# Speed of light and relativity [duplicate]

Suppose there are two observers $A$ & $B$ both are in motion, now $A$ sees $B$ is moving with speed $'u'$. A says that another object $'d'$ is moving with speed $c+u$ w.r.t. A in the same direction as $B$. What stops him from saying this ?

The formula $\frac {u+v}{1+\frac{uv}{c^2}}$ is such that $'v'$ is speed of the object $'d'$ w.r.t.$B$ and not $A.$ and $u$ is the speed of $B$ w.r.t. $A$, now even though $B$ also sees that the object $'d'$ moves faster than the speed of light. He can still always see light to travel with the speed of light and also he can be an inertial observer?

What is the problem here?

## marked as duplicate by jinawee, tpg2114♦, Waffle's Crazy Peanut, John Rennie, Brandon EnrightDec 20 '13 at 18:40

• I don't know if I understood your question correctly, but what I think you are saying is the example where we can see two galaxies moving away from each other at a speed higher than c. There is no problem with that. What we cannot observe is something moving away at a speed higher than c w.r.t us. – cinico Dec 20 '13 at 11:53
• @cinico Can you give me the proof why two bodies can't have relative velocity greater than $c$. – Isomorphic Dec 20 '13 at 11:54
• – jinawee Dec 20 '13 at 11:57
• Your formulation is inconsistent. You start denoting the objects by capital letters A and B, use small quoted italics for speed $'u'$ and, afterwards, say that object $'d'$ has speed $c+u$. Do you ever know what consistency ever is? – Val Dec 20 '13 at 12:00
• @iota: two bodies can have a relative velocity greater than $c$ as viewed from some other inertial frame. They can just never see a velocity greater than $c$ relative to their own inertial frame. – John Rennie Dec 20 '13 at 12:02

The situation in which A observes some object moving at a velocity $c+u$ cannot occur within the framework of special relativity, therefore your scenario is meaningless. The theory puts an upper speed limit on observable motion, which cannot even be reached by massive bodies, but only massless ones. This limit is given precisely by $c$. Hence, assuming that $u$ is positive, nothing can move at $c+u$.

• I just showed that in this case the two axioms of SR aren't violated here – Isomorphic Dec 20 '13 at 11:52
• I wonder how you did that. As a simple example for why it does not make sense: take the factor gamma (which occurs in Lorentz transformations) and insert a velocity greater than c. You will end up with complex expressions, which does not make much sense. – Frederic Brünner Dec 20 '13 at 11:56
• Take $c=2$ in some units and $u=1$ in those units. and take c+u=3. and see if v comes out to be real or complex. – Isomorphic Dec 20 '13 at 12:06
• That is not what I was talking about. The formula for relativistic addition of velocity is derived under the assumption that Lorentz transformations are used to change between different inertial systems. But these transformations are not defined for velocities larger than c. – Frederic Brünner Dec 20 '13 at 12:10
• The concept of an object moving faster than the speed of light is not defined. The question does not make sense. – Frederic Brünner Dec 20 '13 at 12:15

As per your question we'll assume that $u + v > c$, though individually $u$ and $v$ must be less than $c$ because we can never observe speeds faster than light.
Now, you are correct to say that from the perspective of A the relative velocity of B and C is greater than light. But when we talk about $c$ being the fastest speed possible we mean the fastest speed relative to the observer i.e. in this case relative to A.
We know B will see A moving at the speed $u$, but the question is what speed will B see C moving? Will B see C moving at a speed, $w$, faster than light? The answer is no, because this is where the equation for combining relativistic speeds applies. The speed $w$ is given by the formula you quote:
$$w = \frac {u+v}{1+\frac{uv}{c^2}}$$
and this speed is always less than $c$.