Newbie question about relativity.

[StefenRoebke]

So I am relatively (pun intended) new to the world of Special and General relativity and trying to reconsile what I read with what makes sense in my head (apparently probably the biggest hurdle to this whole area.)

I took several physics courses in college, but they were your very standard Newtonian physics classes, and relativity was discussed very little.

One issue I am having is due to the following hypothetical experiment that I thought up the other day. Please correct me if I go astray, and if I do not, please explain how this fits into the STR:

Say you have a train car 10m long. On the cieling and floor of said train car are mirrors. For the sake of the experiment, let's say the mirrors perfectly reflect light, and there is no glass to change the speed of light reflecting off the mirrors. The sides of the train car ar clear so they can be seen through.

On one end of the train car is a laser angled at such an angle that it's x-component (parallel to the track) is 1m/s, and it's z component(non-normal, othogonal to the track) is 0 (a very very small angle, I know.) The end of the car opposite where the laser is being fired is open.


Now on the train is one observer, and off the train is another observer. The train is moving at a speed of 1 m/s in the direction of the x component of the laser, and the outside observer is stationary relative to the train.

Now, since both observers outside and inside the train see the light traveling at c, and c is not effected by the speed at which the train is moving, won't the observer inside the train see the light exit the car after 10 seconds, and the observer outside the car NEVER see the light exit the train? (to him, the light is traveling in the direction of the track at the same speed as the train is moving.)

How is this reconsiled with STR?

 

[Doc Al]

Say you have a train car 10m long. On the cieling and floor of said train car are mirrors. For the sake of the experiment, let's say the mirrors perfectly reflect light, and there is no glass to change the speed of light reflecting off the mirrors. The sides of the train car ar clear so they can be seen through.

OK.

On one end of the train car is a laser angled at such an angle that it's x-component (parallel to the track) is 1m/s, and it's z component(non-normal, othogonal to the track) is 0 (a very very small angle, I know.) The end of the car opposite where the laser is being fired is open.

I don't understand. The speed of light is 3 x 10⁸ m/s, so if it's angled such that its speed parallel to the train direction is 1 m/s (according to folks inside the train) it's basically being shot vertically? How high are the ceilings in this train?

Now on the train is one observer, and off the train is another observer. The train is moving at a speed of 1 m/s in the direction of the x component of the laser, and the outside observer is stationary relative to the train.

Do you mean one observer is on the train and another is on the tracks?

Now, since both observers outside and inside the train see the light traveling at c, and c is not effected by the speed at which the train is moving, won't the observer inside the train see the light exit the car after 10 seconds, and the observer outside the car NEVER see the light exit the train? (to him, the light is traveling in the direction of the track at the same speed as the train is moving.)

Again, I don't really understand the set up. Can you draw a diagram? In any case, if the light beam exits the train as observed in one frame it will exit the train in any frame.

 

[StefenRoebke]

it is almost being shot vertically. The height of the train doesn't matter. What the set up does is creates a psuedo-wave front traveling at 1 m/s along the length of the car. The laser is still traveling at the speed of light, it is just bouncing back and forth on the ceiling and floor.

One observer is on the train, the other is say, sitting on a bench next to the track.

 

[jambaugh]

One point. Remember that since the outside observer sees the source of light moving the angle he sees the light emitted will be more than what the observer inside the train sees.

Picture the light coming from a vertical laser. As a pulse of light travels up the laser the laser is traveling (as seen by the external observer) and so the path he sees while the pulse is inside the laser is diagonal and not vertical. Now given your case is not vertical the observed emission angles will nonetheless differ.

When you take this in consideration along with length "contraction" and time "dilation" (and relative simultaneity) you will see that the two observers agree phenomenologically on what they see.

 

[Doc Al]

it is almost being shot vertically. The height of the train doesn't matter. What the set up does is creates a psuedo-wave front traveling at 1 m/s along the length of the car. The laser is still traveling at the speed of light, it is just bouncing back and forth on the ceiling and floor.

One observer is on the train, the other is say, sitting on a bench next to the track.

Ah... I think see what you're getting at.

You are arranging the light such that, as seen by the track observer, the x-component of the beam's speed exactly matches the speed of the train (1 m/s). Thus from the view of the track observer, the light never exits the train.

The trick here is to realize that while the speed of the light beam is the same for all observers, the angle it makes is different for different observers. (This is called aberration.) In this case, the train observers see the light going perfectly vertical within the train (no angle at all). So both sets of observers see the same thing: The light bounces back and forth without ever leaving the train.

If, as you originally stated, you want the angle of the beam to be slightly tilted towards the direction of the train motion so that the light's speed in the x-direction is 1 m/s according to train observers, then the light's speed in the x-direction according to track observers will be slightly greater than 1 m/s. (The angle of the light is frame dependent--aberration again.) In this case, all observers will see the light exit the train.

 

[StefenRoebke]

But how can the speed be different between the 2 observers, I thought that one of the requirements is that the speed is the same to all observers, regardless of inertial reference frame.

 

[Doc Al]

But how can the speed be different between the 2 observers, I thought that one of the requirements is that the speed is the same to all observers, regardless of inertial reference frame.

The speed is not different, just the x-component. All observers see the beam move at speed c, but with a different angle with the x-axis and thus a different x-component of velocity.

 

[thenewmans]

Umm, I'm not clear on your thought experiment but the speed of light is constant regardless of your speed (inertial frame). What is not constant is time. So to the outside observer, the light in the train is moving at C. He might think that the inside observer sees it move slower. But if he looks carefully, he'd see that time ticks slower in the train too. The inside observer doesn't notice so he sees the light move at C too.

I started an entry on this that might help.
http://bevnap.blogspot.com/2008/02/special-relativity.html
And look for video #42. The Lorentz Transformation (1/2 hr 1985)
http://www.learner.org/resources/series42.html

 

[Nickelodeon]

Now, since both observers outside and inside the train see the light traveling at c, and c is not effected by the speed at which the train is moving, won't the observer inside the train see the light exit the car after 10 seconds, and the observer outside the car NEVER see the light exit the train? (to him, the light is traveling in the direction of the track at the same speed as the train is moving.)

What would happen if you took it to extremes. I think if you got the train going near the speed of light you wouldn't need to point the beam any further forward on the ceiling to get the beam out the door in the same number of bounces.
The observer still on the station platform would, I think, still see the beam coming out the door but wavelength shifted either towards blue or red or neither depending on the relative direction of travel.