2
$\begingroup$

My first question was going to be:

Does EEP hold at all time scales?

I believe this was successfully answered at Einstein's Principle of Equivalence - Infinite Acceleration/Energy?

The answer is no, but it does not claim to. Einstein's equivalence principle says that gravitational and inertial acceleration are locally indistinguishable from each other. The key being "locally" in space and in time.

Nonetheless, while considering this problem, the question arises: what happens to a craft that is accelerated indefinitely? I recognize that this is somewhat physically unrealistic, but imagine using some extraordinarily bright and continuous stationary photon source to impart momentum to the craft. I assume that this constant input of energy would eventually cause the object to collapse into a black hole! Is that true?

$\endgroup$
1
$\begingroup$

Einstein's equivalence principle says that gravitational and inertial acceleration are locally indistinguishable from each other.

Correct. Constant acceleration is completely indistinguishable from a perfectly uniform gravitational field, but no real gravitational field is perfectly uniform, so there are tidal effects if you make measurements over a large enough distance. When we say "locally indistinguishable" it means you confine your measurements to a chunk of spacetime that's small enough in the given gravitational field so that it looks like flat spacetime, and then the tidal effects are negligible.

what happens to a craft that is accelerated indefinitely

If you apply uniform acceleration to a craft, that is, the acceleration feels like constant gravity to the people in the craft, then its speed gets closer and closer to c without ever reaching it. That may seem weird, but it works that way because of the relativistic formula for velocity addition. For further details, please see the classic Usenet Relativistic Rocket page.

I assume that this constant input of energy would eventually cause the object to collapse into a black hole! Is that true?

No, that's not true. Even if the speed of the craft is a tiny fraction just under lightspeed, so the crew are experiencing a huge time dilation relative to the rest frame they took off from, they won't notice anything unusual, although their view of the universe will change due to length contraction and Doppler shift.

However, if such an ultra-relativistic craft crashed into something, that could create a black hole, but the required kinetic energy is incredibly high. Eg, if you could turn all the energy of a dozen Earths colliding with a dozen antimatter Earths into kinetic energy, and used that energy to slam two billiard balls into each other, it could create a black hole, providing the collision fragments stay close enough to the collision point.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ In the high energy billiard ball collision example, I assume this effect occurs because the particles composing the billiard ball experience such a large impulse by the sudden deceleration of collision, so as to overcome the repulsive forces (and quantum mechanical prohibitions of distance) normally keeping the constituent particles of mass at normal-life space-separation distances, resulting in a configuration of elementary particles at the same inter-particle separation as is seen/predicted for a black hole? $\endgroup$ – Thomas Lee Abshier ND Nov 8 '19 at 13:08
  • 1
    $\begingroup$ @Thomas Not really, although that's part of what's going on. The main thing is that gravity (i.e., spacetime curvature) is proportional to the stress-energy-momentum in a region. If we can concentrate sufficient energy into a small enough region we get a black hole. A black hole with a Schwarzschild radius equal to a billiard ball has a mass about a dozen times that of Earth. Merely giving a billiard ball that much kinetic energy won't create a black hole: the billiard ball is still just a billiard ball in its rest frame. $\endgroup$ – PM 2Ring Nov 8 '19 at 13:32
  • 1
    $\begingroup$ But smashing two such balls gives us enough energy in the centre of mass frame of the two balls. However, we lose roughly half of the kinetic energy we pumped in due to the gravitational waves that are emitted as the black hole forms. $\endgroup$ – PM 2Ring Nov 8 '19 at 13:33
  • 1
    $\begingroup$ In theory, we could make a black hole with no matter at all, just using the energy of light. Obtaining such an intense concentration of light is far beyond our technical capabilities, but it has a name: kugelblitz. $\endgroup$ – PM 2Ring Nov 8 '19 at 13:36
  • $\begingroup$ Thanks, you offer an interesting addition to the concept - that the collision produces a shock wave/gravitational wave, and that roughly half of the energy is lost as gravitational radiation. I assume this amount of energy loss is a prediction from GR, and it would vary slightly with the duration-of-impulse? $\endgroup$ – Thomas Lee Abshier ND Nov 8 '19 at 14:05

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.