# Rotating mass increase?

1. Mar 21, 2004

### sazzles

Hi, I know that a moving object has a greater mass due to relativistic effects. This may be a stupid question, but will a rotating or spinning object's mass increase as its angular velocity increases?
Thank you for any help.

2. Mar 21, 2004

### Chen

When the angular velocity increases so does the "normal" velocity of the object ($$v = \omega r$$). So yes, if you change the angular velocity the mass of the object will change as well.

But bear in mind that in constant circular motion, both $$\omega$$ and $$v$$ are constant. So the mass of the object changes, relative to you, only when it first started to rotate and when it stops.

3. Mar 21, 2004

It's actual mass won't increase. It's rest mass is always constant.

The whole concept of "relativistic mass" of "effective mass" and "mass increasing" comes from the relativistic corrected momentum, i.e.

$$p = \gamma m_0 v$$

So an objects mass never increases.

However, an object with angular velocity is related to its linear velocity by

$$\omega = \frac{v}{r}$$

so it will experience relativistic effects for large enough radius.

4. Mar 21, 2004

### sazzles

Thanks, I think I understand now.
Just one small thing, I'm still confused about, does this mean the rest mass of an electron is its mass when it is not spinning, even though it is always spinning?

5. Mar 21, 2004

Although there is a property of electrons called "spin," they are not actually spinning like you would spin a tennis ball. Spin as it applies to subatomic particles is a little bit more abstract.

And yes, an electron's rest mass never changes.

6. Mar 21, 2004

### sazzles

Oh dear, now I am totally lost. If electrons are not actually spinning, how does the Pauli Exclusion principle work?

7. Mar 21, 2004

### Chen

They have a spin, but it's not the same kind of spin as that of a tennis ball or Earth.

Looks like someone thought this was an important enough subject to warrant its very own website:
http://www.electronspin.org/
(Ahh, sorry, apparently it's a site about a book with the same name.)

Last edited: Mar 21, 2004
8. Mar 21, 2004

### sazzles

Damn my text book, it says
.
The link doesn't seem to work, are you sure it is right?

9. Mar 21, 2004

### sazzles

oops, the link's fine, it's just my PC

10. Mar 21, 2004

### pmb_phy

The mass of which you speak is relativistic mass which, as you indicate, is a function of speed.

If you have a body which has a spatial extent (i.e. it is not a point particle) and as such it is meaningful to speak of it in the classical sense of spining then each part of the body, except those parts on the axis of rotation, is moving while the body as a whole stays at rest. The mass of the body is the sum of the masses so you simply integrate (i.e. add up) the contributions of all parts of the body. The result will be a function of the body's angular velocity.

For an explicit example see
http://www.geocities.com/physics_world/sr/rotating_cylinder.htm

sazzles was speaking of a spinning object, not an object which has a translational motion. That means, if I understood him correctly, the center of mass of the object remains at rest while other parts of the body move around the axis of rotation.

sazzles question indicates that she is speaking of relativistic mass and not rest mass (aka proper mass). Which one you call "actual mass" is a matter of taste and the subject of a long going debate in physics. E.g. Many people prefer to think of relativistic mass as "actual mass" and think of "rest mass" as "intrinsic mass."

However since the translational motion of the object is zero it makes no difference which one sazzles is speaking of since, in this case, even when the body has no translational motion it is still spinning and if the body is spinning even its rest mass increases. This can readily be seen since the "rest mass" of a macroscopic body is given by the "total energy"/c2 as measured in the zero momentum frame. Since a spinning body has more energy than the same body which is not spinning then it's "rest mass" increases as well. The mass derived in the above link and given in Eq. (8) is actually the rest mass of the object.

Also the electron's spin is not a classical spin in that different parts of the electron are moving.

Last edited: Mar 21, 2004
11. Mar 21, 2004

### sazzles

Wow, thanks pmb_phy for that detailed reply. It was really useful. One minor point, I'm a 'she' not a 'he', I thought my avatar and user name might have conveyed that.

