Calculating Total Time Dilation in a Falling Photon Experiment

[Josh3to107]

This is my very first post, so here it goes. I've had this idea for a while but this is my first attempt at putting it into words. We've all heard of the hypothetical demonstration that tries to put gravity into perspective by pitting a fired bullet against a simple falling bullet. The idea is that if you fired a bullet parallel to the ground and at the exact same time you dropped a bullet from the same height, they would hit the ground at the same time. This thought experiment is 'similar' and starts with two perfectly parallel, perfectly plumb and perfectly reflective mirrors. Between these two mirrors exists a perfect vacuum and you have the ability to fire, or release, a single photon that is perfectly parallel to the ground, thus making its trajectory absolutely perpendicular to the opposing mirror. Now using classical physics, a single photon fired perfectly perpendicularly between two perfectly parallel and perfectly reflective mirrors should 'bounce' back and forth ad infinitum (I believe that this is precisely how Einstein's 'photon clock' is constructed). All is well and good out side of any gravitational field, but now let's consider gravity and general relativity. Our imaginary mirror apparatus is constructed on the surface of Earth in a very real gravitational field. Here is my question, will the photon 'fall' towards earth? And if so, will it 'accelerate'? If one were to drop an object at the exact moment as the photon was released, negating any air resistance, would the photon and outside object reach the ground at the same time? This is how I imagine the experiment would go because as I understand it gravity and acceleration are indistinguishable experimentally.

 

[Nugatory]

All is well and good out side of any gravitational field, but now let's consider gravity and general relativity. Our imaginary mirror apparatus is constructed on the surface of Earth in a very real gravitational field. Here is my question, will the photon 'fall' towards earth? And if so, will it 'accelerate'? If one were to drop an object at the exact moment as the photon was released, negating any air resistance, would the photon and outside object reach the ground at the same time? This is how I imagine the experiment would go because as I understand it gravity and acceleration are indistinguishable experimentally.

You don't need general relativity with its attendant mathematical complexity to work through this problem. At the scale of a dropped object on the surface of the earth, tidal effects are completely negligible, so you can apply the equivalence principle ("It's the same as performing the experiment in a spaceship accelerating at 1g") and treat it using ordinary special relativity. For that matter, if the times involved are small enough, even a purely classical treatment will work. For example:

You are standing in a spaceship accelerating at 1g (10 meters per second per second because I'm doing the arithmetic in my head and I don't want to mess with the more precise value 9.8). You drop an object from a height of five meters, and it hits the floor in one second (classical $s=\frac{1}{2}at^2$ and actually the spaceship, including the floor, is accelerating upwards until the floor hits the object).

At the same time as you drop the object, you also send a flash of light sideways. You are asking whether this flash of light "falls" along with the object, and the answer is that it does; the light moves sideways while the floor accelerates upwards towards it. The easiest way to visualize this may be to imagine how it looks from the point view of an observer coasting along at a constant speed that just happens to be exactly equal to the speed of the spaceship at the moment that the object is dropped and flash of light emitted - at that moment, everything is at rest relative to that observer although the ship is accelerating from rest.

Although the light does "fall" with the object, the effect is hard to see because the speed of light is so great. If we make the reasonable assumption that the mirrors are three meters apart, then the light moves between them in 10 nanoseconds and "drops" only 100 nanometers with each bounce; it's fully attenuated and disappears long before the guy on ship can see the drop. In the second that it takes for the dropped object to reach the floor, the light would be reflected 100 million times.

 

[DaveC426913]

If one were to drop an object at the exact moment as the photon was released, negating any air resistance, would the photon and outside object reach the ground at the same time?

The short answer is, essentially, yes.

Gravitational lensing shows us that light does bend in the presence of gravity. Light from distant sources passes an intervening galaxy and, like in your experiment, starts to "fall" toward the galaxy.

 

[DaveC426913]

I am curious though, if it would fall at the same rate as any other object, which is darned fast.

If you set this photon bouncing at sixteen feet above the ground, would it literally veer so as to hit the ground after one second?

 

[SlowThinker]

I am curious though, if it would fall at the same rate as any other object, which is darned fast.

If you set this photon bouncing at sixteen feet above the ground, would it literally veer so as to hit the ground after one second?

I'd prefer to say "5 meters" but generally, yes.

Funny thing, the light is going at an angle just before it hits the ground, so it takes more time between the bounces. I'm thinking this is gravitational time dilation for kids
(Edit: Now I don't really know how I arrived at the idea. It seemed obvious just a few minutes ago )
After time $t$, the vertical projection of the photon's speed is $(at)$, horizontal projection is $$v^2=c^2-(at)^2\text{, or }\,\,\,\,\,v/c=\sqrt{1-(at)^2/c^2}$$ and depth is $$h=-\frac{1}{2}at^2$$so we have $$(at)^2=-2ah\text{ and }\,\,\,\,\,v/c=\sqrt{1+2 a h/c^2}$$which, for small height, can be approximated by $$v/c=1+a h/c^2$$
This happens to be the correct (approximate) formula for time dilation at height difference $h$.
We should take into account that the vertical speed is less than $(at)$ after a very long time. It might work but I've never done it.

 

[Nugatory]

I am curious though, if it would fall at the same rate as any other object, which is darned fast.
If you set this photon bouncing at sixteen feet above the ground, would it literally veer so as to hit the ground after one second?

It will have moved $3\times{10}^6$ kilometers horizontally, so the net deflection from horizontal is one part in one hundred million. I don't know that "veer" is quite exactly the word that I would use for that - "almost unimaginably straight" sounds more right to me.

But with that said, yes it will, to the limits of the classical approximation that I made in the previous post.

 

[Nugatory]

Funny thing, the light is going at an angle just before it hits the ground, so it takes more time between the bounces. I'm thinking this is gravitational time dilation for kids

Gravitational time dilation is even easier than that. Place a strobe light that flashes once per second on the floor of a compartment in the accelerating rocket and calculate the rate of arrival of the flashes at the ceiling, remembering that each flash has slightly farther to travel than the previous one. That's how Einstein arrived at the idea in the first place.

You are right that the light is "going at an angle" when it hits the floor as viewed by an observer who is accelerating along with the rocket. However, it's not going at an angle as viewed by the inertial observer who is at rest relative to the rocket at the moment the flash is emitted so the time between bounces doesn't change (which is why the motion of the light and the dropped object is most easily analysed in that frame). When I ignored the "going at an angle" effect for the rocket observer, I was making an approximation: If you have a right triangle, and one side is of length 100,000,000 and the other side is of length 1... the difference between $\sqrt{10^{16}}$ and $\sqrt{10^{16}+1}$ is completely impossible to measure so we can say that the length of the hypotenuse is the length of the long side.

We should take into account that the vertical speed is less than $(at)$ after a very long time. It might work but I've never done it.

I said above "if the times involved are small enough, even a purely classical treatment will work" and that's why I chose a relatively modest drop of five meters and an elapsed time of one second. The change of speed of the rocket (relative to the observer who was at rest relative to the rocket at the start of the experiment) is only 10 m/sec so we don't have to worry about this problem with the vertical speed.

If we're going to run the experiment for a longer time we do have to worry about this effect, so we have to use Rindler coordinates (Google will find the explanation) - that's what they're for.

 

[jartsa]

Funny thing, the light is going at an angle just before it hits the ground, so it takes more time between the bounces. I'm thinking this is gravitational time dilation for kids

It's motion time dilation that occurs because of the falling motion. You found out that it's equal to gravitational time dilation.

I suggest you calculate the total time dilation.