Other Compilation of severe errors in famous textbooks

[Demystifier]

Erm, I did. And they don't cancel. Did you try it? Take two degrees of freedom with different k's and m's, then it is obvious that one term will cancel and the other will be left.

I find it hard to believe, because there is a general theorem telling that energy is conserved whenever the Hamiltonian does not have an explicit time dependence. In our case the Hamiltonian is essentially
$$H=\frac{p_1^2}{2}+\frac{p_2^2}{2}-\alpha_1^2\frac{x_1^2}{2}-\alpha_2^2\frac{x_2^2}{2}$$
where $\alpha_i^2=-\omega_i^2$ are positive constants. Hence the Hamiltonian does not have an explicit time dependence so the theorem should apply. Indeed, since the Hamiltonian is diagonalized, i.e. there is no coupling between $x_1$ and $x_2$, the Hamiltonian is simply
$$H=H_1+H_2$$
so each $H_i$ is conserved separately, as in the case of only one degree of freedom. Since this contradicts your result, can you present the essential steps of your calculation?

Anyway, here is mine calculation. The equations of motion (in the Newton form) are
$$\ddot{x}_i(t)=\alpha_i^2x_i(t)$$
so the general solution is
$$x_i(t)=c_{i+}e^{\alpha_i t}+c_{i-}e^{-\alpha_i t}$$
where $c_{i\pm}$ are arbitrary constants. Hence
$$x_i^2(t)=c_{i+}^2e^{2\alpha_i t}+c_{i-}^2e^{-2\alpha_i t}+c_{i+}c_{i-}$$
$$\dot{x}_i(t)=\alpha_i(c_{i+}e^{\alpha_i t}-c_{i-}e^{-\alpha_i t})$$
$$\dot{x}_i^2(t)=\alpha_i^2(c_{i+}^2e^{2\alpha_i t}+c_{i-}^2e^{-2\alpha_i t}-c_{i+}c_{i-})$$
Therefore
$$\dot{x}_i^2(t)-\alpha_i^2x_i^2(t)=-2\alpha_i^2c_{i+}c_{i-}$$
is time independent, so finally we have the energy
$$H=\sum_{i=1,2} \frac{1}{2}\left( \dot{x}_i^2(t)-\alpha_i^2x_i^2(t) \right) =-\sum_{i=1,2} \alpha_i^2c_{i+}c_{i-}$$
which is time independent, Q.E.D.

 

[martinbn]

What you've written is correct, but irrelevant. You have two different $\omega_j$, as I guessed, and you wagged your finger at me for doing so. L&L do something else. They look for a solution of the form $x_j=A_je^{i\omega t}$. And if you put that into the energy the time dependence will remain if your $\alpha_j$ are not the same.

 

[Demystifier]

What you've written is correct, but irrelevant. You have two different $\omega_j$, as I guessed, and you wagged your finger at me for doing so. L&L do something else. They look for a solution of the form $x_j=A_je^{i\omega t}$. And if you put that into the energy the time dependence will remain if your $\alpha_j$ are not the same.

If they look for a solution in which both modes have the same frequency, then such a solution simply doesn't exist. In fact, the modes which are not solutions don't conserve energy even with real $\omega$. But modes which are not solutions are nonphysical simply because they are not solutions, so it's irrelevant whether they conserve energy or not.

 

[vis_insita]

But you don't make explicitly wrong statements about collapse, because you accept "collapse" at least in the sense of information update and you do not deny the quantum Zeno effect. In that sense you are not like Ballentine.

Ballentine does not deny the Quantum Zeno effect either. I think you misunderstood him. In

Ballentine, L.E. Found Phys (1990) 20: 1329. https://doi.org/10.1007/BF01883489.

he has a short discussion of the Quantum Zeno effect. His conclusion is "Thus the Quantum Zeno effect actually occurs for this system. [This refers to what is reported in https://doi.org/10.1103/PhysRevA.41.2295 ] But it is misleading to explain it as being due to a "collapse of the wave function" caused by measurement. No "collapse" actually occurs; rather the excitation of the atom is impeded by the string perturbation of the optical pulses and the coupling to the radiation field. Moreover [...], the effect occurs regardless of whether or not any measurement of the emitted photons is actually made."

Do you think there is anything wrong with these statements? If not, what other explicitly wrong statements do you have in mind?

Also, I believe there is no reason to accept the collapse even in the sense of "information update". The update of information is properly handled without a collapse by means of conditioning upon past measurement results. (As is argued in his textbook and also in the paper above.)

