Time dilation for the clock in the orbit

[victorvmotti]

Suppose that we want to compute the total time dilation for a clock located in an orbiting satellite relative to the clock in our cell phone on the ground.

Consider two different approaches below.

1. Use special relativity and compute time contraction due to velocity. Use approximation of general relativity in the Newtown limits and compute time expansion due to the less gravity and then find the total time dilation.

2. Don't use special relativity. Stick to the approximation of general relativity based on the symmetry and find Schwarzschild metric and the geodesic for the Earth limits. Find the time dilation assuming a relative velocity in the metric.The question is:

Which of them are more justified and provide a better approximation? Are they equivalent? What happens when the relative velocity of the satellite is zero?

 

[pervect]

Time dilation can be regarded as the ratio of coordinate time to proper time (or was it the other way around?). The choice one makes for the coordinate system implies the choice of some specific metric, which may or may not be the Schwarzschild metric. The Scharzschild metric makes many calculations easier, but people routinely misinterpret the physical significance of the Schwarzschild coordinates, for instance by interpreting the r coordinate as measuring radial distance. It would probably be wisest wise to pick some standard coordinate system (I think the IAU has one) with an associated standard metric if one is unsure.

After one picks the coordinate system (preferably one that one is familiar with so one doesn't make any errors in interpreting the results), the rest is straightforwards. One calculates the orbit - most likely a Newtonian approximation will suffice. Then one compares the proper time interval and the coordinate time interval for the orbit (or section of orbit) and takes the appropriate ratio. To do this one needs to understand what proper time and coordinate time are, of course, and the necessary formula. Feel free to ask if needed.

Note that "time dilation" is not a tensor, it's not a coordinate independent quantity. So the choice of coordinates will matter to the numerical answer, though it won't have any true physical significance.

 

[victorvmotti]

The Scharzschild metric makes many calculations easier, but people routinely misinterpret the physical significance of the Schwarzschild coordinates, for instance by interpreting the r coordinate as measuring radial distance.

Can you explain your point here a little more? What is the correct interpretation of the $r$ coordinates in the Schwarzschild metric?

 

[DrGreg]

What is the correct interpretation of the $r$ coordinates in the Schwarzschild metric?

If you take all the points with the same Schwarzschild $r$ coordinate, they form a spherical surface of area $4 \pi r^2$.

A sphere of that surface area in flat spacetime (i.e. far from any gravity) would have a radius of $r$, but around a star or planet, the distance to the centre would be greater than $r$ due to spacetime curvature. (And around a black hole, "distance to the centre" is an undefined concept. $r$ represents distorted distance only outside the event horizon.)

 

[victorvmotti]

I see your point that we assume an asymptotic flat metric.

What I had in mind was how good is the approximation in the two approaches above.

When we pick the second approach and use the Schwarzschild metric we get this equation:

$dt' = \sqrt{1-\frac{3GM}{c^2r}}dt = \sqrt{1-\frac{3r_s}{2r}}dt$

where $r_s$ is the Schwarzschild radius: $r_s = 2GM/c^2$.

Here we not only assume the asymptotic flat metric to measure $r$ but also switch to Newtown gravity when we want to cancel $v$:$$ v = \sqrt{\frac{GM}{r}} $$

So it appears that in the second approach there are many more approximation assumptions.

 

[m4r35n357]

Suppose that we want to compute the total time dilation for a clock located in an orbiting satellite relative to the clock in our cell phone on the ground.

Your question is answered in detail here.

 

[victorvmotti]

Your question is answered in detail here.

Thanks, this is great, exactly what I was looking for, that is not using $v$.