Understanding Special Relativity: The Relationship Between Force and Velocity

[StephenD420]

Hello everyone

I was wondering why in special relativity that the force when the velocity is parallel to the force the force = dP/dt= \gamma^3 ma, whereas when the velocity is perpendicular to the force the force = dP/dt= \gamma ma. Why is this? Is it because when the velocity is parallel to the force both the velocity of the gamma and v in the momentum, \gammamv, changes, thus the product rule applies, whereas when the velocity is perpendicular only the velocity in the momentum changes not the velocity in the gamma, thus the product rule does not apply?

Thanks for the help in explain this.
Stephen

 

[StephenD420]

bump...

 

[hikaru1221]

Well you're right When force is perpendicular to velocity or momentum, it doesn't change the MAGNITUDE of momentum and thus speed. Because the gamma coefficient only depends on speed (or magnitude of velocity), in that case, gamma doesn't change. But in the case where force is parallel to velocity, it changes both direction and magnitude of velocity, so gamma also changes.

 

[collinsmark]

Hello Stephen

I was wondering why in special relativity that the force when the velocity is parallel to the force the force = dP/dt= \gamma^3 ma, whereas when the velocity is perpendicular to the force the force = dP/dt= \gamma ma. Why is this? Is it because when the velocity is parallel to the force both the velocity of the gamma and v in the momentum, \gammamv, changes, thus the product rule applies, whereas when the velocity is perpendicular only the velocity in the momentum changes not the velocity in the gamma, thus the product rule does not apply?

First, I plead with you to be careful with your terminology. Three momentum is

\vec p = \gamma m \vec v

Gamma is not multiplied by the momentum, gamma actually is part of the momentum! In special relativity, that's a pretty important point.

Semantics/terminology aside, I believe you are asking whether you can simply treat gamma as a constant and simply bring it out of the time derivative. Here is my answer: I wouldn't do that unless you are prepared to back it up with some math to show that it is a constant.* It's not intuitively obvious that gamma remains constant when the 3-force is perpendicular to the 3-velocity. Generally speaking, The v in γmv is the same v as the v within γ itself (unless you work with components individually, but even then, the component in question is still within γ).

But there is an easier way to show that f = γma if the force is perpendicular to the velocity. This is the route I suggest taking if you are ever quizzed on this. It's the easy way. Note that,

\vec v = (v_x, v_y, v_z)

Let's assume that the object is moving along the x-axis only (v_y and v_z are zero), and the 3-force is applied in the y direction. Now evaluate

F_y = m \frac{d}{dt}(\gamma v_y) \left|_{v_y = 0, v_z = 0}

But be careful, make sure you do the evaluation after you take the derivative. This is very important. Just like in Newtonian mechanics, just because the component's velocity happens to be zero doesn't mean the component's acceleration happens to be zero! However, once you come across a v_y that is outside any derivative, then it is okay to to zero that whole term. Go ahead and do this (using the product rule), and you'll immediately see why F = γma if the force is perpendicular to the velocity. That's the easy way.

*Now, as hikaru1221 points out, gamma is only dependent on speed, so a 3-force perpendicular to the 3-velocity should not change gamma. But you should be prepared to back that up mathematically if you ever use that (or risked getting "dinged" by your instructor, who may not have gone through the math himself/herself). But it can be done (I proved it to myself last night on a paper napkin). I can't give you the full thing but here is a hint on how to do it.

If we use the units c = 1, then gamma in longhand form is

\frac{1}{\sqrt{1 - (v_x^2 + v_y^2 + v_z^2)}}

Now solve for d(\gamma)/dt [/tex], with the nonzero velocity only on the x-axis (but again, be careful -- don&#039;t do any evaluations until <i>after</i> taking the derivative!), and noting that the dot product between perpendicular vectors is zero. Then point out that if \dot \gamma is zero, it implies that \gamma is a constant.

 

[hikaru1221]

Hurray to collinsmark for a very detailed, helpful post
@to the OP: You can learn a lot besides your particular problem from collinsmark's post