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The closer I get to a black hole, does time slow down proportionally?

Does time stop near the event horizon?

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    $\begingroup$ From the perspective of a distant observer, your time slows down towards 0 as you approach the event horizon, although time seems normal to you. The exact amount of time dilation depends on your motion, as John explains here: physics.stackexchange.com/a/159359/123208 $\endgroup$ – PM 2Ring May 13 at 14:52
  • $\begingroup$ When you say it depends on my motion what do you mean? $\endgroup$ – Sykhow May 13 at 14:56
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    $\begingroup$ At a given height, you get different time dilation if you're hovering above the BH compared to dropping in free-fall, or in orbit around the BH. $\endgroup$ – PM 2Ring May 13 at 15:02
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This answer is going to be longer than necessary, but I feel you need to grasp a few things about relativity first before I can address your question. I don't know how experienced you are with physics/maths so I shall assume "not very" and will try and explains things conceptually, avoiding equations etc.

First a clarification: the time you experience always occurs at "one second per second". Relativity does not change this. Instead relativity talks about how time appears different to observers if they are moving at different speeds (say one is moving near the speed of light) or if they experience appreciably different gravitational field strengths (important for your question). One thing you should know is that, for light there is no concept of time. This is difficult for me to explain in a way that might seem intuitive to someone used to classical physics, and I could spend a long time trying to explain it, so for now just accept it as fact. However, anything with mass, like you or me, will always experience time. And we will always experience it at one second per second in our own reference frame. In relativity we call this "proper time" because it is the time that is physical for us. However, confusingly, there is also "coordinate time". This is part of time that forms a dimension of spacetime. If we are movingly slowly compared to the speed of light then coordinate time is pretty much equal to proper time. But as we move faster through spacetime the difference between coordinate time and proper time becomes more apparent. This is where the "twin paradox" (not actually a paradox) comes from and if you want to learn more about the difference between proper time and coordinate time I suggest you look into the twin paradox.

So the short answer to your question is that you will continue to experience time in a normal way as you enter a black hole. If the black hole is very large then spacetime isn't even very curved at the event horizon. This is good news for you because curves in spacetime give rise to "tidal forces", which are the same type of forces that occur between the Earth and moon and are responsible for the tides. It is these forces that lead to "spaghettification" where your body will be stretched out due to the difference in gravity experienced by your toes and your head. If the black hole is small then spacetime could be steeply curved around it. This is very bad news for you because the tidal forces may rip you apart before you even reach the event horizon. Assuming you survive, you will continue to experience time in the normal way until you reach the singularity itself. There no one knows what happens. Maybe it's like "Interstellar" and you get to go to your daughters bedroom and have a chat, but in all likelihood you're probably dead from tidal forces by now.

The longer answer is that, although you experience time normally, that is not what someone else who looks at you fall into the black hole will see. Lets say they watch you fall in through a telescope far away from the black hole. In order to see you, light must bounce off you and travel to their eyes. Lets say you are emitting light to send to them. The closer you reach the event horizon, the longer it takes the light to reach them. This is because the path light must take through spacetime is getting more and more curved, hence longer. The light will also appear redder and redder. This is because the light is a wave. Each tiny part of the wave is taking longer to reach you, so the wave "looks" longer. What this means is that the frequency of the light wave is lower. S,o as you move closer to the event horizon the light takes longer to reach someone else, and also has lower energy when it does reach them. Eventually the energy of the light will be so low that it would be very hard to detect.

At the point of the event horizon light cannot escape the black hole. So any light you emit to someone will never leave. In fact, the event horizon is the exact point where light can no longer leave, so if you emitted a light particle on the event horizon it would just stay there. The person observing you fall in will no longer be able to see you, since the light can't reach them. When you hear people talk about a person being "frozen in time" when they pass the event horizon of a black hole this is what they actually mean. The image of you is forever confined to the event horizon. You, however, will continue to experience time just fine, until you reach the singularity or until gravitational forces rip you apart.

Now we'll use some maths. The difference in coordinate time between when you emit two photons and when and observer will observe them is always the same, it is the proper time you experience that is affected by the graviational field. In relativity we use something called the line element. The line element tells us what spacetime geometry looks like, by telling us how distances in spacetime are related. For objects with mass like you and me the distance moved through spacetime is always equal to the negative of the proper time we experience. We write this as $$ds^2=-d\tau^2.$$ Where $\tau$ is proper time and $s$ is distance in spacetime.

For spherically symmetric objects, like a black hole, we can use the Schwarzschild solution. This tells us that the geometry of spacetime around any spherical mass is $$ds^2= -(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2+r^2d\theta^2+sin^2(\theta)d\phi^2,$$ where $G=c=1$ and we are using Schwarzschild coordinates (spherical polar coordinates plus time).

If we assume both you and your observer are standing still then $dr^2$, $d\theta^2$, $d\phi^2$ are all equal to zero. Then $$d\tau^2=(1-2m/r)dt^2.$$ So proper time for you is $$\Delta\tau_A=\sqrt{1-2\frac{m}{r_A}}\Delta t_A,$$ and proper time for your observer is $$\Delta\tau_B=\sqrt{1-2\frac{m}{r_B}}\Delta t_B.$$ The ratio of the two is $$\frac{\Delta\tau_B}{\Delta\tau_A}=\sqrt{\frac{1-2\frac{m}{r_B}}{1-2\frac{m}{r_A}}} \frac{\Delta t_B}{\Delta t_A} = \sqrt{\frac{1-2\frac{m}{r_B}}{1-2\frac{m}{r_A}}} \Leftarrow \Delta t_A=\Delta t_B.$$ We can clearly see that this ratio tends to infinity as $r_A\rightarrow 2m$, which is the event horzion. Thus the proper time experienced for $B$ between each photon arriving will tend to infinity.

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