Definition of Form-Invariant Function: Q&A

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Einj
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Hello everyone. I'm reading Weinberg's 'Gravitation and Cosmology' and I'm having some problems understanding the definition of a 'form-invariat function'. He says:
A metric ##g_{\mu\nu}## is said to be form-invariant under a given coordinate transformation ##x\to x^\prime##, when the transformed metric ##g^\prime_{\mu\nu}(x^\prime)## is the same function of its argument ##x^{\prime\mu}## as the original metric ##g_{\mu\nu}(x)## was of its argument ##x^\mu##, that is,
\begin{equation}
g^\prime_{\mu\nu}(y)=g_{\mu\nu}(y) \; \text{ for all }y.
\end{equation}
If the previous condition was true doesn't this simply mean that ##g_{\mu\nu}^\prime## is the same function as ##g_{\mu\nu}##? Should we ask for:
\begin{equation}
g^\prime_{\mu\nu}(x^\prime)=g_{\mu\nu}(x)
\end{equation}
?

Thanks a lot!
 
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Einj said:
If the previous condition was true doesn't this simply mean that ##g_{\mu\nu}^\prime## is the same function as ##g_{\mu\nu}##?

That is precisely what he means, that they are the same function.
I think that's why he used the symbol "y" instead of x's, so that it wouldn't be confused as having something to do with the coordinates x and x'.
 
Last edited:
The problem with [itex]g'_{\mu\nu}(x')=g_{\mu\nu}(x)[/itex] is that it means the two functions have same value in a single point. But because [itex]x'[/itex] associates different numbers to that point compared to [itex]x[/itex], [itex]g'_{\mu\nu}(x')=g_{\mu\nu}(x)[/itex] necessarily means [itex]g[/itex] and [itex]g'[/itex] are different functions and don't have the same form.
But [itex]g'_{\mu\nu}(y)=g_{\mu\nu}(y)[/itex] means that if we give the two functions, the same numbers, they will give equal results which means they're the same function. Note that y in the right refers to a different point than y in the left because they are same numbers in different coordinate systems.
 
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Shyan said:
Note that y in the right refers to a different point than y in the left because they are same numbers in different coordinate systems.
Oh I see. That is clear. Thanks!