# Light shone in a train bouncing off mirrors

1. May 29, 2007

### JustinTime

I have a question about a thought experiment I read about. I think the question or a similar was asked, and I'm sure others have asked this question, but I'm asking it in this way so it can be answered through the framework of my question. The framework of how I'm understanding it, and so that if there is a flaw, I can be shown what that is. I don't have a math or physics background, but have found what I've read about relativity fascinating. I got a book called Einstein's Relativity by Max Born and a lot of it is over my head but I understand bits. The thought experiment I'm listing wasn't in this book.

The thought experiment is a train is moving forward and on the top and bottom are mirrors which reflect a light up and down. It is said that for the person on the train, they will see the light go straight up. But an an observer positioned outside of the train will see it traverse diagonally. What I don't understand is this: If the speed of light is constant, it moves independent of the train. So when the light is first reflected to the opposite mirror, why wouldn't both observers see the light bounce and hit the other mirror slightly *behind* where it hit the other? I understand that if a person in a train throws up a ball, to him it appears to go straight up but to the person outside it is seen as an arc. But with light, it travels independent of the speed of the train. In order for the person in the train to see the light travel a straight path, wouldn't the beam of light have to shift or travel additionally in the direction he's going?
Thank you in advance to anyone who takes the time to answer.

Justin

2. May 29, 2007

### JesseM

Light's speed is independent of the source, but the direction of a beam of light from a laser or a flashlight is not. If the beam travels parallel to the body of a flashlight when the flashlight is at rest on Earth, as measured in the Earth's rest frame, then the beam must also travel parallel to the body of a a flashlight when the flashlight is on a train, as measured in the rest frame of the train. If it didn't work this way, then the laws of physics would work differently in different frames, and a person in a sealed windowless room could determine their absolute velocity by seeing what angle a beam left a flashlight as measured in the room's rest frame.

3. May 30, 2007

### SpazAttack

The idea behind this is that the laws of physics are exactly the same if you are moving with a constant velocity as opposed to you not moving at all. So although the person on the platform will indeed see the photons direction changed (like the poster above me said, direction is not independent), the person on the train would not, because from his frame of reference, he isnt even moving.

4. May 31, 2007

### JustinTime

Thank you, that is interesting. How do we know that the beam doesn't travel parallel to the body of the flashlight?

5. May 31, 2007

### JesseM

Well, it does travel parallel in the flashlight's rest frame, just not in a frame where the flashlight is moving. You can predict this just based on the fact that the equations of electromagnetism are "Lorentz-invariant", meaning electromagnetic waves and charged particles obey the same equations in different reference frames.

6. Jun 1, 2007

### JustinTime

It seems strange to me that if the speed of light is constant that it would travel in a path parallel to the flashlight. If the flashlight is moving, the light would have to move as well. If the beam is observed moving parallel to the flashlight it would be moving at the same speed as the flashlight, but light moves at a constant speed. I think it would be said that the flashlight is at rest in its own inertial frame. Something just doesn't seem to add up!

7. Feb 11, 2009

### Xieper

I found the theories around what happens if you assum that (in this example) the train is approaching or at the speed of light very confusing (and so I can't remember them). I must look into it again and see if I can grasp the concept of these theories now I'm a little older and now find it MUCH more interesting.....

8. Feb 11, 2009

### ZikZak

What criteria are you using to determine whether "the flashlight is moving?" You must be assuming the presence of an absolute reference frame against which all motion is measured. There is no such reference frame. The person in the train is entirely justified in considering the flashlight to be at rest, and it will behave exactly as if it were, because indeed, relative to the train, it is.

9. Feb 13, 2009

### I. N. Stine

Wow! This is fascinating!

What if the flashlight is placed on the track bed and bound with duct tape to a cross tie so that it is exactly perpendicular to the tracks according to a protractor held against it?

Then, when the train rumbles by, a hobo standing in the embankment and looking in a window can see the beam hit a hole in the train floor and go up to the ceiling, where it hits a spot behind the spot in the floor!

OK, if I understand you right, a rider in the train will see the beam come through the hole in the floor and go straight up to a spot that is not behind the hole in the floor. The spot on the ceiling is the same distance away from the rear wall as the hole in the floor! Right? Wow! I love revatitly! it is so cool!

Thank you for explaining it so good! You are a really wise fellow!

10. Feb 13, 2009

### michelcolman

No, both will see the spot of light behind the hole. To the observer in the train, the light moves diagonally because from his point of view the flashlight is moving backward underneath the train.

