thermodynamics - Entropy generation of constant volume heat addition process

For an ideal gas constant volume heat addition process, change of entropy equation is:

$$\Delta S= c_v \ln\frac{T_2}{T_1}+ R \ln\frac{v_2}{v_1}= \int \frac{dq}{T}+S_{gen}$$

The term $R \ln\frac{v_2}{v_1}$ equals zero, since it’s a constant volume process.

For ideal gas $\int \frac{dq}{T} =c_v \ln\frac{T_2}{T_1}$ .

Then:

$$\Delta S= c_v \ln\frac{T_2}{T_1}= c_v \ln\frac{T_2}{T_1}+S_{gen}$$

Therefore, the $S_{gen}$ term equals zero and the process is reversible.

The question is: why does the $S_{gen}$ term equal zero and the process is reversible when this is a heat addition through a finite temperature difference?

To give a numerical example, imagine that an ideal gas is put in a rigid tank of uniform temperature where its initial temperature is $400$ K, and a hot reservoir at $500$ K. then heat is transferred from the hot reservoir to the rigid tank until the temperature of the rigid tank is $430$ K.

referring to the expressions above $T_1 =400$ K and $T_2 =430$ K.

Answer

The approach is to start from the definition of entropy change for a system $dS$ according to the laws of thermodynamics.

$$ dS \equiv \frac{\delta q}{T} $$

The heat flow $\delta q$ is path dependent and $T$ is the temperature of the system. We take a reversible path to find the reversible entropy change. We include an irreversible term as needed. Therefore, we write the expression below for the entropy change of a system under any process.

$$ \Delta S = \int \frac{\delta q_{rev}}{T} + \Delta S_{irr}$$

Reversible processes are those where the system and surroundings are in exact mechanical (equal pressures), thermal (equal temperatures), and chemical (equal chemical potentials) equilibrium at all points in time during the process. Such processes do not exist in the real world. They are hypothetical processes that allow us to make fundamental insights.

The difference between the entropy change of a system undergoing a reversible process $\Delta S_{rev}$ and the entropy change of a real world process is the irreversible entropy generation $\Delta S_{irr}$ or $S_{gen}$.

Ideal gases are also non-existent in the real world. They are however closely approximated by real gases to the point that we allow for the assumption even in practice.

For an ideal gas that undergoes a reversible change in temperature at constant volume, we obtain the following:

$$ dU^\star_{rev} = C_V dT = \delta q $$

$$ dS^\star_{rev} = C_v \frac{dT}{T} $$

$$ \Delta S^\star_{rev,V} = C_V \ln(T_f/T_i) $$

The last step requires that we assume that heat capacity is constant (the heat capacity of an ideal gases can depend only on temperature).

For an ideal gas that undergoes a reversible change in volume at constant temperature, we can also prove

$$ \Delta S^\star_{rev,T} = R \ln(V_f/V_i) $$

Combining the two expressions, we obtain the entropy change of an ideal gas with constant heat capacity undergoing any reversible change in temperature and volume as

$$ \Delta S^\star_{rev} = C_V \ln(T_f/T_i) + R \ln(V_f/V_i) $$

The total entropy change of the universe is the sum of the system and the surroundings. In a reversible process, the system and surroundings have the same entropy change. The total of the universe is therefore zero.

For an irreversible process, the entropy change of an ideal gas at constant heat capacity will still be the same as above. The irreversible entropy change is included and assigned to the surroundings. We use the term $S_{gen}$ to obtain

$$ \Delta S^\star_{univ} = \Delta S^\star_{sys} + \Delta S^\star_{surr} + S_{gen} = S_{gen}$$

The $S_{gen}$ term accounts for the fact that system and surroundings are not at perfect mechanical, thermal, or chemical equilibrium at all points in time during the process.

The entropy change of the system is found using a reversible path. The irreversibility is assigned to the surroundings. Using the definition of entropy, we can make the comparable statement $dS_{gen} = \delta q/T_{surr}$.

The founding equation starts with $\Delta S$. This is ambiguous. Is it to be $\Delta S_{univ}$, $\Delta S_{sys}$, or $\Delta S_{surr}$? This ambiguity should be clarified first.

The first expression after the equal sign is the entropy change for an ideal gas with constant heat capacity that undergoes a change in temperature and volume. Using only this term, we would intuitively set $\Delta S$ as $\Delta S_{sys}$. We cannot set it to $\Delta S_{univ}$. We can only set $\Delta S$ to $\Delta S_{surr}$ when we make the statement that the surroundings is an ideal gas.

The second expression is the entropy change of any generic irreversible process for any type of material. It could be viewed as $\Delta S_{sys}$ or $\Delta S_{surr}$. In the former case, $T = T_{sys}$. In the latter case, $T = T_{surr}$.

Let's now drop the ambiguous $\Delta S$ to obtain this.

$$C_V \ln(T_f/T_i) + R\ln(V_f/V_i) = \int \frac{\delta q}{T} + S_{gen}$$

The left side is the entropy change for an ideal gas under any process. This is the system. The right side is therefore the surroundings.

The fact that we include $S_{gen} \neq 0$ on the right side means that we are defining an irreversible process in the surroundings. The remaining term $\int \ldots$ must be the reversible entropy change of the surroundings. Therefore, $\delta q = \delta q_{rev,surr}$ and $T = T_{surr}$.

When the system is rigid, the term for $V_f/V_i$ drops from the left side. With temperatures $T_i = 400$ K, $T_f = 430$ K, and $T_{surr} = 500$ K, the expression becomes as below.

$$C_V \ln(430/400) = \int \frac{\delta q_{rev,surr}}{500} + S_{gen}$$

With the assumption that you know the material, you know $C_V$. You have one equation and one two unknowns. You have one of two approaches to an answer. You must state the amount of heat flow out of the surroundings ($\delta q_{rev,surr}$) as a constant. An immediate option is to say that this value is the reversible heat given to the system $\delta q_{rev,surr} = -\delta q_{rev,sys} = - C_V dT$. From this, you obtain

$$C_V \ln(430/400) = C_V \frac{400 - 430}{500} + S_{gen}$$

This allows you to solve for $S_{gen}$.

Alternatively, you must state the amount of irreversible entropy generated $S_{gen}$. When you also say that $\delta q_{rev,surr}$ is constant, you can solve for it.

Finally, with the parameters as given, the process is not reversible because the temperatures of the system and the surroundings are not the same at all stages. You could presume to call the process reversible so that $S_{gen} \equiv 0$. This means, you must ignore the temperature of the surroundings in your considerations. This gives the following:

$$C_V \ln(430/400) = \int \frac{\delta q_{rev,surr}}{T_{surr}}$$

With knowledge of the material, you use this to determine the reversible entropy change in the system or in the surroundings.

quantum field theory - What is the spin-statistics theorem in higher dimensions?

In $d = 3+1$ dimensions, the spin-statistics theorem states that fermionic particles have half-integer spin and bosonic particles have integer spin, in a well-behaved relativistic quantum field theory. We also know that in $d = 2 + 1$ dimensions, there's more freedom; you can have anyons, which have neither bosonic or fermionic exchange statistics.

I'm interested in what the exact statement of the spin-statistics theorem is in higher dimensions. I would assume it would say that "particles are fermionic if and only if they're spinors", where a "spinor" is any representation of the Lorentz group where a $2 \pi$ rotation does not act trivially.

However, I haven't found a definite statement of this anywhere. Moreover, this paper seems to say, very surprisingly, that it totally changes depending on the value of $d \, (\text{mod}\ 8)$, e.g. in $d = 8$ spinors must be bosons.

What's going on here? What's the statement of the spin-statistics theorem in higher dimensions?

Answer

When formulating a physical theory, one usually begins with a set of axioms. The theory itself will be just as useful as its axioms are accurate. In particular, when dealing with a supposedly fundamental theory, the following set of axioms seems natural and well-motivated by experiments:

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  There is a set of fundamental entities, referred to as particles, which cannot be further subdivided. Any set of particles can be combined to form more complex systems.

  
  



• 
  

  This "combining" of particles is commutative (an electron plus a positron is the same as a positron plus an electron), and there is an inverse-operations (i.e., to each particle we can associate an anti-particle).

  
  


• 
  

  A few other technical but natural assumptions regarding how systems of particles compose.

  
  

One can prove (Deligne's theorem on tensor categories, cf. Ref 1) that the most general mathematical structure that describes a physical model with these properties is a super-Lie algebra. More precisely, particles can be thought of as irreducible representations of such an algebra, in the sense of Wigner, and where "composing particles" is just a tensor product. Needless to say, it may very well be the case that the odd part of the algebra is trivial, and indeed no experiment has ever measured a superpartner. (The theorem does not imply that supersymmetry must exist; rather, it says that supersymmetry is the only possible extension of conventional QFT if we want to keep the concept of particle, as opposed to, say, strings).

If one further assume that it makes sense to evolve these particles in time so as to talk about scattering (and that the scattering matrix satisfies some natural assumptions, such being non-trivial), then it follows from a theorem due to Haag, Łopuszański, and Sohnius (Ref. 2), that the super-Lie algebra is a direct sum of the super-Poincaré algebra and a reductive (bosonic) Lie algebra, the algebra of internal symmetries (colour, isospin, etc.). As before, particles are to be thought of as irreducible representations of the symmetry algebra. The details can be found in this PSE post (the odd part of the algebra, if non-trivial, organises particles into supermultiplets).

As described therein, particles can be further organised according to the effect of the fundamental group of the rotation group, to wit, $\mathbb Z_2$. If a particle transforms trivially, it is said to be a boson, and if non-trivially, a fermion. In other words, bosons are invariant under a $2\pi$ rotations, while fermions acquire a negative sign. This is not the content of the spin-statistic theorem, but rather a tautology that follows from the definition of boson/fermion.

