# Rotational kinematics

1. Apr 12, 2004

### FulhamFan3

Everyone that's taken basic physics know that the kinetic energy of a spinning object with no translational velocity is

KE=(1/2)Iw^2.

where I is the moment of mass and w is the angular velocity. I've been trying to find a similiar formula incorporating special relativity but the math is more challenging than I thought. Does anybody have an answer to this?

2. Apr 12, 2004

### turin

Not off the top of my head, but I would assume that you could use a "relativistic" moment of inertia:

I = I(ω).

I'll see if I can come up with anything in the next couple o' whiles or so.

EDIT:
Alrighty then. I did a derivation for the kinetic energy of a rotating disk and got an expression. It agrees to first order with the non-relativistic expression (I think; I'll leave it to you to double check the expansion of the natural log and then let me know). Did you want something that has the same form as the non-relativistic expression? If so, I don't think I can help you.

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Last edited: Apr 12, 2004
3. Apr 12, 2004

### DW

You get such a simple equation from Newtonian mechanics because of the form the Newtonian kinetic energy equation takes. The Kinetic energy of the body in its center of momentum frame is
$$KE = \Sigma (1/2)m_{i}v_{i}^2$$
When all the center of momentum frame motiom is the same angular motion this simplifies as
$$KE = \Sigma (1/2)m_{i}r_{i}^{2}\omega ^2$$
$$KE = (1/2)[\Sigma m_{i}r_{i}^{2}]\omega ^2$$
$$KE = (1/2)I\omega ^2$$
For special relativity it would be
$$KE = \Sigma (\gamma_{i} - 1)m_{i}c^2$$
Even if all the motion is the same angular motion you see that there is no general way to separate the angular velocity terms from the distance terms to yeild a similar simplification. So this as complicated as the sum is must remain is the SR expression for rotational kinetic energy.

4. Apr 13, 2004

### FulhamFan3

instead of a general espression like "I" how about just a disc with uniform density and radius r.

turin: in your formula what is z and p

5. Apr 13, 2004

### DW

If all particles it is comprised of move at the same angular velocity then the integral yields:
$$KE = \sigma \pi r^{2}c^{2}[2\frac{c^{2}}{r^{2}\omega ^{2}}(1 - \sqrt{1 - \frac{r^{2}\omega ^{2}}{c^{2}}}) - 1]$$
Good luck separating the distance information from the angular velocity into a seperate factor you may call the "moment of inertia".

Last edited: Apr 13, 2004
6. Apr 13, 2004

### FulhamFan3

ok, now that we have a forula can I see how you derived it?

7. Apr 13, 2004

### DW

Take the equation I gave you:
$$KE = \Sigma (\gamma_{i} - 1)m_{i}c^2$$
Rewrite as an integral
$$KE = \int_{0}^{r} [\gamma (r) - 1]c^{2}\sigma 2\pi rdr$$
where $$\gamma (r) = \frac{1}{\sqrt{1 - \frac{r^{2}\omega ^{2}}{c^{2}}}}$$
and do the integration.

8. Apr 13, 2004

### turin

The formula is for a uniform, perfectly rigid, extremely flat cylinder (disk). &Delta;z is the height of the cylinder (thickness of disk). &rho; with the ~ over it is the proper mass density of the cylinder (disk).

9. Apr 14, 2004

### DW

Of course since there is no absolute perfect rigidity in special relativity you really had to choose a frame according to which it remained of uniform density. In essence you are doing the same thing as I did except I chose a different behavior for my $$\sigma$$. I chose for my $$\sigma$$ to be uniform with respect to the inertial frame for which it spins without translation.

10. Apr 14, 2004

### turin

DW,
Actually, I don't know how to treat the rigidity issue, so I ignored it. I assumed a T00 = &rho;proper c2 &gamma;2, and then integrated that over the volume (area, actually) of the disk, assuming that the area of the disk would not change. I appologize for not listing this as an assumption. I assumed that, since the motion was entirely transverse, the radius of a given element of the disk should not change. Of course, this probably induces a stress, but I assumed that the induced stress would only show up in the Tij components, and since I considered only the T00 for the energy, I shouldn't have to regard the induced stress.

