Relativistic index notation del-operator

1. Jan 22, 2013

arwright3

Hi,

I've been wondering about this forever and I finally decided to ask on the forums. In relativistic index notation (with c= $\hbar$ =1) with the minkowski metric g$\mu$$\nu$=diag(1,-1,-1,-1), the 4-vector $x^{\mu}=(t,x,y,z)=(x^0,\vec{x})$, and with the del operator defined as $\partial_{\mu}\equiv \frac{\partial}{\partial x^{\mu}}=(\partial_{t},\nabla)$. I should have that:
$\partial^{\mu} x_{\mu}=\partial_{\mu} x^{\mu}=1$
but this is inconsistent with the way I usually think of vector calc because I should have
$\partial_{\mu} x^{\mu}=\partial_{t}t+\nabla\bullet\vec{x}$
and
$\partial_{t}t=1$
$\nabla\bullet\vec{x}=3$
so, with the way I normally think, I should have:
$\partial_{\mu} x^{\mu}=4$
Where am I going wrong here?

P.S. all of this notation is straight out of Peskin and Schroeder's QFT text

2. Jan 22, 2013

Staff: Mentor

You are not going wrong at all. In 4D flat spacetime, it should be 4.

3. Jan 22, 2013

WannabeNewton

Hi arwright3! It's because there is an implied summation involved so $\partial _{\mu }x^{\mu } = \partial _{0}x^{0} + \partial _{1}x^{1} + \partial _{2}x^{2} + \partial _{3}x^{3} = 4$. Don't forget when you have an index on the top and the same index on the bottom there is an implied summation! Cheers!

4. Jan 23, 2013

arwright3

Thanks for your help! One more quick question. Can someone derive this:
$\partial^{\mu} x^{\nu}=g^{\mu \nu}$?

P.S. I'm trying to pick up this 4-vector stuff on my own because I need it for QFT and I never had an undergrad class in special relativity. If anyone has a suggestion for a good resource I'd appreciate it. I've read a few pdfs that I found with various google searches, and I understand how the Lorentz transformations work and everything, but what I'd really like is something that explains what is and is not allowed when working with four-vectors. There are many derivations in my QFT book that are ~2 lines long and I don't know how to work out the intermediate steps. Most of these things have nothing to do with QFT, even the classical E&M field equations/lagrangian are hard for me to understand and work with in their index notation form.

5. Jan 23, 2013

WannabeNewton

$\partial _{\mu }x^{\nu } = \delta _{\mu }^{\nu }$ so $g^{\sigma \mu }\partial _{\mu }x^{\nu } = \partial ^{\sigma }x^{\nu } = g^{\sigma \mu }\delta _{\mu }^{\nu } = g^{\sigma \nu }$ and you can just relabel sigma to nu from there if you would like. I have a good book for you that gives you a ton of practice with 4 - vectors and Einstein summation and tensor calculus and all the other fun stuff which I will pm to you however I have never done QFT so I do not know the depth at which you need to know these things - I am basing it off of what I know from GR.

6. Jan 23, 2013

arwright3

Thanks, that's very nice of you. and BTW I like your profile picture!