- #1

- 3,962

- 20

Looking on the internet, it seems to very difficult to find a simple straightforward statement of the equations for relativistic centripetal force. This is my point of view and hopefully with some feedback we can come to a consensus.

Consider a test mass moving in a circle with radius r. The Newtonian equation for the acceleration of the test mass is:

[tex]a = \frac{v^2}{r} = r \omega^2 = r \frac{d\theta^2}{dt^2} [/tex].

From the point of view of a non rotating inertial observer in flat space at rest with the centre of rotation, the acceleration of the mass in SR terms is identical to the Newtonian equation.

To an observer co-moving with the test mass, the acceleration of the test mass is:

[tex]a' = r \frac{d\theta^2}{d\tau^2} \gamma^2 = r \frac{d\theta^2}{dt^2} \gamma^2 = \frac{v^2}{r}\gamma^2 = a \gamma^2[/tex]

due to time dilation of the co-moving observer's clock. ([itex]\gamma = 1/\sqrt{ (1-v^2/c^2) }[/itex])

To the co-moving observer, no notion of relativistic mass is involved, so the force is simply:

[tex]F' = ma' = m \frac{v^2}{r}\gamma^2 = ma \gamma^2[/tex]

which is the proper force. If the test mass was resting on the inside of a rotating cylinder with radius r and angular velocity v/r, this is the force that would be measured by a set of weighing scales between the test mass and the wall of the cylinder. If the centripetal force is provided by a string of length r and the test mass was attached to the end of a string by a tension gauge, then this is the force that would be directly measured by the tension gauge.

It is known that the Lorentz transformation of transverse force is [itex]F = F ' \gamma^{-1}[/itex], so it follows that the force measured by the non rotating inertial observer at rest with the centre of rotation is:

[tex] F = \frac{F'}{\gamma} = \frac{ma'}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma [/tex]

This is not just a notional calculation of coordinate force, but is what would be measured by a tension gauge at the axis end of the rope. This is surprising, because the tension at one end of the rope is different from the tension at the other, which is not the case in Newtonian mechanics. It is also a measure of the centripetal force that would be provided by a non rotating electromagnetic field to keep the test mass moving in a circle. In both these cases, it is the force provided from the non-rotating frame to the moving test mass in the rotating frame.

Given that the Newtonian equation for angular momentum is [itex]L = mr^2\omega = mrv[/itex] it is not difficult to work out that the relativistic angular momentum of the test mass in the non rotating frame is [itex]m\gamma r v[/itex] from the above equations.

Consider a test mass moving in a circle with radius r. The Newtonian equation for the acceleration of the test mass is:

[tex]a = \frac{v^2}{r} = r \omega^2 = r \frac{d\theta^2}{dt^2} [/tex].

From the point of view of a non rotating inertial observer in flat space at rest with the centre of rotation, the acceleration of the mass in SR terms is identical to the Newtonian equation.

To an observer co-moving with the test mass, the acceleration of the test mass is:

[tex]a' = r \frac{d\theta^2}{d\tau^2} \gamma^2 = r \frac{d\theta^2}{dt^2} \gamma^2 = \frac{v^2}{r}\gamma^2 = a \gamma^2[/tex]

due to time dilation of the co-moving observer's clock. ([itex]\gamma = 1/\sqrt{ (1-v^2/c^2) }[/itex])

To the co-moving observer, no notion of relativistic mass is involved, so the force is simply:

[tex]F' = ma' = m \frac{v^2}{r}\gamma^2 = ma \gamma^2[/tex]

which is the proper force. If the test mass was resting on the inside of a rotating cylinder with radius r and angular velocity v/r, this is the force that would be measured by a set of weighing scales between the test mass and the wall of the cylinder. If the centripetal force is provided by a string of length r and the test mass was attached to the end of a string by a tension gauge, then this is the force that would be directly measured by the tension gauge.

It is known that the Lorentz transformation of transverse force is [itex]F = F ' \gamma^{-1}[/itex], so it follows that the force measured by the non rotating inertial observer at rest with the centre of rotation is:

[tex] F = \frac{F'}{\gamma} = \frac{ma'}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma [/tex]

This is not just a notional calculation of coordinate force, but is what would be measured by a tension gauge at the axis end of the rope. This is surprising, because the tension at one end of the rope is different from the tension at the other, which is not the case in Newtonian mechanics. It is also a measure of the centripetal force that would be provided by a non rotating electromagnetic field to keep the test mass moving in a circle. In both these cases, it is the force provided from the non-rotating frame to the moving test mass in the rotating frame.

Given that the Newtonian equation for angular momentum is [itex]L = mr^2\omega = mrv[/itex] it is not difficult to work out that the relativistic angular momentum of the test mass in the non rotating frame is [itex]m\gamma r v[/itex] from the above equations.

Last edited: