Relativistic centripetal force

1. Sep 25, 2007

Helios

I derived the (magnitude of ) relativistic centripetal force

mv$$^{2}$$/R[1-(v/c)$$^{2}$$]$$^{3/2}$$

but my few books don't even mention relativistic centripetal force. Can someone corroborate this?

2. Sep 25, 2007

pervect

Staff Emeritus
Hmm - can you tell me first what you mean by "relativistic centripetal force"? Specifically, if we have some mass m moving in a circle, are you computing

dp/dt, where t is the time in the lab frame

dp/dtau, where tau is the "proper time" of the moving particle

or perhaps, something else, entirely?

3. Sep 26, 2007

Helios

We know what centripetal force means. We know that relativistic means close to the speed of light. Let me create a word problem.

Superman holds a 1m wire with a 1 kg ball attached to the end. He spins the ball in a circular motion. The speed of the ball was measured at 9/10 the speed of light. What is the centripetal force or the inward directed force?

4. Sep 26, 2007

HallsofIvy

Staff Emeritus
You still haven't answered pervect's questions! "Force" isn't just a specific number. Like (almost) everything else, it is relative to the frame of reference of the observer.

I suspect you want the force from superman's perspective, from the frame of reference in which the center of the circle around which the ball is moving is motionless. Do you see that the motion of the ball is always perpendicular to the radius of the circle in that frame of reference? Do you remember that relativistic effect happen along the direction of motion?

5. Sep 26, 2007

Helios

Let me say it this way.

Superman holds a 1m wire with a 1 kg ball attached to the end. He spins the ball in a circular motion. The speed of the ball was measured at 9/10 the speed of light. The wire was equipt with a TENSION SENSOR. As Superman spun the ball, a digital display could display the magnitude of the tension ( a specific number, mks units ). What was this value?

6. Sep 26, 2007

pervect

Staff Emeritus
I doubt that the above quantity is a single number, i.e. I think it will probably depend on where the tension sensor is mounted (i.e. at the center, or near the ball).

I think it might be instructive to show how to compute the 4-acceleration of the ball, which should be the same as the force on the ball from the perspective of someone on the ball, and also for a tension sensor mounted near the ball.

This is also the reading that an accelerometer on the ball would have, i.e. the "proper acceleration".

The path that ball takes through space time is

x = cos(omega*t)
y = sin(omega*t)

To find the 4-velocity (and hence the 4-acceleration) we need to re-write this as

x(tau), y(tau), t(tau)

where tau is proper time. We will also write as tau as $\tau$ and omega as $\omega$ when we use latex in some of the more complex equations.

If we use geometric units, so that c=1, we can write:

$$x = r \cos \frac{\omega \tau}{\sqrt{1-r^2 \omega^2}}$$

$$y= r \sin \frac{\omega \tau}{\sqrt{1-r^2 \omega^2}}$$

$$t=\frac{\tau}{\sqrt{1-r^2 \omega^2}}$$

We can then compute the 4-velocity as (dt/dtau, dx/dtau, dy/dtau). We can confirm that -(dt/dtau)^2 + (dx/dtau)^2 + (dy/dtau)^2 = -1.

See the wikipedia article on four-velocity] for more information on this approach.

Similarly, we can compute the 4-acceleration. See the wikipedia article on 4-acceleration.

I used maple to calculate the above. I won't give all the components, though I'll mention that dt^2 / dtau^2 = 0.

The final end result is that the magnitude of the 4-acceleration is:

$${\frac {{r}{\omega}^{2}}{ \left( 1 - {r}^{2}{\omega}^2\right) }}$$

i.e. if we let v = r omega

$$\frac{v^2}{r \left(1-v^2\right)}$$

Note that because I've used geometric units, v is a dimensionless number that would be equal to (v/c) in non-geometric units.

This is different than your result, you must have been calculating something else (and unfortunately you haven't explained what you've calculated at all, or how you went about doing it).

Last edited: Sep 26, 2007
7. Sep 26, 2007

Helios

So you're saying that acceleration has the dimensions of reciprocal length??? That's what your last formula indicates. Acceleration has the dimensions LENGTH/TIME SQUARED. You must have been calculating something else.

8. Sep 26, 2007

Staff: Mentor

I suspect you are misreading pervect's tau ($\tau$), which indicates proper time, as an r. Also note the special units being used:

9. Sep 26, 2007

Helios

Well I guess I did! but his formula is STILL wrong. Acceleration does not have the dimensions of reciprocal time either. Wait! that is an "r" implied by his last step.!?

Would pervect use a capital R if he intends to correct his work?

Last edited: Sep 26, 2007
10. Sep 26, 2007

genneth

As pervect said, he's using units where c=1, so the units of time and length are the same. Therefore, displacement has units of length or time, velocity is unitless, and acceleration is reciprocal time or reciprocal length. I haven't gone through his algebra, but it looks correct here. Notice that he said the force measured by a sensor would depend on where the sensor is situated -- this is a crucial phenomenon in this case.

11. Sep 26, 2007

Helios

If you think it's correct then write his last formula in the conventional way, not in "geometrical units". Then we'll see. Call me "old school" but to say that "acceleration is reciprocal time or reciprocal length" is absurd.
Also "force measured by a sensor would depend on where the sensor is situated" is also absurd. It is not "a crucial phenomenon in this case". If the tension differed throughout the wire, points on the wire would not be static with respect to one another. Let's not muddy up this problem by contemplating the composition of the wire. At the start I had to invent the tension sensor just to avoid questions like "what do I mean by force".

