1
$\begingroup$

The total energy $E$ of an object in free fall from rest at an infinite radius is known to be

$$ E = m c^2 $$

As the object falls, its energy doesn't change since the object is merely moving freely through curved spacetime.

For a distant observer, is $E$ the sum of the potential energy $U$ and the kinetic energy $E_k$ such that

$$ U = \left(1 - \frac{r_0}{r}\right) mc^2 $$

and

$$ E_k = \left(\frac{r_0}{r}\right) m c^2 $$

?

$\endgroup$
  • $\begingroup$ From the source: Saying that the universe undergoes energy inflation and that something that looks redshifted actually has constant energy is a more complicated and convoluted way of looking at things. Why are photons that aren't near a massive source of gravitation also redshifted since their energy doesn't change but our global energy scale does? $\endgroup$ – Connor Dolan Mar 15 '18 at 12:14
  • $\begingroup$ As measured by whom? $\endgroup$ – Rob Jeffries Mar 17 '18 at 7:40
  • $\begingroup$ Yes, the energy doesn't change. If you drop a 1kg brick into a black hole, the black hole mass increases by 1kg. $\endgroup$ – John Duffield Mar 18 '18 at 11:02
2
$\begingroup$

I'll assume that the energy is measured by a stationary observer ($dr=0$).

Taking the spherically symmetric Schwarzschild metric as an example and radial infall from infinity at rest, then we know that. $$\left(1 - \frac{r_s}{r}\right) \frac{dt}{d\tau} = 1$$ So this expression gives the time dilation inferred by an observer at infinity for an infalling object, which I assume is what you mean by the clock rate - the proper time interval divided by the coordinate time interval. $$ d\tau = dt (1-r_s/r)$$

This leads to expressions for $dr/dt$, a velocity in coordinate space time $$\frac{dr}{dt}=-\left(1 - \frac{r}{r_s}\right)\left(\frac{r_s}{r}\right)^{1/2}$$ and a velocity according to a stationary observer at $r$ $$v = -\left(\frac{r_s}{r}\right)^{1/2}$$ Where, in each case, $r_s$ is the Schwarzschild radius and $c=1$ units are used.

The kinetic energy measured by the stationary observer is $(\gamma -1)mc^2$ and thus $$ E_K = \left(\frac{1}{(1 - r_s/r)^{1/2}} -1\right)mc^2 $$

According to coordinate space time, the equivalent kinetic energy expression would be $$E_K = \left(\frac{1}{[1 - (1 - r_s/r)^2 (r_s/r)]^{1/2}}-1\right ) mc^2$$ which also doesn't simplify to what you suggest, since you see that this approaches zero at the Schwarzschild radius.

I think that in GR, this simple separation between potential and kinetic energy is not possible.

$\endgroup$
  • $\begingroup$ IMHO it's possible Rob. But it isn't what you find in your textbook. See the 2013 AMPS paper an apologia for firewalls. Tucked away in the conclusion on page 27 is footnote 31, containing a reference 87 to Friedwardt Winterberg’s 2001 paper gamma ray bursters and Lorentzian relativity. The point to note is that falling bodies do not slow down. $\endgroup$ – John Duffield Mar 18 '18 at 11:01

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.