The collapse is certainly not necessary to explain the Zeno effect, but it is useful as a quick and dirty way to obtain it.

This somewhat contradicts your earlier statement that there are experiments that "show that the collapse exists." There would be nothing wrong with criticising a "quick and dirty" derivation of the Quantum Zeno effect if there is a cleaner one available.

 

[Demystifier]

Ballentine does not deny the Quantum Zeno effect either. I think you misunderstood him. In

Ballentine, L.E. Found Phys (1990) 20: 1329. https://doi.org/10.1007/BF01883489.

he has a short discussion of the Quantum Zeno effect. His conclusion is "Thus the Quantum Zeno effect actually occurs for this system. [This refers to what is reported in https://doi.org/10.1103/PhysRevA.41.2295 ] But it is misleading to explain it as being due to a "collapse of the wave function" caused by measurement. No "collapse" actually occurs; rather the excitation of the atom is impeded by the string perturbation of the optical pulses and the coupling to the radiation field. Moreover [...], the effect occurs regardless of whether or not any measurement of the emitted photons is actually made."

Do you think there is anything wrong with these statements? If not, what other explicitly wrong statements do you have in mind?

That's perfectly fine, but that's not how he explained it in the book.

Also, I believe there is no reason to accept the collapse even in the sense of "information update". The update of information is properly handled without a collapse by means of conditioning upon past measurement results. (As is argued in his textbook and also in the paper above.)

It's not necessary to use collapse in the sense of information update, but I think it's also not wrong to use it that way.

This somewhat contradicts your earlier statement that there are experiments that "show that the collapse exists." There would be nothing wrong with criticising a "quick and dirty" derivation of the Quantum Zeno effect if there is a cleaner one available.

Perhaps it was not obvious from the context, but I distinguish a "true" collapse from a FAPP collapse (essentially, an update of information). The latter makes sense even when the former doesn't exist.

 

[vis_insita]

That's perfectly fine, but that's not how he explained it in the book.

What exactly are you referring to? I don't remember any discussion of the Quantum Zeno effect in the book. He criticises the collapse on the grounds that continuous observation cannot completely halt the time evolution of the system (called "the watched-pot-paradox" by him), which seems to be implied by the collapse postulate. But this is not what happens with the Quantum Zeno effect.

It's not necessary to use collapse in the sense of information update, but I think it's also not wrong to use it that way.

The paper I cited discusses some instances in which it is wrong (or ambiguous). For sequential measurements the conditional probability for obtaining the result $a$, given some earlier result $b$ only agrees with the collapse postulate if -- paradoxically -- the first measurement didn't change the state of the system.

Perhaps it was not obvious from the context, but I distinguish a "true" collapse from a FAPP collapse (essentially, an update of information). The latter makes sense even when the former doesn't exist.

True, but if Ballentine's arguments are correct, then the collapse can be used only in some idealized situations, but certainly not FAPP.

 

[martinbn]

If they look for a solution in which both modes have the same frequency, then such a solution simply doesn't exist. In fact, the modes which are not solutions don't conserve energy even with real $\omega$. But modes which are not solutions are nonphysical simply because they are not solutions, so it's irrelevant whether they conserve energy or not.

No, you are still not reading it carefully. The amplitudes are chosen so that you have a solution.

 

[Keith_McClary]

This may still be controversial:

The Casimir Effect and the Quantum Vacuum
R. L. Jaffe (2005)

I have presented an argument that the experimental confirmation of the Casimir effect does not establish the reality of zero point fluctuations. Casimir forces can be calculated without reference to the vacuum and, like any other dynamical effect in QED, vanish as α → 0. The vacuum-to-vacuum graphs (See Fig. 1) that define the zero point energy do not enter the calculation of the Casimir force, which instead only involves graphs with external lines. So the concept of zero point fluctuations is a heuristic and calculational aid in the description of the Casimir effect, but not a necessity

 

[weirdoguy]

This may still be controversial:

Why?

 

[Demystifier]

This may still be controversial:

The Casimir Effect and the Quantum Vacuum
R. L. Jaffe (2005)

It's not a book but since you mentioned it, I think I have resolved the controversy completely. First I proved a no-go theorem showing what the Casimir effect is not
http://de.arxiv.org/abs/1605.04143and then I explained what it really is
http://de.arxiv.org/abs/1702.03291

 

[Demystifier]

No, you are still not reading it carefully. The amplitudes are chosen so that you have a solution.

I have presented my calculation in detail and you said that it's correct but irrelevant. Please present your calculation in detail, because otherwise I will never understand what's your point.

