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Relativistic Sling Forces

Posted April 28, 2010 12:00 AM by Jorrie
Pathfinder Tags: Centripetal force relativity

The Newtonian centripetal force is well known as: FN = m r ω2 = m v2/r , which is the force needed to move a mass m in a circle with radius r at an angular velocity of ω.

Let mass m be at the end of a long string with negligible own mass and length r. Anchor the string at the origin of an inertial coordinate system in free space. Without altering the string length, insert two small force gauges (of negligible mass) inline, one at the 'origin end' and one at the 'mass end'. Sling the mass around the origin. According to Newton, the gauges will both measure the centripetal force as given above. Now sling the mass at a relativistic transverse velocity (v=rω). What happen to the readings on the two gauges (given that the string doesn't break)?

One can half-expect that they will record higher values than given by the above Newtonian formula. What is very surprising is that Special Relativity (SR) predicts that the gauge at the 'mass end' will record a higher force (Fm) than the gauge at the 'origin end' (Fo). This is despite the fact that they measure forces in the same ('massless') string!

I will give the SR formulas first and then discuss how it works.

Fo = m v2/[r √(1 - v2/c2)] and Fm = m v2/[r (1 - v2/c2)]

Note the square root below the line for Fo. That's the normal relativistic velocity time dilation factor. For Fm there are hence two time dilation factors below the line. For readability, replace 1/√(1 - v2/c2) with γ, the Lorentz factor, so that:

Fo = γ m v2/r and Fm = γ2m v2/r

As a numerical example, let's use 1 kg at the end of a 1 km long string and swing it 1% of c, or 3,000,000 m/s, giving γ = 1.00005. If my sums are correct, the forces will be (in kilo-Newton):

FN ≈ 9,000,000 kN

Fo ≈ 9,000,450 kN

Fm ≈ 9,000,900 kN

OK, there are surely no strings that can take these forces, but it illustrates the relative values. The 'hand-waving' argument for the difference between Fo and Fm is simply that the gauge at the mass moves at relativistic speed relative to the gauge at the origin, so it measures a higher force (by the time dilation factor). But, that's not good enough - some clarifications are needed.

To start with, measuring force essentially means measuring acceleration. Unit-wise, acceleration is a distance interval divided by the square of a time interval. As in the famous 'twin-paradox', a clock moving non-inertially (accelerated) is time dilated or slowed down by a factor γ relative to an inertial clock. This means the circling clock measures the said time interval as too short by this factor. Square that and we have the reason why the mass-gauge will experience an acceleration that is a factor γ2 larger than what Newton would have predicted.

Secondly, why would the origin-gauge register a force of only a factor γ larger than the Newton value? We now have to think 'moving mass' for the circling object. As far as the origin observer is concerned, the 1 kg mass now weighs γ kg and would hence need a force γ times what Newton predicts. This is smaller than the γ2 times Newton that the mass-gauge registers. This is still not very rigorous, but it gives the general idea.

Confusing, but true according to SR; and who would wish to argue with Einstein on that?

-J

Notes.

1. The whole thing is so counter-intuitive: if the string had a small, but significant mass, Newton would predict that the gauge at the mass-end will record a smaller force than the one at the origin. A relativistic treatment becomes very complex in such a case, but normally the mass at the end will dictate and the end-gauge will still measure a larger force than the one at the origin.

2. Relativity 4 Engineers appendix C treats relativistic accelerations in a general way, but also leads up to fairly technical discussions.

3. If you have the inclination, here is a paper by Hrvoje Nikolic, with a seriously technical treatment of relativistically rotating frames.

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Anonymous Poster
#1

Re: Relativistic Sling Forces

04/29/2010 3:18 AM

what if the there was no string and there were a chain of gauges(negligible mass) one after the other till the mass at the end? Will the reading in each of the gauges increase as we reach the masss?

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Guru
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#2
In reply to #1

Re: Relativistic Sling Forces

04/29/2010 7:37 AM

Yes, although I have not tried to calculate that, I believe that the force will increase smoothly, but not linearly, from Fo to Fm.

-J

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Guru
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#3

Re: Relativistic Sling Forces

10/07/2010 5:35 AM

1. For human acheivable velocities, values of Fo and Fm are practically same. Where we find application of this negligible difference?

2. When artificial satelite is blasted off with greater than escape velocity, it takes highly elliptic orbit. Is the 'Relativistic Sling Forces' apply in this case? Why not it take exactly circular orbit like sling?

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Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

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#4
In reply to #3

Re: Relativistic Sling Forces

10/07/2010 7:49 AM

Hi pritam,

I know of no practical place where the relativistic sling forces are significant. It is just a theoretical oddity.

BTW, when a probe is blasted off at greater than escape velocity, it takes on a hyperbolic orbit, not a highly elliptical one. You may perhaps be thinking of the "escape transfer orbit", which is highly elliptical. The final energy to escape Earth is added later, near the perigee of this orbit, where it is most cost effective.

-J

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