Other Compilation of severe errors in famous textbooks

[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.

 

[vanhees71]

It's amazing that we have such a long debate about such a simple issue. Maybe we should carefully study LL 1 again. It's information density is very high, i.e., though the text looks pretty short, it takes quite a time to study each chapter carefully and get a full understanding of it.

 

[TSny]

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.

LL give two separate arguments for why $\omega^2$ must be real and positive (see page 67 of their text). The first argument is based on what they call “physical arguments” while the second, independent proof is “mathematical”. Your argument above appears to be similar to their second argument. They provide the mathematical steps to show how the positivity of $k_{ik}$ leads to $\omega^2$ being real and positive. I take it we agree that LL’s reasoning here is good.

LL’s “physical argument” for why $\omega^2$ must be real and positive is a separate argument which I also think is OK. It’s a simple proof by contradiction. If $\omega^2$ is not real and positive, then $\omega$ must have an imaginary part, $i \alpha$, where $\alpha$ is real. Then each $x_k = A_k e^{i \omega t}$ would have the same exponential factor $e^{-\alpha t}$. Thus, each $\dot x_k$ would also have the factor $e^{- \alpha t}$. By inspection of the forms of $T$ and $U$ (and keeping in mind that they are positive definite) you can see that this would imply that the total energy $E$ cannot be constant in time. But $E$ must be conserved for this system as shown earlier in the text in section 6, pages 13-15. So, the assumption that $\omega^2$ is not real and positive leads to a contradiction. Therefore, $\omega^2$ must be real and positive.

I don’t see anything wrong with this line of reasoning.

 

[Demystifier]

If $\omega^2$ is not real and positive, then $\omega$ must have an imaginary part, $i \alpha$, where $\alpha$ is real. Then each $x_k = A_k e^{i \omega t}$ would have the same exponential factor $e^{-\alpha t}$.

I don't think it's a valid argument because it is not clear why each $x_k$ needs to have the same exponential factor. It seems much more natural to consider $\omega_k$ with the imaginary part $i \alpha_k$, so that each $x_k = A_k e^{i \omega_k t}$ has a different exponential factor $e^{-\alpha_k t}$.

 

[TSny]

I don't think it's a valid argument because it is not clear why each $x_k$ needs to have the same exponential factor. It seems much more natural to consider $\omega_k$ with the imaginary part $i \alpha_k$, so that each $x_k = A_k e^{i \omega_k t}$ has a different exponential factor $e^{-\alpha_k t}$.

Seeking particular solutions where each $x_k$ has the same exponential factor $e^{i\omega t}$ is standard fare. (For example, see Goldstein’s text.) It can then be shown that any solution of the equations of motion can be expressed as a linear combination of these particular solutions.

 

[Demystifier]

Seeking particular solutions where each $x_k$ has the same exponential factor $e^{i\omega t}$ is standard fare.

Not really. Consider, for instance, two degrees of freedom ($k=1,2$) without a coupling between $x_1$ and $x_2$. In practice one usually starts with the ansatz $x_k(t)=A_ke^{i\omega t}$. But then from two differential equations for $x_1$ and $x_2$ one obtains a quadratic equation for $\omega^2$ with two solutions $\omega^2_1$ and $\omega^2_2$. Then, when one turns $\omega^2_1$ back into the differential equations, one finds that it is a solution only if $A_2=0$. Similarly, $\omega^2_2$ gives a solution only if $A_1=0$. Hence the actual solution is
$$x_1(t)=A_1e^{i\omega_1 t}, \;\;\; x_2(t)=A_2e^{i\omega_2 t}$$
which means that the two oscillators never oscillate with the same frequency. There is no solution of the form $x_1(t)=e^{i\omega t}$, $x_2(t)=e^{i\omega t}$ with the same $\omega$.

 

[vanhees71]

I'm really puzzled, why there's still so much confusion. This example is now utmost simple. You have in this case the matrix $(k_{ij})=\mathrm{diag}(m \omega_1^2,m \omega_2^2)$ already diagonalized. Thus the EoM reads
$$\mathrm{d}_t^2 \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=-\begin{pmatrix} \omega_1^2 x_1 \\ omega_2^2 \vec{x}_2 \end{pmatrix}.$$
These are simply two-uncoupled harmonic oscillators.

Now you look for single-frequency solutions with the ansatz
$$\vec{x}(t)=\vec{A} \exp(-\mathrm{i} \omega t).$$
Plugging this into the EoM gives after cancelling the common factor $\exp(-\mathrm{i} \omega t)$
$$-\omega^2 \vec{A}=-\begin{pmatrix}\omega_1^2 A_1 \\ \omega_2^2 A_2 \end{pmatrix}.$$
Now if $\omega_1 \neq \omega_2$, you necessarily must have either
$$\omega=\omega_1, \quad A_2=0$$
or
$$\omega=\omega_2, \quad A_1=0.$$
This is not surprising since you simply have the two eigenvectors $(A_1,0)$ and $(0,A_2)$ for the normal modes of the two oscillators. Writing it in terms of the two normalized eigenvectors $\vec{e}_1=(1,0)$ and $\vec{e}_2=(0,1)$ the general solution thus is
$$\vec{x}(t)=\vec{e}_1 (C_{11} \exp(-\mathrm{i} \omega_1 t) +C_{12} \exp(\mathrm{i} \omega_1 t) +\vec{e}_2 (C_{21} \exp(-\mathrm{i} \omega_1 t) +C_{22} \exp(\mathrm{i} \omega_1 t),$$
as you can of course read off directly from the EoM, because the matrix $\hat{k}$ was already given in diagonalized form.

 

[HomogenousCow]

The errata for Schwartz's QFT text is rumored to be longer than the book itself

 

[vanhees71]

I thought that's more true for Peskin&Schroeder's QFT text...

 

[MathematicalPhysicist]

I thought that's more true for Peskin&Schroeder's QFT text...

It's true of every qft...

 

[WWGD]

And the index of the book cannot give both the right term and the right page simultaneously!

 

[WWGD]

Erratas are for minor typos, and OP excluded this out of discussion.

Google is rarely that precise (I've actually had the opposite problem), so that it may get you something reasonably-close to a list of mistakes. worth a try, I'd say.