Friday, 30 November 2018

homework and exercises - Inductance of two parallel wires


I've been asked to calculate the inductance per unit length for two wires or radius $a$ separated by $2d$ where $2d>>a$.


Starting from $\int_{s} B.dS = LI$ Im not sure what surface to take?


For each wire the field at a distance r away is given by $\int_{l} B.dl = \mu_0 I$ and by superposition $B$ in the first integral is their sum. And so $$LI= \int_{s} B_1.dS + \int_{s} B_2.dS $$ where $B_1$ and $B_2$ are the contributions from the two wires.




Answer



my answer


Answer if anyone is interested. In the end the areas outside the inner edges of the wire cancelled by symmetry and so the surface i was looking for was the area enclosed.


general relativity - What are the units of the quantities in the Einstein field equation?


The Einstein field equations (EFE) may be written in the form:



$$R_{\mu\nu}-\frac {1}{2}g_{\mu\nu}R+g_{\mu\nu}\Lambda=\frac {8\pi G}{c^4}T_{\mu\nu}$$ where the units of the gravitational constant $G$ are $\mathrm{\frac{N\,m^2}{kg^2}}$ and the units of the speed of light are $\mathrm{\frac{m}{s}}$.


What are the units of the Ricci curvature tensor $R_{\mu\nu}$, the scalar curvature $R$, the metric tensor $g_{\mu\nu}$, the cosmological constant $\Lambda$ and the stress-energy tensor $T_{\mu\nu}$?




electromagnetism - Why does the electric field of an infinite line depend on the distance, but not on an infinite plane?


I understand that with an infinite plane, as you get closer, the infinitesimal contributions to the electric field become greater in module. The direction of the vectors become less perpendicular to the plane as you get closer, hence reducing the overall electric field in the perpendicular direction. Both effects offset each other such that the electric field is the same at whatever distance.


However, wouldn't this same argument apply to an infinite line? Gauss's law shows that the electric field of an infinite line depends on the distance, but supposedly the same would happen as with the infinite plane, and yet it doesn't.


Does anyone have an explanation for this?



Answer



Well, the explanation really is held in Gauss's Law. That shows that the field of an infinite line is distance dependent, while an infinite plane is not.



But I expect you're looking for a more intuitive answer. So I'll give my best shot at one. Keep in mind, that dealing with an infinite anything tends to be non-intuitive, so this may be slightly "hand-wavy".


Electric field works on an inverse-square law, meaning that the when you view a charge from twice as far away, the field strength is four times weaker. It turns out that visual perception is also (approximately) an inverse square law. The amount of area in your field of view an object takes up follows the same proportionality as electric field does. So a way to get a rough estimate for how an electric field changes strength, is to see how the sources size changes with distance.


If you consider looking at an infinite plane, it would take up your entire field of view, (assuming your view is only 180 degrees). As you got further and further from the plane, it would still take up your entire field of view. So the strength of the field doesn't change.


However, the infinite wire does change size. While it's long dimension stays the same in your view, it get's thinner. So the field strength gets smaller and approaches zero at infinity, since the wire would appear infinitesimally thin as you approach infinity.


Hopefully that made it a little more intuitive!


Thursday, 29 November 2018

terminology - Are quantum mechanics and quantum physics the same field?


What is the difference between quantum mechanics and quantum physics?



Answer



There is one more term one should discuss in this thread, quantum theory.


They're largely equivalent and in most cases, each of them may be replaced by any other. S_H overstates the difference by suggesting that "quantum mechanics" is a quantum version of "mechanics" while "mechanics" is just a subset of physics. However, "quantum mechanics" isn't used just for mechanics. Quantum mechanics is used for the new set of principles that underlie all quantum theories whether or not they could be interpreted as "mechanics" or e.g. "field theory". After all, quantum mechanics shows that particle-like "mechanical" properties of elementary particles or light are always complemented by their field-like "wave" properties. In this sense, quantum mechanics does imply that "mechanics" includes the rest of physics including fundamental fields when it's done properly.


The term "quantum theory" was born around 1900 when Max Planck explained the black body radiation by the light quanta. Quantum mechanics is what was born in the mid 1920s except that later, all these advances would be considered parts of the same "revolution" that may be called either "quantum theory" or "quantum mechanics".


So one could say that all the three terms are synonyma. Physicists would almost always pick "quantum mechanics" as the preferred label for any theory that follows the postulates of quantum mechanics or for the postulates themselves. On the other hand, "quantum physics" and "quantum theory" is more likely to be used by the outsiders or in the non-expert context.


electric circuits - Why doesn't current decrease in series combination?


I know that the question is quite stupid but I want to get an insight of this case. consider 3 resistors connected in series with a battery, after the current passes through resistor 1 it loses some of its energy, the kinetic energy of the charge carriers will definitely also decrease and so does the drift velocity then why doesn't the current decrease? Its quite confusing.




Answer



Current is due to the drift velocity of electrons . In the Transient State when the current sets up there is an accumulation of electrons at places like bends in the wire. There is an electric field ( small) in the wire that drives the current. At any place where there is an accumulation of charges the outflowing current will be less than the inflowing current and the field will act in such a way so as to equalize the current. It so happens at the ends of the resistor. But all this happens in such a quick time that for all practical purposes the current is same in every brach of the circuit.


It is just like the electrons which have reached the end of the first resistor communicate the presence of the resistor to the electrons coming behind them and convey them that they have to come slowly because there is a resistor ahead.


Griffith explains this very efficiently. Also there a certain beautiful answers here concerning these questionS !


quantum mechanics - Is resonance an energy eigenstate?


In the particle physics book of Martin & Shaw, they used QM to derive the decay distribution, namely, breit wigner formula. What confused me was that, here they assumed the resonance state $\psi_0 $ is an energy eigenstate with energy $E_0$.



But in another quesion of mine Does the Breit–Wigner formula indicate "violation" of energy conservation? @anna v suggested that the resonance does not have a definite mass. Hence it can't be an energy eigenstate. This seems to be a contradiction.


enter image description here enter image description here enter image description here enter image description here




Wednesday, 28 November 2018

statistical mechanics - Once a quantum partition function is in path integral form, does it contain any operators?


Once a quantum partition function is in path integral form, does it contain any operators?



I.e. The quantum partition function is $Z=tr(e^{-\beta H})$ where $H$ is an operator, the Hamiltonian of the system.


But if I put this into the path integral formalism so that we have something like $Z= \int D(\bar{\gamma},\gamma) e^{-\int_0^\beta d\tau\,[\frac{i\hbar}{2}(\gamma\partial_t\bar{\gamma}-\bar{\gamma}\partial_t \gamma) + H(\bar{\gamma},\gamma)]}$, is the $H(\bar{\gamma},\gamma)$ an operator?


Thanks!



Answer



Nope, Feynman's path integral formulation of quantum mechanics is a method to directly calculate the complex probability amplitudes and all objects that appear in its formalism - not counting proofs of equivalence with other approaches to quantum mechanics - are $c$-numbers representing classical observables.


In particular, the exponent in the path integral - which should be $iS$ ($i$ times the action i.e. $i$ times the integrated Lagrangian), not the Hamiltonian - is a $c$-number-valued function of the "classical observables", the same function that is relevant for the classical (non-quantum) theory. So the path integral is an infinite-dimensional integral over otherwise "ordinary classical variables" that produces some probability amplitudes - the same ones that may be (but don't have to be) obtained from the operator formalism.


Uncertainty principle in path integral


By the way, some sorts of the path integral include integration over both positions and momenta, $\int Dx(t)\, Dp(t)$. How it is possible that both of them are treated "classically" as $c$-numbers? Doesn't it violate the uncertainty principle?


The answer is that it doesn't violate the uncertainty principle. One may still deduce that $xp-px=i\hbar$ from the path integral as long as she is careful about putting the right values of the time $t$ (the argument). The quantities $x(t)p(t-\epsilon)$ and $p(t)x(t-\epsilon)$ differ. A necessary condition for this difference to exist is the fact that "most" trajectories contributing to the path integral are discontinuous.


General relativity and the conservation of momentum


I'm trying to understand the conservation of momentum in general relativity.


Due to the curvature of space-time by matters and energy, the path of a linear motion appears to be distorted.


Therefore from the frame of reference of an observer who is at a not-so-curved region of space-time, it appears that the velocity vector of an object that is moving along a straight line in a very-curved space-time is changing?


Does that imply that from the observer's frame of reference (in a not-so-curved space-time) momentum of that object (in the very-curved space-time) is not conserved?


In this sense is momentum not always conversed in general relativity or did I misinterpret something?



Answer



When a particle is deflected by gravity the gravitational field will also be modified by the particle. To form a conservation law for momentum you need to take into account the momentum in the gravitational field as well as the particle. This can be done e.g. using pseudo-tensor methods.


This works but remember that momentum is a relative concept. Even in Newtonian dynamics it depends on the velocity of your reference frame. In general relativity it also depends on the frame but a much wider class of frames is valid. This means that momentum conservation depends on the choice of co-ordinates. Locally you can pick an inertial reference frame but over extended regions there is no inertial frame. A momentum in one location cannot be simply added to a momentum vector in another location.


Nevertheless, momentum conservation laws over extended regions do work correctly in general relativity. For an extended description of the formalism and why it works see my article at http://vixra.org/abs/1305.0034



Edit: In the comments below MWT p457 has been cited to support the idea that energy and momentum are only conserved in specific cases. I am adding this to directly refute what has been said there.