11. Apr 14, 2004

### outandbeyond2004

The way I would do the problem is to start with the object at rest in an inertial reference frame, with the origin at the object's center of inertia. (Do I assume correctly that the spin axis is through the center of inertia?) Then impart an angular velocity to the object d$\omega$. Calculate the energy using DW's formula. That would give us an equation relating the incremental energy to the incremental angular velocity. That can be integrated . . .

Last edited: Apr 20, 2004
12. Apr 15, 2004

### turin

If you use DW's formular, I think you would first need to differentiate wrt ω to do what your doing. That would give you dKE/dω. Then, incrementing by dω would give you an increment of dKE. Putting this in an integral would give you integral of dKE = KE. Am I looking at the wrong DW equation?

Last edited: Apr 15, 2004
13. May 14, 2004

### Sammywu

I wonder whether you shall treat this problem separately in a microscopic environment or a macro environment.

In a microscopic environment, I seem to see that a rotational system forms a very stable motion. If we try to assume a set of even distributed matter fields Phi_i and with rotation as Killing vector fields, a rotation can be seen as a symetric motional system, will that lead to a rule such that in this system, it's stable enough so that its motion is conserved just like a linear motion.

In a macroscopic environment, there is no true even-districute matter field. The rotational system is unstable and will continuously tend to reform into a more stable motion system.

14. Nov 23, 2007

### bjschaeffer

There is an important application of the relativistic moment inertia.
Macgregor, in his book, The enigmatic electron, has shown that the moment of inertia of a sphere rotating at its maximum velocity, that is, the speed of light at the equator, is 1/2mr^2. This result explains very simply why the spin of the electron is 1/2.

15. Nov 25, 2007

### Chris Hillman

Warning: necropost!

BJ, resurrecting long-dead threads can cause confusion for frequent posters so it is frowned upon at PF.

Oh wow.... you just reopened a gaping can of big fat wriggling carnivorous flesh-feeding maggots. Earlier this year pervect, myself, and Greg Egan engaged in a long thread on the so-called "Ehrenfest paradox", which concerns the relavistic spinning disk but which is also relevant for any relativistically spinning object. From this thread it should be evident that there are numerous common misconceptions even concerning rotation in flat spacetime, which take time and effort to overcome (not to mention mastery of modern computational tools; see Poisson, A Relativist's Toolkit). I would characterize one theme of this long discussion as "anyone who says 'It's perfectly simple' is talking like a d--n f--l'!", if I might so express it.

Well, there are many claims that of the form "suggestion S explains mysterious datum M". The problem is usually that S conflicts with A, B, ... L.

The book you mentioned, M. H. MacGregor, The Enigmatic Electron, Kluwer, 1992, concerns a classical model of the electron and therefore could be unkindly described as belonging to "the literature of the anachronism". See the https://www.amazon.com/Enigmatic-El...9826/ref=sr_1_1/002-3007882-05&s=books&sr=8-1 from the publisher. It seems fair to say that suggestions that the electron can be well-modeled using Reissner-Nordstrom or Kerr-Newman solutions are somewhat notorious in gtr. Someone has been pushing fringe to cranky views :yuck: for quite some time over at this WP article, which tends to be rather misleading as a result. I see that recent versions mention the book you cited.

Last edited by a moderator: May 3, 2017
16. Nov 25, 2007

### pervect

Staff Emeritus
While I don't have access to that book, from what I have been able to see online at google books, I don't believe the result. But it's good to have a definite source for where this idea came from, even if it is from a necropost.

The problem is that the momentum per unit volume of an object under stress is not equal to the relativistic mass times the velocity. Thus integrating the relativistic mass will not give a correct value for the relativistic moment of inertia tensor, because one cannot compute the relativistic momentum from density * relativistic mass * velocity.

