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Imagine a stationary person (with respect to a spaceship) floating in space and looks at a clock on the spaceship going 0.5c and sees it's clock ticking slower than his own, and concludes that it is moving closer to the speed of light than himself...

I learned recently that people on the spaceship will actually see the 'stationary' person's clock ticking not faster, but slower also, with respect to the spaceship's clock, which would make the spaceship person conclude that his spaceship was stationary and the floating man was moving the opposite direction at 0.5c.

My question is this... when physics says we can't move at the speed of light, I ask, with respect to who? The floating man calculates that he is moving at 0c and can accelerate all the way to 1c. But the person on the spaceship is moving, but concludes that he is stationary, so he calculates that his speed is 0c and he can accelerate all the way to 1c. If the man in the spaceship accelerates all the way to 0.99c with respect to his own reference frame, then the stationary person would see him accelerating from 0.5c to 1.499c, faster than light.

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    $\begingroup$ You can't reach the speed of light relative to everybody. Search for "relativistic velocity addition" on this site. $\endgroup$ – CuriousOne Jun 2 '16 at 11:50
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    $\begingroup$ I believe @curiousone probably meant to say "anybody", not "everybody". $\endgroup$ – WillO Jun 2 '16 at 12:07
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    $\begingroup$ Or, to explain @CuriousOne's pithy and witty "relative to everybody" (LOL I really like it) and what it has to do with his suggestion: relativistic velocity addition is not linear as you suggest (i.e. you don't simply add velocities like you have done) but its nonlinearity is such that $c$ is the same for all observers. You move at $0.5\,c$ relative to me, but we will both conclude from out independent measurements that any light beam moves at $c$ (although we will in general disagree about the direction - this latter phenomenon is called "relativistic aberration"). $\endgroup$ – WetSavannaAnimal Jun 2 '16 at 12:10
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    $\begingroup$ To set the record about my English straight: it's not my first language and I do frequently mistake any- and everybody. :-) In this case I initially wrote "anybody" and then I decided to go with the grammatically incorrect (right?) "everybody" to make a stronger statement. @WetSavannaAnimalakaRodVance guessed my intentions correctly, but I don't mind being called out on it. I appreciate that somebody likes it. :-) $\endgroup$ – CuriousOne Jun 2 '16 at 20:47
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    $\begingroup$ @WillO: Maybe I should have put "everybody" in quotation marks? I think that would have made a proper improper statement, right? $\endgroup$ – CuriousOne Jun 2 '16 at 20:48
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To answer this, You can't add velocities as in Newtonian mechanics. In relativity, If frame B is moving at velocity u and C is moving at velocity v both with respect to an inertial frame A, then relative velocity between B and C is $\frac{u+v}{1+ uv/c^2}$.

So substituting your arguments here would now leave us with relative speed less than c. Just because in Newtonian mechanics we deal with low velocity we ignore the denominator.

Note: Also if you substitute both u and v to a number close to c the expression is still less than c, this might give you an idea that it impossible to reach the speed of light.

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"The stationary person would see him accelerating from 0.5c to 1.499c, faster than light" The statement itself is a large mistake... The person in the space ship knows that he had accelerated from 0.5 to 0.99 time the speed of light, but the person outside would only see him accelerating from 0.5c to 0.99665551839c. It is because relative velocity is given by (v+u)/(1+uv/c^2). This shall be proved in STOR.

With respect to any frame in the universe, maximum time of interaction between two events is distance divided by the speed of light i.e, maximum speed of interaction is speed of light.

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