Modern Physics/Addition of Velocities

In classical physics, velocities simply add. If an object moves with speed u in one reference frame, which is itself moving at v with respect to a second frame, the object moves at speed u+v in that second frame.

This is inconsistant with relativity because it predicts that if the speed of light is c in the first frame it will be v+c in the second.

We need to find an alternative formula for combining velocities. We can do this with the Lorentz transform.

Because the factor v/c will keep recurring we shall call that ratio &beta;.

We are considering three frames; frame O, frame O' which moves at speed u with respect to frame O, and frame O" which moves at speed v with respect to frame O'.

We want to know the speed of O" with respect to frame O,U which would classically be u+v.

The transforms from O to O' and O' to O" can be written as matrix equations,



\begin{pmatrix} x' \\ ct' \end{pmatrix} = \gamma \begin{pmatrix} 1 & - \beta \\ -\beta & 1 \end{pmatrix} \begin{pmatrix} x \\ ct \end{pmatrix} \quad \begin{pmatrix} x \\ t \end{pmatrix} = \gamma' \begin{pmatrix} 1 & - \beta' \\ -\beta' & 1 \end{pmatrix} \begin{pmatrix} x' \\ ct' \end{pmatrix} $$

where we are defining the &beta;'s and &gamma;'s as


 * $$\begin{matrix}

\beta = \frac{u}{c} & \gamma = \frac{1}{\sqrt{1-\beta^2 }} \\ \beta^\prime = \frac{v}{c} & \gamma^\prime = \frac{1}{\sqrt{1-{\beta^\prime}^2 }} \end{matrix} $$

We can combine these to get the relationship between the O and O" coordinates simply by multiplying the matrices, giving



\begin{pmatrix} x \\ ct \end{pmatrix} = \gamma \gamma^\prime \begin{pmatrix} 1+\beta \beta' & - (\beta + \beta') \\ - (\beta + \beta') & 1+\beta \beta'  \end{pmatrix} \begin{pmatrix} x \\ ct \end{pmatrix} \quad (1)$$

This should be the same as the Lorentz transform between the two frames,



\begin{pmatrix} x \\ ct \end{pmatrix} = \gamma \begin{pmatrix} 1 & - \beta \\ -\beta'' & 1 \end{pmatrix} \begin{pmatrix} x \\ ct \end{pmatrix} \quad (2) \mbox{ where } \begin{matrix} \beta'' & = & \frac{U}{c} \\ \gamma & = & \frac{1}{\sqrt{1-{\beta}^2 }} \end{matrix} $$

These two sets of equations do look similar. We can make them look more similar still by taking a factor of 1+&beta;&beta;' out of the matrix in (1) giving#



\begin{pmatrix} x \\ ct \end{pmatrix} = \gamma \gamma' (1+\beta \beta') \begin{pmatrix} 1 & - \frac{\beta + \beta'}{1+\beta \beta'} \\ - \frac{\beta + \beta'}{1+\beta \beta'} & 1+  \end{pmatrix} \begin{pmatrix} x \\ ct \end{pmatrix} $$

This will be identical with equation 2 if


 * $$\beta''=\frac{\beta + \beta'}{1+\beta \beta'} \mbox{ (3a)  and }

\gamma'' = \gamma \gamma' (1+\beta \beta') \mbox{ (3b)}$$

Since the two equations must give identical results, we know these conditions must be true.

Writing the &beta;'s in terms of the velocities equation 3a becomes


 * $$\frac{U}{c}=\frac{\frac{u}{c} + \frac{v}{c}}{1+\frac{uv}{c^2}}$$

which tells us U in terms of u and v.

A little algebra shows that this implies equation 3b is also true

Multiplying by c we can finally write.

$$U = \frac{u+v}{1+\frac{uv}{c^2}} $$

Notice that if u or v is much smaller than c the denominator is approximately 1, and the velocities approximately add but if either u or v is c then so is U, just as we expected.