Timeline for Einstein's equivalence principle (EEP) and indefinite acceleration
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Nov 21, 2019 at 20:22 | vote | accept | burgerking | ||
| Nov 19, 2019 at 9:09 | comment | added | PM 2Ring | @burgerking OTOH, I've been assuming that the rocket is accelerating in a perfect vacuum, but even deep intergalactic space contains some atoms (and photons). Eventually, the ultra-relativistic rocket will be going so fast that the momentum of the collisions with those atoms & photons will become significant, and that friction will make it very difficult for the rocket to continue accelerating, no matter what sort of propulsion mechanism we use. | |
| Nov 19, 2019 at 8:58 | comment | added | PM 2Ring | @burgerking No, it isn't possible to create a black hole by just accelerating an object for a very long time. Those relatively small collisions in the rocket engine don't have a lot of energy or momentum in the centre of mass frame of the rocket + exhaust. | |
| Nov 19, 2019 at 8:47 | comment | added | burgerking | @PM 2Ring. Thank you for your answers. I'm not sure if I am making myself clear or if I fully understanding what you are saying. Based on your last comment, it seems like it is indeed possible to create a black hole by accelerating an object for a very long time. After all, the acceleration itself is nothing but a series of many many collisions that impart energy. And in some frame, that looks like a catastrophic, nearly instantaneous event. | |
| Nov 18, 2019 at 7:57 | comment | added | PM 2Ring | @burgerking The rate of acceleration involved in the collision isn't really relevant. The ship & the rock have different velocities in different frames, but all observers will calculate the same energy of impact involved in the collision. If the energy density in a region of space (strictly speaking, the stress-energy-momentum) is greater than or equal to the Schwarzschild limit, then a black hole forms. | |
| Nov 18, 2019 at 7:22 | comment | added | burgerking | But a collision is just a fast acceleration into another frame. My question is why that differs from an acceleration spread out in time. I'm not a relativity expert, but I imagine there is a frame where the collision looks like a billion years of slow acceleration. | |
| Nov 17, 2019 at 14:18 | comment | added | PM 2Ring | @burgerking Maybe I'm misunderstanding your question, but it doesn't matter how quickly the ship and the rock acquire their speeds. All that matters is what their relative speed is when they collide. | |
| Nov 17, 2019 at 13:07 | comment | added | burgerking | Thank you for the discussion and link, I am still confused. Let us imagine a rock in frame A. In frame B, a spacecraft, having been accelerated for a billion years, is travelling at relative speed very close to light speed. The spacecraft collides with the rock, and as discussed (hypothetically at least) this could create a black hole. In this picture, the spacecraft is rapidly accelerated from one frame into another. How is this fundamentally different from a spacecraft that is slowly accelerated from one frame to another over a billion years of acceleration at a few g? | |
| Nov 10, 2019 at 5:41 | comment | added | PM 2Ring | @burgerking You may find this article by John Baez on torsors helpful. | |
| Nov 10, 2019 at 5:39 | comment | added | PM 2Ring | @burgerking I agree that it can seem hard to reconcile. The kinetic energy of a body doesn't mean much when the body is isolated, since any body is traveling at 0.999999c in some frame. If a body is a black hole in some frame, it must be a black hole in every frame, you can't create (or destroy) a BH by mere mathematical manipulation. So kinetic energy is relevant when you have 2 or more objects with different speeds, then you can talk about the kinetic energy of one body relative to another, or the KE of each body relative to the centre of mass of the collection of bodies. | |
| Nov 10, 2019 at 5:24 | comment | added | PM 2Ring | @Thomas Yes, I don't know the fine details (such calculations are beyond my skills), I got that figure of 50% in a related conversation some time ago. The basic idea is that any changes to the gravitational field corresponding to the quadrupole moment or higher moments must emit gravitational waves. Forming a black hole by smashing 2 bodies together causes very intense changes to spacetime curvature, so a lot of energy is dissipated as gravitational waves. | |
| Nov 8, 2019 at 15:42 | comment | added | burgerking | "If we can concentrate sufficient energy into a small enough region we get a black hole." Isn't that exactly what we are doing when we accelerate something indefinitely. I agree with your point that the billiard ball is still just a billiard ball in its own rest frame. What I find confusing is that in MY rest frame, eventually its relativistic mass will exceed that required to form a black hole. How can I reconcile these two very different outcomes? | |
| Nov 8, 2019 at 14:05 | comment | added | Thomas Abshier | 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? | |
| Nov 8, 2019 at 13:36 | comment | added | PM 2Ring | 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. | |
| Nov 8, 2019 at 13:33 | comment | added | PM 2Ring | 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. | |
| Nov 8, 2019 at 13:32 | comment | added | PM 2Ring | @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. | |
| Nov 8, 2019 at 13:08 | comment | added | Thomas Abshier | 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? | |
| Nov 8, 2019 at 12:43 | history | edited | PM 2Ring | CC BY-SA 4.0 |
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| Nov 8, 2019 at 11:11 | history | answered | PM 2Ring | CC BY-SA 4.0 |