MWT begin by saying that there is no such thing as energy or momentum for a closed universe because "To weigh something one needs a platform on which to stand to do the weighing" This is pure Wheeler rhetoric of the type for which he is greatly admired, but in this case it is simply misleading. Weight is a Newtonian term with no useful counterpart in general relativity except in the specific case of an isolated system in an asymptotically flat spacetime. For other situations such as the closed cosmology energy and momentum conservation take a different but equally valid form.


They go on to say that in a closed universe total energy or momentum or charge is "trivially zero" They justify that it is zero because you can use Gauss's divergence theorem to write the charge, energy or momenta as a boundary integral. For a closed universe the boundary disappears making the result zero. This is of course correct, but they give no justification for calling this answer trivial. Energy and momenta are initially defined as a sum of volume integral contributions from each physical field including electromagnetic fields, fermionic fields, gravitational field etc. It is in this sense that we understand that conserved quantities can move and can transform from one form to another but the total remains constant. It is a property of gauge fields that when the dynamical field equations are used the conserved quantity is the divergence of a flux from just the gauge field so that it can be integrated over a volume and be calculated as a boundary surface integral. This gives charge/energy/momenta a holographic nature where they can be considered either as a volume integral over contributions from different fields of a surface integral over the gauge field flux. The important thing to understand is that to go from the volume to the surface form the field equations must be used. This means that the total charge/energy/momenta in a closed universe is zero but that this is not in any sense a trivial result. If you calculate total energy as a volume integral for a configuration of fields that do not satisfy the equations of motion the answer will not necessarily be zero. Stating that it is zero is therefore making a non-trivial assertion about the dynamics. This is what conservation laws are all about.


MWT go on to explain why it would make no sense to have an energy-momentum 4-vector globally. The invalid assumption they are making is that energy and momenta need to form a 4-vector. A 4-vector is a representation of the Poincare group and is the natural form that energy-momentum takes in special relativity where Poincare invariance is the global spacetime symmetry. In general relativity the global spacetime symmetry is diffeomorphism invariance so the correct expectation is that all quantities should take the form of a representation of the diffeomorphism group for the manifold. This is what happens. If you demand an energy-momentum 4-vector then of course you will only get an answer locally, and also for an asymptotically flat spacetime where Poincare symmetry is valid at spatial infinity, but demanding such a 4-vector is simply the wrong thing to do in general relativity.


In the fully general case of any spacetime we can apply Noether's theorem using invariance under diffeomorphism generated by any contravariant transport vector field (Observe that it is invariance of the equations that is required, not invariance of the solutions. Some people like to confuse the two) The result is a conserved current with a linear dependence on the transport vector field. This is the correct form for a representation of the diffeomorphism group. I refer to my cited paper for the mathematical details.


This current gives conservation laws for energy, and momenta including generalizations of angular momenta as well as linear momenta depending on the transport vector field chosen. If it transports space-like hypersurfaces in a timelike direction it will give an energy conservation law and if it transports in spacelike directions it gives momenta conservation laws. These energy and momenta do not normally form 4-vectors but they can be integrated to give non-trivially conserved quantities. The global form a conservation law must take is that the total energy and momenta in a volume must change at a rate which is the negative of the flux of the quantity over the boundary, and this is what you get with the currents derived from Noether's theorem.


It may be that other people will want to add comments here that dispute the validity of energy conservation in other ways. I refer once again to my new article at http://vixra.org/abs/1305.0034 where I refute all the objections that I have heard. Triviality is dealt with in item (6) and 4-momentum is dealt with in item (8). Unless someone comes up with a novel objection I will just refer to the numbered objections in this paper in future.


Remember, there are no authorities in science and any expert may be shown to be wrong either by reasoning or by experiment.


general relativity - Why isn't an infinite, flat, nonexpanding universe filled with a uniform matter distribution a solution to Einstein's equation?


In Newtonian gravity, an infinite volume filled with a uniform distribution of mass would be in perfect equilibrium. At every point, the gravitational forces contributed by masses in one direction would be exactly counterbalanced by those in the opposite direction.


But when Einstein tried to apply General Relativity to possible cosmologies, he found it necessary to include the cosmological constant in order to get a static universe.


In qualitative terms, it seems to me that the gravitational stresses that the masses would impose on the spacetime should all cancel out, and likewise, that the resulting flat spacetime should have no effect on the motion of the masses.


However, the math of the situation is beyond my current skills, so I'm asking how it produces the nonequilibrium condition?


(I realize that such an equilibrium solution might not be stable, and that there are many other very good reasons to believe in an expanding universe, so I'm not trying to promote any alternative theories. I'm just curious about this particular point. )



Answer



Nice question!



Here's a possible statement of the logic in the newtonian case. (1) In newtonian mechanics, we assume that inertial reference frames exist (this is one popular modern way of restating Newton's first law), we assume that such frames are global, and we assume that we can always find such a frame by observing a test particle that is not acted on by any force. (2) In the newtonian homogeneous cosmology, we could assume that the force on a chosen test particle P can be found by some limiting process, and that the result is unique. (This is basically a bogus assumption, but I don't think that ends up being the issue here.) (3) Given that the result is unique, it must be zero by symmetry. (4) By assumptions 1 and 2, P defines an inertial frame, and by assumption 1, that frame can be extended to cover the entire universe. Therefore all other particles in the universe must have zero acceleration relative to P.


In general relativity, assumption 1 fails. Test particles P and Q can both be inertial (i.e., no nongravitational forces act on them), but it can be false that they are not accelerated relative to one another. For example, we can make an FRW cosmology in which, at some initial time, $\dot{a}=0$, but then it will have $\ddot{a}\ne0$ (in order to satisfy the Einstein field equations for a uniform dust). (In this situation, the Einstein field equations can be reduced to the Friedmann equations, one of which is $\ddot{a}/a=-(4\pi/3)\rho$.)


This shows that the newtonian argument (or at least one version of it) fails. It does not prove that there is no other semi-newtonian plausibility argument that explains why an initially static universe collapses. However, I'm not sure what criteria we would be able to agree on as to what constitutes an acceptable semi-newtonian plausibility argument. Some people have developed these semi-newtonian descriptions of cosmology at great length, but to me they appear to lack any logical foundations that would allow one to tell a correct argument from an incorrect one.


electromagnetism - Definition of Ampere



On Wikipedia it says:



This force is used in the formal definition of the ampere, which states that it is "the constant current that will produce an attractive force of $2 × 10^{-7}$ newton per metre of length between two straight, parallel conductors of infinite length and negligible circular cross section placed one metre apart in a vacuum."



In reference to the definition of an Ampere, why was $2 × 10^{–7}$ chosen?



Answer



The definition of Ampere is obtained by the below equation of force between two infinitely long parallel current carrying conductors.
enter image description here


Where $F$ is force, $\triangle{L}$ is small length element, $\mu_0$ is absolute permeability of vaccum or free space, $I_1, I_2$ are current flowing through two conductors.


By calculation we can obtain that $\frac{\mu_0}{4\pi}=10^{-7} T A^{-1} m$



When $\triangle{L}=1m, I_1=I_2=1A, r=1m$. By substituting the values in the above equation given in the figure, we obtain $\frac{F}{\triangle{L}}$ to be equal to $2X10 ^{-7}N m{-1}$.


Thus we have the definition of one ampere as: One ampere is that current, which when flowing through each of the two parallel conductors of infinite length and placed in free space of one metre from each other, produces between them a force of $2X10^{-7}$ newton per metre of their lengths.


Tuesday, 27 November 2018

homework and exercises - Why can't a single photon produce an electron-positron pair?


In reading through old course material, I found the assignment (my translation):




Show that a single photon cannot produce an electron-positron pair, but needs additional matter or light quanta.



My idea was to calculate the wavelength required to contain the required energy ($1.02$ MeV), which turned out to be $1.2\times 10^{-3}$ nm, but I don't know about any minimum wavelength of electromagnetic waves. I can't motivate it with the conservation laws for momentum or energy either.


How to solve this task?



Answer



Another way of solving such problems is to go to another reference frame, where you obviously don't have enough energy.


For example you've got a $5 MeV$ photon, so you think that there is plenty of energy to make $e^-e^+$ pair. Now you make a boost along the direction of the photon momentum with $v=0.99\,c$ and you get a $0.35 MeV$ photon. That is not enough even for one electron.


standard model - Reasons for choosing $SU(3)$ as the color group vs. $SO(4)$


What are the reasons that $SU(3)$ is used for QCD?


Why wouldn't the simpler & smaller group $SO(4)$ make a better candidate?




cosmology - How far has a 13.7 billion year old photon travelled


I've read that the size of the observable Universe is thought to be around ~46 billion light years, and that the light we see from the most distant galaxies were emitted ~13.7 billion years ago as a result of the expansion. So is the actual distance the photons have travelled 46 billion light years, or 13.7 billion light years?



Answer



Age of universe is roughly 13.8 billion years, so anything with longest traveling record can only travel 13.8 billion light years at most (in any direction; not necessarily to Earth) because nothing can travel faster than light.


Now, we are at the center of the observable universe (which is spherical with visible radius 13.8 billion years), so longest traveling record holders are the oldest ones and are at the edge of the observable universe and they have traveled 13.8 billion light years so far.


The actual calculated radius 46 billion light years is different from the visible radius because the universe has expanded.


