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I'm trying figure out a way to describe bodies in a motion over a Einsteinian Universe. My question is: how is it possible? Is there some formula to describe the gravitational interaction in terms of curvature of space? Or simply this not exist yet? Thanks

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    $\begingroup$ Objects moving under gravitational force move along geodesics in the curved spacetime. $\endgroup$ – G. Smith Jun 10 at 3:47
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    $\begingroup$ Geodesic equation $\endgroup$ – John Rennie Jun 10 at 3:55
  • $\begingroup$ Einsteinian Universe is a static Universe proposed by Einstein in his paper Kosmologische Betrachtungen zur allgemeinen Relativitdtstheorie. Sitzungsber. preuss. Akad. Wiss., 1917, 1, 142—152, where the cosmological constant was first introduced. The term Einstein manifold is also used, see en.wikipedia.org/wiki/Einstein_manifold. $\endgroup$ – Alex Trounev Jun 10 at 8:00
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Yes there is, although to be particular it is not the curvature of space, but rather the curvature of spacetime.

The equations you are looking to start off with are the einstein field equations, which are $$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R+Λg_{\mu\nu}=\frac{8πG}{c^4}T_{\mu\nu}$$

This is a system of ten or so differential equations packed into one. It’s not exactly trivial to solve most of the time, but you want to solve for $g_{\mu\nu}$, an object known as the metric tensor. The metric tensor is essentially a correction for the pythagorean theorem in curved spaces.

Using the metric you can get the full form of the christoffel symbols, which need to be used in something called the geodesic equation.

The geodesic equation is an equation that when solved will give you the path of the straightest line in spacetime. If you can solve for that straightest path, you’ll know how objects will move in a gravitational field, like in orbits or in radial freefall. In some cases such as the schwarschild spacetime, which is the spacetime often used for non-rotating black holes or planets moving around the sun, you can solve for the path of motion just from the corrected pythagorean theorem using some of the considerations taken from the geodesic equations or from mathematical objects known as killing vectors.

The geodesic equation is this: $$\frac{d^{2}x^{\mu}}{d\lambda^2}+Γ^{\mu}_{\sigma\nu}\frac {dx^{\sigma}}{d\lambda}\frac {dx^{\nu}}{d\lambda} = 0 $$

Where $Γ^{\mu}_{\sigma\nu}$ are the christoffel symbols.

This is the crux of general relativity, that the einstein field equations describe how mass and energy curve spacetime, and how that spacetime tells objects in it how to move, and how objects in that spacetime will follow the straightest paths in it, which creates gravity.

EDIT: Another commenter here has suggested that i explain some of the other terms in the field equations, which i will.

I’ll start off by giving the equation for the christoffel symbols in terms of the metric and it’s derivatives.

The equation is: $$Γ^{\mu}_{\sigma\nu}=\frac{1}{2}g^{\mu\lambda}(g_{\lambda\sigma ,\nu}+g_{\lambda\nu , \sigma}-g_{\sigma\nu , \lambda})$$

The commas represent partial derivatives of the metric terms with respect to the indices following them, and $g^{\mu\lambda}$ is the inverse metric tensor, just like how you can have the inverse matrix of a matrix, which the metric tensor can be represented as.

Now that we have the christoffel symbols we can get something known as the Riemann curvature tensor $R^{\beta}_{\mu\sigma\nu}$ which is a measure of the curvature of our space and is quite useful as in the field equations you will notice a similar term $R_{\mu\nu}$ which is what’s known as a contraction of the riemann tensor.

The equation for the riemann tensor in terms of the christoffel symbols is: $$R^{\beta}_{\mu\sigma\nu}= \Gamma^{\beta}_{\nu\mu , \sigma}- \Gamma^{\beta}_{\sigma\mu , \nu}+\Gamma^{\beta}_{\sigma\lambda}\Gamma^{\lambda}_{\nu\mu}-\Gamma^{\beta}_{\nu\lambda}\Gamma^{\lambda}_{\sigma\mu}$$

The ricci tensor, as a contraction of the riemann tensor is $R^{\beta}_{\mu\beta\nu}=R_{\mu\nu}$

The equation for the ricci tensor is: $$R_{\mu\nu}=R^{\beta}_{\mu\beta\nu}= \Gamma^{\beta}_{\nu\mu , \beta}- \Gamma^{\beta}_{\beta\mu , \nu}+\Gamma^{\beta}_{\beta\lambda}\Gamma^{\lambda}_{\nu\mu}-\Gamma^{\beta}_{\nu\lambda}\Gamma^{\lambda}_{\beta\mu}$$

Now the ricci scalar is like the name suggests, a scalar and is a contraction of the ricci tensor. The equation for that is $$R=g^{\mu\nu}R_{\mu\nu}$$

That’s the equations for all the curvature terms. The last term in the field equations (besides the cosmological constant) is the stress energy tensor, $T_{\mu\nu}$. That just essentially represents the amount of mass-energy at points in spacetime. For cases like the schwarschild spacetime mentioned before, the stress-energy tensor is zero so in general for the simple cases you don’t have to worry about it too much. Not that it isn’t important, but still.

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  • $\begingroup$ Please use latex – replace the unicode symbols for greek letters with \mu, etc. $\endgroup$ – Prof. Legolasov Jun 10 at 4:22
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    $\begingroup$ Also, if you're aiming to give a short overview of the mathematics of GR, I suggest adding the two missing equations – the formula for the Cristoffels in terms of the metric, and the formula for the curvature tensors. Otherwise, nice answer – upvoted :) $\endgroup$ – Prof. Legolasov Jun 10 at 4:27
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    $\begingroup$ @Solenodon My intent was to show the general field equations and explain the old “spacetime tells matter how move” quote, and the general idea that objects follow geodesics, but tbh i got a bit lazy with the other ones. I was just kind of hoping OP would research that themselves. But very well, i’ll add the riemann and ricci tensor in terms of the christoffel symbols, the ricci scalar and the christoffel symbols in terms of the metric. $\endgroup$ – Thatpotatoisaspy Jun 10 at 4:33
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    $\begingroup$ Oh I don't expect you to give the whole overview of course, but just quoting an equation that uses undefined symbols can be kind of cruel to OP, at least in my view. $\endgroup$ – Prof. Legolasov Jun 10 at 4:36
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    $\begingroup$ @StephenG I actually meant Mathjax (I'm aware of the difference), but I said latex anyway... I'm in desperate need of a cup of coffee. $\endgroup$ – Prof. Legolasov Jun 10 at 5:46

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