# Is light slowed down when passing through a tube that exerts gravity?

## Main Question or Discussion Point

Bill and Ben are a fixed distance from each other.
Bill sends a laser pulse to Ben and times how long it takes to get there and bounce back.

Now a very heavy tube is placed in between Bill and Ben. The light pulse is sent again so it passes through the tube. The tube is along the same axis that the laser travels so its gravity does not alter the path of the light.

Does the light take longer this time?

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A.T.
Does the light take longer this time?
Yes, because the distance is now longer.

Bill_K
I'd say yes, but not because of a distance effect. Time passes more slowly in a region of higher gravitational potential. In fact, this is a straight-line analog of the Shapiro delay.

A.T.
I'd say yes, but not because of a distance effect. Time passes more slowly in a region of higher gravitational potential.
But spatial geometry would be distorted too. Are these two cumulative effects?

Bill_K
But spatial geometry would be distorted too. Are these two cumulative effects?
Oops I take it back. In the weak field approximation h00 and hij are the same order of magnitude. For a slowly moving test particle only h00 counts, since in the equations of motion hij gets multiplied by vivj/c2, but for a light ray both space and time distortions are equally important.

A.T.
For a slowly moving test particle only h00 counts.
If the test particle is much slower than c, then it needs less time with the massive tube in place.

I gather(for light) that the time as measured by Bill or Ben will increase with the presence of the tube, and this is due to a combination of increased space within the gravitational field and time also running slower within the field.

Are the effects exactly inversely proportional to each other?
(i.e. a field such that doubles the distance of the journey would half the speed of a clock running in it)

A.T.
(i.e. a field such that doubles the distance of the journey would half the speed of a clock running in it)
The clocks will have different rates at different points in the field. What you can say, is that the local light speed, measured with a local ruler and a local clock is the same everywhere.

This is a rough analogy with the Schwartzchild metric considering horizontal motion only.

Using this version of the Schwarzschild metric where p represents the Newtonian gravitational potential (-GM/r) and the radial axis is represented by x:

## d\tau^2 = (1+ 2p)dt^2 - (1+2p)^{-1}dx^2 - dy^2##

Moving horizontally, dx=0 and the equation can be rearranged to:

##(dy/dt)^2 = (1+2p) - (d\tau/dt)^2##

If we represent the local horizontal velocity measured by a static observer at x by (dy'/dt') then we can rewrite the equation as:

##(dy/dt)^2 = (1+ 2p) - (1+2p)(1-(dy'/dt')^2)##

##(dy/dt)^2 = (1+2p)(dy'/dt')^2##

##dy/dt = \sqrt{(1+2p)}(dy'/dt')##

In other words, whatever the local horizontal velocity is measured to be, the velocity according to the observer at infinity is will be slower by a factor of ##\sqrt{1+2p}##.

Now if a photon is sent down the tube, a local observer inside the tube will of course measure the velocity of photon as c as it passes close by. Similarly, a local observer anywhere within the tube will measure the local velocity of the slow test particle to be 0.1c as it passes close by, i.e. the same as when the tube was not there.

It is worth noting that Bill and Ben wil have there clocks affected by the tube as well so I am assuming they are sufficiently far from the ends of the tube, that the error can be considered negligible.

<P.S.> Had a crash when editing this and lost some of the original.

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A.T.
Similarly, a local observer anywhere within the tube will measure the local velocity of the slow test particle to be 0.1c as it passes close by.
Why should the slow particle stay at 0.1c in a gravitational field?

Bill_K
In this problem, for motion along the z axis,

ds2 = gtt dt2 - gzz dz2

where gtt and gzz are functions of z alone. The path of a light ray, ds = 0 and

dt/dz = √(gzz/gtt)

In the weak field approximation, gtt = 1 - 2φ and gzz = 1 + 2φ, where φ is the Newtonian potential (e.g. φ = M/r for Schwarzschild)

Thus the elapsed time is

t = ∫ √((1 + 2φ)/(1 - 2φ)) dz

The integrand is greater than 1, so the elapsed time is always increased by the presence of the tube. Distortions in both time and space contribute to the delay.