12. Mar 21, 2004

### pmb_phy

Oops! Sorry. I corrected it. And you're quite welcome. This topic came up elsewhere in this forum. I think someone was curious about this "mass of a moving body" thing compared to "mass = rest mass" and in investigating the difference they asked a similar question. So I expected this question to arise again so I did out an explicit example and placed in on my web site. I'm glad to see that it was useful.

13. Nov 27, 2006

### Chris Hillman

When we spin up a CD, does its mass increase?

Hi, sazzles, you raised an interesting question way back in 2004:

Consider the question I raised in the title of this post: "When we spin up a CD, does its mass increase?" This question is analgous to "if we heat up a saucepan, does its mass increase"?

The best short answer to both questions is: yes! You might well ask: "doesn't this violate conservation of mass?" The answer is "no, because we did work when we spun up the CD and when we heated up the saucepan, and the energy we added slightly increased the effective gravitational mass".

Needless to say, these effects are very small when you use the CD drive in your computer or cook your dinner.

Particle by particle, that is true. In these examples, however, if we think of the saucepan and CD as composed of atoms (cautiously trying to avoid quantum mechanics as far as possible, however), then we can see that when we heat the pan, we make the individual atoms move faster, even though the pan as a whole is sitting on our burner. The increased kinetic energy however adds to the gravitational mass we assign to the pan. Similar remarks apply to the CD.

In short, I agree that it is best to think of the "rest mass" (or better, the "mass", unqualified) of a particle as an invariant, and the so-called "relativistic mass" as representing the mass plus the relativistic kinetic energy, but in gtr (and in many competing relativistic classical field theories of gravitation), all forms of mass-energy gravitate, so it should not be terribly surprising that simply making the constituents move in any way we can without making the body as a whole move will increase its effective gravitational mass.

Chris Hillman

14. Nov 28, 2006

### tehno

Gravitational field of the rotating object will increase.
Speaking about its' mass you have to accept or not to accept definition of a relativistic mass.

15. Nov 28, 2006

### pervect

Staff Emeritus
Measured in a frame in which the average momentum of the CD is zero, there is no difference between the invariant mass and the relativistic mass. Both of them increase.

Look at the defintion of invariant mass:

m = sqrt(E^2 - (pc)^2) / c^2

This defintion applies to systems in special relativity as well as point particles, if one just adds up the total energy E, and takes the magniutde of the total linear momentum vector, p.

(There are some unpleasant wrinkles that result from this when the system in question is not isolated, but we don't need to go into that, they don't affect the argument).

When you spin up the CD, you add energy to the system, while the momentum stays constant (zero), thus the invariant mass of the CD has increased.

Last edited: Nov 28, 2006
16. Nov 28, 2006

### Staff: Mentor

What avatar?

And your user name doesn't have any female connotation to me, which may simply mean that I'm old enough to be totally out of touch with your generation's vocabulary or culture. :uhh:

You don't need the concept of relativistic mass in order to say that a rotating object has a larger mass than a non-rotating one. An extended object is a bound system of particles. The invariant mass (a.k.a. "rest mass") of each individual particle is related to its energy and momentum by

$$E^2 = (pc)^2 + (m_0 c^2)^2$$

which gives

$$m_0 c^2 = \sqrt { E^2 - (pc)^2 }$$

The invariant mass of the system is related to the total energy and total momentum in the same way:

$$m_0 c^2 = \sqrt { E_{total}^2 - (p_{total} c)^2 }$$

In the reference frame in which the object's motion is purely rotational, ${\vec p}_{total} = 0$ (remember momentum is a vector), so $m_0 c^2 = E_{total}$ in that frame. But you get the same value for $m_0 c^2$ in any inertial reference frame; it's simply easiest to calculate in the frame in which the object has no translational motion.

A rotating object has more non-translational energy (in the form of rotational kinetic energy) than a non-rotating one, so it has a larger invariant mass.

