Relativistic centripetal force

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  • #26
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genneth, are you implying that Superman can get the ball going at the speed of light just by creating the force ( tension )

F = mc^2 / R ?

If not then why not? I would think a relativistic formula is needed.
Yes, as long as m means the relativistic mass, i.e. [tex]\gamma m_0[/tex]. Notice that I calculated the acceleration, not the force. To be perfectly rigorous, you should actually work out the rate of change of 4-momentum.
 
  • #27
Jorrie
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Pervect, your formula for acceleration, even in standard units, is still coming up with dimensions of reciprocal length.
You mean this one from pervect?
[tex]\frac{\beta^2 c^2}{R \left(1-\beta^2 \right)}[/tex]

If so and given that [itex]\beta[/itex] is dimensionless, how on Earth do you get this to have reciprocal length dimensions in standard (SI) units?
 
  • #28
malawi_glenn
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Iam trying to derive this formula:

[tex] p = const. \cdot r [/tex]

For a charged particle moving in a magnetic field, perpendicular to the field lines. And r is the radius of the curvature.

Now I needed the relativistic centripetal force, in Lab-frame (shall measure momenta of relativistic particles tracks in a bubble chamber). All sources I have consulted (wanted to check my derivation), tells me that:

[tex] F = \dfrac{\gamma m v^2}{r} = \dfrac{pv}{r} [/tex]

Just so you guys might have something more to talk about:)
 
  • #29
pervect
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Iam trying to derive this formula:

[tex] p = const. \cdot r [/tex]
Perhaps you made a typo, and are looking for dp/dt?
 
  • #30
malawi_glenn
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the constant is 0.3*B, where B is measured in T, and p in GeV/c

Where is my suggested typo?

That forumula is widely used in particle physics.

So what I am saying is that the formula you might be looking for is the one I posted:
[tex] F = \dfrac{\gamma m v^2}{r} = \dfrac{pv}{r} [/tex]
(The Centripetal force)

If one uses this expression for relativistic centripetal force, and F = qvB (when particle is coming in perpendicular to field lines), you'll get:

p = 0.3*B*r, if p is measured in GeV/c
 
  • #31
cornfall
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Rotation?

Do you remember that relativistic effect happen along the direction of motion?
What relativistic effect are you referring to?
 
  • #32
pervect
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the constant is 0.3*B, where B is measured in T, and p in GeV/c

Where is my suggested typo?

That forumula is widely used in particle physics.
Sorry, I misunderstood the question. It seems to me that you have two separate questions, though, if I'm understanding it correctly this time.

1) How do you compute the momentum of a particle, given its observed radius of curvature in a magnetic field of known strength. (You've supplied your own answer to this one, it looks right though I haven't double checked it)

2) What is the force (i.e. dp/dt) on that particle in the lab frame?

The answer gamma m v^2 / r is correct here as well with these assumptions.

Note that you have to specify a frame in order to ask what the 3-force is. One of the advantages of 4-forces is that they are geometric, frame-independent objects , but the magnitude of the 3-force definitely depends on the frame in which it is measured.
 
  • #33
malawi_glenn
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Sorry, I misunderstood the question. It seems to me that you have two separate questions, though, if I'm understanding it correctly this time.

1) How do you compute the momentum of a particle, given its observed radius of curvature in a magnetic field of known strength. (You've supplied your own answer to this one, it looks right though I haven't double checked it)

2) What is the force (i.e. dp/dt) on that particle in the lab frame?

The answer gamma m v^2 / r is correct here as well with these assumptions.

Note that you have to specify a frame in order to ask what the 3-force is. One of the advantages of 4-forces is that they are geometric, frame-independent objects , but the magnitude of the 3-force definitely depends on the frame in which it is measured.
The thing is that I dont have a question, I just wanted to add what I obtained to this discussion.

Maybe I should have written "I WAS trying to derive this forumula.." And then I played a bit, and also searched in some books etc to werify my steps, thats all. And I wanted to contribute with my "research" in this thread, where to OP seemed to find the expression i posted.
 
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