Once classified the most general particle content of a theory, the next step is to construct specific models for them. The most general technique we have is known as quantum field theory, where one embeds these particles into fields, that is, objects with simple transformations under the (super-)Poincaré group. This allows us to construct models in a straightforward manner (Lagrangian field theory). Without fields, it is very complicated to construct interactions that satisfy the required assumptions (unitarity, covariance, etc.).

Let me stress the following: particles are described by representations of (the universal conver of) $\mathrm{SO}(d-1)$, while fields are described by representations of (the universal cover of) $\mathrm{SO}(1,d-1)$. In both cases it makes sense to speak of bosons/fermions, but the spin-statistics theorem concerns the latter, not the former. What's more, the spin-statistics theorem does not say that bosonic fields must commute, and fermionic fields anti-commute. Rather, it says the following: if bosonic fields anti-commute for spacelike separations, then the field must be trivial (and, similarly, so must be fermionic fields if they commute). In other words, we do not say that (fermions) bosons must (anti-) commute; instead, we say that the other possibility is forbidden. If we further assume that these are the only two possibilities, then we do get the quoted result: bosons must commute and fermions anti-commute.

It is important to note that other options besides $ab\pm ba$ have been considered in the literature, and while they make sense from an algebraic point of view, they do not tend to work well from a physics point of view. I do like to think that the fundamental reason these two possibilities are the only ones allowed by nature is the aforementioned Deligne theorem. That being said, the canonical proof of the spin-statistics theorem as quoted above can be found in Ref.3 (which addresses the $d=4$ case only, but the argument admits a natural extension to arbitrary $d$), and in Ref.4 (this time for arbitrary $d$, but using a more involved argument). In short, the higher-dimensional spin-statistics theorem is identical to the $d=4$ case: bosonic fields cannot anti-commute, and fermionic fields cannot commute.

References.

1. 
   

   Supersymmetry and Deligne's Theorem, Urs Schreiber, https://www.physicsforums.com/insights/supersymmetry-delignes-theorem/

   
   


2. 
   

   Weinberg's QFT, Vol.III.

   
   
   


3. 
   

   Wightman's PCT, spin-statistics, and all that.

   
   


4. 
   

   Causality, antiparticles, and the spin-statistics connection in higher dimensions, S. Weinberg, https://doi.org/10.1016/0370-2693(84)90812-8 and Causal fields and spin-statistics connection for massless particles in higher dimensions, N. Ohta, https://doi.org/10.1103/PhysRevD.31.442.

   
   

black holes - Space-time and gravitational fields

I am reading a book by Carlo Rovelli, Seven Brief Lessons On Physics, and would like to check if I have understood something. Apologies if my question is badly phrased, feel free to edit where appropriate.

I am not a physicist, just an enthusiast.

It was this excerpt that made me think gravitational waves had something to do with time.

The heat of the black holes is like Rosetta Stone of physics, written in a combination of three languages-Quantum, Gravitational and Thermodynamic - still awaiting decipherment in order to reveal the true nature of time

And the following that made we wonder if they were to do with space as well? My understanding is that space and time are synonymous?

The heat of black holes is a quantum effect upon an object, the black hole, which is gravitational in nature...

It was the next line, following that inspired my question was this,

... It is the individual quanta of space, the elementary grains of space, the vibrating 'molecules' that heat the surface of black holes

Talking about the gravitational field being space-time

The gravitational field, as we saw in the first lesson, is space itself, in effect space-time

Text referred to in the 'First Lesson'

Einstein had been fascinated by this electromagnetic field and how it worked... soon came to understand that gravity, like electricity must be conveyed by a field as well... the gravitational field is not diffused through space; the gravitational field is that space itself

Question: Is space-time 'made of' gravitational waves? Is that field it's fundamental building block?

It seems to me from all of this that space-time is indeed

Answer

I'll try to boil down several of your questions and answer what I think is most fundamental, and hopefully clarify things in the process:

Gravity is completely synonymous with the shape of spacetime across all 4 dimensions (3 space, 1 of time). The reason we speak of spacetime is thus: When you (having negligent mass) stand in a "gravity field" such as that caused by a massive object such as the Earth, you notice 2 things:

First, that space seems to have a direction, i.e. objects will "fall" towards the center of the dominant mass

Second, that your watch will tick somewhat more slowly than it did when you were far away from this gravity well.

These phenomena of acceleration (a gradient in space) and changes in the rate of your watch ticking (so a gradient in time) is why we don't speak of space and time as separate entities - they are inextricably linked, and movement through one affects your movement through the other. The stronger the gravity field you're exposed to (so the more massive the object you're near), the slower your watch will tick when compared to someone standing safely outside the field. Spacetime is the fabric of the universe - so far as we know, there's nothing "under" it, nothing that it is "made up of" - and GR treats it as such. So when you think of spacetime, think of it as a landscape of hills and valleys in both 3D space and in time, all caused by the various masses that reside there, and understand that those hills and valleys are gravity.

Now, gravity waves, then, are ripples in this landscape that can be caused by an accelerating massive object. So if two black holes accelerate and smash into each other, some ripples in spacetime will travel out from the disturbance and we may be able to detect it.

With our understanding of what gravity is above, what would a gravity wave be expected to 'look like?'

We'd expect to be able to pick it up one of two ways: Either by a change in space (seen as a distance fluctuation between 2 points in our detector) or a change in time (an aberration in the rate a very reliable clock is ticking). Which now makes perfect sense, because gravity is simply the shape of space and time, bound up together.

differential geometry - Is partial derivative a vector or dual vector?

The textbook(Introduction to the Classical Theory of Particles and Fields, by Boris Kosyakov) defines a hypersurface by $$F(x)~=~c,$$ where $F\in C^\infty[\mathbb M_4,\mathbb R]$. Differentiating gives $$dF~=~(\partial_\mu F)dx^\mu~=~0.$$ The text then says $dx^\mu$ is a covector and $\partial_\mu F$ a vector. I learnt from another book that $dx^\mu$ are 4 dual vectors(in Minkowski space), $\mu$ indexes dual vector themselves, not components of a single dual vector. So I think $\partial_\mu F$ should also be 4 vectors, each being the directional derivative along a coordinate axis. But this book later states that $(\partial_\mu F)dx^\mu=0$ describes a hyperplane $\Sigma$ with normal $\partial_\mu F$ spanned by vectors $dx^\mu$, and calls $\Sigma$ a tangent plane (page 33-34). This time, it seems to treat $\partial_\mu F$ as a single vector and $dx^\mu$ as vectors. But I think $dx^\mu$ should span a cotangent space.

I need some help to clarify these things.

[edit by Ben Crowell] The following appears to be the text the question refers to, from Appendix A (which Amazon let me see through its peephole):

Elie Cartan proposed to use differential coordinates $dx^i$ as a convenient basis of 1-forms. The differentials $dx^i$ transform like covectors [...] Furthermore, when used in the directional derivative $dx^i \partial F/\partial x^i$, $dx^i$ may be viewed as a linear functional which takes real values on vectors $\partial F/\partial x^i$. The line elements $dx^i$ are called [...] 1-forms.

quantum mechanics - Momentum as Generator of Translations

I understand from some studies in mathematics, that the generator of translations is given by the operator $\frac{d}{dx}$.

Similarly, I know from quantum mechanics that the momentum operator is $-i\hbar\frac{d}{dx}$.

Therefore, we can see that the momentum operator is the generator of translations, multiplied by $-i\hbar$.

I however, am interested in whether an argument can be made along the lines of "since $\frac{d}{dx}$ is the generator of translations, then the momentum operator must be proportional to $\frac{d}{dx}$". If you could outline such an argument, I believe this will help me understand the physical connection between the generator of translations and the momentum operator in quantum mechanics.

Answer

In the position representation, the matrix elements (wavefunction) of a momentum eigenstate are $$\langle x | p\rangle = \psi_p(x) = e^{ipx}$$ The wavefunction shifted by a constant finite translation $a$ is $$\psi(x+a)$$ Now the momentum operator is the thing which, acting on the momentum eigenstates, returns the value of the momentum in these states, this is clearly $-i\frac{d}{dx}$.

For our momentum eigenstate, if I spatially shift it by an infinitesimal amount $\epsilon$, it becomes $$\psi(x+\epsilon) = e^{ip(x+\epsilon)} = e^{ip\epsilon}e^{ipx} = (1+i\epsilon p + ...)e^{ipx}$$ i.e. the shift modifies it by an expansion in its momentum value. But if I Taylor expand $\psi(x+\epsilon)$, I get $$\psi(x+\epsilon)= \psi(x)+\epsilon \frac{d}{dx} \psi(x)+... = \psi(x)+i\epsilon(-i\frac{d}{dx})\psi(x)+... $$ So this is consistent since the infinitesimal spatial shift operator $-i\frac{d}{dx}$ is precisely the operator which is pulling out the momentum eigenvalue.

homework and exercises - How much energy in form of heat does a human body emit?

How much energy in form of heat does a human body emit at rest level?

Answer

That's easy. The energy requirement of an average man is 2,500 Calories per day, and one Calorie is 4184J. Therefore he emits about 10.5MJ/day or about 120W.

An average woman requires 2,000 Calories per day, so she emits about 97W.

homework and exercises - Calculate acceleration and time given initial speed, final speed, and travelling distance?

A motorcycle is known to accelerate from rest to 190km/h in 402m.

Considering the rate of acceleration is constant, how should I go about calculating the acceleration rate and the time it took the bike to complete the distance?

