- #1
ionm
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My question is about comparing the time dilation of a clock on a spinning disk versus a clock in the vicinity of a massive object. It seems there should be a connection between the two, because of the equivalence principle, but I'm missing something because I don't quite get the answer I would expect.
These are my reasoning steps:
1. The relation between the proper time of a clock rotating uniformly with [itex]\omega [/itex] at a distance ##r ## from the center, and a clock at rest at the center is:
[tex]\Delta t_r = \sqrt{1-r^2\omega^2/c^2}\Delta t_\text{center}[/tex]
2. The relation between the proper time of a clock inside a spherical gravitational field and the proper time of a clock far away is:
[tex]\Delta t_r = \sqrt{1-\dfrac{2GM}{rc^2}}\Delta t_0[/tex]
3. Now let's say we want to rotate our clock such that the centrifugal acceleration reproduces the effect of the gravitational acceleration, so we impose
$$ r\omega^2 = \dfrac{GM}{r^2} $$
but then we get two different relations for the time. Why?
[tex]\Delta t_r = \sqrt{1-r^2\omega^2/c^2}\Delta t_\text{center}[/tex]
[tex]\Delta t_r = \sqrt{1-2r^2\omega^2/c^2}\Delta t_0[/tex]
4. Now of course, there is no reason why the radius should be the same, so let me rephrase the question. We can always pick ##r## and ##\omega## such that the centrifugal acceleration reproduces the effect of gravity:
$$ r\omega^2 = \dfrac{GM}{R^2} $$
From the equivalence principle we would expect that these two cases should be equivalent, so the proper times of the two clocks should be the same. We can fix ## \Delta t_0=\Delta t_\text{center} ## if they are not moving with respect to each other. However, the relations we get are not quite the same, but still depend on the radius:
[tex]\Delta t_r = \sqrt{1-r^2\omega^2/c^2}\Delta t_\text{center}[/tex]
[tex]\Delta t_r = \sqrt{1-2rR\omega^2/c^2}\Delta t_0[/tex]
Why is that? Isn't this result contradicting the principle of equivalence? Or is there a mistake in my reasoning?
These are my reasoning steps:
1. The relation between the proper time of a clock rotating uniformly with [itex]\omega [/itex] at a distance ##r ## from the center, and a clock at rest at the center is:
[tex]\Delta t_r = \sqrt{1-r^2\omega^2/c^2}\Delta t_\text{center}[/tex]
2. The relation between the proper time of a clock inside a spherical gravitational field and the proper time of a clock far away is:
[tex]\Delta t_r = \sqrt{1-\dfrac{2GM}{rc^2}}\Delta t_0[/tex]
3. Now let's say we want to rotate our clock such that the centrifugal acceleration reproduces the effect of the gravitational acceleration, so we impose
$$ r\omega^2 = \dfrac{GM}{r^2} $$
but then we get two different relations for the time. Why?
[tex]\Delta t_r = \sqrt{1-r^2\omega^2/c^2}\Delta t_\text{center}[/tex]
[tex]\Delta t_r = \sqrt{1-2r^2\omega^2/c^2}\Delta t_0[/tex]
4. Now of course, there is no reason why the radius should be the same, so let me rephrase the question. We can always pick ##r## and ##\omega## such that the centrifugal acceleration reproduces the effect of gravity:
$$ r\omega^2 = \dfrac{GM}{R^2} $$
From the equivalence principle we would expect that these two cases should be equivalent, so the proper times of the two clocks should be the same. We can fix ## \Delta t_0=\Delta t_\text{center} ## if they are not moving with respect to each other. However, the relations we get are not quite the same, but still depend on the radius:
[tex]\Delta t_r = \sqrt{1-r^2\omega^2/c^2}\Delta t_\text{center}[/tex]
[tex]\Delta t_r = \sqrt{1-2rR\omega^2/c^2}\Delta t_0[/tex]
Why is that? Isn't this result contradicting the principle of equivalence? Or is there a mistake in my reasoning?