Recently I posted a question about variation of metric. I thought I understood it and talked with my friend about it. After that he said he's not convinced because he can't prove variation of metric are tensor as well as orginal one. I considered a lot but I can't get an answer, so plz let me ask a question.

He says: Where $D^{\alpha}_{\beta} = ∂x^{\alpha}/∂x'_{\beta}$, $$\delta g_{\mu \nu} = \delta (g_{\mu \nu})$$ and under a Lorenz transformation $$\delta g’_{\mu \nu} = \delta(D^{\rho}_{\mu} D^{\sigma}_{\nu}g_{\rho \sigma}) \neq D^{\rho}_{\mu} D^{\sigma}_{\nu} \delta g_{\rho \sigma} $$ in general so $\delta g_{\mu \nu}$ is not a tensor anymore. And he suggests some examples:

For flat Minkowski metric, $$\delta g = g(x+\delta x) - g(x) = 0$$ but for radial coordinates $\delta g \neq 0$ so it's not a tensor (indices are omitted here).

However, I think:

  1. We can't define sum of fields that have different properties under the transformation so the varation should be tensor as well. e.g. scalar + vector = ?

  2. We should treat $\delta g_{\mu \nu}$ as $(\delta g)_{\mu \nu}$ (i.e. ($\mu, \nu$)-component of tensor $\delta g$), in other words it's tensor if this condition $$(\delta g)'_{\mu \nu} = D^{\rho}_{\mu} D^{\sigma}_{\nu} (\delta g)_{\rho \sigma}$$ is satisfied. However, I can't give him explanation about this point anymore.

So my question is: $\delta g$ is a tensor again? If so/not, please tell me what is wrong in our discussion.


  • I think the problem is whether $\delta g $ is equal to $g(x+\delta x) - g(x)$ or just a arbitrary tensor .

  • I think function is also functional when $x$ of $g(x)$ is a particular point. In other words, $$g_{\delta}[g] \equiv g_{\mu\nu}(x_0)= \int_{M} \delta(x-x_0) g_{\mu\nu}(x) $$ so we can take arbitrary 2-rank tensor as its variation, but if we think $x$ as a function of proper time $\tau$ on world-line, then we can think $g$ is a functional of $x$: $g_{\mu \nu}[x]$ so $\delta x$ resurges.

  • Corrected some mistakes in my question.


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