# 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!