Tag Info

New answers tagged

0

Note that the property "black hole" is relative to an observer. A freely falling observer will not notice anything strange like an event horizon etc. Those observers will not have an information paradox. But similar, an observer at rest relative to the black hole will have no information paradox too, since nothing ever falls into it. For a calculation ...


0

If you consider a spaceship as v → c then someone in a stationary frame observing the spaceship will see ∆ t → 0 as well (on the spaceship). The thing to note is that the spaceship is still moving forwards at close to c. Time is only frozen for things moving on the spaceship (as observed by someone in a stationary frame) but the spaceship itself is still ...


0

The thing to consider is not $t$ the 'coordinate time', but $\tau$ the 'proper time' measured by the clock of the in-falling particle. $$ \Delta\tau = \int \sqrt{-ds^2}, $$ where $ds^2$ is the spacetime interval along a time-like path. All in-falling particles take finite proper time to cross the horizon. Further, all in-falling, massive particles will ...


0

From Wikipedia, the free encyclopedia A sonic black hole (sometimes called a dumb hole) is a phenomenon in which phonons (sound perturbations) are unable to escape from a fluid that is flowing more quickly than the local speed of sound. They are called sonic, or acoustic, black holes because these trapped phonons are analogous to light in astrophysical ...


1

Whether you see the incoming background radiation as blueshifted depends on your relative motion compared to the large scale Hubble flow. Even in the absence of a black hole you can accelerate in a direction and see blue shift in the forward direction. And that's fundamentally what is happening in the black hole, to stay at a fixed distance from the black ...


1

No observer will ever observe a true horizon because it takes an infinite time to form. All we real observers will ever see is an apparent horizon, though for typical black holes the difference between the real and apparent horizons is extremely small. The Hawking radiation does depend on the motion of the observer. For example a freely falling observer ...


0

This might not be right, but as I understand it, the particle and anti particle both have mass. The tidal force of the black hole is able to separate the 2 particles, one of them, flying off in to space, the other, flying towards the black hole. So from inside the event horizon you'd see both particles, well, you'd have to look very close cause particles ...


1

Unfortunately, A complete theory of Hawking radiation does not exist at the moment. Hawking radiation is described by a sort of WKB approximation far from the balck hole horizon. In order to describe the phenomenon in the whole space time, in fact we would need a complete theory of Quantum gravity, or at least, a consistent theory (that means with a generic ...


0

Doesn't the Schwarzschild metric combined with Hawking radiation imply that nothing ever gets past the event horizon of a black hole? No. First of all, let's set Hawking radiation aside, we don't need it. Let's look at this: An observer looking from the outside will never see the particle cross the event horizon, even if he/she looked for an arbitrarily ...


0

If, as is canonically accepted, the black hole evaporates within a finite time, then the freely falling particle will be released from its gravitational attraction. By the premise of the question, in the observer frame the particle can not have passed the event horizon during the intervening period. This would also solve the quantum information paradox, as ...


0

Actually I think heuristics are very important and understanding them correctly is critical. I would point point out a Feynman statement that we really don't understand the Pauli Exclusion Principle because we don't have a good heuristic for it. Hell go ask several physicists to describe the proof starting with the Wightman Axioms and they can't do it! In ...



Top 50 recent answers are included