Finding a solution for Relativistic Acceleration

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Michio Cuckoo
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I tried all my mathcad software such as Maple and I can't seem to find a solution to this time based differential equation.

m}\left%20(%201%20-%20\frac{\left%20[%20f(t)%20\right%20]^{2}}{c^{2}}%20\right%20)^{\frac{3}{2}}.gif
 
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Michio Cuckoo said:
I tried all my mathcad software such as Maple and I can't seem to find a solution to this time based differential equation.

m}\left%20(%201%20-%20\frac{\left%20[%20f(t)%20\right%20]^{2}}{c^{2}}%20\right%20)^{\frac{3}{2}}.gif

Solving differential equations is often a matter of guessing and checking.
This equation comes up in Rindler coordinates, so I actually know the solution.

Let g = F/m.
Switch to a new independent variable T that is related to t through
t = c/g sinh(gT/c).

Then try the solution:

f = c tanh(gT/c)

In terms of the original t, we can rewrite f as

f = c (gt/c)/√(1 + (gt/c)2)

or

f = c (ct)/√((ct)2 + c2/g)
 
Solving differential equations is often a matter of guessing and checking.
Michio, It's important part of everyone's physics education to pick up some standard techniques for solving DEs, so you don't have to depend entirely on the math software to do it for you, which often does an imperfect job. And this equation is a very easy one to solve. Setting the constant parameters to one to simplify the discussion,

df/dt = (1 - f2)3/2

Separate the variables (t on one side, f on the other) and integrate immediately:

t = ∫(1 - f2)-3/2 df

To do the integral, the factor 1 - f2 suggests making a trig substitution. Let f = sin θ.

t = ∫(cos θ)-3 cos θ dθ = ∫ sec2 θ dθ = tan θ

So we have f = sin θ, t = tan θ, giving us an algebraic relationship and the solution:

t = f/√(1 - f2) or f = t/√(1 + t2)
 
Bill_K said:
t = ∫(1 - f2)-3/2 df

To do the integral, the factor 1 - f2 suggests making a trig substitution. Let f = sin θ.

Interestingly, it works to try the substitution f = sin(θ), then you get t = tan(θ), but it also works to try the substitution f = tanh(θ), then you get t = sinh(θ). I like the latter better, because that gets you to the Rindler coordinates.