 

[Demystifier]

What exactly are you referring to? I don't remember any discussion of the Quantum Zeno effect in the book. He criticises the collapse on the grounds that continuous observation cannot completely halt the time evolution of the system (called "the watched-pot-paradox" by him), which seems to be implied by the collapse postulate. But this is not what happens with the Quantum Zeno effect.

Watched pot and quantum Zeno are two names for the same effect.

 

[martinbn]

Then their was his scathing rebuke of the Dirac Delta function. Rather than say we need further developments in math to make sense of it, which has now been done (admittedly requiring mathematicians like Grothendieck whose mathematical reputation is the equal of Von-Neumann himself) he simply dismissed it as a fiction. Von-Neumann is one of my heroes, being more that just a great mathematician, but that even rarer beast, a polymath, however like all human beings perfect he was not.

Thanks
Bill

That sounds interesting. Where can we read more about it? By the way, I am not sure about the history, but I think that Grothendick was not involved here. You need to credit Schwartz.

 

[martinbn]

I have presented my calculation in detail and you said that it's correct but irrelevant. Please present your calculation in detail, because otherwise I will never understand what's your point.

For this there is no need for any calculation other than what is written in the book. They look for a solution of the form $x_k(t)=A_ke^{i\omega t}$. They plug it in the Euler-Lagrange equations, the exponentials cancel because they are the same. This give linear equations for the $A_k$'s, which will have a non-zero solution if $\omega$ is chosen so that the corresponding determinant is zero. It is all written in the book!

 

[vis_insita]

Watched pot and quantum Zeno are two names for the same effect.

The watched-pot-paradox discussed by Ballentine is not the same as the Quantum Zeno effect. The paradox is an absurd consequence of the "quick-and-dirty"-explanation of the Quantum Zeno effect as due to a "collapse" caused by measurement. Abandoning the collapse avoids the absurd consequence and resolves the paradox, without denying the real Quantum Zeno effect.

 

[Demystifier]

For this there is no need for any calculation other than what is written in the book. They look for a solution of the form $x_k(t)=A_ke^{i\omega t}$. They plug it in the Euler-Lagrange equations, the exponentials cancel because they are the same. This give linear equations for the $A_k$'s, which will have a non-zero solution if $\omega$ is chosen so that the corresponding determinant is zero. It is all written in the book!

What you just said has absolutely nothing to do with violation of energy conservation. If you cannot (or do not want to) present an explicit calculation showing violation of energy conservation, then I see no point in further discussion.

 

[Demystifier]

The watched-pot-paradox discussed by Ballentine is not the same as the Quantum Zeno effect. The paradox is an absurd consequence of the "quick-and-dirty"-explanation of the Quantum Zeno effect as due to a "collapse" caused by measurement. Abandoning the collapse avoids the absurd consequence and resolves the paradox, without denying the real Quantum Zeno effect.

If we agree that (according to quantum Zeno) frequent measurements slow down the decay, then I see nothing absurd in the idea that quantum Zeno in the continuous measurement limit (that is, the watched pot) can, in principle, stop the decay completely.

 

[martinbn]

What you just said has absolutely nothing to do with violation of energy conservation. If you cannot (or do not want to) present an explicit calculation showing violation of energy conservation, then I see no point in further discussion.

No, this was just a response to your claim that you don't get a solution. With this solution you plug in the energy and you can see that if $\omega$ is not real you will have an overall exponential factor that makes the energy to either decay or increase, and because you have only one $\omega$ you cannot get cancellations for more than one degree of freedom. I thought we cleared that, and your only objection was that it is not a solution, hence my comment.

 

[vanhees71]

I'm a bit puzzled about this long debate. It's enough to consider only the one-dimensional case. You have a Hamiltonian system (I'll add the mass to get the dimensions correct) with a Hamiltonian that is not expclicitly time dependent. Thanks to Noether the total energy thus must be conserved. Indeed we have
$$H=\frac{p^2}{2m}-\frac{\alpha^2}{2} q^2.$$
Then the EoM. reads
$$\dot{p}=-\partial_q H=\alpha^2 q, \quad \dot{q}=\partial_{p} H=\frac{p}{m}.$$
This gives
$$m \ddot{q}=\alpha^2 q$$
with the general solution
$$q(t)=A \exp(\lambda t) + B \exp(-\lambda t), \quad \lambda=\frac{\alpha}{\sqrt{m}}.$$
Then
$$p=m \dot{q}=m A \lambda \exp(\lambda t) - m B \lambda \exp(-\lambda t).$$
Finally
$$H=\frac{p^2}{2m}-\frac{\alpha^2}{2} q^2=-2 A B \alpha^2=\frac{p_0^2}{2m}-\frac{\alpha^2}{2} q_0^2 = \text{const}.$$

