- #1
eep
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Hi, I'm working out of Hartle's book Gravity and had a question about Ch. 5 problem 7. The problem states that a particle has constant acceleration along the x-axis, that is, the acceleration when measured in it's instantaneous rest frame is always the same constant. They want me to get the parametric equations for x and t as functions of proper time [itex]\tau[/itex], assuming at t=0 that x=0 and v=0.
I assumed that since the acceleration is constant, the three-force is also constant. I can then use
[tex]
\frac{dp}{dt} = F = ma
[/tex]
for some constant a. Since F is constant I can integrate once to get
[tex]
{\gamma}m\frac{dx}{dt} = mat
[/tex]
Or, solving for [itex]\frac{dx}{dt}[/itex]
[tex]
\frac{dx}{dt} = \frac{at}{\sqrt{1+a^2t^2}}
[/tex]
Using the relationship [itex]{d\tau}^2 = dt^2 - dx^2[/itex] I can arrive at
[tex]
d\tau = dt\sqrt{(\frac{1}{1+a^2t^2})}
[/tex]
Integrating and solving for t and letting [itex]\tau[/itex] = 0 when t=0 I get
[tex]
t(\tau) = \frac{{\sin}h(a\tau)}{a}
[/tex]
I can then solve for x and get
[tex]
x(\tau) = \frac{{\cos}h(a\tau)}{a} - \frac{1}{a}
[/tex]
due to the initial conditions. Did I do this right?
EDIT: I really just need to know if my first couple steps are right, as everything follows from that.
I assumed that since the acceleration is constant, the three-force is also constant. I can then use
[tex]
\frac{dp}{dt} = F = ma
[/tex]
for some constant a. Since F is constant I can integrate once to get
[tex]
{\gamma}m\frac{dx}{dt} = mat
[/tex]
Or, solving for [itex]\frac{dx}{dt}[/itex]
[tex]
\frac{dx}{dt} = \frac{at}{\sqrt{1+a^2t^2}}
[/tex]
Using the relationship [itex]{d\tau}^2 = dt^2 - dx^2[/itex] I can arrive at
[tex]
d\tau = dt\sqrt{(\frac{1}{1+a^2t^2})}
[/tex]
Integrating and solving for t and letting [itex]\tau[/itex] = 0 when t=0 I get
[tex]
t(\tau) = \frac{{\sin}h(a\tau)}{a}
[/tex]
I can then solve for x and get
[tex]
x(\tau) = \frac{{\cos}h(a\tau)}{a} - \frac{1}{a}
[/tex]
due to the initial conditions. Did I do this right?
EDIT: I really just need to know if my first couple steps are right, as everything follows from that.
Last edited: