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I have read this:

What is the general relativity explanation for why objects at the center of the Earth are weightless?

And this:

Gravitational time dilation at the earth's center

And this:

http://en.wikipedia.org/wiki/Riemann_curvature_tensor

Where it says:

" Converting to the [[tensor index notation]], the Riemann curvature tensor is given by: $$R^\rho{}_{\sigma\mu\nu} = dx^\rho(R(\partial_{\mu},\partial_{\nu})\partial_{\sigma})$$ where
$$\partial_{\mu} = \partial/\partial x^{\mu}$$ are the coordinate vector fields. The above expression can be written using [[Christoffel symbols]]: $$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma} - \partial_\nu\Gamma^\rho{}_{\mu\sigma} + \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma} - \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$$ "

And this:

Time dilation without curvature at the center of mass, how is that possible?

Where John Rennie says in the comments:

"a stationary particle is weightless if all the Christoffel symbols ΓαttΓαtt are zero. However this does not mean the Riemann tensor has to be zero as well. It's possible for the Riemann tensor to be non-zero, i.e. there is curvature, and the Christoffel symbols to be zero. This is what happens at the centre of the Earth."

" the curvature is described by the Riemann tensor not by the Christoffel symbols (though they can be derived from each other). The Riemann tensor is not zero at the centre of the sphere."

What I do not understand is, he says

"the Riemann tensor can be non-zero, even if the Christoffel symbols are all zero."

But this seems to me to contradict the formula:

"$$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma} - \partial_\nu\Gamma^\rho{}_{\mu\sigma} + \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma} - \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$$ "

Because if all the Christoffel symbols are zero, then the Riemann tensor seems to be zero here.

The curvature if I understand correctly is just calculated with the formula of the Riemann tensor as I understand.

Question:

  1. Can somebody please help me understand the math formula of the curvature (which is just the Riemann tensor as I understand) at the center, so that I see it is not zero when r=0?
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    $\begingroup$ I think the simple answer is: $\Gamma$ is zero at that point, but not at points nearby, so the derivatives do not vanish. $\endgroup$ – user121664 Dec 15 '16 at 21:50
  • $\begingroup$ OK , but what is the formula like at r=0? and how can I see it is not 0 there? $\endgroup$ – Árpád Szendrei Dec 15 '16 at 22:05
  • $\begingroup$ Hi @ÁrpádSzendrei if you don't mind, I'd like to give you a piece of advice: next time, try to keep your post more focused, and remove all the unnecessary stuff. You should not begin every post with "I've read this and that". Just ask the question, and at the beginning of the post if possible. One or two paragraphs at most. Remove the superfluous quotes. I think this will lead to more upvotes and more answers for you :-) Cheers $\endgroup$ – AccidentalFourierTransform Dec 15 '16 at 22:50
  • $\begingroup$ This is, in the end, simply an exercise in doing elementary GR computations: Take the interior Schwarzschild metric and compute the Christoffel symbols as well as the Riemann tensor. It's tedious, but I'm not sure what the conceptual question is. $\endgroup$ – ACuriousMind Dec 15 '16 at 23:05
  • $\begingroup$ Yes I understand, thank you. But the question is more simple then that. I really looked at this link en.wikipedia.org/wiki/Riemann_curvature_tensor. And I somehow cannot 'find' the radius itself in the Riemann tensor. I simply don't know where to use r=0 since I do not see the radius in the metric. $\endgroup$ – Árpád Szendrei Dec 16 '16 at 2:17
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Lets just calculate all the Riemann tensors inside a spherical symmetric body using the metric: \begin{equation} ds^2=g_{\mu\nu}dx^\mu dx^\nu=-\mathrm{e}^{\nu(r)}dt^2+\mathrm{e}^{\lambda(r)}dr^2\tag{1}+r^2(d\theta^2+\sin^2(\theta)d\phi^2). \end{equation} The non-vanishing components of $R_{\alpha{}\beta\gamma\delta}$ are:

