1
$\begingroup$

I came across statements such as "the acceleration observed in a weak gravitational field is mainly due to curvature in the time coordinate. " I want to know how we can explicitly find the curvature in a particular coordinate and what would it physically signify. I.e. I want to know the relation between the curvature in a particular coordinate and the Riemann tensor, (Since the Ricci scalar is a scalar so it should not depend on choice of coordinates which leaves the Riemann tensor.)

PS. There are other questions asking about acceleration and curvature of t coordinate but I want to know the mathematical relation between the curvature in one coordinate and the Riemann tensor. A mathematical answer with some physical intuition would be greatly appreciated.

$\endgroup$
2
  • 2
    $\begingroup$ I'm not sure what the question here is. The Riemann tensor is the curvature, and if you know what a tensor is you also know what its coordinate expressions are. $\endgroup$ – ACuriousMind Jun 18 '16 at 9:08
  • $\begingroup$ Here is brief explanation of what the Riemann tensor means. You cannot find "the curvature in a particular coordinate", that doesn't make any sense (i.e. the Riemann curvature of a 1D space is always zero). I get the feeling you're imagining one dimension "bending" but (1) that's extrinsic curvature, not intrinsic, and (2) curvature is much more complicated than that. $\endgroup$ – knzhou Jun 19 '16 at 4:30
1
$\begingroup$

The geodesic equation is given by, \begin{equation} \frac{\mathrm d^2}{\mathrm d\tau^2}x^{\mu}+\Gamma^{\mu}_{\lambda\sigma}\frac{\mathrm dx^{\lambda}}{\mathrm d\tau}\frac{\mathrm dx^{\sigma}}{\mathrm d\tau}=0 \end{equation} which is a set of 4 equations for $x^{i}$.

$\Gamma^{i}_{jk}$ tells us about the curvature of the space time which can be written in terms of derivatives of the metric: \begin{equation} \Gamma^{\mu}_{\alpha\beta}=\frac{1}{2}g^{\mu\nu}(\partial_{\alpha}g_{\gamma\beta}+\partial_{\beta}g_{\gamma\alpha}-\partial_{\gamma}g_{\alpha\beta}) \end{equation} The question is can we get the standard gravitational law from this, For this let's take the following conditions: $c=1$ and $v^i\ll 1$ which implies that $\mathrm dx^i\ll \mathrm dt$. Now the expansion of the second term in the geodesic equation yields,

$$ \Gamma^{\mu}_{\lambda\sigma}\frac{\mathrm dx^{\lambda}}{\mathrm d\tau}\frac{\mathrm dx^{\sigma}}{\mathrm d\tau}~\approx~ \Gamma^{\mu}_{00}\left(\frac{\mathrm dt}{\mathrm d\tau}\right)^2 + 0+\ldots$$ as $\mathrm dx\ll \mathrm dt$. and also $\Gamma^{\mu}_{00}=\frac{1}{2}g^{\mu\sigma}(\partial_0g_{00}+\partial_0 g_{\sigma 0}-\partial_{\sigma}g_{00})$. Now $\partial_0$ acting on everything will be 0, thus giving;

$$ \frac{\mathrm d^2}{\mathrm d\tau^2}x^i~=~\frac{1}{2}g^{i\sigma}\partial_{\sigma}g_{00}\left(\frac{\mathrm dt}{\mathrm d\tau}\right)^2\;.$$

Now we can take, $g_{00}=\eta_{00}+h_{00}=-1-2\Phi$, where $\Phi$ is the usual gravitational potential. After this we have to put in some approximations regarding the potential strength and the usual Newtonian gravity appears.

Coming to the other part of the question as to how the Riemann tensor depends on the metric, lets look at the following.

we have the following relation: \begin{equation} R^{\sigma}_{\lambda\nu\mu}=\partial_{\mu}\Gamma^{\sigma}_{\lambda\nu}-\partial_{\nu}\Gamma^{\sigma}_{\lambda\mu}+\Gamma^{\sigma}_{\mu\rho}\Gamma^{\rho}_{\lambda\nu}-\Gamma^{\sigma}_{\nu\rho}\Gamma^{\rho}_{\lambda\mu}. \end{equation}

Now we go to a frame of reference in which the christoffel connection vanishes, i.e., the first derivative of the metric is 0, in which case the Riemann tensor becomes: \begin{equation} R^{\sigma}_{\lambda\nu\mu}=g^{\sigma\rho}(\partial_{\mu}\partial_{\lambda}g_{\rho\nu}-\partial_{\mu}\partial_{\rho}g_{\nu\lambda}-\partial_{\nu}\partial_{\lambda}g_{\rho\mu}+\partial_{\nu}\partial_{\rho}g_{\mu\lambda}) \end{equation} This gives the relation between the metric and the Riemann tensor.

$\endgroup$
1
  • $\begingroup$ You don't have to use \begin{equation} \end{equation}; just use $$ $$ - it would render the same result here. $\endgroup$ – user36790 Jun 18 '16 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.