So, the answer to your question is 13.8 billion light years.



general relativity - Another layman blackhole question, pulling one end of a string out from behind the event horizon


No long explanation is needed,


What would happen if I were to allow one end of a rope to fall past the event horizon of a black hole while I held the other end?



Would I be able to pull it out? Would the rope feel extremely (infinitely?) heavy?




general relativity - Does time expand with space? (or contract)


Einstein's big revelation was that time and space are inseparable components of the same fabric. Physical observation tells us that distant galaxies are moving away from us at an accelerated rate, and because of the difficulty (impossibility?) of defining a coordinate system where things have well defined coordinates while also moving away from each other without changing the metric on the space, we interpret this to mean that space itself is expanding.



Because space and time are so directly intertwined is it possible that time too is expanding? Or perhaps it could be contracting?



Answer



The simple answer is that no, time is not expanding or contracting.


The complicated answer is that when we're describing the universe we start with the assumption that time isn't expanding or contracting. That is, we choose our coordinate system to make the time dimension non-changing.


You don't say whether you're at school or college or whatever, but I'm guessing you've heard of Pythagoras' theorem for calculating the distance, $s$, between two points $(0, 0, 0)$ and $(x, y, z)$:


$$ s^2 = x^2 + y^2 + z^2 $$


Well in special relativity we have to include time in the equation to get a spacetime distance:


$$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$


and in general relativity the equation becomes even more complicated because we have to multiply the $dt^2$, $dx^2$, etc by factors determined by a quantity called the metric, and usually denoted by $g$:


$$ ds^2 = g_{00}dt^2 + g_{11}dx^2 + g_{22}dy^2 + ... etc $$



where the $... etc$ can include cross terms like $g_{01}dtdx$, so it can all get very hairy. To be able to do the calculations we normally look for ways to simplify the expression, and in the particular case of the expanding universe we assume that the equation has the form:


$$ ds^2 = -dt^2 + a(t)^2 d\Sigma^2 $$


where the $d\Sigma$ includes all the spatial terms. The function $a(t)$ is a scale factor i.e. it scales up or down the contribution from the $dx$, $dy$ and $dz$, and it's a function of time so the scale factor changes with time. And this is where we get the expanding universe. It's because when you solve the Einstein equations for a homogenous isotropic universe you can calculate $a(t)$ and you find it increases with time, and that's what we mean by the expansion.


However the $dt$ term is not scaled, so time is not expanding (or contracting).


Monday, 26 November 2018

cosmology - What do the names "E mode" and "B mode" mean? Where do they come from?


This has been bugging me a bit since the BICEP announcement, but if there are any resources that answer my question in a simple way, they've been buried in a slew of over-technical or over-popularized articles; Wikipedia isn't much help either.


It is clear to me that when the CMB is described as having "E modes" and "B modes", some reference is being made to electric and magnetic fields. What is the precise nature of this reference? I suspect it is simply an appeal to the fact that the polarization can be split into a curl-free component, which is the gradient of something, and a divergence-free component, which is the curl of something else, and these are formally analogous to electric and magnetic fields. Calling them that way certainly brings to bear all our intuition from electrostatics and magnetostatics into how such modes can look. Is this suspicion correct? Or is there some actual electric or magnetic field involved (as, for example, in TE modes in a waveguide)?


Secondly, how exactly does one split the polarization field into these components? It's not quite the sort of 3D vector field to which Helmholtz's theorem applies:





  • It is a vector field over a sphere instead of all space. This sphere can be seen as the celestial sphere, or equivalently as the surface of last scattering.




  • It does not automatically have a magnitude on top of its direction, though I suspect one can happily use some measure of the degree of polarization for this. (Is that correct? If so, exactly what measure is used?)




How exactly is the polarization field defined, over what space, and exactly what mathematical machinery is used to split it into E modes and B modes? Are there analogues to the scalar and vector potential? If so, what do they physically represent?



Answer



Planck, BICEP, et al are all detecting electromagnetic radiation, but the "E-modes" and "B-modes" refer to polarization characteristics of this radiation, not the actual electric and magnetic fields. As you surmised, the names derive from an analogy to the decomposition of a vector field into curl-less (here "E" for electric or "G" for gradient) and divergence-less ("B" for magnetic or "C" for curl) components, as follows...


The first step is the measurement of the standard Stokes parameters $Q$ and $U$. In general, the polarization of monochromatic light is completely described via four Stokes parameters, which form a (non-orthonormal) vector space when the various waves are incoherent. For light propagating in the $z$ direction, with electric field:



$$ E_x = a_x(t) \cos(\omega_0 t - \theta_x (t)) \, \, , \quad E_y = a_y(t) \cos(\omega_0 t - \theta_y (t)) $$


the Stokes parameters are:



  • $ I = + $ , intensity

  • $ Q = - $ , polarization along $x$ (Q>0) or $y$ (Q<0) axes

  • $ U = < 2 a_x a_y \cos(\theta_x - \theta_y) > $ , polarization at $\pm 45$ degrees

  • $ V = < 2 a_x a_y \sin(\theta_x - \theta_y) > $ , left- or right-hand circular polarization


In cosmology, no circular polarization is expected, so $V$ is not considered. In addition, normalization of $Q$ and $U$ is traditionally with respect to the mean temperature $T_0$ instead of intensity $I$.


The definitions of $Q$ and $U$ imply that they transform under a rotation $\alpha$ around the $z$-axis according to: $$ Q' = Q \cos (2 \alpha) + U \sin (2 \alpha) $$ $$ U' = -Q \sin (2 \alpha) + U \cos (2 \alpha) $$



These parameters transform, not like a vector, but like a two-dimensional, second rank symmetric trace-free (STF) polarization tensor $\mathcal{P}_{ab}$. In spherical polar coordinates $(\theta, \phi)$, the metric tensor $g$ and polarization tensor are:


$$ g_{ab} = \left( \begin{array}{cc} 1 & 0 \\ 0 & \sin^2 \theta \end{array} \right) $$ $$ \mathcal{P}_{ab}(\mathbf{\hat{n}}) =\frac{1}{2} \left( \begin{array}{cc} Q(\mathbf{\hat{n}}) & -U(\mathbf{\hat{n}}) \sin \theta \\ -U(\mathbf{\hat{n}}) \sin \theta & -Q(\mathbf{\hat{n}})\sin^2 \theta \end{array} \right) $$


As advertised, this matrix is symmetric and trace-free (recall the trace is $g^{ab} \mathcal{P}_{ab}$).


Now, just as a scalar function can be expanded in terms of spherical harmonics $Y_{lm}(\mathbf{\hat{n}})$, the polarization tensor (with its two independent parameters $Q$ and $U$) can be expanded in terms of two sets of orthonormal tensor harmonics:


$$ \frac{\mathcal{P}_{ab}(\mathbf{\hat{n}})}{T_0} = \sum_{l=2}^{\infty} \sum_{m=-l}^{l} \left[ a_{(lm)}^G Y_{(lm)ab}^G(\mathbf{\hat{n}}) + a_{(lm)}^C Y_{(lm)ab}^C(\mathbf{\hat{n}}) \right]$$


where it turns out that:


$$ Y_{(lm)ab}^G = N_l \left( Y_{(lm):ab} - \frac{1}{2} g_{ab} {Y_{(lm):c}}^c\right) $$ $$ Y_{(lm)ab}^C = \frac{N_l}{2} \left( Y_{(lm):ac} {\epsilon^c}_b + Y_{(lm):bc} {\epsilon^c}_a \right)$$


where $\epsilon_{ab}$ is the completely antisymmetric tensor, "$:$" denotes covariant differentiation on the 2-sphere, and


$$ N_l = \sqrt{\frac{2(l-2)!}{(l+2)!}} $$


The "G" ("E") basis tensors are "like" gradients, and the "C" ("B") like curls.



It appears that cosmological perturbations are either scalar (e.g. energy density perturbations) or tensor (gravitational waves). Crucially, scalar perturbations produce only E-mode (G-type) polarization, so evidence of a cosmological B-mode is (Nobel-worthy) evidence of gravitational waves. (Note, however, that Milky-Way "dust" polarization (the "foreground" to cosmologists) can produce B-modes, so it must be well-understood and subtracted to obtain the cosmological signal.)


An excellent reference is Kamionkowski. See also Hu.


Diagram-like perturbation theory in quantum mechanics


There seems to be a formalism of quantum mechanics perturbation that involve something like Feynman diagrams. The advantage is that contrary to the complicated formulas in standard texts, this formalism is intuitive and takes almost zero effort to remember (to arbitrary orders).


For example, consider a two level atom $\{|g\rangle, |e\rangle\}$ coupled to an external ac electric field of frequency $\omega$. Denote the perturbation by $\hat V$, with nonzero matrix element $\langle e|\hat V |g\rangle$.


Then the second order energy correction reads $$E^{(2)} = \langle e|\hat V |g\rangle\frac{1}{\omega_g - \omega_e +\omega} \langle g|\hat V |e\rangle + \langle e|\hat V |g\rangle\frac{1}{\omega_g - \omega_e -\omega} \langle g|\hat V |e\rangle $$ where the first term corresponds to the process absorb a photon then emit a photon while the second process is emit a photon then absorb a photon.


Does anybody know the name of this formalism? And why it is equivalent to the formalism found in standard texts?




homework and exercises - An Electric Potential Glued to a Cube-Shaped Insulator to Replicate a Point Charge: Charge Distribution


I have been going back over this problem with a friend for the better part of a day:


A potential is glued to a cube-shaped insulator so that outside of the insulator the field is the same as a point particle. How can we calculate the surface charge distribution and the volume charge distribution?



