Directional Acceleration at Relativistic Speeds

[PhysicsLord]

Homework Statement

A particle flies along in the positive +x direction. It has a constant force F applied 30º clockwise to the x-axis.
It is moving at .6 c. What is the angle of acceleration?

Homework Equations

a = F/(mγ³)

The Attempt at a Solution

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I'm pretty sure I know how to do this problem.
I expect that arctan_-1(a_y/a_x) should give me the answer.

Now for a_x, I'm pretty sure the equation above works fine. It's what my textbook gave me. I also agree with the derivation. Nonetheless, it only applies to particles accelerated parallel to the direction of the velocity.
So a_x = Fsin(30)/mγ³)

Now, I'm almost certain that the equation should be different for a_y. Otherwise, a_y/a_x would just have a lot of cancellation and be sin(30)/cos(30) = tan(30) --> angle of acceleration = tan^-1(tan(30)) = 30, which I don't think is right (otherwise the question would be a useless one to ask).

I tried a couple of different things to find a_y, but I couldn't find anything anywhere (be it online or in my head) that told me what the effect of velocity is on a acceleration when it is parallel to the force.

 

[PeroK]

How is Force defined?

 

[PhysicsLord]

How is Force defined?

Is it not the time derivative of momentum?

 

[PeroK]

Is it not the time derivative of momentum?

Yes, although you may want to be more precise about what time you are talking about. In any case, that allows you to relate force and acceleration for 2D motion.

 

[PhysicsLord]

Yes, although you may want to be more precise about what time you are talking about. In any case, that allows you to relate force and acceleration for 2D motion.

Ok. So I'm not sure what You mean by which time. I assume that it's the proper time for the object on which the force is acting. In any case, any difference in parallel vs. perpendicular force shouldn't change anything when we consider the time since they would both be acting on the same reference frame.

I figure then that I could integrate p = γmv = 1/√(1-v²)•mv -> dp/dt = F = -1/2•(1-v²/c²)^-3/2•(0-a²/c²)•mv+ maγ = a²mγ³/c² + maγ

That's my best attempt, but I'm pretty sure it's wrong. I'm not sure what to do about trying to differentiate for the perpendicular direction. My understanding is that it speed should only matter in the direction (or antiparallel to) of motion.

 

[PeroK]

Ok. So I'm not sure what You mean by which time. I assume that it's the proper time for the object on which the force is acting. In any case, any difference in parallel vs. perpendicular force shouldn't change anything when we consider the time since they would both be acting on the same reference frame.

I figure then that I could integrate p = γmv = 1/√(1-v²)•mv -> dp/dt = F = -1/2•(1-v²/c²)^-3/2•(0-a²/c²)•mv+ maγ = a²mγ³/c² + maγ

That's my best attempt, but I'm pretty sure it's wrong. I'm not sure what to do about trying to differentiate for the perpendicular direction. My understanding is that it speed should only matter in the direction (or antiparallel to) of motion.

The force and acceleration in this case are derivatives with respect to coordinate time. If you differentiate momentum with respect to proper time of the particle you get the four-force. That said, you seem to be using coordinate time in any case.

For 2D motion, momentum is a vector. $\vec{F} = \frac{d\vec{p}}{dt}$

Also, you have incorrectly differentiated $\gamma$ - in particular, the $v^2$ term.

$\frac{d}{dt}v^2 \ne a^2$

Note also that velocity is a vector and $v^2 = \vec{v} \cdot \vec{v}$

 

[PeroK]

PS I'm going offline now, but perhaps someone else can help.