Can a Spinning Object Increase its Mass through Acceleration?

[mpolo]

This is a just for fun question. I saw the movie "Contact" and they built a machine that generated gravity by spinning rapidly. I assume this comes form general relativity. Is this correct? My question then is this. Is there a simple formula that can be used in which something moving in an orbit / accelerating can have its mass increase as a result of increase in acceleration. I would love to have that equation and put it into my graphing calculator. If you could show example that would be great.

 

[PeterDonis]

I saw the movie "Contact" and they built a machine that generated gravity by spinning rapidly.

I think you are referring to generating a wormhole by spinning rapidly, correct?

As far as the physics involved in that scene in "Contact", Kip Thorne in Black Holes and Time Warps describes how Sagan consulted him about that, and Thorne suggested using a wormhole, which would have to be held open by some form of exotic matter (i.e., matter that violates technical conditions called "energy conditions" that all ordinary matter obeys--it is not known whether such exotic matter can actually exist). As I understand it, the machine portrayed in the movie was supposed to be a creating the wormhole, but I don't know that the spinning part was based on any actual physics; certainly the wormhole solution that Thorne describes is not spinning and is not produced by spinning anything rapidly.

 

[Ibix]

The book suggests that The Machine creates a target for an alien wormhole generator to lock on to, which is why it never produces a wormhole again. We never see an actual wormhole generator.

As Peter says, it's just science fiction.

 

[cosmik debris]

This spinning thing seems to have been around in SciFi for a number of years, the engine rooms of many of the ships contain spinning things. I suspect that the idea goes back to old discussions about the contraction of the rim of a spinning disc in SR and some people concluded from this that the spinning and the resultant contraction meant that spacetime had to be curved by this apparatus.

Cheers

 

[mpolo]

@cosmik debris

Yes, you hit on it, except for its not Special Relativity its General Relativity. That is what gave Einstein the idea for General Relativity. I read that in a book God's Equation. A spinning disc causes the curvature of space time. Apparently this is not a well known or accepted reason for the cause of gravity. Has anyone ever weighed a spinning sphere or disk and see if it gets heavier?

 

[Nugatory]

]es, you hit on it, except for its not Special Relativity its General Relativity. That is what gave Einstein the idea for General Relativity.

The spinning disk problem that @cosmik debris mentions most certainly does not involve general relativity. Google for "Ehrenfest disk" for more information.

I read that in a book God's Equation.

As with all popularizations, you have to be careful about taking it too seriously. In this case, it seems to have rather seriously misled you.

A spinning disc causes the curvature of space time. Apparently this is not a well known or accepted reason for the cause of gravity. Has anyone ever weighed a spinning sphere or disk and see if it gets heavier?

I doubt that anyone has ever actually performed this measurement because the effect would be small and the practical difficulties in performing the measurements to the required level of accuracy would be enormous. However, there is no doubt whatsoever that a rotating disk will weight very slightly more than the same disk when not rotating (it would be good exercise to calculate approximately how much - the classical expression for kinetic energy and $E=mc^2$ will suffice), as many related experiments have been done and yield results that match the predictions of special relativity. This effect has nothing to do with the rotation causing curvature of spacetime - you could get the same effect by adding energy to the disk by heating it.

 

[mpolo]

@Nugatory You made my day. Thanks for the info. I was hoping for a good strong statement about the subject and an idea about how physicists nowadays view this concept of rotation as it relates to mass. So are you saying that the rotating disk weighs more simply because we have added energy to the system via energy of increased motion? More energy, equals more mass because mass and energy are basically equivalent. I want to make sure I understand what you are saying Thanks.

 

[Nugatory]

So are you saying that the rotating disk weighs more simply because we have added energy to the system via energy of increased motion? More energy, equals more mass because mass and energy are basically equivalent.

That's pretty much it.

 

[mpolo]

Excellent Thanks...

 

[DrStupid]

So are you saying that the rotating disk weighs more simply because we have added energy to the system via energy of increased motion?

It has at least more mass due to the additional energy. The resulting weight is not as easy to preditc.

 

[mpolo]

So what is the formula for predicting the extra weight or mass of a spinning disk?

 

[Nugatory]

So what is the formula for predicting the extra weight or mass of a spinning disk?

I gave you a very big hint in #6, so big that you're not going to get another one until you've at least thought about it on your own for a day or so.

