0
$\begingroup$

I have recently started studying quantum field theory from the book Quantum Field Theory and the Standard Model by Schwartz. In chapter 2 it is said that, contrary to GR, one can ignore the index position, because we are working in Minkowski. I do understand this when the indices are contracted, nevertheless in chapter 3, we arrive to the following equality

\begin{equation} \partial_\mu \left(\sum_n \frac{\partial L}{\partial(\partial_\mu \phi_n)}\partial_\nu \phi_n - g_{\mu \nu}L\right)=0 \end{equation}

and from here the energy-momentum tensor for a classical field theory is defined as

\begin{equation} T_{\mu \nu}=\sum_n \frac{\partial L}{\partial(\partial_\mu \phi_n)}\partial_\nu \phi_n - g_{\mu \nu}L\,. \end{equation}

Yet in Tong's notes and some other references (where the position of the indices is taken into account as supposed - http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf), one finds that

\begin{equation} T'^\mu_ \nu=\sum_n \frac{\partial L}{\partial(\partial_\mu \phi_n)}\partial_\nu \phi_n - \delta^\mu_\nu L\,. \end{equation}

Now, I can lower the index from this definition and find

\begin{equation} T'_{\mu \nu}=\sum_n \frac{\partial L}{\partial(\partial^\mu \phi_n)}\partial_\nu \phi_n - g_{\mu \nu}L\,. \end{equation}

Now it seems to me that considering each convention $T'_{\mu \nu} \neq T_{\mu \nu} $. Of course, if I contract the first index the equality is restored, yet I don't see how these components are equal alone. Anyone who can help? Is this a typo or am I missing something?

$\endgroup$
  • 1
    $\begingroup$ The first author is being careless with indices. $\endgroup$ – Jahan Claes Jul 8 '16 at 20:39
  • $\begingroup$ It's not a typo. In Schwartz, $F^{\mu\nu} = F_{\mu\nu} = F_{\mu}^{\ \nu}$, and so on. $\endgroup$ – knzhou Jul 8 '16 at 20:44
  • $\begingroup$ I still don't understand what that means... for example, if I want to compute $T_{11}$, I would get $T_{11}=\sum_n \frac{\partial L}{\partial(\partial_x \phi_n)}\partial_x \phi_n +L$. Now I could do the same for the other definition and I get $T_{11}=\sum_n -\frac{\partial L}{\partial(\partial_x \phi_n)}\partial_x \phi_n +L$ . How are they the same? $\endgroup$ – blackhole1511 Jul 8 '16 at 21:10
  • 1
    $\begingroup$ @blackhole1511 You need to figure out where the indices go before plugging in specific components. It'll take some extra work, but you can always recover the positioning uniquely. $\endgroup$ – knzhou Jul 8 '16 at 21:23
  • 1
    $\begingroup$ As a simpler case, if you wanted $A^0$ and Schwartz wrote $A^\mu = B_\mu$, you would have to raise the index to $B^\mu$ first, then extract $B^0$. $\endgroup$ – knzhou Jul 8 '16 at 21:24
2
$\begingroup$

The stress-energy tensor is not symmetric by its definition. It is symmetric only if we require the system to be rotationally invariant and to have no intrinsic spin (all representations of the internal Lorentz algebra must vanish).

In particular one can show that: $$ M^{\mu}_{\alpha\beta} = (x_{\alpha}T^{\mu}_{\beta} - x_{\beta}T^{\mu}_{\alpha}) + \textrm{(Irrep)}^{\mu}_{\alpha\beta} $$ where the left hand side is the representation of a generator of Lorentz transformations and the last term on the right hand side is proportional to some sort of "internal current fields" that have to transform according to the Lorentz group as well.

Taking derivatives of the left hand side and demanding it to vanish (for rotational invariance) proves the statement if the last term on the right hand side is zero too.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.