\begin{align} R_{trtr}&=\frac{1}{4} e^{\nu (r)} \left(-\lambda '(r) \nu '(r)+2 \nu ''(r)+\nu '(r)^2\right),\\ R_{t\theta t\theta}&=\frac{1}{2} r e^{\nu (r)-\lambda (r)} \nu '(r),\\ R_{t\phi t\phi}&=\frac{1}{2}e^{\nu (r)-\lambda (r)}r\sin ^2(\theta )\nu '(r),\\ R_{r\theta r\theta}&=\frac{1}{2} r \lambda '(r),\\ R_{r\phi r\phi}&=\frac{1}{2} r \sin ^2(\theta ) \lambda '(r),\\ R_{\theta\phi\theta\phi}&=r^2 \sin ^2(\theta ) e^{-\lambda (r)} \left(e^{\lambda (r)}-1\right), \end{align} and the ones related to those by the symmetries of $R_{\alpha{}\beta\gamma\delta}$. To get to $R^\alpha{}{}_{\beta\gamma\delta}$ one has to raise with $g^{\alpha\alpha}$. The last five are zero at $r=0$ for non-singular metric potentials $\nu$ and $\lambda$. So the only remaining interesting one is $R_{trtr}$ or if you whish \begin{equation} R^t{}_{rtr}=g^{tt}R_{trtr}=\frac{1}{4} \left(-\lambda '(r) \nu '(r)+2 \nu ''(r)+\nu '(r)^2\right)\tag{2}. \end{equation} This one only depends on derivatives of the metric potentials. To get those one would need to solve the field equations for a specific source term. If we assume an ideal fluid we are in the setting of the Tolman–Oppenheimer–Volkoff equations or if we assume a constant density sphere we recover the interior Schwarzschild solution. I will not go into detail but one can expand both systems around $r=0$ to get series expansions of the metric potentials and their derivatives. For $R^t{}_{rtr}$ follows in both cases \begin{equation} R^t{}_{rtr}\rightarrow \frac{4 \pi }{3}\left(3P_c+\rho_c\right)+16 \pi^2\left(P_c^2-\frac{\rho_c^2}{9}\right)r^2 +\mathrm{O}(r^4),\tag{3} \end{equation} and therefore \begin{equation} R^t{}_{rtr}(r=0)= \frac{4 \pi }{3}\left(3P_c+\rho_c\right).\tag{4} \end{equation} $P_c$ is the central pressure and $\rho_c$ the central internal energy density. In case of the interior Schwarzschild solution one could relate both quantities to stellar mass and radius.

So yes in short: some components of the Riemann tensor do not vanish because the derivatives of some metric potentials do not vanish at $r=0$ but in fact the metric potential $\nu(r)$ itself does not vanish at the center. $\lambda(0)=0\Leftrightarrow \mathrm{e}^{\lambda(0)}=1$ but $\nu(0)$ is finite both for the TOV and internal Schwarzschild solution. So $R_{trtr}$ is unequal zero too.

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Yes, I sympathize. You are reading an article in wiki about the Riemann tensor that is fairly mathematical, with just no examples. Not a good way to learn.

Look up the Schwarzschild solution in wiki, it might have the Riemann and/or the Ricci tensor components calculated, and you'll see there the coordinate r or R for the radial distance. Still, be careful, you'll need to find it for the internal Schwarzschild solution, the external one will give you a singularity at 0, the angularity of the black hole. I am not sure I've seen it in Wikipedia. You migh find it elsewhere if you google it, or you may need to calculate it.

The metric, Christophel symbols and Reimman tensor components, or some of them, depend on the radial distance r, so you can set r=0 after you have the expressions.

Although the weight is zero, just an epsilon away it is not, and that difference between accelerations at two close points is the tidal force, and the derivative of the Christophel symbols (or some of them) and why all the Riemann tensor components are not zero

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The notation:

$$ \partial_\mu\Gamma^\rho{}_{\nu\sigma} $$

means:

$$ \frac{\partial}{\partial x^\mu}\left(\Gamma^\rho{}_{\nu\sigma}\right) $$

so we are taking the derivative of $\Gamma^\rho{}_{\nu\sigma}$ with respect to the coordinates $t$, $r$, $\theta$ and $\phi$. The fact a function has the value zero at some point does not necessarily mean its derivative is zero at that point, and indeed this is the case at the centre of a sphere. That's why the Riemann tensor can be non-zero at $r=0$.

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Sorry that this will appear as an answer and not a comment but I don't have enough rep. Be careful in setting $r=0$ in the Christoffel symbols and some of them are equal to $1/r$ which is clearly undefined and doesn't make any sense. With the Scwarzschild solution you can get a rough rough rough idea of curvature in this space time by contracting the Riemann Tensor twice to determine the Ricci Scalar. You will see as $r$ tends to infinity the space time curvature will tend to zero.

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  • $\begingroup$ some of them are equal to $1/r$ - this of course a result of using polar coordinates and it's true even in flat spacetime (or indeed in Euclidean space). $\endgroup$ – John Rennie Dec 16 '16 at 7:36
  • $\begingroup$ I should have mentioned I was considering spherical polar coordinates $\endgroup$ – Rumplestillskin Dec 16 '16 at 7:45

protected by Qmechanic Dec 16 '16 at 8:56

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