 

[DrStupid]

I gave you a very big hint in #6, so big that you're not going to get another one until you've at least thought about it on your own for a day or so.

In #6 you told him how to estimate the mass but you excluded the additional effect on the weight.

 

[mpolo]

@Nugatory E=mc^2 is not the right formula in my humble opinion. I assume that is your hint. E=mc^2 does not account for acceleration factor. No gamma factor either. A spinning disk is a motion of acceleration this fact should not be ignored. A better hint would be F=ma. or Lorentz formula for the increase in mass. Although Lorentz formula does not account for acceleration. I was just hoping there existed another formula besides Newton's formula and Einstein's formula.

 

[DrStupid]

E=mc^2 is not the right formula in my humble opinion.

For the mass it is. You just need to know the energy to use it. With the classical estimation (as suggested by Nugatory) it is quite easy but it might become tricky in relativity.

 

[PAllen]

For the mass it is. You just need to know the energy to use it. With the classical estimation (as suggested by Nugatory) it is quite easy but it might become tricky in relativity.

Why, unless you get beyond the domain of weak field GR? As long as the disk is in equilibrium (thus producing a stationary metric) and itself gravity is not significant, the theory of Komar mass would mandate that the invariant mass is the weight for the disk as a test body in some overall filed.

 

[DrStupid]

Why, unless you get beyond the domain of weak field GR?

Because

$E_{rot} = {\textstyle{1 \over 2}}J \cdot \omega ^2$

doesn't work in relativity. You need to assume the disc to be sufficiently (but not infinite) rigid and than to integrate the relativistic kinetic energy.

 

[mpolo]

@DrStupid Hey that formula looks real interesting. Let's say we have a disk that has a 1 meter diameter and it weighs a 1 kilogram and it is spinning at 10000 rpm.
How much heavier would it be?
Can you show me how to use your formula above so I can put it in my TI Calculator. I would really love to see this worked out don't skip any steps. what is the w^2 represent? What is the E unit and the J is that Joules? I am just off on summer vacation and am learning how to put equations into my TI calculator.

 

[DrStupid]

J is the moment of inertia. In case of a disc it is

$J_{disc} = {\textstyle{1 \over 2}}m \cdot r^2$

And $\omega$ is the angular velocity. It is proportional to the frequency f according to

$\omega = 2 \cdot \pi \cdot f$

With

$m = 1\;kg$
$r = 1\;m$
$\omega = 2 \cdot \pi \cdot 10000\;\min ^{ - 1} = 2 \cdot \pi \cdot \frac{{10000}}{{60}}\;s^{ - 1}$

you get

$E_{kin} = {\rm 274156}\;{\rm J}$

and therefore

$\Delta m = {\rm 3}{\rm .05} \cdot {\rm 10}^{{\rm - 12}} \;kg$

That's quite hard to measure.

 

[jbriggs444]

@DrStupidwhat is the w^2 represent? What is the E unit and the J is that Joules?

That "w" is actually the greek letter omega ("$\omega$"). It is the angular rotation rate, typically in radians per second. There are 2pi radians in a circle.

The E is energy and would typically be measured in Joules. (*)

The J is the "moment of inertia" of the spinning disk. (I am more used to seeing it as "I" rather than "J"). The moment of inertia is computed based on the mass of the spinning disk and its radius. If you are familiar with integral calculus, it is the integral over the disk of an incremental mass element multiplied by the square of the distance of that mass element from the center of rotation. If you are not familiar with integral calculus, you can look up a formula.

For a disk of uniform density, mass m and radius r spinning around its center, the moment of inertia is $\frac{1}{2}mr^2$ The mass would typically be measured in kilograms and the radius in meters. The moment of inertia would then be in units of kilogram meter².

That should be enough information so that you can first calculate the moment of inertia of a 1 kg disk 1 meter in diameter.
You should also be able to convert 10000 rpm to radians per second.
Finally, you should be able to use that information to compute the energy of the spinning disk.

Here on these forums, we do not believe in solving problems for you. We believe in helping you solve them for yourself.

(*) The basic formula works with any coherent system of units. Centimeters, grams, seconds and ergs works just as well as meters, kilograms, seconds and Joules. Feet, slugs, seconds and foot-pounds works too.

