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Question about distance...by jumpin'jack
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#1
Jan612, 06:06 PM

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If two objects move away from each other both at the speed of light, after 1 second will they be approximately 372,000 miles apart?



#2
Jan612, 10:06 PM

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I have two rocket launchers. I point one of them to my left and the other one to my right and fire them at the same time. Each rocket leaves its launcher at (almost) the speed of light relative to me. After one second, I will see one rocket about 186,000 miles to my left and the other about 186,000 miles to my right. So I will see them about 372,000 miles apart, but I won't have seen anything moving faster than light. An observer on one of the rockets won't see anything moving faster than light either. If you're on one of the rockets, after one second you'll see me and my rocket launcher about 186,000 miles behind you  but you won't see the other rocket 186,000 miles farther beyond me. You'll see it leaving the rocket launcher in slow motion because of time dilation (my clock is running slow relative to you) so it won't have moved very far at all. 


#3
Jan612, 11:04 PM

Sci Advisor
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P: 5,079

This question gets at two key aspects of special relativity: separation speed and velocity addition. Correcting the impossible bodies traveling at c, let's say A is moving .99c to the west from B, and C is moving at .99c to the east from B. B sees A and C separating at 1.98 c.
Now let's ask what C sees. Using velocity addition: (u + v)/ (1 + uv/c^2) C sees B moving west at .99c, and A moving west at .99995c (approx.). So, according to C, the separation speed between A and B is quite small. It is, of course, possible to describe how C interprets B's measurements, but I think it is more important to directly understand the two concepts of separation speed and velocity addition directly. 


#4
Jan712, 08:56 AM

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Question about distance...



#5
Jan812, 07:53 AM

P: 2

thanks for the responses and sorry for the obvious mistake.
Taking the ABC example, what if you took out the rocket launcher and it was just two rockets passing in opposite directions at the same speed, say 0.9c. If each rocket was observing the other, they'd be separating at 1.8c. How is this explained with relativity? I'll check out the separartion speeds and addition, very interesting, thanks again. 


#6
Jan1012, 12:07 AM

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That's 0.9c relative to WHAT? You need that third point in the middle, the guy with the rocket launcher, to define the point that they're each moving away from at .9c. Without that point, you don't have two speeds to add; you just have one speed, the speed of the spaceships relative to each other. If the rocket launcher bothers you, we can think about it differently. I am standing still, and I see see a rocket off to my left and about 162,000 miles away, screaming towards me at .9c. And off to my right, there's another rocket also traveling towards me at .9c and about 162,000 miles away. Because something moving at .9c covers about 162,000 miles in a second, the two rockets will pass each other in opposite directions right under my nose in about one second and then apart in opposite directions, each moving at .9c relative to me. But that's .9c relative to me. Someone on one of the rockets will see me moving away at .9c and the other rocket moving away at .99995c (thanks to PAllen for doing the math). And if you want to take me out of the problem completely, then there's no .9c anywhere, just two rockets moving apart from each other at .99995c. 


#7
Jan1012, 12:32 AM

P: 863

A slightly different version of Nugatory's rocket launchers is for the rockets to leave at say 100,000 miles/sec, such that the combined speed is 200,000 miles/sec, which is faster than the speed of light.
Only somebody riding the rocket will only see the other rocket moving at 155,000 miles/sec. At very near the speed of light, that Nugatory used, the man on the rocket still only sees the other rocket going very near the speed of light, not more. That's how the addition of velocities work. 


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