Derivation of Velocity from Acceleration in Special Relativity

[ecastro]

I got stuck in deriving the velocity of the particle from the acceleration equation. Here are the details of the problem.

The acceleration of a particle with a relativistic momentum is

\vec{a} = \frac{\vec{F}}{\gamma m} - \frac{\vec{v}}{\gamma m c^2}\left(\vec{F} \cdot \vec{v}\right).

By integrating the above equation, the velocity of the particle can be calculated if the force applied and the mass of the particle are given. The greek letter \gamma in this case is \left(1 - \frac{v^2}{c^2}\right)^{-1/2}

Considering that a constant force \it{F} is applied to a mass \it{m} initially at rest, the velocity of the particle is

\int \vec{a} dt = \int \left[\frac{\vec{F}}{\gamma m} - \frac{\vec{v}}{\gamma m c^2}\left(\vec{F} \cdot \vec{v}\right)\right] dt \\<br /> <br /> \vec{v} = \int \left[\frac{\vec{F}}{\gamma m} - \frac{\vec{v^2}}{\gamma m c^2}\left(\vec{F} \right)\right] dt.

Since F and m are constants in time, they are removed from the integral. Gamma is taken out from the two terms. Then,

\vec{v} = \frac{\vec{F}}{m} \int \frac{1}{\gamma} \left(1 - \frac{v^2}{c^2}\right) dt.

Using the definition of gamma, the integrand is

\vec{v} = \frac{\vec{F}}{m} \int \left(1 - \frac{v^2}{c^2}\right)^{3/2} dt.

This is where I get stuck. I don't know how to integrate it since the integrand is not a direct function of t but of v which is a function of t. I tried having v = at, but I don't know if this is correct.

 

[Dale]

You cannot substitute v=at because that only works if a is constant, which it is not here. Personally, I would start with your last equation, differentiate both sides with respect to t, and solve the resulting differential equation. Are you familiar with differential equations, or do you have access to a math package like Mathematica?

 

[DrStupid]

Using the definition of gamma, the integrand is

\vec{v} = \frac{\vec{F}}{m} \int \left(1 - \frac{v^2}{c^2}\right)^{3/2} dt.

This is where I get stuck. I don't know how to integrate it since the integrand is not a direct function of t but of v which is a function of t.

You have the differential equation
\frac{{dv}}{{dt}} = \frac{F}{m} \cdot \left( {1 - \frac{{v^2 }}{{c^2 }}} \right)^{\frac{3}{2}}
that can be solved by separation of variables.

 

[Meir Achuz]

You have to know F(x,v,a,t). If F is constant, or a function of only t, then m\gamma^3 dv=Fdt
can be integrated.

 

[ecastro]

You cannot substitute v=at because that only works if a is constant, which it is not here. Personally, I would start with your last equation, differentiate both sides with respect to t, and solve the resulting differential equation. Are you familiar with differential equations, or do you have access to a math package like Mathematica?

Yes. I am familiar with the mathematics. The resulting differential equation will be the one shown by DrStupid. By separating the variables,

\frac{dv}{dt} = \frac{F}{m} \cdot \left(\frac{c^2 - v^2}{c^2}\right)^{3/2} = \frac{F}{m c^2} \cdot \left(c^2 - v^2\right)^{3/2} \\<br /> \left(c^2 - v^2\right)^{-3/2} dv = \frac{F}{m c^2} dt \\<br /> \int \left(c^2 - v^2\right)^{-3/2} dv= \int \frac{F}{m c^2} dt \\<br /> \frac{v}{c^2 \sqrt{c^2 - v^2}} = \frac{F t}{m c^2}.

Could you please check the mathematics? After this, I just need to solve for v, right?

You have to know F(x,v,a,t). If F is constant, or a function of only t, then m γ^3 dv = F dt can be integrated.

Would you be suggesting that dt = \frac{m \gamma^3}{F} dv? If I were to substitute this to my last equation, wouldn't be the constants F and m cancel, so does \gamma? I would be left out with the integral of dv which will result into v.

 

[DrStupid]

\frac{dv}{dt} = \frac{F}{m} \cdot \left(\frac{c^2 - v^2}{c^2}\right)^{3/2} = \frac{F}{m c^2} \cdot \left(c^2 - v^2\right)^{3/2}

\frac{dv}{dt} = \frac{F}{m} \cdot \left(\frac{c^2 - v^2}{c^2}\right)^{3/2} = \frac{F}{m c^3} \cdot \left(c^2 - v^2\right)^{3/2}

should be more accurate

 

[ecastro]

Oh, yes, I forgot about the 3rd-power-exponent. Thanks a lot!