Answer

For constant acceleration you have two equations that you need to solve for time $t$ and acceleration $a$

$$ x = \frac{1}{2} a t^2 $$ $$ v = a t $$

It is up to you from here.

$$ \left(402\, {\rm m}\right) = \frac{1}{2} \, a \,t^2 $$ $$ \left( 190 {\rm \frac{km}{hr}}\right) \left( \frac{1000 {\rm \frac{m}{km}}}{3600 {\rm \frac{sec}{hr}}} \right) = \left(52.777 {\rm \frac{m}{sec}} \right)\, a \,t $$

electrostatics - Gauss's Law of Electric Field how it actually works? & How Gauss derived it?

1. 
   

   I want to know how Gauss derived his equation of Electric Field. Did he derive it from Coulomb's law? I don't think so.

   
   


2. 
   

   Please tell me some details about how this law works?

   
   



3. 
   

   inside a Gaussian surface neat $E = 0$ and why it doesn't depend on the shape of the Gaussian surface? (I know, the electric field goes in the Gaussian surface is canceled by those going out of it, but how come it doesn't depend on the shape of that surface?)

   
   

homework and exercises - Calculating the generating functional for the free scalar field explicitly by completing the square

I'm trying to reproduce the calculation resulting in equation (3.12) in the following script:

http://www.desy.de/~jlouis/Vorlesungen/QFTII11/QFTIIscript.pdf

The only difference is that in my notes, the Feynman propagator was defined with an extra minus sign. Let $A_x = (\Box_x + m^2) $, then

$$A_xD(x-y) = -i\delta(x-y).$$

The generating functional is given by:

$$Z[J] = \int D \phi \exp{i\int d^4x(-\frac{1}{2}\phi A \phi +J\phi)}$$

Let $I = \int d^4x(-\frac{1}{2}\phi A \phi +J\phi)$ and change variables

$$\phi(x) = \phi'(x) + \phi_0 (x) \equiv \phi'(x) + i\int d^4y D(x-y)J(y)$$

so that $D\phi'=D\phi$ and

$$A_x\phi_0(x) = J(x)$$

Then

$$I = \int d^4x \left( -\frac{1}{2} \left[ \phi'A\phi' + \phi'A\phi_0 + \phi_0A\phi' + \phi_0A\phi_0 \right] +J\phi' + J\phi_0\right)$$

this simplifies to

$$I = \int d^4x \left( -\frac{1}{2} \phi'A\phi' -\frac{1}{2} \phi_0A\phi' +\frac{1}{2} J\phi' + \frac{1}{2} J\phi_0\right)$$

Now, we just want to keep the $-\frac{1}{2} \phi'A\phi' $ and $\frac{1}{2} J\phi_0= \frac{i}{2} \int d^4y J(x)D(x-y)J(y)$ parts. The first one is independent of $J$ and provides the normalization, and the second, upon multiplication by $i$ and exponentiation, leads to the correct answer:

$$Z[J] = N e^{-(1/2)\iint d^4x d^4y J(x)D(x-y)J(y)}.$$

However, I absolutely can't see why the other two parts of the expression for $I$ need to cancel. That is, why the following holds:

$$-\frac{1}{2} \phi_0(x) A\phi'(x) +\frac{1}{2} J(x)\phi'(x) = 0?$$

Answer

The cancellation occurs by integrating the term $-\frac12 \phi_0(x)A\phi'(x)$ twice by parts.

Ignoring surface terms and the $-\frac12$ coefficient, this is:

\begin{aligned} \int d^4x ~\phi_0(x)A\phi'(x) &= \int d^4x ~\phi_0(x)\partial^2\phi'(x) + \phi_0(x)m^2\phi'(x)\\ &= -\int d^4x ~\partial\phi_0(x)\partial\phi'(x) + \int d^4x ~\phi_0(x)m^2\phi'(x)\\ &= \int d^4x ~\partial^2\phi_0(x)\phi'(x) + \int d^4x ~\phi_0(x)m^2\phi'(x)\\ &= \int d^4x ~A\phi_0(x)\phi'(x)\\ &= \int d^4x J(x)\phi'(x) \end{aligned}

electrostatics - Wimshurst machine - How does it work?

Look at this VIDEO to see what is supposedly happening:

A Wimshurst machine is a seemingly simple device consisting of two plastic wheels with embedded metal plates on the rim. The wheels spin in opposite direction, facing each other. Two metal bars, one on each side, span the device: They are at an angle of 90°, but separated by the two discs spinning between them. At the end of the bars, metal brushes touch the plates spinning underneath them. A slight initial charge imbalance on one of the plates (which necessarily exists) causes the bar brushing against it to polarize along its length, so the metal plate on the other end of the same disc is charged oppositely and retains that charge when the brush disconnects during the spinning. As this two-plate "dipole" spins past the bar located behind the other disc, it induces an opposite dipole in its opposing plates on the other disc, which, in turn, polarizes the bar brushing against them. The charged metal plates give off their charges to two Leyden jars which connect to the plates with brushes of their own.

However, why do the Leyden-jar-brushes collect the charges? They tend to get very negative/positive, so why don't they push the electrons/holes away instead of collecting them? Wouldn't the brushes charge the passing plates which then travel around to the other Leyden brush, and get discharged there?

Answer

Ignore the collecting combs and Leiden jars for the moment. As you noted, the Wimhurst disk and 45° shorting bars form a powerful electrostatic pump such that the left side of both disks gets charged one way and the right side of both disks gets charged the other way.

This charge feeds back through the shorting bars and builds up until something leaks the charge, and that's where the combs and Leiden jars come in. Under one pair of combs, the disk sectors are charged highly negative, and since the two adjacent sectors on the two disks are charged the same way, the electrons wants to escape. The combs provide such an escape route, and the electrons jump to the combs and accumulate in the Leiden jar. Yes, the charge already in the jar repels the incoming electrons, but the pump is strong enough so that (to a point) the electrons keep coming.

I found an interesting article in the 1888 Proceedings of the Institution of Electrical Engineers called "The Influence Machine from 1788 to 1888". It includes an interesting explanation of why the Wimhurst Machine needs both combs and brushes to operate properly.

Brushes directly contact the underlying sectors, and so only an infinitesimal potential is required to push electrons across them. Combs are above the sectors, with an air gap, so that a significant voltage must be present before the electrons cross the gap.

The shorting bars should end in brushes, otherwise the machine won't be self-starting. With brushes, the inevitable imbalance of charge will start being amplified right away; if combs were used then there'd have to be a significant initial charge before the electrons could cross the gap and start the amplification process.

The Leiden jar collection points should end in combs, otherwise a discharge may reverse the polarity of the machine. Were brushes used, the charge on the sectors under the brushes could be completely neutralized, returning the machine to its initial state awaiting a random imbalance. With combs, however, not all of the charge on the sector can be removed by the discharge, and thus the machine will maintain its polarity and not have to start from scratch.

(Nineteenth century technology is SO cool...)

vacuum - Does the density of virtual particles decrease when space expands due to dark energy?

I understand that vacuum (i.e. QED vacuum) is theorized to contain spontaneously appearing and annihilating pairs of virtual particles. Suppose the average density of those particles at a given moment in time and a particular location in space is Dinit.

Suppose we wait an amount of time such that dark energy expanded the space at that location and the density of virtual particles is measured to be Dexp.

Will the density at which the virtual particles appear at that location be the same or different? In other words, will Dinit = Dexp or Dinit > Dexp?

Edit: Maybe my understanding is incomplete, this is what I would like to know:

Suppose at some time tinit, the universe has a volume Vinit, and some experiment is performed that estimates that, on average, in some window of time, 2 electron-positron pairs appear and annihilate. The "pair density" Dinit is computed to be 2/Vinit.

Then time tlong passes, dark energy expands space, and the universe now has volume Vnew. The same measurement is performed, estimating the average number of virtual particle pairs appearing and disappearing in the new expanded universe.

Question, would A or B be expected to be observed:

A: New avg number of pairs conserved at ~2, thus pair density Dexp=2/Vnew, and thus Dexp < Dinit?

OR

B: New avg pairs >2, thus pair density Dexp= >2/Vnew, and thus pair density conserved and Dexp ~= Dinit?

astronomy - What does the sky look like to human eyes from orbit?

There are numerous pictures, obviously, of the blackness of space from the shuttle, the space station, and even the moon. But they all suffer from being from the perspective of a camera, which is not sensitive enough to pick up the stars in the background when compared to the bright foreground objects (the limb of the Earth, the station, moon, etc). I've seen some photos that show a few of the brightest stars, but nothing special.

Are there any photos or eye witness accounts from astronauts of what it looks like to a human with night-adjusted vision? If I were in an orbit similar to the space station and looked away from the Earth, would I be able to see more stars than I ever could on Earth, or would it only be marginally better than the best terrestrial night viewing?

Answer

I haven't been to space :( and don't know of any accounts to point you to but I suspect that the view would be marginally better than that on earth.

First, the "black of space" would be really black i.e. no light. Even in dark skies there is a bit of scattered light in the atmosphere (even if just from scattered starlight) so you'd have higher contrast.

Additionally, you wouldn't have any "seeing" effects, the bluring of the starlight by the shifting atmosphere so the stellar images would be more concentrated. Not that you'd probably be able to consciously notice. The net effect would be to sharpen the stellar image on you eyes and increase the contrast a bit more.

I suspsect you'd see a few more stars for the above reasons, especially the latter as you eyes would have a better chance to detect light from the sharper stellar images. However, it wouldn't be a huge amount more as the stars are intrinsically faint and your eyes are only so senstive.

wavefunction - Physical intepretation of nodes in quantum mechanics

I am taking my second course in QM, and my head is starting to spin as it probably should. But I would very much like to clear up my head about a few details regarding the wave function. As I know it is impossible to predict where particles are and one can only give a probability of where it should be.

