# Another flashlight from spaceship question

• dilletante
In summary, the observer on the spaceship sees the beam move away from the ship at c. The observer on the ship does not think he is moving.

#### dilletante

When a spaceship is moving at 150,000 km/sec and you shine a flashlight forward, both an outside observer and the observer on the ship perceive the beam as traveling at c. I think I understand the reason for that when the light is shined forward, but I am confused in the case where he shines it backwards from the rear of the ship.

When he shines it forward, I assume the outside observer sees the ship moving at .5c and the beam moving at c, which means he sees the beam move away from the ship a distance of about 150,000 km after 1 second.

The observer in the ship has a shortened ruler and a slower clock compared to the outside observer, so for him the beam has traveled more than 150,000 km away and taken less time to do it because of his slower clock. The combination of these two things assures him that the beam is traveling at speed c.

But if he shines the light out the back of the spaceship, the outside observer will see that the light has separated from the ship by 450,000 km after one second. If the observer on the ship has a shorter ruler and slower clock, it seems like he would perceive that the light has traveled further than 450,000 km in less than 1 second.

I know this is obviously the wrong way to look at it, since the guy on the ship has to see it as traveling at speed c regardless of which direction he shines it.

The guy on the ship does not think he is moving. There is no change for him in ruler size or time passed. He sees the light moving away from his stationary position at c.

You're leaving out relativity of simultaneity. Whenever you're analyzing these types of scenarios, you have to include that, because it works together with time dilation and length contraction. Leaving out anyone of the three can lead to erroneous conclusions.

To see how it works, consider two sets of three observers. One set corresponds to the "outside" observer: he now has two companions, one 300,000 km away to the left and one 300,000 km away to the right, and all three are at rest relative to each other (and the distances I just gave are as measured in their common rest frame).

The other set corresponds to the spaceship; it now has two companion spaceships, one in front and one in back, and all three are moving at the same speed relative to the first set of observers, so they are all three at rest relative to each other. They are separated by 300,000 km as measured in their common rest frame (so the distance between them is shorter in the other frame).

Just as the spaceship in the middle passes the outside observer in the middle, the spaceship emits two light pulses, one forward and one backward. In the spaceship frame, both pulses take one second to travel to the next spaceship (the one in front and the one in back), and both pulses cover 300,000 km, so the speed of both pulses is 300,000 km per second.

In the outside observer frame, the forward-moving pulse has to catch up with the forward spaceship, since they are both moving the same direction. If all three ships are moving at 0.5c in this frame, the distance between them is length contracted to about 260,000 km. This distance will decrease at 0.5c, or 150,000 km/sec, because that is the difference in speeds between the ships and the light, in the outside observer frame. So it will take about 1.73 seconds for the light pulse to reach the forward spaceship. In this time the light will cover 260,000 km (the distance between the ships) plus another 260,000 km (the distance the forward ship travels in 1.73 seconds), or 520,000 km. Divide 520,000 by 1.73 and you get 300,000 km/sec.

Now let's look at the backward-moving light pulse in the outside observer frame. In this frame, the pulse is moving towards the spaceship, so the distance between them decreases at 1.5c, or 450,000 km/sec. That means it will take about 0.578 seconds for the pulse to reach the ship. Notice that this is a shorter time than the forward pulse takes: in other words, in the outside observer frame, the forward and backward pulses do not reach their respective ships at the same time, although they do in the spaceship frame. This is relativity of simultaneity. And to calculate the speed of the backward light pulse, we simply note that the pulse has to travel only 173,333 km (the 260,000 km distance between the ships, minus 86,667 km, the distance the ship in back travels in 0.578 sec at 0.5c). And if you divide 173,333 by 0.578, you get 300,000 km/sec. So both light pulses travel at c in the outside observer frame; the rear one covers a shorter distance, but it also takes a shorter time because of relativity of simultaneity.

PeterDonis said:
You're leaving out relativity of simultaneity. Whenever you're analyzing these types of scenarios, you have to include that, because it works together with time dilation and length contraction. Leaving out anyone of the three can lead to erroneous conclusions.

To see how it works, consider two sets of three observers. One set corresponds to the "outside" observer: he now has two companions, one 300,000 km away to the left and one 300,000 km away to the right, and all three are at rest relative to each other (and the distances I just gave are as measured in their common rest frame).

The other set corresponds to the spaceship; it now has two companion spaceships, one in front and one in back, and all three are moving at the same speed relative to the first set of observers, so they are all three at rest relative to each other. They are separated by 300,000 km as measured in their common rest frame (so the distance between them is shorter in the other frame).

I appreciate both answers, and agree with Phinds that the guy on the ship doesn't think he is moving. I also agree with the calculations Peter has provided, which explain how things look to the outside observer. I am not sure it explains how things look to the guy on the spaceship, however, but I assume the answer is in there somewhere and I am being dense.

The guy on the middle ship could think he is at rest with the other ships, as Phinds suggests. He could also synchronize his clock with theirs since they are at rest with respect to each other. From his perspective they are always at rest, 300,000 km apart.

So clocks have to come into it somehow, because the outside observer sees the beam hit the rear ship quite a bit before it hits the forward ship. However, if the guys in the front and rear ships both stop their clocks at the moment the beam arrives, their clocks should read the same, not the difference between 1.73 seconds and .578 seconds. Therefore they can radio back their clock reading to the middle guy, and he can deduce that the beams took the same amount of time to travel each way.