Answer



Introduction


$\def\ph{\varphi}\def\vr{\vec{r}}\def\eps{\varepsilon}\def\rmr{{\rm r}}\def\pd{\partial}\def\l{\left}\def\r{\right}\def\ltag#1{\tag{#1}\label{#1}}\def\div{\operatorname{div}}\def\grad{\operatorname{grad}}\def\nR{{\mathbb{R}}}$ You do not need a combination of a surface charge and a volume charge to re-produce a point-charge field outside a cube.


The surface-charge distribution is uniquely determined if you just use that for your purpose. The construction of this surface charge distribution is described in the first part of this answer.


If you just use a volume charge distribution this is not uniquely determined. An example for that case is given in the second part of the answer.



Surface charge distribution


Let $V$ be the the open interior of the cube containing the origin. For points $\vr$ outside the quader we set the potential $\ph(\vr)$ equal to the potential $$ \ph_Q(\vr) =\frac Q{4\pi\eps_0|r|} $$ of a point charge $Q$ at the origin. This gives you also the boundary condition $$ \ph(\vr) = \ph_Q(\vr)\text{ for }\vr\in\partial V $$ for the potential inside the quader. Solve the Dirichlet boundary value problem with the Laplacian equation $\Delta\ph(\vr)=0$ and with this boundary condition for the potential $\ph(r)$ at points $\vr\in V$ inside of the quader.


Let $\partial^+_\nu\ph$ and $\partial^-_\nu\ph$ be the limits of the outer normal derivatives of $\ph$ on $\partial V$ from the outside and the inside of $V$, respectively. The surface charge will be $\sigma(\vr) = -\eps_0(\partial^+_\nu\ph-\eps_{\rm r}\partial^-_\nu\ph)$. Why? Try to explain this with an integral over the surface of a small volume sitting on the surface.


Here we assume that the insulator has a homogeneous and isotropic relative permittivity $\eps_\rmr$.




The following picture shows a section of the solution for the Dirichlet problem in the cube $[-1,1]^3$. Only the part $\bigcup_{z\in[-1,0]}[0,-z]^2\times\{z\}$ of the cube has been modelled that generates the full cube through a sequence of reflections. Natural boundary conditions $\partial_\nu \ph=0$ were used at the symmetry planes. The Dirichtlet boundary condition at $z=-1$ is $$\ph(x,y,z) = \frac1{\sqrt{x^2+y^2+1}}.$$


potential inside the cube


The domain has been modelled with gmsh and solution is computed with getdp.


The limit of the outer normal derivative of $\ph$ from the outside at the boundary $z=-1$ is \begin{align} \pd^+_\nu \ph(x,y,-1) &= -\pd_z \ph(x,y,-1) = \l.\frac{z}{\l(x^2+y^2+z^2\r)^{3/2}}\r|_{z=-1}\\ \ltag{derPotOutside} &=\frac{-1}{\l(x^2+y^2+1^2\r)^{3/2}} \end{align} The following picture shows the field $\sigma:=-(\pd^+_\nu\ph-\pd^-_\nu\ph)$ in the boundary area $[0,1]^2\times\{-1\}$. Thereby, $\pd^+_\nu\ph$ is given by \eqref{derPotOutside} and $\pd^-_\nu\ph$ results from the numerical solution of the boundary value problem. Note, that this corresponds to a normalized solution of the original problem with $\eps_\rmr=1$.


sigma



One can check the numerical result $\sigma$ with the integral $\int_{(x,y)\in[0,1]^2} \sigma(x,y,-1) dA$ which must give $\frac\pi6$. This works within the bounds of numerical precision.


For reproduction of the results I link here the geometry definition and the problem definition for the FEM solver.




The normal component of the field strength is discontinuous at the charged surface but the potential is continuous.


This can be demonstrated with a uniformly charged circular disk in the (x,y)-plane with center at the origin and radius $R$. We denote the charge density with $\sigma$. The complement of the disk is free space with permitivity $\eps_0$. For the demonstration we calculate the potential on the z-axis: \begin{align} \ph(z) &= \frac\sigma{4\pi\eps_0}\int_{r=0}^R \frac{ 2\pi r dr}{\sqrt{r^2 + z^2}}\\ &=\frac\sigma{2\eps_0} \left[\sqrt{r^2+z^2}\right]_{r=0}^R\\ &=\frac\sigma{2\eps_0} \left(\sqrt{R^2+z^2}-|z|\right) \end{align} The potential can be continuously extended at $z=0$ with the value \begin{align} \ph(0) = \frac {\sigma R}{2\eps_0}. \end{align}


Volume charge distribution


The volume charge distribution is not uniquely determined. For an instance if you have $\eps_\rmr=1$ you can put a ball with uniform volume charge into the cube. That is $$ \rho(\vr) = \begin{cases} \frac{Q}{\frac34\pi R^3}&\text{ for }|\vr|

Another possible space charge distribution is \begin{align} \rho(\vr) = \begin{cases} \frac{\frac87 Q}{\frac34\pi R^3}&\text{ for } \frac12 R<|\vr|

If the insulator is a quader with constant permitivity $\eps_\rmr\neq1$ then you have given:




  1. the potential $\ph$ on the boundary $\partial V$ of the quader by the outer potential field $\ph^+(\vr)=\frac{Q}{4\pi\eps_0|\vr|}$ of the point charge

  2. the normal derivative $\partial^-_\nu\ph$ at the boundary of the quader through the normal component of the charge displacement density $$\partial^-_\nu\ph(\vr) = \frac1{\eps_\rmr}\partial^+_\nu\ph(\vr).$$


The taks is now to find an at least two times differentiable scalar function $\ph:V\rightarrow\nR$ interpolating the required potential and the required normal derivative at $\partial V$. The Volume charge density can then be calculated by $$ \rho(\vr) = -\eps_\rmr\eps_0 \Delta \ph(\vr). $$




The actual problem for varying $\eps_r$ is \begin{align} \div\l(\eps(\vr) \grad\ph(\vr)\r) = -\rho. \end{align} In your case $\eps(\vr)$ is even discontinous such that you need the weak form of $\eps(\vr)$ and you end up with the problem: \begin{align} \int_{\nR^3} (\grad\delta\ph(\vr))\eps(\vr)\grad\ph(\vr) d V = \int_{\nR^3} \delta\ph(\vr)\rho(\vr) dV \end{align} for all test functions $\delta\ph$. This can be solved numerically.


I do not know if there are known Green functions for that problem.


particle physics - Are oscillations of electron chirality experimentally observable?



Is there any plausible experiment by which chirality oscillations in electrons could be observed experimentally, such as through some analogy to neutrino oscillation experiments?




Sunday, 25 November 2018

Quantum entanglement definition






  1. How can we define Quantum entanglement (in QFT)?




  2. What are the known mathematical settings and special physical (or logical) conditions of QE applied to Quantum computing?





Answer



Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated or interact in ways such that the quantum state of each particle cannot be described independently — instead, a quantum state must be described for the system as a whole.



Measurements of physical properties such as position, momentum, spin, polarization, etc., performed on entangled particles are found to be appropriately correlated. For example, if a pair of particles are generated in such a way that their total spin is known to be zero, and one particle is found to have clockwise spin on a certain axis, then the spin of the other particle, measured on the same axis, will be found to be counterclockwise, as to be expected due to their entanglement. However, this behavior gives rise to paradoxical effects: any measurement of a property of a particle can be seen as acting on that particle (e.g., by collapsing a number of superposed states) and will change the original quantum property by some unknown amount; and in the case of entangled particles, such a measurement will be on the entangled system as a whole. It thus appears that one particle of an entangled pair "knows" what measurement has been performed on the other, and with what outcome, even though there is no known means for such information to be communicated between the particles, which at the time of measurement may be separated by arbitrarily large distances.


tensor calculus - Levi Civita covariance and contravariance


I read some older posts about this question, but I don't know if I'm getting it. I'm working with a Lagrangian involving some Levi Civita symbols, and when I calculate a term containing $\epsilon^{ijk}$ I obtain the contrary sign using $\epsilon_{ijk}$. I always apply the normal rules: $\epsilon_ {ijk}=\epsilon^{ijk}=1$; $\epsilon_ {jik}=\epsilon^{jik}=-1$ etc. I believed that there is no difference between covariant and contravariant Levi-Civita symbol. What do you know about this?




special relativity - Dirac spinor and Weyl spinor


How can it be shown that the Dirac spinor is the direct sum of a right-handed Weyl spinor and a left-handed Weyl spinor?


EDIT: - Let $\psi_L$ and $\psi_R$ be 2 component left-handed and right-handed Weyl spinors. Their transformation properties are known. When I put these two spinors in a column and construct a four-component column which is a direct sum of $\psi_L$ and $\psi_R$ i.e., $\psi_D=\psi_L\oplus\psi_R$. This I defined to be the Dirac spinor. Right? Since it is a direct sum under Lorentz transformation, the corresponding Lorentz transformation matrix is diagonal. Right? Then it is easy to show that, it satisfies the Dirac equation in Chiral basis. Right? This is possible because we started from the definition of left-handed and right-handed Weyl spinors and their transformation properties are known. Right? This is explicitly carried out in the book by Lewis Ryder. But suppose I start the other way around. I solve the Dirac equation in chiral basis. Then no one tells me that the upper two components are really left-handed and lower two are really right-handed. Suppose I take this chiral basis solution of Dirac equation and now take that to be my definition of Dirac spinor. Then how can I show the opposite, that it is made up of two irreps of Lorentz group i.e., $\psi_D=\psi_L\oplus\psi_R$ ?



