Extra Dimensions (in String Theory) - What does it mean?

I have been reading a lot about string theory and the necessity of extra dimensions (particularly as visualized in Calabi-Yau spaces), as "curling-ups" in our apparently 3-dimensional (or 4-dimensional, including time) world. I do not understand, though, how all these "extra dimensions" are actually adding dimensions to our world.

Even if these little multi-dimensional curls are everywhere and too tiny for us to see, how is it still adding extra dimensions? I mean, if our eyes were sharp enough, couldn't one still specify their location according to an $x,y,z$ axis? They will be a certain amount to the "left," a certain amount "high," and a certain amount front or backward (etc). It's still length, width and depth.

What am I missing here?

electricity - Why is the charge transferred by electrons and not by protons?

Charges are transferred by electrons which we all know. But why can't it be transferred by protons? Well, I searched on Google where I found similar questions already being asked on many sites. However, I didn't find any answers. They were saying that electrons are mobile and similarly protons are busy being nucleons. Also, someone was saying, "By the simple mechanism of separating two things, electrons are pulled out of their Fermi energy levels and into a conduction band to effect charge transference and the creation of a static charge" which I didn't understand at all.

Answer

You can totally transfer charge using protons. Or using Na+ or any kind of charged particle. It happens all the time - if you look at how a wet cell battery works, you'll find that while charge across the wire is carried by electrons, current flows along the salt bridge via charged ions (theoretically protons could be among these). I bet if you could somehow make a superfluid out of He+ or something you could use that to carry charge like a superconducting wire.

I think our intuition that electrons are always the charge carriers comes from the fact that:

1.) electrons are very light compared to protons, so if you imagine putting a proton in an electric field and an electron in the same electric field, the electron will be accelerated 1000x more (same charge, 1/1000th the mass). This means that when there is a charge imbalance and either proton or electron flow could alleviate it, the electrons will flow way before the protons are impelled to move.

2.) electrons are the mobile charged particles in a solid - all the protons are bound in nuclei. Since almost all of our intuition about current flow comes from wires or other solid conductors, we're almost exclusively worrying about electron flow.

kinematics - Stellar outflows in the Milky Way

There appears to be a lot of evidence that gas flows outward in Galaxies. I've been trying to parse through the available data and am unable to get a clear answer to this question: is there any evidence for an average flow of stellar material in our galaxy? That is, do we know if stars are falling in, falling out, in a perfectly stable orbit (or do we just not have enough data yet)?

general relativity - Stress-energy tensor for a fermionic Lagrangian in curved spacetime - which one appears in the EFE?

So, suppose I have an action of the type: $$ S =\int \text{d}^4 x\sqrt{-g}( \frac{i}{2} (\bar{\psi} \gamma_\mu \nabla^\mu\psi - \nabla^\mu\bar{\psi} \gamma_\mu \psi) +\alpha \bar{\psi} \gamma_\mu \psi\bar{\psi}\gamma_\nu \psi g^{\mu\nu})$$ Where $\psi$ is a fermionic field and the rest has the usual meaning ($\alpha$ is a coupling constant). Now, if I write down the Canonical energy momentum tensor, i find $$ \tilde{T}_{\mu\nu}= \frac{\delta L}{\delta \nabla^\mu\psi} \nabla_\nu\psi+ \nabla_\nu\bar{\psi} \frac{\delta L}{\delta \nabla^\mu\bar{\psi}}- g_{\mu\nu} L = 2i\bar{\psi} \gamma_{(\mu}\nabla_{\nu)}\psi -g_{\mu\nu} L $$

But, if I write the Einstein's tensor in general relativity i get $$T_{\mu\nu}=\frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu\nu}}=2 i\bar{\psi} \gamma_{(\mu}\nabla_{\nu)}\psi + 2 g \bar{\psi} \gamma_\mu \psi\bar{\psi}\gamma_\nu \psi- g_{\mu\nu} L$$

The two are obviously different. So, which one should i use in the Einstein's equations? The problem comes when you write an interaction term of the type $A_\mu A^\mu$, where $A$ is some current. Because otherwise the two tensor coincide. The first energy momentum is the one invariant under translations, so it is the one satisfying $$\nabla_\mu \tilde{T}^{\mu\nu} = 0$$ While the second satisfy the same identity only if $$\nabla_\mu A^\mu = 0$$ Basically my question is, which one of the two should be used in the Einstein's equations? $G_{\mu\nu} = \kappa \overset{?}{T}_{\mu\nu}$ Or am i doing something wrong and the two tensor do actually coincide?

Answer

As @Holographer has mentioned in a comment, the correct formula for the stress-tensor that enters the EFE is $$ T_{\mu\nu} = \frac{2}{\sqrt{-g}} \frac{\delta S_{\text{matter}}}{ \delta g^{\mu\nu} } $$ whereas what you are computing is the canonical stress energy tensor. However, there is a subtle relation between the two, which I will elaborate upon here.

Apart from a theory that contains only scalars, the canonical stress tensor is never the one that enters the EFE. This is because, in general, the canonical stress tensor is not symmetric and therefore cannot possibly be the same stress tensor that enters in the EFE. For instance the canonical stress tensor for electromagnetism is $$ (T^{EM}_{\mu\nu})_{\text{canonical}} = F^\rho{}_\mu \partial_\nu A_\rho + \frac{1}{4} g_{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} $$ which is not only not symmetric, but also not gauge invariant. PS - The non-symmetry is due to the spin of the field involved and is closely related to the angular momentum tensor.

However, there is an ambiguity in the construction of the stress tensor (the ambiguity does not change the conserved charges which are physical quantities). This ambiguity allows construction of an improved stress tensor (often known as the Belinfante tensor) that is symmetric and conserved. It is this improved tensor that enters the EFE. (ref. this book)

To see the equivalence, let us recall the standard construction of the stress-tensor. Consider a coordinate transformation $$ x^\mu \to x^\mu + a^\mu (x) $$ Since the original Lagrangian is invariant under translations (where $a^\mu$ is constant), the change in the action under such a coordinate transformation is $$ \delta S = \int d^d x \sqrt{-g} \nabla_\mu a_\nu T_B^{\mu\nu} $$ Now, if the stress-tensor is symmetric then we can write $$ \delta S = \frac{1}{2} \int d^d x \sqrt{-g} \left( \nabla_\mu a_\nu + \nabla_\nu a_\mu \right) T_B^{\mu\nu} $$ Note that the term in the parenthesis is precisely the change in the metric under the coordinate transformation. Thus, $$ \delta S = \frac{1}{2} \int d^d x \sqrt{-g} \delta g_{\mu\nu} T_B^{\mu\nu} \implies \frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g_{\mu\nu}} = T_B^{\mu\nu} $$

Thus, we see that the symmetric Belinfante stress tensor is precisely the gravitational stress tensor. Note of course that what I've said holds specifically in a Minkowskian background, since the construction of $T_B^{\mu\nu}$ assumes Lorentz invariance.

Extra energy in quantum tunneling

In quantum tunneling, the probability of finding an electron inside the potential barrier is non zero . So we can actually find an electron which had an energy $E$ in a place where classically it should have an energy bigger than $E$.

So if we find an electron in this potential barrier what will its energy be ?

• 
  

  The energy of the electron is conserved so its $E$ and the fact that its $KE \lt 0$ is just a weird fact of quantum mechanics. (I don't like this answer)

  
  


• 
  

  Since we know the electron is there we can treat it as a classical particle and because of that it's energy has rose to a value bigger than the potential of the potential barrier. The energy of the electron is not conserved unless the electron manages to pick up energy from nowhere.

  
  

These are the answers I have in mind, I'm practically sure none of them is correct ...

So how can we interpret this fact from an energy point of view ? The question may be silly but I'm just beginning quantum mechanics and I'm having a difficult time trying to understand it.

I hope someone can help me.

(Sorry for my bad english)

Answer

Short answer: Position and energy are not compatible observables, meaning you can not determine them both at the same time, much like position and momentum are non-compatible observables.

Long answer: If you know the energy of your particle, that means it's wavefunction is an eigenfunction of the Hamiltonian (a solution to the time-independent Schrödinger equation). This wavefunction will be spread out over the system with non-zero components in the classically forbidden region, i.e. there is a finite probability to find the particle in this region.

To actually find it there, you must perform a measurement. The measurement will collapse the wavefunction to one which is localized around the point where you happen to find it (let's assume we do find it in the classically forbidden region). The new wavefunction now is no longer an eigenfunction of the Hamiltonian, and the particle therefore does not have a well-defined energy. To determine the energy of the particle you would have to perform an energy measurement. This measurement would collapse the wavefunction into an eigenstate of the Hamiltonian, which would again be spread out over the system, i.e. the particles position would now be undetermined. Furthermore, the energy you would measure would likely be different from the original energy of the particle (before position and energy measurements).

As for energy conservation: When you introduce a measurement apparatus the system is no longer closed, and energy conservation does not apply unless you consider the total system, including the measurement apparatus.

cosmology - Number $g(T)$ of relativistic degrees of freedom as a function of temperature $T$

Let us consider the total number of relativistic degrees of freedom $g(T)$ for particle species in our universe:

$$g(T)=\left(\sum_Bg_B\right)+\frac{7}{8}\left(\sum_Fg_F\right)$$

Where the sums are over the degrees of freedom for bosons ($B$) and and fermions ($F$) which are relativistic when the universe has temperature $T$ (meaning $T$ > their mass energy). For example the photon contributes a $g_{ph}=2$ for the two polarization degrees of freedom it has.

Now, I heard of the following rough estimates for $g(T)$:

When $T_1\geq 1GeV$ we have $g(T_1)\approx 100$.

When $100MeV\geq T_2\geq 1MeV$ we have $g(T_2)\approx 10$.

When $ 0.1MeV\geq T_3 $ we have $g(T_3)\approx 3$.

I am trying to reproduce these estimates by counting all relativistic particles at the specific $T$ values and summing up their degrees of freedom. However, there seem to be contradictions and unclarities here.

For example, the lowest of the three values $g(T_3)\approx3$ is supposedly due to the 2 polarizations of photons and 1 spin degree of freedom of the electron neutrino. However, shouldn't we also count the spin degree of freedom of the electron anti-neutrino? And what about the other two neutrino species? Why include electron neutrino but leave out the others?

Similarly, for $g(T_2)\approx 10$ I would expect to count 2 photon polarizations, 1 spin d.o.f. for neutrinos and anti-neutrinos (6 d.o.f.s in total), 2 spins for electrons and anti-electrons and muons and anti-muons (8 in total), again 2 spins for up, down and strange quark particle anti-particle pairs (12 in total). I am not sure if I missed any particle species here, but we already have $g(T_2)\approx30$ instead of $10$.

Could someone explain to me how to do this counting properly and why some species appear to be missing from consideration even though they should count as relativistic?

Answer

First, note that the equation you use is only valid when all relativistic particles are in thermal equilibrium. The more general equation, which allows for particles with different temperatures, is $$ g(T) = \sum_B g_B\left(\frac{T_B}{T}\right)^4 + \frac{7}{8}\sum_F g_F\left(\frac{T_F}{T}\right)^4 $$ where $T$ is the photon temperature and $T_B$, $T_F$ are the temperatures of each boson and fermion.

The degrees of freedom for all Standard Model particles are listed in the table below (source: http://www.helsinki.fi/~hkurkisu/cosmology/Cosmo6.pdf):

At temperatures $T\sim 200\;\text{GeV}$, all particles are present, relativistic, and in thermal equilibrium, so we find $$ g(T) = 28 + \frac{7}{8}\cdot 90 = 106.75 $$ When $T\sim 1\;\text{GeV}$, the temperature has dropped below the rest energy of the $t$, $b$, $c$, $\tau$, $W^+$, $W^-$, $Z^0$, and $H^0$ particles, therefore these are no longer relativistic (and will have annihilated) and we have to take them out of the equation. We are left with $$ g(T) = 18 + \frac{7}{8}\cdot 50 = 61.75 $$ When $T$ drops below $100\;\text{MeV}$, the remaining quarks and gluons are locked up in non-relativistic hadrons, and the muons have annihilated. All that's left are photons, electrons, positrons, neutrinos and anti-neutrinos, so that $$ g(T) = 2 + \frac{7}{8}\cdot 10 = 10.75 $$ So far, all relativistic particles were in thermal equilibrium. However, as the temperature drops to $1\;\text{MeV}$, the neutrinos decouple and move freely, which means their temperature will start to diverge from the photon temperature. At $T < 500\;\text{keV}$, the electrons and positrons are no longer relativistic, so only the photons and neutrinos remain, and $$ g(T) = 2 + \frac{7}{8}\cdot 6\left(\frac{T_\nu}{T}\right)^4, $$ where $T_\nu$ is the neutrino temperature. Calculating this requires a bit of work.

