# Theory of relativity/Special relativity/momentum

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This article will derive the relativity-correct formula for momentum on theoretical grounds, using the Lorentz transform of Special Relativity and the requirement that momentum be conserved in all frames of reference.

The thought experiment is a simple collision between two identical particles, A and B. The collision is perfectly elastic, that is, energy is conserved. The particles have the same mass, and their speeds, before and after the collision, are the same. By setting up the collision in this way, we don't have to be concerned with issues such as mass change or energy change, that is, E = mc².

The collision as seen in the primed frame.

The collision takes place at x' = y' = t' = 0 in the primed frame of reference. The x and y components of the speed are ±r and ±s, respectively. The equations of the particles' motion in the primed frame are:

A before collision (t' < 0): ${\displaystyle x'=rt'\qquad y'=st'}$
A after collision (t' > 0): ${\displaystyle x'=rt'\qquad \ y'=-st'}$
B before collision (t' < 0): ${\displaystyle x'=-rt'\quad \ y'=-st'}$
B after collision (t' > 0): ${\displaystyle x'=-rt'\qquad y'=st'}$

Now choose an unprimed frame of reference moving to the left with speed r. This will cancel out the horizontal motion of particle B. The Lorentz transforms both primed-to-unprimed and unprimed-to-primed, are

${\displaystyle x={\frac {x'+rt'}{\sqrt {1-r^{2}/c^{2}}}}\qquad \qquad x'={\frac {x-rt}{\sqrt {1-r^{2}/c^{2}}}}}$
${\displaystyle y=y'\qquad \qquad \qquad \qquad y'=y}$
${\displaystyle t={\frac {t'+rx'/c^{2}}{\sqrt {1-r^{2}/c^{2}}}}\qquad \qquad t'={\frac {t-rx/c^{2}}{\sqrt {1-r^{2}/c^{2}}}}}$

Solving these, we get, in the unprimed frame:

A before collision (t < 0): ${\displaystyle x={\frac {2rt}{1+r^{2}/c^{2}}}\qquad y={\frac {st{\sqrt {1-r^{2}/c^{2}}}}{1+r^{2}/c^{2}}}}$
A after collision (t > 0): ${\displaystyle x={\frac {2rt}{1+r^{2}/c^{2}}}\qquad \ y={\frac {-st{\sqrt {1-r^{2}/c^{2}}}}{1+r^{2}/c^{2}}}}$
B before collision (t < 0): ${\displaystyle x=0\qquad y={\frac {-st}{1+r^{2}/c^{2}}}}$
B after collision (t > 0): ${\displaystyle x=0\qquad \ y={\frac {st}{1+r^{2}/c^{2}}}}$

We now analyze the collision in the unprimed frame. In this frame, B has no horizontal speed. It goes straight down, collides, and comes straight back up.

The collision as seen in the unprimed frame.

A's speed is the square root of the sum of the squares of the x and y components of its speed:

${\displaystyle A_{\textrm {speed}}={\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}}$

B's speed is simpler:

${\displaystyle B_{\textrm {speed}}={\frac {s}{\sqrt {1-r^{2}/c^{2}}}}}$

We know that the momentum vector must point in the same direction as the particle's motion, and its magnitude must be some function of the mass and speed. It must also be exactly proportional to the mass—two protons have twice the momentum of one. Also, in the non-relativistic limit, the momentum is given by:

${\displaystyle {\vec {p}}=m{\vec {v}}}$

So we set the momentum formula to:

${\displaystyle {\vec {p}}=m{\vec {v}}f(v)}$

where f is some function of the (magnitude of the) momentum.

Sneak preview: Those who have seen the formula know that ${\displaystyle f(x)={\frac {1}{\sqrt {1-x^{2}/c^{2}}}}}$. The goal of this article is to derive that.

Using this formula with the unknown function f, we have:

${\displaystyle A_{\textrm {x\ momentum,\ before\ or\ after}}={\frac {2mr}{1+r^{2}/c^{2}}}\ \ f\left({\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}\right)}$
${\displaystyle A_{\textrm {y\ momentum,\ before}}={\frac {ms{\sqrt {1-r^{2}/c^{2}}}}{1+r^{2}/c^{2}}}\ \ f\left({\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}\right)}$
${\displaystyle A_{\textrm {y\ momentum,\ after}}={\frac {-ms{\sqrt {1-r^{2}/c^{2}}}}{1+r^{2}/c^{2}}}\ \ f\left({\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}\right)}$
${\displaystyle B_{\textrm {x\ momentum,\ before\ or\ after}}=0\,}$
${\displaystyle B_{\textrm {y\ momentum,\ before}}={\frac {-ms}{\sqrt {1-r^{2}/c^{2}}}}\ \ f\left({\frac {s}{\sqrt {1-r^{2}/c^{2}}}}\right)}$
${\displaystyle B_{\textrm {y\ momentum,\ after}}={\frac {ms}{\sqrt {1-r^{2}/c^{2}}}}\ \ f\left({\frac {s}{\sqrt {1-r^{2}/c^{2}}}}\right)}$