Answer



From the relativistic covariance of the Dirac equation (see Section 2.1.3 in the QFT book of Itzykson and Zuber for a derivation. I also more or less follow their notation.), you know how a Dirac spinor transforms. One has $$\psi'(x')=S(\Lambda)\ \psi(x)$$ under the Lorentz transformation $$x'^\mu= {\Lambda^\mu}_\nu\ x^\nu= {\exp(\lambda)^\mu}_\nu\ x^\nu=(I + {\lambda^\mu}_\nu+\cdots)\ x^\nu\ .$$ Explicitly, one has $S(\Lambda)=\exp\left(\tfrac{1}8[\gamma_\mu,\gamma_\nu]\ \lambda^{\mu\nu}\right)$.


To show reducibility, all you need is to find a basis for the gamma matrices (as well as Dirac spinors) such that $[\gamma_\mu,\gamma_\nu]$ is block diagonal with two $2\times 2$ blocks. Once this is shown, it proves the reducibility of Dirac spinors under Lorentz transformations since $S(\Lambda)$ is also block-diagonal. Such a basis is called the chiral basis. It is also important to note that a mass term in the Dirac term mixes the Weyl spinors in the Dirac equation but that is not an issue for reducibility.


While this derivation does not directly use representation theory of the Lorentz group, it does use the Lorentz covariance of the Dirac equation. I don't know if this is what you wanted.



(I am not interested in your bounty -- please don't award me anything.)


Saturday, 24 November 2018

special relativity - “Relativistic Baseball”


On Randall Munroe’s new blog “what if”, he answers the question:



“What would happen if you tried to hit a baseball pitched at 90% the speed of light?”



http://what-if.xkcd.com/1/


He concludes:



… the air molecules in front of this ball don’t have time to be jostled out of the way. The ball smacks into them hard that the atoms in the air molecules actually fuse with the atoms in the ball’s surface. Each collision releases a burst of gamma rays and scattered particles.



These gamma rays and debris expand outward in a bubble centered on the pitcher’s mound. They start to tear apart the molecules in the air, ripping the electrons from the nuclei and turning the air in the stadium into an expanding bubble of incandescent plasma.


… Suppose you’re watching from a hilltop outside the city. The first thing you see is a blinding light, far outshining the sun. This gradually fades over the course of a few seconds, and a growing fireball rises into a mushroom cloud. Then, with a great roar, the blast wave arrives, tearing up trees and shredding houses.


Everything within roughly a mile of the park is leveled, and a firestorm engulfs the surrounding city. The baseball diamond is now a sizable crater, centered a few hundred feet behind the former location of the backstop.



Is that all actually reasonable, or is it all just hyperbole?



Answer



I found the narrative to be consistent with my view of physics. Let me address this major point about the integrity of the ball as it travels through the air.


The distance from the pitcher's mound to home plate is $18.39 m$, and the diameter of the ball is about $7.4 cm$. Since the ball is sufficiently fast, we are comfortably out of normal fluid mechanics and know that collisions will happen based entirely on the trajectory of the ball. Let me be clear that the ball holds together mostly due to the fact that all the atoms in the ball have the same momentum vector - NOT because molecular forces are making much of a difference.


The density of air is about $1.2 kg/m^3$. Multiply this by the area of the ball times the distance it travels to get that the ball collides with $92.4 g$ in its path to home plate. The weight of a major league baseball is about $145 g$. In the absence of fusion, it would appear correct to say that the speed of the ball changes very little through that distance, due to the simple reasoning that the ball is heavier than the air.


At the point that the ball hits the bat, it really shouldn't matter much to us what happens because the conservation of energy mandates absolutely that such an explosion happens. I would maintain, however, that the center could be considerably past home plate, since the nuclei in the ball will have to go through a good number of collisions before the momentum is dissipated. Keep in mind, the Earth is only normally supporting the full weight of the stadium directly downward. The weight of Wembly Stadium is in the neighborhood of $(2.2 g/cm^3) \times 90,000 m^3 + 23,000 \text{tonnes} = 2.2 \times 10^{8} kg$, which leads to a normal downward force of about 2 billion Newtons. The momentum of the ball is about 89 million Newton-seconds.



I think a fraction of the stadium itself will head down the street a little bit while exploding. In fact, it might travel quite a good distance, since the shine from the ball plasma is highly directional and will hit the bleachers which have less of the high mass concrete (mostly foundations). The ball has the momentum of about 90 fully loaded 18-wheelers traveling at 60 mph. That is a lot of momentum, and it would likely break through a few walls, but it may still dissipate sufficiently (the momentum, not the energy) within the stadium but not the field. That's the one major point where I take issue with Munroe.


Why is 3D stress tensor acting only on three surfaces?


I'm trying to learn about the stress tensor (in 3D)


enter image description here


The tensors are said to have directions (the first subindex $i$ in $\sigma_{ij}$) and specify the surface upon which they act (the second subindex $j$ in $\sigma_{ij}$).


What confuses me is why is it defined only to act on three surfaces, even when the cube has six surfaces?




electromagnetism - Conversion of Moving coil galvanometer to ammeter


A galvanometer can be converted into an ammeter by connecting a low resistance (called shunt resistance) in parallel to the galvanometer.


enter image description here


Firstly, why do we need to connect the resistance?


If a resistance is connected in parallel, then some of the current will flow through resistance. So how can the current in the circuit be determined? Won't it be inaccurate, as some of the current flows through resistance?



Answer



First things first, the resistance is connected in the circuit parallel to the galvanometer to minimise the total resistance of the ammeter. Whenever 2 resistances are connected in parallel, the equivalent resistance is always less than the lowest resistance.


You know that:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$


$$\implies R_{eq} = \frac{R_1R_2}{R_1 + R_2}$$



eg: Try plugging in values. If $R_1 = 5\Omega$ and $R_2 = 0.2\Omega$:
$$R = \frac{5\times0.2}{5 + 0.2} = \frac{1}{5.2} = 0.19$$ which is less than 0.2, thus reducing the equivalent resistance.



This is done so that the ammeter has the least resistance possible and it doesn't interfere with the potential drop across the circuit.




Now, the ammeter is callibrated in such a way that it gives the correct reading, even if the current through G isn't what it says.
It's like calculating your weight while holding a 20kg bag. You can still get the correct weight by shifting the scale of your weighing machine down by 20kg! ;)



quantum mechanics - Treating matter waves as light waves?


Is it valid to treat a matter wave as a light wave with wavelength equal to the de Broglie wavelength of the matter wave? Either way please can you explain why?



Answer



It's not valid to treat light and matter waves alike. Why? Apart from the obvious reason that they are not the same (we can after all see light distinctly from matter), the two have different equations of motion - the (non-relativistic) "matter wave" obeys the famous Schrodinger equation, while classical light waves go around according to Maxwell's equations (or the standard wave equation).


There are many differences because of this:




  1. A 'matter wave' is (unavoidably) complex valued, unlike a light wave, where the electric and magnetic fields are always real valued.

  2. The frequency of a matter wave goes as the inverse square of its wavelength, and that of a light wave goes as the inverse of its wavelength. This means that while light of all wavelengths has the same speed $c = \nu\lambda$, the speed of a matter wave depends essentially on its wavelength/frequency. (In addition to John Rennie's point - the phase velocity of a matter wave is half its group velocity, while they are the same for light).

  3. There can always be a time when there is no light in the universe, as light waves can be absorbed and emitted; but the total "intensity" of a matter wave must always be a constant for all time (this follows from the probability interpretation).

  4. A rather famous example is the case of refraction: when light goes into a region where it slows down, it bends towards the normal; when matter goes into a region of higher potential (where it slows down), it bends away from the normal to the surface of the region: this is because in the latter case, only the component of velocity along the normal reduces. This is once again because the wavelength increases with increasing speed for a light wave, but decreases with increasing speed for a matter wave. In fact, as the dependence of wavelength on speed is reciprocal between the two cases, matter waves obey the "inverse" of Snell's law.


Finally, there is only ever one electromagnetic field in space at any point of time, and all these waves are disturbances on that field. These disturbances all add up. When there are many ("entangled") particles, however, the wave equation itself is different (a many-particle Schrodinger equation) and it is incredibly hard to express the wave function as a simple "sum" of many matter waves, one for each particle. Instead, matter waves are not waves in real space anymore, but simply a function of various properties of a material system.


Friday, 23 November 2018

special relativity - Relativistic mass or rest mass


If we have two particles having same rest mass say $m$ and equal velocity approaching $c$ and they do a head on collision and stick together, then the mass of both particles after collision should be equal to 2(relativistic mass) . My view is it should be 2(rest mass) because they are no longer in motion so their masses should be again equal to rest mass. But by energy conservation it should be equal to relativistic mass. What do actually happen here?




Integers powers of fields in a QFT Lagrangian


Why can we not have non-integer powers of fields in a QFT Lagrangian, eg. $\phi^{5/2}$? Or if we wanted a charged $\phi^3$ theory, could we not have a $(\phi^*\phi)^{3/2}$ term?




Answer



This is a property of renormalization group fixed points---- in usual circumstances, with local Lagrangians, noninteger powers get smoothed out to integer powers. To understand qualitatively why, consider the simplest example of renormalization--- the central limit theorem.


