# 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,

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

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

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

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

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

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

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

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

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

This will be identical with equation 2 if

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

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

Writing the β's in terms of the velocities equation 3a becomes

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

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.

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

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.