Hi,(adsbygoogle = window.adsbygoogle || []).push({});

I've been wondering about this forever and I finally decided to ask on the forums. In relativistic index notation (with c= [itex]\hbar[/itex] =1) with the minkowski metric g_{[itex]\mu[/itex]}_{[itex]\nu[/itex]}=diag(1,-1,-1,-1), the 4-vector [itex]x^{\mu}=(t,x,y,z)=(x^0,\vec{x})[/itex], and with the del operator defined as [itex]\partial_{\mu}\equiv \frac{\partial}{\partial x^{\mu}}=(\partial_{t},\nabla)[/itex]. I should have that:

[itex]\partial^{\mu} x_{\mu}=\partial_{\mu} x^{\mu}=1[/itex]

but this is inconsistent with the way I usually think of vector calc because I should have

[itex]\partial_{\mu} x^{\mu}=\partial_{t}t+\nabla\bullet\vec{x}[/itex]

and

[itex]\partial_{t}t=1[/itex]

[itex]\nabla\bullet\vec{x}=3[/itex]

so, with the way I normally think, I should have:

[itex]\partial_{\mu} x^{\mu}=4[/itex]

Where am I going wrong here?

-Adam

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Relativistic index notation del-operator

**Physics Forums | Science Articles, Homework Help, Discussion**