 

[martinbn]

@vanhees71 The discussion is about the argument of L&L. What they do is to look for a solution of the form $x_k=A_ke^{i\omega t}$. The $\omega$ has to be a root of some polynomial. Can it be non-real? The argument is that it has to be real, otherwise there the energy will have an overall factor of the form $e^{\lambda t}$ with $\lambda$ real, which contradicts conservation of energy. @Demystifier said that this argument is erroneous because the time dependence will cancel out. To support his claim he showed and example. The problem was that in his example the solution was of the form $x_k=A_ke^{i\omega_k t}$, where the omegas are different and you get a cancellation. That is incorrect, more accurately irrelevant, in L&L the solution has the same $\omega$. Then he said that with the same $\omega$ you cannot get a solution, which is also wrong. Of course you cannot get a solution with non-real $\omega$, but that is their claim as well. So I am still unconvinced that they have done anything wrong.

 

[vanhees71]

A solution what for? I don't quite understand the problem obviously. Of course, I always can make the ansatz with $x=A \exp(\mathrm{i} \omega t)$, but if you solve for a problem with a negative oscillator postential, $V(x)=-\alpha^2 x^2/2$, you get the EoM
$$m \ddot{x}=\alpha^2 x$$
and Plugging in your ansatz you get
$$-m \omega^2=\alpha^2 \; \Rightarrow \; \omega=\pm \mathrm{i} \alpha/\sqrt{m},$$
and again we get the general solution indicated above
$$x(t)=A_1 \exp(\alpha t/\sqrt{m}) + A_2 \exp(-\alpha t/\sqrt{m}).$$
Energy is of course conserved.

The trajectory is always unbound (the potential has no minima, and the energy is not bounded from below). So there's no surprise here at all.

 

[andresB]

he has a short discussion of the Quantum Zeno effect. His conclusion is "Thus the Quantum Zeno effect actually occurs for this system. [This refers to what is reported in https://doi.org/10.1103/PhysRevA.41.2295 ] But it is misleading to explain it as being due to a "collapse of the wave function" caused by measurement. No "collapse" actually occurs; rather the excitation of the atom is impeded by the string perturbation of the optical pulses and the coupling to the radiation field. Moreover [...], the effect occurs regardless of whether or not any measurement of the emitted photons is actually made."

Do you think there is anything wrong with these statements? If not, what other explicitly wrong statements do you have in mind?

What about interaction-free measurements?
https://www.nature.com/articles/ncomms7811

 

[Demystifier]

The problem was that in his example the solution was of the form $x_k=A_ke^{i\omega_k t}$, where the omegas are different and you get a cancellation.

No, my example is sufficiently general to include the case in which the omegas are equal (which happens when the alphas are equal), and the cancellation happens even in this case.

 

[martinbn]

No, my example is sufficiently general to include the case in which the omegas are equal (which happens when the alphas are equal), and the cancellation happens even in this case.

Not general enough to claim that their argument is wrong.

 

[Demystifier]

Then he said that with the same $\omega$ you cannot get a solution, which is also wrong.

It's not wrong, the omegas in the solution are the same as the omegas in the Hamiltonian (because the Hamiltonian is diagonalized i.e. there is no coupling between the two degrees of freedom), so if the omegas in the Hamiltonian are different then so are the omegas in the solution.

 

[Demystifier]

Not general enough to claim that their argument is wrong.

Even if that was true (which was not), you missed that I mentioned also a much more general argument (actually a theorem) that the energy is conserved whenever the Hamiltonian does not have an explicit dependence on time, which in our case is true because the omegas in the Hamiltonian are time independent.

 

[bhobba]

That sounds interesting. Where can we read more about it? By the way, I am not sure about the history, but I think that Grothendick was not involved here. You need to credit Schwartz.

Its in the beginning - you can't miss it.

Grothendick worked on Nuclear spaces that are used in Rigged Hilbert spaces eg Schwartz spaces or the space of good functions as Littlewood called it:
https://en.wikipedia.org/wiki/Nuclear_space
Thanks
Bill

 

[vanhees71]

Even if that was true (which was not), you missed that I mentioned also a much more general argument (actually a theorem) that the energy is conserved whenever the Hamiltonian does not have an explicit dependence on time, which in our case is true because the omegas in the Hamiltonian are time independent.

