Time for light to go 100m, vertically

[DeiRenDopa]

In Newton's world, the time taken for light to go 100 meters is 100 divided by the speed of light, c. No if's, no but's.

In Einstein's SR (special relativity) world, relative motion, who has what clocks where, etc are important. If we have two observers (A and B), each with an identical clock, at rest with respect to each other, and 100 m apart, we can work out the time it takes light to go between them as follows:

A starts her clock when her laser fires, and stops it when she receives a laser signal from a mirror (retroreflector) surrounding B's clock. She calculates the time it takes light to go 100 m as the time interval between the starting and stopping of her clock, divided by two.

For B it's the same (with A and B swapped).

Comparing notes later, A and B agree that the time it takes light to go 100 m is the same. And its value is 100 divided by c.

What about in Einstein's GR (general relativity) world?

What if C (Chris) is on the surface of a perfectly spherical, iso-dense, non-rotating, airless planet, of mass M and radius r. And P (Pat) is on a platform on a vertical tower, directly above Chris (the platform and tower are ultra-strong, and have a mass as close to zero as never mind). They have each measured the height of the platform, using their own standard rulers, and agree that height is 100 m. They have identical lasers, clocks, and retroreflectors.

Chris starts her clock when her laser fires, and stops it when she receives a return laser signal from the retroreflector surrounding Pat's clock. She calculates the time it takes light to go 100 m as the time interval between the starting and stopping of her clock, divided by two.

For Pat it's the same: Pat starts his clock when his laser fires, and stops it when he receives a return laser signal from the retroreflector surrounding Chris's clock. He calculates the time it takes light to go 100 m as the time interval between the starting and stopping of his clock, divided by two.

First question: When they compare notes, do Chris and Pat agree on how long it takes light to go 100m? If not, why not?

Second question: If the answer to the first question is "yes", is that time equal to 100 divided by c? If not, what is it? And why?

If something more concrete would be helpful, assume the planet has a mass of 10^25 kg, and a radius of 5,000 km. If that mass is too wimpy, take 10^30 kg.

 

[psionl0]

Since Chris and Pat are stationary relative to one another I gather that the only variable is their relative closeness to the centre of gravity of the planet. Since time in this instance runs more slowly for one of the players than the other and since they would still measure the speed of light from the source as being the same, they would have to disagree on the separation between them.

 

[sol invictus]

First question: When they compare notes, do Chris and Pat agree on how long it takes light to go 100m? If not, why not?

No. The standard answer for the reason is, well, gravity. A ray of light passing through a region with a gravitational field takes longer than one passing through the same size (measured from the outside) region without one. For closely related reasons, clocks run slower farther down in the gravity field.

Second question: If the answer to the first question is "yes", is that time equal to 100 divided by c? If not, what is it? And why?

According to Pat it will take longer than 100m/c; according to Chris, less time. I can tex the formula for it if you like. The difference will be very small - I guess (and it's really a guess, I didn't check) for realistic numbers the fractional change will be of the order of 10^-14, give or take a few orders of magnitude.

 

[sol invictus]

They have each measured the height of the platform, using their own standard rulers, and agree that height is 100 m. They have identical lasers, clocks, and retroreflectors.

Incidentally this is a bit ambiguous, since you didn't specify how they measured the height. Clearly there are ways they could do it - like by naive light timing - where they would disagree. On the other hand if they do it by climbing up or down the ladder, counting meter sticks stacked end to end, they will obviously agree.

No matter how they do it, the answer in my previous post is correct.

 

[DeiRenDopa]

Since Chris and Pat are stationary relative to one another I gather that the only variable is their relative closeness to the centre of gravity of the planet.

Yes.

I also tried to remove any other complications, due to an atmosphere, or the planet's rotation, or deviations from a perfectly (spherically) symmetric gravity field, ...

Since time in this instance runs more slowly for one of the players than the other

Yes.

The gravitational time dilation, for Chris, is:

[latex]T=\frac{T_0}{\sqrt{1-\frac{2GM}{Rc^2}}}[/latex]

Where T₀ is the time Chris's clock measures, T the time a clock very far away from the planet measures, G is the gravitational constant (big G, not little g), M the mass of the planet, c the speed of light, and R the radius of the planet.

For Pat it's the same, except that R is now the radius of the planet plus 100 m.

and since they would still measure the speed of light from the source as being the same

I'm not sure of that.

The speed of light is constant (c) only locally. Using her clocks and rulers, Chris measures the speed of light at her location to be c; using his clocks and rulers, Pat measures the speed of light at his location to be c.

Measuring the speed of light non-locally (i.e. 'from a distance') is tricky; you have to be extremely careful to accurately describe how you perform the measurement!

, they would have to disagree on the separation between them.

As sol has noted, I was not entirely unambiguous in my description of how Chris and Pat measure the distance between them: "They have each measured the height of the platform, using their own standard rulers, and agree that height is 100 m."

What I meant was the second of the methods sol described, "they do it by climbing up or down the ladder, counting meter sticks stacked end to end". And, as he notes, in this case Chris and Pat agree that the distance is 100 m.