One dimensional fields--- Central Limit Theorem


Suppose I have a quantity X(t) which is determined by the sum of many random quantities over time. The quantities have some mean, which leads X to drift up or down, but subtract out a linear quantity $X'(t) = X(t) - mt$ and fix m by requiring that X(t) have the same mean now as at a later time (in finance/probability language, make X a Martingale). Then all that is left are the fluctuations.


It is well known that these fluctuations average out by the central limit theorem, so that X at a later time $t+\epsilon$ is a gaussian centered around the value of X(t), as long as $\epsilon$ is large enough to include many independent quantities which add up to a gaussian. So the probability density of $X(t+\epsilon)$ is:


$$ \rho(X(t+\epsilon)|X(t)) = e^{-a(x(t+\epsilon)-x(t))^2}$$


Further, the increments are independent at different times, so that the total probability of a path is the product of these independent increments


$$ P(X(t)) = e^{-\int \dot{X}^2}$$


Where you need to choose the scaling of the fluctuations of X appropriately with $\epsilon$ to get to the continuum limit (a trivial case of renormalization), and this P(X) is the weight function in a path integral. Notice that you get an exactly quadratic Lagrangian without any fine-tuning. This is the generic situation. Further, the analytic continuation of this is the particle action for nonrelativistic quantum mechanics, and if you reinterpret time as proper time, this quadratic action is the relativistic propagator as well.


If you make a Lagrangian which has the wrong power, not two, then it will renormalize to the quadratic Lagrangian under very general conditions. For instance, suppose



$$ P(X(t+\epsilon)|X(t) ) = e^{ -|X(t+\epsilon)-X(t)|^{16.7} }$$


This leads to a boxy short-scale motion, but if you average over a few $\epsilon$'s, you recover a quadratic Lagrangian again.


To see why, there is a useful heuristic. Suppose I add a term to the action of the form


$$ \dot X^2 + |(\dot X)|^n $$


If I define the scale of fluctuations in X by setting the $\dot X$ term to 1 (meaning that I normalize the scale of X by the random-walk fluctuations left over when the mean drift is zero), then X has scale dimension -1/2. The new term, integrated in time, has scale dimension which is determined by dimensional analysis to be -n/2 + n-1, so that if n>2, the scale dimension is negative. This is obvious, because the n=2 term sets the scaling, so any higher power must be more negative.


The scale dimension tells you how important the perturbation is at long scales, because it determines how fast the correlations in the quantity fall off. So for integer n, all the correlations disappear at long distances, and you are free to conclude that the theory is perfectly Brownian at long distances.


This is true for integer n. For noninteger n, there are subtleties. If you have exponentials of powers, the resulting distribution for $X(t+\epsilon)$ always has a finite variance, and you always recover the central limit theorem at long distances. But if you make the distribution Levy, you can get new central limit fixed points. These are not exponentials of fractional powers, but the distributions themselves have fractional power tails (the Lagrangian has logarithms of fractional powers).


Since there is no rigorous mathematical theory of quantum fields, these heuristics have to be the guide. You can see them work with lattice models, so they are certainly correct, but proving them is just out of reach of current mathematical methods.


Free field Lagrangians are (free) Central Limit Fixed Points


The central limit points are given by free field Lagrangians. These are quadratic, so they are central-limit stable. If you average a free field at short distances to get a longer distance description, you get the same free field, except that you can recover certain symmetries, like rotational or Lorentz invariance, which might not be there originally. But ignore this. The fundamental renormalization criterion is that you start from a renormalization fixed point, and if you want interactions, you perturb away from this free fixed point by adding nonlinearities.



These fixed points are not all stable to nonlinear interactions, they get perturbed by generic polynomial interactions. If you start with a scalar field


$$\int (\partial\phi)^2 + P(\phi) d^dx$$


The scale dimension of $\phi$ is (d-2)/2 (zero in 2d, 1/2 in 3d, 1 in 4d), which is found by dimensional analysis. This dimensional analysis is assuming that the coefficient of the gradient term is 1, which is exactly for the same reason as in 1d, the coefficient of the gradient term is analogous to the coefficient of the time-derivative term in Brownian motion--- setting it to one normalizes the fluctuations of the field to be those of the free field at long distances. You should be confident that any free field action which is reasonably close to this one will converge to this when you look at it at long distances. This perhaps requires a few discrete symmetries to ensure that you recover rotational invariance, but the intuition is that it is a central limit theorem again).


With this scaling, higher derivative terms have negative scale dimension, so they should disappear when you look at long distances--- their contribution only alters the long-distance central-limit fluctuations, and this just changes the normalization of the field.


Each polynomial interaction also has a scale dimension, which gives you a first order view of how important it is. You can check that in 4d, the naive scale dimension of a term of the form $\phi^n$ is positive only for n=1,2,3, it is zero for n=4, and it is negative for higher n.


This means that there are three coefficients which alter the fluctuations at long distances, which are the linear, quadratic, cubic and quartic terms. The linear term is analogous to the drift in the Brownian motion, and it can be absorbed into a field redefinition. The quadratic term is the mass, and must be tuned to close to zero to approach a continuum limit. The cubic term is a relevant interaction, and the quartic term is a marginal interaction. The space of scalar theories is defined by the quadratic/cubic/quartic couplings.


This is not a theorem, but it should be, and it will be one day. For now, it suffices to say that if you consider such theories, and you look at them at longer distance scales, these coefficients are the only ones that blow up or stay stable. For any other polynomial, the coefficients fall away to zero (I am neglecting polynomial terms of the form $\phi|\nabla\phi|$, or which break rotational invariance, but have positive scale dimension)


If you impose $\phi$ to $-\phi$ symmetry, cubic terms are forbidden, and you only get quadratic/quartic terms. I will assume your model treats positive and negative values symmetrically from now on, although the general case is just slightly more complicated.


The expectation is that adding a term of the form $(|\phi|)^{3/2}$ should renormalize at long distances to a quadratic/quartic coupling of some sort. The reason is that the short distance fluctuations of the field are wild, the $|\phi|^{3/2}$ will get averaged over many neighbors, and if you just look at the correlations of fields at relatively long distances, you will see a RG fixed point, and we know the polynomial ones, and we conjecture that they exhaust the space.


If central limit analogies are not persuasive, one can justify this conjecture using the particle path formulation of field theory. Each polynomial term is a particle point-interaction, of k-particles with each other, and we assume that at long distances, the particles are far apart and free. Whatever the short distance interaction, even if it is a complicated thing, at long distances it should look like the scattering of asymptotic particles, 2->2, 2->3 1->2, and these are all polynomial terms. If there is a complicated superposition of many-particle states which looks like a 2->3.5 interaction, at long distances, you expect the particles to separate, and you recover a polynomial interaction.



None of this is rigorous, and it is very likely that there are counterexamples. In two dimensions, this is completely false, because the scale dimension of phi is zero, so every function of $\phi$ is as good as any other. The condition of being a RG fixed point can be fruitfully analyzed by assuming conformal invariance, and then there is a whole world of examples which differ from the higher dimensional intuition. But in 2d (one space one time dimension) particles don't separate with time, they move together forever, so the idea that you should recover a decomposed scattering picture of interactions at long-distance fails.


Fine Tunings


The scalar example above required fine-tuning of the quadratic term to a special value to reach a nontrivially flucutating limit, because there were terms with positive mass dimension. This is a problem for physics, because to tune things precisely is not physically plausible.


The systems which do not require fine tuning with no space-time symmetry beyond Lorentz invariance exclude scalars, and only have chiral fermions and spin-1 gauge fields, where the fixed points are self-tuning--- there are no terms which have positive scale dimensions for these guys. The standard model is build entirely out of these, with one scalar Higgs field which has a fine tuned.


If you allow supersymmetry, you get a whole world of fluctuating scalar limits, which are protected. These theories are studied widely.


Fractional powers


It is possible to make Levy fields in higher dimensions, by having a fractional power propagator. Such Levy fields have a Lagrangian which has fractional powers of momentum, but not fractional powers of the fields. As far as I know, these nonlocal fields are not studied in the literature, but they appear in a formal way (no physics) in the program of "analytic regularization" of the 1960s, which became dimensional regularization in the 1970s, and in Banks-Zaks fixed points or "Unparticle" theories. There is a recent paper of Haba which analyzes these theories.


There is no full classification of renormalization group fixed points, and it is likely that there are many more that we have missed. But these are not likely to be Lorentz invariant theories.


Once you leave the world of unitary quantum theories, you can get statistical field theories with all sorts of potentials. In particular, the Lifschitz statistical field theory


$$ \int |\nabla^2\phi|^2 + Z|\nabla\phi|^2 + V(\phi) d^4x$$



Gives a dimensionless $\phi$, and should be as rich a statistical theory in 4d for choices of V as conformal theories in 2d. But this is not a unitary theory, and it has been only studied in limited ways. Near eight dimensions, it has an $\epsilon$ expansion which was worked out by Mukhamel in the late 1970s, and which is very close to the normal scalar $\epsilon$ expansion. But in 4d, where it is most interesting, nothing is known.


Even studying this numerically is a challenge, because tuning Z to zero requires a lot of work. These type of Lifschitz points are in vogue right now, because Horava has speculatively connected theories of this sort to quantum gravity. But they are interesting in their own right, even for the nonunitary purely statistical case, because every symmetric renormalization group fixed point is mathematically special and should be understood exhaustively.


popular science - Is there a proof of existence of time?