The Second Law of Thermodynamics implies that the entropy density $s(T)$ is given by $$ s(T) = \frac{\rho(T) + P(T)}{T}, $$ where $\rho$ is the energy density and $P$ the pressure. Using the Fermi-Dirac and Bose-Einstein distributions, one finds that for relativistic particles $$ \rho(T) = \begin{cases} \dfrac{g_B}{2}a_B\, T^4&\text{bosons}\\ \dfrac{7g_F}{16}a_B\, T^4&\text{fermions} \end{cases} $$ and $P = \rho/3$, so that $s(T) = 4\rho(T)/3T$. Let us now consider the entropy density of the photons and the electrons and positrons at high temperatures, when they are still relativistic: $$ s(T_\text{high}) = \frac{2}{3}a_B\,T_\text{high}^3\left(2 + \frac{7}{8}\cdot 4\right) = \frac{4}{3}a_B\,T_\text{high}^3\left(\frac{11}{4}\right). $$ At low temperatures, the electrons and positrons become non-relativistic, most annihilate and the remaining particles have negligible contribution to the entropy, therefore $$ s(T_\text{low}) = \frac{4}{3}a_B\,T_\text{low}^3\,. $$ However, thermal equilibrium implies that the entropy in a comoving volume remains constant: $$ s(T)a^3 = \text{constant}. $$ Also, the temperature of the neutrinos drops off as $T_\nu \sim 1/a$ after they decouple. Combining these results, we find $$ \left(\frac{T_\text{low}}{T_{\nu,\text{low}}}\right)^3 = \frac{11}{4} \left(\frac{T_\text{high}}{T_{\nu,\text{high}}}\right)^3. $$ At high temperatures, the neutrinos are still in thermal equilibrium with the photons, i.e. $T_{\nu,\text{high}}=T_\text{high}\,$, so finally we obtain $$ T_\nu = \left(\frac{4}{11}\right)^{1/3}T $$ at low temperatures. Therefore, $$ g(T) = 2 + \frac{7}{8}\cdot 6\left(\frac{4}{11}\right)^{4/3} = 3.36. $$ A more detailed treatment is given in the same link from where I took the table.

classical mechanics - Euler-Lagrange equations and friction forces

We can derive Lagrange equations supposing that the virtual work of a system is zero.

$$\delta W=\sum_i (\mathbf{F}_i-\dot {\mathbf{p}_i})\delta \mathbf{r}_i=\sum_i (\mathbf{F}^{(a)}_i+\mathbf{f}_i-\dot {\mathbf{p}_i})\delta \mathbf{r}_i=0$$

Where $\mathbf{f}_i$ are the constrainded forces and are supposed to do no work, which it's true in most cases. Quoting Goldstein:

[The principle of virtual work] is no longer true if sliding friction forces are present [in the tally of constraint forces], ...

So I understand that we should exclude friction forces of our treatmeant. After some manipulations we arrive to:

$$\frac{d}{dt}\frac {\partial T}{\partial \dot q_i}-\frac{\partial T}{\partial q_i}=Q_i$$

Further in the book, the Rayleigh dissipation function is introduced to include friction forces. So given that $Q_i=-\frac {\partial \mathcal{F}}{\partial \dot q_i}$ and $L=T-U$, we get:

$$\frac{d}{dt}\frac {\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}+\frac {\partial \mathcal{F}}{\partial \dot q_i}=0$$

Question: Isn't this an inconsistency of our proof, how do we know the equation holds? Or is it just an educated guess which turns out to be true?

Answer

The main point is that Goldstein is not saying we must exclude friction forces in our treatment, but we must place them in the tally of applied forces (that we keep track of in D'Alembert's principle) and not in the other bin of the remaining forces, see this and this Phys.SE posts.

Of course, there does not exist a generalized potential $U$ for the friction forces ${\bf F}=-k {\bf v}$, only the Rayleigh dissipation function, see this Phys.SE post and this mathoverflow post.

electromagnetism - Is there a "special" inertial frame determined by the value of E and B at a point in an EM field?

Given the facts that $(E^2 - B^2)$ and $(E\cdot B)$ are Lorentz invariants of the EM field, and that the energy density $(E^2 + B^2)$ is not invariant, it seems that at each point in an EM field there should be unique inertial frame in which the field energy is minimum. Can that minimum value, and that inertial frame, be considered Lorentz invariants?

Answer

Since $E^2-B^2$ is invariant under boosts, any boost must change both $E^2$ and $B^2$ by the same increment $\epsilon$, if it changes them at all. In other words, a boost either increases both $E^2$ and $B^2$ by the same amount or decreases them both by the same amount, if it changes them at all.

There are two cases two consider: $E\cdot B=0$ and $E\cdot B\neq 0$. Throughout this answer, only a single point is considered, and "energy density" means the energy density at that point.

First suppose $E\cdot B=0$. In this case, there is a frame in which either $E$ or $B$ is zero. (See the appendix for an outline of a proof.) This must be the frame in which the energy density is a minimum, because any boost away from that frame will increase both $E^2$ and $B^2$. (It can't decrease them, because then one of them would end up being negative, but $E^2$ and $B^2$ cannot be negative.)

Now consider the case $E\cdot B\neq 0$. In this case, there is a frame in which the electric and magnetic field vectors are parallel to each other. (See the appendix for an outline of a proof.) Denote these fields by $E_0$ and $B_0$. In such a frame, we have $$ |E_0\cdot B_0| = \sqrt{E_0^2B_0^2}. \tag{1} $$ After a boost, we have $$ E\cdot B = \sqrt{E^2B^2}\cos\theta = \sqrt{(E_0^2+\epsilon)(B_0^2+\epsilon)}\,\cos\theta, \tag{2} $$ and since $E\cdot B$ is invariant, this gives $$ \sqrt{E_0^2B_0^2} = \sqrt{(E_0^2+\epsilon)(B_0^2+\epsilon)}\,\cos\theta. \tag{3} $$ This shows that we must have $\epsilon\geq 0$. Therefore, if we start in a frame in which the vectors $E$ and $B$ are parallel, a boost cannot decrease the value of $E^2+B^2$, so it cannot decrease the energy density at the given point. This shows that the energy at any given point is minimized in any frame where $E$ and $B$ are parallel to each other.

The frame that minimizes the energy density is not unique. For example, if we start in a frame where $E$ and $B$ are parallel to each other (or if one of them is zero) and then apply a parallel boost, the fields $E$ and $B$ are unchanged.

Summary: There is always a frame in which (at the given point) either $E$ and $B$ are parallel to each other or one of them is zero. The frame that does this is not unique. Any such frame minimizes the energy density (at that point).

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Appendix

The claim is that if $E\cdot B=0$, then there is a frame in which either $E$ or $B$ is zero; and if $E\cdot B\neq 0$, then there is a frame in which they are parallel. Here's an outline of the proof, using the Clifford-algebra approach that was described in my answer to

Wigner Rotation of static E & M fields is dizzying

Let $\gamma^0$, $\gamma^1$, $\gamma^2$, $\gamma^3$ be mutually orthogonal basis vectors, with $\gamma^0$ being timelike and the others being spacelike. In Clifford algebra, vectors can be multiplied, and the product is associative. I'll use $I$ to denote the identity element of the algebra, and I'll use $\Gamma$ to denote the special element $$ \Gamma\equiv \gamma^0\gamma^1\gamma^2\gamma^3. \tag{4} $$ Mutually orthogonal vectors anticommute with each other, and the product of parallel vectors is proportional to $I$.

The electric and magnetic fields are components of the Faraday tensor $F_{ab}$, which are the components of a bivector $$ F = \sum_{a In the frame defined by the given basis, the electric and magnetic parts of $F$, which I'll denote $F_E$ and $F_B$ (so $F=F_E+F_B$), are the parts that do and do not involve a factor of the timelike basis vector $\gamma^0$, respectively. The quantity $F$ satisfies $$ F^2=(F^2)_I + (F^2)_\Gamma \tag{6} $$ where subscripts $I$ and $\Gamma$ denote terms proportional to $I$ and $\Gamma$, respectively. Both parts, $(F^2)_I$ and $(F^2)_\Gamma$, are invariant under proper Lorentz transformations. The part $(F^2)_I$ is proportional to $E^2-B^2$, and the part $(F^2)_\Gamma$ is proportional to $E\cdot B$.

The starting point for the proof is that in four-dimensional spacetime, any bivector may be written in the form $$ F = (\alpha+\beta\Gamma)uv \tag{7} $$ where $\alpha,\beta$ are scalars and where $u$ and $v$ are mutually orthogonal vectors. If neither vector is null (the null case wil be treated last), then we can always find a frame in which they are both proportional to basis vectors in a canonical basis. The factor $\alpha+\beta\Gamma$ is the same in all frames (only $u$ and $v$ are affected by Lorentz transformations, because $\Gamma$ is invariant), so in the latter frame, $F_E$ is proportional to the product of two basis vectors (one of which is timelike), and $F_B$ is proportional to the product of the other two. After reverting to the traditional formulation in which the electric and magnetic fields are represented by vectors in 3-d space rather than by bivectors in 4-d spacetime, this is equivalent to the condition that $E$ abd $B$ both be proportional to a single vector.

If one of the coefficients $\alpha,\beta$ in (7) is zero (so that $E\cdot B=0$), then this proves the existence of a frame in which one of them is zero. If both coefficients $\alpha,\beta$ are non-zero (so that $E\cdot B\neq 0$), then then this proves the existence of a frame in which they are parallel.

Finally, if one of the vectors $u,v$ in (7) is null, then $E$ and $B$ are already parallel to each other and have the same magnitude.

Relativistic constant acceleration in the framework of general relativity

Regarding the constant acceleration of a spacecraft towards relativistic velocities, the general equations are found at the following link: Baez

However, if the distance covered is in excess of 1 billion light-years, general relativity would have to be used to compute the distance.

So an attempt to find the corresponding equations was made in this post: post for the framework of GR

Can someone actually give a numerical example of the calculations involved in the latter link?

Also no description was given for the calculation of $a(t)$.

Thanks

general relativity - Significance for LQG of Sen's result on entropy of black holes?

Sen 2013 says,

...we apply Euclidean gravity to compute logarithmic corrections to the entropy of various non-extremal black holes in different dimensions [...] For Schwarzschild black holes in four space-time dimensions the macroscopic result seems to disagree with the existing result in loop quantum gravity.

How serious a problem is this for LQG? Does this mean that LQG doesn't have GR as its semiclassical limit? Does that mean it's a dead theory, or maybe just that it needs to be modified? Is the technique using Euclidean gravity reliable?

Since I'm not a specialist, I'd be interested in a hand-wavy explanation of what the Euclidean gravity technique is about.

Sen, http://arxiv.org/abs/arXiv:1205.0971

newtonian mechanics - Understanding which string breaks when one pulls on a hanging block from below

Intro:

In completing Walter Lewin's 6th lecture on Newton's Laws, he presents an experiment (go to 42:44) which leaves me baffled.

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Experiment:

(I recommend watching the video; see link above.)

• There is a $2$ kg block with 2 identical strings attached to it: one at the top, the other at the bottom.


• The top string is attached to a "ceiling", and the bottom to a "floor".


• Professor Lewin "stretches" the system (by pulling on the bottom string) with the block not accelerating.


• One string snaps.

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Prediction:

• Initially, the top string has a tension of approximately $20$ N, to counter the force of gravity. The bottom string has no tension at all.


• Then, when Lewin pulls the bottom string, it gains some tension $n$ N. To counter act the force exerted by the bottom string, the top string exerts now $20 + n$ N.


• I assume that the string with more force will give out sooner, leading me to conclude that the top string will break.

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Results:

(This was conducted by Lewin, not me; see link above.)

• Trial 1: Bottom string breaks.


