Relativistic Momentum

The page focuses on momentum when traveling close to the speed of light

The Main Idea

Momentum is a property of a moving body; it can be narrowed down to simply mass and velocity (hence, [math]\displaystyle{ \overrightarrow{p} = m*\overrightarrow{v} }[/math]). However, the usual momentum equation does not always apply. When traveling near the speed of light, a new equation must be used. This equation was discovered by Albert Einstein in the early 1900's. This discovery revolutionized physics and introduced a new constant, gamma or [math]\displaystyle{ \gamma }[/math], a quantity relating velocity and momentum.

A Mathematical Model

The relative equation for momentum is as follows:
[math]\displaystyle{ \overrightarrow{p} = \gamma * m * \overrightarrow{v} }[/math]
where p is the momentum of the system, m is mass, and v is the velocity. The new constant [math]\displaystyle{ \gamma }[/math] is a bit more complicated.

The equation for [math]\displaystyle{ \gamma }[/math] is as follows:
[math]\displaystyle{ \gamma = \sqrt{\frac{1}{1-\frac{\left\vert \overrightarrow{v} \right\vert^2}{c^2}}} }[/math]
where v is again the velocity, and c is the speed of light or [math]\displaystyle{ 3 * 10^8 }[/math]

Lastly, momentum is most practical in the case of predicting position using Iterative Prediction.

Iterative prediction normally uses the position update equation: [math]\displaystyle{ {\vec{r}_{f} = \vec{r}_{i} + \vec{v}_{avg}{Δt}} }[/math]

This equation still applies relative to the speed of light, but appears in a slightly different form seen below.

[math]\displaystyle{ {\vec{r}_{f} = \vec{r}_{i} + \frac{1}{\sqrt{1-\frac{\left\vert\vec{v}_{avg}\right\vert^2}{c^2}}}\vec{v}_{avg}{Δt}} }[/math]

Visual Model

Again, this formula should only be used when traveling close to the speed of light. As you can see in the following chart, momentum is only noticeably affected around [math]\displaystyle{ 10^7 }[/math].

The difference in Newtonian momentum [math]\displaystyle{ \overrightarrow{p} = m*\overrightarrow{v} }[/math] and Einstein's relativistic momentum [math]\displaystyle{ \overrightarrow{p} = \gamma * m * \overrightarrow{v} }[/math] can be easily visualized in the following graph.

Notice the graph only goes up to [math]\displaystyle{ 2.9 * 10^8 }[/math] or .96c. The relativistic curve continues to grow extremely exponentially to the point where the Newtonian momentum graph is not visible. Although it is important to emphasize this difference, the graph was cut short for practical purposes.

Examples

Simple

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

What is the momentum of the proton?

[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] kgm/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?

[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] kgm/s

Difficult

Suppose that a proton (mass = [math]\displaystyle{ 1.7 * 10^{-27} }[/math] kg) is moving in a straight line with a constant velocity [math]\displaystyle{ \lt 0 , 1.8 * 10^8 , 0\gt }[/math] m/s. The proton begins at a point <4.2, 7.8, -1.3>.

Where will the proton be 5 ns later? (ns = nanosecond = 1 * 10^{-9} s)

[math]\displaystyle{ \left\vert \overrightarrow{v} \right\vert = \sqrt{(0)^2 + (1.8*10^8)^2 +(0)^2} }[/math] m/s [math]\displaystyle{ = 1.8 * 10^8 }[/math] m/s.

[math]\displaystyle{ \frac{\left\vert \overrightarrow{v} \right\vert}{c} = \frac{1.8 * 10^8 m/s}{3 * 10^8 m/s} }[/math] [math]\displaystyle{ = .6 }[/math]

[math]\displaystyle{ \gamma = \frac{1}{\sqrt{1-(.6)^2}} = 1.25 }[/math]

Plug values in

[math]\displaystyle{ {\vec{r}_{f} = \vec{r}_{i} + \gamma\vec{v}_{avg}{&Delta;t}} }[/math]

[math]\displaystyle{ \vec{r}_{f} = \lt 4.2, 7.8, -1.3 \gt m + (1.25)*\lt 0,1.8*10^8,0 \gt m/s*(5*10^{-9})s = \lt 4.2, 7.8, -1.3\gt m + \lt 0,1.125,0\gt m = \lt 4.2, 8.925, -1.3 \gt m }[/math]

Therefore the proton is at the position < 4.2,8.925,-1.3 > meters after 5 nanoseconds.

Connectedness

This concept of relativistic momentum affects several different majors. For example, on a quantum physics level, relativistic momentum aids in accurately predicting the position of a particle through a certain period of time.

Relativistic momentum is applicable throughout majors. Hopefully, one day, you'll find its use for you.

History

In 1905, Albert Einstein released his Special Theory of Relativity to the public. This theory set the "speed limit" for the universe at the speed of light. When objects, came closer to the speed of light, many entities are drastically changed (one being momentum). This changed the direction of physics immensely. After this discovery, physicists were better able to predict and calculate momentum on the microscopic scale of fast-moving particles.

See also

Links to different uses of momentum

Further reading

Momentum Principle

Impulse Momentum

Momentum with Respect to External Forces

Iterative Prediction and Position Update