Re: Circular motion in SR

On Mar 13, 7:57 pm, "Pmb" <some...@xxxxxxxxxxxxx> wrote:
"Dono" <sa...@xxxxxxxxxxx> wrote in message

news:10e6e8d6-50b2-41f1-b1c7-d9e46892e548@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx



On Mar 13, 7:08 pm, "Pmb" <some...@xxxxxxxxxxxxx> wrote:
"Dono" <sa...@xxxxxxxxxxx> wrote in message

news:47a8a70a-073a-40d0-b133-ae9d8856f002@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Mar 13, 1:18 pm, "Pmb" <some...@xxxxxxxxxxxxx> wrote:

<ram.rac...@xxxxxxxxx> wrote in message

news:a37dc3a4-7094-4691-8a3a-eb2f830b015d@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

I'm trying to write the equations for circular motion according to SR
laws of motion. I'm doing some kind of mistakes, because I'm not
getting real solutions to the equations.

This is what's going on: You have an body of mass m0 circling around
a
stationary center point that is a distance of r from the body. There
is a force F that attracts the body towards the center point, making
it move in a circle around it. What's the velocity v of the object?

I've used these equations:

v=sqrt(a*r)
a=F/(m0*gamma^3)
gamma=1/sqrt(1-(v/c)^2)

I keep getting only imaginary solutions for v. What am I doing wrong?

Thanks, Ram.

I've used this in solving the equation of motion for a charged particle
moving in a uniform magnetic field. In such a case the force is
constant
in
magnitude and the direction is radially inward towards the center of
the
circle the particle moves in. This is used in cyclotrons.
Seehttp://www.geocities.com/physics_world/sr/cyclotron.htm

There is an error in your equations, namely a=F/(m0*gamma^3). If a is
the
radial acceleration, and there is no tangential acceleration, then F =
ma

Pete,

I'm afraid that there is another error.
In eq(5), when you calculate d/dt (\gamma*m_0....)
you forget to calculate d \gamma/dt.
After all, \gamma is a function of v and v is a function of t. You
calculate d v_x/dt and d v_y/dt but you forgot to calculte d \gamma/
dt.

Gamma is a function of the speed (i.e. magnitude of velocity) only. Since
the speed of the particle is a constant of motion it follows that d
\gamma/dt = 0.

How could this be?

v^2=v_x^2+v_y^2

You calculate d v_x/dt and you don't calculate dv/dt?

Actually I did calculate dv/dt. I just didn't show it explicitly. See Eq.
(5). Note that I wrote

F = dp/dt = d/dt[gamma*m_0*(v_x e_x + v_y e_y)]

This is identical to the expression

F = m dv/dt = gamma*m_0*d/dtv_x e_x + v_y e_y)

where m = gamma*m_0 and v = v_x e_x + v_y e_y. The value of dv/dt is dv/dt =
d(v_x)/dt e_x + d(v_y)/dt e_y). This is then set equal to Eq. (4) to obtain
Eq. (5).

Best wishes

Pete



What I am tellling you is that you missed calculating d \gamma/dt .
This should have triggered a term in gamma^3*dv/dt , clearly missing
in your computations.
Instead, you factored \gamma out of the derivation, an error.
.

Relevant Pages

  • Re: Circular motion in SR
    ... getting real solutions to the equations. ... circle the particle moves in. ... radial acceleration, and there is no tangential acceleration, then F ... the speed of the particle is a constant of motion it follows that d ...
    (sci.physics.relativity)
  • Re: Circular motion in SR
    ... circle the particle moves in. ... radial acceleration, and there is no tangential acceleration, then F = ... the speed of the particle is a constant of motion it follows that d ... Actually I did calculate dv/dt. ...
    (sci.physics.relativity)
  • Re: Circular motion in SR
    ... I've used this in solving the equation of motion for a charged particle ... magnitude and the direction is radially inward towards the center of the ... radial acceleration, and there is no tangential acceleration, then F = ma ...
    (sci.physics.relativity)
  • Re: Circular motion in SR
    ... magnitude and the direction is radially inward towards the center of the ... circle the particle moves in. ... radial acceleration, and there is no tangential acceleration, then F = ma ... the speed of the particle is a constant of motion it follows that d ...
    (sci.physics.relativity)
  • Re: Circular motion in SR
    ... I've used this in solving the equation of motion for a charged particle ... in magnitude and the direction is radially inward towards the center of ... radial acceleration, and there is no tangential acceleration, then F = ma ...
    (sci.physics.relativity)