I'm still not understanding all these arguments. I guess one should move this discussion to the mechanics forum, but isn't LL discussing the case of small oscillations around a minimum of a potential. Then the eigenfrequencies are of course all real.

That's easy to see. Without loss of generality we can shift the coordinates such that the minimum of the potential is at $\vec{x}=0$. Then we can approximate the potential in a neibourhood of the origin by
$$V(\vec{x})=\frac{1}{2} A_{jk} x_j x_k,$$
where $A_{jk}$ is a symmetric positive definite matrix, i.e., there's a orthogonal transformation, $\vec{x}'=\hat{O} \vec{x}$ such that $A_{jk}' x_j' x_k'=m \sum_{j} \omega_j^2 x_j^{\prime 2}$ with all $\omega_j^2>0$, and then the solutions for the normal coordinates $x_k'$ are just
$$x_k'(t)=A_k \exp(\mathrm{i} \omega_k t) + B_k \exp(-\mathrm{i} \omega_k t).$$
So for this case LL seems right to me.

You can of course solve everything in the original coordinates. The EoMs read
$$m \ddot{x}_j=-\partial_{j} V=A_{jk} x_k \; \Rightarrow \; m\ddot{\vec{x}}=-\hat{A} \vec{x}.$$
This you can solve by the ansatz
$$\vec{x}(t)=\vec{a} \exp(\mathrm{i} \omega t).$$
This gives
$$-m \omega^2 \vec{a} = -\hat{A} \vec{a},$$
which means that $\omega$ must be chosen such that $m \omega^2$ is an eigenvalue of $\hat{A}$ and $\vec{a}$ the corresponding eigenvector.

Since $\hat{A}=\hat{A}^{\text{T}}$ is positive definite there are $3$ positive eigenvalues (which may all be different from each other or not) you have $\omega \in \mathbb{R}$, and one can choose the eigenvectors $\vec{a}_k$ real and orthonormal. Then the general solution reads
$$\vec{x}(t) = \sum_{j=1}^3 \vec{a}_k [A_k \exp(\mathrm{i} \omega_k t)+B_k \exp(-\mathrm{i} \omega k t)].$$
This can be written in real form as
$$\vec{x}(t)=\sum_{j=1}^3 \vec{a}_k [C_k \cos(\omega_k t) + D_k \sin(\omega_k) t],$$
where $C_k,D_k \in \mathbb{R}$.

If $\hat{A}$ is not positive definite, the approximation doesn't make sense, because then there are unbound solutions as discussed above.

 

[Demystifier]

I'm still not understanding all these arguments.

What you say completely agrees with my statements and I don't understand the arguments by @martinbn either.

 

[vis_insita]

If we agree that (according to quantum Zeno) frequent measurements slow down the decay, then I see nothing absurd in the idea that quantum Zeno in the continuous measurement limit (that is, the watched pot) can, in principle, stop the decay completely.

We don't agree. The effect is caused by perturbing the system into an unstable state which almost instantly decays back into the ground state via spontaneous emission. The repeated excitation and the coupling to the EM-field cause the delayed decay. Since the metastable state isn't affected, the result of the disturbance is still a superposition and not a collapsed state. All this has nothing to do with frequent measurement or an induced collapse at all. In particular the effect is independent of whether any emitted photons are measured or not.

As far as I understand it the decay could only be halted in the unphysical limit of infinite coupling. Thus the collapse is at best an imperfect model for the real effect, and cannot be true exactly. Another way of putting it is that the collapse implies a strictly exponential decay law. But for short times the time evolution necessarily deviates from the exponential law. Thus the collapse should be expected to give wrong results in the limit of continuous observation.

 

[vis_insita]

What about interaction-free measurements?
https://www.nature.com/articles/ncomms7811

I don't know. What about them? They say they employ the Quantum Zeno effect. I'm not sure if they claim to have shown the existence of a collapse, athough this is probably their interpretation of the cause of that effect. Do you think something in their setup conflicts with the analysis of Ballentine?

 

[Keith_McClary]

what the Casimir effect is not

Is there anyone still arguing that vacuum energy is needed to explain the Casimir effect (or that the Casimir effect proves the existence of vacuum energy)?

 

[TSny]

Regarding the LL discussion, I don't see any errors in the LL argument. As some others have pointed out, it is important to keep in mind that their argument is based on the assumption that $T$ and $U$ are always non-negative. Here is how I would summarize their argument:

1. $T$ is a positive definite quadratic form $T = \frac{1}{2}\sum m_{ik}\dot x_i \dot x_k$.