• Trial 2: Top string breaks.


• Trial 3: Bottom string breaks.

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Additional Notes:

The results don't seem consistent. If I was right, I'd expect all 3 experiments to be right; conversely, if I was wrong, I'd expect all 3 experiments wrong, with one exception: the results are more-less random and one result isn't preferred over the other.

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Question:

• Why was my prediction incorrect?


• Was there a flaw in my logic?


• Why were the results inconsistent?

Answer

Your predictions of the forces adding up is correct, if nothing accelerates. Because, think about it... you are adding up forces, right? That is what you do in Newton's 1st law. Which is the law that only applies when nothing accelerates.

What if you were told that you can't use Newton's 1st law in the second case? Is something accelerating in the second case?

Or in other words, is the string trying to accelerate something in the second case?

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Solution

If something should accelerate, we are in Newton's 2nd law. If not, Newton's 1st law. Let's write it out with the forces from each string and weight $w$ present:

$$- F_{up} +F_{down} + w=0\qquad \qquad - F_{up} +F_{down} + w=ma$$

(I hope it's okay I've put the y-direction downwards.)

• 
  

  If you pull slowly down, no significant speeding up happens of the box. $F_{down}$ has some constant value. It all balances out. The 1st law.

  
  



• 
  

  If you pull fast down, the box tries to speed up fast to follow along. That means large $a$. That requires large force to cause it. And the force, that tries to cause is the $F_{down}$.

  
  

Look at those two equations again. In the first case $F_{up}=F_{down}+w$, so the upper string breaks. In the second case $F_{up}=F_{down}+w-ma$. Hmm, here is being subtracted the part $ma$...

So, is $F_{up}$ becoming smaller? No, of course not, it has it's tension and only grows as you pull downwards. Rather $F_{down}$ becomes larger. Because it tries to cause the $a$.

And as you see, it tries to but simply can't apply enough force to cause that acceleration. The necessary force in the lower string is more than the strength of the string, so it breaks.

electromagnetism - Vector Potential Oscillating E Field of the "Null" Field of a Hertzian Dipole?

The vector potential of a Hertzian dipole falls off spherically as $1/r$. The polar axis of the dipole is a "Null" field -- meaning no electric and magnetic field. The absence of magnetic field is clear enough since there is no curl in the vector potential there. However, the potentials description of the E field involves the equation:

$$\mathbf E = -∇Φ - ∂\mathbf A/∂t$$

which seems to indicate there will be an oscillating electric field in the far field of the Hertzian dipole -- even in the "Null" of the dipole, where the where there should be no electric field.

Clearly, as there is no poynting vector in the "Null", an electron placed there can not oscillate in response to an oscillating electric field as there is no energy locally available, otherwise we'd encounter a violation of conservation of energy or a violation of locality.

How is this conundrum resolved?

Answer

The vector potential of an oscillating dipole (using the usual electric dipole approximation) can be written as $$ \vec{A} =\frac{\mu_0 I_0 l}{4\pi r} \cos \omega(t-r/c)\ \hat{z},$$ where the dipole is of length $l$, with a current $I_0 \cos \omega t$ and $\hat{z}$ is a unit vector along the z-axis of the dipole.

Using the Lorenz gauge one can then calculate a corresponding scalar potential $$\phi = \frac{\mu_0 I_0 l c^2\cos \theta}{4\pi} \left( \frac{\cos \omega(t-r/c)}{cr} + \frac{\sin \omega(t-r/c)}{\omega r^2} \right),$$ where $\theta$ is the usual polar angle to the z-axis.

So now the electric field is given by $$\vec{E} = -\frac{\partial \vec{A}}{\partial t} - \nabla \phi$$

If you discard all the terms that have a denominator of $r^2$ or higher from the right hand term (i.e. just consider the far-field), you find the gradient of the scalar potential only has a radial component $$\vec{E} \simeq \frac{\mu_0 I_0 l \omega}{4\pi r} \sin \omega(t-r/c) \hat{z} - \frac{\mu_0 I_0 l \omega \cos \theta}{4\pi r} \sin \omega(t-r/c) \hat{r} $$

But when $\theta =0$ (or $\pi$) along the z-axis, then $\hat{r} \equiv \hat{z}$ and the two terms cancel leaving no far-field.

I am still pondering whether there is a more qualitative way of explaining this.

electrostatics - How to find the distribution of charge on two spheres connected by a conducting wire?

A solid metal sphere of radius $R$ has charge $+2Q$. A hollow spherical shell of radius $3R$, concentric with the first sphere, has net charge $-Q$.

What would be the final distribution of the charge if the spheres were joined by a conducting wire?

Answer: $+Q$ on outer shell, $0$ on inner sphere.

Why is that? I would assume that the total charge should just evenly spread between the spheres.

Answer

Using Gauss' law and the condition that there can be no electric field inside a conductor the initial charge distribution is as follows:

• 
  

  $+2Q$ charge on the outside of the inner sphere.

  
  


• 
  

  $-2Q$ charge on the inside of the outer sphere ($-Q$ original charge and $-Q$ induced charge)

  
  
  


• $+Q$ induced charge on the outside of the outer sphere

There is an electric field between the two sphere and so there must be a potential difference between the two spheres.
Joining the two spheres with a conducting wire means that charges will flow until the potential difference between the two spheres becomes zero ie there is no electric field between the two conductors and inside the conducting wire.
So no charge can reside on the outside of the inner sphere and the inside of the outer sphere and this is achieved by the $-2Q$ on the inside of the outer sphere and the $+2Q$ on the outside of the inner sphere neutralising one another leaving charge $+Q$ on the outside of the outer sphere.

soft matter - What happens when a bacteria tumble? Is it an active or a passive process?

I was wondering what actually happens during the tumbling process of a "run and tumble" process. When the bacteria stops to tumble and reorient itself, is that an active process, or is the reorientation due to thermal rotationnal diffusion?

Cheers,

Nico

electromagnetism - Is neutron decay a purely electromagnetic phenomenon?

Until reading the Phys.SE post here about the neutron decay I never feel strange the fact about the antisymmetricity of this decay. But indeed why this decay is antisymmetric. The neutron is his own antiparticle, and this is without any restriction.

Could I suppose that the phenomena has a purely electromagnetic cause? Put a neutron into an environment of positrons, will the result of the neutron decay be an antiproton, a positron and a neutrino? And if this is right, will an ordinary positive charged environment give the same result?

Comment: The question is not about the process of the decay which is explainable with the weak nuclear force. It's about the hidden parameters, which bring the neutron to the decay.

Answer

Neutron decay is not an electromagnetic phenomenon at all. It is governed by the weak nuclear force. This is well supported by fact that the lifetime of the neutron fits neatly into weak universality, by the energy distribution of neutron decay products, and by the fact that anti-neutrino initiated "inverse beta decay" can be shown to be proportional to the power of nearby nuclear reactors or to beam intensity in accelerator experiments.

Secondly the neutron is not it's own antiparticle. The anti-neutron is a distinct particle.

newtonian mechanics - Is a causal relationship implied by Newton's 2nd Law?

Throughout my time learning physics I have been imbued with the notion that forces cause accelerations, period. Accelerations don't cause forces, and they aren't merely correlated phenomena. By causality, I am content with the following definition:

Connection between two events or states such that one produces or brings about the other; where one is the cause and the other its effect.

That is to say, an object experiences an acceleration because it is exposed to a net force; the force does not arise because of the acceleration. However, some philosophical thinking on Venturis has shaken my confidence in this idea. If the acceleration of the fluid through the constriction is caused by an unbalanced force, what causes the unbalanced force in the first place? Another way of asking the question is, how is the bounding geometry causally linked to the pressure distribution of the flow? My only answer as yet is that there's no other way to satisfy mass, momentum, and energy conservation simultaneously, but that seems decidedly unsatisfying. Is there any causality implied by Newton's 2nd Law?

Answer

One of my favorite quotes, and I think this complements Ján Lalinský's answer:

"Does the engineer ever predict the acceleration of a given body from a knowledge of its mass and of the forces acting upon it? Of course. Does the chemist ever measure the mass of an atom by measuring its acceleration in a given field of force? Yes. Does the physicist ever determine the strength of a field by measuring the acceleration of a known mass in that field? Certainly. Why then, should any one of these roles be singled out as the role of Newton's second law of motion? The fact is that it has a variety of roles." - Brian Ellis, The Origin and Nature of Newton's Laws of Motion (1961), as cited by A P French, Newtonian Mechanics. (a fantastic book)

$F=m\ddot{x}$ isn't a definition of force or a definition of mass, it's a relationship.

As for your specific example, I can't help on the dynamics, but from the fact that the water accelerates, it must be pushed from behind (or pulled from the front, but you know with pressure the two are equivalent). This image on wikipedia:

http://en.wikipedia.org/wiki/File:Venturi.gif

in which high pressure is indicated by a dark blue color, gives you a pressure gradient. Clearly the change in pressure is enough to explain the acceleration/deceleration. Why is there a pressure gradient? That is deserving of its own question*.

*I'm answering the question "Is a causal relationship implied by Newton's 2nd Law?"

How can a moving frame transmit energy (a fighter plane shoots a bullet)?

This question shows that a bullet shot from a flying plane receives a large amount of energy from the plane, even much more than the energy from the propellant of the bullet.

But I wonder how can the plane transfer additional energy to the bullet? For example, the propellant gas expands and creates the pressure that pushes the bullet forward, this energy is limited by the propellant itself. While the plane only recoils backward, how can the plane transfer force, or energy to the bullet, in what method since it even does not touching the bullet? If the barrel is a part of the plane, the plane even pulls the bullet backward by friction.

quantum mechanics - Time reversal symmetry and real symmetric Hamiltonian matrix

In the literature (like those in quantum chaos), it seems that time-reversal symmetry implies that the Hamiltonian of the system is a real symmetric one, instead of just being complex Hermitian.

Is there any rigorous justification about this link?

special relativity - Is causality a formalised concept in physics?

I have never seen a “causality operator” in physics. When people invoke the informal concept of causality aren’t they really talking about consistency (perhaps in a temporal context)?

For example, if you allow material object velocities > c in SR you will be able to prove that at a definite space-time location the physical state of an object is undefined (for example, a light might be shown to be both on and off). This merely shows that SR is formally inconsistent if the v <= c boundary condition is violated, doesn’t it; despite there being a narrative saying FTL travel violates causality?

Note: this is a spinoff from the question: The transactional interpretation of quantum mechanics.

electromagnetism - How can relativistic effects be the cause of magnetic forces when the normal speeds of carrier charges are so low?

I had read here that magnetism arises from a current because of the special relativistic effect associated with the speed of the moving charges in that current. However that speed is only on the order of 0.024 cm/s, essentially zero compared to the speed of light, so there cannot be a relativistic effect to consider. Is special relativity the origin of magnetism?

entropy - How does Landauer's Principle apply in quantum (and generally reversible) computing

I understand that a reversible computer does not dissipate heat through the Landauer's principle whilst running - the memory state at all times is a bijective function of the state at any other time.

However, I have been thinking about what happens when a reversible computer initializes. Consider the state of the physical system that the memory is built from just before power up and initialization. By dint of the reversibility of underlying microscopic physical laws, this state stay encoded in the overall system's state when it is effectively "wiped out" as the computer initialized and replaces it with a state representing the initialized memory (set to, say, all noughts).

So it seems to me that if $M$ bits is the maximum memory a reversible algorithm will need to call on throughout its working, by the reasoning of Landauer's principle, ultimately we shall need do work $M\, k\,T\,\log 2$ to "throw the excess entropy out of the initialized system".

Question 1: Is my reasoning so far right? If not, please say why.

Now, specializing to quantum computers, this seems to imply some enormous initialization energy figures. Suppose we have a system with $N$ qubits, so the quantum state space has $2^N$ basis states. Suppose further that, for the sake of argument, the physics and engineering of the system is such that the system state throughout the running of the system only assumes "digitized" superpositions, $i.e.$ sums of the form:

$$\frac{1}{\sqrt{\cal N}}\sum_{j\in 1\cdots 2^N} x_j \,\left|\left.p_{1,j},p_{2,j},\cdots\right>\right.$$

where $x_j, \;p_{k,j}\in{0,1}$ and ${\cal N}$ the appropriate normalization. To encode the beginning state that is wiped out at power up and initialization, it seems to me that the Landauer-principle-behested work needed is $2^N \,k\,T\,\log 2$. This figure reaches 6 000kg of energy (about humanity's yearly energy consumption) at around 140 qubits, assuming we build our computer in deep space to take advantage of, say, 10K system working temperature.

