- #1
arwright3
- 6
- 0
Hi,
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
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