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## How to add velocities in special relativity

As I mentioned in this blogpost, in special relativity any observer will measure the speed of light in a vacuum to be $c$, irrespective of whether the observer is moving towards or away from the source of light. We can think of the speed of light as a cosmic speed limit, nothing can travel faster than it.

But, let us suppose that we have two reference frames $S$ and $S^{\prime}$ moving relative to each other with a speed of $v=0.9c$, 90% of the speed of light. Surely, if someone in frame $S^{\prime}$ fires a high-speed bullet at a speed of $u^{\prime}= 0.6c$, an observer in frame $S$ will think that the bullet is moving away from him at a speed of $u = v + u^{\prime} = 0.9c + 0.6c = 1.5c$, which seemingly violates the comic speed limit.

What have we done wrong?

We cannot simply add velocities, as we would do in Newtonian mechanics. In special relativity we have to use the Lorentz transformations to add velocities. How do we do this? Let us remind ourselves that the Lorentz transformations can be written as The Lorentz transformations to go either from reference frame $S \text{ to } S^{\prime}$, or to go from $S^{\prime} \text{ to } S$.

## Calculating a velocity in two different reference frames

To calculate the velocity $u$ of some object moving with a velocity $u^{\prime}$ in reference frame $S^{\prime}$ we need to use these Lorentz transformations.

We start off by writing $x = \gamma \left( x^{\prime} + vt^{\prime} \right) \text{ (1) }$

and $t = \gamma \left( t^{\prime} + \frac{x^{\prime}v}{c^{2} } \right) \text{ (2) }$

We will now take the derivative of each term, so we have $dx = \gamma \left( dx^{\prime} + vdt^{\prime} \right)$

and $dt = \gamma \left( dt^{\prime} + \frac{dx^{\prime}v}{c^{2} } \right)$

We can now write $dx/dt = u$ (the velocity of the object as seen in frame $S$) as $\frac{dx}{dt} = \frac{ \gamma \left( dx^{\prime} + vdt^{\prime} \right) }{ \gamma \left( dt^{\prime} + \frac{dx^{\prime}v}{c^{2} } \right) }$

The $\gamma$ terms cancel, and dividing each term on the right hand side by $dt^{\prime}$ gives $\frac{dx}{dt} = \frac{ \left( dx^{\prime}/dt^{\prime} + vdt^{\prime}/dt^{\prime} \right) }{ \left( dt^{\prime}/dt^{\prime} + \frac{dx^{\prime}v}{c^{2} dt^{\prime} } \right) } = \frac{ \left( u^{\prime} + v \right) }{ \left( 1 + \frac{ u^{\prime} v}{c^{2} } \right) }$ $\boxed{ u = \frac{ \left( u^{\prime} + v \right) }{ \left( 1 + \frac{ u^{\prime} v}{c^{2} } \right) } }$

where $u^{\prime}$ was the velocity of the object in reference frame $S^{\prime}$.

Going back to our example of $v = 0.9c$ and $u^{\prime} = 0.6c$, we can see that the velocity $u$ as measured by an observer in reference frame $S$ will be $u = \frac{ 0.6c + 0.9c }{ \left( 1 + \frac{ (0.6c \times 0.9c) }{ c^{2} } \right) } = \frac{ 1.5c }{ 1 + 0.54 } = \frac{ 1.5c }{ 1.54} = \boxed {0.974c}$, not $1.5c$ as we naively calculated.

## The constancy of the speed of light

What happens if a person in reference frame $S^{\prime}$ shines a light in the same direction as $S^{\prime}$ is moving away from $S$? In this case, $u^{\prime}=1.0c$. Putting this into our equation for $u$ we get $u = \frac{ 0.6c + 1.0c }{ \left( 1 + \frac{ (0.6c \times 1.0c) }{ c^{2} } \right) } = \frac{ 1.6c }{ 1 + 0.6 } = \frac{ 1.6c }{ 1.6} = 1.0c$

So they both agree that the light is moving away from them with the same speed $c$!

### 9 Responses

1. on 20/12/2018 at 02:54 | Reply hiroji kurihara

Speed of Light

On everything of a light ray before incoming (on wavelength, amplitude, waveform, etc and on these varying), observer’s motion has not any effect. So, in the formula c = f λ, f and c vary.

http://www.geocities.co.jp/Technopolis/2561/eng.html

Sorry, I cannot receive E-mail. I do not have PC.

3. on 01/05/2019 at 06:08 | Reply Hiroji kurihara

Lorentz contraction

Plain waves of light (wavelength is constant) are coming from the upper right 45 degrees. Two bars of the same length are moving to the right and the left at the same speed. The number of waves hitting the bars is the same. Lorentz contraction is unthinkable. And also, time dilation will be denied.

• on 20/07/2019 at 05:01 | Reply Hiroji Kurihara

Constancy of speed of light he

4. on 11/07/2019 at 20:50 | Reply Hiroji Kurihara

Lorentz contraction

In a moving passenger car, MM experiment is being done. Between two light pathes diverged by a half mirror, there is considerable difference in length. Lorentz contraction will not stand up.

5. on 11/07/2019 at 21:06 | Reply Hiroji Kurihara

Sorry, I made a mistake. URL of my web site is below.

http://lifeafterdeath.vip/eng.html

6. on 20/07/2019 at 05:04 | Reply Hiroji Kurihara

Constancy of speed of light

They say, it stands up on an observer in every inertial frame. Yes, when the light source shines in that frame, it is true.

Some man mistook this fact natural for a great discovery. And it is believed widely.

7. on 07/11/2019 at 06:18 | Reply Hiroji kurihara

Basis of special relativity

We seem to measure c by the light source situated on the same inertial frame. A web site says, reasonable basis of the cconstancy of c cannot be found in web (with three words).

8. on 15/11/2019 at 20:34 | Reply Hiroji kurihara

Reexamination of propagation of light

In outer space, a mirror is reflecting star light ray. Speed of reflected light relative the mirror is constant. Speed of incident light relative to the mirror is not constant (the latter is constant relative to the aether).