The simplest case is a "frictionless" particle "bouncing" back and forth inside a infinite square well. Eg a particle in the following potential

$$ V(x) = \left( \begin{array}{cc} 0 \ , & \text{for} \ 0 \leq x \leq a \\ \infty \ , & \text{elsewhere} \end{array} \right) $$ Which gives rise to the following normalized solution $$ \psi_n(x) = \sqrt{\frac{2}{a}} \sin\left( \frac{\pi n}{a}x \right) $$ My problem is what the nodes in the square function represents. If I draw $|\Psi_2(x,0)|^2=|\psi_2(x)|^2$ I obtain a graph similar to the one below.

What is the physical explenation that finding the particle around a small region around $a/2$ is close to zero? Or why is it so much less likely to find it near $a/2$ than $a/4$? Eg why is $$ P(a/2 -\varepsilon \leq X \leq a/2+\varepsilon) = \int_{a/2-\varepsilon}^{a/2+\varepsilon} \left| \psi_2(x) \right|^2 \,\mathrm{d}x \sim 0 $$ for small $\varepsilon$

electromagnetism - Motional EMF and EMF?

What is the difference between motional EMF = $-vBL$ , and Faraday's law of induction $\displaystyle\mathcal{E} = \left|\frac{d\Phi_B}{dt}\right|$? Aren't they the same? What is the relation of Lorentz force to motional EMF?

Answer

Faraday's law $\mathcal{E}=-d\Phi/dt$ can be used in a variety of situations, including ones where the phrase "motional EMF" is appropriate.

Your particular expression $-vBL$ is applicable only for a very particular situation. Probably a sliding bar, which is part of a circuit, in a uniform magnetic field. That expression can be derived using Faraday's law, and is a one- or two-liner if you go through it. I believe you can derive it using other methods ($\vec{F}=q\vec{v}\times\vec{B}$ and such), but Faraday's law is applicable here too, and in so many other situations where that force law would give misleading answers.

So I suppose the phrase motional EMF is used when there is physical movement of a conductor. The term Faraday's law is typically used to indicate the method one uses to calculate what the EMF is.

To address your Lorentz force law question more explicitly: Faraday's law is used especially in situations where $F=q(\vec{E} + \vec{v}\times\vec{B})$ might yield a misleading answer since an induced electric field causes by a changing magnetic field causes the force, rather than a magnetic force as one might expect. (Well, that's the usual interpretation. SR grumble grumble.) You don't run into this situation with the sliding bar example, but if you have a stationary conducting loop immersed in a changing magnetic field, one might ignore the electric field in the Lorentz force law since you're not actively creating such a field. But actually there is an electric field; Faraday's law tells you what the path integral of that electric field is, which is useful.

classical mechanics - Boundary conditions for calculus of variations in phase space and under canonical transformations

This might be a stupid question, but I just don't get it. In Hamiltonian mechanics when examining conditions for a $(\boldsymbol{q},\boldsymbol{p})\rightarrow(\boldsymbol{Q},\boldsymbol{P})$ transformation to be canonical one starts with $$ \dot{q}_ip^i-H(\boldsymbol{q},\boldsymbol{p},t)= \dot{Q}_iP^i-\bar{H}(\boldsymbol{Q},\boldsymbol{P},t)+\frac{d}{dt}W(\boldsymbol{q},\boldsymbol{Q},t)$$ where $\bar{H}$ is the transformed Hamiltonian, and $W$ is the generating function (now a function of $\boldsymbol{q}$ and $\boldsymbol{Q}$). This term shouldn't break Hamilton's principle, since $$ \delta\int_{t_1}^{t_2} dt\frac{d}{dt}W(\boldsymbol{q},\boldsymbol{Q},t)=\delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_2}-\delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_1}=0-0=0 .$$ But I don't see why the variation of $W$ should disappear at the endpoints (say at $t_1$). Expanding leads to: $$ \delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_1}=\left(\frac{\partial W}{\partial q_i}\right)_{t_1}\underbrace{\delta q_i(t_1)}_{=0}+ \left(\frac{\partial W}{\partial Q_i}\right)_{t_1}\delta Q_i(t_1)=\left(\frac{\partial W}{\partial Q_i}\right)_{t_1}\delta Q_i(t_1).$$ $\boldsymbol{Q}$ is itself a function of $\boldsymbol{q}$ and $\boldsymbol{p}$, so $$ \delta Q_i(t_1)=\left(\frac{\partial Q_i}{\partial q_k}\right)_{t_1}\underbrace{\delta q_k(t_1)}_{=0}+\left(\frac{\partial Q_i}{\partial p_k}\right)_{t_1}\delta p_k(t_1)=\left(\frac{\partial Q_i}{\partial p_k}\right)_{t_1}\delta p_k(t_1). $$ It seems as if we also needed the variation of $\boldsymbol{p}$ to vanish at the endpoints, and I don't get this because (at least in cartesian coordinates) $\boldsymbol{p}=m\dot{\boldsymbol{q}}$ and the velocity can be different along the original and the varied orbitals even at the endpoints (they can point in totally different directions), so in general $\delta \dot{\boldsymbol{q}}(t_1)\neq 0$. What am I doing wrong? Can someone help me with this, please?

Answer

These are very good questions.

1. 
   

   Let us start with the old phase space variables $(q^k,p_{\ell})$. The Hamiltonian action is $$S_H~=~\int_{t_i}^{t_f} \! dt ~L_H, \qquad L_H~:=~\dot{q}^j p_j - H(q,p,t).\tag{A}$$ Its infinitesimal variation reads $$ \delta S_H ~=~ \text{bulk-terms} ~+~ \text{boundary-terms},\tag{B}$$ where $$\text{bulk-terms}~=~\int_{t_i}^{t_f} \! dt \left(\frac{\delta S_H}{\delta q^j}\delta q^j + \frac{\delta S_H}{\delta p_j}\delta p_j \right)\tag{C}$$ yield Hamilton's equations, and where $$\text{boundary-terms}~=~\left[p_j\underbrace{\delta q^j}_{=0} \right]_{t=t_i}^{t=t_f}~=~0\tag{D}$$ vanish as they should because of, say$^1$, essential/Dirichlet boundary conditions (BCs) $$ q^j(t_i)~=~0\qquad\text{and}\qquad q^j(t_f)~=~0. \tag{D}$$ Notice that the momenta$^2$ $p_j$ are unconstrained at the boundary.

   
   


2. 
   

   Next let us consider new phase space variables $(Q^k,P_{\ell})$. The action of type 1 reads$^3$ $$S_1~:=~\int_{t_i}^{t_f} \! dt ~L_1~=~S_K+\left[ F_1(q,Q,t) \right]_{t=t_i}^{t=t_f}, \qquad S_K~:=~\int_{t_i}^{t_f} \! dt ~L_K, $$ $$ L_1~:=~L_K+\frac{dF_1(q,Q,t)}{dt}, \qquad L_K~:=~ \dot{Q}^j P_j - K(Q,P,t),\tag{F}$$ where the old positions $q^j=q^j(Q,P,t)$ are implicit functions of the new phase space variables $(Q^k,P_{\ell})$. Its infinitesimal variation reads $$ \delta S_1 ~=~ \text{bulk-terms} ~+~ \text{boundary-terms},\tag{G}$$ where $$\text{bulk-terms}~=~\int_{t_i}^{t_f} \! dt \left(\frac{\delta S_1}{\delta Q^j}\delta Q^j + \frac{\delta S_1}{\delta P_j}\delta P_j \right)\tag{H}$$ yield Kamilton's equations, and where $$\text{boundary-terms}~=~\left[\underbrace{\left(P_j+\frac{\partial F_1}{\partial Q^j}\right)}_{=0}\delta Q^j +\frac{\partial F_1}{\partial q^i}\underbrace{\delta q^j}_{=0} \right]_{t=t_i}^{t=t_f}~=~0\tag{I}$$ vanish as they should.

   
   

--

$^1$ Alternatively, one could impose natural BCs, or perhaps some mixture thereof.

$^2$ Note that in QM it would conflict with the HUP to simultaneously impose BCs on a canonical conjugate pair.

$^3$ Notation conventions: Kamiltonian $K\equiv\bar{H}$ and type 1 generating function $F_1\equiv G_1\equiv W$.

thermodynamics - Why do energy transfers always result some heat loss?

explain that energy transfers and transformations in mechanical systems always result in some heat loss to the environment. Like why is it necessary for a heat loss?

Answer

This is a hard topic but I'll try to give a brief reply.

"Heat" consists of random motion of atoms vibrating about in all directions at a variety of speeds. A heat engine strives to catch that random motion and turn it into a moving piston or a spinning turbine wheel. A simple measure of how much heat we have in (for example) a sample of hot gas is its temperature, and a perfect capture of heat energy from hot gas by an engine would mean the exhaust of that engine would be at zero temperature.

But the environment surrounding that engine is not at "zero temperature", it is at 270 degrees celsius above zero, which means that the exhaust of the engine will leave it with heat that the engine could not capture. This is one reason why heat engines cannot in principle extract all the heat energy out of a sample of hot gas.

Another reason is that no engine is perfect is that all its bearings and sliding and spinning parts have friction trying to slow them down as they move, and that friction steals work from the engine and turns it into heat which leaves the engine while it is running.