Why should the slow particle stay at 0.1c in a gravitational field?
First, I must stress that I mean the local velocity will remain at 0.1c, but that is difficult to prove. I am just assuming that, because locally, everything measured in a gravitational field changes in such a way that it appears nothing has changed. Also, there is a danger that with an extreme gravitational field, of slow particles traversing the distance in less time than light particles, if slow particles speed up and light particles slow down. (Maybe the tube will collapse into a black hole before that happens)

<EDIT>Then again, the slow test particle is passing from high potential to low potential and then back to high potential, so for the first part of the journey it is effectively falling and local observers will measure its speed to be increasing, so I retract the claim that they always measure the velocity to be 0.1c. On the way back out, the particle is climbing to higher potential so its local speed slows down. The difficulty is not having a ready metric for a finite hollow tube.

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mfb
Mentor
For slow initial speeds and if the mass of the tube is not too large, the tube will certainly speed up the process.

Just make a tube out of earth, and let the test mass start above its surface with a velocity of 1m/s. The slow test mass will need something of the order of 1 hour with earth, but half a year without (2 hours and a year including the way back).

A.T.
The difficulty is not having a ready metric for a finite hollow tube.
Consider the extreme case: Bill releases the particle at rest. How do the travel times with and without the tube compare?

Bill_K
The difficulty is not having a ready metric for a finite hollow tube.
It's perfectly reasonable to use the weak field limit, in which the field of the tube is described by its Newtonian potential.

In post #9 I did rough analogy with the Schwartzchild metric considering horizontal motion only, but on reflection vertical motion in the metric, is a better (but still rough) analogy.

Using this version of the Schwarzschild metric where p represents the Newtonian gravitational potential (-GM/r) and the radial axis represented by x:

## d\tau^2 = (1+ 2p)dt^2 - (1+2p)^{-1}dx^2 - dy^2##

Moving vertically, dy=0 and the equation can be rearranged to:

##(dx/dt)^2 = (1+2p)^2 - (1+2p)(d\tau/dt)^2##

If we represent the local vertical velocity measured by a static observer at x by (dx'/dt') then we can rewrite the equation as:

##(dx/dt)^2 = (1+ 2p)^2 - (1+2p)^2(1-(dx'/dt')^2)##

##(dx/dt)^2 = (1+2p)^2(dx'/dt')^2##

##dx/dt = (1+2p)(dx'/dt')##

In other words, whatever the local horizontal velocity is measured to be, the velocity according to the observer at infinity is will be slower by a factor of ##(1+2p)##, in rough agreement with the result obtained by Bill_W.

However, A.T. is right that the local velocity of the photon remains constant, but the local velocity of the slow particle does not and is on average higher than when the tube is not there. This means the ratio between the travel time of the photon and the slow particle as measured by Bill and Ben depends on the mass of the tube.

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For slow initial speeds and if the mass of the tube is not too large, the tube will certainly speed up the process.

Just make a tube out of earth, and let the test mass start above its surface with a velocity of 1m/s. The slow test mass will need something of the order of 1 hour with earth, but half a year without (2 hours and a year including the way back).
Agree.

If I recall correctly, in Newtonian physics, if we have various straight tunnels passing through the Earth, the travel time from entry to exit of a falling particle is the same for all the tunnels, independent of length.

Consider the extreme case: Bill releases the particle at rest. How do the travel times with and without the tube compare?
Yep, I messed up :)

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mfb
Mentor
Agree.

If I recall correctly, in Newtonian physics, if we have various straight tunnels passing through the Earth, the travel time from entry to exit of a falling particle is the same for all the tunnels, independent of length.
Right, approximately 42 minutes if you assume a homogeneous density (otherwise the time depends on the tunnel) and neglect the rotation.

A.T.