# Relativistic Momentum

This page gives the relativistic definition of linear momentum and compares it to the traditional definition of linear momentum.

## The Main Idea

The Linear Momentum of an object is traditionally defined as [math]\displaystyle{ \vec{p} = m \vec{v} }[/math]. Momentum is a useful quantity because many other concepts relate to it. For example, forces cause the momentum of a system to change according to Newton's Second Law: the Momentum Principle, and in certain situations, the momentum of a system is conserved. However, for systems containing objects moving at speeds comparable to the speed of light, both Newton's second law and conservation of momentum appear to be violated. As it turns out, if the definition of the momentum of a particle is adjusted to accommodate high-speed objects, these relations hold true after all. The adjusted definition of momentum is called the relativistic definition and defines momentum as follows:

[math]\displaystyle{ \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} }[/math]

where [math]\displaystyle{ \vec{p} }[/math] is the momentum of the particle, [math]\displaystyle{ m }[/math] is mass, [math]\displaystyle{ \vec{v} }[/math] is the velocity of the particle, [math]\displaystyle{ v }[/math] is the magnitude of the velocity (the speed of the particle), and [math]\displaystyle{ c }[/math] is the speed of light (about [math]\displaystyle{ 3 * 10^8 }[/math] m/s).

The factor [math]\displaystyle{ \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} }[/math] appears in many other relativistic formulas, so it has been given a name (the Lorentz factor) and a symbol ([math]\displaystyle{ \gamma }[/math]). This turns the relativistic momentum formula into

[math]\displaystyle{ \vec{p} = \gamma m \vec{v} }[/math].

The Lorentz factor is a ratio and therefore has no units.

Note: sometimes, particularly in older textbooks, the factor [math]\displaystyle{ \gamma * m }[/math] is considered the "relativistic mass" of the particle. This would reduce the relativistic momentum formula to [math]\displaystyle{ \vec{p} = m_{rel} * \vec{v} }[/math]. However, this way of thinking is no longer favored, since the mass of a moving particle cannot be measured independently from its momentum, and there is therefore no evidence to suggest that a particle's mass increases as its speed approaches the speed of light. It is recommended that you use the formula [math]\displaystyle{ \vec{p} = \gamma m \vec{v} }[/math], keeping [math]\displaystyle{ \gamma }[/math] and [math]\displaystyle{ m }[/math] separate and using the particle's rest mass (traditional notion of mass) for m.

### A Mathematical Model

Let us mathematically analyze the equations

[math]\displaystyle{ \vec{p} = \gamma m \vec{v} }[/math]

and

[math]\displaystyle{ \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} }[/math].

One observation worth noting is that in the limit where the speed of the particle [math]\displaystyle{ v }[/math] is much smaller than the speed of light [math]\displaystyle{ c }[/math], the fraction [math]\displaystyle{ \frac{v^2}{c^2} }[/math] is approximately 0 and so [math]\displaystyle{ \gamma }[/math] is approximately 1. This explains why the traditional equation [math]\displaystyle{ \vec{p} = m \vec{v} }[/math] works very well for objects moving at everyday speeds; the relativistic definition and the traditional definition agree when [math]\displaystyle{ \gamma }[/math] is 1. The relativistic definition should be used if [math]\displaystyle{ v \gt \frac{1}{10} c }[/math]; for slower particles, the error caused by using the traditional formula is negligible.

Another observation worth noting is that in the limit where the speed of the particle [math]\displaystyle{ v }[/math] approaches the speed of light [math]\displaystyle{ c }[/math], the fraction [math]\displaystyle{ \frac{v^2}{c^2} }[/math] approaches 1 and so [math]\displaystyle{ \gamma }[/math] approaches infinity. This has far-reaching implications. An object of non-zero mass traveling at the speed of light would have infinite momentum according to the relativistic definition, so an infinite impulse must be applied to a particle at rest to change its speed to the speed of light. As real forces are always finite in both magnitude and duration, no particle of nonzero mass can ever achieve the speed of light. Let us analyze what happens if a constant force is applied to a particle at rest for a long time. The particle's momentum will increase linearly over time as a result of the force, as described by Newton's second law. At first, this means its speed increases linearly over time as well. However, as the particle enters the relativistic regime, [math]\displaystyle{ \gamma }[/math] begins to increase as well, meaning that the particle's speed doesn't have to increase as much to sustain the steady increase in momentum. Finally, when the particle's speed is nearly the speed of light, even the tiniest increases in [math]\displaystyle{ v }[/math] cause [math]\displaystyle{ \gamma }[/math] to skyrocket, meaning that the particle hardly speeds up while the particle's momentum continues to steadily increase. The speed of the particle would asymptotically approach the speed of light over time if the force continues to be applied.

## Examples

### Simple

Suppose that a proton (mass = [math]\displaystyle{ 1.7 * 10^{-27} }[/math] kg) is moving with a speed of [math]\displaystyle{ .97c }[/math]

What is the magnitude of the momentum of the proton?

