# Why is gravitational time dilation not symmetric

## Summary:

Why does the derivation of gravitational time dilation use classical mechanics' definition of kinetic energy as opposed to the relativistic definition of kinetic energy?

## Main Question or Discussion Point

Why do we use the equation ##\frac {1}{2}mv^2 = \frac {GmM}{r}## to derive potential velocity, and then put that in the Lorentz factor in order to derive gravitational time dilation? Shouldn't we be using the relativistic definition of kinetic energy -> ##mc^2(\gamma - 1)## to derive the gravitational time dilation? Using the first approach we get $$\gamma = \frac 1 {\sqrt {1 - \frac {2GM} {rc^2}}},$$ but using the relativistic definition of kinetic energy we get $$\gamma = \frac {GM} {rc^2}+1$$, so clearly one of these approaches is wrong.

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Why do we use the equation ##\frac {1}{2}mv^2 = \frac {GmM}{r}## to derive potential velocity, and then put that in the Lorentz factor in order to derive gravitational time dilation?
Where have you seen gravitational time dilation derived this way? Can you give a reference?
Why do we use the equation ##\frac {1}{2}mv^2 = \frac {GmM}{r}## to derive potential velocity, and then put that in the Lorentz factor in order to derive gravitational time dilation?
As Peter is hinting, your derivation is incorrect. ##v## in your calculation is the Newtonian velocity of something dropped from rest at infinity to radius ##r##. But the gravitational time dilation formula you quote applies between hovering observers - who would regard such infalling objects as moving, so a different time dilation factor would apply. Even if the Newtonian calculation of the velocity were correct, which it's not.

Furthermore, ##\gamma## is the kinematic time dilation factor between inertial frames, so can only be used locally in general relativity. It makes no sense to apply it to comparing rates of distant clocks in curved spacetime because ##v## cannot be defined the way it requires.

Peter asked for a reference. If you can provide one, we'd be interested because it's a source to avoid. If, on the other hand, it's something you came up with yourself, you've simply found an amusing coincidence, not a derivation.
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Why does the derivation of gravitational time dilation use classical mechanics' definition of kinetic energy as opposed to the relativistic definition of kinetic energy?
That is interesting. I know of two different derivations, and neither uses KE at all. The ones I am familiar with either use the equivalence principle or directly from the metric.

If you don’t like that derivation then I would just find another one that you do like
Where have you seen gravitational time dilation derived this way? Can you give a reference?
I did not find it online, someone told me about it.
I knew this derivation made no sense because the energy principle in this scenario is applying to an energy conserved system consisting of an infinite amount of space between the observer and the object and it makes so many assumptions that I can't even comprehend. However it is quite insane that the approximation for kinetic energy in the system can be used in for the velocity in the Lorentz factor and one can get the expression for gravitational time dilation.
Did this someone give you an actual reference? Or were they just making it up?

As you can see from the responses you have gotten in this thread, there is no such derivation.

