# Observeving Black Hole and Time Dilation Question

• B
Hello all,

Thank you for taking time to read my question.

I am curious, what happens if following criteria is met;

1. Place a indestructable stopwatch inside the black hole set it to 5 second
2. Place another stop watch outside of black hole and set it to 5 second
3. Place a indestructable metal bar (does not flex, bend) that goes across both in and outside of black hole with a button pusher to activate both stop watch simultaneously.

Question,

1. Will the timer end at the same time? If not, why?
2. If light speed is not involved in the equation, does the time continues normally?

From what I understood, time slows down to compensate for light, because light speed cannot change, so if the light were to travel the same distance when affected by gravity, time has to slow down to compensate. But what if we just take light speed out of question and place a mechanical switch that pushs both timer's button at the same time from point A to point B. Will the timer still be affected by the time dilation?

Thanks for taking time to read my noob question.

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If you mean by 'inside the black hole' any object that passed the event horizon then it is impossible for those objects to remain stationary.

tionis
Gold Member
3. Place a indestructable metal bar (does not flex, bend) that goes across both in and outside of black hole with a button pusher to activate both stop watch simultaneously.
Hi, Seiten. I think special relativity forbids perfect, rigid objects because they violate causality. In other words, the signal would travel faster than light, I think. So, no.

As MeJennifer has written, spacetime geometry is rather different past the event horizon. Also we can't tell what is going on inside the horizon using general relativity, loosely put. There is not a mean of communication from inside to outside the event horizon.

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If you mean by 'inside the black hole' any object that passed the event horizon then it is impossible for those objects to remain stationary.

Hi MeJennifer, Thanks for replying, my main question is, if there is a way to activate both timer in/out of black hole simultaneously, will they reach 0 at the same time.

Hi, Seiten. I think special relativity forbids perfect, rigid objects because they violate causality. In other words, the signal would travel faster than light, I think. So, no.
Hi Tionis, thanks, so basically, a perfect rigid object were to exist it will be traveling both from past into future in order to push the two buttons simultaneously because it violates causality (aka time travel).

As MeJennifer has written, spacetime geometry is rather different past the event horizon. Also we can't tell what is going on inside the horizon using general relativity, loosely put. The is not a mean of communication from inside to outside the event horizon.

Mr-R, Thanks

I see, I am pretty sure it is unrealistic.

My main question really boils down to why do 1 second inside black hole would mean a lot more time outside. Does time literally slow down? Or is just perceived to be slowed.

If count down from 5 second starts simultaneously both inside and outside of black hole, shouldn't they reach 0 at the same time?

Or, make it a lot more realistic,

If count down from Earth and Moon starts at the same time, shouldn't they reach 0 at the same time as well?

My main question really boils down to why do 1 second inside black hole would mean a lot more time outside. Does time literally slow down? Or is just perceived to be slowed.
Doesn't have to be inside a black hole for that effect to happen. Lets say one is close to the black hole and the other is far away. "Really or perceived" doesn't really make sense. According to relativity it depends on the observer. Type time dilation on Wikipedia.

If count down from 5 second starts simultaneously both inside and outside of black hole, shouldn't they reach 0 at the same time?
See above.

If count down from Earth and Moon starts at the same time, shouldn't they reach 0 at the same time as well?
Nope. Depends on the observer. The person on the Moon would count something and the person on Earth would count something else.
Just read a bit on Special relativity mathematically and then have a look at gravitational time dilation as well. Should give you an idea about how different frame references work.

If count down from Earth and Moon starts at the same time, shouldn't they reach 0 at the same time as well?
The concept of "the same time" is a tricky thing in curved spacetime.

Time dilation is a real, physical effect with real, physical consequences - it is not just some trick of perception. With that being said, it is best to regard time dilation as a relationship between observers, rather than something that "happens" to a clock; as such, the amount of time dilation will depend on who measures it.

Will the timer end at the same time?

The notion of simultaneity is ill defined across regions of curved space-time; the notion of time, in a sense, is a purely local phenomenon that may not be shared between observers.

Mr-R
Ibix
From what I understood, time slows down to compensate for light, because light speed cannot change, so if the light were to travel the same distance when affected by gravity, time has to slow down to compensate. But what if we just take light speed out of question and place a mechanical switch that pushs both timer's button at the same time from point A to point B.
The rod is held together by electromagnetic forces, which propagate at lightspeed. So the push will travel along the rod slower than light - in fact at the speed of sound in the material which will be a lot slower.

It's not that things change to keep lightspeed constant, it's that the way the universe is set up defines a finite invariant speed that nothing can exceed. Light travels at that speed, but has nothing directly to do with why it's invariant.

tionis
Gold Member
Hi Tionis, thanks, so basically, a perfect rigid object were to exist it will be traveling both from past into future in order to push the two buttons simultaneously because it violates causality (aka time travel).

Yes. The rigid rod would behave like a tachyon.

Orodruin
Staff Emeritus
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Hi MeJennifer, Thanks for replying, my main question is, if there is a way to activate both timer in/out of black hole simultaneously, will they reach 0 at the same time.
As others have said, "simultaneous" is a very tricky concept already in special relativity as it is frame (and convention) dependent. This gets even worse in GR. There is not even a proper way of defining your observer inside the event horizon to be stationary with respect to the outside observer. Furthermore, the "radius" you are thinking of no longer has the meaning you think it has. Unfortunately, those concepts will be difficult to grasp at B-level.

If count down from 5 second starts simultaneously both inside and outside of black hole, shouldn't they reach 0 at the same time?
Again, the "same time" loses its meaning in GR and is tricky already in SR. This renders this question moot.

PeterDonis
Mentor
what happens if following criteria is met

They can't be met; it's physically impossible. Not only is it impossible to have an infinitely rigid bar, as others have pointed out; and not only is there no well-defined notion of simultaneity in this scenario, as others have pointed out; it is also impossible to place an object inside the hole's horizon so that it stays in the same place. Nothing can stay in the same place inside the horizon.

time slows down to compensate for light, because light speed cannot change

This is a (heuristic) way of understanding how time dilation due to relative motion arises in special relativity. However, it is a very limited heuristic, and doesn't work in the presence of gravity. There is a different concept of "time dilation" in the presence of gravity, but it only works outside the hole's horizon; inside the horizon this concept is not well-defined.

Orodruin
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Nothing can stay in the same place inside the horizon.
If we are in that mood.
Not only that, it is not well defined what "the same place" would mean inside the horizon.

PeterDonis
Mentor
Not only that, it is not well defined what "the same place" would mean inside the horizon.

Yes, agreed.