Situation 1: flashlight on board
Observer on train: flashlight is stationary, so it shines straight up
Observer on ground: flashlight is moving forward, so it shines diagonally forward

Situation 2: flashlight on ground
Observer on train: flashlight is moving backward, so it shines diagonally backward
Observer on ground: flashlight is stationary, so it shines straight up

There can never be a situation where two observers really see different things (like a spot of light in two different locations). They will disagree on many other things, like at what time the spot hit the ceiling, how long it took the light beam to travel from bottom to top, how long the train is, whether two events happen at the same time, whether two clocks in different parts of the train are synchronised or not, etc... But they would never see a spot of light hitting different parts of the train.

If they would see the spot of light in two different locations, imagine tying a light-sensitive explosive charge to the ceiling of the train exactly above the hole. As seen from the train, it would explode. But as seen from outside, it would not. This is clearly impossible.

11. Feb 13, 2009

### I. N. Stine

Thanks for explaining it to me! I love retalivity but every time I think about I seem to get something backward.

I believe I read somewhere that Einstein claimed that light did not get a change in speed from the speed of a light bulb or something like that. I think Einstein said that light does not get momentum from the thing that gives off the light. Does that sound like I am remembering it right?

What is confusing me when I study your good explanation is this. If the flashlight is in the moving train, and the hobo is sitting in the embankment watching in a window, when a piece of light comes out of the flash light, it should go straight up at its regular speed. And, if Einstein is right, and I remember him right, the piece of light will not have any speed in the direction the train is going. So, while the piece of light is going from the flash light up to the ceiling, the ceiling is moving forward but the piece of light is not moving forward. It seems that the piece of light will hit a spot on the ceiling further back than the location of the flash light. The hobo sitting outside should see it that way. Come to think of it, the rider in the train should see it that way too. Am I right at all about this?

In the first post in this thread, the writer was concerned about the light beam going straight up when the train was sitting still. Whether the rider or the hobo was watching it. Then when the train is moving, the first writer was saying ( I believe I remember it right ) that the light would get left behind and hit further back on the ceiling. Whether, again, the rider or the hobo watched it.

One time I borrowed a college physics book and was reading about relavility. It had two pictures. One picture showed light going straight up and down. I guess that was what the rider was supposed to see. The other picture showed light going in a diagonal zig zag and moving along with the train. That must have been what the hobo was supposed to see. It not have a third picture. So I don't know what anybody was supposed to see when the train was still.

I am confused about why the book showed different people seeing two different ways that light moved but you are explaining that everybody must see the same thing.

Wher am i getting mixed up on this? Does light get left behind or does light get speed from the train and go forward too?

I know all of you figured all this out a long time ago. But it gets mixed up for me. Thanks!

12. Feb 13, 2009

### altonhare

The guy on the ground sees the light take a diagonal for the following reason. The atom of the source excites, sends a signal to your eye, and your brain says "there's the flashlight". Your brain decides it's *exactly* 4 feet from a nearby stop sign and 1 foot off the ground. The train then moves a finite distance (delta x) to the right before an atom in the ceiling (1 foot above the source) absorbs the signal, then retransmits it to your eye. Your brain then says "There's a spot right there". This spot will be 4 feet - delta(x) from the stop sign and 1 foot above the flash light. You perceived the source to be 4 feet to the left of the stop sign and the target (ceiling) to be 4-dx from the stop sign. (4,0) to (4-dx,1) is diagonal to the two coordinate axes considered

The guy in the train sees the source, decides its 4 feet from the wall, then the train, wall, flashlight, guy, and the signal itself move dx to the right, and the guy sees the spot exactly 4 feet from the wall again, just 1 foot higher. (4,0) to (4,1) is parallel to an axis, call it vertical or horizontal.

One way to think about light is as if the atoms between the source and the target are *physically connected* by a 2 strand entwined rope or something. When an atom in the source "excites" it torques the rope, which propagates straight up to the next atom, exciting it. Both atoms torque a rope connected to your eye, too. Just as if you took two shoe strings and wrap them around each other, then at one end take the two strings and pull them laterally, you'll torque it and cause the other end to "spin". This is the idea behind the principle of ray reversibility. The light signal propagates along this permanent physical interconnection that travels along with the atoms it connects i.e. the atoms in the flash light and the ceiling.

So if you're watching the train go by the rope connecting the two atoms is still "vertical", you just see the flash light and the ceiling at two different x-axis locations which makes it appear that light traveled both "up" and along the x axis, which makes it seem like it traveled diagonally. The guy on the train sees the flash light and the ceiling at the same location on the x axis, which makes it look like the light traveled parallel to the y axis. It's just an optical illusion, a result of light's finite propagation speed.