With a system of units where the unit of energy is not given by unit mass times unit distance squared per unit time, the formula is almost correct, but needs a constant added to account for the unit choice.

 

[PAllen]

Because

$E_{rot} = {\textstyle{1 \over 2}}J \cdot \omega ^2$

doesn't work in relativity. You need to assume the disc to be sufficiently (but not infinite) rigid and than to integrate the relativistic kinetic energy.

But the suggestion was to integrate kinetic energy! If the rim speed is less than e.g. .1 c, you don't need relativistic KE. Anything more than that, there is no plausible way to hold the disc together. Even if you pretend, you just need to integrate relativistic KE instead.

What I was wondering about was your distinction between mass and weight in your post #13. Until you get a really massive disc, such that self gravitation matters, you should just be able to compute total KE and add its contribution to mass. Even then (when self gravitation matters), the mass also would be affected. Unless you get a disk too big or massive to behave like a test object in the background geometry, the mass and weight must be the same.

 

[DrStupid]

If the rim speed is less than e.g. .1 c, you don't need relativistic KE.

That depends on the error you are willing to accept.

Until you get a really massive disc, such that self gravitation matters

Same again: "matters" depends on the acceptable error.

 

[PAllen]

That depends on the error you are willing to accept.
Same again: "matters" depends on the acceptable error.

Ok, but I was wondering also about the distinction you made between weight and mass. Even Jupiter is treated as a point mass in the latest and most precise ephemeris. You would need something like a brown dwarf not too far from the sun before the effects encapsulated in the misataqua equation become relevant even for very high precision work.

 

[DrStupid]

Then you also need to wonder about the distinction between the mass of rotating and non-rotating objects.

 

[PAllen]

Then you also need to wonder about the distinction between the mass of rotating and non-rotating objects.

You add energy to an object to make it rotate, increasing both its mass and weight. To speak of mass and weight being different, you have to have a situation where you have gone outside the bounds in which the weak equivalence principle is true. Even frame dragging effects are not relevant because they affect orientation of spin axis over time, not the trajectory of the body. Note, I speak here of a broad notion of WEP where the issue is when an n body system's behavior over time is dependent on something other than the mass of each body (noting mass includes all forms of internal energy and binding energy), along with initial positions and velocities. Even gravitational radiation per se is not relevant. Only effects such that two seemingly equivalent n body systems radiate differently would entail a distinction between mass and weight.

 

[DrStupid]

To speak of mass and weight being different, you have to have a situation where you have gone outside the bounds in which the weak equivalence principle is true.

We are already out of the scope of classical mechanics. Why should the weak equivalence principle remain valid?

Even frame dragging effects are not relevant because they affect orientation of spin axis over time, not the trajectory of the body.

Really? Trajectories aroud a black hole are identical for rotating and non-rotating black holes?

 

[PAllen]

We are already out of the scope of classical mechanics. Why should the weak equivalence principle remain valid?
Really? Trajectories aroud a black hole are identical for rotating and non-rotating black holes?

1) The WEP has broad applicability in GR.

2) Ok, you are right, This is a counter example, where a two body system of spinning BH and small body behaves differently than one with equal mass nonspinning BH even if there is no significant GW. So I agree you are right for a sufficiently extreme case, which I guess was your point.

 

[mpolo]

@DrStupid and @jbriggs444

Thank you both so much for this formula and the additional detailed explanations. This formula is so cool! I am going to have a great time putting it into my calculator and playing with it. This is better than video games! The additional info and unit details really helps me understand what is going onwith this equation.

 

[PAllen]

1) The WEP has broad applicability in GR.

2) Ok, you are right, This is a counter example, where a two body system of spinning BH and small body behaves differently than one with equal mass nonspinning BH even if there is no significant GW. So I agree you are right for a sufficiently extreme case, which I guess was your point.

Now I am not so sure. The following source reiterates my prior claim:

https://arxiv.org/abs/gr-qc/0006075

The correct question to ask for a Kerr BH would be whether the "center of mass" (by some reasonable definition) of a small (there is no lower size limit) Kerr BH orbiting e.g. a star moves differently than a small nonrotating BH (making them small removes GW issues from the picture). If this happened, you would have a violation of the WEP and weight would different from mass. The literature claims there are no known predicted violations of the WEP in GR (except charge couplings between a charged particle and its own field, which are excluded in careful definitions). I have not been able to locate any literature on the motion of a Kerr BH in a background geometry, but would now expect that it would not violate the WEP and thus there is no sense in which weight of spinning object (even extreme) is different from its mass (to the extent classical GR is true).

