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 β.

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,

$\left(\begin{array}{c}{x}^{\prime }\\ c{t}^{\prime }\end{array}\right)=\gamma \left(\begin{array}{cc}1& -\beta \\ -\beta & 1\end{array}\right)\left(\begin{array}{c}x\\ ct\end{array}\right)\left(\begin{array}{c}{x}^{″}\\ t^{″}\end{array}\right)={\gamma }^{\prime }\left(\begin{array}{cc}1& -{\beta }^{\prime }\\ -{\beta }^{\prime }& 1\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ c{t}^{\prime }\end{array}\right)$

where we are defining the β's and γ's as

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

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

$\left(\begin{array}{c}{x}^{″}\\ c{t}^{″}\end{array}\right)=\gamma {\gamma }^{\prime }\left(\begin{array}{cc}1+\beta {\beta }^{\prime }& -(\beta +{\beta }^{\prime })\\ -(\beta +{\beta }^{\prime })& 1+\beta {\beta }^{\prime }\end{array}\right)\left(\begin{array}{c}x\\ ct\end{array}\right)(1)$

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

$\left(\begin{array}{c}{x}^{″}\\ c{t}^{″}\end{array}\right)={\gamma }^{″}\left(\begin{array}{cc}1& -{\beta }^{″}\\ -{\beta }^{″}& 1\end{array}\right)\left(\begin{array}{c}x\\ ct\end{array}\right)(2)\text{ where }\begin{array}{ccc}{\beta }^{″}& =& \frac{U}{c}\\ {\gamma }^{″}& =& \frac{1}{\sqrt{1-{{\beta }^{″}}^2}}\end{array}$

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

$\left(\begin{array}{c}{x}^{″}\\ c{t}^{″}\end{array}\right)=\gamma {\gamma }^{\prime }(1+\beta {\beta }^{\prime })\left(\begin{array}{cc}1& -\frac{\beta +{\beta }^{\prime }}{1+\beta {\beta }^{\prime }}\\ -\frac{\beta +{\beta }^{\prime }}{1+\beta {\beta }^{\prime }}& 1+\end{array}\right)\left(\begin{array}{c}x\\ ct\end{array}\right)$

This will be identical with equation 2 if

${\beta }^{″}=\frac{\beta +{\beta }^{\prime }}{1+\beta {\beta }^{\prime }}\text{ (3a) and }{\gamma }^{″}=\gamma {\gamma }^{\prime }(1+\beta {\beta }^{\prime })\text{ (3b)}$

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

Writing the β'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.