3
$\begingroup$

According to Wikipedia, this is the formula for the Bekenstein bound $$ S \le \frac{2\pi k RE}{\hbar c}$$ where $E$ is the total mass-energy. That seems to imply that the presence of a black hole in the region is dependent on an observer's frame of reference. Yet, my understanding is that the Bekenstein bound is the maximum entropy that any area can withstand before collapsing into a black hole.

Does this mean that the existence of black holes is observer dependent? Or that even if an observer does not report a black hole in their frame, one is guaranteed to form there in the future?

$\endgroup$

1 Answer 1

3
$\begingroup$

Individually, $E$ and $R$ are both frame-dependent, but at least in inertial reference frames, the product $ER$ is actually invariant.

Suppose we have a system with rest energy $E_0$ and proper length $R_0$. When we boost to a frame with Lorentz factor $\gamma$, in that frame the system has $E=\gamma E_0$ and $R=\frac{R_0}{\gamma}$, so $ER=E_0R_0$.

I don't know how to prove this in the general case, but the above should demonstrate the possibility that a similar argument might hold (i.e. the energy and the characteristic length of a system transform in opposing ways).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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