Last edited: Nov 28, 2006
17. Nov 28, 2006

Staff Emeritus
From this reply and Chris's I take it that the "increase in gravitational mass due to the added kinetic energy of the parts" is accepted terminology. But I was taught to make a careful distinction between the gravitating mass and the total mass-energy tensor, which includes other forms of energy, not to mention momentum-stress. In this terminology what would increase would be the energy, and that would indeed contribute to the curvature. Am I out of date? Too inflexible? What?

On this same topic, my copy of Diagrammatica arrived yesterday, and I discovered that Veltman, writing in 1994, uses ict in his Minkowski metric. Yeee Gods! He devotes part of an appendix to the issue, but here's what he says in the text:
It occurs to me that this attitude, rigorously enforced, would shorten an awful lot of threads on thei forum.

18. Nov 28, 2006

### Staff: Mentor

Chris can address the applicability of this to gravitational mass... I was thinking purely in terms of SR.

19. Nov 29, 2006

### RandallB

I disagree:
If your last formula is correct then the first could be revised to:

$$E^2 = (p_{total} c)^2 + (m_0 c^2)^2$$

BUT this leaves you in the same position as Newton’s error in debating Leibniz on “Vis-viva”. In effect your arguing the Energy disappears as the momentum does. This was resolved by the French (du Chatelet) in the mid 18th Cent.
Basically you cannot sum the vectors of the many momentums and then square them to obtain Energy (vis-viva) you must square each momentum and then sum all the energies (note what happens to the momentum vector sign when squaring).

Then you can see that $m_0 c^2 = E_{total}$ is Not True!

Rest Mass, Invariant Mass, and IMO Gravitational Mass as well; all remain the same no matter how fast the object spins, or how much Kinetic Energy is stored in the rotation.

So I agree with selfAdjoint that mass remains unchanged and therefore does not affect the gravitational mass. But I’m not convinced that accounting for a ‘spin energy’ will produce any additional gravitation or gravitational curvature from some form of an energy tensor. I would need to see some experimental evidence to be convinced of that.

Last edited: Nov 29, 2006
20. Nov 29, 2006

### Chris Hillman

Gravitational energy or mass?

I've read a lot of papers (especially recent ones) but I'm not a physicist by training, so take my reading for what it is worth. A notorious difficulty with constructing thought experiments in gtr is that all forms of mass-energy gravitate, so to model a CD which is originally nonspinning and which is then spun up to some new spinning equilibrium state, one would presumably have to model something like a laser pulse coming in from infinity, striking the rim, and transfering momentum, which would be rather awful. Even in this case, since we are discussing changing the geometry of spacetime (i.e. changing the gravitational field), it can be tricky to compare the initial and final states "location by location". Nonetheless, I suspect that most researchers would speak of "augmented active gravitational mass" of the CD rather than "augmented gravitation due to the augmented kinetic energy of the bits of matter in the CD".

Hmm... here's a new conundrum: is the inertial mass of an operating CD player greater, due to the effect just mentioned? (I am thinking of a battery-operated model.)

Ye gods indeed! This particular choice of notation is now so nonstandard that it is likely to annoy many of his readers. History shows that otherwise valuable books which insist on employing highly nonstandard notation have less impact than they would have otherwise, so I feel his choice may prove to be self-defeating. This is actually quite a serious issue for any student who might plan to adopt his notation and who might later try to read some gtr literature. I have in mind important techniques like NP tetrads, where the notation is notoriously delicate, and I can think of many other places where huge changes in notation would make it much harder to avoid sign errors in "translating" from a paper by Chandrasekhar, say, to the notation prefered by Veltmann. It's hard enough to avoid errors of this kind even if one does not make changes of the kind he advocates!

I happen to feel that there are many good reasons for prefering real variables here, quite apart from the issue of consistency with the huge literature on gtr, almost all of which uses the terminology and notation of semi-Riemannian geometry. I admit that the standard terminology does lead to some common student misconceptions, e.g. the "metric" structure we are imposing to form a Lorentzian manifold (by bundling an indefinite but nondegenerate quadratic form) certainly does not play well with "metric topology" as in elementary courses on general topology, unlike Riemannian manifolds (obtained by bundling a positive definite quadratic form), but these misconceptions are easily cleared up.

Chris Hillman