Now the issue of the two observers measuring light's speed is "another can of worms" that has to do with the observation that a clock belonging to the guy on the train will tick fewer times than the guy on the ground. Let's analyze the situation. The guy on the ground will measure a diagonal distance. Taking the velocity of light to be irrespective of the source like any wave, we would naively expect specific x and y components of the velocity, which of course have to add up to c. Let's say the train's moving along at ~100 km/sec (fast train). We naively expect 0.999999944*c in the y direction and 0.0003 in the x direction. So we expect the observer on the ground, watching, to perceive the light to take a little longer to go from the flash light to the ceiling than the guy on the train. The guy on the train doesn't measure a diagonal, he only has a single component, 1*c in the y direction. We'd expect the guy on the ground to report that the light took 1/.999999944 longer to get to the ceiling than the guy on the train. But in reality this isn't what we see. The guy on the ground measures the same time. It seems that the clock belonging to the guy on the ground ticks 1/.999999944 faster than the guy's clock on the train, exactly making up for the extra distance he perceives due to the "diagonal". Nature is tricky like that.

Why do clocks slow down? Who knows, all we know is that they do so just enough so that we always *measure* the speed of light to be the same, no matter what situation we find ourselves in.

Yous said you "haven't had any math" but if you're interested in relativity you should, at a *minimum* learn enough math to follow the derivation of the Lorentz transformations (if you haven't done so already).

Last edited: Feb 13, 2009
13. Feb 13, 2009

### matheinste

Hello antonhare

Quote:-

---One way to think about light is as if the atoms between the source and the target are *physically connected* by a 2 strand entwined rope or something. When an atom in the source "excites" it torques the rope, which propagates straight up to the next atom, exciting it. Both atoms torque a rope connected to your eye, too. Just as if you took two shoe strings and wrap them around each other, then at one end take the two strings and pull them laterally, you'll torque it and cause the other end to "spin". -----

As visualizations go this is pretty bizarre and also misleading.

Matheinste

14. Feb 13, 2009

### altonhare

How is it misleading? The atom can only take in or release an integral number of links of this rope, which justifies quantization. The signal can only propagate rectilinear, justifying this observation. It simulates propagating perpendicular plane waves of classical EM. It's a pretty useful visualization.

Edit: Additionally, granting that it may not be perfect, is a stream of bullets or abstract 2D plane waves somehow superior?

15. Feb 13, 2009

### matheinste

Hello antonhare.

In my opinion the most fundamentally misleading part is describing the difference in "views" between the two observers as an "optical illusion". The remark about quantization is quite irrelevant to the understanding of the geometry of the situation and just throws in a level of complication into the description of a simple effect that can, and has already been, adequately explained earlier in this thread.

Matheinste

16. Feb 13, 2009

### altonhare

How is it not an optical illusion? The guy on the train sees (x1,0) ; (x1,1) and the guy on the ground sees (x2,0) ; (x2-dx, 1) where dx is just the distance-traveled by the train relative to the guy on the ground. One infers a vertical path and the other infers a diagonal path, purely because light's propagation speed is finite, so the guy on the ground must wait while the train moves to the side before he can see the second photon.

The guy on the ground infers a longer distance. The distance between A and B, though, is static. When A releases the photon it travels a distance D to B, no matter who's watching, whether they're moving or not, or whether they're doing cartwheels or not. This static separation doesn't change because of testimonials or calculations.

The remark on quantization was in response to your general dislike of the visualization, to point out other reasons I like it.

Last edited: Feb 13, 2009
17. Feb 13, 2009

### matheinste

It is not an optical illusion because what either/both observers see is real for them. Both are correct and neither sees an illusion. Perspective might be a better word, but even this is inaccurate.

Matheinste.

18. Feb 13, 2009

### altonhare

Are you saying we could barely squeeze a brick between A and B but then also stick 2 bricks between A and B?

Edit: The problem here is that the "distance" calculated in SR is actually "distance traveled" (i.e. a distance-traveled by the light signal). Distance traveled is a dynamic concept whereas distance is a static concept. An observer that assumes the two are the same makes a gross logical error.

Last edited: Feb 13, 2009
19. Feb 13, 2009

### matheinste

Quote:-

---Are you saying we could barely squeeze a brick between A and B but then also stick 2 bricks between A and B? ----

I'm afraid i do not undeerstand what you are asking.

Matheinste

20. Feb 13, 2009

### altonhare

Like I said, there are two concepts here, the static concept distance and the dynamic concept "distance traveled". The latter is what we measure, not the former. .