Question

I read somewhere online that the formula for time dilatation for a static object at a distance \(R\) away from the center of a planet of mass \(M\) under the theory of General Relativity is given as follows.\[\Delta t =\frac{\Delta t_0}{\sqrt{1 - \dfrac{2GM}{Rc^2}}}\]I noticed that this is equivalent to the time dilitation formula from the theory of Special Relativity with the value \(v=v_e = \sqrt{\frac{2GM}{R}}\) substituted in, which is simply the escape velocity.\[\Delta t = \frac{\Delta t_0}{\sqrt{1 - \dfrac{\color{red}{v_e}^2}{c^2}}}\]Any explanations or proofs of this?

Answer 1

good observation.

Answer 2

the first equation is a just the velocity gained at the radius, they are essential the same it is just the v one in for any velocity and the g one is for the velocity gained by gravity.

Answer 3

Yes, I know that, but the time dilatation is for a particle static at this point. This is the time slowing down due to gravity, not relative speed.

Answer 4

I guess time dilates due to energy. from gravitational potential or kinetic energy

Answer 5

After some thinking, what do you guys think of this reasoning? @Jemurray3, opinions?
1. Lets have a particle fall from infinity to a distance \(R\) from the center of the planet. Before it starts falling, it "looks" at the planet, and notices that they do not have any relative velocity, so it is then to assume by Special Relativity that there is no time dilatation present between the two objects. Now, the object closes its eyes, and begins it approach towards the planet. During this fall, it senses no acceleration, as locally all the laws of physics work exactly as one would expect, even though it is being pulled towards the planet. Once it reaches a distance \(R\), it will have reached a velocity \(v = \sqrt{\frac{2GM}{R}}\) relative to the planet, though for all it knows, it hasn't accelerated at all.
2. Now, let us use an infinitely long rope to slowly lower down the object from infinity to the distance \(R\) in such a way that it ends up at rest once it has reached its destination. This time, it will feel the tension force from the rope, so it knows it must be accelerating (though it senses this going away from the planet...remember that it isn't actually looking at the planet...only interpreting what it feels). By integrating the force with respect to time, the object could in theory calculate its final velocity. I believe this would then come out to \(v=-\sqrt{\frac{2GM}{R}}\), which is same in magnitude, though opposite in direction. Thus, it will think the time dilatation between it and the planet is \[\Delta t =\frac{\Delta t_0}{\sqrt{1 - \dfrac{2GM}{Rc^2}}}.\]My feeling is that this interpretation would then be valid because of the equivalence principle? Thoughts?

Answer 6

I still think kit has to do with the enrgy because if you calculate the potential energy and the kinetic energy they are the same

Answer 7

That is an intriguing notion, though I have never seen that proposed in other contexts though. I'll keep it in mind.

Answer 8

This is a good observation, and one that I've never noticed. I also noticed just now that the time dilation can be expressed as \[\Delta t = {{\Delta t_0} \over {\sqrt{1-{R_s \over R}}}}\]where R_s is the Schwarzschild radius.\[R_s = {2GM \over c^2}\]

Answer 9

Yeah, I've seen that before as well...it's a cool relationship...and clearly shows that when \(R = R_s\), we have infinite time dilatation.

Answer 10

Really interesting findings.

Answer 11

@jwuphysics
Yep, that's usually the way this time ratio is given.

@yakeyglee
That might seem naive, but is this not logical as Einstein stated that an acceleration field or a gravity field act in the same way?

Your falling clock should feel no difference being at rest in the gravity field of the mass or travelling at speed ve, after being accelerated from 0-speed and infinite distance to that same place (without the mass present).

Answer 12

@Vincent-Lyon.Fr - I know that...that is my basis of part 1. The falling is undetectable. Part 2 is if we were being lowered by a rope... which would then be detectable...and thus the clock could interpret that as being accelerated in the opposite direction, which would make a relative velocity with the planet, which then could be predicted as a discrepancy in time measurements from the planet and the clock.

Answer 13

what about some kind of - conservation of energy
If we suggest that traveling through time ("your speed in the time axis) can be exspressed in energy, just like the speed in x/y/z axis.
my problem is that by getting closer to the planet u lost potential energy (it's size is the escape velocity) and in the second equation you gaind kinetic energy.
Tnx! I had this idea for a few years but never got into it

Answer 14

@or_dobkowski - It's not that you gain kinetic energy...it's rather that it "feels" that way...the rope has to exert a force on the object in order to lower it such that when it gets to the desired distance from the planet, it is at zero velocity..

Answer 15

I didn't read your comment in deep yet, I'm reffering only to your first post.
I don't think you get my intention.
Looking from the aspect of energy.
The system doesn't mind from what kind of energy - potential, kinetick or electric.
there is equivilent time deviation for systems with equivelent energy

The loss of potentail energy when aprroching from infinty to a distance R from the planet is the same amount of energy for a body with escape velocity has (that's how u calculate the escape velocity)
Do u see my intention? there is equivilent time deviation for systems with equivelent energy
(sorry for bad spelling)
but I still see a problem as I mentioned earlier

Answer 16

I'm still a little confused as to what your problem is. Whether you gain or lose the energy, the square of it is going to be the same number: \((+v_e)^2=(-v_e)^2=v_e^2\)

Also, are you suggesting something kinda like this?\[\Delta t = \frac{\Delta t_0}{\sqrt{1-\dfrac{\color{red}{\left(\dfrac{2 \Delta E}{m}\right)}}{c^2}}}\]

Answer 17

I didn't have an equation in my head but this looks correct for this two situations at least

I was confussed about where u choose potential energy to be zero.
not very important maybe.
But it should be choosen to be zero at infinity, and than it is -infinnity at R=0
and there will be a problem of sings with your suggustion

so the relative energy due to gravity will be
\[-Gmm/R\]

so the correct form off this equation should have the absolute value of the relative energy
Maybe we are on to something