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 (see Conservation of Momentum). However, for systems containing objects moving at speeds near 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].
A Mathematical Model
The relative equation for momentum is as follows:
[math]\displaystyle{ \vec{p} = \gamma * m * \vec{v} }[/math]
where [math]\displaystyle{ \vec }[/math] is the momentum of the system, [math]\displaystyle{ m }[/math] is mass, and [math]\displaystyle{ \vec{p} }[/math] 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}{Δt}} }[/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