It is clear that the x component of the momentum is conserved:

${\displaystyle A_{\textrm {x,\ before}}+B_{\textrm {x,\ before}}=A_{\textrm {x,\ after}}+B_{\textrm {x,\ after}}}$

The y component needs to be solved carefully:

${\displaystyle A_{\textrm {y,\ before}}+B_{\textrm {y,\ before}}=A_{\textrm {y,\ after}}+B_{\textrm {y,\ after}}}$

Collecting terms in the above line, and dividing by 2ms:

${\displaystyle {\frac {\sqrt {1-r^{2}/c^{2}}}{1+r^{2}/c^{2}}}\ \ f\left({\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}\right)\ =\ {\frac {1}{\sqrt {1-r^{2}/c^{2}}}}\ \ f\left({\frac {s}{\sqrt {1-r^{2}/c^{2}}}}\right)}$

or:

${\displaystyle f\left({\frac {\sqrt {4r^{2}+s^{2}-r^{2}s^{2}/c^{2}}}{1+r^{2}/c^{2}}}\right)\ =\ {\frac {1+r^{2}/c^{2}}{1-r^{2}/c^{2}}}\ \ f\left({\frac {s}{\sqrt {1-r^{2}/c^{2}}}}\right)}$

This is a rather formidable equation, involving two parameters, r and s, from which to deduce the function f.

We can simplify this by letting s approach zero:

${\displaystyle f\left({\frac {2r}{1+r^{2}/c^{2}}}\right)\ =\ {\frac {1+r^{2}/c^{2}}{1-r^{2}/c^{2}}}\ \ f(0)}$

Since we have hypothesized that ${\displaystyle f(0)=1\,}$, that is, the momentum approaches ${\displaystyle {\vec {p}}=m{\vec {v}}}$ in the classical limit:

${\displaystyle f\left({\frac {2r}{1+r^{2}/c^{2}}}\right)\ =\ {\frac {1+r^{2}/c^{2}}{1-r^{2}/c^{2}}}}$

Now let:

${\displaystyle z={\frac {2r}{1+r^{2}/c^{2}}}}$

${\displaystyle r={\frac {c^{2}}{z}}(1-{\sqrt {1-z^{2}/c^{2}}})}$

Then:

${\displaystyle 1+{\frac {r^{2}}{c^{2}}}={\frac {2c^{2}}{z^{2}}}(1-{\sqrt {1-z^{2}/c^{2}}})}$
${\displaystyle 1-{\frac {r^{2}}{c^{2}}}={\frac {2c^{2}}{z^{2}}}(z^{2}/c^{2}-1+{\sqrt {1-z^{2}/c^{2}}})}$
${\displaystyle f(z)={\frac {1-{\sqrt {1-z^{2}/c^{2}}}}{{\sqrt {1-z^{2}/c^{2}}}\ -\ (1-z^{2}/c^{2})}}}$
${\displaystyle \ \ ={\frac {1-{\sqrt {1-z^{2}/c^{2}}}}{{\sqrt {1-z^{2}/c^{2}}}\ (1-{\sqrt {1-z^{2}/c^{2}}})}}}$
${\displaystyle \ \ ={\frac {1}{\sqrt {1-z^{2}/c^{2}}}}}$

So the relativistically correct equation for momentum is:

${\displaystyle {\vec {p}}={\frac {m{\vec {v}}}{\sqrt {1-v^{2}/c^{2}}}}}$
The Lorentz factor as a function of speed, from 0 to the speed of light.

The correction factor that we have derived is called the Lorentz factor, typically denoted with a lowercase gamma.

${\displaystyle \gamma ={\frac {1}{\sqrt {1-v^{2}/c^{2}}}}}$

This same function appears prominently in the Lorentz transform. It determines the length contraction and time dilation. For non-relativistic speeds, it is nearly 1. As the speed approaches the speed of light, it grows without limit. At v=0.99c, it is about 7.

This means that an object's momentum grows unboundedly large as its speed approaches c.

The next article in this series is Special relativity/energy.