## Relativistic Version of F=ma

Notes provided by forum experts as FAQ and reference material. This is not a wiki, but rather a small repository for material relevant to common discussions.

### Relativistic Version of F=ma

Most are familiar with Newton's second law, $\vec F=m \vec a$, in which the mass $m$ of an object relates the force $\vec F$ and the acceleration $\vec a$ vectors. In Newton's theory, the acceleration is always parallel to the force, since $m$ is a scalar (just a number). In relativity, however, our concept of velocity is radically different, and the corresponding equation takes a significantly different form. As I will demonstrate, acceleration is not always parallel to the force. But, before doing this, let's look at the relativistic version of $f=ma$ in one dimension.

In one dimension...

To begin, we will use Newton's second law in it's original form, where ordinary force on an object is defined as the rate of change of momentum.

$\fs3 f = dp / dt$

The relativistic version of momentum is

$\fs3 p = \gamma m v$

where $v$ is the speed of a particle with non-zero rest mass $m$ and $\gamma$ is the Lorentz factor given by

$\fs3 \gamma = \frac{1}{\sqrt{1-v^2/c^2}}$.

Note that for simplicity, I will be letting the speed of light $\fs3 c = 1$ for the remainder of this post. Since the the Lorentz factor $\fs3 \gamma$ is changing with time, we need to find its derivative.

$\fs3 \frac{d \gamma}{dt} = \frac{d}{dt}\left( \frac{1}{\sqrt{1-v^2}} \right) = \frac{v \dot{v}}{\left( \sqrt{1-v^2}\right)^3} = \gamma^3 v a$

where $\fs3 a = \dot{v}$ is the acceleration. Now we can calculate the force.

$\fs3 f = m \left(\frac{d \gamma}{dt} v + \gamma \frac{dv}{dt} \right) = ma \left( \gamma^3 v^2 +\gamma \right) = \gamma^3 m a$

In two dimensions...

In two dimensions, the Lorentz factor becomes

$\fs3 \gamma = \frac{1}{\sqrt{1-v_x^2-v_y^2}}$

Taking the derivative, we find that

$\fs3 \frac{d \gamma}{dt} = \frac{d}{dt}\left( \frac{1}{\sqrt{1-v_x^2-v_y^2}} \right) =
\frac{v_x \dot{v_x} + v_y \dot{v_y}}{\left( \sqrt{1-v_x^2-v_y^2}\right)^3}$

In two dimensions, force is a vector given by

$\fs3 \left[\begin{array}{GC+40}f_x \\ f_y \end{array}\right] =
\left[\begin{array} \frac{d p_x}{dt} \\ \frac{d p_y}{dt} \end{array}\right] =
m\left[\begin{array} \frac{d}{dt} \left( \gamma v_x \right) \\ \frac{d}{dt}\left( \gamma v_y \right) \end{array}\right] =
m\left[\begin{array}\frac{d \gamma}{dt}v_x + \gamma \frac{d v_x}{dt} \\
\frac{d \gamma}{dt} v_y +\gamma \frac{d v_y}{dt}\end{array}\right]$

For a particle traveling in the $\fs3 x$ direction, $\fs3 v_y$ is initially zero. So the Lorentz factor derivative becomes

$\fs3 \left\ \frac{d \gamma}{dt}\right|_{v_y=0} =
\frac{v_x \dot{v_x}}{\left( \sqrt{1-v^2}\right)^3}$

The force vector is then given by

$\fs3 \left[\begin{array}{GC+30}f_x \\ f_y \end{array}\right] =
m\left[\begin{array}\gamma^3 a_x \\
\gamma a_y \end{array}\right]$

It should be clear that we could do the same thing in three dimensions and find that

$\fs3 \left[\begin{array}{GC+30}f_x \\ f_y \\ f_z \end{array}\right] =
m\left[\begin{array}\gamma^3 a_x \\
\gamma a_y \\ \gamma a_z \end{array}\right]$

Clearly, force and acceleration need not be parallel in relativity.

Transverse and Longitudinal Mass

In the early days, around 1900, Hendrik Lorentz defined mass as the ratio of force to acceleration rather than the ratio of momentum to velocity. Using such a definition, we see from the results above that mass would directional! That is, an object would have more mass in the direction of travel (longitudinal) than it would perpendicular to travel (transverse).

$\fs3 m_L = \gamma^3 m_0$

$\fs3 m_T = \gamma m_0$

where m_0 is the rest mass. Today, these concepts are no longer used. Instead, mass $m$ without a subscript is always assumed to be the rest mass unless otherwise stated. For more on the subject of the concept of mass in relativity, see Relativistic mass [Fact or Fiction].

Sparky
Active Member

Posts: 1250
Joined: 26 Jul 2007
Location: United States
Likes received:2

Return to Expert Notes

### Who is online

Users browsing this forum: No registered users and 2 guests