We have settled on a standard way of expressing mass at relativistic speeds, as per Matt Strassler's excellent blog.

That is no problem for linear momentum, mass is invariant.

All I want to know is does the same principle apply to angular momentum, for every object, from golf balls to neutron stars? Or have I pushed the invariance idea into the wrong area?

EDIT I just discovered this post Does rotation increase mass and I really want to avoid semantics about the word mass, but it seems to say mass does increase END EDIT

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    $\begingroup$ Read Ben's answer to the earlier question more carefully. He uses exactly the same example system I used and get the same result (though expressed in slightly different language). $\endgroup$ – dmckee --- ex-moderator kitten Oct 29 '16 at 3:58

The mass of a extended body in rotation is larger than the mass of the same body not rotating, but that increase in mass should not be attributed to any change in the mass of the constituent particles.

Here I am defining the mass of an object (in the usual manner of modern treatments) as the square of that object's energy-momentum four-vector: $$ m \equiv \frac{1}{c^2}|\mathbf{p}|^2 = \frac{1}{c^2}\sqrt{E^2 - (\vec{p}c)^2} \;.$$ Such masses are Lorentz invariants: they do not change under a Lorentz boost (change in inertial velocity). This is the only meaning of "mass" in modern treatments and in particular the phrase "relativistic mass" is nowhere to be seen.

There is kinetic energy in rotation and that energy contributed to the time-like component of the four momentum (i.e. the $E$ above).

To expand on how the overall body can grow more massive while the particle that make it up do not, we have to examine the four-momentum of a compound system.

Like other vectors, you can add four-vectors component-wise. So trivial compound body consisting of a symmetric rigid1 rotor of total mass $M_\textrm{rot} = 2m$ when not rotating and separation $r$. At a given moment each sub-mass has three-momentum $\pm\vec{p}$ in the rotor's COM frame and therefore energy $E = \sqrt{(\vec{p}c)^2 + (mc^2)^2}$ in the same frame. That makes their total four-momentum $$ \mathbf{P}_\textrm{tot} = \left( 2\sqrt{(\vec{p}c)^2 + (mc^2)^2}, \vec{0} \right) \;.$$ The mass of the rotor is \begin{align*} M_\textrm{rot} &= \frac{1}{c^2} \left| \mathbf{p}_\textrm{tot} \right| \\ &= \frac{1}{c^2}E^2 \\ &= 2\sqrt{\left(\frac{\vec{p}}{c}\right)^2 + m^2} \;. \end{align*} This obviously exceeds $2m$ for any non-zero $\vec{p}$.

The basic lesson here is that the mass of systems is not automatically the sum of the mass of the parts in relativity; a matter often inexplicably left off of lists of things that are different in Einstein's world. One of the most interesting games you can play with this fact is showing that two photons (each of exactly zero mass using the definition above) can none-the-less form a system of non-zero mass.

A word of caution: I've not considered the binding energy of this system which would not be acceptable in a serious treatment as any rotation swift enough to make an appreciable difference to the mass of the system would also necessarily alter the binding energy by an appreciable amount (that is, the assumption of rigidity would be violated).

1 I know. No truly rigid bodies in relativity. We can assume this thing isn't actually rigid but that we spin it up slowly enough to pretend.

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    $\begingroup$ Thank you very much for your time. I know about half of what you have written, I also now know much more about how subtle it can be. This is a 3 beer, ( or possibly aspirin), read through answer you have given me . Much appreciated though. $\endgroup$ – user108787 Oct 29 '16 at 4:08
  • $\begingroup$ In general, in baryons with the same quark content but different $J$, the one with higher $J$ is heavier, right? Obviously your caveat about binding energy applies strongly since it's the strong interaction, though. $\endgroup$ – Robin Ekman Oct 29 '16 at 9:58
  • $\begingroup$ So what mass M do you use when you apply a force to the center of mass of your rotating body (angular velocity $\omega$) and want to know its change of linear momentum $\vec P=M\vec v$, in Newton's 2nd law $\vec F=\frac {d(M \vec v)}{dt}$? $\endgroup$ – freecharly Oct 29 '16 at 13:31
  • $\begingroup$ And what mass do you use for the rotating body to calculate the gravitational force according to Newton's law of gravitation? $$F=G\frac{m_1m_2}{r^2}$$ $\endgroup$ – freecharly Oct 29 '16 at 15:44
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    $\begingroup$ @freecharly you use $M_{rot}$ in both cases. $\endgroup$ – Virgo Oct 29 '16 at 17:50