Question 2: Given that we could build a quantum computer with 140 qubits in "digital" superpositons as above, do we indeed need such initialization energies?

One can see where arguments like this might go. For example, Paul Davies thinks that similar complexity calculations limit the lower size of future quantum computers because their complexity (information content) will have to respect the Bekestein Bound. P.C.W. Davies, "The implications of a holographic universe for quantum information science and the nature of physical law", Fluctuation and Noise Lett 7, no. 04, 2007 (see also http://arxiv.org/abs/quantph/0703041)

Davies points out that it is the Kolmogorov complexity that will be relevant, and so takes this as an indication that only certain "small" subspaces of the full quantum space spanned by a high number of qubits will be accessible by real quantum computers. Likewise, in my example, I assumed this kind of limitation to be the "digitization" of the superposition weights, but I assumed that all of the qubits could be superposed independently. Maybe there would be needfully be correlations between the superpositions co-efficients in real quantum computers.

I think we would hit the Landauer constraint as I reason likewise, but at a considerably lower number of qubits.

Last Question: Am I applying Landauer's principle to the quantum computer in the right way? Why do my arguments fail if they do?*

gravity - The theory of moon creation when a Mars size planet hit Earth

As we know the predominant theory where does the moon come from is that a Mars size planet hit the earth and took a chunk out of it which eventually materialized into moon.

My question is that if a Mars size object were to hit Earth, wouldn't it knock it off the orbit all together? What kind of collision is required to knock a planet of its orbit. By 'knock off' I mean it would alter the orbit of Earth and possibly speed so that it will not have stable orbit anymore so it will either (gradually) leave solar system or (gradually) collapse into sun.

thermodynamics - Reversible cycle approximated by Carnot cycles

My textbook, W.E. Gettys, F.J. Keller, M.J. Skove, Physics 1, gives the definition of a reversible transformation as a transformation that can be inverted by effectuating only infinitesimal changes in the surroundings. I admit that I have no idea of what infinitesimal means in a rigourous mathematical language (if we exclude non-standard analysis, but I do not think that it is used in elementary thermodynamics), therefore the definition is quite confusing to me.

Then the book proves that the entropy of a Carnot cycle is zero and, since a reversible cycle can be approximated by the sum of many Carnot cycles, the entropy of a reversible cycle is zero too.

I suppose that we can mathematically formalize the meaning of such an approximation by saying that for any reversible cycle there is a sequence $\{\{C_{1,n},\ldots,C_{n,n}\}\}_n$ of sets of Carnot cycles $C_{1,n},\ldots,C_{n,n}$ (say that the $n$-th set of Carnot cycles used to approximate is composed by $n$ cycles) such that $$\oint\frac{\delta Q}{T}=\lim_{n\to\infty} \sum_{k=1}^n \frac{Q_{C,k,n}}{T_{C,k,n}}+\frac{Q_{H,k,n}}{T_{C,k,n}}.$$As a side note, I realise that, for any Carnot cycle $C_{k,n}$, we have $\frac{Q_{C,k,n}}{T_{C,k,n}}+\frac{Q_{H,k,n}}{T_{C,k,n}}=0$ and therefore, provided that the approximation holds, ${\displaystyle\oint}\frac{\delta Q}{T}=0$.

But, how can we see that a reversible cycle can be arbitrarily approximated by Carnot cycles (i.e. that ${\displaystyle\oint}\frac{\delta Q}{T}=\lim_{n\to\infty} \sum_{k=1}^n \frac{Q_{C,k,n}}{T_{C,k,n}}+\frac{Q_{H,k,n}}{T_{C,k,n}}$, if we use the notation I have introduced and if I have correctly understood the meaning of the approximation)?

Answer

I am not familiar with the textbook you mentioned but there is a conventional "proof" of the equality ${\displaystyle\oint} _\textrm{rev}\frac{\delta Q}{T}=0$ for states that can be described by a thermal parameter, usually temperature $T$ and by a mechanical parameter, such as $V$. The "proof" goes by approximating the area enclosed by the $T,V$ cycle with Carnot cycles, that is consisting of isothermal-adiabatic-isothermal-adiabatic segments. The reason why I used the quotation marks around the word proof because this is not really a proof in a mathematical sense. There are many loose ends, among them the funny way of rectifying the cycle with isothermal pieces that do not join.

It was never obvious to me that the process converges. We have a lot more complicated case when there are several mechanical/deformation variables. The rectification of an arbitrary curve in higher dimensions with non-joining segments must be even more tenuous than in 2D. Note too that the existence of adiabatic surfaces that are explicitly postulated here is equivalent mathematically with the very existence of the entropy function that we wish to prove, so this not really a proof at all.

This whole problem can be side-stepped by a trick that goes back at least to Fermi (see his book), namely the use of an auxiliary Carnot cycle engine tied to a reservoir whose temperature $T_0$ is higher than any one in the system whose cycle you are interested in, and then the system exchanges heat only through this auxiliary Carnot engine at all temperatures. One now only postulates the existence of isothermal heat exchanges, not adiabatics. Using Kelvin's principle it follows that $ T_0 {\displaystyle\oint} \frac{\delta Q}{T} \le 0$ with $T_0 >T$, and from $T_0 >0$ we get ${\displaystyle\oint} \frac{\delta Q}{T}\le 0$ for all cycles. Now if the process is reversible then reversing it we get both ${\displaystyle\oint}_\textrm{rev} \frac{\delta Q}{T}\ge 0$ and ${\displaystyle\oint} _\textrm{rev}\frac{\delta Q}{T}\ge0$ so then ${\displaystyle\oint} _\textrm{rev}\frac{\delta Q}{T}=0$.

Is thermodynamic equilibrium static or dynamic?

Suppose a system has two parts, $A$ & $B$. They were initially at different temperatures and hadn't achieved mechanical equilibrium. After attainting thermodynamic equilibrium, do they continue to exchange heat energy? ie. after attaining thermal equilibrium, ie. Is the heat that $A$ gains from $B$ the same as the heat gained by $B$ from $A$ so that the temperature remains constant ? Meaning to say, is thermodynamic equilibrium dynamic like chemical equilibrium? Or is it static? What is the reason for any of the two?

Answer

It is a dynamic equilibrium. To see this notice that the Boltzmann distribution law is statistical in nature, for instance, the molecules that have energies between $E_1$ and $E_1+\Delta E_1$ at time $t_1$ are not necessarily the same than the ones in between those same energies at time $t_2$. Particles collide (and thus exchange kinetic energy) all the time, and while the distribution stays the same, the particles that contribute to different parts of the distribution keep changing.

Update: It is dynamic, like chemical equilibrium, because the molecules do not all have the same kinetic energy. If, at equilibrium, you randomly pick a molecule, there some probability (from the Boltzmann distribution) that it will have its energy between $E_1+\Delta E_1$. If after some time $\Delta t$ you pick those same molecules again, there is again the same probability to find those molecules with an energy between $E_1$ and $E_1+\Delta E_1$ is the same, but those same molecules will likely have a different energy than the one you measured earlier. this is because they molecules collide all the time and exchange energy with each other. From a macroscopic point of view the two states, at times $t_1$ and $t_1+\Delta t$, are the same; from a microscopic point of view, the two states are different. Thus it is a macroscopic equilibrium not a microscopic one. Same when have ice and water at equilibrium. The molecules that form part of the ice or part of the water keep changing all the time. At equilibrium, you keep the same amounts of water and ice, but the molecules that make up the ice and the water keep being exchanged between the ice and the water.

electromagnetism - Nature of electromagnetic waves

EM waves are defined as a propagating disturbance of the EM field. But aren't the waves simply a mathematical concept, i.e., analogous to the oscillating (increasing and decreasing) strength of the field as opposed to actually consisting of a physical thing moving up an down?

quantum mechanics - Is normal-ordering useful in supersymmetric theories?

As far as I know, the most immediate effects of normal-ordering is the elimination of the zero-point energy (and the zero-point background of any conserved charge) and the elimination of tadpoles in perturbation theory.

In SUSY the zero-point background of all charges is zero, and (in the theories I've studied) there are not tadpoles either. So it seems that normal-ordering is essentially effectless. Is this true to all practical purposes, or is there some point where normal-ordering is useful?

How do you determine the value of the degeneracy factor in the partition function?

In the partition function, expressed as $$Z = \sum_j g_je^{-\beta E_j}$$ I'm wondering what determines the $g_j$ factor. I've been trying to look around the internet for an explanation of it but I can't find one. I guess it is the number of degenerate states in a given energy level? How do you determine how many degenerate states there are? A simple example involving how to determine the $g_j$ factor would be great. Thanks for the help!

Answer

Consider a quantum system with state (Hilbert) space $\mathcal H$. For simplicity, let the Hamiltonian $H$ of the system have discrete spectrum so that there exists a basis $|n\rangle$ with $n=0,1,2,\dots$ for the state space consisting of eigenvectors of the Hamiltonian. Let $\epsilon_n$ denote the energy corresponding to each eigenvector $|n\rangle$, namely \begin{align} H|n\rangle = \epsilon_n|n\rangle \end{align} Now, it may happen that one or more of the energy eigenvectors $|n\rangle$ have the same energy. In this case, we say that their corresponding shared energy eigenvalue is degenerate. It is therefore often convenient to have a the concept of the energy levels $E_j$ of the system which are simply defined as the sequence of distinct energy eigenvalues in the spectrum of the Hamiltonian. So, whereas one can have $\epsilon_n = \epsilon_m$ if $n\neq m$, one cannot have $E_n = E_m$ if $n\neq m$. Moreover, it is often convenient to label the energy levels in increasing index order so that $E_m < E_n$ whenever $m

The degeneracy $g_n$ of the energy level $E_n$ is defined as the number of distinct energy eigenvalues $\epsilon_m$ for which $\epsilon_m=E_n$. For simplicity, we assume that none of the levels is infinitely degenerate so that $g_n\geq 1$ is integer for all $n$.

The partition function of a system in the canonical ensemble is given by \begin{align} Z = \sum_n e^{-\beta\epsilon_n} \end{align} In other words, the sum is over the state labels, not over the energy levels. However, noting that whenever there is degeneracy, sum of the terms in the sum will be the same, we can rewrite the partition function as a sum over levels \begin{align} Z = \sum_{n} g_n e^{-\beta E_n} \end{align} The degeneracy factor is precisely what counts the number of terms in the sum that have the same energy.

As for a simple example, consider a system consisting of two, noninteracting one-dimensional quantum harmonic oscillators. The eigenstates of this system are $|n_1, n_2\rangle$ where $n_1,n_2 = 0, 1, 2, \dots$ and the corresponding energies are \begin{align} \epsilon_{n_1,n_2} = (n_1+n_2+1)\hbar\omega. \end{align} The canonical partition function is given by \begin{align} Z = \sum_{n_1,n_2=0}^\infty e^{-\beta \epsilon_{n_1, n_2}} = \underbrace{e^{-\beta\hbar\omega}}_{n_1=0,n_2=0} + \underbrace{e^{-\beta(2\hbar\omega)}}_{n_1=1,n_2=0} + \underbrace{e^{-\beta(2\hbar\omega)}}_{n_1=0,n_2=1} + \cdots \end{align} If you think about it for a moment, you'll notice that, in fact, the energy levels of this composite system are \begin{align} E_n = n\hbar\omega, \qquad n_1+n_2 = n \end{align} and that the degeneracy of the $n^\mathrm{th}$ energy level is \begin{align} g_n = n \end{align} so that the partition function can also be written in the form that uses energy levels and degeneracies as follows: \begin{align} Z = \sum_{n=1}^\infty g_n e^{-\beta E_n} = \underbrace{e^{-\beta\hbar\omega}}_{n=1} + \underbrace{2e^{-\beta(2\hbar\omega)}}_{n=2} + \underbrace{3e^{-\beta(3\hbar\omega)}}_{n=3} + \cdots \end{align}

fourier transform - Localized field quanta?

