0
$\begingroup$

If I understand correctly the equation for kinetic energy in relativity is $$ E_k= mc^2 \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right) \;,$$ and the equation for escape velocity in General Relativity is the same as in Newtonian Physics so $v_e=\sqrt{\frac{2GM}{r}}$ and when something is moving at escape velocity the kinetic energy must be equal to the opposite of the Gravitational Potential Energy, so $U=-E_k$. If I I substitute in $\sqrt{\frac{2GM}{r}}$ for the escape velocity, I get $$U=-mc^2\left(\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}-1\right)\;.$$ So is this the correct equation for Gravitational Potential Energy in General Relativity?

$\endgroup$
2
$\begingroup$

This has some correct elements, but there are flaws in your reasoning. You can't get correct results in GR by plugging relativistic correction factors into Newtonian equations.

So is this the correct equation for Gravitational Potential Energy in General Relativity?

There isn't really "the" equation for gravitational energy in GR. It's not an instantaneous action at a distance theory like Newtonian gravity. What you've cooked up resembles the square root of the time-time component of the metric in the Schwarzschild spacetime. The metric is what plays the role of a potential in GR, although we can define a scalar potential in a static spacetime. The earth's field, for example, is not static, because the earth is rotating.

One way to see that your equation can't really be valid in GR is that it predicts $U=0$ when $m=0$, but we know that light rays do interact gravitationally.

the equation for escape velocity in General Relativity is the same as in Newtonian Physics so $v_e=\sqrt{\frac{2GM}{r}}$

This logic doesn't make sense. The derivation of this equation uses $K=(1/2)mv^2$, which is false in relativity.

$\endgroup$
1
$\begingroup$

In GR, in general, the definition of a gravitational potential energy is meaningless. GR is a geometric theory, so you have to find the analogy with the concept of "potential" in a geometric object. It turns out that the geometrical equivalent of the gravitational potential is the metric tensor.

The metric $g$ is a symmetric tensor of second rank which basically tells you how to "measure" space-time interval. Here the link if you want to deepen the subject: https://en.wikipedia.org/wiki/Metric_tensor.

In GR particles move along specific curves called geodesics, the equation of geodesics is: $$\ddot x^\mu = -\Gamma_{\sigma\nu}^\mu\dot x^\sigma \dot x^\nu$$ where the gamma factors are called Christoffel symbols, here the link: https://en.wikipedia.org/wiki/Christoffel_symbols.

You see that there is an analogy of equation of motion in newtonian mechanics where $$m\ddot{\textbf{x}}=\textbf{F}$$ and thus you can think of Christoffel symbols as something regarding gravitational force.

Since for the Levi-Civita connection Christoffel symbols are completely defined by the metric and, in particular, by the derivatives of the metric in this way $$\Gamma^\mu_{\sigma\nu}=\frac{1}{2}g^{\mu\tau}(\partial_{\sigma}{ g_{\nu\tau}} +\partial_{\nu}{ g_{\sigma\tau}} - \partial_{\tau}{ g_{\sigma\nu}})$$ then you can think of the metric as the gravitational potential (the force is the derivative of the potential).

This is just an analogy and of course it has nothing to do with rigorous definitions (except for the geometric part)...

$\endgroup$

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.