Yet another reason no heat engine is perfect is that it is impossible to completely prevent heat from leaking out through the walls of the engine while it is running. That leakage also contains heat energy that the engine obviously cannot use to do work for us.

quantum mechanics - Violation of Pauli exclusion principle

From hyperphysics (emphasis mine):

Neutron degeneracy is a stellar application of the Pauli Exclusion Principle, as is electron degeneracy. No two neutrons can occupy identical states, even under the pressure of a collapsing star of several solar masses. For stellar masses less than about 1.44 solar masses (the Chandrasekhar limit), the energy from the gravitational collapse is not sufficient to produce the neutrons of a neutron star, so the collapse is halted by electron degeneracy to form white dwarfs. Above 1.44 solar masses, enough energy is available from the gravitational collapse to force the combination of electrons and protons to form neutrons. As the star contracts further, all the lowest neutron energy levels are filled and the neutrons are forced into higher and higher energy levels, filling the lowest unoccupied energy levels. This creates an effective pressure which prevents further gravitational collapse, forming a neutron star. However, for masses greater than 2 to 3 solar masses, even neutron degeneracy can't prevent further collapse and it continues toward the black hole state.

How then, can they collapse, without violating the Pauli Exclusion Principle? At a certain point does it no longer apply?

Answer

The Pauli exclusion principle is being applied here to FREE neutrons. There are always free energy/momentum states for the neutrons to fill, even if they are compressed to ultra-high densities; these free states just have higher and higher energies (and momenta).

One way of thinking about this is in terms of the uncertainty principle. Each quantum state occupies approximately $h^3$ of position-momentum phase-space. i.e. $(\Delta p)^3 \times (\Delta x)^3 \geq h^3$. (Actually each momentum state can accomodate 2 fermions - spin up and spin down).

If you increase the density, $\Delta x$ becomes small, so $\Delta p$ has to become large. i.e. the neutrons will occupy higher and higher momentum states as the Fermi energy is increased. You can make the density as high as you like and the PEP is not violated because the particles gain higher momenta.

The increasing momenta of the neutrons supplies an increasing degeneracy pressure. However, there is a "saturation", because eventually all the neutrons become ultrarelativistic and so an increase in density does not lead to such a big increase in pressure. Technically - $P \propto \rho$ at extremely high densities.

It is then a bit of standard textbook astrophysics to show that a star supported by such an equation of state is not stable and will collapse given the slightest perturbation.

In reality neutron stars are not supported by ideal degeneracy pressure - there is a strong repulsive force when they are compressed beyond the nuclear saturation density, with something like $P \propto \rho^2$. Yet even here, an instability is reached at finite density because in General Relativity, the increasing pressure contributes (in addition to the density) to the extreme curvature of space and ultimately means that the star collapses at a finite density and pressure.

electromagnetism - Why can coercivity be used in place of magnetization when modelling permanent magnets?

I'm trying to understand how permanent magnets can be modeled in finite element methods. FEMM's manual, here, states that it treats a PM as a solenoid with the same surface current as the PM, and then solves the field from there, which I understand. However, they use the coercivity $H_c$ in place of the magnetization $M$ (I might be dropping factors of $\mu$ here.) to calculate this surface current, i.e., $H_c (\hat{m} \times \hat{n})$.

The argument given is that this is the amount of current you would need to surround the magnet in (in opposite direction) to drive the field to zero (this is the definition of coercivity), which is the same field you get with no current at all. Therefore, the surface current of the original PM can be physically cancelled out with $-H_c (\hat{m} \times \hat{n})$, so this must be it's actual value.

My problem is that when we model a permanent magnet in absence of any external currents, we are at the $H=0$ point of the $B-H$ diagram (i.e., $B = B_r$), and this is not the same physical point. Intuitively I would want to use $B_r$ in place of $M$, modulo some factors of $\mu$ again.

I know that this makes assumptions about linearity of the magnet, but I'm not exactly sure where they creep in. Thanks in advance for any input.

Answer

I think the answer is that you can't directly measure $B$ inside the material, but you can figure out $H$. So you need to know how $B$ depends on $H$ for calculation purposes. For so-called linear demagnetization curve permanent magnets, we can linearize the $B-H$ curve in the second quadrant, and write

$H(B) = \nu B - H_c$

where $\nu = \frac{H_c}{B_r}$ is the reluctivity (inverse of permeability). Note that if this were a linear material without hysteresis (e.g. a soft iron core), you would just have

$H(B) = \nu B$.

If you apply Ampere's law you get $\nabla \times H = J_f$. For our permanent magnet, the extra term $-\nabla \times H_c$ in the first equation is equivalent to adding an extra bound current term to the second equation. The bound current only has support on the surface, so we treat the permanent magnet like a soft iron core surrounded by a surface current of value $-H_C \hat {n}\times \hat{m}$.

I found the following documents useful: http://www.permagsoft.com/english/assets/applets/DemagnetisingCurve.pdf http://www.permagsoft.com/english/assets/applets/CompEngl.pdf

astronomy - Can the axis of rotation of a celestial body point in any arbitrary direction?

I am developing a small computer program that involves moderately simple simulation of elliptical Kepler orbits for fictional, generated star systems. I'm doing this without much prior knowledge of orbits (apart from some basic equations) or astrophysics in general.

I'm attempting to create loosely realistic orbits in the sense that they are not all in one plane and that they are elliptical with the parent body at one focus. The program assumes there are only interactions between a body and the one it orbits. Planets do not affect other planets' orbits. All other forces are explicitly ignored with orbits following only Kepler's laws.

The axis of rotation of each body in this simulation will be static (ie. without precession) and the axial tilt will be pseudorandomly generated. I wish to align an orbit with zero inclination with the equatorial plane of the parent body.

The real question, then, is the following: are there are any important constraints to the direction of the axis of rotation of an arbitrary hypothetical body orbiting another hypothetical body of significantly greater mass that I should take into consideration when determining the axis and axial tilt?

Which is to say, can the axis of rotation of, say, a planet, "point" in any arbitrary direction?

(The axis can be assumed to be a vector in the direction of the north pole. The north pole is here simply the pole that is "above" the orbital plane when axial tilt is zero.)

Answer

Uranus has an axial tilt of 97.77 degrees (it's on its side). So we have axial tilts ranging from 23.5 degrees (earth) to uranus's axis to venus's retrograde rotation. I think its safe to say that the axis of rotation can point in any direction. Remember that the axis of rotation is in fact the direction of the angular momentum, which can be changed by torque, i.e. collisions that are not directed right at a planet's center-of-gravity. Considering that the early solar system featured many of these types of collisions, I imagine that over time the axial tilt of a plant may be in any conceivable direction. For more re: Uranus's tilt, check this.

quantum mechanics - Particle hole symmetry in 2nd quantization

In second quantization one the Particle hole trasnformation is defined as \begin{align} \hat{\mathcal{C}} \hat{\psi}_A \hat{\mathcal{C}}^{-1} &= \sum_B U^{*\dagger}_{A,B} \hat{\psi}^{\dagger}_B \\ \hat{\mathcal{C}} \hat{\psi}_A^{\dagger} \hat{\mathcal{C}}^{-1} &= \sum_B \hat{\psi}_B U^{*}_{B,A} \\ \hat{\mathcal{C}} i \hat{\mathcal{C}}^{-1} &= +i \end{align} And if in a 2nd quantized Fermionic Hamiltonian ($\hat{\mathcal{H}} $) Particle Hole symmetry is present then $$ \hat{\mathcal{C}} \hat{\mathcal{H}} \hat{\mathcal{C}}^{-1} = \hat{\mathcal{H}} $$ I want to see what this equation means in single particle basis. In single particle basis I can write the 2nd quantized Hamiltonian ($\hat{\mathcal{H}}$) as $$ \hat{\mathcal{H}}=\sum_{A,B}\hat{\psi}^\dagger_{A}H_{A,B}\hat{\psi}_B $$ Here the matrix $H$ is the Hamiltonian in single particle basis. Now, with the transformation rules on should get $$ U H^{*} U^{\dagger} = - H $$ In the single-particle basis. But what I am getting using the transformation rules is $$ U^* H U^{*\dagger} = -H $$ Now I have started to think whether the transformation rules given here are right or not. I wanted to know if the transformation rule or my calculation is wrong.
Source: Topological phases: Classification of topological insulators and superconductors of non-interacting fermions, and beyond Equation 17

electric circuits - What causes a resistor to heat up?

In the following video clip at 2:10,

http://www.youtube.com/v/YslOUw5oueQ ,

Professor Walter Lewin talks about a misconception people have that the energy going through a wire to a resistor is in the form of kinetic energy of electrons. He proves this cannot be so as follows. The current density is J = I/A = Vne where V is the drift velocity (or average velocity), n is the number of electrons per volume, and e is the charge of an electron. A is a cross-sectional area of the wire.

We can make A as large as we want (keeping the current constant), and therefore V will have to become very small, and the electrons will have very little kinetic energy. Yet the resistor (say a light bulb) dissipates the same amount of power P=(I^2)*R. Therefore, it must be that the form of energy is not the kinetic energy of electrons.

My first question is, if we make A larger why does it have to be that V goes down? Perhaps n goes down - we increased the volume (by increasing the cross-sectional area), so there should be fewer electrons per volume?

My second question, my main question is, if the energy is not the kinetic energy of the electrons, what does in fact bring energy to the resistor and how does it heat up?

Answer

my main question is, if the energy is not the kinetic energy of the electrons, what does in fact bring energy to the resistor and how does it heat up?

We assume steady state operation.

The drift velocity of the electrons entering the resistor must equal the drift velocity of the electrons leaving the resistor. This follows from the fact that the current into the resistor equals the current out of the resistor.

However, the electrons leaving the resistor have less potential energy than those entering the resistor. This follows from the fact that there is an electric field through the structure of the resistor and, thus, there is a potential difference between the ends of the resistor.

The electric field within the resistor structure accelerates (does work on) the electrons increasing their kinetic energy however, this energy is quickly given up to the structure of the resistor via collisions; the resistor gets hotter.