So, a couple of questions... one observer sees the beams hit at two distinctly different times, yet the other on the middle ship confirms via the clocks that they hit at exactly the same time. The outside observer would see all the ships' clocks as running at the same speed, although a little slow compared to his own. But if he sees the beams arriving at different times, he should also be able to see their clocks and he would say that their times read differently when the beams hit, and that they stopped at different times. This must not be the case, but why?

The second question is, if the outside observer thinks it takes 1.73 seconds for the beam to reach the forward ship, the middle ship guy must think it takes 1 second, because to him it has traveled 300,000 km as he measures it. But clocks do not slow that much at .5c, I don't think. But it seems the guy in the forward ship must have stopped his clock at one second past the time it was emitted according to the middle guy's clock. That could be a long second compared to the outside observer, but not 1.73 times as long, could it? Maybe it could, but confusion reigns in my brain.

And I am more confused by the rear ship's clock. His clock should read one second later than the middle guy's emmission time also. But that is longer than the perceived time of the outside observer, and I am quite sure his clock doesn't run faster to make up the difference.

Sorry for the overly long reply.

dilletante said:
I ... agree with Phinds that the guy on the ship doesn't think he is moving. I am not sure it explains how things look to the guy on the spaceship, however
I didn't check the rest of the post, but do you see the contradiction in these two statements?

Well, "contradiction" may not be quite the right word. I can't contradict that you see what you see, but I'm asking why it is that you don't think "the guy on the spaceship doesn't think he is moving" explains why he sees what he sees.

dilletante said:
I am not sure it explains how things look to the guy on the spaceship, however, but I assume the answer is in there somewhere and I am being dense.

You give it in your very next paragraph:

dilletante said:
The guy on the middle ship could think he is at rest with the other ships, as Phinds suggests. He could also synchronize his clock with theirs since they are at rest with respect to each other. From his perspective they are always at rest, 300,000 km apart.

The observers in the three spaceships all do think they are at rest relative to the other ships; moreover, they can verify that this is the case by exchanging round-trip light signals and verifying that the round-trip travel time of light beams, by each of their clocks, is unchanging. (They can also verify that the light signals they receive from the other spaceships are neither redshifted nor blueshifted.)

dilletante said:
clocks have to come into it somehow, because the outside observer sees the beam hit the rear ship quite a bit before it hits the forward ship.

Yes, because of relativity of simultaneity. The beams hitting the rear and front ships are simultaneous in the rest frame of the ships, but not in the rest frame of the outside observer.

dilletante said:
if the guys in the front and rear ships both stop their clocks at the moment the beam arrives, their clocks should read the same, not the difference between 1.73 seconds and .578 seconds.

Yes, they do read the same. The ships' clocks are all synchronized with respect to the rest frame of the ships. That means the times they read are according to that rest frame. The 1.73 seconds and 0.578 seconds are times according to the rest frame of the outside observer; that frame's time is different from the time in the rest frame of the ships.

dilletante said:
if he sees the beams arriving at different times, he should also be able to see their clocks and he would say that their times read differently when the beams hit,

No, he wouldn't, because those clocks don't read the time according to his rest frame. See above.

To see the difference more concretely, suppose that there are two other outside observers, stationed at the points, in the outside observer's frame, where the two light beams will reach the ships. That is, one is stationed 520,000 km in front of the outside observer (i.e., in the direction the ships are moving), and the other is stationed 173,333 km in the rear of the outside observer (i.e., in the opposite direction from the one in which the ships are moving). Then, when the front observer sees the front light beam hit the front ship, his clock will read 1.73 seconds, while the clock in the front ship, just passing him at that instant, will read 1 second. And when the rear observer sees the rear light beam hit the rear ship, his clock will read 0.578 seconds, but the clock in the rear ship, just passing him at that instant, will read 1 second. This is a concrete manifestation of the relativity of simultaneity.

dilletante said:
if the outside observer thinks it takes 1.73 seconds for the beam to reach the forward ship, the middle ship guy must think it takes 1 second, because to him it has traveled 300,000 km as he measures it.

Yes.

dilletante said:
But clocks do not slow that much at .5c

No, they don't. But you're leaving out relativity of simultaneity again. Consider the front outside observer, the one who is 520,000 km in front of the original outside observer (as measured in the outside observer's rest frame). When the light beam is emitted from the middle ship, according to the outside observer, this front observer's clock reads zero. But according to the middle ship, when the light beam is emitted, the front outside observer's clock reads 0.866 seconds, because of relativity of simultaneity. So according to the middle ship, while the light beam is traveling, only 0.866 seconds elapse on the front outside observer's clock, while 1 second elapses on the middle ship's clock, for a time dilation factor of about 1.15, which is the correct value for a relative velocity of 0.5c.

The best way to see all these relationships in context is to draw a spacetime diagram. Unfortunately I don't have a handy way of doing that, but it's a good exercise to do it yourself.

PeterDonis said:
The best way to see all these relationships in context is to draw a spacetime diagram. Unfortunately I don't have a handy way of doing that, but it's a good exercise to do it yourself.
I will certainly do that, and thank you for the explanation. It is much clearer now. Thanks to Phinds as well.