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Suppose I am in a spaceship. In this spaceship there is also my friend. So my friend and my relative velocity is zero. So no matter how fast my ship goes my friend's velocity relative to me is zero($0$). So in this logic my ship can go as fast as I like (even greater than the velocity of light).

But a viewer from outside the ship will see the ship going in a velocity. So we won't be able to go in a velocity greater than light. This is of course wrong. Where am I wrong?

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    $\begingroup$ It's not exactly clear what your issue is. In your frame, the ship has zero velocity. And no inertial observer will measure the ship to have a speed $\ge c$. $\endgroup$ – PM 2Ring Jan 7 at 9:38
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So in this logic my ship can go as fast as I like (even greater than the velocity of light)

First of all, nothing goes faster than the speed of light. You should measure the speed of the spaceship to be zero, because you are in it. Everything outside the ship goes at $c$.

In relativity, there is a unique way of calculating relative velocities. The person (in frame $S$ observing you and your spaceship will say that you are moving at the speed of light (the same speed as your spaceship - and he/she would also say that your relative velocity with respect to the spaceship is zero.

$ Vrel = v+c/(1+ vc/c^2)$ = $c$. You (in frame $S'$) may say that you are moving with respect to the spaceship with velocity $v$. But the observer will disagree and say that you are not moving at all with respect to the spaceship.

Get familiar with the Lorentz transformations. They are a great way to find solutions to these kind of problems.

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    $\begingroup$ "Everything outside the ship goes at $c$". What?! $\endgroup$ – PM 2Ring Jan 7 at 11:07
  • $\begingroup$ @PM2Ring in his example, the spaceship goes at velocity $c$. $\endgroup$ – KV18 Jan 7 at 12:38
  • $\begingroup$ Not exactly. They haven't stated a specific speed. The OP says: "my ship can go as fast as I like (even greater than the velocity of light)". $\endgroup$ – PM 2Ring Jan 7 at 12:47
  • $\begingroup$ Ya. Well,I know that specific line did not make sense. I guess it's because he confused between his and the observer's frame of reference. $\endgroup$ – KV18 Jan 7 at 12:53

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