On a more fundamental level, the energy flows from the battery to the resistor through the space around the conductors via the electromagnetic field. See, for example, William Beaty's description of energy flow in a simple circuit here.

mathematics - Does Gödel preclude a workable ToE?

Gödel's incompleteness theorem prevents a universal axiomatic system for math. Is there any reason to believe that it also prevents a theory of everything for physics?

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Edit:

I haven't before seen a formulation of Gödel that included time. The formulation I've seen is that any axiomatic systems capable of doing arithmetic can express statements that will be either 1) impossible to prove true or false or 2) possible to prove both true and false.

This leads to the question: Are theories of (nearly) everything, axiomatic systems capable of doing arithmetic? (Given they are able to describe a digital computer, I think it's safe to say they are.) If so, it follows that such a theory will be able to describe something that the theory will be either unable to analyse or will result in an ambiguous result. (Might this be what forces things like the Heisenberg uncertainty principle?)

Answer

If a "Theory of Everything" means a computational method of describing any situation, and true arithmetic formulas exist (as Gödel has shown) which cannot be proven, true arithmetic formulas exist which are necessary to describe some situation which cannot be discovered computationally, or if discovered incidentally, can not be proven true. So for example this computational method to be complete, would need to be able to prove the validity of math and logic, without using math and logic, since math and logic are separate from physics.

The definition above that "Gödel's theorem is a statement that it is impossible to predict the infinite time behavior of a computer program." is both incorrect, and anachronistic (at first Gödel rejected the Church-Turing definition of 'computability', but later (meaning by 1946) had to eventually discover it on his own). Besides Gödel wasn't a Computer Scientist even if his logic would be useful to them at some later date. The problem described above is a specific application of Gödel's theorem called the 'Halting Problem', but his theorem is much broader than that and it's implications much greater. What Gödel's first theorem basically states is that:

Any effectively generated axiomatic system $S$ cannot be both consistent and complete. In particular, for any effectively generated axiomatic system $S$ that is consistent which proves certain basic conclusions true, there are some basic true conclusion that is not provable within that system $S$.

For any formal effectively generated axiomatic system $S$, if $S$ includes a statement of its own consistency then S is inconsistent.

One of the answers above noted that:

1. Gödel's theorem only applies to formal axiomatic systems (which is true)

However went on to suggest that "Almost no useful, real-world physical theories have ever been stated as formal axiomatic systems". This is completely false given how Gödel defined formal axiomatic systems. By formal axiomatic systems Gödel meant 'computable' meaning any system able to derive results (conclusions) through functions (or logic) that algorithmically computable. Physics completely relies on two such systems - Mathematics and Logic, which means Physics also is.

Is it really being suggested Physics is not computable? Physics makes predictions using math and logic, both of which are formal axiomatic systems. Physics also describes it's observed behaviour using the same systems. Physics is nothing less than a formal axiomatic system used to describe nature, though it does presuppose these other systems. Even if some of its axioms are observed or measured it derives results from these, or laws about them through functions that computable ($E=MC^2, F=MA$), therefore Gödel absolutely applies.

This means a Theory of Everything and indeed physics must either be internally consistent, but incomplete, meaning not actually able to describe every possible situation, or it must be complete but inconsistent, meaning able to describe every possible situation, but contain inconsistencies (self-contradictions). That physics requires mathematics to prove its own truths shows that physics is incomplete (since it needs to presuppose the consistency of Maths as an axiomatic system) just as mathematics requires logic to prove its theorems (for the same reason, Maths cannot prove logic, but must simply presuppose it). This is direct evidence of Gödel's claim that no axiomatic system can prove its own consistency, and so, is incomplete. Additionally, people have also shown that the Incompleteness theorem even holds in Quantum Mechanics (which is also consistent, but not complete).

Any 'Theory of Everything' cannot be complete since it cannot explain math, or logic, and there will be physical phenomenon whose behaviour cannot be computed. Just as physics itself, the physics of a TOE, in addition to physical observation, requires math and logic, showing how incomplete physics by itself is (though it is consistent).

Why gravity decreases as we go under ground?

We all know that gravity decreases as we go upward, we also know that gravity decreases as we go inside the earth? I don't know why gravity decreases as we go downward or inside the Earth? Please explain?

solid state physics - In the diode equation, why the exponential $exp$ and the ideality factor $n$ are there? What do they represent & what is their significance?

In the Shockley diode equation, why the exponential $\exp$ and the ideality factor $n$ are there? What do they represent & what is their significance?

I have to work on Solar Photovoltaics, and I need to understand the Shockley diode equation clearly.

Could quantum mechanics work without the Born rule?

Slightly inspired by this question about the historical origins of the Born rule, I wondered whether quantum mechanics could still work without the Born rule. I realize it's one of the most fundamental concepts in QM as we understand it (in the Copenhagen interpretation) and I know why it was adopted as a calculated and extremely successful guess, really. That's not what my question is about.

I do suspect my question is probably part of an entire field of active research, despite the fact that the theory seems to work just fine as it is. So have there been any (perhaps even seemingly promising) results with other interpretations/calculations of probability in QM? And if so, where and why do they fail? I've gained some insight on the Wikipages of the probability amplitude and the Born rule itself, but there is no mention of other possibilities that may have been explored.

quantum mechanics - Does Feynman's derivation of Maxwell's equations have a physical interpretation?

There are so many times that something leaves you stumped. I was recently reading the paper "Feynman's derivation of Maxwell's equations and extra dimensions" and the derivation of the Maxwell's equations from just Newton's second law and the quantum mechanical commutation relations really intrigued me. They only derived the Bianchi set, yet with slight tweakings with relativity, the other two can be derived.

Awesome as it is, does this even have a physical interpretation? How is it possible to mix together classical and quantum equations for a single particle, which aren't even compatible, and produce a description of the electromagnetic field?

Answer

Feynman's derivation is wonderful, and I want to sketch why we would expect it to work, and what implicit assumptions it's really making. The real issue is that by switching back and forth between quantum and classical notation, Feynman sneaks in physical assumptions that are sufficiently restrictive to determine Maxwell's equations uniquely.

To show this, I'll give a similar proof in fully classical, relativistic notation. By locality, we expect the force on a particle at position $x^\mu$ with momentum $p^\mu$ depends solely on $p^\mu$ and $F(x^\mu$). (This is Eq. 1 in the paper.) Then the most general possible expression for the relativistic four-force is $$\frac{d p^\mu}{d\tau}= F_1^\mu(x^\mu) + F_2^{\mu\nu}(x^\mu)\, p_\nu + F_3^{\mu\nu\rho}(x^\mu)\, p_\nu p_\rho + \ldots$$ where we have an infinite series of $F_i$ tensors representing the field $F$. (Of course, we already implicitly used rotational invariance to get this.) I'll suppress the $x^\mu$ argument to save space.

It's clear that we need more physical assumptions at this point since the $F_i$ are much too general. The next step is to assume that the Lagrangian $L(x^\mu, \dot{x}^\mu, t)$ is quadratic in velocity. Differentiating, this implies that the force must be at most linear in momentum, so we have $$\frac{d p^\mu}{d\tau}= F_1^\mu + F_2^{\mu\nu}\, p_\nu.$$ This is a rather strong assumption, so how did Feynman slip it in? It's in equation 2, $$[x_i, v_j] = i \frac{\hbar}{m} \delta_{ij}.$$ Now, to go from classical Hamiltonian mechanics to quantum mechanics, we perform Dirac's prescription of replacing Poisson brackets with commutators, which yields the canonical commutation relations $[x_i, p_j] = i \hbar \delta_{ij}$ where $x_i$ and $p_i$ are classically canonically conjugate. Thus, Feynman's Eq. 2 implicitly uses the innocuous-looking equation $$\mathbf{p} = m \mathbf{v}.$$ However, since the momentum is defined as $$p \equiv \frac{\partial L}{\partial \dot{x}}$$ this is really a statement that the Lagrangian is quadratic in velocity, so the force is at most linear in velocity. Thus we get a strong mathematical constraint by using a familiar, intuitive physical result.

The next physical assumption is that the force does not change the mass of the particle. Feynman does this implicitly when moving from Eq. 2 to Eq. 4 by not including a $dm/dt$ term. On the other hand, since $p^\mu p_\mu = m^2$, in our notation $dm/dt = 0$ is equivalent to the nontrivial constraint $$0 = p_\mu \frac{dp^\mu}{d\tau} = F_1^\mu p_\mu + F_2^{\mu\nu} p_\mu p_\nu.$$ For this to always hold, we need $F_1 = 0$ and $F_2$ (hereafter called $F$) to be an antisymmetric tensor and hence a rank two differential form. We've now recovered the Lorentz force law $$\frac{d p^\mu}{d\tau} = F^{\mu\nu} p_\nu.$$

Our next task is to restore Maxwell's equations. That seems impossible because we don't know anything about the field's dynamics, but again the simplicity of the Hamiltonian helps. Since it is at most quadratic in momentum, the most general form is $$H = \frac{p^2}{2m} + \mathbf{A}_1 \cdot \mathbf{p} + A_2.$$ Collecting $\mathbf{A}_1$ and $A_2$ into a four-vector $A^\mu$, Hamilton's equations are $$\frac{dp^\mu}{d\tau} = (dA)^{\mu\nu} p_\nu$$ where $d$ is the exterior derivative. That is, the simplicity of the Hamiltonian forces the field $F$ to be described in terms of a potential, $F = dA$. Since $d^2 = 0$ we conclude $$dF = 0$$ which contains two of Maxwell's equations, specifically Gauss's law for magnetism and Faraday's law. So far we haven't actually used relativity, just worked in relativistic notation, and indeed this is where our derivation and Feynman's run out of steam. To get the other two equations, we need relativity proper.

