0
$\begingroup$

I am doing a problem from Schutz, Introduction to general relativity.The question asks you to find a coordinate transformation to a local inertial frame from a weak field newtonian metric tensor $$ds^2=-(1+2\phi)dt^2+(1-2\phi)(dx^2+dy^2+dz^2).$$ I looked at the solution from a manual and it has the following equation,$$x^{\alpha '} = (\delta^\alpha_\beta + L^\alpha_\beta)x^\beta$$ $L^\alpha_\beta$ is a function of the Newtonian potential $\phi$.

My question is: Is this a valid tensor equation?

The transformation is motivated by the idea that when $\phi$ is zero then you already have a locally inertial frame hence $L^\alpha_\beta$ are all zero and $x^{\alpha '} = x^{\alpha}$ which is very understandable. But is the equation, $x^{\alpha '} = (\delta^\alpha_\beta + L^\alpha_\beta)x^\beta$, symbolically correct (because the superscripts don't balance out like they normally do in tensor calculus)? But may be if $L^\alpha_\beta$ is not a tensor then it does not have to obey those principles of tensor calculus.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Firstly, yes, the indices do work properly, although there is confusion about primes. I normally put primes on the object itself (ie I would write $x'^\alpha$, not on the index), but this is just notation and I don't actually know what the normal convention is (and my notation is even more confusing in this case I think). If you want to prime the indices then the expression should really be:

$$x^{\alpha'} = \left(\delta^{\alpha'}{}_\beta + L^{\alpha'}{}_\beta\right)x^\beta$$

And this is fine.

However, there is an important point here, which I think you have realised. The things that drive coordinate / basis transformations are not tensors: they are, rather, just nonsingular matrices, and what this is is just matrix multiplication. It's easy to see that they are not (the components of) tensors because they have indices in two different bases, and that makes no sense at all. And if you do a suitable change of basis to get their indices all in one basis, you find that the result is always $\delta^\alpha{}_\beta$!

Improve this answer
$\endgroup$
3
  • $\begingroup$ But the issue is this: If I let $L^\alpha_\beta =0$ then by the nature of the problem $x^{\alpha '} = x^{\alpha}$ which makes a lot of sense. But if I put primes like you have then I get $x^{\alpha '} = x^{\alpha '}$ which is just confusing (or atleast I cant make sense out of them). $\endgroup$
    – Shaz
    Commented Jul 22, 2018 at 12:03
  • $\begingroup$ @Shaz really what you mean is $x^{\alpha'} = \delta^{\alpha'}{}_\beta x^\beta$: the change-of-basis matrix (which here is the identity matrix) is the thing that adds or removes primes on indices. This is why I prefer primes on the things themselves, as I find $x'^\alpha = x^\alpha$ much easier to understand: it's just saying that the primed coordinates are the same as the unprimed ones. However I don't know how common (or how good) my notation is. $\endgroup$
    – user107153
    Commented Jul 22, 2018 at 12:20
  • $\begingroup$ Ahhhh. I should have seen that :D . Thanks a bunch for pointing that out. $\endgroup$
    – Shaz
    Commented Jul 22, 2018 at 12:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.