See for instance Rindler, "Intro to special relativity", second edition, pg 132, eq (45.9)

The x component of the momentum density for an object under stress (with a stress energy tensor in its rest frame of $$t^{ij}_0$$ is

$$g^1 = \mu \gamma^2(\mu) \left( \rho_0 + t^{11}_{0}/c^2 \right)$$

Here $\rho_0 c^2 = t^{00}_0$ and the object is moving in the x direction with v/c = $\mu$.

Thus we see that stress in the rest frame of the object affects its momentum density, an example of how energy and momentum intermix in SR.

While some detailed work has been done on the relativistic ring and hoop that I believe is more likely to be correct, it unfortunately has *not* been peer reviewed. I'll send you a link via PM to the webpage that I believe is most likely to be approximately correct even though it hasn't been peer reviewed. (I don't want to be mysterious - we've recently tightened up our policy on links to webpages, which is why I'm not posting it).

Last edited: Nov 25, 2007
17. Nov 26, 2007

### bjschaeffer

I will see for instance Rindler later. Please find below the complete calculation and let me know what is false in it.
MacGregor assumes that the velocity at the equator of the spherical electron is the velocity c of light. Let us compute the observed mass of a filled sphere of radius R in relativistic rotation with its axis at rest. $\rho_0$ is the specific mass on the axis where the matter is at rest. The distance to the axis is r. The height of an elementary cylinder of thickness dr is h. The linear velocity of rotation is v=ωr=(c/R)r (=c on the equator). We have then

$$m=\int_0^R\rho_0(r)2\pi(2h)rdr =2\pi\rho_0 R^2 \int_0^R\frac{2\sqrt{R^2-r^2}}{\sqrt{1-\frac{(\omega r)^2}{c^2}}}d\frac{r^2}{2R^2} =2\pi\rho_0 R^3 \int_0^R\frac{\sqrt{1-\frac{r^2}{R^2}}}{\sqrt{1-\frac{c^2 r^2}{R^2 c^2}}}d\frac{r^2}{R^2} =2\pi\rho_0 R^3 \int_0^Rd\frac{r^2}{R^2}=2\pi\rho_0 R^3$$
The same calculation may be done for the moment of inertia of the sphere:

$$I=\int_0^R r^2 dm=\int_0^R r^2\rho_0(r)2\pi(2h)rdr =2\pi\rho_0 R^3 \int_0^R\frac{2\sqrt{1-\frac{r^2}{R^2}}}{\sqrt{1-\frac{r^2}{R^2}}}d\frac{r^4}{4R^4} =\pi\rho_0 R^3 \int_0^R d\frac{r^4}{R^4}=\pi\rho_0 R^5=\frac{1}{2}mR^2$$

The Einstein-Planck relationship gives the proper rotational frequency of the electron. This implies that the Einstein-Planck frequency be the same as the rotational frequency of the electron. It is of course a very simple hypothesis:

$$\omega=2\pi\nu=2\pi\frac{mc^2}{h}$$

Knowing the rotational frequency, we are able to compute the radius of the electron:

$$R= \frac{c}{\omega}=\frac{\hbar}{mc}=R_C$$

$R_C\$ is the Compton radius, the same appearing in the Klein-Gordon equation, and, of course in the Compton effect. MacGregor obtains then the intrinsic angular momentum of the electron :

$$L=I\omega=\frac{1}{2}mR_C^2\omega=\frac{1}{2}mvR_C=\frac{1}{2}mc\frac{\hbar}{mc}= \frac{1}{2}\hbar$$

in accord with the observation.
Usual quantum mechanics is unable to get such a result with understandable calculations.

18. Nov 26, 2007

### pervect

Staff Emeritus
This equation appears to be based on finding the relativistic mass, which if I've understood the book's symbology correctly, the author calls $\rho_0(r)$ as a function of r.