Last edited: Sep 26, 2007
12. Sep 26, 2007

genneth

You know, physicist have been using geometrical units since special relativity became widely used. It's only elementary physics courses that even bother to not use them. Furthermore, it's actually popular to use units where c=1 and hbar=1, as it avoids a lot of wasted ink. In this case, I'll oblige and do the obvious translation:

$$a = \frac{v^2}{R\left(1 - \frac{v^2}{c^2}\right)}$$

Now, we said that the tension measured by the sensor is different depending on where the sensor is -- but that doesn't imply the wire is moving. If I have a thick, heavy wire, just dangling freely, the tension on different parts of the wire is different, but the wire is still stationary. However, that example is simply meant to illustrate that a varying tension doesn't imply motion, NOT to suggest a reason for the non-uniform tension. The precise reason for the non-uniform tension is simply that as measurement of force inevitably involve measuring time, you have to contend with time dilation. Sensors at different points along the wire would be subject to different amounts of time dilation.

Last point: pervect is pretty good at relativity related stuff. I suggest you trust him, and his analysis. We're not exactly here just to make up random stuff to confuse people.

13. Sep 26, 2007

Helios

What do you mean "the obvious translation" ? You got rid of a c-squared with a sleight of hand. Where did it go? It's not obvious to me.
Also you say "measurement of force inevitably involve measuring time". Oh? Why is this true? If I have a sensor with a spring or counter-weight, there's no time measure needed. I don't need a watch to weigh myself.
Let's just say the sensor is a microscopic hookian transducer. Please resist mentioning the composition of the wire.

Last edited: Sep 26, 2007
14. Sep 26, 2007

genneth

Perhaps I was speaking too loosely. Obvious is only ever obvious when you already know the answer; my bad. But I hope it's clear that in units where c=1, the c just disappears. To know where to put it back just requires a bit of dimensional analysis -- 1-v^2 certainly has the wrong units, but a substitution of v->v/c fixes it.

So far no-one has mentioned the composition of the wire. I don't know why you keep bringing it up... In any case, what makes you think the modulus of a wire isn't subject to relativistic effects?

15. Sep 26, 2007

Helios

My point is made -- pervect's formula for acceleration is flawed by dimensional analysis.

You ask--"what makes (me) think the modulus of a wire isn't subject to relativistic effects?"

answer--because there is no motion in the radial direction. Just suppose of the sensor is a hookian spring. Measure the spring and know the force. There is no "length contraction" of the sensor. There is no "time dilation" of the sensor. It doesn't matter where the sensor is on the wire. It doesn't matter what company made the sensor. In fact, let's just say Superman has the abilty to tell us the force! What would he tell us? What would he tell us is the CENTRIPETAL FORCE?

Last edited: Sep 26, 2007
16. Sep 26, 2007

genneth

Well, let's do the analysis again, this time in his frame. The position of the particle is still (x,y) = (R cos wt, R sin wt). In this frame, everything is in fact pretty simple, as the times all match up: a = rw^2 = v^2/r where things are measured in his frame. As you can see, it's not the same as the acceleration as measured in the moving frame.

17. Sep 26, 2007

pervect

Staff Emeritus
Acceleration has units of reciprocal length in geometric units. See for instance http://en.wikipedia.org/wiki/Geometrized_unit_system

But I'll try and redo the post in non-geometric units since you don't seem to like geometric units.

Last edited: Sep 26, 2007
18. Sep 26, 2007

pervect

Staff Emeritus
This is an actual error (typo). It should be

x = R*cos(omega*t) and y= R*sin(omega*t)

In standard units, that's

$$x = R \cos \frac{\omega \tau}{\sqrt{1-R^2 \omega^2 / c^2}}$$

$$y= R \sin \frac{\omega \tau}{\sqrt{1-R^2 \omega^2/c^2}}$$

$$t=\frac{\tau}{\sqrt{1-R^2 \omega^2}}$$

In standard units, the square-norm is -c^2 (dt/dtau)^2 + (dx/dtau)^2 + (dy/dtau^2), and it should be equal to -c^2.

In standard units, let $\beta = R \omega / c$ Then we can write the acceleration as

$$\frac{\beta^2 c^2}{R \left(1-\beta^2 \right)}$$

Last edited: Sep 26, 2007
19. Sep 26, 2007

Helios

Would pervect write the formula for centripetal force that he think's is correct in a conventional form with the International System of Units so that we could compute the force in Newtons?

20. Sep 26, 2007

pervect

Staff Emeritus
See above. Reload the page if you've just looked at it, I've edited it several times to get the latex right.

Note that I'm not the one calling this the "centripetal force". What I'd call what I'm computing is the proper acceleration of the body, i.e. the acceleration that someone on the body would measure with an accelerometer.

The force on the string as measured by the person holding the string will be different. What I"ve computed above would be the force as measured by a force gauge mounted on the far end of the string, near the body. This is because it's easy and standard to compute, and there shouldn't be much argument about the computation.

Last edited: Sep 26, 2007