# What is the source of curvature in an accelerating frame?

PeterDonis
Mentor
2019 Award
The Ricci tensor is a contraction of the Riemann tensor and may be zero while allowing some curvature that isn't described by the Ricci tensor (this is called Weyl curvature, I believe).
Yes. For example, all of the curvature in any vacuum solution of the EFE, such as the Schwarzschild solution, is Weyl curvature.

It's possible that you are still confusing "vanishing Ricci tensor" with "spacetime is flat". As I said before, they are not the same; you can have a non-flat spacetime (or region of spacetime) with a vanishing Ricci tensor.
If the field is measured at distance $r$ from the center of the mass, then it should satisfy or be approximated to Poisson equation.
$$\bigtriangledown ^2 \Phi=4 \pi G \rho$$
Now $$\Phi=\frac{1}{2}g_{00}$$
So, given that $T_{ij}$ is a representative of $\rho$, $g_{00}$ not Ricci tensor should be the representative of $\Phi$.
But Ricci tensor is a function of second derivatives of $g_{00}$, which makes me wondering why the field dynamical variable is treated differently in both classical and relativity theories. In classical theory the dynamical variable is the field potential for which its second derivative is not vanishing at $r$ from the source, but in GR, the dynamic variable is the metric of the space time for which its second derivatives in the form of Ricci Tensor vanishes at the same distance from the source!

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PeterDonis
Mentor
2019 Award
$$\Phi = \frac{1}{2} g_{00}$$
Actually, in the weak field approximation, $g_{00} = 1 + 2 \Phi$.

given that $T_{ij}$ is a representative of $\rho$, $g_{00}$ not Ricci tensor should be the representative of $\Phi$.
And it is; $g_{00}$ corresponds to $\Phi$ (with the correction I gave above), and the Einstein tensor corresponds to $\nabla^2 \Phi$.

In classical theory the dynamical variable is the field potential for which its second derivative is not vanishing at $r$ from the source
Yes, it does. The equation you wrote down is $\nabla^2 \Phi = 4 \pi \rho$. But outside the source, $\rho = 0$, so $\nabla^2 \Phi$ is also zero.

Yes, it does. The equation you wrote down is $\nabla^2 \Phi = 4 \pi \rho$. But outside the source, $\rho = 0$, so $\nabla^2 \Phi$ is also zero.
Just a bit of calculus, the equation of classical potential at a distance $r$ from the source is given by,
$$\Phi=\frac{-1}{r}$$
Taking the second derivatives with the respect of $r$ will give $\frac{-2}{r^3}$, why should it be zero?

PeterDonis
Mentor
2019 Award
Taking the second derivatives with the respect of $r$ will give $\frac{-2}{r^3}$, why should it be zero?
I'll respond to this with another question: how can $\nabla^2 \Phi$ be nonzero when $\rho$ is zero, which it is everywhere outside the source? Doesn't that violate Poisson's equation?

(Hint: you might want to check the actual definition of the Laplacian operator $\nabla^2$ in spherical coordinates.)

I'll respond to this with another question: how can $\nabla^2 \Phi$ be nonzero when $\rho$ is zero, which it is everywhere outside the source? Doesn't that violate Poisson's equation?

(Hint: you might want to check the actual definition of the Laplacian operator $\nabla^2$ in spherical coordinates.)
This is cool, yes it should violate it unless $\nabla^2\Phi$ is zero which it is according to the defintion of Lapalacisn operator. Thank you.