Directional Acceleration at Relativistic Speeds

In summary, the conversation discusses the calculation of the angle of acceleration for a particle moving at .6 c with a constant force of 30º clockwise to the x-axis. The homework equations used are a = F/(mγ3) and arctan-1(ay/ax), and the attempt at a solution involves integrating p = γmv = 1/√(1-v2)•mv and differentiating with respect to coordinate time. The conversation also touches on the concept of four-force and the differentiation of velocity and momentum.
  • #1
PhysicsLord
4
0

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


[/B]
I'm pretty sure I know how to do this problem.
I expect that arctan-1(ay/ax) should give me the answer.

Now for ax, 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 ax = Fsin(30)/mγ3)

Now, I'm almost certain that the equation should be different for ay. Otherwise, ay/ax 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 ay, 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.
 
Physics news on Phys.org
  • #2
How is Force defined?
 
  • #3
PeroK said:
How is Force defined?
Is it not the time derivative of momentum?
 
  • #4
PhysicsLord said:
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.
 
  • #5
PeroK said:
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 = a23/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.
 
  • #6
PhysicsLord said:
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 = a23/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}##
 
Last edited:
  • #7
PS I'm going offline now, but perhaps someone else can help.
 

FAQ: Directional Acceleration at Relativistic Speeds

What is directional acceleration at relativistic speeds?

Directional acceleration at relativistic speeds refers to the change in velocity of an object as it travels at speeds close to the speed of light. It is a measure of how quickly an object's direction of motion is changing.

What is the difference between directional acceleration and regular acceleration?

Regular acceleration refers to the change in speed of an object, while directional acceleration also takes into account the change in direction of motion. Regular acceleration can be calculated using the formula a = Δv / Δt, while directional acceleration is calculated using the formula a = Δv / Δt - Δθ / Δt.

Can directional acceleration occur in a straight line?

Yes, directional acceleration can occur in a straight line as long as the object is changing its direction of motion. For example, a car going around a circular track at a constant speed is experiencing directional acceleration, even though it is moving in a straight line.

How is directional acceleration affected by relativistic speeds?

At relativistic speeds, directional acceleration becomes more significant because the object is moving at speeds close to the speed of light. This means that even small changes in direction can result in large changes in velocity, leading to a high directional acceleration.

What are the real-world applications of directional acceleration at relativistic speeds?

Understanding directional acceleration at relativistic speeds is crucial for many fields of science, including astrophysics and particle physics. It also has practical applications in space travel and the development of high-speed transportation systems.

Similar threads

Replies
4
Views
1K
Replies
3
Views
3K
Replies
9
Views
599
Replies
1
Views
1K
Replies
13
Views
2K
Back
Top