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This is intuitively very simple problem but I am unable to complete it with Mathematical rigor. Here is the deal:

A coordinate system $(u,v,w,p)$ in which the metric tensor has the following non-zero components, $g_{uv}= g_{ww}=g_{pp}=1$. Find the coordinate transformation between $(u,v,w,p)$ to the regular coordinates $(t,x,y,z)$.

So in the first part of the same question you are supposed to prove that coordinate space $(u,v,w,p)$ is flat which is easy enough to see because the metric components are constant, all Riemann tensor components are zero. Hence the space is Minkowski. Which, intuitively, implies that the transformation between the coordinates will be linear because standard $(t,x,y,z)$ form a Minkowski space. But I am unable to prove it Mathematically.

Here is what I did. The question had a hint to calculate $e_u.e_u$ and $e_v.e_v$ (e represents the basis in either coordinate system and subscript implies which coordinates they belong to. I am removing the vector heads or the circumflex). From the metric tensor in this coordinate they are going to be zero.

So, I did write the conversion between the basis of the two systems,

$$

e_{\alpha '} = \Lambda^\alpha_{\alpha '} = \frac{\partial x^\alpha}{\partial x^{\alpha '}}e_\alpha

$$

Here the primed indices represent the $(u,v,w,p)$ coordinates and unprimed represents $(t,x,y,z)$. This gives the following dot product,

$$

e_{\alpha '}.e_{\beta '}= \Lambda^\alpha_{\alpha '}\Lambda^\beta_{\beta '}e_\alpha.e_\beta = \frac{\partial x^\alpha}{\partial x^{\alpha '}}\frac{\partial x^\beta}{\partial x^{\beta '}}e_\alpha.e_\beta

$$

And I can use the orthonormality of the $(t,x,y,z)$ coordinates to simplify $e_\alpha.e_\beta$ but even then using the metric all,

$$

e_{\alpha '}.e_{\beta '} = \frac{\partial x^\alpha}{\partial x^{\alpha '}}\frac{\partial x^\beta}{\partial x^{\beta '}}e_\alpha.e_\beta

$$

gives me is a bunch of (actually 16) partial differential equations but I do not know where to go from there.

How do I prove that the partial differential equations have a solution,

$$x^\alpha = A^\alpha_{\alpha '}x^{\alpha '}$$

where $$A^\alpha_{\alpha '}$$ are constants? Surely it can be seen that it is ONE of the solution of the system of PDEs but is that the only solution(?). I arrived at the answer intuitively but want to be able to SEE it mathematically being a solution. Also, how do I end up calculating the values of $$A^\alpha_{\alpha '}$$?

A coordinate system $(u,v,w,p)$ in which the metric tensor has the following non-zero components, $g_{uv}= g_{ww}=g_{pp}=1$. Find the coordinate transformation between $(u,v,w,p)$ to the regular coordinates $(t,x,y,z)$.

So in the first part of the same question you are supposed to prove that coordinate space $(u,v,w,p)$ is flat which is easy enough to see because the metric components are constant, all Riemann tensor components are zero. Hence the space is Minkowski. Which, intuitively, implies that the transformation between the coordinates will be linear because standard $(t,x,y,z)$ form a Minkowski space. But I am unable to prove it Mathematically.

Here is what I did. The question had a hint to calculate $e_u.e_u$ and $e_v.e_v$ (e represents the basis in either coordinate system and subscript implies which coordinates they belong to. I am removing the vector heads or the circumflex). From the metric tensor in this coordinate they are going to be zero.

So, I did write the conversion between the basis of the two systems,

$$

e_{\alpha '} = \Lambda^\alpha_{\alpha '} = \frac{\partial x^\alpha}{\partial x^{\alpha '}}e_\alpha

$$

Here the primed indices represent the $(u,v,w,p)$ coordinates and unprimed represents $(t,x,y,z)$. This gives the following dot product,

$$

e_{\alpha '}.e_{\beta '}= \Lambda^\alpha_{\alpha '}\Lambda^\beta_{\beta '}e_\alpha.e_\beta = \frac{\partial x^\alpha}{\partial x^{\alpha '}}\frac{\partial x^\beta}{\partial x^{\beta '}}e_\alpha.e_\beta

$$

And I can use the orthonormality of the $(t,x,y,z)$ coordinates to simplify $e_\alpha.e_\beta$ but even then using the metric all,

$$

e_{\alpha '}.e_{\beta '} = \frac{\partial x^\alpha}{\partial x^{\alpha '}}\frac{\partial x^\beta}{\partial x^{\beta '}}e_\alpha.e_\beta

$$

gives me is a bunch of (actually 16) partial differential equations but I do not know where to go from there.

How do I prove that the partial differential equations have a solution,

$$x^\alpha = A^\alpha_{\alpha '}x^{\alpha '}$$

where $$A^\alpha_{\alpha '}$$ are constants? Surely it can be seen that it is ONE of the solution of the system of PDEs but is that the only solution(?). I arrived at the answer intuitively but want to be able to SEE it mathematically being a solution. Also, how do I end up calculating the values of $$A^\alpha_{\alpha '}$$?

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