# I Why are coordinates independent in GR?

#### TomServo

Summary
Looking for fundamental reason why we only look at independent coordinates in GR
I can see that by the tensor transformation law of the Kronecker delta that

$\frac{\partial x^a}{\partial x^b}=\delta^a_b$

And thus coordinates must be independent of eachother.

But is there a more straightforward and fundamental reason why we don’t consider dependent coordinates? Is it just built-in to the transformation law?

Obviously if we could have dependent coordinates, then for a metric like this plane wave one:

$ds^2=-2dudv+f(u)g(u)dy^2+f(u)/g(u)dz^2$

...I could rescale my y and z coordinates by:

$dy’^2=f(u)g(u)dy^2$ and $dz’^2=f(u)/g(u)dz^2$

And have

$ds^2=-2dudv+dy’^2+dz’^2$

And thus turn the metric (or any metric by such a rescaling) to flat spacetime. So I see why it’s good to have linearly independent coordinates to forbid transformations like the above.

But why? Is this requirement the very motivation for the tensor transformation laws? What’s the underlying reason?

Thanks.

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#### PeroK

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In general, the number of independent coordinates is determined by the dimension of the space. If, for example, in the 2D plane you have $y$ dependent on $x$, then you have a 1D curve.

You could, I guess, have more than 2 coordinates for the x-y plane, but one of them would be redundant.

If you have the minimum number of coordinates then they must be independent.

#### TomServo

In general, the number of independent coordinates is determined by the dimension of the space. If, for example, in the 2D plane you have $y$ dependent on $x$, then you have a 1D curve.

You could, I guess, have more than 2 coordinates for the x-y plane, but one of them would be redundant.

If you have the minimum number of coordinates then they must be independent.

#### PeroK

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2018 Award
Independent does not mean orthogonal.

#### TomServo

Independent does not mean orthogonal.
That's true, but perhaps you can help me to see something. Let's say I have the 1+1 Minkowski space, with coordinates x and t with basis vectors $\hat{x}=(0,1)$ and $\hat{t}=(1,0)$. Now I define a new coordinate u=t+x to replace t, and so my coordinates are u and x with basis vectors $\hat{u}=(1,1)$ and $\hat{x}=(0,1)$.

These are independent basis vectors, so in that sense they are independent, but based on the definition of u it seems that $\frac{\partial u}{\partial u}=1$ and $\frac{\partial u}{\partial x}=1\neq 0$, in conflict with the notion that $\frac{\partial x^a}{\partial x^b}=\delta^a_b$, so it seems they aren't independent after all.

Where have I gone wrong?

Is it that we cannot introduce just one new coordinate, but when we say we are introducing a new coordinate we are implicitly introducing D new coordinates, but the new special coordinate and D-1 coordinates such that $x'^a=x^a$? And thus what I should have said is that I'm introducing new coordinates u and x', such that u=t+x and x'=x, and thus $\frac{\partial u}{\partial u}=1$ but $\frac{\partial u}{\partial x'}=0$?

#### PeroK

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2018 Award
Sorry, will be offline for a bit. You're really differentiating a curve there. There's a subtle difference between that and a new coordinate system.

Hopefully someone else can explain in full.

#### martinbn

That's true, but perhaps you can help me to see something. Let's say I have the 1+1 Minkowski space, with coordinates x and t with basis vectors $\hat{x}=(0,1)$ and $\hat{t}=(1,0)$. Now I define a new coordinate u=t+x to replace t, and so my coordinates are u and x with basis vectors $\hat{u}=(1,1)$ and $\hat{x}=(0,1)$.

These are independent basis vectors, so in that sense they are independent, but based on the definition of u it seems that $\frac{\partial u}{\partial u}=1$ and $\frac{\partial u}{\partial x}=1\neq 0$, in conflict with the notion that $\frac{\partial x^a}{\partial x^b}=\delta^a_b$, so it seems they aren't independent after all.

Where have I gone wrong?

Is it that we cannot introduce just one new coordinate, but when we say we are introducing a new coordinate we are implicitly introducing D new coordinates, but the new special coordinate and D-1 coordinates such that $x'^a=x^a$? And thus what I should have said is that I'm introducing new coordinates u and x', such that u=t+x and x'=x, and thus $\frac{\partial u}{\partial u}=1$ but $\frac{\partial u}{\partial x'}=0$?
This is because of abuse of notations. When you switch from the coordinates $(t,x)$ to $(u,x)$ by the relation $u=t+x$, what is really meant is that you are considering new coordiantes $(u,\tilde{x})$ with $u=f(t,x)=t+x$ and $\tilde{x}=g(t,x)=x$.

#### TomServo

This is because of abuse of notations. When you switch from the coordinates $(t,x)$ to $(u,x)$ by the relation $u=t+x$, what is really meant is that you are considering new coordiantes $(u,\tilde{x})$ with $u=f(t,x)=t+x$ and $\tilde{x}=g(t,x)=x$.
Okay, so my hypothesis at the end was correct?

I don't like abuse of notations because of the confusion they cause, I'm grateful to you for point it out.

"Why are coordinates independent in GR?"

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