High School 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?