# Circular orbits in SR

In CM there is this equation of motion for ciruclar orbits:
F=k/r^2=m*v^2/r, where k depends on the force involved.

I know this gets complicated in GR, but is there a simple expression for circular orbits in SR, what happends to the m*v^2/r ?

If the forces are only electromagnetic, does k/r^2 change in SR ?

Also, is the quantity L=mrv conserved in SR, or some modification ?

I want to know what is the equation(s) for circular orbits in SR, in A) electromagnetic force k/r^2 and B) gravitational force if it is possible. Any reference ?

Integral
Staff Emeritus
Gold Member
I could be wrong, but SR applies to unaccelerated systems. Orbits, by definition, are accelerated systems; therefore you will not find orbits in SR.

Further for SR to be of any interest you need to have a substantial fraction of c involved. Orbits at that speed would be in the world of QM. Once again, no SR orbits.

JesseM
I could be wrong, but SR applies to unaccelerated systems. Orbits, by definition, are accelerated systems; therefore you will not find orbits in SR.
As it turns out that is wrong, in SR one can easily analyze the behavior of accelerating systems from the perspective of an inertial reference frame, and the modern view is that even non-inertial frames are considered part of SR if the curvature of spacetime is zero. See Can special relativity handle acceleration? from the Usenet Physics FAQ for more info.

If you meant GR, on the other hand, there exists such an expression. If I remember correctly, it is

$$r \left( \frac{d\phi}{dt} \right)^2=v=\sqrt{\frac{GM}{r-\frac{2GM}{c^2}}}$$

Where $$t$$ is the time measured by a comoving observer.

As you can see, you don't need the general relativistic correction when $$r>>\frac{2GM}{c^2}$$.

Note also that stable circular orbits only exist when $$r>\frac{3GM}{c^2}$$ because of the speed limit of $$v<c$$.

Dale
Mentor
2020 Award
In CM there is this equation of motion for ciruclar orbits:
F=k/r^2=m*v^2/r, where k depends on the force involved.

I know this gets complicated in GR, but is there a simple expression for circular orbits in SR, what happends to the m*v^2/r ?
In units where c=1 f you start with uniform circular motion, we have an equation for the worldline (parameterized by coordinate time):
$$s=(t,r\;cos(t\omega),r\;sin(t\omega),0)$$

which gives a four-velocity:
$$u=\frac{ds}{d\tau}=\left(\frac{1}{\sqrt{1-r^2\omega^2}},-\frac{r\omega\;sin(t\omega)}{\sqrt{1-r^2\omega^2}} ,\frac{r\omega\;cos(t\omega)}{\sqrt{1-r^2\omega^2}} ,0 \right)$$

and therefore a four-force:
$$f=\frac{d(mu)}{d\tau}=\left( 0, -\frac{mr\omega^2\;cos(t\omega)}{1-r^2\omega^2} , -\frac{mr\omega^2\;sin(t\omega)}{1-r^2\omega^2}, 0\right)$$

So the magnitude of the four-force is:
$$|f|=\frac{m r\omega^2}{1-r^2\omega^2}$$

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In relativistic physics, the Born coordinate chart is a coordinate chart for (part of) Minkowski spacetime, the flat spacetime of special relativity. It is often used to analyze the physical experience of observers who ride on a rigidly rotating ring or disk.

http://en.wikipedia.org/wiki/Born_coordinates

I think there's a typo above, and the force should be

$$f=\frac{d(mu)}{d\tau}=\left( 0, -\frac{mr\omega^2\.cos(t\omega)}{1-r^2\omega^2} , -\frac{mr\omega^2\.sin(t\omega)}{1-r^2\omega^2}, 0\right)$$

Easy to do with so much Tex. Last edited:
Dale
Mentor
2020 Award
I think there's a typo above
Oops! thanks for the heads-up. I have fixed it.

Oops! thanks for the heads-up. I have fixed it.

Doesnt look like it's corrected, two dots missing, is m rest mass of the orbiting object? And f points radially inwards ?

Dale
Mentor
2020 Award
It is correct, the dots were also mistakes. Yes, m is the invariant mass of the orbiting object, and yes, f points radially inwards.

Awesome, also, if w=v/r, then this force is just gamma^2 * mv^2/r, so all we need to do when going from CM to SR is v->gamma*v ?