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The basic conclusion is that Feynman's derivation is great, but not completely mysterious. In particular, it isn't really mixing classical and quantum mechanics at all -- the quantum equations that Feynman uses are equivalent to classical ones derived from Hamilton's equations, because he is using the Dirac quantization procedure, so the only real purpose of the quantum mechanics is to slip in $\mathbf{p} = m \mathbf{v}$, and by extension, the fact that the Hamiltonian is very simple, i.e. quadratic in $\mathbf{p}$. The other assumptions are locality and mass conservation.

It's not surprising that electromagnetism pops out almost 'for free', because the space of possible theories really is quite constrained. In the more general framework of quantum field theory, we can get Maxwell's equations by assuming locality, parity symmetry, Lorentz invariance, and that there exists a long-range force mediated by a spin 1 particle, as explained elsewhere on this site. This has consequences for classical physics, because the only classical physics we can observe are those quantum fields which have a sensible classical limit.

special relativity - Does the (relativistic) mass change? Why?

I learned recently that when an object moves with a velocity comparable to the velocity of light the (relativistic) mass changes. How does this alteration take place?

electromagnetism - Is Electromagnetic Mass Possible?

If the sinusoidal electric component of a light wave were off-set to one side of the magnetic component and then the smaller "lobe" were to cancel out with much of the larger side, then where would the energy go? Would it not form a closed loop much like a mass-bearing string? Could the electrical energy not be converted into a gravitational field to bend space-time over the length of the wave to form a 1-D "string" along the junction of the perpendicular E and B fields? This would be much like folding a sheet of paper in half so that one edge protrudes past the other. The protruding edge being the electric component and the rest being electrical energy converted into mass/gravity (comparable to a mass-bearing string). The magnetic component could then arise from Lorentz symmetry as described by Lubos:

If you only start with the $E_z$ electric field, the component $F_{03}$ is nonzero. However, when you boost the system in the $x$-direction, you mix the time coordinate $0$ with the spatial $x$-coordinate $1$. Consequently, a part of the $F_{03}$ field is transformed into the component $F_{13}$ which is interpreted as the magnetic field $B_y$, up to a sign.

Is the described "string wave" model possible?

I realize that I haven't provided a mechanism to explain the proposed model. However "I do not reject food because I do not understand digestion." -Oliver Heavenside

From here on the rest of this post is merely a compilation of the "evidence" that circumstantially supports the proposed idea. Please do not feel required to address these topics. They are here because I think electromagnetic mass is real. I realize this borders on "promoting unaccepted theories," but it is merely my way of assessing the possibility. I promise not to bring this up again on SE if it is refuted.

The concept seems compatible with the standard model:

a.) an origin of charge has been proposed.

b.) eliminating one lobe has reduced the "spin" or magnetic field by 1/2 relative to a light wave.

c.) Lepton number 1 applies due to the "strong interaction" of the remaining electric component

d.) a mechanism of mass has been proposed.

Maxwells equations seem to fit:

The Laws of Electromagnetism have geometric relations incorporated into them that naturally arise from the proposed electron-as-an-EM-wave model: the dot product and the cross product specifically.

Ampere's Law

Ampere’s Law describes the magnetic field produced by the flow of electrons along a wire. The negative components of electrons flowing along the wire should repel each other, which would mean that the negative components should protrude from the wire like the dorsal fins of sharks swimming parallel and breaking the surface. Since the electrons are travelling in the same direction along the wire the magnetic components (in the direction of the side fins of the shark analogy) should be tangent to the surface of the wire, which results in a circular magnetic field around the wire just as Ampere’s Law and the Biot-Savart Law predict.

Faraday's Law

Faraday’s Law describes the electric voltage produced in a coil of wire as a magnetic field through it changes. The voltage is proportional to the number of loops of wire, which is counter-intuitive/non-conservative. Why should the voltage depend on the number of loops? An explanation naturally arises from the proposed model. Electrons inherently have velocity in the form of a "Poynting-like" vector. When the loop of wire encounters a changing magnetic field the "poyinting vectors" align and under the right orientations they align with the wire loops and thereby form electric current.

In the bottom orientation the magnet produces no EMF along the wire. In the top orientation the EMF is along the wire as seen in generators.

Schrodinger's statistical model won because Maxwells equations had already divorced physics from first causes. Saying that the surface covered area of a loop was the cause of EMF instead of the poynting vectors of electrons adding over the distance of the circumference.

Gauss's Laws

Gauss’ laws of electricity and magnetism are easily integrated with the proposed model as the net magnetic flux is zero and the net charge is unchanged. However Coulombs Law does not hold true at the subatomic level. The force between two charged electrons is modified by the presence of other electrons if electrons are not point charges, but electric components of an EM-wave.

Coulomb's law

Coulomb's law for point charges:

$$F=k\frac{q_1 q_2}{r^2}=\frac{1}{4\pi\epsilon_0}\frac{qq_0}{r^2}\hat{u}$$

Does not hold at the subatomic level for all directions if a blip of negative charge sticks off one side of a mass-bearing electromagnetic wave. This naturally provides a classical explanation for "quantum tunnelling:" certain orientations of subatomic particles behave unlike point charges.

Atomic Orbitals

Take one electron and one proton and place them near each-other. The proposed model suggests an intrinsic and intuative reason why electrons don’t fall into the nucleus, which current theories lack. The velocity of an imbalanced EM wave is perpendicular to both the electric and magnetic components, which means that the radial acceleration of attraction towards the nucleus experienced by the electric component of an electron is always perpendicular to its velocity or the “Poynting vector” of an electron. All that is left for the electron to do is to set up the lowest energy standing wave possible. This also suggests that the proton rotates with the electron's orbit so that the "fins" constantly point toward eachother. Spherical harmonics should arise naturally from this arrangement and approximate to the schrodinger equation.

Relativistic Explanation of the Lorentz Force

Using relativistic tensors: "If you only start with the $E_z$ electric field, the component $F_{03}$ is nonzero. However, when you boost the system in the $x$-direction, you mix the time coordinate $0$ with the spatial $x$-coordinate $1$. Consequently, a part of the $F_{03}$ field is transformed into the component $F_{13}$ which is interpreted as the magnetic field $B_y$, up to a sign\cite{Lubos}."

If the electron is an E-M wave as proposed and the magnetic components align with the extern magnetic field, then performing the reverse of the above transformation should convert the external magnetic field into an electric component in the $E_z$ direction so that the electron feels the equivalent to a charge perpendicular to its direction of motion, which explains the Lorentz force.

Numerous Other Explanations

There are many other phenomena that seem to fit with the proposed model. I'm just running out of steam!

A Little History

Electrons were viewed as "matter waves" in DeBrogli's model. "Matter waves" is a polite way of saying: something is waving but we don't know what. Heisenberg came along and said that despite not knowing what is waving we can assume that the wave doesn't have any undiscovered properties that would allow for knowing both position and momentum (an arrogant assertion!). Schrodinger then came along and noticed that tweaking spherical harmonics provided a reasonable model of atomic orbitals (statistically). That all led to the Copenhagen interpretation, which Einstein called the "Born-Heisenberg tranquilizing philosophy, or religion." Electrons have continued to be treated as point charges or matter waves as is convenient for interpreting experimental results ever since.

QED and QCD have introduced "virtual photons" in explaining the interactions of "point charges" and light, etc. Surely "virtual photons" would be unnecessary if the point charges were instead modelled as EM waves themselves.

Particle physicists have invented a "higgs field" which they propose vibrates to create mass. The higgs field seems like adding epicycles: sorta unnecessary when there is a simpler and better alternative (that I have not really understood yet!). It has been obvious for decades that physics has stalled while trying to model hadrons as point charges. String theory has been a lonely success story waiting to happen. Electromagnetic string-waves are the future! (unless someone refutes me :-).

lagrangian formalism - Why do we require local gauge invariance

My thought on this are somewhat scattered so I apologise in advance.

Maxwell's equations are gauge invariant. The physical Electric and Magnetic fields don't depend on whether we use $A_\mu$ or $A_\mu+\partial_\mu\Lambda$. We have the free EM Lagrangian $\mathcal{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ from which we can derive Maxwell's equations. It thus makes sense that the action stemming from this Lagrangian also be gauge invariant.

My issue arises when we couple to matter.

Let's take the Lagrangian for EM fields coupled to matter;

$$\mathcal{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+j^\mu A_\mu +\bar{\Psi}(i\gamma^\mu\partial_\mu-m)\Psi$$

Under $A_\mu\rightarrow A_\mu+\partial_\mu\Lambda$ the interaction term transforms as

$$\delta(j^\mu A_\mu)=j^\mu \partial_\mu\Lambda=\partial_\mu(\Lambda j^\mu)-\partial_\mu j^\mu\Lambda$$

Ignoring the total derivative term the invariance of the action then requires $\partial_\mu j^\mu=0$.

This is generally the point where textbooks say that we can find a $j^\mu$ made up of the matter fields using Noether's theorem; for spinor fields under global $U(1)$ transformations we would have $j^\mu=\bar{\Psi}\gamma^\mu\Psi$.

However, this conserved current is only obtained when the matter fields are on-shell, but we (presumably) require gauge invariance of the action even when the fields are off-shell. (Please correct me if this is incorrect). If the fields are off-shell we can cancel out the extra term we obtain by requiring that the matter fields transform in an appropriate way under local $U(1)$ transformation. We can then update our definition of gauge invariance to a local $U(1)$ transformation of the matter fields and the usual transformation of the EM fields.

My question; Why should we want to do this anyway? Even if we have the extra term $j^\mu\partial_\mu\Lambda$ Maxwell's equations are unaffected which was the reason we wanted gauge invarance in the first place. From this perspective, the extra term is not an issue.

