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Travelling faster than the speed of light
Double light speed

Lets say that

  • an airplane can fly at 4/5 of the speed of light, and
  • I can run at 2/5 of the speed of light, and
  • I'm in the airplane.

Suddenly I start running towards the cockpit down the middle lane. Now what?

From what I know, the speed of light is not the speed of light just because it is. The speed of light is instead the maximum speed limit because light moves as fast as possible and therefore equals the max speed.
But when it's the maximum speed limit, nothing can ever move faster.

So, I'm running in the airplane... now what?

  • $\begingroup$ The question is very similar to physics.stackexchange.com/q/11398/373 . And the first answer should give you the answer. To make a long story short, speeds don't add up intuitively when close to the speed of light. $\endgroup$ – Frédéric Grosshans Jun 21 '12 at 10:53

This must have been asked before, but the only near duplicate I can find is How to derive addition of velocities without the Lorentz transformation? and this isn't an exact duplicate.

The point is that relativistic velocities can't just be added. In your example let u be the plane's speed, 0.8c, and v be your speed, 0.4c, then the speed a stationary observer sees, w, is given by:

$$ w~=~\dfrac{u+v}{1+uv/c^2} $$

which is about 0.91c.

The other passengers in the plane will see you running at 0.4c, but remember that the aeroplane's time is running more slowly than a stationary observers time. So what looks like 0.4c to the passengers on the plane looks slower to me standing still on the ground.

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  • $\begingroup$ But me in the plane, I don't feel the slowdown or how? Only the stationary man can see it? $\endgroup$ – Steeven Jun 21 '12 at 10:50
  • $\begingroup$ I've updated my answer to address your comment $\endgroup$ – John Rennie Jun 21 '12 at 10:52
  • $\begingroup$ Thanks. That means that it is the time that is experienced differently for each person. $\endgroup$ – Steeven Jun 21 '12 at 10:56
  • $\begingroup$ It is not just time that slows down, but simultaneity that fails, and you can't just talk about it in terms of time slowing down--- this doesn't distinguish between running backwards and running forwards. $\endgroup$ – Ron Maimon Jun 21 '12 at 16:47
  • $\begingroup$ Agreed, but given the question was a fairly basic one I thought trying to introduce the Lorentz transforms would be an equation too far. $\endgroup$ – John Rennie Jun 21 '12 at 17:04

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