it is quite insane that the approximation for kinetic energy in the system can be used in for the velocity in the Lorentz factor and one can get the expression for gravitational time dilation
No, it's not "insane", it's just a coincidence, as @Ibix has pointed out. What's more, the coincidence is even less impressive than it might seem at first glance. Consider: the kinetic energy expression includes a velocity ##v##. But what object has this velocity? Certainly not the object whose gravitational time dilation is equal to the expression you give, because that object is static--that expression for gravitational time dilation is for an object which is at rest relative to the gravitating mass, at radial coordinate ##r## (which is not quite the same as the actual proper distance of the object from the center of the gravitating mass). So the coincidence is the result of mixing together two formulas that don't even apply to the same object.
I know it is a coincidence I'm just saying that it is funny that you can incorrectly come to a correct conclusion by failing to be precise. And no this person didn't have a source he just attempted to oversimplify the derivation then I asked why he used the approximation for kinetic energy and he said you just assume you're not going that fast and then I got even more confused.
this person didn't have a source he just attempted to oversimplify the derivation
He wasn't oversimplifying a valid derivation to begin with: the entire reasoning is invalid because, as I said, the ##v## in the kinetic energy formula can't possibly apply to the object whose gravitational time dilation is given by the formula that is derived. The same would be true if you tried to use the relativistic formula for kinetic energy.
I know it is a coincidence I'm just saying that it is funny that you can incorrectly come to a correct conclusion by failing to be precise.
This is essentially numerology, just with algebra instead of arithmetic. Like numerology, coincidences come up more often than you might naively think. Like numerology, there's no meaning to the coincidences.
But why is a mass traveling at a velocity relative to a distant "stationary" observer seen to have a slower ticking clock and why is this the same situation observed for a mass resting (v=0) on the top of a mountain on Earth and the same situation when the object begins to free fall (v=0+ ) off of that mountain top? What is common to these three situations? What is the mechanism at play?
This is essentially numerology, just with algebra instead of arithmetic. Like numerology, coincidences come up more often than you might naively think. Like numerology, there's no meaning to the coincidences.
This may be more than coincidence. Perhaps it is a hint that there is an unrecognized, underlying mechanism at work.
What is common to these three situations?
The last two aren't different, because the mass resting on the mountain and the mass just starting to free-fall aren't different with respect to time dilation (since the mass just starting to free-fall still has negligible velocity and is still at the same altitude).

The first situation has nothing significant in common with the second except that, like any phenomena in General Relativity, they are both ultimately due to spacetime geometry (see below). Time dilation due to relative velocity is symmetric: each observer sees the other's clock running slow. Time dilation due to different altitudes in a stationary gravitational field is not symmetric: both observers agree that the observer at the higher altitude has a faster running clock.

What is the mechanism at play?
The underlying "mechanism" in general is spacetime geometry. But see below.

This may be more than coincidence. Perhaps it is a hint that there is an unrecognized, underlying mechanism at work.
As above, the underlying "mechanism" is spacetime geometry. But nothing in the spacetime geometry requires that there must always be some kind of match between "velocity time dilation" and "gravitational time dilation". The match in this particular case is, as has already been said, just a coincidence.

I strongly recommend that you review the PF rules on personal speculation before posting further on this subject. Your posts are already close to the line on that.
But why is a mass traveling at a velocity relative to a distant "stationary" observer seen to have a slower ticking clock
Ticks of a clock turn out to mark regular intervals in spacetime in the same way that marks on a ruler mark regular distances in space. The two clocks' paths through spacetime are not parallel so, just as the tick marks on rulers don't align if they aren't parallel, the clocks' ticks don't align.
why is this the same situation observed for a mass resting (v=0) on the top of a mountain on Earth and the same situation when the object begins to free fall (v=0+ ) off of that mountain top?
It isn't. Two objects at the same height with no relative speed between them tick at the same rate. If the clocks are at different altitudes then they tick at different rates (in some sense). The reason is the same - the "distances" through spacetime indicated by the clocks are along paths that are not parallel, with the added complication that even defining "parallel" in curved spacetime is complicated.
This may be more than coincidence. Perhaps it is a hint that there is an unrecognized, underlying mechanism at work.
It really isn't. How Newtonian gravity emerges as an approximate form of GR is well understood. And how SR emerges as a special case of GR is well understood. Ripping bits out of both and jamming them together is meaningless, even if there are some coincidences when you do. It's like taking a piece of a jigsaw with sky on it and another piece with grass on it and noticing that the edges happen to fit. You can put them together, but you're rather missing the point if you do.
But why does both the clock traveling at a relative velocity tick slower AND so does the clock located at a deeper position in a gravity field? Inertial-gravitational equivalence .
But why does both the clock traveling at a relative velocity tick slower AND so does the clock located at a deeper position in a gravity field?
Their worldlines aren't parallel and/or spacetime is curved so "parallel" is a complicated concept. Thus times along their worldlines are like distances along non-parallel rulers or distances along different latitude lines - they don't align. As I already said.
Inertial-gravitational equivalence .