0
$\begingroup$

Question #1:

Lost as to how the second equality in the following equation holds —

$$\frac{\partial}{\partial \tau} (A^2) = \frac{\partial}{\partial \tau} (\eta_{\mu\nu}A^\mu A^\nu) = 2\eta_{\mu\nu}A^\mu \frac{\partial A^\nu}{\partial \tau}.$$

Seems to me like there is a similar trick to like when you use the Euler Lagrange equation to get the Klein Gordan equation,

$$(\partial_\mu \phi)^2=(\partial_\mu \phi)(\partial^\mu \phi)=\eta^{\mu\nu}\partial_\mu \partial_\nu \phi.$$

But I can't put my finger on it.

Answer to question #1 taking eranreches and G. Smith's tips into account:

$$ \begin{eqnarray}\frac{\partial}{\partial \tau} (A^2) &=& \frac{\partial}{\partial \tau} (\eta_{\mu\nu}A^\mu A^\nu) \\ &=& \frac{\partial}{\partial \tau} (\eta_{\mu\nu}A^\mu) A^\nu + \eta_{\mu\nu}A^\mu \frac{\partial}{\partial \tau}(A^\nu)\\ &=& \eta_{\mu\nu}\frac{\partial}{\partial \tau} (A^\mu) A^\nu + \eta_{\mu\nu}A^\mu \frac{\partial}{\partial \tau}(A^\nu) \\ &=& \eta_{\nu\mu}\frac{\partial}{\partial \tau} (A^\nu) A^\mu + \eta_{\mu\nu}A^\mu \frac{\partial}{\partial \tau}(A^\nu) \\ &=& 2\eta_{\mu\nu}A^\mu \frac{\partial}{\partial \tau} (A^\nu) \\ \end{eqnarray}$$

where in the 2nd to last line the index $\mu$ has been renamed to $\nu$, and in the last line, we used the fact that $\eta_{\mu\nu}$ is symmetric.

Question #2:

I have the following "rules" for index-notation "transformations":

  • $A'^\mu = \Lambda^\mu_\nu A^\nu$
  • $A^\mu = \delta^\mu_\nu A^\nu$
  • $\partial^\mu = \eta^{\mu\nu} \partial_\nu$

I am perfectly comfortable with the top equation, but my question is, what is the difference between the 2nd and 3rd? I know that the $\delta^\mu_\nu$ and the $\eta^{\mu\nu}$ are difference (one is the 4x4 identity matrix, and the other the metric tensor with the tt component -1, or vice-versa depending on your metric preference), but I'm not sure in what circumstances to use each when they seem to be used in similar circumstances, i.e. to raise of lower an index.

My only guess is that, one needs to use $\delta^\mu_\nu$ when working with 4-vectors like $A^\mu$ to keep all the components the same. However, I'm confused why then (by implication) we would WANT to change one of the components of the derivative as with using the metric tensor to raise/lower such as with the 3rd equation.

$\endgroup$
  • $\begingroup$ for the second question, notice that $\delta^\mu_\nu$ doesn't lower the index at all, it just "selects" the index $\mu$. You can think of $A^\nu$ and $A_\nu$ loosely as you'd think $|\psi\rangle$ and $\langle \psi|$ in QM, and of the metric as something that makes you go back and forth, combining them gives you something similar to a scalar product. I hope no mathematician reads this but it helped my intuition at the time. For a more formal explanation look for tangent and cotangent spaces and something called musical isomorphism $\endgroup$ – user2723984 Jun 5 at 20:57
1
$\begingroup$

Q1: The second equality is just $\frac{{\rm d}}{{\rm d}\tau}\left(fg\right)=\frac{{\rm d}f}{{\rm d}\tau}g+f\frac{{\rm d}g}{{\rm d}\tau}$ plus re-indexing.

Q2: I would suggest you to post each of your questions separately.

$\endgroup$
  • $\begingroup$ credited with answer above $\endgroup$ – Lopey Tall Jun 6 at 16:14
1
$\begingroup$

Re #1: Use the fact that $\eta^{\mu\nu}$ is symmetric.

Re #2: The first equation transforms the vector between two different inertial frames. The second and third equations are identities within one reference frame.

$\endgroup$
  • $\begingroup$ credited with answer above $\endgroup$ – Lopey Tall Jun 6 at 16:14

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.