Two objects moving away from eachother at c - Special relativity

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Two rockets moving away from each other at 0.95c do not exceed the speed of light in any reference frame due to the principles of special relativity. When calculating the speed of one rocket from the reference frame of the other, the relativistic velocity addition formula must be used, resulting in a speed of 0.9987c, which is still below c. Time dilation affects the perception of time for each rocket, but neither rocket experiences time travel or changes in size in the way described. Light traveling at c in a vacuum means that the rockets cannot see each other if they are moving away at relativistic speeds. Understanding these concepts clarifies the behavior of objects in high-speed motion according to special relativity.
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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