Yes, the mass increases according to the energy you've added by $E=mc^2$.

  • $\begingroup$ Thanks for the quick response, if you don't mind I will hold off accepting your answer, obviously we can call mass whatever we like for the particular problem, but the inconsistency, if there is one, bothers me. $\endgroup$ – user108787 Oct 28 '16 at 21:55
  • $\begingroup$ Yes this is the easy answer, which always holds. How much energy will you have added relativistically when you bring a body with classical inertial moment $I$ to an angular speed $\omega$? $\endgroup$ – freecharly Oct 28 '16 at 22:23
  • $\begingroup$ Additional question: What happens to the circumference if the rotating body is a circular disk? Does $\pi= circumference/diameter$ change its value? $\endgroup$ – freecharly Oct 28 '16 at 22:44
  • $\begingroup$ I think you are right, linear momentum is treated different than angular momentum and invariance is used only when needed. It's a choice we make, nature could not care less, so I am wrong to extend the invariance idea further than linear. $\endgroup$ – user108787 Oct 28 '16 at 22:55
  • $\begingroup$ There is certainly a little extra energy held in the stress caused by the centripetal force. The stress-energy tensor, if used right, will include that energy. Also, there's energy (in the form of stress) in the different relative velocities (inner material wants to foreshorten less than external material). So the actual angular velocity calculated from the applied energy is less than a more naive guess at it. So a precise calculation almost needs to be done numerically rather than analytically because those stresses are likely non-linear. @freecharly-it wants to shrink. $\endgroup$ – Digiproc Oct 28 '16 at 23:09

I am only considering Special Relativity. When the rotating object is composed of particles with rest mass $m_i$ with rotational speed $v_i=r_i\omega$, where $r_i$ are the distances to the rotational axis and $\omega$ is the angular velocity, the total rest mass of the object is $M= \sum_i m_i$ and the relativistic mass is $$M_\textrm{rel}=\sum_i \frac{m_i}{{\sqrt{1-(v_i/c)^2}}}=\sum_i \frac{m_i}{{\sqrt{1-(r_i\omega/c)^2}}}$$ This doesn't include additional potential energy due to internal tension in the rotating object. The sum can also written as an integral over mass elements dm.

  • $\begingroup$ Thank you, this should be an obvious thing, but since I joined this site, lots of things that I assumed to be true, are not. Regards $\endgroup$ – user108787 Oct 28 '16 at 22:23
  • $\begingroup$ @CountTo10- You are right. I have had similar experiences. Greetings $\endgroup$ – freecharly Oct 28 '16 at 22:39
  • $\begingroup$ -1: The answer is incorrect because the op is asking "Does the inertia of a body depend upon its energy content?" and the answer to that question is yes; and the concept of relativistic mass needs to die. $\endgroup$ – Robin Ekman Oct 29 '16 at 7:45
  • $\begingroup$ @Robin Ekman - The question of the OP is "Does a spinning object acquire mass due to its rotation?". You can easily check this by reading the title. If "the relativistic mass needs to die" then also Einsteins formula $E=mc^2$ needs to die? $\endgroup$ – freecharly Oct 29 '16 at 12:44
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    $\begingroup$ That question is a special case of the one I quoted, which is the title of Einstein's paper. And yes $E = mc^2$ needs to die in favor of the correct $E^2 - p^2 = m^2$. $\endgroup$ – Robin Ekman Oct 29 '16 at 13:23