After the canonical quantization of the Klein-Gordon field (for example), we interpret the quantum of excitations of the fields with definite energy and momentum as particles. But our mental image of a particle is that of an entity which is localized at some point in space. But are they also localized in space?

However, in quantum mechanics a particle is represented by a wavefunction. If it's not in a position eigenstate (dirac-delta function), and no measurement of position has been made on it, as I understand, position is not a defining attribute of a particle in quantum mechanics. The best representation of a particle with resonably well-known position is described by a wave-packet peaked about some point is space. This is possible only when we have made sure that the particle is trapped in a small finite region.

I've read in Ryder's book that "quantum field theory is a theory of point particles." Are the field quanta always point particles by construction or only under some controlled situations they behave as point particles with definite locations? If yes, how can we write such a state mathematically? In a nutshell, are the field quanta necessarily localized in space?

Answer

To represent the states of a single particle in the $3$-space you should exploit the so called Newton-Wigner position representation.

The point is that the so obtained wavefunction, for a KG field, is a complex function defined in the rest space of the reference frame you are adopting and it has a strongly non-local behavior under the action of Lorentz transformation. This is different from what happens for quantum fields that locally transform under Lorentz transformations.

Position-defined states are again formally represented by Dirac deltas in that representation. If $\psi_{x_0}= \delta(x-x_0)$ is the wavefuntion of a KG particle in NW representation spatially localized at $x_0$ for $t=0$, the field configuration $\phi_{x_0}(t,x)$ associated to it is a solution of KG equation such that, for $t=0$, is localized around $x_0$ just up to the Compton length of the particle.

ADDENDUM

Let us focus on the momentum representation space for a real scalar KG particle with mass $m>0$. There are several, unitarily equivalent, possibilities to define that (one-particle) Hilbert space $\cal H$ used to build up the Fock space of the field. The most elementary one is thath referring to the standard measure $d^3p$ (instead of the covariant one $d^3p/E$).

In this representation ${\cal H}= L^2({\mathbb R}^3, d^3p)$ and states are represented by normalized wavefunctions $\psi =\psi(\vec{p})$ belonging to that functional space.

The momentum operator is therefore defined as (I adopt here the signature $-+++$ and $c=\hbar=1$): $$\left(P^i \psi\right)(\vec{p}):= p^i \psi(\vec{p})\:, \quad \left(P^0 \psi\right)(\vec{p}):= E(\vec{p})\psi(\vec{p})=\sqrt{\vec{p}^2 +m^2}\psi(\vec{p})\:,\quad i=1,2,3\:,$$ for those wavefunctions such that the right-hand sides still are elements of $L^2(\mathbb R^3, d^3p)$. These operators are automatically self-adjoint and the $\mathbb R^3$ in $L^2(\mathbb R^3, d^3p)$ turns out to coincide with the joint spectrum of $P_1,P_2,P_3$ as usual.

Let us come to the NW position operators and the corresponding representation of $\cal H$. In the Hilbert space $L^2(\mathbb R^3, d^3p)$, the three components of the three Newton-Wigner position operators are simply defined as the unique self-adjoint extensions of the symmetric operators:
$$\left(X^i\psi\right)(\vec{p}) := - i\frac{\partial}{\partial p^i}\psi(\vec{p})\qquad \psi \in C_0^\infty(\mathbb R^3)\:.$$ As usual, the (Newton-Wigner) position representation is the representation of the formalism in an unitarily equivalent Hilbert space where the $X^i$s are multiplicative. So the $L^2$ space is the one constructed upon joint spectrum of these operators: $L^2(\mathbb R^3, d^3x)$. In practice, also introducing time evolution, the wavefunctions in this space are related with those in the momentum representation by the usual Fourier-Plancherel transform. For $\psi \in L^1(\mathbb R^3, d^3p)\cap L^2(\mathbb R^3, d^3p)$: $$\phi_{NW}(t,\vec{x}):= \int_{\mathbb R^3} \frac{e^{i \vec{p}\cdot \vec{x}}}{(2\pi)^{3/2}} e^{-i tE(\vec{p})}\psi(\vec{p}) d^3p\:.$$ Conversely, the field representation is: $$\phi(t,\vec{x}):= \int_{\mathbb R^3} \frac{e^{i \vec{p}\cdot \vec{x}}}{(2\pi)^{3/2}\sqrt{2E(\vec{p})}} e^{-i tE(\vec{p})}\psi(\vec{p}) d^3p\:.$$ When acting on $\psi$ with the usual unitary action of the Lorentz group (henceforth $x= (t,\vec{x})$ and $p=(E(\vec{p}),\vec{p})$ are $4$-vectors and I define $E(p):= E(\vec{p})$), i.e.: $$\left(U_\Lambda\psi\right)(\vec{p}) = \sqrt{\frac{E(p)}{E(\Lambda^{-1} p )}}\psi\left( \vec{\Lambda^{-1} p}\right) \:,$$ the field representation transforms locally $$\phi \to \phi'\:,\quad \phi'(x):=\phi(\Lambda^{-1}x)\:,$$ whereas the transformation law of the Newton-Wigner representation, $\phi_{NW}$, is given in terms of a quite complicated integral formula and the action turns out to be highly non-local, as the transformed $\phi'_{NW}(x)$ also depends on the values of the functions $\phi_{NW}(x)$ at events different for $x$ itself.

Finally, consider a particle localized at $\vec{x}_0$ for $t=0$. Therefore, at least formally, we can write: $$\phi_{NW}(0,\vec{x}) = \delta(\vec{x}-\vec{x}_0)\:.$$ The covariant representation instead reads: $$\phi(0,\vec{x}):= \int_{\mathbb R^3} \frac{e^{i \vec{p}\cdot (\vec{x}-\vec{x}_0)}}{(2\pi)^{3}\sqrt{2E(\vec{p})}} d^3p\:.$$ It is possible to prove that the shape of that function is concentrated in a set including $\vec{x}_0$ with linear extension of the order of $1/m$, namely the Compton length of the particle.

There is a wide, quite technical, literature on these topics. Some results can be extended to QFT in curved spacetime for static spacetimes. An elementary discussion in that context can be found in Fulling's textbook on QFT in curved spacetime. A general discussion though confined to Minkowski spacetime, but even including fields with spin, appears in chapter 20 of Barut-Raczka's textbook on the theory of group representations. Several attempts to encompass a physically sensible notion of time observable exists in the literature especially making use of POVM.

cosmology - How is CMB related to the temperature of the universe

As I understand it, CMB (cosmic microwave background) is the radiation emitted when matter decoupled at the early stages of the big bang. The thing I don't understand is do all stars emit this kind of radiation or just the ones formed at the early stages of the big bang and are those stars travelling at the speed of light away from us also how is the fact that light is redshift related to the temperature of empty space.

Answer

Today, stars don't emit the CMB – they emit much more energetic (hotter, higher-frequency, shorter-wavelength) radiation in general. CMB was emitted by "everything" that could emit radiation at the relevant time, around 400,000 years after the Big Bang. At that time, the radiation was in thermal equilibrium with everything else, so its distribution to frequencies was fully described by the Planck black body curve.

The universe was expanding and the wavelengths were just being expanded linearly with the size of the universe. The temperature was dropping inversely proportionally to the size of the Universe. At some moment, the radiation got decoupled so it started to live its own life, with the temperature no longer linked to the temperature of the matter around. The temperature continued to drop inversely proportionally to the size of the universe while the temperature of the stars etc. stayed higher and independent because the two subsystems no longer interacted.

None of these things has anything to do with the speed of divergence of mutually distant galaxies. The temperature is a local quantity and it's determined by the local physics while the mutual speed of distant galaxies is not a local effect. It is a global effect and an illusion of a sort, too. The increasing distance between the galaxies is due to the expansion of the space in between them. However, all the calculations of the temperature and its evolution may be applied locally at each place of the universe, independently of distant places of the universe, and this calculation of the temperature and its evolution is moreover the same at every place because the universe is uniform at cosmological scales.

electromagnetism - Does there exist electric field around all the substances?

• A system of two equal and opposite charges separated by a certain distance is called an electric dipole.


• Electric dipole moment ($p$) is defined as the product of either charge ($q$) and the length ($2a$)of the electric dipole.



• Magnitude of electric field ($E$) due to an electric dipole at a distance $r$ from its centre in a direction making an angle $\theta$ with the dipole is given by the equation,$$E=\frac{1}{4\pi\epsilon}.\frac{p\sqrt{3\cos^2\theta+1}}{r^3}$$
  where, $p=2aq$ ($2a$ is the distance of separation of the charges $q$).

From the above equation, I thought that, electric field around a neutral body won't be zero. Electric field around a neutral body would be zero if and only if the distance of separation between the dipole charges is zero (from the equation, we can notice that $E$ tends to zero as $p$ tends to zero or $2a$ tends to zero). So, even if a body is neutral, I thought that electric field need not be zero.

I got a doubt here. We know that, in a neutral atom, electron and proton have equal and opposite charges, and even they are separated by a certain distance. So, I assumed a pair of electron and proton to behave as a dipole, so that I could fit them to the above dipole equation. I thought, as the distance between electron and proton is not zero, there must exist electric field around them. As a result, we can expect electric field around all the atoms, thus producing a vector added field around the substance. But, we don't feel electric field around all the substance. Is it that electric field around the substances negligible? I don't know whether I am correct or wrong, or whether I have misunderstood the concept. If any is the case, please explain.

[My book mentions one related example: "The molecules of water, ammonia, etc behave as electric dipoles. It is because, the centres of positive and negative charges in these molecules lie at a small distance from each other" It is to be noted that, molecules are considered here and atoms are neglected.]

quantum field theory - Why the Hamiltonian and the Lagrangian are used interchangeably in QFT perturbation calculations

Whenever one needs to calculate correlation functions in QFT using perturbations one encounters the following expression:

$\langle 0| some\ operators \times \exp(iS_{(t)}) |0\rangle$

where, depending on the textbook S is either (up to a sign)

1. 
   

   $\int \mathcal{L}dt$ where $\mathcal{L}$ is the interaction Lagrangian

   
   

   or

   
   


2. 
   
   

   $-\int \mathcal{H}dt$ where $\mathcal{H}$ is the interaction Hamiltonian.

   
   

It is straightforward to prove that if you do not have time derivatives in the interaction terms these two expressions are equivalent. However these expressions are derived through different approaches and I can not explain from first principles why (and when) are they giving the same answer.

Result 1 comes from the path-integral approach where we start with a Lagrangian and do perturbation with respect to the action which is the integral of the Lagrangian. Roughly, the exponential is the probability amplitude of the trajectory.

Result 2 comes from the approach taught in QFT 101: Starting from the Schrödinger equation, we guess relativistic generalizations (Dirac and Klein-Gordon) and we guess the commutation relations to be used for second quantization. Then we proceed to the usual perturbation theory in the interaction picture. Roughly, the exponential is the time evolution operator.

Why and when are the results the same? Why and when the probability amplitude from the path integral approach is roughly the same thing as the time evolution operator?

Or restated one more time: Why the point of view where the exponential is a probability amplitude and the point of view where the exponential is the evolution operator give the same results?

Answer

Starting from the Hamiltonian formulation of QM one can derive the path-integral formalism (see chapter 9 in Weinberg's QFT volume 1), where the Hamiltonian action is found to be proportional to $\int \mathrm{d}t (pv - H)$.

For a subclass of theories with "a Hamiltonian that is quadratic in the momenta" (see section "9.3 Lagrangian Version of the Path-Integral formula" in above textbook), the term $(pv - H)$ can be transformed into a Lagrangian $L_H = (pv - H)$. Then the Lagrangian action is proportional to $\int \mathrm{d}t L_H$. Both actions give the same results because one is exactly equivalent to (and derived from) the other.

$$ \int \mathrm{d}t (pv - H) = \int \mathrm{d}t L_H$$

Moreover, when working in the interaction representation you do not use the total Hamiltonian but only the interaction. The derivation of the Hamiltonian action is the same, except that now the total Hamiltonian is substituted by the interaction Hamiltonian $V$. Again you have two equivalent forms of write the action either in Hamiltonian or Lagrangian form.

If you consider Hamiltonians whose interaction $V$ does not depend on the momenta, then the $pv$ term vanishes and the above equivalence between the actions reduces to

$$ - \int \mathrm{d}t V = \int \mathrm{d}t L_V$$

where, evidently, the interaction Lagrangian is $L_V = -V$

This is what happens for instance in QED, where the interaction $V$ depends on both position and Dirac $\alpha$ but not on momenta.

Note: There is a sign mistake in your post. I cannot edit because is less than 10 characters and I have noticed the mistake in a comment to you above, but it remains.

classical mechanics - What will happen if a plane trys to take off whilst on a treadmill?

So this has puzzled me for many a year... I still am no closer to coming to a conclusion, after many arguments that is. I don't think it can, others 100% think it will.

If you have a plane trying to take off whilst on a tread mill which will run at the same speed as whatever the planes tires rotation speed is will it take off?

[edited to be more clear]

The question is simple. Will a plane take off if you put this plane onto a treadmill that will match whatever speed the plane wheels are moving at. So the plane should not be able to move.

This is a hypothetical situation of course. But I am very interested.

Answer

Idealizing the plane's wheels as frictionless, the thrust from the propeller accelerates the plane through the air regardless of the treadmill. The thrust comes from the prop, and the wheels, being frictionless, do not hold the plane back in any way.

If the treadmill is too short, the plane just runs of the end of it and then continues rolling towards take off.

If the treadmill is long enough for a normal takeoff roll, the plane accelerates through the air and rotates off of the treadmill.

UPDATE: Don't take Alfred's word for it. Mythbusters has actually done the experiment.

UPDATE 2: I've been thinking about how the problem is posed (for now as I'm typing this) and it occurred to me that the constraint "run at the same speed as whatever the planes tyres rotation speed" actually means run such that the plane doesn't move with respect to the ground.

Consider a wheel of radius $R$ on a treadmill. The treadmill surface has a linear speed $v_T$ to the right. The center of the wheel has a linear speed $v_P$ to the left. The CCW angular speed of the wheel is:

$\omega = \dfrac{v_T + v_P}{R}$

If "run at the same speed as whatever the planes tyres rotation speed" means:

$\omega = \dfrac{v_T}{R}$

then the constraint requires $v_P = 0$. That is, the question, as posed, is:

If the treadmill is run such that the plane doesn't move, will the plane take off?

Obviously, the answer is no. The plane must move to take off. Looking at mwengler's long answer, we see what is happening. The rotational speed of the tires and treadmill are not the key, it is the acceleration of the treadmill that imparts a force on the wheel axles (ignoring friction for simplicity here).

So, it is in fact the case that it is possible, in principle, ( don't think it is possible in practice though) to control the treadmill in such a way that it imparts a holding force on the plane, preventing it from moving. But, once again, this force is not proportional to the wheels rotational speed, but to the wheel's angular acceleration (note that in the idealized case of massless wheels, it isn't even possible in principle as the lower the moment of inertia of the wheels, the greater the required angular acceleration).

particle physics - What is the significance of the QCD scale parameter $Lambda$?

I see that it appears as a constant in the relation for the running of the strong coupling constant. What is its significance? Does it have to be established by experiment? Is it somehow a scale for quark confinement? If yes, how? I ask because I saw this in Perkins' Particle Astrophysics

After kT fell below the strong quantum chromodynamics (QCD) scale parameter ∼ 200 MeV, the remaining quarks, antiquarks, and gluons would no longer exist as separate components of a plasma but as quark bound states, forming the lighter hadrons such as pions and nucleons.

Answer

Dear dbrane, $\Lambda_{\rm QCD}$ is the only dimensionful parameter of pure QCD (pure means without extra matter).

It is dimensionful and replaces the dimensionless parameter $g_{\rm QCD}$, the QCD coupling constant. The process in which a dimensionless constant such as $g$ is replaced by a dimensionful one such as $\Lambda$ is called the dimensional transmutation:

http://en.wikipedia.org/wiki/Dimensional_transmutation

The constant $g$ isn't quite constant but it depends on the characteristic energy scale of the processes - essentially logarithmically. Morally speaking, $$ \frac{1}{g^2(E)} = \frac{1}{g^2(E_0)} + K \cdot \ln (E/E_0), $$ at least in the leading approximation. Because $g$ depends on the scale, it is pretty much true that every value of $g$ is realized for some value of the energy scale $E$. Instead of talking about the values of $g$ for many specific values of $E$, one may talk about the value of $E$ where $g$ gets as big as one or so, and this value of $E$ is known as $\Lambda_{\rm QCD}$ although one must be a bit more careful to define it so that it is 150 MeV and not twice as much, for example.

Yes, it is the characteristic scale of confinement and all other typical processes of pure QCD - those that don't depend on the current quark masses etc. In most sentences about the QCD scale, including your quote, the detailed numerical constant is not too important and the sentences are valid as order-of-magnitude estimates. However, given a proper definition, the exact value of $\Lambda_{\rm QCD}$ may be experimentally determined. With this knowledge and given the known Lagrangian of QCD - and the methods to calculate its quantum effects - one may reconstruct the full function $g(E)$.

Magnetic poles in Halbach array?

I am a bit confused by the description of Halbach arrays. It is said that the line of magnets aligned in certain way results in cancellation of magnetic field on one side of the array, and amplification on the other side. And there is a schematic distribution of magnetic flux -

I don't understand what happens to the poles of magnets, or the direction of magnetic flux? Does it change its direction in each next or third element of array? Or is it possible to make something like a magnetic monopole, so that the direcition of magnetic flux will be uniform on one side of the array?

There are some other images showing different arrays with arrows which, I suppose, represent the direction of the magnetic flux.

Diagram 1 Diagram 2

The Diagram 2 looks like a monopole, because there is the same direction of flux all the time? Is it true, or am I misunderstanding something?

Answer

No, the Halbach array is not related to the monopole at all. It is just an special arrangement of magnet.

Any arrangement of a "dipole" magnet would not result in any monopole. It can be easily understood as follows: If each individual magnet contains no magnetic monopole, i.e., $\nabla\cdot\mathbf{B}_i=0$ true for whole space, then the resulting magnetic field is $\mathbf{B} = \mathbf{B}_1 + \mathbf{B}_2$ when you put them together. Hence the resulting divergence is also $\nabla\cdot\mathbf{B}=0$.

I don't understand what happens to the poles of magnets, or the direction of magnetic flux? Does it change its direction in each next or third element of array?

You should know that magnetic pole is not a very well defined concept as the "boundary" of the magnetic pole is not fixed. The flux line does change the direction across the boundary. But the most important things (in the first figure) is that the magnetic field line is continuous and closed, which means that there is no magnetic monopole.

All the field lines go out have to come back at the next nearest neighbor. The effect of Halbach array is simply to push the closed magnetic field line on one side, hence the resulting magnetic field was enhanced on one side.

The Diagram 2 looks like a monopole, because there is the same direction of flux all the time? Is it true, or am I misunderstanding something?

No, the flux direction is not in the same direction over the whole region, they are alternating instead. You can think of it by putting two array in diagram one (flipped one of them) together. Each of them forming a small field line loop. At the center of two up arrows ($\uparrow$) the magnetic field is pointing up. Also, at the center of two down arrows ($\downarrow$), the magnetic field line is pointing down. Hence, it is clear that the field line is alternating after each next nearest neighbor. The wiggler of the electron path is the result of the alternating magnetic field.

I think there is a problem for the Diagram 2, the oscillating frequency is double of what it is supposed to be. It should be the same as the period of the array.

spacetime - When it is said that the Higgs field is a scalar, do they mean Lorentz scalar?

We often hear that Higgs boson is a scalar boson, and that Higgs field is a scalar field. I was always thinking that this means "4-scalar". In other words, it is space-time invariant, .i.e. it's properties are not changing if move relative to it. Unlike photon which changes it's frequency, polarization and other properties, if we look at it from moving frame, Higgs is always the same.

Is above correct? If not then what is correct and what is not?

Answer

You are right, in this case, scalar means Lorentz invariant field. But it is not invariant under the transformations of SU(2)xU(1) of the electroweak model. And it is a scalar under the SU(3) of QCD.

So the four real components of the Higgs are indeed invariant under space-time transformations.

Physicists are usually not very clear in these distinctions, and you have to guess under which transformation the field is a scalar.

cosmology - Could the universe's antimatter be hiding in black holes?

As far as we know almost all the mass in the universe is matter, not antimatter. There are three scenarios:

• The universe started off with more matter than antimatter and the small amount of antimatter has mostly been converted to energy via collisions with matter


• The universe started off with equal amounts of matter and antimatter and some unknown process caused the situation we see today


• The universe started off with less matter than antimatter and somehow the ratio reversed (I've never heard this discussed as it seems so unlikely)

We reason that mass is matter because if there were large amounts of antimatter they should produce antimatter galaxies, which would give off huge amounts of energy when they interact with matter galaxies. This has never been observed so we conclude that all galaxies, therefore all mass in the universe is matter.

As I understand it there are two main types of black holes, stellar remnants (the collapsed remnants of dead stars with more than two solar masses) and super-massive black holes at the centre of many galaxies, including one thought to be at the centre of our own galaxy.

Stellar black holes are obviously originally composed of matter, for the same reason that all galaxies are composed of matter.

We don't know how super-massive black holes formed. Is there enough mass in them to contain all the antimatter in the universe if matter and antimatter were originally created in equal quantities? I realise this shifts the unknown mechanism from "why was more matter created than antimatter" to "why is antimatter more likely to form super-massive black holes" but is this hypothetical scenario possible?

quantum mechanics - One particle states in an interacting theory

Question:

What is the general definition of one particle states $|\vec p\rangle$ in an interacting QFT? By general I mean non-perturbative and non-asymptotic.

---

Context. 1)

For example, in Weigand's notes, page 42, we can read (edited by me):

As $[H,\vec P]=0$, these operators can be diagonalised simultaneously. Write the eigenstates as $$\begin{aligned} H|\lambda(\vec p)\rangle&=E(\lambda)|\lambda(\vec p)\rangle\\ \vec P|\lambda(\vec p)\rangle&=\vec p|\lambda(\vec p)\rangle \end{aligned}\tag{1}$$ One particle states are those with $E(\vec p)=\sqrt{\vec p^2+m^2}$.

For me, this is on the right track as a general definition, because we are defining $|\vec p\rangle$ as the eigenstates of operators that commute, which is fine. But there are some issues with this definition:

• 
  

  Why is $E$ a function of $\vec p$? why is $E$ not an independent label on $|\vec p\rangle$? Just as $p_x,p_y,p_z$ are independent, why is $p_0$ not independent as well$^{[1]}$?

  
  
  


• 
  

  Once we know that $E=E(\vec p)$, how do we know that $E^2-\vec p^2$ is a constant? (we define mass as that constant, but only once we know that it is actually a constant). By covariance, I can see that $p^2$ has to be a scalar, but I'm not sure what's the reason that this scalar has to be independent of $p$ (for example, in principle we could have $p^2=m^2+p^4/m^2$).

  
  


• 
  

  How do we know that these $|\vec p\rangle$ are non-degenerate? Why couldn't we have, for example, $H|\vec p,n\rangle=E^n(\vec p)|\vec p,n\rangle$, with $E^n\sim \alpha/n^2$, as in the hydrogen atom?

  
  

These properties are essential to prove, for example, the Källén-Lehmann spectral representation, which is itself a non-perturbative, non-asymptotic result.

Context. 2)

In Srednicki's book, page 53, we can read (edited by me):

Let us now consider $\langle p|\phi(x)|0\rangle$, where $|p\rangle$ is a one-particle state with four-momentum $p$. Using $\phi(x)=\mathrm e^{iPx}\phi(0)\mathrm e^{-iPx}$, we can see that $$ \langle p|\phi(x)|0\rangle=\mathrm e^{ipx} \tag{2} $$

This only works if $P^\mu|p\rangle=p^\mu|p\rangle$, which we could take as the definition of $|p\rangle$. Yet again, I don't know why we would expect $p^2$ to be a constant (that we could identify with mass). Or why are the eigenstates of $P^\mu$ non-degenerate.

The expression $(2)$ is essential to scattering theory, but the result itself is supposedly non-asymptotic (right?). Also, it fixes the residue of the two-point function, which is one of the renormalisation conditions on the self-energy ($\Pi'(m^2)=0$). This means that $(2)$ is a very important result, so I'd like to have a formal definition of $|p\rangle$.