 

[DeiRenDopa]

No. The standard answer for the reason is, well, gravity. A ray of light passing through a region with a gravitational field takes longer than one passing through the same size (measured from the outside) region without one. For closely related reasons, clocks run slower farther down in the gravity field.

Thanks very much sol!

This is what I suspected. I already knew about gravitational time dilation, but didn't know how it would apply in the case of Chris and Pat measuring the time light takes to go 100 m, vertically.

According to Pat it will take longer than 100m/c; according to Chris, less time. I can tex the formula for it if you like.

Yes please!

The difference will be very small - I guess (and it's really a guess, I didn't check) for realistic numbers the fractional change will be of the order of 10^-14, give or take a few orders of magnitude.

The difference in time dilation - using the above formula, and the lower mass planet - is ~3x10^-13, if I have done my sums correctly.

How do you relate that time dilation difference to the difference between what Chris observes and what Pat observes?

 

[sol invictus]

Yes please!

OK. First, note that the spacetime metric in the situation you described is the Schwarzschild metric for radii greater than the radius of the planet. Since there's no atmosphere, the motion of light will be along null geodesics of that metric:

[latex]$ds^2=-f(r)c^2dt^2+dr^2/f(r)+r^2d\Omega^2$[/latex],
where
[latex]$f(r)=1-{2GM \over r c^2} = 1-r_s/r.$[/latex]

Null geodesics in the radial direction are easily found by setting ds²=dOmega²=0:
[latex]$c dt=\pm dr/f(r)$[/latex]

The solution is
[latex]$c(t_f-t_i) = r_f-r_i + r_s \log{r_f-r_s \over r_i - r_s},$[/latex]
for a radial null geodesic that begins at (t_i, r_i) and ends at (t_f, r_f).

That tells you the coordinate time it takes for light to travel between Chris and Pat. To get the proper time - the time measured by Chris or Pat's clock - just multiply that time interval by the square root of f(r) at their location.

So the ratio of the times they measure is given by the ratio (f(R+100m)/f(R))^1/2. The actual measured time is given by the above formula times f(R)^1/2 or f(R+100m)^1/2.

 

[DeiRenDopa]

Thank you very much sol!
{insert dance smilie here}

 

[oody]

I'm no expert, but I was just reading about the Pound-Rebka Experiment^WP which in 1959 measured the gravitational redshift over a vertical difference of 22.5 meters.

 

[DeiRenDopa]

I'm no expert, but I was just reading about the Pound-Rebka Experiment^WP which in 1959 measured the gravitational redshift over a vertical difference of 22.5 meters.

It's closely related.

I started a thread on this general topic a while ago (and must get back to it): Gravitational Time Dilation, Shapiro, GPS, Gravitational Redshift, etc.

What I'm after here is quite specific, and am very grateful that sol has already given me everything I wanted (in this thread).

 

[DeiRenDopa]

OK. First, note that the spacetime metric in the situation you described is the Schwarzschild metric for radii greater than the radius of the planet. Since there's no atmosphere, the motion of light will be along null geodesics of that metric:

[latex]$ds^2=-f(r)c^2dt^2+dr^2/f(r)+r^2d\Omega^2$[/latex],
where
[latex]$f(r)=1-{2GM \over r c^2} = 1-r_s/r.$[/latex]

Null geodesics in the radial direction are easily found by setting ds²=dOmega²=0:
[latex]$c dt=\pm dr/f(r)$[/latex]

The solution is
[latex]$c(t_f-t_i) = r_f-r_i + r_s \log{r_f-r_s \over r_i - r_s},$[/latex]
for a radial null geodesic that begins at (t_i, r_i) and ends at (t_f, r_f).

That tells you the coordinate time it takes for light to travel between Chris and Pat. To get the proper time - the time measured by Chris or Pat's clock - just multiply that time interval by the square root of f(r) at their location.

So the ratio of the times they measure is given by the ratio (f(R+100m)/f(R))^1/2. The actual measured time is given by the above formula times f(R)^1/2 or f(R+100m)^1/2.

r_s is the Schwartzchild radius.

For the lower mass planet (i.e. M = 10^25 kg), r_s = ~1.5 mm (! ) ; for the higher mass one, ~1.5 km.

The coordinate time (for light to travel between Chris and Pat) is ~3.3 x 10^-7 s (i.e. ~0.33 µs), for both cases. The difference between 100/c and the coordinate time is ~4.3x10^-16 s and ~4.3x10^-11 s, respectively.

The square root of f(r) is almost the same at the two locations, and differs from 1 by < 1 part per thousand. So the difference between what Chris's and Pat's clocks read is ~10^-20 s and ~10^-15 s, respectively.

(Someone please check my sums)

 

[DeiRenDopa]

The "lower mass" planet is not dissimilar to the Earth (aside from being perfectly spherically symmetric, isodense, non-rotating, and airless! ): radius 5,000 km (vs ~6,370 for the real Earth), mass 10^25 kg (vs ~6x10^24).