It seems to me that there is no such thing as time. There is only movement in the universe and we compare our own movement to a different object to have a sense of time. It can be a clock or a atomic vibration.


Does this view of time work within the current framework of phsics?



Do physicists have an explanation/proof about time's existence?



Answer



It's easy to get mixed up between time and the flow of time, and I think you've done this in your question.


Take time first: since 1905 we describe any event as a spacetime point and label it with four co-ordinates ($t$, $x$, $y$, $z$). Saying that time doesn't exist means we can ignore the time co-ordinate and label everything by just it's spatial co-ordinates ($x$, $y$, $z$), which is contradiction with observations. The time co-ordinate obviously exists and be used to distinguish events that happen at the same place but at different times.


Now consider the flow of time: actually this is a tough concept, and relativity makes it tougher. We all think we know what we mean by the flow of time because we experience time passing. To take your example of movement, we describe this as the change of position with time, $d\vec{r}/dt$, where we regard time as somehow fundamental. I'm guessing that this is what you're questioning i.e. whether the flow of time is somehow fundamental.


I don't think there is a good answer to this. To talk about the flow of time you'd have to ask what it was flowing relative to. In relativity we can define the flow of co-ordinate time relative to proper time, $dt/d\tau$, and indeed you find that this is variable depending on the observer and in some circumstances (e.g. at black hole event horizons) co-ordinate time can stop altogether. But then you'd have to ask whether proper time was flowing. You could argue that proper time is just a way of parameterising a trajectory and doesn't flow in the way we think time flows.


At this point I'm kind of stuck for anything further to say. If I interpret your question correctly then you do have a point that just because we observe change of position with time (i.e. movement) this doesn't necessarily mean time is flowing in the way we think. However I'm not sure this conclusion is terribly useful, and possibly it's just semantics.


gravity - Gravitational effects of a single human body on the motion of planets


(This is going to be a strange question.)


How big a difference does the existence (or positioning) of a single human body make on the motion of planets in our solar system, millions of years in the future? I know we can't predict what the difference will be, but do we have reason to think that there likely will be a non-negligible difference?


Why might there be a non-negligible difference? Well, figuring out the motion of the planets in our solar system is an n-body problem. So that motion is supposed to be chaotic - highly sensitive to changes in initial conditions. At least, on a timescale of 5 million years, the positions of planets should be highly sensitive to conditions now. But just how sensitive is an n-body system to tiny perturbations in the gravitational field?


A single human body exerts some small amount gravitational force on nearby planets. So, if I add a single human to Earth's population now, or I move them from one position to another, how much would that change the motion of planets in the future? And over what timescale?


Bonus questions:




  • Would the differences continue to grow over time, or would they eventually diminish to nothing? (I figure that in a sufficiently chaotic system they'd just keep growing, but would be interested to hear otherwise.)





  • Would the effects be similar on the scale of a galaxy, or beyond?





Answer



Lasker published a well-known result in 1989 showing that the solar system is chaotic, the inner planets more so than the outer planets. Quoting the Scholarpedia article (written by Lasker himself):



An integration over 200 million years showed that the solar system, and more particularly the system of inner planets (Mercury, Venus, Earth, and Mars), is chaotic, with a Lyapunov time of 5 million years (Laskar, 1989). An error of 15 m in the Earth's initial position gives rise to an error of about 150 m after 10 Ma; but this same error grows to 150 million km after 100 Ma. It is thus possible to construct ephemerides over a 10 million year period, but it becomes essentially impossible to predict the motion of the planets with precision beyond 100 million years.




So one approach to your question, to get at least a qualitative answer, would be to compare a 15-m error in the Earth's initial position to a 70-kg error in its mass. Let's start with the Earth-Sun gravitational potential energy, which depends on the mass of the Earth $\left(m\right)$ and its orbital radius $\left(r\right)$:


$$U\left(r, m\right) = -mr^{-1},$$


in units where $GM_\textrm{Sun} = 1.$ The errors in $U$, one due to the error in $m$ and the other due to the error in $r$ will be


$$\delta U_{m} = r^{-1}\delta m \textrm{ }\textrm{ (magnitude), and}$$


$$\delta U_{r} = mr^{-2}\delta r.$$


The ratio of the errors is


$$\frac{r^{-1}\delta m}{mr^{-2}\delta r} = \frac{\delta m}{m} \frac{r}{\delta r} \approx 6 \times 10^{-14}.$$


You can use SI units to get the numerical result, but you don't have to plug in any values to see what is going on. Because the units cancel, we can just compare the ratio of the errors to the ratio of the values. The ratio of the errors, $\delta m / \delta r$, is approximately 5. But the ratio of the values, $r/m$, is $\approx 10^{-13}.$


So, if the system's sensitivity to the mass error scales in a similar way to its sensitivity to the position error, it seems the mass error will have a much smaller effect than the position error for calculations covering 10 million years. Calculations that cover a longer period are not reliable regardless of the source of error.


thermodynamics - Why doesn't the entropy increase when two similar gases mix with each other?


Entropy increases when two substances mix with each other.


For example, the entropy of mixing of two different gases are given by $$\Delta S= 2Nk\ln\frac{V_f}{V_i}\;.$$


But, the entropy doesn't increase when the two gases mixing are same.



This is pointed by Daniel V Schroeder:



It's important to note that this result applies only if the two gases are different like helium and argon. If you start with the same gas on both sides, the entropy doesn't increase at all when you remove the partition.



Now, why is this so? Both the gases, though the same, increase their individual entropies when they expand, don't they?


So, why did Schroeder say there is no entropy of mixing? What actually happens such that there is no change in entropy when the gases are the same?




Thursday, 22 November 2018

astronomy - Could the earth have another moon?


First, to clarify: I'm not asking if perhaps there's a moon that we haven't found yet. The question is, theoretically, would the earth be able to have another stable moon in addition to the current one? Or, if the orbit couldn't be stable, why not? How large/small of a moon would it be able to have? And how do we know all this?



Answer



Jupiter has over 60 moons, and the dozens of man-made satellites (along with the thousands of pieces of space-trash) orbiting the Earth could be considered tiny moons.


Earth's main Moon would disrupt the orbit of anything smaller at certain radii. The particular disrupted orbits are called resonances, and occur where the orbital period in question divided by the period of the Moon reduces to a small fraction, like 1/2, 2/3, etc. It comes from the Moon giving a predictable tug on objects at those resonance orbits, similar to pushing a child on a swing higher and higher so that you eventually pitch him right out onto the ground. (Do not try this at home.)


The principle is spectacularly illustrated by Saturn's rings. (Some of*) the dozens of dark tracks indicate where one of the moons has a resonance. http://upload.wikimedia.org/wikipedia/commons/thumb/b/b1/Saturn%27s_rings_dark_side_mosaic.jpg/2200px-Saturn%27s_rings_dark_side_mosaic.jpg


[EDIT: To answer more of your questions, there are lots of stable orbits, and the Moon can also act to stabilize, not just destabilize orbits- see http://en.wikipedia.org/wiki/Trojan_(astronomy) . Any stable orbit would have to be above the top of the atmosphere, about 100 km, or else friction with atmospheric gas would drag it down to Earth. Moons could range from specks of dust up to potentially extremely large, even as large as the first Moon, but at great size the options for stability become much more limited. See http://en.wikipedia.org/wiki/Euler%27s_three-body_problem. All of this arises from classical, Newtonian mechanics. To greatly simplify history, Tycho Brahe took a lot of data on planet positions in the sky, Kepler took those observations and developed his purely empirical laws of orbital motion, and Newton took those empirical laws and developed a theoretical framework to account for them.]


Further reading-


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


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



*Some gaps are due to other or unknown factors, but for the most part...


vacuum - Why does Hauksbee's electrostatic machine produce light?


I'm reading on the history of the discovery of electricity and the electron, and I've went from reading about Rutherford's gold leaf experiment all the way back to Francis Hauksbee's spinning glass machine.


Hauksbee was essentially the first person to preform scientific study on the effects of electrostatics in vacuum. He observed that when spinning a glass orb evacuated of air (vacuum inside the orb) and while placing his hand on the spinning orb, a charge was created such that blueish glow was seen inside the orb, where his hands were placed, and on the opposite side of the orb.


As seen here:


https://youtu.be/iWmpBpzvIfY



I understand the glow of fluorescent tubes due to an electric current driven through a gas -filled tube, i understand the effect of the release of a photon when an electron decreases an energy level, and that gives the glow.


What I don't understand is why is the glow inside of the orb instead of outside, and why does the glow "stick" to Hauksbee's hand? and furthermore, what does the vacuum have to do with the glowing effect? Can I preform the same results with an orb that inside is a nearly complete vacuum or no air at all?


I'm intrigued because the advancement in our scientifically and technological abilities was due to the "combination" of vacuum and electrostatics (tribo-electric effect) and I want to understand the thought process that brought Hauksbee to build this machine.


Thanks in advance! :)




thermodynamics - How light causes increase in temperature


Temperature is the measure of movements of atoms. So if something is said to have high temperature it means that its atoms are moving fast or have high KE energy.


There are basically two ways heat can be travels from on object to another. 1) conduction. 2) Radiation.


My question is how does radiation causes increase in KE energy of atoms, when photons only interact with electrons? Or how does electrons going up in energy state translates to atom having more KE energy.