2. Given: $U$ is a positive definite quadratic form $U = \frac{1}{2}\sum k_{ik} x_i x_k$.

3. The total energy $E = T+V$ is conserved. This was discussed back in chapter II of LL.

Claim: If a solution of the equations of motion has the form $x_k = A_ke^{i \omega t}$, then $\omega$ must be real. (Note that $\omega$ is assumed to be the same for all $x_k$.) We don't need to consider the trivial solution where all of the $A_k$'s are zero.

LL's "physical argument”:

If $\omega$ has negative imaginary part, then the magnitudes of all the $x_k$’s and all the $\dot x_k$’s for which $A_k \neq 0$ will grow exponentially with time. Since $T$ and $U$ are positive definite forms, $E$ must necessarily grow exponentially with time. If $\omega$ has positive imaginary part, then all the $x_k$’s and all the $\dot x_k$’s for which $A_k \neq 0$ will decay exponentially with time. $E$ must then decay exponentially with time. But we know that $E$ must be a constant. So, $\omega$ cannot have an imaginary part.

 

[vanhees71]

Is there anyone still arguing that vacuum energy is needed to explain the Casimir effect (or that the Casimir effect proves the existence of vacuum energy)?

Nobody is arguing that the usual introductory-textbook treatment (using simply a boundary condition for two plates) is the infinite-charge-value limit. It's an effective theory to describe the manybody system, but the Casimir effect is not some vacuum fluctuation (since the vacuum is the one state which does not fluctuate at all) but due to the fluctuations of the charges in the plates and the em. field (the quantum pendant of van der Waals forces). See

R. L. Jaffe, The Casimir effect and the quantum vacuum,
Phys. Rev. D 72 (2005) 021301.
https://dx.doi.org/10.1103/PhysRevD.72.021301
https://arxiv.org/abs/hep-th/0503158v1

Also @Demystifier has written a PLB (open access):

https://doi.org/10.1016/j.physletb.2016.08.036
and a very nice pedagogical paper:

https://doi.org/10.1016/j.aop.2017.05.013https://arxiv.org/abs/1702.03291v2

 

[atyy]

Well, one severe conceptual mistake is in some textbooks (if I remember right even in the Feynman Lectures vol. II and in Berkeley Physics Course vol. II, which shows that also Nobel Laureates make mistakes ;-)) when treating the magnetostatics of a wire relativistically. The mistake lies in the assumption that the wire is uncharged in the rest frame of the wire. In fact it's uncharged in the rest frame of the electrons. The correct treatment has to take into account the "self-induced" Hall effect though it's academic for house-hold currents, where the drift velocities are of the order of 1mm/s, but if you want to treat it fully relativistically you must take into account the correct relativistic version of Ohm's Law, $\vec{j}=\sigma \gamma (\vec{E} + \vec{v} \times \vec{B}/c)$ with $\sigma$ the usual electric conductivity (a scalar as any transport coefficient), $\gamma=(1-v^2/c^2)^{-1/2}$ (using Heaviside-Lorentz units).

Oh no, is Schroeder (of Peskin and Schroeder) wrong too in this presentation of Purcell?
http://physics.weber.edu/schroeder/mrr/MRRtalk.html

I never realized there was a relativistic correction to Ohm's law ...

 

[Demystifier]

Is there anyone still arguing that vacuum energy is needed to explain the Casimir effect (or that the Casimir effect proves the existence of vacuum energy)?

Most still compute Casimir effect in terms of vacuum energy, without thinking whether it is really necessary to do it that way.

 

[fluidistic]

So, can we conclude that TSny's post (#83) finally settles once and for all that Landafshitz's textbook is fine on that part? I am constantly swinging my lighter beneath my physical copy of that textbook, I need to know whether I set it on fire or not!

 

[TSny]

I am constantly swinging my lighter beneath my physical copy of that textbook, I need to know whether I set it on fire or not!

NO NO NO

 

[vanhees71]

Oh no, is Schroeder (of Peskin and Schroeder) wrong too in this presentation of Purcell?
http://physics.weber.edu/schroeder/mrr/MRRtalk.html

I never realized there was a relativistic correction to Ohm's law ...

Yes it's wrong too, because he assumes the wire having $\rho=0$ in the lab frame. Of course it's the conduction electrons that are moving in the lab frame while the positive ions are at rest, and then there's the Hall effect. Taking into account the Hall effect leads to the correct relativistic Ohm's Law and to the correct conclusion that $\rho'=0$ in the rest frame of the conduction electrons. You find this treatment here:

https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/https://itp.uni-frankfurt.de/~hees/pf-faq/relativistic-dc.pdf
That there's a relativistic correction to Ohm's Law goes back to the seminal work by Minkowski and can be found in many textbooks on relativistic classical electrodynamics.

