2
$\begingroup$

When you pick Riemann normal coordinates at a point, then the Christoffel symbols vanish and the metric is flat, but the Riemann curvature tensor does not necessarily vanish. Since Einstein said that locally the laws of special relativity hold, how does that reconcile with the Riemann curvature tensor not being zero (which means that spacetime doesn't appear to be flat in that local neighborhood)?

$\endgroup$
5
$\begingroup$

Click here for reference.

At a specific space-time point , you could always find a inertial frame, such that the metrics is Minkowski at this point (say at $x=0$ in this inertial frame), but this does not mean, that the metrics is Minkowski for $x \neq 0$, in fact, you may made a Taylor series around $x=0$ and you find :

$g_{\mu\nu}(x) = n_{\mu\nu} + \frac{1}{3} R_{\mu \alpha\nu\beta} x^\alpha x^\beta + O(x^3)$

$\Gamma_{\alpha \beta}^\mu(x) = -\frac{1}{3} (R^\mu_{\alpha\beta\nu} + R^\mu_{\beta\alpha\nu})x^\nu + O(x^2)$

And finally, the curvature tensor, being a tensor, cannot vanish completely in the new (inertial) frame, by the laws of transformations of tensors.

So, if you made an experiment, and a experiment need always some interval of space, and some interval or time, so, in some sense, no experiment is stricly "locally", you will find differences with a flat spacetime.

$\endgroup$

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.