# Relativistic velocity derivation under constant force

I've been attempting to derive the relativistic velocity equation (for an object accelerating with a constant force- $$\frac{Fct}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$) for near a month without any solution.

I've derived an equation form the F=ma relation which includes a Sin function, so I know it's wrong. However, I haven't been able to determine why. Could someone point me in the right direction?

My derivation:

Using $$a=\frac{F}{m}$$ and $$a\equiv\frac{dv}{dt}$$

$$\frac{dv}{dt}=\frac{F}{m}$$

$$m dv= F dt$$ where $$m\equiv\frac{m_{o}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

$$\frac{m_{o} dv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}= F dt$$

Integrate both sides

$$m_{o}c Sin^{-1}(\frac{v}{c})=Ft + constant$$

Which gives an answer that doesn't make sense and obviously doesn't represent v(t)

$$\frac{v}{c}=Sin(\frac{Ft + const}{m_{o}c})$$

Any idea where the error is, or how to get the correct expression? Any help is appreciated.

Related Special and General Relativity News on Phys.org
...My derivation:

Using $$a=\frac{F}{m}$$ and $$a\equiv\frac{dv}{dt}$$ here is the mistake

$$\frac{dv}{dt}=\frac{F}{m}$$

$$m dv= F dt$$ where $$m\equiv\frac{m_{o}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

$$\frac{m_{o} dv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}= F dt$$

...
You have to write

$$\vec{F}=\frac{d\,\vec{p}}{d\,t},\quad \vec{p}=m\,\gamma\,\vec{v}, \quad \gamma=\frac{1}{\sqrt{1-v^2/c^2}}$$

Thus first integrate

$$\vec{F}=\frac{d\,\vec{p}}{d\,t}\Rightarrow \vec{p}}=\vec{F}\,t+\vec{p}_o$$

plug in $$\vec{p}=m\,\gamma\,\vec{v}$$ and solve for the velocity.

Jorrie
Gold Member
$$\vec{F}=\frac{d\,\vec{p}}{d\,t}\Rightarrow \vec{p}}=\vec{F}\,t+\vec{p}_o$$

plug in $$\vec{p}=m\,\gamma\,\vec{v}$$ and solve for the velocity.
I assume you are talking coordinate values here, not "proper values", meaning the constant force will produce decreasing coordinate acceleration with time.

Question: will such a constant (coordinate) force produce a constant proper acceleration, i.e., as measured by an accelerometer on board the spacecraft?

Last edited:
I assume you are talking coordinate values here, not "proper values", meaning the coordinate acceleration will decrease with time.
I don't really understand the terminology here! It is not you by my English!

The constant force $\vec{F}$ I mean is like the spacecraft is like a charge accelerating in a constant eletrical field. Is that what you mean by constant coordinate force?

Wow... Thanks much, RC. I'm still not sure where my original math went wrong, but I'm okay with pondering that until I get the answer. My urgent task was to know how to derive the equation, which you have generously provided.

Jorrie- Yes: this applies to accelerating charges in a constant electric field (very good guess RC- you're on fire tonight), so no proper measurements required.

Regards to all,
Carey

Newton's 2nd law must be written in the form

$$\vec{F}=\frac{d\,\vec{p}}{d\,t}\Rightarrow \vec{F}=\gamma\,m\,\vec{a}+\gamma^3\,m\,\frac{\vec{v}\cdot\vec{a}}{c^2}\,\vec{v}$$

for the spatial compontets of the four-force.

But this is a horrible point to start!

This link gives the equations of a rocket with constant acceleration (as felt by the occupants of the rocket)

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

It does not show the derivations but it gives you something to cross check your answers against.

This link gives a derivation of velocity from force. I’m not sure if the derivation is correct but the final answer agrees with the equations given by Baez.

http://www.usna.edu/Users/physics/mungan/Scholarship/ConstantAcceleration.pdf [Broken]

The author makes the observation “The fact that the passengers feel one gee means that the force F measured in frame R is mg. However, as proved by French Eq. (7-20), an observer in frame E measures the same (longitudinal) force, despite the fact that she measures neither the same (relativistic) mass nor acceleration of the rocket. “ which simplifies things a bit.

Substituting F/m for a in the Baez equations(where m is the proper mass) and dividing by c we get:

$$\frac{v}{c}= \frac{1}{\sqrt{1+(m c)^2/(Ft)^2}}$$

and

$$\frac{v}{c}= tanh\left(\frac{F T}{m c}\right)$$

The last equation is in terms of the proper time and becomes:

$$\frac{v}{c}= tanh\left( sinh^{-1}\left(\frac{F t}{m c}\right)\right)$$ in terms of coordinate time.

I hope that helps.

Last edited by a moderator: