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Relativistic mass question

  1. Feb 19, 2005 #1
    Suppose that nothing in the universe existed, except for one solitary billiard ball. For the sake of definiteness, let it be an eight ball.

    So the center of mass of the universe is the center of mass of the eightball. Now, the eight ball has an inerital mass m, which is either a function of its speed v in a frame or not. Suppose that m(v), therefore m=m(x,y,z,t) in general. OK so now here is my question. Suppose that in a frame of reference where the center of mass is at rest, and the location of the number 8 on the surface of the solid spherical billiard ball is at rest, the rest mass is exactly m0. What would be the total inertial mass of the billiard ball in a reference frame in which the center of mass was at rest, but the number 8 was orbiting the center of mass with tangential speed v?

    For clarity, let the radius of the billiard ball be R. You can presume that R(t) or not, thats your choice depending on how you want to handle temperature, and radiation, etc.

    Regards,

    Guru

    PS: This is supposed to be an ordinary 8 ball. You can presume that the density is constant if you want. You can presume the net electric charge is zero if you want or not. And you can start off by assuming the formula for relativistic mass is correct.

    If you want a model for the interior structure of the eightball, pretend that its basically a lattice structure, and that the total number of molecules is N.



    [tex] m = \frac{m_0}{\sqrt{1-v^2/c^2}} [/tex]
     
    Last edited: Feb 19, 2005
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  3. Feb 19, 2005 #2

    JesseM

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    Only one of these can be an inertial frame in which the laws of SR apply, the other must be an accelerating frame in which they don't (I'm not sure what laws apply to rotating frames in relativity). Which one are you assuming is the inertial frame?
     
  4. Feb 19, 2005 #3
    If the "orbiting" frame is artificial then it isn't an IRF, so in order to get the number 8 to rotate about an axis, somehow energy got put into the system to get it to spin.

    In other words start off with an IRF being the one in which the center of mass is at rest, and all the parts are at rest, so therefore the number 8 isn't spinning in this frame. Since all the parts are at rest, and remaining at rest in this frame, this frame is surely an inertial frame. Now, we want to get the number 8 to start orbiting an axis of rotation in this frame, which we know is inertial. The only way to do that is to have an external source spin the eight ball. By stipulation there is nothing external to the eightball, so there can be no external source of energy to get it to spin, so lets modify the problem somewhat.

    Assume that there is one and only one net external force which briefly acts upon this eight ball, to get it to spin, and that after this external force acts upon the eight ball, the number 8 now has a tangential speed of V, in this IRF.

    What is the inertial mass of the eight ball now, after this external energy entered the eightball system?

    Regards,

    Guru
     
  5. Feb 19, 2005 #4

    pervect

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    For starters, I would suggest re-writing this question to use standard terminology. see the following link

    relativistic mass

    for the correct usage of "relativistic mass" and "invariant mass". It would appear that you probably meant to say "relativistic mass", but if that is indeed what you meant, why not say it? "Inertial mass" is ambiguous at best in this context - the generally preferred notion of mass nowadays is invariant mass, which does not depend on the velocity of the ball.

    Laymen word questions this poorly and ambiguously all the time, which usually entials the resultant of a very long explanation of some very basic physics. Someone aspiring to be a "physicsguru" should make some effort to follow the standard conventions, especially on such basic concepts as mass.

    Next, you need to (as another poster has already mentioned) which coordinate system is non-rotating if you want an answer as to what general relativity predicts. General relativity distinguishes rotating coordinate systems from non-rotating ones - so does Newtonian physics for that matter. Only a non-rotating coordinate system can qualify as an "inertial frame" in Newtonian physics, or have a locally Lorentzian metric in General relativity.

    For your own benefit, it might be helpful if you imagine what sort of experiments you would do to measure the mass of the billiard ball. Are you going to measure it's resistance to acceleration? Are you going to put a test particle into orbit around it and measure the orbital period as a functionof distance?

    You specify the billard ball as being "alone in the universe". If you want an answer according to General relativity, a little more precise phrasing of the question would also be very helpful. The usual assumption would be that the billiard ball is in an asymptotically flat space-time. This is also a requirement for many of the definitions of "system mass" to apply in General relativity.
     
  6. Feb 19, 2005 #5

    JesseM

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    Wait, so you're saying these aren't two different frames looking at the same physical system, but a single inertial frame looking at two distinct physical situations, one where the ball isn't rotating and one where it has been given some energy that causes it to spin?
     
  7. Feb 19, 2005 #6
    Exactly, yes.

    As you rightly said, the laws of SR don't apply in a non-inertial frame of reference, so we start off in an IRF and remain there. I had to modify the question of course.

    Regards,

    Guru
     
  8. Feb 19, 2005 #7

    JesseM

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    Well, one complication here is that there's no such thing as a rigid body in relativity, so if you have a nonrigid sphere, it will get squashed along its poles and become oblate when it rotates. And with relativity the change in shape would be complicated further still by the fact that volume elements further from the axis of rotation would be moving faster and thus would experience greater Lorentz contraction. But I imagine that whatever velocity we choose to spin it, there would be some non-spherical shape it could have when not spinning that would lead it to become perfectly spherical when it is spinning, so let's assume we pick such a shape.

    In that case, to find the total relativistic mass of a spinning sphere you should just be able to integrate the relativistic mass of each volume element--this is probably easiest in spherical coordinates, where a volume element is defined by [tex]dV = R^2 \, sin(\phi) \, dR \, d\phi \, d\theta[/tex]. If the sphere makes one revolution in time [tex]t_0[/tex], then a volume element at coordinates [tex](R, \phi, \theta)[/tex] should be moving at speed [tex]\pi R^2 cos^2 (\phi) / t_0[/tex], so its relativistic mass should be [tex]1 / \sqrt{1 - \pi^2 R^4 cos^4 (\phi) / {t_0}^2 c^2}[/tex] times its rest mass. So if the sphere has rest mass M and radius [tex]R_0[/tex], then I think the relativistic mass would be [tex]M \int_{0}^{2\pi} d\theta \, \int_{0}^{\pi} sin(\phi) \, d\phi \, \int_{0}^{R_0} R^2 / \sqrt{1 - \pi^2 R^4 cos^4 (\phi) / {t_0}^2 c^2} \, dR[/tex].
     
    Last edited: Feb 20, 2005
  9. Feb 20, 2005 #8

    pervect

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    Careful - even in the weak field limit, when the body has negligible gravitational potential energy, the intergal for the mass will be

    [tex]
    \int T^{00} d^3x
    [/tex]

    and T00, (the density component of the stress-energy tensor) will probably not transform the way you describe. For an ideal fluid, T00 would transform by a factor of [tex]\gamma^2[/tex] rather than [tex]\gamma[/tex]. Of course an ideal fluid has no tensile strength so it's not really a great model to use for a spinning body which has to be a solid to hold together - unless it's bound together by gravity, which opens another (different) can of worms ,involving the proper handling of gravitational self-energy.

    At this point I'm not aware of any simple realistic model for T00 for an actual substance (as you point out, there's no such thing as a rigid body in relativity). This leaves the messy problem of figuring out what T00 would do for some specific actual substance. Probably there is some way to work the problem with an idealized solid with an idealized Young's modulus, but it looks to be a messy problem. And probably not one that has any real--world applications.

    The calculation of the mass of a body when it has significant gravitational self energy is considerably more complicated, hopefully we don't have to go into it though because it involves lots of Killing [vectors :-)].

    See for instance "Gravitation" (MTW), pg 448 for a discussion of the case described above where there is no significant gravitational binding energy.

    For a semi-intuitive explanation of the gamma^2 transformation property of T00 think of a bunch of cubes strung together by wire. The moving cubes will have a relativistic increase in mass by a factor of gamma due to their velocity, but will also have a smaller volume due to the fact that they Lorentz contract, so their density will increase by a factor of gamma^2, not just gamma.

    Any physical substance will of course deform, the non-Euclidiean geometry of a spinning body just makes life more interesting (and confusing). A physical substance won't have any voids, but exactly how it will deform under the combination of pressure and tension (due the the non-Euclidean geometry) will depend on the properties of the solid. The total number of atoms in the rotating object will of course remain constant if it's made out of atoms - but how they are distributed will be different in detail than the simple calcuations you [Jesse] give above. I should also point out that contributions to the mass of the body are made not only by summing the mass of the atoms - one has to include the enregy in the interactions between atoms (at least in principle). In practice, of course, the energy in the chemical bonds is not large enough to appreciably change the mass. But this also means that the body must fly apart before one can measurably change it's mass if it's made out of normal matter, just because the energy in the chemical bonds is not a significant fraction of the rest mass of normal matter.

    Something like a neutron star might be a different story, however. But note that a neutron star will act far more like a fluid than a rigid body. Also note that the gravitational self- energy won't be able to be neglected.

    I do think it's true that the total energy of a body at rest that is "spun up" must be equal to the rest mass of the original body plus the energy used to spin it up (at least in an asymptotically flat space-time where the notion of the mass of a body makes sense in GR). But I haven't attempted to work out the formulas in detail. I just thought I had better post this cautionary note that the problem is considerably more complicated than it appears on the surface.
     
  10. Feb 20, 2005 #9

    JesseM

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    Well, I don't know very much about GR--for an object with negligible gravitational potential energy, would the "density component of the stress-energy tensor" be identical to the "density" of relativistic mass? For example, if you imagine that the rotating sphere is made up of atoms of equal rest mass [tex]m[/tex] which are uniformly distributed through the volume of the sphere in the spinning state (which, due to the squashing effect, would mean they wouldn't be uniformly distributed in the object's non-spinning state), then wouldn't each atom's relativistic mass when spinning be [tex]m \pi R^2 cos^2 (\phi) / t_0[/tex], since each atom is travelling along a circle of radius [tex]R cos(\phi)[/tex] and making a revolution in time [tex]t_0[/tex]? And wouldn't the total relativistic mass just be the sum of each atom's relativistic mass? This may not be the same as the integral of the density component of the stress-energy tensor over the volume of the sphere, of course, in which case the "total relativistic mass" might not be a useful physical quantity (I don't know if the resistance to acceleration would be proportional to the relativistic mass for a spinning object, for example).

    edit: I messed up the relativistic mass of an atom there, I should have said "wouldn't each atom's relativistic mass when spinning be [tex]m / \sqrt{1 - (\pi R^2 cos^2 (\phi) / t_0)^2 / c^2}[/tex], since each atom is travelling along a circle of radius [tex]R cos(\phi)[/tex] and making a revolution in time [tex]t_0[/tex]?"
     
    Last edited: Feb 20, 2005
  11. Feb 20, 2005 #10

    pervect

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    Yes.

    There are two possibilities. The first possibility is what you describe, and it would correspond to the body undergoing elastic deformation (no "plastic flow"). Another possiblility is that the atoms might actually permanently "flow".

    Let's consider the elastic deformation case - no permanent position changes , so the object resumes its original spherical shape when it stops rotating. What I think should happen in this case is that you will have the same number of atoms, but they will not be at the usual interatomic spacing when you go to a "comoving frame". They will be further apart in the comoving frame than they will be in the inertial frame.

    This represents a form of energy storage, like the energy stored in a spring. This energy will contribute to the mass of the system - as I mentioned in my previous long response, the system energy includes the energy of the interaction between atoms, as well as the energy of the moving atoms themselves.

    [edit - included factor of 1/2]

    I think the best semi-practical approach is to make the assumption that the speeds obtainable by a roating body made of matter are much less than relativistic, look at the newtonian energy .5*I*w^2 and say that the total energy of the rotating sphere is esentially m0*c^2 + .5*I*w^2.

    [edit-clarify]
    Perhaps it would be more consistent to express this in terms of 'relativistic mass'. Personally, I have a dislike of the term "relativistic mass", I feel it leads to confusion. But for the sake of consistency, I'll say that an energy of m0*c^2 + .5*I*w^2 is equivalent to a relativistic mass of m0 + .5*I*w^2/c^2
    [end edit]

    This should give suitable accuracy for all practical purposes, and it can be justified from first principles. It should basically be good enough to show that the mass change is not detectable experimentally. And it avoids a lot of headaches in dealing with the stress-energy tensor.

    There's a little bit about this problem in the sci.physics.faq on the rigid rotating disk, but it doesn't have a numerical solution to it (the amount of energy stored in the stresses) either. In fact, the author of the FAQ has asked anyone who has done the calculation to let him know (there is no mention of any response to his plea in the FAQ).

    the specific FAQ entry is:
    http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html
     
    Last edited: Feb 20, 2005
  12. Feb 20, 2005 #11
    I don't follow. Are you choosing a frame in which the ball is always at rest? If so then the proper mass is the same as it is in the rest frame. The rest mass will vary according to the potential energy of the body.
    That is true when a body (object with non-zero size) is not under stress. If a body is under stress then its possible that the rest mass does not equal relativistic mass.
     
  13. Feb 20, 2005 #12
    This is a wonderful attempt at an answer, I am going to need time to look at it obviously but when I'm done I will make my comments.

    Very Kind regards,

    Guru

    PS (when I say attempt i mean nothing bad. Each answer has the assumptions built into it, so to obtain any answer is excellent. Again thank you I will inspect this)
     
  14. Feb 20, 2005 #13
    In this post I will keep track of my thoughts as they come.


    Yes that is a complication, and one which your solution deals with. :)

    Once you input energy into a billiard ball to get it to spin, it will buldge at the equator just like our real earth. This will be true regardless of relativity.


    This is what I was most concerned about in the solution of this problem. I am well aware that if the radius is large enough, then a certain particular value of [tex] \omega [/tex] can be chosen so that points on the equator (where the buldge is hence the distance from the center is largest) come out to be equal to the speed of light c. And I want to see what happens at the end of the problem, when we let v=c, where v is the tangential speed of a point at the equator.

