I have two questions, and I'll address them while explaining my calculation and my, probably banal, uncertainties.

We're basically deriving the Energy-Momentum tensor for a scalar field from Noether theorem in a similar way of the one exposed in Weinberg, The Quantum Theory of Fields, Vol I, pag. 311.

Consider the lagrangian density of a complex scalar field:

$$ \mathscr{L} \, =\, -\partial^{\mu} \phi^* \partial_{\mu} \phi - m^2 \phi \phi^* $$

and the following transformation

$$ \phi \rightarrow \phi(x) - a^\mu (x) \partial_\mu\phi(x) \\ \phi^* \rightarrow \phi^*(x) - a_\mu (x) \partial^\mu\phi^*(x) $$

My professor writes that, beside the terms proportional to $a$ (which I have still to prove to myself give $\eta^{\mu \nu} \mathscr{L}$, with $\eta$ the metric), the variation of the Lagrangian is

$$ (\partial_\mu a_\nu)(\partial^{\mu} \phi^* \partial^{\nu} \phi + \partial^{\nu} \phi^* \partial^{\mu} \phi) $$

While I obtain

$$ (\partial_\mu a^\nu)(\partial^{\mu} \phi^* \partial_{\nu} \phi )\, + \, (\partial^\mu a_\nu) (\partial^{\nu} \phi^* \partial_{\mu} \phi) $$

I strongly feel that they could be the same but I don't know how to play with indices to reach the same result. Can someone help me with ths? That was my first question.

The second one is about I derived that result.

Following Weinberg, The Quantum Theory of Fields, Vol I, pag. 311, I have that the variation of the lagrangian, under the transformation written above, is (varying $\phi$ and $\phi^*$ independently)

$$ \frac{\partial \mathscr{L}}{\partial \phi} (- a^\nu \partial_\nu \phi) - \frac{\partial \mathscr{L}}{\partial \phi^*} (a_\nu \partial^\nu \phi) - \frac{\partial \mathscr{L}}{\partial (\partial_\mu \phi)} \partial_\mu (a^\nu \partial_\nu \phi) - \frac{\partial \mathscr{L}}{\partial (\partial^\mu \phi^*)} \partial^\mu (a_\nu \partial^\nu \phi^*) $$

Now I have a problem with the last 2 terms, why introducing the second and different index $\mu$? I just don't get it why it has to be a different index, can somebody explain that to me?


3 Answers 3


First answer: if $\eta$ is Minkowski metric, then the coordinate derivative of it vanishes, i.e. $$ \partial_\rho\eta_{\mu\nu}=0. $$ Hence, we can raise and lower the same indices, e.g. $$ \begin{align} (\partial_\mu a^\nu)(\partial^\mu\phi^\ast \partial_\nu\phi)&=(\partial_\mu (\eta^{\nu\rho} a_\rho))(\partial^\mu\phi^\ast \partial_\nu\phi)\\ &=(\partial_\mu a_\rho)\eta^{\nu\rho}(\partial^\mu\phi^\ast \partial_\nu\phi)\\ &=(\partial_\mu a_\rho)(\partial^\mu\phi^\ast \eta^{\nu\rho} \partial_\nu\phi)\\ &=(\partial_\mu a_\rho)(\partial^\mu\phi^\ast \partial^\rho\phi), \end{align} $$ which is after renaming the repeated index $\rho$ to $\nu$ we get the same expression.

Second answer: the index of derivation must be always different from that of Lagrangian, because otherwise we lose information.

  • $\begingroup$ Thank you, now that I notice, using the metric was the obvious thing to do. In which sense we lose information? I don't think I'm really getting it both physically and mathematically $\endgroup$ Nov 29, 2016 at 20:40
  • 1
    $\begingroup$ @Runlikehell Suppose following example: $\frac{\partial}{\partial(\partial_\mu \phi)}(\partial_\mu\phi)=1$. This is just one possible result. If instead we calculate this with different index, we get $\frac{\partial}{\partial(\partial_\nu \phi)}(\partial_\mu\phi)=\delta^\nu_\mu$, which is more general that the first case. It seems this is not so important, but it may play crucial role in many cases. $\endgroup$
    – Immanuel
    Nov 29, 2016 at 20:58

As for your first question, per definition $a_\mu b^\mu = a_\mu b_\nu \eta^{\mu \nu} = a^\mu b_\mu$, where $\eta$ is the metric -> you can hence always shift indices up/and down if you sum over them.

Regarding your second question, expressions of the form $a_\nu \partial^\nu$ can be regarded as a scalar. You just need to introduce a further index if you want to take more directional derivatives …

  • $\begingroup$ Thank you, now that I notice, using the metric was the obvious thing to do, could you elaborate the second answer more? I'm not so sure I got it, instead I'm pretty sure I'm not getting it. $\endgroup$ Nov 29, 2016 at 20:39

May I also add that $\eta^{\nu\rho}\eta_{\nu\sigma}=\eta^{\rho\nu}\eta_{\nu\sigma}=\delta^\rho_\sigma$ and e.g $\delta^\rho_\sigma a^\sigma=a^\rho$ plus that $\partial_\mu a^\nu=\eta^{\nu\sigma}\partial_\mu a_\sigma$. They may (or may not) come in handy.


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