Gravitational Time Dilation & Blueshift

[nutgeb]

Consider a theoretical nonrotating solid massive ball of constant density. A shaft is drilled from a point on the surface down to the center. A light source on the surface is aimed down the shaft. Observers at each end of the shaft carry clocks which initially were synchronized before separation.

1. The gravitational potential (or spacetime curvature) at the exact center of the sphere is zero, as calculated by the shell theorem, etc.

2. The clock at the surface will become time-dilated (i.e., the clock runs slower) compared to the clock at the center. For this reason, the spectrum of the light beam from the surface will appear redshifted when viewed by the observer at the center.

3. The light beam descending from the surface to the center will experience gravitational acceleration toward the center of the sphere during its journey. Gravitational acceleration at any instant is linearly proportional to distance from the center (up to and including the surface). Due to this the gravitational acceleration, the light beam from the surface will appear to be blueshifted to the observer at the center.

I would like to confirm that each of the above 3 points is correct. Also, is there a single equation for calculating the net gravitational redshift/blueshift resulting from these two causes combined, or is it easiest just to calculate each separately and then offset them against each other?

 

[JesseM]

2. The clock at the surface will become time-dilated (i.e., the clock runs slower) compared to the clock at the center. For this reason, the spectrum of the light beam from the surface will appear redshifted when viewed by the observer at the center.

The clock at the center is deeper in a potential well, so it's the one that runs slower relative to a clock at the surface. Gravitational potential is not the same thing as local spacetime curvature. The clock of an observer inside a massive hollow sphere will run slower than the clock of an observer outside even though spacetime in the hollow region is totally flat, as discussed in this paper.

 

[nutgeb]

Thanks Jesse, but I don't think so. As far as I can tell, the Weyl-Majumdar-Papapetrou (WMP) spacetime solution applies only if a massive sphere is electromagnetically charged with the charge being equal to the mass. Without an EM charge, there is no gravitational effect and no time dilation inside the sphere.

It makes no sense that the "gravitational potential" is something different from the local spacetime curvature. They are one and the same; if spacetime is flat there is no gravitational effect at all.

The paper you cited doesn't seem very serious, even if it is peer reviewed. They are speculating about warp drives and time machines. The interrelation between the massive shell and the EM charge isn't clearly spelled out. There are a number of analytical papers available on WMP spacetime solutions, e.g. http://arxiv.org/abs/gr-qc/0502047"

 

[A.T.]

Thanks Jesse, but I don't think so. As far as I can tell, the Weyl-Majumdar-Papapetrou (WMP) spacetime solution applies only if a massive sphere is electromagnetically charged with the charge being equal to the mass. Without an EM charge, there is no gravitational effect and no time dilation inside the sphere.

With no charge, all you need is the interior Schwarzschild solution (search this forum), it shows you that the clock in the center is the slowest.

It makes no sense that the "gravitational potential" is something different from the local spacetime curvature. They are one and the same; if spacetime is flat there is no gravitational effect at all.

Gravitational time dilation is not a local effect, but a clock rate ratio between two distant points in space. It is determined by the spacetime curvature between the two points, and can occur even if spacetime is locally flat at both points.

 

[George Jones]

2. The clock at the surface will become time-dilated (i.e., the clock runs slower) compared to the clock at the center. For this reason, the spectrum of the light beam from the surface will appear redshifted when viewed by the observer at the center.

3. The light beam descending from the surface to the center will experience gravitational acceleration toward the center of the sphere during its journey. Gravitational acceleration at any instant is linearly proportional to distance from the center (up to and including the surface). Due to this the gravitational acceleration, the light beam from the surface will appear to be blueshifted to the observer at the center.

See

https://www.physicsforums.com/showthread.php?p=2147975#post2147975

for my (somewhat non-rigourous) calculations and A.T.'s animation.

 

[nutgeb]

OK, got it! Thank you all for the very helpful references.

Follow-on question: If the surface clock is thrown upward (from just below the surface) at exactly the Newtonian escape velocity of the massive ball [planet] at its surface, then at the instant when that clock crosses the surface, will the gravitational time dilation of the center clock (as compared to the surface clock) be exactly canceled out by the SR time dilation resulting from the motion of the surface clock (as calculated in the center clock's reference frame)?

My guess is yes, but I'm hoping someone has already worked the problem.

 

[A.T.]

Follow-on question: If the surface clock is thrown upward (from just below the surface) at exactly the Newtonian escape velocity of the massive ball [planet] at its surface, then at the instant when that clock crosses the surface, will the gravitational time dilation of the center clock (as compared to the surface clock) be exactly canceled out by the SR time dilation resulting from the motion of the surface clock (as calculated in the center clock's reference frame)?

Should be easy to compute.

What I find curious: Do a clock swinging in tunnel around the center and a clock at rest in the center experience the same proper time intervals, between their meetings? This is more complex, as it involves integration. IIRC George Jones computed it and said the answer is "no". But a book that provides no math, stated "yes".

 

[nutgeb]

Well the math for the SR time dilation of the surface clock traveling outward at escape velocity, as compared to the center clock, is straightforward. It's actually the same as the gravitational time dilation equation for the external Schwarzschild solution. But I can't show it now because Latex isn't working right.

The calculation of gravitational time dilation for the internal Schwarzschild solution is more complex. George Jones shows an equation for this calculation in the Forum thread he cited above.

However, before tackling the equation, I have a basic question about how gravitational blueshift works.

It is often said that gravitational blueshift corresponds to the difference between the underlying clock rates at the emitter and at the observer. Which means that the wavelength at the instant of emission is shorter as measured by the observer's clock than when in the emitter's frame, because the observer's clock ticks more slowly.

On the other hand, without reference to clock rate differentials, it is also said that the light beam is "accelerated" as it falls deeper into the gravity well, and this acceleration increases the (kinetic) energy of the falling light, which also equates to a shortening of the wavelength and blueshift.

I don't understand why these two complementary explanations are not cumulative, rather than alternatives to each other, because they occur at different stages in the light's journey.

It seems logical that lightbeam's wavelength is longer (as considered at the observer's clock rate) at the instant of emission; and then subsequently over the full course of its journey, the light beam is also progressively accelerated, causing additional blueshift. In other words, the amount of gravitational blueshift ought to be something like TWICE the amount of gravitational time dilation.

Why is that logic wrong? I understand that if you bring the observer's clock to the emitter's location, it will no longer tick slower than the emitter's clock. But that's not a very satisfying explanation for the 2-stage process as I described it.