0
$\begingroup$

Sorry if this is obvious, but I've searched and cannot find this answered.

taking a minor change to the light clock experiment on a traveling spaceship.
putting the clock in the horizontal position, does the light appear (to the outside observer) to move faster in one direction than the other? It must, correct? assuming you have two light clocks that are both horizontal, one on top of the other. but the light starts from opposite sides. the light in the clocks must appear to leave at the same moment, and arrive at the other side at the same moment, but both photons are covering different distances.

I understand that length contraction affects the space, but I don't understand how it would affect it to account for this difference. wouldn't the length be the same? contracted in the direction of the ships movement, but still, it's the same space being traveled?

$\endgroup$
2
  • 1
    $\begingroup$ related: physics.stackexchange.com/questions/276574/… $\endgroup$
    – Paul T.
    Commented Nov 19, 2018 at 18:41
  • $\begingroup$ Hello, I'm having trouble understanding the question. Can you please describe the "light clock experiment" you reference? You mention how you want to change it, but I don't know what you are starting from $\endgroup$
    – Paul T.
    Commented Nov 19, 2018 at 18:48

3 Answers 3

0
$\begingroup$

First, one of the postulates of special relativity is the invariance of the speed of light: the speed of light is constant in every reference frame. By this postulate both observers see the light move the same speed. Any length contraction or time dilation works to preserve this fact.

You say that the inside observer sends light the same distance in opposite directions. This observer says both lights turned on simultaneously and were detected simultaneously.

The outside observer sees the spaceship moving. As such, they would observe the length of the spaceship to be contracted and the spaceship's clock to run at a different rate. More important for this experiment is that they would not agree that the two lights turned on simultaneously. This is because another consequence of special relativity: the relativity of simultaneity.

To understand your question, I recommend you draw a Minkowski diagram of the situation. You should mark the trajectories of the "front" and "back" of the spaceship; the events of two lights turning on; the path that each light takes; and the events where they are detected.

$\endgroup$
5
  • $\begingroup$ Thanks Paul. This helped. The biggest thing I was missing was that the front and rear of the spaceship would be observed at different time lags. It's very confusing, of course, but at least that helps to make sense of it. $\endgroup$
    – MD13
    Commented Nov 19, 2018 at 19:44
  • $\begingroup$ I had another thought experiment that this lag seems to help with - $\endgroup$
    – MD13
    Commented Nov 19, 2018 at 19:47
  • $\begingroup$ instead of time clocks, they are flashlights at opposite ends. we have 4 light sensors. each flashlight has a sensor right in front of it, and one at opposite end of the spaceship. there are also 4 beams, two at each end, that are tilted towards the center, all secured by cables. when the light sensors immediately in front of the flashlights are triggered, two beams are released and fall towards each other. long enough so if they are released at the same time, they meet at a point and support each other, not falling. but if they are released at different times, they fall and overlap. $\endgroup$
    – MD13
    Commented Nov 19, 2018 at 19:56
  • $\begingroup$ the same thing happens with the sensors at the destinations. In this way, only if the sensors release simultaneously at both the beginning and end would all 4 beams be supported. so how could they be supported for both observers? but the fact that the time lag changes over the distance of the ship answers this. I think. $\endgroup$
    – MD13
    Commented Nov 19, 2018 at 19:59
  • $\begingroup$ * 4 beams = steel or wood or whatever material. not light beams $\endgroup$
    – MD13
    Commented Nov 19, 2018 at 20:05
0
$\begingroup$

A still standing Einstein light clock ticks like this:

tic..toc..tic..toc..tick

An identical fast moving 'vertical' Einstein light clock ticks like this:

tic..........toc..........tic..........toc..........tick

An equally fast moving 'horizontal' Einstein light clock ticks something like this:

tic..................toc..tic..................toc..tick

Now let us transform the above diagram to a verbal description, let's say each point is one second of time:

Light moves at speed of light to the right for 18 seconds, then light moves at speed of light to the left for 2 seconds, then light moves at speed of light to the right for 18 seconds, then light moves at speed of light to the left for 2 seconds.

An exercise for the original poster: Convert the two other diagrams to verbal descriptions.

$\endgroup$
0
$\begingroup$

Imagine that we are stationary, and a train is about to pass us traveling close to the speed of light. Suppose we see lightning hit the front and the back of the train simultaneously just as the train passes us. How does that appear to a passenger sitting in the center of the train?

In our reference frame, the passenger would be traveling toward the lightning strike at the front of the train and away from the lightning strike at the back of the train. Since the speed of light is constant in our reference frame, we would see the light from the strike in front of the train reach the passenger first, followed by the light from the strike in the back of the train.

But since the speed of light is also constant in the passenger’s frame of reference, the passenger would see the strike at the front before he sees the strike at the back. The passenger would say that the strike at the front of the train occurred before the strike at the back of the train. Even though the two events were simultaneous in one reference frame, they are not simultaneous in another.

Now let’s try to measure the length of the train. In our stationary reference frame, we measure the location of the front and the back of the train simultaneously and subtract to get the length. The passenger will see us measuring the location of the front before we measure the location of the back. The passenger would say, ”No wonder you got a shorter length as I was moving past you. The back is moving toward the front, and you measured the front before you measured the back.” The passenger would perceive the full length of the train, but in our reference frame we would measure a shorter length.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.