Solving Tensor Calculus Question from Schutz Intro to GR

[shahbaznihal]

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.

 

[Orodruin]

It is a linear change of coordinates, not a tensor equation. Coordinates are not tensors.

 

[shahbaznihal]

It is a linear change of coordinates, not a tensor equation. Coordinates are not tensors.

Hence the rules of tensor calculus do not apply. This is just some equation where symbols do not have tensorial meanings.

But then if you want to transform from one coordinate system to another, how do you know how the coordinates transform? Given you only know the metric tensors in the two frame? Or it can't be done that way and you need some additional information about how the two frames are related? Perhaps some physical intuition like in this problem where the related them by the idea that if $\phi=0$ then you already have a locally inertial frame. Which was then used to derive the equation $x^{\alpha '} = (\delta^\alpha_\beta + L^\alpha_\beta)x^\beta$.

 

[Orodruin]

But then if you want to transform from one coordinate system to another, how do you know how the coordinates transform? Given you only know the metric tensors in the two frame?

Just knowing the components of the metric is generally not sufficient because there can be several coordinate systems satisfying the same requirement. However, note that with your ansatz of a linear relationship between the coordinates, you have
$$
\frac{\partial x'^{\alpha'}}{\partial x^\beta} = \delta^\alpha_\beta + L^\alpha_\beta.
$$
This should let you find some restrictions on $L$ based on the transformation rules for the metric tensor.

 

[shahbaznihal]

That is how the solution proceeds as well. Actually, since $L^\alpha_\beta$ is a function of Newtonian potential $\phi (x,y,z)$ then it also needs to be differentiated which gives, $$\frac{\partial x^{\alpha '}}{\partial x^\gamma}=\delta^{\alpha}_\gamma + L^{\alpha}_\gamma + L^\alpha_{\beta, \gamma} x^\beta$$ But yeah, that is the point of the solution.

 

[shahbaznihal]

Just knowing the components of the metric is generally not sufficient because there can be several coordinate systems satisfying the same requirement. However, note that with your ansatz of a linear relationship between the coordinates, you have
$$
\frac{\partial x'^{\alpha'}}{\partial x^\beta} = \delta^\alpha_\beta + L^\alpha_\beta.
$$
This should let you find some restrictions on $L$ based on the transformation rules for the metric tensor.

But in deriving the above equations, the idea of contraction is used. So the coordinates are not tensors but their derivatives are. Correct?

Is the idea of contraction a tensor operation? Or is it just a way of compactly writing a bunch of equations i.e. $$f^\alpha_\beta = T^\alpha_\gamma T^\gamma_\beta$$ is just a set of functions with terms quadratic is some function $T$?

 

[Orodruin]

But in deriving the above equations, the idea of contraction is used. So the coordinates are not tensors but their derivatives are. Correct?

No, this is not correct. Summing over two indices does not necessarily mean that you have tensor expressions. Consider the Lorentz transformation $x'^{\alpha'} = \Lambda^{\alpha'}_\beta x^\beta$. The transformation coefficients $\Lambda^{\alpha'}_\beta$ are not tensors, they are coefficients that tell you how the coordinates are related.

The derivative of a coordinate with respect to some curve parameter is (by definition) the components of a tangent vector. The partial derivatives of some set of coordinates with respect to another set of coordinates are the transformation coefficients that tell you how tensor components transform between the systems. They are not by themselves tensors.

 

[shahbaznihal]

Thanks a bunch for the greatly useful discussion :)