Other Compilation of severe errors in famous textbooks

[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.