Solution:

- [math]\displaystyle{ \frac{v}{c} = \frac{.97c}{c} }[/math] [math]\displaystyle{ = .97 }[/math]
- [math]\displaystyle{ \gamma = \frac{1}{\sqrt{1-(.97)^2}} = 4.1135 }[/math]
- Plug values in

- [math]\displaystyle{ p = \gamma * m*v = (4.1135)*(1.7*10^{-27})*(.97*(3*10^8)) = 2.035 * 10^{-18} }[/math] kg*m/s

### Middling

Suppose that a proton (mass = [math]\displaystyle{ 1.7 * 10^{-27} }[/math] kg) is moving with a velocity [math]\displaystyle{ \lt 1 * 10^7 , 2 * 10^7 , 3 * 10^7\gt }[/math] m/s.

What is the momentum of the proton?

Solution:

- [math]\displaystyle{ \left\vert \overrightarrow{v} \right\vert = \sqrt{(1*10^7)^2 + (2*10^7)^2 +(3*10^7)^2} }[/math] m/s [math]\displaystyle{ = 3.7 * 10^7 }[/math] m/s.
- [math]\displaystyle{ \frac{\left\vert \overrightarrow{v} \right\vert}{c} = \frac{3.7 * 10^7 m/s}{3 * 10^8 m/s} }[/math] [math]\displaystyle{ = .12 }[/math]
- [math]\displaystyle{ \gamma = \frac{1}{\sqrt{1-(.12)^2}} = 1.007 }[/math]
- [math]\displaystyle{ \overrightarrow{p} = \gamma * m * \overrightarrow{v} }[/math]
- Plug values in

- [math]\displaystyle{ \overrightarrow{p} = (1.007)*(1.7*10^{-27})*\lt 1 * 10^7 , 2 * 10^7 , 3 * 10^7\gt = \lt 1.7 * 10^{-20}, 3.4 * 10^{-20}, 5.1 *10^{-20}\gt }[/math] kg*m/s

### Difficult

Suppose that a proton (mass = [math]\displaystyle{ 1.7 * 10^{-27} }[/math] kg) is moving at a speed of [math]\displaystyle{ .95c }[/math]. A force of magnitude [math]\displaystyle{ 2 * 10^{-17} }[/math]N acts on it for .5s in the direction of the proton's velocity, causing it to speed up. What is its new speed at the end of the .5s time interval? Use [math]\displaystyle{ 3*10^8 }[/math] for [math]\displaystyle{ c }[/math].

Solution:

Let us first find the initial momentum of the particle:

[math]\displaystyle{ \vec{p} = \gamma m \vec{v} }[/math]

[math]\displaystyle{ \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-.95^2}} = 3.203 }[/math]

[math]\displaystyle{ \vec{p} = 3.203 * 1.7 * 10^{-27} * (.95 * 3 * 10^8) = 1.55 * 10^{-18} }[/math] kg*m/s

Now let us find the momentum at the end of the impulse using the impulse-momentum theorem.

[math]\displaystyle{ \vec{p}_f = \vec{p}_i + \vec{J} }[/math]

Since the impulse is in the same direction as the velocity, this vector addition becomes scalar (regular) addition.

[math]\displaystyle{ \vec{p}_f = 1.55 * 10^{-18} + .5 * 2 * 10^{-17} = 1.155 * 10^{-17} }[/math] kg*m/s

Now let us solve for the final speed using

[math]\displaystyle{ \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} }[/math]

[math]\displaystyle{ 1.155 * 10^{-17} = \frac{1}{\sqrt{1-\frac{v^2}{(3 * 10^8)^2}}} * 1.7 * 10^{-27} * v }[/math]

It is impossible to algebraically solve for [math]\displaystyle{ v }[/math], but using an equation solver yields

[math]\displaystyle{ v = 2.997 * 10^8 }[/math] m/s.

This demonstrates that when a particle is moving at nearly the speed of light, even a large impulse cannot cause it to meet or exceed the speed of light. An impulse of [math]\displaystyle{ 10^{-17} }[/math] N*s may not seem very large, but it increased the particle's momentum by approximately a factor of 10.

## Connectedness

Relativistic momentum has applications in quantum physics, particle physics, and astrophysics. In these branches of physics, particles routinely travel at relativistic speeds, and often, their momenta are directly measurable by the particles' interactions with matter. In industry, relativistic momentum is mostly used by space flight institutions.

## History

In 1905, Albert Einstein published his Special Theory of Relativity. This theory set the "speed limit" for the universe at the speed of light and redefined many properties such as momentum and energy for fast-moving objects. As particles' speeds approach the speed of light, the laws of physics change even for concepts as fundamental as space and time. After this discovery, physicists were better able to understand the motion of fast-moving particles. Einstein published the more comprehensive General Theory of Relativity in 1915, which allowed differently accelerating frames of reference to be reconciled. Some aspects of relativity have since been used in a variety of technologies, including GPS.