 

[DrStupid]

I have not been able to locate any literature on the motion of a Kerr BH in a background geometry, but would now expect that it would not violate the WEP and thus there is no sense in which weight of spinning object (even extreme) is different from its mass (to the extent classical GR is true).

That is quite surprising. To my knowledge trajectories around a BH depend on its angular momentum. In order to make this happen without affecting the trajectory of the BH itself, the differences in energy and momentum would need to be emitted as gravitational waves. Is that the case?

Another point is the relativistic velovity dependence of the gravitational force. The gravitational mass is different from rest mass for relativistic velocities. In "Measuring the active gravitational mass of a moving object" [American Journal of Physics 53, 661 (1985)] DW Olson and RC Guarino derived a factor of $\gamma \cdot \left( {1 + \beta ^2 } \right)$ for hyperbolic trajectories (just to give an example). Keeping the same weight for rotating and non-rotating bodies would require that this effect cancels always out over all parts of a rotating body (at least with center of mass at rest) - independend from geometry and angular velocity. Is that the case?

 

[PAllen]

That is quite surprising. To my knowledge trajectories around a BH depend on its angular momentum. In order to make this happen without affecting the trajectory of the BH itself, the differences in energy and momentum would need to be emitted as gravitational waves. Is that the case?

Another point is the relativistic velovity dependence of the gravitational force. The gravitational mass is different from rest mass for relativistic velocities. In "Measuring the active gravitational mass of a moving object" [American Journal of Physics 53, 661 (1985)] DW Olson and RC Guarino derived a factor of $\gamma \cdot \left( {1 + \beta ^2 } \right)$ for hyperbolic trajectories (just to give an example). Keeping the same weight for rotating and non-rotating bodies would require that this effect cancels always out over all parts of a rotating body (at least with center of mass at rest) - independend from geometry and angular velocity. Is that the case?

This raises a number of question I have given some thought to.

To understand these issues, we need to separate notions:

1) A body's mass (within domain of GR where one can talk of a body with adequately defined mass), however it arises (rotating or not, to any degree). We don't care how internal components may add non-linearly to constitute mass, just whether we can speak of a reasonably isolated body whose mass can be measured. Then we can define effective passive gravitational mass by how it responds to the background metric of a much more massive body (of any nature, rotating or not). This same body's effects on much tinier test particles, far away, can be used to define an active gravitational mass. Far away, because we want the net influence of the body, not issues of energy/pressure distribution within it producing complex near field. One may also similarly define inertial mass.

2) Velocity dependent interaction (and even acceleration dependent interactions). The latter are discussed in Steve Carlip's well known paper "Aberration and the Speed of Gravity", where in he shows that (analyzed from a force point of view), a moving body generates both velocity and acceleration dependent gravitational forces, this being how the 'apparent speed' of gravity may appear to be near infinite. That a moving charge produces velocity dependent force on another test charge in no way is said to change the charge of the moving charge. Similarly, that a moving body produces velocity and acceleration dependent gravitational interactions with a test body doesn't change what the effective mass of the moving body is - it is the mass as in (1) measured in a quasilocal rest frame of the body. Guarino is discussing the velocity dependent effect.

3) Then, the question of how one accumulates component information to explain empirical mass (1), is a separate question. In the case of pressure-less, non interacting, uncharged, dust body, with insignificant self gravitation, then even if internal motions are relativistic, the invariant mass (= sum of particle total energies in the center of momentum frame) will be equal to the mass in (1). Beyond this, you would need to use either a quasilocal mass integrating the stress energy tensor in a nontrivial way (for example, Bartnik mass), or, more commonly, treat the body as embedded in isolation in asymptotically flat spacetime and compute its ADM mass. Another option is komar mass if the body may be treated as producing a stationary metric, which is true for a uniformly spinning disk in equilibrium. Generally, whenever several of these are valid, they come out the same.

A further issue is that the question of orbits near a Kerr BH is probing near field of an extended body. Nothing about the equivalence principle says probing the near field of different same mass bodies with different SET distributions will fail to detect differences. While a BH is all vaccuum, it may be considered the remnant of different collapses for different BH parameters (angular momentum, for our purposes), and its having different near fields is irrelevant to the EP.

