# 广义相对论|PHYS3203 General Relativity代写

Recognising tensors. One way to prove that something is a vector or tensor is to show explicitly that it satisfies the coordinate transformation laws. This can be a long and arduous procedure if the tensor is complicated, like $R_{\mu \nu \kappa^{*}}^{\lambda}$. There is another way, usually much better! Show that if $V_{\nu}$ is an arbitrary covariant vector and the combination $T^{\mu \nu} V_{\nu}$ is known to be a contravariant vector (note the free index $\mu$ ), then
$$\left(T^{\prime \mu \nu}-T^{\lambda \sigma} \frac{\partial x^{\prime \mu}}{\partial x^{\lambda}} \frac{\partial x^{\prime \nu}}{\partial x^{\sigma}}\right) V_{\nu}^{\prime}=0$$
Why does this prove that $T_{\mu \nu}$ is a tensor? Does your proof actually depend on the rank of the tensots involved?

Solution:

If $T^{\mu \nu} V_{\nu}$ and $V_{\nu}$ are known to be tensors (vectors), we can write
$$T^{\prime \mu \nu} V_{\nu}^{\prime}=\frac{\partial x^{\prime \mu}}{\partial x^{\lambda}} T^{\lambda \kappa} V_{\kappa}=\frac{\partial x^{\prime \mu}}{\partial x^{\lambda}} \frac{\partial x^{n \nu}}{\partial x^{\kappa}} T^{\lambda \kappa} V_{\nu}^{\prime}$$
which implies
$$\left(T^{\prime \mu \nu}-T^{\lambda \sigma} \frac{\partial x^{\prime \mu}}{\partial x^{\lambda}} \frac{\partial x^{\prime \nu}}{\partial x^{\sigma}}\right) V_{\nu}^{\prime}=0$$
and
$$T^{\prime \mu \nu}=T^{\lambda \sigma} \frac{\partial x^{\prime \mu}}{\partial x^{\lambda}} \frac{\partial x^{\prime \nu}}{\partial x^{\sigma}},$$
since $V_{\nu}$ is arbitrary. The same reasoning applies to tensors of arbitrary rank.

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