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A solid sphere of mass M is rotating along an axis.

We can consider it as a collection of large number of point masses, every point mass is moving with respect to center of mass with velocity which depends on its radius from rotating axis. Then, according to relativity, the mass of every point increases and consequently the mass of the sphere increases.

But if we consider the overall sphere, it is not moving at all and its mass remains the same. which produces a contradiction.

Please tell me where I am wrong.

Kyle Kanos
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vats dimri
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    Hint: $K_\text{ball}=K_\text{CoM}+K_\text{rotation}$, where $K$ denotes kinetic energy. Your text seems to equate $K_\text{ball}$ with $K_\text{CoM}$ instead. – Danu Mar 12 '15 at 13:08
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    Mass of the sphere does not increase – Sean Mar 12 '15 at 13:16
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    Note that the notion of "relativistic mass" is misleading and outdated, which appears to be causing the confusion here. – Kyle Kanos Mar 12 '15 at 13:29
  • The relativistic mass does increase, and I don't think this is a relativistic mass vs. rest mass confusion. It's more subtle, having to do with the difference between stationary and static mass distributions. –  Mar 12 '15 at 14:37

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Theoretically, a sphere rotating at the speed of light (on its outer edge) would have an infinite mass due to the mass portions on the outer edge. A rotating sphere IS moving and has momentum (that is why flywheels can store energy).

Practically, any such sphere would fly apart far before any kind of relativistic speeds would be reached.

Jiminion
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The mass of a moving body increases only for the observer in a rest frame, not for the moving body, as it feels itself as being in rest.

The rotating sphere is a different story, there are other forces and the point on moving sphere feels them, you cannot play a simple paradox game here. The situation is more difficult. I suppose you dont have any gravitation forces in the example.

However, as @Danu commented, you put a kinetic energy $K_{rotation}$ to the sphere to rotate it, so your total energy must increase by this. I expect that $E=mc^2$ contains all types of internal energies, so the rotation should come there also and your mass is increased.

jaromrax
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  • Can't we apply the mass increase formula to center of mass of the sphere? – vats dimri Mar 12 '15 at 13:31
  • I think, we are in CMS of the sphere all the time. Or do you want to rotate also your coordinate system around the sphere to see it in rest? But there would be very strange forces in your new system...making difficult considerations about the sphere points. – jaromrax Mar 12 '15 at 14:06
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If the sphere is not moving, than is not even rotating on its axis.

I don't really get what you're trying to say. Relativistic mass increases with speed, so if you're considering a moving sphere and a standing one they're not gonna have the same mass, but in the formula of relativistic mass m0 comes in, which is the rest mass (when the object is not moving).

user3211966
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In relativity (but often not Newtonian physics) there is a huge difference between a static mass distribution and a stationary one. A static distribution is one in which there is no velocity, whereas a stationary one is defined by looking the same at any given time. All static distributions are stationary, but your rotating sphere is an example of a stationary distribution that is not static.

The fact is, the rotating sphere has additional angular momentum and energy. Thus it does have a greater "relativistic mass" (i.e. total energy minus rest mass energy). In fact, there's nothing terribly relativistic about the misconception here; a rotating Newtonian sphere has a kinetic energy, even though the mass isn't going anywhere.

For concreteness, your approach of looking at each point mass works (ignoring material stresses). If a uniform sphere has rest mass $M$ and radius $R$, its rest mass density will be $\rho = 3M/4\pi R^3$. If it is rotating with angular velocity $\omega < c/R$, then each point with colatitude $\theta$ has velocity $R\omega \sin\theta$, and so the total relativistic mass is $$ M_\mathrm{rel} = \int\limits_\text{sphere} \frac{\rho}{\sqrt{1-(R\omega/c)^2\sin^2\!\theta}} \, \mathrm{d}V = \frac{Mc}{2R\omega} \log\left(\frac{1+R\omega/c}{1-R\omega/c}\right) = M \left(1 + \mathcal{O}\left((R\omega/c)^2\right)\right). $$ This diverges as $R\omega \to c$.