Unfortunately, as I noted earlier:

The correct formulation for the i'th component of momentum in flat space-time is

$$P^i = \int T^{0i} dV$$

where $$T^{ij}$$ is the stress-energy tensor. The 0'th component of the momentum is the energy.

This is based on the fact that one of the interpretations of the stress-energy tensor is that $T \cdot u$ gives the density of energy-momentum in a region, where T is the stress-energy tensor and u is the 4-velocity, see for instance MTW's "Gravitation". (You'll probably find similar remarks in Rindler as well with different notation.)

To deal with spherical coordinates in flat-space-time it might be useful to note that one would more generally write:

$$P^i = \int \sqrt{\left| T^{0i}\,T_{0i} \right| }\, dV$$

(I believe this is correct, but it's from fallible memory, not directly from a textbook).

The issue of how to deal with non-flat space-times gets much more involved, but hopefully we don't need to get into that. I'll just point out that one needs certain preconditions to even *define* the total energy or total momentum in a non-flat space-time.

Because the stress terms in the stress-energy tensor do affect the momentum, one would need a complete material model, including the stresses, to find the answer. The issue is complicated by the non-existence of a Born-rigid spherical spinning body, see for instance the sci.physics.faq, so it's not clear that the notion of a rigid sphere spinning at light speed at its periphery makes any physical sense.

See The rigid rotating disk in relativity for a popular discussion and some references. The result for the non-existence of a spinning Born-rigid sphere is due to Ehrenfest, and is the well-known Ehrenfest paradox which Chris Hillman talked about.

The associated faq entry is:

Last edited: Nov 26, 2007
19. Nov 28, 2007

### bjschaeffer

The formula by Rindler is applicable only if the mechanical properties of the material making the electron are known. This is not the case.
The centrifugal acceleration, $\omega c\simeq 10^{28}$ is an enormous quantity. No macroscopic material resists such stresses. The strength of elementary particles needs a different approach and the point is that the electron is not broken simply because it exists. The use of elasticity theory is a nonsense for the electron sphere.

The calculation of MacGregor is stationary, not transitory and purely geometric. The process of creation of the spin is unknown. It is possible to compute without using the mass, with only the volume depending of the radius. The calculation is valid only for an electron at low linear velocity so that there is only one frame which is the frame centered at the center of the sphere. The Ehrenfest paradox concerns the problem of accelerating an electron from zero to a given rotational velocity. This problem is not only very complicated, but unrealistic because it will probably never be possible to rotate at will an electron at a given rotational velocity.

20. Nov 28, 2007

### pervect

Staff Emeritus
I agree that one needs to know the mechanical properties of the material to use the formulas that I've given, and that the mechanical properties of the electron (if the electron can even be understood in this manner) are not known.

I doubt that the formula given by MacGregor and the one given by the references I quoted can both be correct if applied to a material whose mechanical properties are known, because one formula has a result which depends on the state of stress, and the other formula ignores stress.

I would tend to agree that the electron couldn't possibly be made out of anything resembling a normal material, which makes it hard for me personally to get very comfortable with MacGregor's theories.

However, addressing MacGregor's theories as a whole is not something that I'm personally particularly interested in. In any event, discussion of MacGregor's theories would seem to be best done in some other forum, as it doesn't seem particularly relevant to special or general relativity or to the now departed OP's question.

What I can talk about is not MacGregor's theories, but rather the mechanical properties of continuous media according to relativity, which I have done.

I'll add that if one applied MacGregor's formula to calculate the moment of inertia of a star, I would expect that one would not come up with the standard answers given by the literature, for instance (from a google search)

http://arxiv.org/abs/gr-qc/0504020v1 (to first order only)
http://www.aanda.org/index.php?option=article&access=standard&Itemid=129&url=/articles/aa/full/2001/23/aa10321/aa10321.right.html [Broken]

Last edited by a moderator: May 3, 2017