 

[vanhees71]

Regarding the LL discussion, I don't see any errors in the LL argument. As some others have pointed out, it is important to keep in mind that their argument is based on the assumption that $T$ and $U$ are always non-negative. Here is how I would summarize their argument:

1. $T$ is a positive definite quadratic form $T = \frac{1}{2}\sum m_{ik}\dot x_i \dot x_k$.

2. Given: $U$ is a positive definite quadratic form $U = \frac{1}{2}\sum k_{ik} x_i x_k$.

3. The total energy $E = T+V$ is conserved. This was discussed back in chapter II of LL.

Claim: If a solution of the equations of motion has the form $x_k = A_ke^{i \omega t}$, then $\omega$ must be real. (Note that $\omega$ is assumed to be the same for all $x_k$.) We don't need to consider the trivial solution where all of the $A_k$'s are zero.

LL's "physical argument”:

If $\omega$ has negative imaginary part, then the magnitudes of all the $x_k$’s and all the $\dot x_k$’s for which $A_k \neq 0$ will grow exponentially with time. Since $T$ and $U$ are positive definite forms, $E$ must necessarily grow exponentially with time. If $\omega$ has positive imaginary part, then all the $x_k$’s and all the $\dot x_k$’s for which $A_k \neq 0$ will decay exponentially with time. $E$ must then decay exponentially with time. But we know that $E$ must be a constant. So, $\omega$ cannot have an imaginary part.

That doesn't make sense indeed. No matter which signature the matrix $k_{ik}$ has, energy is always conserved. Noether's theorem applied to time-translation invariance tells you that energy is conserved as long as the Hamiltonian (Lagrangian) is not explicitly time dependent, and that's obviously the case here.

Of course, if $k_{ik}$ is not positive definite, you have unbound (exponentially growing) solutions, but that doesn't mean that energy conservation could be violated.

In LL 1 is a much more severe mistake. Somewhere they claim all Hamiltonian systems were integrable, which is for sure wrong. There are more non-integrable Hamiltonian systems than integrable ones in fact.

Nevertheless, the LL series is among the best textbooks written, as are the Feynman Lectures though there are indeed some severe errors and some typos in there. I wish there'd be some recipe to ensure error-free scientific texts. If you have one, please tell us ;-)).

 

[TSny]

No matter which signature the matrix $k_{ik}$ has, energy is always conserved. Noether's theorem applied to time-translation invariance tells you that energy is conserved as long as the Hamiltonian (Lagrangian) is not explicitly time dependent, and that's obviously the case here.

Agreed. Any solution of the equations of motion will conserve energy whether or not $U$ is positive definite . LL are simply arguing that if $k_{ik}$ is positive definite, then there can't be solutions of the equations of motion which are exponentially growing or decaying because energy would not conserved.

Of course, if $k_{ik}$ is not positive definite, you have unbound (exponentially growing) solutions, but that doesn't mean that energy conservation could be violated.

I know. But LL are assuming that $k_{ik}$ is positive definite. This seems to be the important feature of LL's argument that some people here are overlooking.

 

[atyy]

Yes it's wrong too, because he assumes the wire having $\rho=0$ in the lab frame. Of course it's the conduction electrons that are moving in the lab frame while the positive ions are at rest, and then there's the Hall effect. Taking into account the Hall effect leads to the correct relativistic Ohm's Law and to the correct conclusion that $\rho'=0$ in the rest frame of the conduction electrons. You find this treatment here:

https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/https://itp.uni-frankfurt.de/~hees/pf-faq/relativistic-dc.pdf
That there's a relativistic correction to Ohm's Law goes back to the seminal work by Minkowski and can be found in many textbooks on relativistic classical electrodynamics.

So in the "standard" Purcell-Feynman-Schroeder presentation, the correct transformation is obtained because all the errors cancel out?

 

[vanhees71]

I've to check, whether at the end they get it right, but then the derivation would be self-contradictive, because in the beginning they assume the wire's charge density vanishes in the restframe in the wire to deduce that in fact that this is not the case (because of the Hall effect). I think the correct treatment is simple enough, at least not more complicated than the wrong one, to present it at the end of the introductory E&M theory lecture (usualy the 3rd semester of the theory course in German universities).

 

[Demystifier]

their argument is based on the assumption that T and U are always non-negative.