    I'm not sure about this step. But I do realize you are attempting to make some kind of stipulation to facilitate mathematical solution. I suppose now is the time to invoke the (nothing is rigid) stipulation into the argument, and deal with interatomic forces at the equatorial surface.

    I want to be able to eventually see what happens when v approaches c. Also keep in mind that what I really am interested in is the "resistance to acceleration" the body has, although nowhere have I stated it.

    I will think about this region of the problem later, but lets continue.


    I am used to using theta for the angle that goes from zero to ninety, and phi as the angle going from zero to two pi, but i guess it doesnt matter.

    One thing that must be clearly addressed is how to interpret dm, since a lattice has a quantized number of atoms. In other words, what exactly does a differential volume element mean. Density matters when?

    I will think about the remainder of the post later.

    Regards,

    Guru
     
    Last edited: Feb 20, 2005
  15. Feb 20, 2005 #14

    JesseM

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    Well, see pervect's responses above--I think the problem with my calculation is that it doesn't taken into account the forces between the atoms which hold the solid object together, which will change when the object rotates and which also contribute to the relativistic mass.
     
  16. Feb 20, 2005 #15

    Garth

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    My 'Machian' twopenny-worth; "What is the 8-ball rotating with respect to?"

    Garth
     
  17. Feb 20, 2005 #16
    Its rotating in an inertial reference frame. The problem starts off with a real billiard ball being the only thing in existence, and being at rest in an inertial reference frame. And by that is meant only that the center of inertia isn't moving in the frame. But of course either the billiard ball is spinning or not. So what I do to ensure that the frame we watch everything in is always inertial, I stipulate that nothing in the interior of the object is moving in the frame initially, so that I can be assured that the frame is inertial. Then, to get to the point where we have an object which is spinning in an inertial frame, someone has to spin it.

    So its spinning in an inertial reference frame. The material in the body is to be thought of as moving through the coordinates of an inertial reference frame with speed v.

    Regards,

    Guru
     
    Last edited: Feb 20, 2005
  18. Feb 20, 2005 #17
    What do you mean by "a physical substance wont have any voids."

    You say "summing the mass of the atoms." Supposing relativity to be right, the relativistic mass of each atom increases as its speed v in the frame increases. If we are summing the masses, aren't we already taking into account that some external source is inputting energy into the system?

    Also, how in the world do you intend to account for the "energy" in the interactions. What I am interested in here, is the final relativistic mass of the object, assuming relativity to be correct.



    And another question.

    Do you see any way that non-conservation of electric charge could enter this problem?

    Regards,

    Guru
     
    Last edited: Feb 20, 2005
  19. Feb 20, 2005 #18

    pervect

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    There's a little bit about the classical and relativistic stress-energy tensors at the following two links

    http://math.ucr.edu/home/baez/gr/stress.energy.html
    http://people.hofstra.edu/faculty/Stefan_Waner/diff_geom/Sec12.html

    (The later is the only link to cover the classical stress-energy tensor, the first link describes only the relativistic version).

    It's an essential tool in general relativity - it appears on the right hand side of Einstein's field equations. I usually think of it (the stress-energy tensor) as the density of energy of energy and momentum per unit volume, where volume is represented by a vector. (It's possible to represent volume as a vector by taking the time vector that's orthogonal to the volume and scaling it appropriately). Unfortunately, this particlar way of thinking about it does not explain why it's called the stress-energy tensor very well. The second link above does a better job of explaining the name and the classical origins of this tensor (though it's a bit long, and sometimes digresses).

    Multiplying a volume element in any frame (expressed as a vector) by the stress-energy tensor gives the energy-momentum 4-vector of the contents of that volume.

    I'll assume that most readers of this thread are already familiar with the energy-momentum 4-vectors. For those who may not be, or need a refresher

    http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html

    talks about energy-momentum 4-vectors. Understanding the energy-momentum 4-vector is a prerequisite for trying to understand the more advanced stress-energy tensor.

    In the rest frame of an ideal fluid, T00 just gives the density of energy / unit volume, as the first component of the energy-momentum 4-vector is the energy, and the unit vector normal to a non-moving volume is time.

    Note that I use energy rather than relativistic mass by preference - the notion of energy used in this definition includes the rest energy (mc^2) of matter, just as the energy in the energy-momentum 4-vector does.
     
  20. Feb 20, 2005 #19
    That's not quite a good idea since, in general, they are not proportional.

    Pete
     
  21. Feb 20, 2005 #20

    Garth

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    Guru
    And, in an otherwise empty universe, what is it that picks out one particular frame as the inertial frame?

    That's why I used the adjective "Machian".
    regards,
    Garth
     
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