Answer



The answer to your question is very wide and include many phenomena. There is no single mechanism that converts absorbed energy into heat. I will give you a general overview on this topic.



Heat is transferred by radiation which is in general an electromagnetic wave (not light only). For example, IR radiation alone provides 49% of the heat provided to earth from sun radiation. Have a look at Heat section of this page.


In electromagnetic spectrum, the wavelength varies significantly, which means that materials respond differently to different ranges of wavelengths. For example (listing from long wavelength to short wavelength):




  1. Radio waves: The wavelength here is very long such that the materials constituents (atoms and molecules) don't sense the waves. Thus most solids are transparent to radio waves, which means there is no wave-material interaction. There is no heat generated.The transparency is clear as you can get cellphone reception in your home as the wave simply travel through walls and our bodies.




  2. Microwaves: The wavelength here is shorter, the photons energy in this level are comparable to the energy levels of the molecular rotational levels, which means that a molecule could absorb a microwave photon and start to rotate. This rotation is a form of kinetic energy of molecules which is translated as heat when you look at the whole ensemble of molecules. This mechanism is the mechanism known for microwave heating in your kitchen.





  3. IR: The photon energy in this range is comparable to molecular vibrational levels. So if a molecule absorbed an IR photon it will vibrate, which is translated into heat when you look at the whole ensemble of molecules.




  4. Visible light: things get complicated here, the photon energy here is comparable to electronic levels within an atom. There is no simple direct explanation of how photon energy is transformed into heat.




One can argue for visible light that an electron can absorb a photon and radiate it without increasing kinetic energy of atoms (no increase oh heat). That is true but that is a very simplistic picture of what happens in reality. First because the visible light is continuous spectrum, while the photons that can be absorbed by electrons are discrete. So there are many visible light photons with wavelengths that electrons can't absorb, which could be absorbed by ions or affect polar bonds somehow if they existed in the material. Second, some materials like solids (where heat by radiation is efficient) have energy bands rather than the simplistic electron levels concept. So the reaction of materials to visible light in this case is complex. But one of the mechanism of transforming photon energy to heat that I can think of here is the absorption of visible light by ion lattice in some crystals.




  1. UV: The photons in this range have enough energy to ionize the atoms they hit (by liberating an electron from its orbital). Again, there is no simple direct explanation of how photon energy is transformed into heat. One possible route is the inelastic scattering of electrons with materials atoms. For example, have a look at page 5 of this report where it is mentioned that inelastic electron scattering can induce phonons (lattice vibration, which is a form of kinetic energy of atoms)





  2. X rays: The photon energy in this range is much larger than the energy required to move electron from one level to other. X rays photons can ionize an atom and become lower energy photon through Compton scattering. There is no simple direct explanation of how photon energy is transformed into heat.




Briefly, heating by radiation in general can't be attributed to a simple explanation. The mechanisms through which photons energy is transformed into heat depend on the energy of the photon and the properties of the material. The simplistic picture of a photon being absorbed by electron is narrow to explain conversion to heat because it only describes a single electron in a free atom. That is a narrow view of material properties.


Have a look here and the nice figure of Non-ionizing radiation section of this page


Hopefully that was helpful


Does string theory provide quantitative experimental predictions?





Possible Duplicate:
What experiment would disprove string theory?



We carefully observe things, observe patterns and then build theories that predict.


String theory is frequently criticized for not providing quantitative experimental prediction.


What are the problems that prevent this theory from producing quantitative experimental predictions?


Is there no experiment suggested by string theorists to verify validity of their theory? Is the problem mathematical? (or it just requires many dimensional equipments...just kidding).


I am not criticizing the theory because to do that I should understand it first, but I haven't studied it. I just want to know.




electromagnetism - How are magnetic and electric fields transmitted?


This may be a duplicate, but how are (ideally) constant E- and B-/H-fields transmitted thru space? In this situation I like think of EM radiation as a note from a particle informing the universe of a change, but as the fields are constant, there aren't any notes/photons to send. For example, how will a bar magnet affect a cosmic ray passing by earth, and by what mechanism? Is this the province of virtual photons, or something else?



Answer



One has to define in what framework one is talking of electric and magnetic fields.


In the classical framework the field is defined, for simplicity lets take a point charge, as proportional to 1/r^2 and exists up to infinity. Thus classically there is no transmission for a static charge, it just is. When a charge is moving, i.e. changing its (x,y,z,t) dependence radiation is emitted, if there exists acceleration, as electromagnetic waves. The effect of the classical motion of these charges and consequent motion of field lines can be seen in this simulation . A video of constant motion ( non radiating) is here . It is a subject of interest for plasma physics. Another way of looking at static fields is that they are built up of electromagnetic fields of wavelength approaching infinity.


This brings us to the quantum framework, which as far as we have discovered with our experiments, is the fundamental underlying level of physics from which all classical theories emerge.



For example, how will a bar magnet affect a cosmic ray passing by earth, and by what mechanism? Is this the province of virtual photons, or something else?




Photons exist in the quantum mechanical domain and the classical electromagnetic field emerges from a large ensemble of photons. In this framework all information is transmitted by particles . So virtual photons will be transmitted between the bar of magnet and the cosmic ray with the velocity of light , though the effect will be infinitesimal. Virtual means that the exchange happens between known input and output particles and the exchanged photon is not on mass shell. Your example can be written up as such an interaction and the probability of the bar of magnet to change to path of the cosmic ray can be computed but it is not worth the trouble because it will be very very small ( 1/r^2 may be disguised in the solutions of the quantum mechanical problem but it is still there).


Wednesday, 21 November 2018

Forces as vectors in Newtonian mechanics


I seem to be confused about the nature of forces as vectors, in the basic Newtonian mechanics framework.


I know what a vector is as a mathematical object, an element of $R^3$. I understand that if a vector is drawn in a usual physical way as an arrow in space, it can be seen as a mathematical vector by translating it to begin at 0 and seeing where the arrow tip ends up. Generally it seems the word "vector" is used in such a way that a vector remains the same vector if it's translated arbitrarily in space (always corresponding to the same mathematical vector).


But now let's say I have a solid object, maybe a metal cube, with some forces acting on it: I push at it with a stick in the center of one facet, it's held by a rope in a different corner, etc. To specify each force that is acting on the cube it doesn't seem enough to specify the vector: I also need to specify the place of application. The cube behaves differently if I push it in the center as opposed in the corner etc.



I'm reading through J.P.Den Hartog's Mechanics that teaches me how to find the resultant force on the cube. I need to sum forces one by one using the parallelogram law, but I should always be careful to slide each force along its line of application, until two forces meet. I could just translate them all to start at the same point and add, but then I won't find the right resultant force, only its direction and magnitude; I will still need to find its line of application (maybe using moments etc.)


So let's say I'm calculating the resultant force "the right way": by sliding arrows along their lines until tails meet, adding, repeating. What am I doing mathematically? (it's not vector addition, that would correspond to just translating them all to 0 and adding) What mathematical objects am I working with? They seem to be specified with 4 free parameters: 3 for direction/magnitude of the vector and 1 more to displace it to the correct line of application; the location at the line of application seems irrelevant according to the laws of statics.




black holes - Why doesn't Hawking radiation cancel itself out?



Hawking Radiation is formed when particle, anti particle pairs formed by the uncertainty principle are separated by the event horizon of a black hole. It seems like an equal amount of particles and anti-particles should end up on each side of the event horizon. So why don't the particles and anti-particles annihilate with a new partner once separated from their original partner by the event horizon? Thus canceling out any radiation released.



Answer



This is covered by a number of existing questions, but I think it does no harm to present a fresh summary.


Firstly the analogy of the black hole absorbing one member of a pair of virtual particles is just an analogy, and actually a rather poor one, but let's go with it for now. In this analogy you're quite correct that equal numbers of particles and particles would be produced, but they wouldn't cancel each other out. An electron and positron don't just disappear when they meet, they annihilate into two 511keV photons. So a black hole producing equal numbers of particles and antiparticles would be a net emitter of photons.


But this isn't what happens. If the above analogy were correct you'd expect the Hawking radiation from a black hole to be sharply peaked. For example you'd have a peak at 511kev due to electron-positron annihilations, but no radiation below that until you got to the peaks for the neutrino annihilations (whatever mass they are). But what Hawking actually predicted is that the radiation would have a black body spectrum with no sharp peaks at all. This is the origin of the infamous information paradox.


The effect is actually due to the curvature of spacetime near the black hole. You may have heard that the curvature causes odd effects, for example time runs slower for observers hovering near the black hole, and distances measured radially change. These changes are what cause gravitational lensing, so we have real experimental evidence that they happen - this isn't just the wild rantings of theoreticians.


Anyhow, the curvature also means that a quantum field theory vacuum measured near to the black hole looks different from a quantum field theory vacuum measured far from the black hole. If you're interested, the two are related by a Bogoliubov transformation, though the Wikipedia article I've linked will be utterly incomprehensible to non-nerds. The result is that what the observer near the black hole considers to be a vacuum looks to an observer far from the black hole to be a black body glowing at a temperature of:


$$ T = \frac{\hbar c^2}{8\pi GMk_B} $$


There is a more detailed description of this in the Wikipedia article on Hawking radiation. This is fairly mathematical, but armed with my general description above you should be able to follow the main points.


Hawking's argument is quite general and doesn't rely on any specific mechanisms, like particle-antiparticle annihilation, and that's one of the reasons it predicts a featureless black body spectrum.



Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...