The EP would come into play when we ask, even for the near field of an extreme Kerr BH, if the motion of a tiny (very low mass) non-rotating BH would be different from the motion of a tiny Kerr BH. Per numerous claims (see for example, Clifford Will's living review article on testing GR), GR is the unique theory that predicts no violations of even the strong equivelence principle, ever. Experiment so far is consistent with this, but coverage of extreme regimes is very limited. Note that a charged body is not considered valid for the EP because it is never isolated from its own field. However, collections of balanced extreme charges, with extreme motion, such there is no external field, are perfectly ok for the EP (this, of course, is ordinary matter).

Thus, I return to may claim that there is no sense in which the weight (passive gravitational mas) and mass (inertial or active gravitational) will differ, when they are definable at all. As for computing the mass of a spinning body, integrating relativistic energy in COM frame will only be valid when pressure and stress (in natural units) and self gravity are all insignificant. But that is a separate issue from weight and mass being different.

 

[PeterDonis]

To my knowledge trajectories around a BH depend on its angular momentum. In order to make this happen without affecting the trajectory of the BH itself

The trajectories in question are trajectories of test bodies, which by definition contribute no mass, angular momentum, stress-energy, etc. to the system as a whole. Obviously this is an idealization, but the point is that the EP, strictly speaking, only applies to test bodies in this idealized limit. So gravitational spin-spin coupling between two bodies, for example, does not violate the EP, because in order for it to be present at all both bodies must have non-negligible mass and angular momentum, hence neither one is a test body.

In practical terms, a body can be treated as a test body to a good approximation if its mass, angular momentum, stress-energy, etc. are sufficiently small compared to those of the BH (or whatever other central body is producing the overall spacetime geometry that determines the test body trajectories). For example, this approximation works very well for computing orbits in the solar system, even for planets which themselves obviously have non-negligible mass, angular momentum, etc. But it's still an approximation.

the differences in energy and momentum would need to be emitted as gravitational waves. Is that the case?

GR certainly predicts that this is the case if we drop the idealization of one body being a test body. For example, GR calculations of GW emissions from binary pulsar systems match observations very well. Pulsars spin rapidly, so the GR calculations would in principle include effects of spin-spin coupling between them, and any resulting effects on GW emission and changes in orbital parameters. (I suspect that these effects are very small compared to the primary source of GW emission, which is simply the orbital motion of the pulsars, but I have not looked up the calculations to see.) But in this case neither body is being treated as a test body.

The gravitational mass is different from rest mass for relativistic velocities. In "Measuring the active gravitational mass of a moving object" [American Journal of Physics 53, 661 (1985)] DW Olson and RC Guarino derived a factor of $\gamma \cdot \left( {1 + \beta ^2 } \right)$ for hyperbolic trajectories (just to give an example).

These calculations all involve the entire orbit, not just a small patch of spacetime, so they are irrelevant to the EP. The EP only applies within a small patch of spacetime. In such a patch, the velocity dependent effects you mention are not observable. They are only observable when you look at the entire trajectory, which cannot be covered by a single small patch.

For this reason, I think the language "gravitational mass different from rest mass" is misleading, although unfortunately it does appear to be commonly used in the literature.

 

[DrStupid]

weight (passive gravitational mas)

It seems we are talking cross purposis. I am talking about this: https://en.wikipedia.org/wiki/Weight

 

[PeterDonis]

I am talking about this

In a GR context, this definition is already problematic, because in GR gravity is not a force. Even the operational definition given in that article is technically incomplete, since it doesn't specify the state of motion of the object being weighed relative to the source of gravity.

The best quick definition I can come up with would be something like: the force that must be applied to an object to make it follow a worldline that is an integral curve of the timelike Killing vector field of the spacetime.

 

[DrStupid]

The trajectories in question are trajectories of test bodies, which by definition contribute no mass, angular momentum, stress-energy, etc. to the system as a whole.

This is not very helpful because the weight in the gravitational fiel of a massless body is always zero.

These calculations all involve the entire orbit, not just a small patch of spacetime, so they are irrelevant to the EP.

But they are relevant fort the weight. In a locally free falling frame of reference, there is no weight at all. That's the basis for GR.