Relativistic velocity derivation under constant force

[cbriggs]

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.

 

[Rainbow Child]

...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]

\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 ?

 

[Rainbow Child]

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 electrical field. Is that what you mean by constant coordinate force?

 

[cbriggs]

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

 

[Rainbow Child]

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!

 

[yuiop]

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

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

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.