- #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= \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?
-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= \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?
-Adam
P.S. all of this notation is straight out of Peskin and Schroeder's QFT text