Time dilation thought experiment clarification

[localrob]

Sorry for yet another time dilation question.
I have a couple example problems that I can't seem to explain.

1) A ship leaves the Sun and heads for Earth at 0.5c. An observer on Earth would see the the light from the sun take 8 min, and the ship take 16 min.
The pilot's clock would show it took 16 min, but once he lands and they compare clocks, wouldn't the pilot's clock be 16min and the observer's clock be greater than 16 min?


2) If I'm driving a car at 0.5c, then someone turns a light on behind me(shooting at me), because the speed of light is the same for all frames of reference, I would see the light approaching me at c. Right? Light can't go slower than c.

 

[Matterwave]

1) The Pilot's clock should not show 16 minutes, it should show something shorter than that.

2) Yes, the light would approach you at c, even in your reference frame.

 

[localrob]

1) During his flight, his clock would tick normally though, right?
So wouldn't it appear to him that he has traveled for 16min but to observers it appears he's traveled for longer?
I mean, both clocks in their own reference frames would be ticking normally. Once he decelerates and lands they would break symmetry. So who's clock would be "right"? (I'm not sure if "right" is even the correct way to think about it)

 

[Matterwave]

To the pilot, his clock ticks "regularly"; however, he thinks the distance from the Sun to the Earth is shorter than 8 light minutes due to length contraction.

 

[localrob]

Yes that makes sense. Thanks.

 

[ghwellsjr]

Sorry for yet another time dilation question.
I have a couple example problems that I can't seem to explain.

1) A ship leaves the Sun and heads for Earth at 0.5c. An observer on Earth would see the the light from the sun take 8 min, and the ship take 16 min.
The pilot's clock would show it took 16 min, but once he lands and they compare clocks, wouldn't the pilot's clock be 16min and the observer's clock be greater than 16 min?

Just remember that in the Sun-Earth frame, it will take 8 minutes after the ship leaves the Sun for the Earth observer to see both the light from the Sun at the instant of the launch and the launch of the ship. By that time, the ship is already half way to Earth. Then in the next 8 minutes, the Earth observer will see the ship make its trip to Earth and meanwhile, he will watch another 8 minutes of light from the Sun reach the earth.

When you say that an Earth observer sees that the light from the sun takes 8 minutes, what you really mean is that he sees the sun as it was 8 minutes earlier, he doesn't actually observe the light progressing from the Sun to the Earth but he can see the ship's progress from the Sun to the Earth and it takes half the time as it really took because he doesn't see the start of the trip until 8 minutes later.

Now, because the pilot is traveling at 0.5c, his clock will be running slow according to the Sun-Earth frame by a factor of one over gamma which is 0.866 so his clock will advance by 13.86 minutes during the trip but because he is traveling towards the Earth observer, it will appear to the Earth observer that the clock is running fast by the Relativistic Doppler factor of 1.732. In other words, this factor is 13.86 (the time interval on the pilot's clock) divided by 8 (the time interval on the Earth's clock).

Does that answer your question?

2) If I'm driving a car at 0.5c, then someone turns a light on behind me(shooting at me), because the speed of light is the same for all frames of reference, I would see the light approaching me at c. Right? Light can't go slower than c.

Just like you can't watch the progress of light coming from the sun, you cannot observe the progress of any other light coming at you so you cannot tell how fast it is traveling. But in Special Relativity, we define the speed of light's propagation to be c.

 

[ghwellsjr]

To the pilot, his clock ticks "regularly"; however, he thinks the distance from the Sun to the Earth is shorter than 8 light minutes due to length contraction.

Yes, once the ship gets going, according to a frame in which it is at rest, the distance between the Sun and the Earth is divided by gamma which is 6.928 light minutes.

Since his clock will advance by 13.86 minutes and the distance is 6.928 light minutes, he can calculate his speed as 6.928 divided by 13.86 which is 0.5c.

Meanwhile, the Earth observer will calculate the speed as 8/16=0.5c.

 

[localrob]

Maybe looking at this with the ship moving toward the observer isn't the best way. Let me propose a modified problem.
Lets say a stationary observer watches a ship fly from the sun to the Earth at 0.5c.
a) Since it would normally take light 8 min to get to Earth, the observer would see the shop take 16 min to reach earth.

b) If the stationary observer could see the pilot's clock, he would see it moving slower than his own. If the pilot could see the observer's clock, he would see it moving more slowly.

c) The observer would be measuring a longer time frame than the pilot would and the pilot's time frame would be shorter than the observer's. Yet both know that the trip should have taken 16 min. Is it true then that the trip would take less than 16 min because of length contraction?

 

[ghwellsjr]

Maybe looking at this with the ship moving toward the observer isn't the best way. Let me propose a modified problem.
Lets say a stationary observer watches a ship fly from the sun to the Earth at 0.5c.

It sounds like what you really want to do is describe the scenario from the point of view of a Frame of Reference where we don't have to take into account light travel times but you should be aware that this not what any observer can see because they are always subject to light travel times. When you discuss a Frame of Reference, you and I are the observers because we take on an omnipresent role being able to see simultaneously everything that is happening.

a) Since it would normally take light 8 min to get to Earth, the observer would see the ship take 16 min to reach earth.

Yes, in an Earth-Sun FoR, we can see that it will take 16 min for a ship to travel 8 light min at 0.5c.

b) If the stationary observer could see the pilot's clock, he would see it moving slower than his own.

Yes, in this same FoR, the pilot's clock is running slower by a factor of 0.866 (the reciprocal of gamma).

If the pilot could see the observer's clock, he would see it moving more slowly.

More precisely, in another FoR in which the ship is stationary, all the clocks that are moving at 0.5c (those that are at rest with respect to the Earth and sun) are ticking more slowly by the same factor.

c) The observer would be measuring a longer time frame than the pilot would and the pilot's time frame would be shorter than the observer's.

I'm not sure what you mean by "time frame". It's not a normal relativistic term and can be confused with the standard "Frame of Reference" term which involves both time and distances. Maybe you mean time interval. If so, then, yes, in the Earth-Sun FoR, the interval for the trip is 16 minutes but the pilot's clock only advances by 13.86 minutes

Yet both know that the trip should have taken 16 min.

They both know that in the Earth-Sun FoR, the trip takes 16 minutes but if they understand SR, then they also both know that it takes 16 minutes because in that FoR, time is defined in such a way that it takes 8 minutes for light to make the trip from the Sun to the Earth. This isn't something that nature demands or that we can determine by measurement, observation, or logical reasoning. It's arbitrary and in other Frames of Reference, the time that it takes for light to get from the Sun to Earth can be other values. In the same way, the time it takes for the ship to get from the Sun to the Earth can be a wide range of values. It all depends on the chosen FoR and its underlying definition of remote time. No Frame is privileged, not even an Earth-Sun frame.

Is it true then that the trip would take less than 16 min because of length contraction?

In a FoR in which the ship is at rest, yes, the trip takes less time because the distance between the Sun and the Earth is length contracted.

 

[localrob]

That really cleared things up for me, thanks a million.