Two objects moving away from eachother at c - Special relativity

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SUMMARY

This discussion clarifies the relativistic effects of two rockets moving away from each other at speeds approaching the speed of light (0.95c). It emphasizes that, in the reference frame of one rocket, the other rocket does not exceed the speed of light; instead, its speed is calculated using the relativistic velocity addition formula: v' = (u + v) / (1 + (uv/c²)), resulting in a speed of 0.9987c. The conversation also touches on concepts like time dilation and the implications of observing each other at such high speeds, confirming that neither rocket experiences time travel or size alteration in the conventional sense.

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  • Understanding of special relativity principles
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vorcil
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While I've got all the concepts of time dilation,the twin paradox and length contraction in my head, I still can't get my head around the fact of the following scenario,

Just imagine two rockets in space, moving away from each other in opposite directions at 0.95c or even just c,

in the reference frame of one rocket, wouldn't the other rocket be moving at a speed greater than the speed of light?

what is happening to the opposite moving rocket in one rockets reference frame?
is it getting younger? because of time dilation, wouldn't it turn out to be going back in time? and getting larger instead of contracting(or is it stretching negatively?)

help thanks
 
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another thing, light travels at c in a vaccum, so assuming the rockets are traveling in a vacuum also, they won't be able to see each other because in all of the reference frames, they are traveling faster than c"?
 
One thing at a time:
vorcil said:
Just imagine two rockets in space, moving away from each other in opposite directions at 0.95c or even just c,

in the reference frame of one rocket, wouldn't the other rocket be moving at a speed greater than the speed of light?
No, it wouldn't, and here's why: first, those .95c speeds you're talking about are measured from a certain reference frame. Let's say that's the Earth's reference frame. But to find the speed of rocket 2 in the reference frame of rocket 1 (which is what you're asking about), you have to "boost" yourself from the Earth's reference frame to the rocket 1 reference frame, and when you do that, the velocity of rocket 2 changes like so:
v' = \frac{u+v}{1+\frac{uv}{c^2}}
It's not just simple addition, like you might think. So .95c + .95c is wrong. When you plug into the proper formula, you get
v' = \frac{.95c+.95c}{1+\frac{(.95c)(.95c)}{c^2}} = 0.9987c
So in the reference frame of rocket 1, rocket 2 is moving at 0.9987c, which is close to but still less than the speed of light.
 
diazona said:
One thing at a time:

No, it wouldn't, and here's why: first, those .95c speeds you're talking about are measured from a certain reference frame. Let's say that's the Earth's reference frame. But to find the speed of rocket 2 in the reference frame of rocket 1 (which is what you're asking about), you have to "boost" yourself from the Earth's reference frame to the rocket 1 reference frame, and when you do that, the velocity of rocket 2 changes like so:
v' = \frac{u+v}{1+\frac{uv}{c^2}}
It's not just simple addition, like you might think. So .95c + .95c is wrong. When you plug into the proper formula, you get
v' = \frac{.95c+.95c}{1+\frac{(.95c)(.95c)}{c^2}} = 0.9987c
So in the reference frame of rocket 1, rocket 2 is moving at 0.9987c, which is close to but still less than the speed of light.

Oh,

cheers
 

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