Now I must admit my classical EM knowledge is a bit rusty so bear with me on this next point. Obviously the spinor equations of motion will not be invariant under the transformation of $A_\mu$ but is there any particular reason why the should be? Just because the 4-potential is not an observable in the pure EM sector is it necessarily like that in the spinor sector?

A thought has also occurred to me when writing this post; if we allow the term $j^\mu\partial_\mu\Lambda$ to appear in the Lagrangian then $\partial_\mu\Lambda$ itself becomes a dynamical field. I can imagine this could produce issues with regard to renormalizability (guessing) but a more immediate consequence I can see of this is the impossibility of interactions; the field equation for $\partial_\mu\Lambda$ is $j^\mu=0$, so we have no interaction. We could though claim that the field is $\Lambda$ instead of $\partial_\mu\Lambda$, in which case the field equation is $\partial_\mu j^\mu=0$.

I admit this post is a bit all over the place. My questions essentially boil down to;

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  Since the current is only conserved on-shell, why do we use Neother's theorem to create the current which we couple to the EM potential?

  
  


• 
  

  Even if we don't banish the extra term, how screwed are we? Does the extra term produce issues with regards to renormalizability or the resulting spinor equations have properties which disagree with experiments? We could just go from global to local transformations of the spinors and everything works out nicely, but why would we want to do this, apart from local transformations seeming like a more general alternative to global ones?

  
  
  

homework and exercises - The computation of the propagator in two dimensions

I did the computation of the propagator in two dimensions at (19.26) in Peskin & Shroeder as follows.
First I performed a Wick rotation.

\begin{alignat}{2} \int\frac{d^2 k}{(2\pi)^2}e^{-ik\cdot (y-z)}\frac{ik^{\mu}\gamma_{\mu}}{k^2} &=&& -\partial^{\mu}\gamma_{\mu} \left( i\int \frac{d^2 k_E}{(2\pi)^2}e^{ik_E\cdot (y_E-z_E)}\frac{1}{-k_E^2} \right) \\ &=&& \frac{i}{4\pi^2} \partial^{\mu}\gamma_{\mu} \int_0^{\infty}dk_E k_E \frac{1}{k_E^2}\int_0^{2\pi}d\theta e^{ik_E|y_E-z_E|\cos \theta} \\ &=&& \frac{i}{4\pi^2} \partial^{\mu}\gamma_{\mu} \int_0^{\infty}dk_E k_E \frac{1}{k_E^2} 2\pi J_0(k_E|y_E-z_E|) \\ \end{alignat}

where $J_0(s)$ is a bessel function and I made use of Hansen-Bessel Formula.

Setting $s\equiv k_E|y_E-z_E|$

\begin{alignat}{2} &=&& \frac{i}{2\pi} \partial^{\mu}\gamma_{\mu} \int_0^{\infty} ds\frac{1}{s} J_0(s) \\ &=&& 0 \end{alignat} But in the book, $$ \int\frac{d^2 k}{(2\pi)^2}e^{-ik\cdot (y-z)}\frac{ik^{\mu}\gamma_{\mu}}{k^2}= -\partial^{\mu}\gamma_{\mu} \left( \frac{i}{4\pi}\log (y-z)^2 \right) \tag{19.26}$$ Where did I make a mistake?

Answer

After you take $\gamma_\mu \partial^\mu$ out of the integral sign, the integral becomes divergent, so this step is not guaranteed by theorems from calculus and apparently illegal. But I think your calculation can be continued in the following (heuristic) way.

Let $\vec{k}$ denote the momentum in Euclidean space and $k$ its norm, and $r=|\vec{y}-\vec{z}|$. In order to take $\partial^\mu$ out, we can introduce a cut off $\epsilon$, which is a positive small number, then the integration becomes $$ \lim_{\epsilon\to 0}\quad\frac{i}{2\pi}\int^\infty_\epsilon dk (\gamma_\mu \partial^\mu)\frac{J_0(kr)}{k}= \lim_{\epsilon\to 0}\quad\frac{i}{2\pi}(\gamma_\mu \partial^\mu)\int^\infty_{\epsilon\cdot r} ds \frac{J_0(s)}{s} $$

In the expansion of $J_0(s)$: $$ J_0(s)=\frac{sin(s)}{s}=1-\frac{1}{3!}s^2+... $$ the terms of order larger than 1 will contribute to the integral terms like $\epsilon^2 r^2$ which vanishes as $\epsilon$ goes to zero. So these terms as well as other constants are irrelevant to the result. The only relevant one is the zeroth order term: $$ \lim_{\epsilon\to 0}\quad\frac{i}{2\pi}(\gamma_\mu \partial^\mu)\int^\infty_{\epsilon\cdot r} ds \frac{1}{s}= \lim_{\epsilon\to 0}\quad -\frac{i}{2\pi}(\gamma_\mu \partial^\mu)\left[ln(r)+ln(\epsilon)+constant\right] $$ which turns out to be the right answer:$$ -\frac{i}{2\pi}(\gamma_\mu \partial^\mu)ln(r) $$

estimation - Do you round uncertainties in intermediate calculations?

Let's say I have an experimental uncertainty of ±0.03134087786 and I perform many uncertainty calculations using this value. Should I round the uncertainty to 1 significant figure at this stage or leave it unrounded until my final answer?

Answer

tl;dr- No, rounding numbers introduces quantization error and should be avoided except in cases in which it's known known to not cause problems, e.g. in short calculations or quick estimations.

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Rounding values introduces quantization error (e.g., round-off error). Controlling for the harmful effects of quantization error is a major topic in some fields, e.g. computational fluid dynamics (CFD), since it can cause problems like numerical instability. However, if you're just doing a quick estimation with a calculator or for a quick lab experiment, quantization error can be acceptable. But, to stress it – it's merely acceptable in some cases; it's never a good thing that we want.

This can be confusing because many intro-level classes teach the method of significant figures, which calls for rounding, as a basic method for tracking uncertainty. And in non-technical fields, there's often a rule that estimated values should be rounded, e.g. a best guess of "103 days" might be stated as "100 days". In both cases, the issue is that a reader might mistake the apparent precision of an estimate to imply a certainty that doesn't exist.

Such problems are purely communication issues; the math itself isn't served by such rounding. For example, if a best guess is truly "103 days", then presumably it'd be best to actually use that number rather than arbitrarily biasing it; sure, we might want to adjust an estimate up-or-down for other reasons, but making an intermediate value look pretty doesn't make any sense.

Getting digits back after rounding

Often, publications use a lot of rounding for largely cosmetic reasons. Sometimes these rounded values reflect an approximate level of precision; in others, they're almost arbitrarily selected to look pretty.

While these cosmetic reasons might make sense in a publication, if you're doing sensitive work based on another author's reported values, it can make sense to email them to request the additional digits or/and a finer qualification of their precision.

For example, if another researcher measures a value as "$1.235237$" and then publishes it as $``1.2"$ because their uncertainty is on-the-order-of $0.1$, then presumably the best guess one can make is that the "real" value is distributed around $1.235237$; using $1.2$ on the basis of it looking pretty doesn't make any sense.

Uncertainties aren't special values

The above explanations apply to not just a base measurement, but also to a measurement's uncertainty. The math doesn't care for a distinction between them.

So for grammatical reasons, it's common to write up an uncertainty like ${\pm}0.03134087786$ as $``{\pm}0.03"$; however, no one should be using ${\pm}0.03$ in any of their calculations unless they're just trying to do a quick estimate or otherwise aren't too concerned with accuracy.

In summary, no, intermediate values shouldn't be rounded. Rounding is best understood as a grammatical convention to make writing look pretty rather than being a mathematical tool.

Examples of places in which rounding is problematic

A general phenomena is loss of significance:

Loss of significance is an undesirable effect in calculations using finite-precision arithmetic such as floating-point arithmetic. It occurs when an operation on two numbers increases relative error substantially more than it increases absolute error, for example in subtracting two nearly equal numbers (known as catastrophic cancellation). The effect is that the number of significant digits in the result is reduced unacceptably. Ways to avoid this effect are studied in numerical analysis.

–"Loss of significance", Wikipedia

The obvious workaround is then to increase precision when possible:

Workarounds

It is possible to do computations using an exact fractional representation of rational numbers and keep all significant digits, but this is often prohibitively slower than floating-point arithmetic. Furthermore, it usually only postpones the problem: What if the data are accurate to only ten digits? The same effect will occur.

One of the most important parts of numerical analysis is to avoid or minimize loss of significance in calculations. If the underlying problem is well-posed, there should be a stable algorithm for solving it.

–"Loss of significance", Wikipedia

A specific example is in Gaussian elimination, which has a lot of precision-based problems:

One possible problem is numerical instability, caused by the possibility of dividing by very small numbers. If, for example, the leading coefficient of one of the rows is very close to zero, then to row reduce the matrix one would need to divide by that number so the leading coefficient is 1. This means any error that existed for the number which was close to zero would be amplified. Gaussian elimination is numerically stable for diagonally dominant or positive-definite matrices. For general matrices, Gaussian elimination is usually considered to be stable, when using partial pivoting, even though there are examples of stable matrices for which it is unstable.

–"Gaussian elimination", Wikipedia [references omitted]

Besides simply increasing all of the values' precision, another workaround technique is pivoting:

Partial and complete pivoting

In partial pivoting, the algorithm selects the entry with largest absolute value from the column of the matrix that is currently being considered as the pivot element. Partial pivoting is generally sufficient to adequately reduce round-off error. However, for certain systems and algorithms, complete pivoting (or maximal pivoting) may be required for acceptable accuracy.

–"Pivot element", Wikipedia