Relativistic velocity derivation under constant force

  • Thread starter cbriggs
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  • #1
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I've been attempting to derive the relativistic velocity equation (for an object accelerating with a constant force- [tex]\frac{Fct}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]) 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 [tex]a=\frac{F}{m}[/tex] and [tex]a\equiv\frac{dv}{dt}[/tex]

[tex]\frac{dv}{dt}=\frac{F}{m}[/tex]

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


[tex]\frac{m_{o} dv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}= F dt[/tex]

Integrate both sides

[tex]m_{o}c Sin^{-1}(\frac{v}{c})=Ft + constant[/tex]

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

[tex]\frac{v}{c}=Sin(\frac{Ft + const}{m_{o}c})[/tex]

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

Answers and Replies

  • #2
...My derivation:

Using [tex]a=\frac{F}{m}[/tex] and [tex]a\equiv\frac{dv}{dt}[/tex] here is the mistake

[tex]\frac{dv}{dt}=\frac{F}{m}[/tex]

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


[tex]\frac{m_{o} dv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}= F dt[/tex]

...
You have to write

[tex]\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}} [/tex]

Thus first integrate

[tex]\vec{F}=\frac{d\,\vec{p}}{d\,t}\Rightarrow \vec{p}}=\vec{F}\,t+\vec{p}_o[/tex]

plug in [tex] \vec{p}=m\,\gamma\,\vec{v}[/tex] and solve for the velocity.
 
  • #3
Jorrie
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[tex]\vec{F}=\frac{d\,\vec{p}}{d\,t}\Rightarrow \vec{p}}=\vec{F}\,t+\vec{p}_o[/tex]

plug in [tex] \vec{p}=m\,\gamma\,\vec{v}[/tex] 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?
 
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  • #4
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! :smile:

The constant force [itex] \vec{F}[/itex] 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?
 
  • #5
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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
 
  • #6
Newton's 2nd law must be written in the form

[tex]\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}[/tex]

for the spatial compontets of the four-force.

But this is a horrible point to start! :smile:
 
  • #7
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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:

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

and

[tex]\frac{v}{c}= tanh\left(\frac{F T}{m c}\right)[/tex]

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

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

I hope that helps.
 
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