---

$^{[1]}$ in standard non-relativistic QM, $E$ is a function of $\vec p$ because $H$ is a function of $\vec P$. But in QFT, $H$ is not a function of $\vec P$, it is a different operator.

Answer

I'll address your issues with definition (1):

$E$ is a function of $\vec p$ because $\lvert \lambda_{\vec p}\rangle\sim\lvert \lambda_0\rangle$ where by $\sim$ I mean that they are related by a Lorentz boost. That is, to "construct" these states, you actually first sort out all the states $\lvert \lambda_0\rangle,\lambda_0\in \Lambda$ ($\Lambda$ now denotes the index set from which the $\lambda_0$ are drawn) with $\vec P \lvert \lambda_0\rangle = 0$ and $H\lvert \lambda_0\rangle = E_0(\lambda)\lvert\lambda_0\rangle$ and then you obtain the state $\lambda_{\vec p}$ by applying the Lorentz boost associated to $\vec v = \vec p/E_0(\lambda)$ to $\lvert\lambda_0\rangle$. Since $H$ and $\vec P$ area components of the same four-vector, that means that $E_{\vec p}(\lambda)$ is a function of $E_0(\lambda)$ and $\vec p$ - and $E_0(\lambda)$ is determined by $\lambda$, so $E_{\vec p}(\lambda)$ is really a function of $\vec p$ and $\lambda$.

Conversely, every state of momentum $\vec p$ can be boosted to a state with zero momentum. We examine this zero-momentum state for what $\lambda_0$ it is and then name the state we started with $\lambda_p$, so this really constructs all states.

From the above it also directly follows that $E_{\vec p}(\lambda)^2-\vec p^2 = E_0(\lambda)^2$. I'm not sure what your issue with degeneracy is - degeneracy in $P$ and $H$ is not forbidden - nowhere is imposed that $E_{\vec p}(\lambda) \neq E_{\vec p}(\lambda')$ should hold for $\lambda\neq\lambda'$.

electromagnetism - At an atomic level, what happens when you connect two batteries in series so that their voltages are added?

I can't for the life of me figure this out. I feel like i'm missing some crucial detail about how batteries work.

Imagine two batteries connected in series, like this:

Circuit <= -(Battery A)+ <= -(Battery B)+ <= Circuit

As far as I've studied, this is what happens:

1. The negative side of Battery B has a surplus of electrons, while the positive side of Battery A has a "normal" concentration of electrons. Since electrons repel each other, they kinda push themselves to Battery A. (I am imagining a tube with one side filled with "elastic" balls tightly packed and the other side with elastic balls but not pressed together. The balls get pushed until they are all at an equal "pressure").


2. A number of electrons constantly reach the positive end of Battery A with a "force" of x Volts. In practical terms, the electrons are colliding with the battery with a higher intensity. For some reason, this causes the electrons coming out of the negative end of Battery A to have a "force" of 2x Volts. I don't understand why.

Is it because there is now two times more free electrons at the negative end of Battery A? And thus the difference between this end and Battery B is two times the normal?

3. The electrons get pushed again to the positive end of Battery B. From what I understand, the material in the battery "absorbs" them and releases a free electron on the negative end of Battery B.

Please avoid using water analogies. I specifically want to know what happens at the atomic, per-electron level.

I don't care much about the differences between each battery. If possible, pick a battery type you prefer or just talk about batteries in general (characteristics shared by all batteries).

Answer

Chemistry and Physics of Batteries in Series

What happens at the atomic level inside a cell when putting two batteries in series?

Short Answer:

Characteristics of a single cell vs. two cells in series: (i.e., compare the voltage, current, $E$ field, stored energy, power, charge, and operating time in a circuit with the same resistive load.):

• Twice the voltage, due to the summation of voltages with two cells in series (i.e., twice the $E$ field acting on a resistive load through the conductor).
  


• Twice the current with the same load, due to the doubled voltage.
  


• Twice the reaction rate at both the anode and cathode in all cells in series.


• Twice the rate of $H^+$ ion migration from the anode to cathode to maintain charge neutrality at both anode and cathode.
  


• Twice the rate of battery charge depletion. When the current doubles and the charge capacity of each cell is unchanged, the duration of flow is reduced to half.
  


• Within the circuit with series batteries: the same current flows through all the current elements. (Electron current flows through the circuit elements: originating in the outer anode, travels through the conductor, load, outer cathode, inner anode, and finally neutralized at by the inner cathode reaction).


• Within the electrolytes of the cells in series, the magnitude of ion flow rate is the same as the current flow rate through the electrolytes of both cells, ($cell_1$ and $cell_2$).
  

What force drives current through the electrolyte and conductor between the cells in series?

• Ultimately, the sum of the battery cell voltages drives current through the cells, but its action is indirect. The terminal battery voltages drive current through the load, but ion migration and reaction potentials are responsible for driving current within the batteries.


• The doubled voltage of two batteries in series drive twice the current through the load.


• This increased current draws more electrons off the outer anode and conducts it through the load to the outer cathode. (Note: the outer anode and outer cathode refer to the external electrodes of the two cells in series. And obviously, the terms "inner anode" and "inner cathode" refer to the terminals connected between batteries in series. .)


• The increased rate of removal of electrons from the outer anode, and delivery to the outer cathode, disturbs the equilibrium state of the reactions in both the outer anode and outer cathode, resulting in the production of more electrons by the anode reaction, and the consumption of more electrons by the cathode reaction.


• The disturbance of equilibrium state of the outer anode and outer cathode is communicated to the inner anode and inner cathode of the series cells by charge accumulation around the terminals. The charge attraction between ions in the electrolyte results in an ion migration between terminals. The increased current and ion flow results in the adjustment of the rate of reaction. Eventually, the current transient dampens, and the same current flows through the whole circuit (conductors, load, and cells in series), thereby meeting and matching the current flow through the load.

Long Answer:

Battery Chemistry Example: (single cell)

• Circuit: resistive load, one cell, Lead-Acid battery, open circuit.


• The voltage of a single cell is $2.05V = 1.60V_{anode} + .36V_{cathode}$


• The anode half-cell oxidation reaction is: $Pb + SO_4^{2-} \to PbSO_4 + 2e^-$


• The cathode half-cell reduction reaction is: $PbO_2 + SO_4^{2-} + 2e^- + 4H^+ \to PbSO_4 + H_2O$


• Anode electrolyte Reaction: $H^+$ and $SO_4^{2-}$ ions migrate as needed to maintain charge neutrality. $SO_4^{2-}$ ions are removed from the electrolyte by the anode reaction with $Pb$. Two electrons are left on the anode, resulting in a net positive charge in the electrolyte around the anode. The excess $H^+$ ions migrate toward the cathode, where $O_2^{2-}$ ions react with $H^+$ ions to produce $H_2O$, which neutralizes the net charge in the electrolyte.
  


• Why does the anode reaction stop in an open circuit? The anode reaction spontaneously proceeds and liberates electrons, which accumulate on the surface of the anode. $H^+$ ions in solution are attracted to the electrons on the anode, effectively creating a net positively charged ion layer covering the entire anode. This positive layer of $H^+$ ions attracts $SO_4^{2-}$ ions from the electrolyte, but they cannot penetrate the $H^+$ ion layer to react with the $Pb$. When the battery circuit is closed, current flows and the excess electrons on the plate move to the cathode, the $H^+$ ion layer disperses, and the $SO_4^{2-}$ ions migrate to the $Pb$ plate and react.


• Cathode electrolyte reaction: the overall reaction is: $4H^+ O^{2-} \to 2H_2O$. This reaction neutralizes, a) $2H^+$ from the anode, b) $2H^+$ from the cathode, with c) $O^{2-}$ from the cathode, d) resulting in H2O.


• Electrolyte current: Electrons flow from anode to cathode through a conductor, but there is also charge movement in the electrolyte by ion migration where $H^+$ ions migrate from anode to cathode. A big picture view of battery charge movement is 1) electrons move from anode to cathode through a conductor, and 2) those electrons combine with $H^+$ ions at the cathode. The battery maintains overall net electric charge neutrality, but internally the anode and cathode are reservoirs of separated positive and negative charge, both on the terminals and in the electrolyte. The charge differential between terminals produces the associated $E$ field potential energy gradient, which is utilized to perform work. The battery converts the electrostatic-bond energy of one atomic/molecular species into another lower energy species. The energy is converted into an $E$ field by creating a charge separation scenario, which produces electric charge neutrality in the electrolyte. However, the reaction sequence is more complex than that. The cathode releases $O^{2-}$ from the $PbO_2 + SO_4^{2-} + 2e^-$ reaction, the $H^+$ ions react with the $O^{2-}$ and convert it to water. If the $H^+$ had reacted with a free electron at the cathode, it would have formed $H_2$ gas.



• Series Cell Electrostatics and Electrochemistry: The same chemical reactions occur in the individual cells whether in a circuit as a single cell or in series. The difference is the amount of current drawn by the series batteries vs. the single cell through the same resistive load. An electrical potential is generated by the half-cell reactions composing a battery.
  The anode and cathode reactions generate an open circuit potential, which results in a current flow proportional to the voltage when the circuit is closed. When cells are placed in series, the voltage of the two cells adds, resulting in a doubled $E$ field in the space surrounding the batteries. As a result, the higher voltage across the load draws twice the current. This current/rate of electron removal doubles the rate of reaction in the cells. Thus, because the voltage is (essentially) constant from a single battery the controlling factor in current flow is the load (lower resistance, higher current). But, if batteries were placed in series and the load was kept constant, the current would double and the reaction rate would double because the voltage had doubled.

Single Cell Example: Lead-Acid battery, $10$ amp-hr charge

• Load $= R = 4.1 \Omega$


• Voltage $= V = 2.05V$


• Total Stored Charge = $q_{total} = 10amp-hr =10coul/sec \times 3600 sec/hr \times 1hr= 36000 C$


• Current flow $=I = V/R = 2.05V/4.1 \Omega = .5A$


• Operating time = $q_{total}/I = 36000/.5 = 72000 sec = 72000/3600 = 20 hrs$



• Power consumption: $P=IV = .5A \times 2.05V = 1.025W$


• Total Energy Storage: $\Delta E=P \times \Delta t = 1.025W \times 20hr = 20.5 Wh$


• Total Energy storage: $\Delta E = IV \times \Delta t = V \times (I \cdot \Delta t) = 2.05V \times 10Ah = 20.5Wh$

Series Cell Example: Lead-Acid battery, $10AH$/cell, same load

• Load = R = $4.1 \Omega$ (i.e. same circuit, but with two cells in series)


• Voltage = V = $2.05V + 2.05V = 4.1V$


• Total Charge Storage = $q_{total} = 10$ amp-hr $= 36,000 amp-sec= 36,000 coulombs$ (Note: Full discharge after delivering 36000 coulombs of charge. Both batteries have stored 36,000 coulombs each, but the discharge of one cell passes its electrons from the anode of one the the cathode of the second. Thus, the discharge of 36K coulombs depletes both batteries at the same time.)


• Current flow: $I = V/R = 4.1V/4.1 \Omega = 1A$



• Operating time = $q_{total}/I = 36000/1 = 36000/3600 = 10Hr$


• Power delivery/consumption: $P=IV = 1A \times 4.1V = 4.1W$


• Total Energy Storage: $\Delta E=P \times \Delta t = 4.1V \times 10Ah = 41.0Wh$


• Total Energy Storage: $\Delta E = IV \times \Delta t = V \times (I \cdot \Delta t) = 4.1V \times 10Ah = 41.0Wh$

Battery Principles:

Conservation of Charge: The above single battery contains 10Amp-hours of charge, which equals 36000 coulombs. This is the total amount of charge that the battery can deliver as current. Batteries in series, both deplete their electron stores at the same time because all current flows through both batteries. In a closed circuit, at the anode, the reactants react and generate excess electrons. At the cathode, the reactants react with electrons to create products. This proceeds spontaneously and generates a state of electron deficiency at the cathode. The reaction cannot proceed without an outside source of electrons. When electrons conduct from the anode to the cathode, they satisfy the cathode's need for electrons, allowing the cathode reaction to proceed as fast as the electrons are supplied (with limits at upper limits that compete with the maximum reaction rate). When all the reactants at the anode or cathode have reacted to become products is spent, as the battery can no longer supply electrons to, or accept electrons from, the load. The increased current from a series battery source depletes the battery's charge-store twice as fast. The benefit of a higher voltage is a hotter resistor, brighter incandescent lamp, faster DC motor...

Conservation of Energy The total energy available to deliver by cells in series is the sum of the stored energy of the individual cells. As per the above examples: a single cell with 10Amp-hours of charge has $2.05V \times 10Ah = 20.5Wh$ of stored electrical potential energy. Two identical cells in series have $41Wh$ of stored electrical potential energy. Comparing the two circuits: a series battery doubles the voltage and current, resulting in 4 times the power consumption. Thus, even though a circuit with two cells in series holds two times the stored energy, the power consumption increases by four times, resulting in half the battery life. Conversion into any other form of energy satisfies the principle of energy conservation.

Energy Conversion: As a battery discharges, it converts electrical potential energy into various types of energy in the load. Common conversions of electrical energy include heat, kinetic energy, or gravitational, electrical, and magnetic potential energy. Examples of the types of circuit loads include a resistor (thermal energy), motor (kinetic energy, gravitational potential, electrical potential, magnetic potential), an inductor/electromagnet ($B$ field potential energy), or a capacitor ($E$ field potential energy).

Battery Terminal Reactions: In the Lead-Acid battery, before reaction, energy is stored as potential energy by the charge attraction between the chemical reactants at each of the terminals.

• Anode $e^-$ liberation: $Pb_{(s)}$ and $SO_4{^{2-}}_{(aq)}$ are in metal-electrolyte contact, and will spontaneously react to form $PbS{O_4}_{(s)}$ and $2$ extra electrons. The positive oxidation potential of this half-cell reaction is $+1.69V$. The positive potential indicates the reaction proceeds spontaneously at room temperature. The reactants attract and react because of electrostatic forces, such as energetically favorable orbital bonding.
  


• Anode $H^+$ Shielding: After $Pb$ and $SO_4^{2-}$ react, the liberated electrons accumulate on the anode, creating a negative $E$ field. The surface electrons attract a layer of $H^+$ ions. This stops the anode $Pb + SO_4^{2-}$ reaction very quickly. The $H^+$ ion layer bonds with the $SO_4^{2-}$ ions preventing them from migrating close enough to the $Pb$ to react. For the reaction to proceed on the anode, the electrons must be removed. When a conductor connects anode and cathode, electrons flow from a place of excess to a place of deficiency. As the electrons leave the anode, the anode reaction resumes.


• Anode Energy Transfer Sequence: All energy begins as reactant bond-energy and is converted into another form of energy by the load:
  1) Batteries store energy as the bond energy of reactants. The reactants have more bonding energy than the products, so breaking the stronger bonds and remaking weaker bonds mobilizes energy. That energy can be used to convert into other types of energy. An electrostatic force between reactants draws them to react.
  2) In an open circuit battery, a small number of the anode half-cell reactions proceed spontaneously to completion at any moment, resulting in the production of excess electrons, which give the battery its characteristic voltage (in combination with its other half-reaction at the cathode.
  3) At the anode, the energy differential between the old bonds within the reactants and new bonds within the products is converted to a concentration of free electrons and an associated $E$ field.
  4) The anode's $E$ field adds to the cathode's $E$ field (with its electron deficiency), creating a net total $E$ field which permeates the space around the battery. The $E$ field drives electrons as a current through the high permittivity conductor to a load.
  5) Passage of current through a load converts current into another form of energy.
  


• Cathode Reaction: the equivalent but opposite process of reaction proceeds at the cathode (i.e., reduction of reactants). The free energy is positive for the bonding of $PbO_2 + SO_4^{2-} + H^+ + 2e^-$ to produce $PbSO_4 + H_2O$, which means this reaction will proceed spontaneously at STP. The reactants require electrons to proceed, and it does so to a small extent, creating a positive $E$ field since the cathode is now electron deficient after the reactant conversion to products consumed electrons. Note, this reaction requires electrons to complete. Even when the circuit is open, a small number of reactants scavenge loosely bound electrons from other atoms or ions. Reaction sequesters those electrons in neutral end products, which results in an electron deficit, a net positive charge on the cathode, and an associated positive $E$ field. The electron deficiency at the cathode is the other half of the battery, which creates a net positive charge and adds to the $E$ field created by the negative charge on the anode.
  

Current Flow through the Series Batteries: Batteries in series increase the voltage and current through the load. The current passing through the load also passes through every cell in series. Each cell contributes an $E$ field to the total $E$ field of the batteries in series. And, the sum of the individual $E$ fields drives the current through the load. The outer anode and cathode initially supply the electron source and sink. But, after a transient period, the rate of reaction at every anode and cathode in series equilibrates, and the same current flows through all the cells in series.

Ion migration in the electrolyte of a Lead-Acid battery: In the Lead-Acid battery, both anode and cathode reactions produce the same neutral compound, $PbSO_4$. In the process, both anode and cathode consume $SO_4^{2-}$ ions from solution. At the anode, the reaction leaves a net positive charge in the electrolyte, which is carried by the $H^+$ ions.
- At the cathode, the consumption/bonding of $SO_4^{2-}$ with $PbO_2$ results in the accumulation of a negative charge in the electrolyte. This is because the reaction of $PbO_2$ with $SO_4^{2-}$ and $2e^-$ results in the liberation of $O_2^{2-}$. The negative charge in the electrolyte around the cathode attracts the $H^+$ ions, from the anode, resulting in the reaction of $4H^+$ ions with one $O_2^{2-}$ ion to produce $2H_2O$. Thus, in lead-acid batteries, charge neutrality in the electrolyte solution surrounding the solution is maintained by the migration of unpaired $H^+$ ions from the anode to cathode.

Battery as Capacitor: The battery acts as a capacitor when in an open circuit. Opposite charges populate the anode and cathode, and this charge differential is separated by a thin dielectric layer of $H^+$ and $SO_4^{2-}$ ions surrounding the anode. This dielectric layer disperses when current flows through a conductor from the anode to the cathode. Conduction removes the excess electrons from the anode, thereby breaking the ionic bond between the $H^+$ and $SO_4^{2-}$ ions. Dispersal of the $H^+$ layer around the anode enables $SO_4^{2-}$ ions to flow past the previously impermeable $H^+$ ion layer.

• Difference between a battery and capacitor: The comparison between a capacitor and battery is incomplete. When in an open circuit, the two behave as charge-storing devices, with both storing charge on either side of a dielectric. The battery has two modes: 1) open circuit: the capacitor and battery configuration are comparable, and 2) closed circuit, the dielectric disperses, and the battery becomes a charge differential generator, using bond-energy liberation from reactants to create products with a differential concentration of charge.
  


• Capacitive, Field, and Power Flow effects: in the air around a battery and conductor. The open circuit voltage across a battery's terminals acts as an air dielectric capacitor. The charge between terminals polarizes the dielectric of the air/space, producing a small displacement current until charged. However, a conductor is also a capacitor (in the most general sense of the word) since a conductor has a dielectric constant. Placing a conductor between a battery' terminals produces a large displacement current. The Electric Displacement Field is a capacitive effect, causing polarization of the metallic dielectric. The current flowing through the conductor is an attempt by the battery voltage to charge the dielectric of the metal. Since the dielectric constant is so large, the Electric Displacement Field is essentially unlimited. Therefore, the current in the circuit is limited by the resistance of the load, rather than the electric permittivity of the conductor.
  


• Electric Permittivity and Conductivity: The conductor provides a high electric permittivity $\epsilon_{conductor}$ path for the $E$ field to act. The equivalent circuit to the conductor is a huge capacitor in series with a tiny resistor. The massive electric permittivity $\epsilon_{conductor}$ of the metallic conductor provides insignificant capacitive impedance. The current flow of a conductor is not similar to the high-velocity trajectory of particle accelerator. Rather, current flow through a conductor is like an atomic executive pendulum, with the microscopic motion of individual electrons at the anode transmitting the repulsive force of their incremental change in proximity from electron to electron from the anode to the cathode. The conductor is a capacitor with a dielectric constant so high that the size of the capacitor is infinite (for practical purposes), as the capacitive effects of a conductor are significant only at high frequencies. At low frequencies/DC, the equivalent size of the conductor-capacitor is so large that it will never fully charge.


• Electric Displacement Field $D$: A conductor provides a high electrical permittivity path in which the Electric Field $E$ acts (units = Volts/m = newtons/coulomb) act on charge to exert a force on electrical charge. The $E$ field of a battery polarizes the dielectric between the plates of a capacitor. The Electric Permittivity of the vacuum of space is a measure of the capacitance of space and is denoted as $\epsilon_o$ and is = 8.85 \times 10^{-12} Farads/meter ). Vacuum is the smallest possible Electric Permittivity and all other materials are expressed as a ratio, a multiple of $\epsilon_o$. An $E$ field acting in a space with $\epsilon_o)$ will have a Electric Displacement Field $D=\epsilon_o E$, which has units of $D=coulombs/m^2$. The $\epsilon_{air}$ is approximately equal to vacuum permittivity, and that permittivity is very low, so minimal charge is stored on the terminals. However, in a conductor, the $\epsilon_{conductor}$ is large, which means that for any given $E$ field the Electric Displacement Field (coulombs/m2) will be large, and the closed circuit current through a conductor alone is very high. In a series connection of two cells, the $E_{series}$ field is double the Electric Potential of a single cell, hence drives and double the current.

Forces acting inside series cells: Placing two Lead-Acid cells in series doubles the voltage, and hence doubles the current across the load. That same current will pass through the conductor between the two cells in series, and the ions in the electrolyte of both cells will likewise conduct charge at the same rate. All of the cathodes accept electrons to complete the $PbO_2 + S0_4^{2-} +2e^- +4H^+ \to PbSO_4 + 2H_2O$ reaction. And, all the anodes generate electrons in the $Pb + S0_4^{2-} \to PbSO_4 + 2e^-$ reaction.

• Two batteries in series will draw twice the current through the same load. The higher current increases the demand for electrons from the anode and increases the supply of electrons to the cathode.


• The increased demand for electrons at the anode removes electrons from the products side of the reaction, accelerating the anode reaction, causing it to liberate more electrons at a rate commensurate with demand. The anode supplies current at the rate demanded by the energy consumption rate at the anode for the reaction voltage.


• The increased supply of electrons at the cathode supplies more electrons to the reactant side of the reaction, allowing it to increase its reaction rate by producing more products. The increased current causes the cathode reaction to increase its rate of consumption.


• Consider the chemical reactions acting at the inner anode and inner cathode of the batteries in series. (Note: the terms "inner anode" and "inner cathode" refer to the terminals which make contact between the cells.)


• Placing a second cell in series, the voltage across the load increases, which doubles the current through the load. The increased current draws away the excess electrons, which disperses the $H^+$ ion layer, which allows $SO_4^{2-}$ to react with the Pb at a greater rate, and increases the concentration of $H^+$ ions. The positive charge around the anode draws $O_2^{2-}$ ions to migrate from the inner cathode to the outer anode. By increasing the concentration of reactants, by Le Chatlier's principle, the cathode reaction is driven toward products, making the inner cathode more positive, and drawing electrons from the inner anode, which increases its reaction rate, and draws $O_2^{2-}$ ions from the outer cathode in the same way as the outer anode.


• And, at the outer cathode, electrons arrive from the outer anode through the load, resulting in a current of electrons which supplies the outer cathode with electrons to for its reaction.


• In sequence, the increased voltage increases the current flow through the load, which draws off electrons from the outer anode and supplies them to the cathode, which increases the rate of reaction at both the outer anode and outer cathode. In turn, the outer anode affects the inner cathode reaction, and the outer cathode affects the inner anode, both via the modification of the migration rate of ions, which in turn changes the reaction rate at the inner anode and cathode.

Summary: With the higher voltage delivered by cells in series, the load draws an increased current. The reaction rate at the anode increases the current supplied in response. As current arrives at the cathode, the cathode reaction rate increases. The inner anode and inner cathode follow the lead of the outer anode and cathode, increasing their reaction rates - influenced by ion migration in the electrolytes of the two cells. After a short transient, the systems reach a steady state where the reaction rates at all anodes and cathodes match the current demand at the load.