- #1
deuteron
- 64
- 14
Hi! I am trying to understand the object ##\partial^\mu##, and wanted to check if the result I am getting below is true.
The definition of ##\partial_\mu## is:
$$\partial_\mu = ( \frac \partial {\partial x^0} , \frac \partial {\partial x^1}, \frac \partial {\partial x^2},\frac \partial {\partial x^3})$$
We define ##\partial^\mu## as:
$$\partial^\mu = \eta^{\mu\nu}\partial_\nu = (\frac \partial {\partial x^0} , -\frac\partial {\partial x^1} , -\frac \partial {\partial x^2}, -\frac \partial {\partial x^3})$$
In this case, if we were to apply ##\partial^\mu## to ##x_\nu##, we would get:
$$\partial^\mu x_\nu = (\eta^{\mu\rho}\partial_\rho)x_\nu=\eta^{\mu\rho} \partial_\rho (\eta_{\nu\sigma} x^\sigma)$$
Since ##\eta##'s are just number, I rearrange the equation to be:
$$=\eta^{\mu\rho} \eta_{\nu\sigma} \partial_\rho x^\sigma$$
Here, my knowledge of indices and contraction ends and I try to think of the objects like matrices, in which case I write:
$$= \begin{bmatrix} 1&0&0&0\\ 0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}\begin{bmatrix} 1&0&0&0\\ 0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} \begin{bmatrix} \frac {\partial x^0}{\partial x^0} & \frac {\partial x^1}{\partial x^0} & \frac {\partial x^2}{\partial x^0} & \frac {\partial x^3}{\partial x^0} \\ \frac {\partial x^0}{\partial x^1} & \frac {\partial x^1}{\partial x^1} & \frac {\partial x^2}{\partial x^1} & \frac {\partial x^3}{\partial x^1}\\ \frac {\partial x^0}{\partial x^2} & \frac {\partial x^1}{\partial x^2} & \frac {\partial x^2}{\partial x^2} & \frac {\partial x^3}{\partial x^2} \\ \frac {\partial x^0}{\partial x^3} & \frac {\partial x^1}{\partial x^3} & \frac {\partial x^2}{\partial x^3} & \frac {\partial x^3}{\partial x^3}\end{bmatrix} $$
Here, one can observe the last matrix reduces to the identity matrix, and the multiplication of two ##\eta##'s also reduces to the identity matrix, thus I get:
$$=\mathbb I$$
Are these steps correct?
The definition of ##\partial_\mu## is:
$$\partial_\mu = ( \frac \partial {\partial x^0} , \frac \partial {\partial x^1}, \frac \partial {\partial x^2},\frac \partial {\partial x^3})$$
We define ##\partial^\mu## as:
$$\partial^\mu = \eta^{\mu\nu}\partial_\nu = (\frac \partial {\partial x^0} , -\frac\partial {\partial x^1} , -\frac \partial {\partial x^2}, -\frac \partial {\partial x^3})$$
In this case, if we were to apply ##\partial^\mu## to ##x_\nu##, we would get:
$$\partial^\mu x_\nu = (\eta^{\mu\rho}\partial_\rho)x_\nu=\eta^{\mu\rho} \partial_\rho (\eta_{\nu\sigma} x^\sigma)$$
Since ##\eta##'s are just number, I rearrange the equation to be:
$$=\eta^{\mu\rho} \eta_{\nu\sigma} \partial_\rho x^\sigma$$
Here, my knowledge of indices and contraction ends and I try to think of the objects like matrices, in which case I write:
$$= \begin{bmatrix} 1&0&0&0\\ 0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}\begin{bmatrix} 1&0&0&0\\ 0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} \begin{bmatrix} \frac {\partial x^0}{\partial x^0} & \frac {\partial x^1}{\partial x^0} & \frac {\partial x^2}{\partial x^0} & \frac {\partial x^3}{\partial x^0} \\ \frac {\partial x^0}{\partial x^1} & \frac {\partial x^1}{\partial x^1} & \frac {\partial x^2}{\partial x^1} & \frac {\partial x^3}{\partial x^1}\\ \frac {\partial x^0}{\partial x^2} & \frac {\partial x^1}{\partial x^2} & \frac {\partial x^2}{\partial x^2} & \frac {\partial x^3}{\partial x^2} \\ \frac {\partial x^0}{\partial x^3} & \frac {\partial x^1}{\partial x^3} & \frac {\partial x^2}{\partial x^3} & \frac {\partial x^3}{\partial x^3}\end{bmatrix} $$
Here, one can observe the last matrix reduces to the identity matrix, and the multiplication of two ##\eta##'s also reduces to the identity matrix, thus I get:
$$=\mathbb I$$
Are these steps correct?