0
$\begingroup$

I'm trying to prove that the divergence of the energy-momentum-tensor is zero by expressing it in terms of the field strength tensor: $\partial_\mu T^{\mu\nu}=0$.

In doing this, letting the derivative act on the second part of the tensor (the Lagrangian part) one obtains the following term (as a part of the full term) after applying the product rule: $(\partial_\mu F_{\alpha\beta})F^{\alpha\beta}+F_{\alpha\beta}\partial_\mu(F^{\alpha\beta})$. Here the solution asserts that this is equal to simply equal to twice the first part of the term, implying $(\partial_\mu F_{\alpha\beta})F^{\alpha\beta}=F_{\alpha\beta}\partial_\mu(F^{\alpha\beta})$. Can anyone help me in understanding why this is the case? I have not been able to find many resources on this and those (or asking AI lol) have not proven enlightning to me (so far).

Improve this question
$\endgroup$
2
  • $\begingroup$ It will be a million times more obvious that they are the same if you realise that the two terms are both $F_{\alpha\beta}g^{\alpha\gamma}g^{\beta\delta}\partial_\mu F_{\gamma\delta}$ $\endgroup$
    – naturallyInconsistent
    Commented Jul 19 at 1:41
  • $\begingroup$ I have realized the error in my line of thinking, see comment on other reply. $\endgroup$
    – user410662
    Commented Jul 19 at 1:58

2 Answers 2

Reset to default
1
$\begingroup$

Here the solution asserts that this is equal to simply equal to twice the first part of the term, implying $(\partial_\mu F_{\alpha\beta})F^{\alpha\beta}=F_{\alpha\beta}\partial_\mu(F^{\alpha\beta})$. Can anyone help me in understanding why this is the case?

Per your explicit parentheses, your first expression above means the same thing as: $$ F^{\alpha\beta}\partial_\mu F_{\alpha\beta}\;. $$

Since the metric is constant (I assume), you can raise and lower the dummy index pairs however you please: $$ F^{\alpha\beta}\partial_\mu F_{\alpha\beta} = F_{\alpha}^{\;\beta}\partial_\mu F^{\alpha}_{\;\,\beta} = F_{\alpha\beta}\partial_\mu F^{\alpha\beta} $$

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ I think I may have found my error: I implicity assumed that the multiplication was non-commutative (at least not trivially commu.), thinking that I was multiplying two rank-2 tensors, when in reality we are working with components of tensors, so this is not an issue. Funny how it's the small things that trip you up sometimes. Thank you for your answer ^^ $\endgroup$
    – user410662
    Commented Jul 19 at 1:55
1
$\begingroup$

Having realized my error of thinking the multiplication was non-commutative, it becomes clear: $$(\partial_\mu F_{\alpha\beta})F^{\alpha\beta}=F^{\alpha\beta}(\partial_\mu F_{\alpha\beta})=\eta_{\alpha\gamma}\eta_{\beta\lambda}F^{\alpha\beta}(\partial_\mu F^{\gamma\lambda})= F_{\gamma\lambda}(\partial_\mu F^{\gamma\lambda})=F_{\alpha\beta}(\partial_\mu F^{\alpha\beta})$$

Thank you for your answers :)

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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