# Acceleration under a constant force

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 4, Problem 4.15.

Under a constant force ${F}$, an object undergoes hyperbolic motion. In one dimension for a constant force we have

 $\displaystyle \frac{dp}{dt}$ $\displaystyle =$ $\displaystyle F\ \ \ \ \ (1)$ $\displaystyle p$ $\displaystyle =$ $\displaystyle Ft+C \ \ \ \ \ (2)$

where ${C}$ is a constant of integration. If the object starts at ${t=0}$ at rest (in the lab frame), then ${C=0}$, and

$\displaystyle p=\frac{mu}{\sqrt{1-u^{2}/c^{2}}}=Ft \ \ \ \ \ (3)$

which can be solved for the velocity ${u}$ to give

$\displaystyle u=\frac{F}{m}\frac{t}{\sqrt{1+\left(Ft/mc\right)^{2}}} \ \ \ \ \ (4)$

We can also get this formula by integrating the expression for the acceleration

$\displaystyle \mathbf{a}=\frac{\mathbf{F}}{\gamma m}-\frac{\mathbf{u}}{\gamma mc^{2}}\left(\mathbf{u}\cdot\mathbf{F}\right) \ \ \ \ \ (5)$

If ${\mathbf{F}}$ is constant and acts on an object initially at rest, then ${\mathbf{F}\parallel\mathbf{u}}$ and

 $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{du}{dt}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{F}{\gamma m}\left(1-\frac{u^{2}}{c^{2}}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{F}{m}\left(1-\frac{u^{2}}{c^{2}}\right)^{3/2} \ \ \ \ \ (8)$

To find ${u\left(t\right)}$ we integrate:

 $\displaystyle \int\left(1-\frac{u^{2}}{c^{2}}\right)^{-3/2}du$ $\displaystyle =$ $\displaystyle \frac{F}{m}\int dt\ \ \ \ \ (9)$ $\displaystyle \frac{u}{\sqrt{1-u^{2}/c^{2}}}$ $\displaystyle =$ $\displaystyle \frac{F}{m}\left(t+t_{0}\right) \ \ \ \ \ (10)$

If ${u=0}$ at ${t=0}$, the constant of integration is ${t_{0}=0}$ and we get 3 again.

From 3 we can get the inverse function

$\displaystyle t=\frac{mu}{F\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (11)$

Thus as ${u\rightarrow c}$, ${t\rightarrow\infty}$ so the object never quite reaches the speed of light.

Because of the velocity-dependent term in 5, the acceleration due to a constant force is not constant, but rather decreases as ${u}$ increases. If we start with ${a_{0}=F/m=g=9.8\mbox{ m s}^{-2}}$, then the times required to reach various speeds are found from

$\displaystyle t=\frac{u}{9.8\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (12)$

 ${\frac{u}{c}}$ ${t}$ 0.9 ${6.3\times10^{7}\mbox{ s}=2\mbox{ years}}$ 0.99 ${2.14\times10^{8}\mbox{ s}=6.78\mbox{ years}}$ 0.999 ${6.82\times10^{8}\mbox{ s}=21.6\mbox{ years}}$ 0.9999 ${2.16\times10^{9}\mbox{ s}=68.4\mbox{ years}}$ 1.0 ${\infty}$

# Relativistic acceleration in terms of force

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 4, Problem 4.14.

Ordinary force in relativity is given by

 $\displaystyle \mathbf{F}$ $\displaystyle =$ $\displaystyle \frac{m}{\sqrt{1-u^{2}/c^{2}}}\left[\mathbf{a}+\frac{\left(\mathbf{u}\cdot\mathbf{a}\right)\mathbf{u}}{c^{2}-u^{2}}\right] \ \ \ \ \ (1)$

To get a general expression for the acceleration ${\mathbf{a}}$ in terms of the force, we take the dot product of both sides with the velocity ${\mathbf{u}}$:

 $\displaystyle \mathbf{u}\cdot\mathbf{F}$ $\displaystyle =$ $\displaystyle \gamma m\left(\mathbf{u}\cdot\mathbf{a}\right)\left(1+\frac{u^{2}}{c^{2}-u^{2}}\right)\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\gamma mc^{2}\left(\mathbf{u}\cdot\mathbf{a}\right)}{c^{2}-u^{2}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \gamma^{3}m\left(\mathbf{u}\cdot\mathbf{a}\right)\ \ \ \ \ (4)$ $\displaystyle \mathbf{u}\cdot\mathbf{a}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{u}\cdot\mathbf{F}}{\gamma^{3}m} \ \ \ \ \ (5)$

Substituting back into 1, we get

 $\displaystyle \mathbf{a}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{F}}{\gamma m}-\frac{\mathbf{u}}{c^{2}-u^{2}}\frac{\mathbf{u}\cdot\mathbf{F}}{\gamma^{3}m}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mathbf{F}}{\gamma m}-\frac{\mathbf{u}}{\gamma mc^{2}}\left(\mathbf{u}\cdot\mathbf{F}\right) \ \ \ \ \ (7)$

In the limit of small ${\mathbf{u}}$, this reduces to the familiar Newton’s law ${\mathbf{F}=m\mathbf{a}}$, but in the relativistic region, the acceleration depends on the object’s velocity. As a result, the acceleration isn’t parallel to the force unless ${\mathbf{F}}$ is either parallel to ${\mathbf{u}}$ or ${\mathbf{F}\perp\mathbf{u}}$; in the latter case ${\mathbf{u}\cdot\mathbf{F}=0}$ and the second term is zero.