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## Main Question or Discussion Point

Let's say the gravity around a planet of mass M is accurately modeled by the Schwarzschild metric. Then an observer (A) at infinity sees a stationary clock (B) at distance R from the center of the planet being gravitationally time dilated by the factor ##\sqrt{1-\frac{2GM}{Rc^2}}##, i.e. if B sends a light pulse every second in its own time towards A, A will receive them every ##\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}## second in its own time.

More formally, if t

[tex]t_A = t_B\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}[/tex]

If I'm not mistaken we arrive at this result by solving the Schwarzschild metric when dr, dθ and dφ = 0:

[tex] c^2dt_B^2 = \left(1-\frac{2GM}{Rc^2}\right)c^2dt_A^2 - \frac{dr^2}{\left(1-\frac{2GM}{Rc^2}\right)} - r^2 \left(d\theta^2+\sin^2(\theta)d\phi^2\right)[/tex]

And if I'm not mistaken there is a reciprocity in what B observes, i.e. if A sends a light pulse every second in its own time towards B, B will receive them every ##\sqrt{1-\frac{2GM}{Rc^2}}## second in its own time. And to see that all we have to do is go from ##t_A = t_B\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}## to ##t_B = t_A\sqrt{1-\frac{2GM}{Rc^2}}##, am I right?

Now assuming the above is true, let's say we have another stationary clock (C) at distance R

And now the crux of my question: from the point of view of C, how much would B appear to be time dilated? And from the point of view of B, how much would C appear to be time dilated? I guess a naive answer would be to say that since ##t_B = t_A\sqrt{1-\frac{2GM}{Rc^2}}## and ##t_C = t_A\sqrt{1-\frac{2GM}{R_Cc^2}}##, then: [tex]\frac{t_B}{t_C} = \frac{\sqrt{1-\frac{2GM}{Rc^2}}}{\sqrt{1-\frac{2GM}{R_Cc^2}}}[/tex]

But aren't these time dilations only valid in the reference frame of A? We have arrived at the relations between t

I have many more questions about how time dilation would be seen by other observers in motion, for instance I've heard that from the reference frame of an observer orbiting the planet, the planet is the one orbiting the observer and so that the Schwarzschild metric is not the right one to model the situation and calculate the time dilation experienced by other people as seen by that observer.

Also I think it is useful to mention that the above is not homework, I haven't been enrolled in an educational institution for a while now, this is pure curiosity. Ultimately the reason I am asking is that I wonder how a clock on the surface of the Earth would appear to be time dilated from the point of view of a GPS satellite in orbit. From the point of view of an observer at infinity a clock on the surface of the Earth is more time dilated than a GPS satellite by about 0.447ns every second, and the clocks on the satellites are slowed down accordingly, but since a GPS satellite is not in the same frame as an observer at infinity then a clock on Earth shouldn't appear to be dilated by the same amount, and I just do not understand why this isn't taken into account.

More formally, if t

_{A}indicates A's time and t_{B}indicates B time, and their clocks are initially synchronized, then[tex]t_A = t_B\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}[/tex]

If I'm not mistaken we arrive at this result by solving the Schwarzschild metric when dr, dθ and dφ = 0:

[tex] c^2dt_B^2 = \left(1-\frac{2GM}{Rc^2}\right)c^2dt_A^2 - \frac{dr^2}{\left(1-\frac{2GM}{Rc^2}\right)} - r^2 \left(d\theta^2+\sin^2(\theta)d\phi^2\right)[/tex]

And if I'm not mistaken there is a reciprocity in what B observes, i.e. if A sends a light pulse every second in its own time towards B, B will receive them every ##\sqrt{1-\frac{2GM}{Rc^2}}## second in its own time. And to see that all we have to do is go from ##t_A = t_B\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}## to ##t_B = t_A\sqrt{1-\frac{2GM}{Rc^2}}##, am I right?

Now assuming the above is true, let's say we have another stationary clock (C) at distance R

_{C}from the center of the planet. In a similar way we can say that if A sends a light pulse every second towards C, then C will receive them every ##\sqrt{1-\frac{2GM}{R_Cc^2}}## in its own time, or ##t_C = t_A\sqrt{1-\frac{2GM}{R_Cc^2}}##.And now the crux of my question: from the point of view of C, how much would B appear to be time dilated? And from the point of view of B, how much would C appear to be time dilated? I guess a naive answer would be to say that since ##t_B = t_A\sqrt{1-\frac{2GM}{Rc^2}}## and ##t_C = t_A\sqrt{1-\frac{2GM}{R_Cc^2}}##, then: [tex]\frac{t_B}{t_C} = \frac{\sqrt{1-\frac{2GM}{Rc^2}}}{\sqrt{1-\frac{2GM}{R_Cc^2}}}[/tex]

But aren't these time dilations only valid in the reference frame of A? We have arrived at the relations between t

_{A}and t_{B}and t_{A}and t_{C}by solving the Schwartzschild metric when the time coordinate is t_{A}at infinity. And so I would guess that it is not correct to simply replace t_{A}in the metric by t_{B}or t_{C}, because in the Schwarzschild metric the time coordinate on the right hand side of the equation is always implicitly the time experienced at infinity by an observer ticking faster than all other observers? Is this right or is this just my misunderstanding and we can use the metric from whatever point of view we want?I have many more questions about how time dilation would be seen by other observers in motion, for instance I've heard that from the reference frame of an observer orbiting the planet, the planet is the one orbiting the observer and so that the Schwarzschild metric is not the right one to model the situation and calculate the time dilation experienced by other people as seen by that observer.

Also I think it is useful to mention that the above is not homework, I haven't been enrolled in an educational institution for a while now, this is pure curiosity. Ultimately the reason I am asking is that I wonder how a clock on the surface of the Earth would appear to be time dilated from the point of view of a GPS satellite in orbit. From the point of view of an observer at infinity a clock on the surface of the Earth is more time dilated than a GPS satellite by about 0.447ns every second, and the clocks on the satellites are slowed down accordingly, but since a GPS satellite is not in the same frame as an observer at infinity then a clock on Earth shouldn't appear to be dilated by the same amount, and I just do not understand why this isn't taken into account.