If two bodies are travelling at speed $0.9 \, c$ in opposite directions, what will be the speed of one, as observed by another? Newtonian mechanics won't apply at such speeds. As such, how will we calculate the actual speed? The speeds can obviously not cross $c$. How to intuitively understand this?
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1http://en.wikipedia.org/wiki/Velocity-addition_formula – pfnuesel Nov 05 '14 at 10:38
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2There are many related questions already on the site. I've suggested what looks the most obvious duplicate. – John Rennie Nov 05 '14 at 10:42
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Related: http://physics.stackexchange.com/q/23625/2451 – Qmechanic Nov 05 '14 at 11:46
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The theory of relativity tells us what is the answer to this question. See Wikipedia, as John Rennie recommends,
http://en.wikipedia.org/wiki/Velocity-addition_formula#Special_case:_parallel_velocities
If in the formula
$$v_{rel} = \frac{v_1 + v_2}{1 + v_1 v_2/c^2}$$
you set $v_1 = v_2 = 0.9 \, c$, you get $v_{rel} = 1.8/1.81$, i.e. slightly less than $c$. But if you set $v_1 = v_2 = c$, you get $v_{rel} = c$.
To your attention, recent works on the gravity theory show that the velocity of light is a function of the configuration of the universe. In the early universe the value of c was probably different of what is today, (see in the arXiv quant-ph the works of Asher Peres and/or Danny Terno). Still, whatever is its value, the light velocity cannot be exceeded.
With pleasure, Sofia
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1"In the early universe the value of c was probably different of what is today" - is this widely accepted? – pfnuesel Nov 05 '14 at 14:14
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@pfnuesel: No, but there are some theories on the variable speed of light. – Kyle Kanos Nov 05 '14 at 14:24