The "higher mass" planet is kinda like the Sun squished down to the size of the Earth; it has a mass of 10^30 kg (vs ~2x10^30 for the real Sun). There may actually be white dwarf stars with much the same mass and radius.

The most extreme physical object we can imagine - limited by the need for a solid surface - would be a neutron star. They have a mass limit of ~3 sols (i.e. three times the mass of the Sun), and estimates of the radii are rather uncertain (although it's certain the radii are greater than the corresponding Schwartzchild radii!).

Let's see what the time on Chris's and Pat's clocks are, for a Gedanken neutron star. Assume a mass of 6x10^30 kg, and a radius of 9 km. In this case, r_s is ~8.89 km, and the coordinate time ~8.8 µs. Wow! That's pretty dramatically different from 0.33µs (which is 100/c)!!

The time on Chris's clock would be ~0.96 µs; on Pat's 1.33 µs.

But perhaps not. In the OP I said the start and stop events, on each clock, were the firing of the laser and receipt of the return pulse. What the formulae sol provided give are the (coordinate and proper) time it takes for light to travel between Chris and Pat. So are the actual clock readings merely double the above? Or does the return null geodesic require its own special calculation?

 

[sol invictus]

So are the actual clock readings merely double the above? Or does the return null geodesic require its own special calculation?

It's just double. The ingoing and outgoing solutions differ by the +/- sign choice in one of the above equations, which doesn't affect the length of the time interval (to go from ingoing to outgoing pick the other sign and swap r_i and r_f; as you can see the time delta is unchanged).

There is one subtlety here that's connected to the definition of "100 meters". If Chris and Pat measure the height of the platform using meter sticks calibrated at infinity, brought down to the surface of the planet, and stacked up, the top of the tower would not necessarily correspond to the coordinate r=R+100m. For one thing, that would depend on the material the meter sticks are made of (how stiff it is).

If on the other hand the meter is defined using the speed of light, one has to be very careful precisely where and how that calibration is done. For example, the meter could be calibrated independently at every point going up the tower using radially moving light. Or, it could be defined at either the top or the bottom and then propagated up or down in some fashion, or calibrated using non-radial light rays.

None of that invalidates the general result above, but it might affect what r_i and r_f are.

 

[DeiRenDopa]

It's just double. The ingoing and outgoing solutions differ by the +/- sign choice in one of the above equations, which doesn't affect the length of the time interval (to go from ingoing to outgoing pick the other sign and swap r_i and r_f; as you can see the time delta is unchanged).

Thanks.

That's what I guessed, both the conclusion and the reason.

There is one subtlety here that's connected to the definition of "100 meters". If Chris and Pat measure the height of the platform using meter sticks calibrated at infinity, brought down to the surface of the planet, and stacked up, the top of the tower would not necessarily correspond to the coordinate r=R+100m. For one thing, that would depend on the material the meter sticks are made of (how stiff it is).

If on the other hand the meter is defined using the speed of light, one has to be very careful precisely where and how that calibration is done. For example, the meter could be calibrated independently at every point going up the tower using radially moving light. Or, it could be defined at either the top or the bottom and then propagated up or down in some fashion, or calibrated using non-radial light rays.

None of that invalidates the general result above, but it might affect what r_i and r_f are.

That's for sure!

Even for the 'white dwarf' planet we're talking about materials science far in advance of anything we have today, not to mention that 'Chris' and 'Pat' would be robots, not super-fit humans. And there's a bunch of other things I've left out too; for example, the 'neutron star planet' needs to be neutral (real neutron stars will have some net electric charge, possibly a rather high one), and have no magnetic field (real neutron stars likely have horrendously large magnetic fields).

Whatever the vertical distance between Chris and Pat, as long as it's fixed for the duration of the experiment, the outcome (i.e. the two, locally-measured, time durations) is unambiguous and objective.

Are there indirect ways to estimate the vertical distance that separates Chris and Pat? Methods which are both objective and independent of their local rulers? And, as an added bonus, do not depend on 'light from afar'?

Worth putting your thinking cap on for; here's one: estimate the mass of the planet from the orbital periods of test masses a very great distance from the planet (using a 'Kepler's law approximation', which should be good enough). Have Pat drop a ball, of unit mass, on Chris's head; the energy delivered by the blow is some (objective, fixed) function of elevation (and the planet's mass). The unit mass could be something universal, say a carbon-12 atom.

Lurkers, what do you think?

 

[sol invictus]

Are there indirect ways to estimate the vertical distance that separates Chris and Pat? Methods which are both objective and independent of their local rulers? And, as an added bonus, do not depend on 'light from afar'?

Well, ultimately it doesn't matter what units you use, just so long as the definition is well-posed (so you always know how to "count" with your units in every experiment). So it's mostly a question of choosing a convenient definition, and one that can be accurately applied.

The reason for changing from that bar of platinum or whatever that used to define the meter to a definition based on the speed of light is that the speed of light can be more easily and accurately measured. But in a case like this, it's not necessarily the right choice.