Inside Black Holes | Leonard Susskind

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Recommends Yukterez · 2015-02 · Physics Stack Exchange · on “The ever increasing pull of a black hole”

There is no limit in acceleration, but since acceleration is the rate of change of velocity per time, and the time dilatation of the falling particle relative to an outer observer goes to infinity at the event horizon (where the escape velocity would be the speed of light), the falling object never makes it through that border from the viewpoint of an outer observer, because everything freezes at the horizon. From the perspective of the falling particle everything is normal since it is just falling freely and doesn't feel the time dilatation it suffers. For a simple relativistic explanation see Link [LINK] , for further details on the quantum perspective I recommend Link 2 [LINK] and Link 3

Recommends Yukterez · 2022-12 · Physics Stack Exchange · on “Simultaneous events”

@WillO - nevertheless, he should indeed try to draw the diagram, if he doesn't know yet how to do it he can always learn [LINK] it (it isn't that different [this video] to SR anyway)

Recommends Yukterez · 2024-06 · Physics Stack Exchange · on “When do we talk about spaghettification or pancakification in black holes?”

Susskind explains it here [this video] : the pancakification due to gravitational time dilation is only in the reference frame of Bob. In her own frame Alice gets spaghettified, not pancaked, since the local tidal forces are positive all the way. On the left you have an infalling stream of particles with different initial velocities in ingoing Eddington-Finkelstein coordinates where you can see the local spaghettification, and on the right the same particles in the classic Schwarzschild-Droste coordinates where you can see the pancakification in the frame of the stationary coordinate bookkeeper: [LINK]

Cites Yukterez · 2016-02 · Physics Stack Exchange · on “Can we detect gravitational waves generated from inside the event horizon of a black hole?”

Even if all the mass behind the horizon would magically disappear in one moment, you would not notice that from outside the horizon. One could assume that in this case there was no mass left that could bend spacetime anymore and the gravitational field would disappear with c, but on the other side you have to take time dilation into account: From the perspective of the outside observer everything that makes up the black hole is stacked up at the horizon and asymptotically approaches it as time goes to infinity, simply because the factor for the time dilation approaches 0 as an object approaches the horizon. Therefore, what happens inside a black hole at a given proper time of an infalling ob

Cites Yukterez · 2016-03 · Physics Stack Exchange · on “Interior material and spacetime of a black hole described from outside”

This question is mostly about general relativity. Answers regarding the quantum physics behind it are also appreciated, but first I would prefer solutions from a relativistic viewpoint. If you fall into a black hole you reach the singularity at $\text{r}=0$ in a finite proper time. In contrast, for a stationary observer on the outside, the infalling object needs an infinite amount of coordinate time to reach the horizon at $\text{r}=\text{r}_s=2\text{G}\text{M}/ \text{c}^2$, in other words, everything that falls into a black hole gets stuck on the horizon and never crosses it. Relative to a stationary observer outside the black hole the gravitational time dilation on the event horizon $\text

Cites Yukterez · 2016-04 · Physics Stack Exchange · on “Where does the energy of a photon trying to escape a black hole go?”

Since due gravitational time dilation something takes forever to fall inside a black hole from the perspective of an outside observer, there is nothing in it [this video] yet that could come out. The light which is emitted from outside the horizon does reach an outside observer, but since it hasn't yet fallen in, it technically doesn't escape from inside the black hole because it was always on the outside. The energy stays conserved, just the power of the light gets reduced, but therefore the duration of the received signal is longer. Since energy = power x time, there is no energy problem. If you don't throw photons but stones, the sum of kinetic and potential energy also stay conserved: if you

Cites Yukterez · 2016-05 · Physics Stack Exchange · on “Two Black Holes held stationary by EM forces”

If the event horizons overlap you get one big horizon. EM forces can not counteract gravity if the curvature is too large since the force required to counteract gravity becomes infinite at the horizon. You can see this in the equation $$F=\frac{G\cdot M\cdot m}{r^2\cdot\sqrt{1-r_s/r}} $$ which becomes infinite at the horizon $r_s$. Since from the outside perspective everthing that fell into the black hole got stuck on the horizon because of time dilation, the matter that makes up the black hole is layered outside [this video] the horizon, so if both horizons would touch, the two black holes would have to merge. Also see [LINK] : "For a Schwarzschild (non-rotating, uncharged) black hole, the river

Cites Yukterez · 2016-08 · Physics Stack Exchange · on “Time Dilation inside a massive body”

From the perspective of an outside observer there is neither space nor time inside the black hole, since the hole is pinched into the fabric of spacetime (compare Flamm's paraboloid [LINK] to Newtonian space [LINK] ). Nothing ever falls into the black hole since it takes an infinite amount of coordinate time to reach the event horizon, so it can never fall through (see Susskind 1 [this video] and 2 [this video] ). Because of this real black holes do not form in finite coordinate time, so in the system of an outside observer they will always stay collapsars which converge to a black hole, but never really become one because of the time dilation slowing down the process. For the infalling observer who c

Cites ann marie cœur · 2022-01 · Physics Stack Exchange · on “Why does Leonard Susskind draw constant time slices around a black hole as lines passing through the origin at zero?”

In this video Inside Black Holes by Leonard Susskind [this video] , why does he draw the constant time slice as lines passing through the origin at zero? Something seems to be contradicting to have constant time slice meeting at the same point. [LINK] Is that effect due to being around blackhole? How precisely it is?

Cites Matrix001 · 2022-10 · Physics Stack Exchange · on “Why does a photon with wavelength equal to the radius of the blackhole it's falling into have 1 bit of information?”

In this lecture: [this video] Susskind derives the entropy of a black hole by assuming a bit of information could be added by a photon with wavelength equal to that of the black hole's radius. I'm asking for the reasoning behind that.