Gravity ~ Acceleration ~ Centrifuge & GR

Ich
How would you define the path lengths in this scenario?
The length of the respective world lines, as caculated http://en.wikipedia.org/wiki/Proper_time#In_general_relativity". Their path in spacetime, not in space.

On twin descends at a controlled rate deep into a gravitational well. The second twin waits one hundred years and then descends down at the same controlled rate to meet his sibling that is biologically twenty years old. The relative time difference is very physically manifested here.
Just as a visualization, have a look at the drawing. It symbolizes how same synchronous coordinate time (the angle) still means different proper time (path length), depending on the potential.

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Just as a visualization, have a look at the drawing. It symbolizes how same synchronous coordinate time (the angle) still means different proper time (path length), depending on the potential.
Nice demonstration.
However we agree that when the twins get back together again in this gravitational version, they have aged differently and for most people that would suggest gravitational time dilation has a physical "reality". No?

Ich
However we agree that when the twins get back together again in this gravitational version, they have aged differently and for most people that would suggest gravitational time dilation has a physical "reality".
Did I dispute that?
I don't know exactly what part of my answers you interpreted as saying otherwise.
I'm just saying that, in a dynamic spacetime, the very notion of "gravitational time dilation" gets a bit fuzzy, so you can't use it to create 'instantaneous' relativity at a distance. That's all.

Could it be that escape velocity is the essential ingredient for determining the time dilation at a given point?

All we have to do now, is figure out what the escape velocity on the rim of a rotating wheel is.
It cost me a warning to ask that question in a sub forum, but it turns out the "escape velocity" $v_e$ on the rim is equal to the tangential velocity of the rim of the wheel and that time dilation in the gravitational context and in the rotational context can equally and equivalently be expressed by the single equation:

$$\frac{\tau}{t} = \sqrt{1-\frac{v_{e}^2}{c^2}}$$

In the context of a wheel the "escape velocity" is the minimum velocity that a particle would require when launched "upward" (inward) orthogonal to the rim up a spoke in order to arrive at the centre of the wheel, where the "potential" in the artificial gravity field is zero.

Using an analogy of how Newtonian potential is calculated in a gravitational field, the kinetic energy aquired by an object falling from zero potential (infinity in the gravity case and the centre of the wheel in the rotational case) to the point under consideration is equal to the potential energy at that point.

The terminal kinetic energy of a particle falling outward on a wheel is :

$$KE = \int_0^r \, {mr \omega^2} \, dr = \frac{mr^2 \omega^2}{2}= \frac{mv^2}{2}$$

Since this is the same as the kinetic energy of a particle that remains on the rim it can be concluded that the radial "escape velocity" ($v_e$) of a location on the wheel, is equal in magnitude to the tangential velocity ($v_t$) of that point relative to the non rotating frame and so in the case of a wheel these two expressions for the time dilation are equal:

$$\sqrt{1-\frac{v_{e}^2}{c^2}} = \sqrt{1-\frac{v_{t}^2}{c^2}}$$

turin
Homework Helper
... up a spoke ...
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... the kinetic energy aquired by an object falling from zero potential ... to the point under consideration is equal to the potential energy at that point.
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$$KE = \int_0^r \, {mr \omega^2} \, dr$$
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Thanks, kev. That seems straightforward and correct. Basically, your are using dT=dW=F.dr and F=-dV/dr, so that dT=-dV. (Note the inclusion of the minus sign, a minor detail.)

However, something still bothers me: What is the gravitational analogy of the constraint force due to the spoke?

Apparently, the mass on the wheel spoke only allows a 1-D gravitational equivalent. It doesn't seem to work if you allow, for instance, a tangential degree of freedom. So, this doesn't apply to, for instance, space stations that generate their artificial gravity from rotation, does it? To put it another way, there is no concept of artificial gravitational (scalar) potential energy if you allow for a tangential degree of freedom, because there would be a velocity-dependent (tangential) force (in the rotating frame).

I'm not saying that any of this is wrong, I just wonder if anyone else is bothered by this. Anyway, the end result must be true, since it agrees with the more fundamental calculation based on proper time.

Let's two centrifuges, They have different length arms, and spun such that a clock at the end experiences each feels 1g. The clock on the centrifuge on the longer arm will run slower, even though it feels the same g-force as the other clock. Or you could arrange it so that the speed of the ends of the arms are the same for each centrifuge. In this case both clocks will run at the same rate, even though one will feel a greater g0force than the other.
Thanks Janus. Just reiterating what I think you've explained. Correct me if wrong.
Are you saying that the time dilation for an object in a centrifuge is based purely on the speed it is travelling (SR) and that there is no gravitational (GR) time dilation to be counted?
And so the time dilation is independent of the radius of the centrifuge?

Jonathan Scott
Gold Member
Thanks Janus. Just reiterating what I think you've explained. Correct me if wrong.
Are you saying that the time dilation for an object in a centrifuge is based purely on the speed it is travelling (SR) and that there is no gravitational (GR) time dilation to be counted?
And so the time dilation is independent of the radius of the centrifuge?
That's one correct way of looking at it.

An alternative way of reaching the same mathematical result is to integrate the centripetal acceleration as if it were a gravitational field, from the center out to the relevant radius, giving the equivalent gravitational potential difference between those points, and hence derive the corresponding time dilation.

Regardless of whether you're using the SR velocity or the equivalent gravitational potential, the time dilation does not depend on the acceleration. As Janus pointed out, you can have the same time dilation for different accelerations, or different time dilations for the same acceleration.