But the assumption that U is non-negative is violated by negative $\omega^2$. So the true reason why $\omega^2$ needs to be positive is positivity of U, not conservation of energy.

 

[Demystifier]

I wish there'd be some recipe to ensure error-free scientific texts. If you have one, please tell us

There is a recipe, write only about trivial stuff and don't try to be original.

 

[Demystifier]

So, can we conclude that TSny's post (#83) finally settles once and for all that Landafshitz's textbook is fine on that part?

No we can't.

 

[TSny]

But the assumption that U is non-negative is violated by negative $\omega^2$.

I don’t see how the non-negativity of U can be violated by having a negative value of $\omega^2$.

The assumption that $U = \frac{1}{2} \sum k_{ik} x_i x_k$ is a positive definite quadratic form means that U cannot be negative for any choice of real values of the $x_k$’s. So, no matter whether $\omega^2$ is positive or negative, U cannot be negative.

 

[Demystifier]

I don’t see how the non-negativity of U can be violated by having a negative value of $\omega^2$.

The assumption that $U = \frac{1}{2} \sum k_{ik} x_i x_k$ is a positive definite quadratic form means that U cannot be negative for any choice of real values of the $x_k$’s. So, no matter whether $\omega^2$ is positive or negative, U cannot be negative.

You are missing the point. $\omega^2$ are the eigenvalues of the matrix $k_{ik}/m$ (for simplicity I take all masses to be equal) so positivity of $k_{ik}$ is equivalent to positivity of $\omega^2$. So if $k_{ik}$ is positive, then $\omega^2$ cannot be negative. It is a total nonsense to consider negative $\omega^2$ if one has already decided that $k_{ik}$ is positive.

So the correct chain of reasoning is the following:
- Why is $\omega^2$ positive?
- Because $k_{ik}$ is positive.
- Fine, but why is $k_{ik}$ positive?
- Because we want potential energy to be positive.
- Are those positivity requirements related to conservation of energy?
- No, the energy is conserved for any sign of $k_{ik}$, or equivalently, for any sign of $\omega^2$.

 

[vanhees71]

It has nothing to do with kinetic energy but with the question whether the Hamiltonian is bounded from below or not. Only if the matrix $k_{ik}$ is positive (semi-)definite that's the case.

Now make the ansatz
$$\vec{x}(t)=\vec{x}_0 \exp(-\mathrm{i} \omega t).$$
The EoM reads
$$\ddot{\vec{x}}=-\hat{K} \vec{x},$$
where $\hat{K}=(k_{ij}/m)$. Plugging in the ansatz leads to
$$\hat{K} \vec{x}_0 = \omega^2 \vec{x}_0,$$
i.e., $\vec{x}_0$ must be an eigenvector of $\hat{K}$ with eigenvalue $\omega^2$. Any symmetric matrix, which we have here since we can always choose $k_{ik}=k_{ki}$ (or $\hat{K}^{\text{T}} =\hat{K}$), can be diagonalized with an orthogonal transformation. If we choose the appropriate basis we thus have
$$\hat{K}'=\hat{O} \hat{K} \hat{O}^{\text{T}}=\mathrm{diag}(\omega_1^2,\ldots,\omega_d^2),$$
where $d$ is the dimension of the system.

Now it's clear that for any vector
$$\vec{x}^{\text{T}} \hat{K} \vec{x}^{\text{T}}=\vec{x}^{\text{T}} \hat{O}^{\text{T}} \hat{K}' \hat{O} \vec{x} \geq 0$$
if and only if $\omega_j^2 \geq 0$. If all $\omega_j^2 >0$, all motions are bounded oscillations. If one or more eigenvalues are 0 you have directions, given by the eigenvectors, where the particle is unbound and can move as a free particle though the Hamiltonian is still bounded from below.

If one or more eigenvalues are negative, the motion in these directions can be unbound and the particle is accelerated exponentially with time.

In any case the total energy is conserved since the Hamiltonian is not explicitly time-dependent.

If you consider the harmonic (or pseudoharmonic if there are negative eigenvalues) potential as approximation of some other more complicated potential the approximation is only good for the bound oscillatory motion, for which the particle always stays near the equilibrium value, and that's only the case if the potential has a true minimum, and that's where the Hesse matrix of the potential $\hat{K}$ is positive definite.

 

[Demystifier]

It has nothing to do with kinetic energy but with the question whether the Hamiltonian is bounded from below or not.

Of course, but it was @TSny who first framed it in terms of positivity of kinetic and potential energy, so I was replying to him.