Relativistic Mass of an Orbiting Billiard Ball

[Physicsguru]

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, that's 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.



$$m = \frac{m_0}{\sqrt{1-v^2/c^2}}$$

 

[JesseM]

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?

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?

 

[Physicsguru]

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?

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 let's 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

 

[pervect]

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.

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.

 

[JesseM]

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.

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?

 

[Physicsguru]

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?

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

 

[JesseM]

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

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 $$dV = R^2 \, sin(\phi) \, dR \, d\phi \, d\theta$$. If the sphere makes one revolution in time $$t_0$$, then a volume element at coordinates $$(R, \phi, \theta)$$ should be moving at speed $$\pi R^2 cos^2 (\phi) / t_0$$, so its relativistic mass should be $$1 / \sqrt{1 - \pi^2 R^4 cos^4 (\phi) / {t_0}^2 c^2}$$ times its rest mass. So if the sphere has rest mass M and radius $$R_0$$, then I think the relativistic mass would be $$M \int_{0}^{2\pi} d\theta \, \int_{0}^{\pi} sin(\phi) \, d\phi \, \int_{0}^{R_0} R^2 / \sqrt{1 - \pi^2 R^4 cos^4 (\phi) / {t_0}^2 c^2} \, dR$$.

 

[pervect]

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

$$\int T^{00} d^3x$$

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 $$\gamma^2$$ rather than $$\gamma$$. 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.

 

[JesseM]

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

$$\int T^{00} d^3x$$

and T⁰⁰, (the density component of the stress-energy tensor) will probably not transform the way you describe.

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 $$m$$ 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 $$m \pi R^2 cos^2 (\phi) / t_0$$, since each atom is traveling along a circle of radius $$R cos(\phi)$$ and making a revolution in time $$t_0$$? 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 $$m / \sqrt{1 - (\pi R^2 cos^2 (\phi) / t_0)^2 / c^2}$$, since each atom is traveling along a circle of radius $$R cos(\phi)$$ and making a revolution in time $$t_0$$?"

 

[pervect]

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?

Yes.

For example, if you imagine that the rotating sphere is made up of atoms of equal rest mass $$m$$ 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 $$m \pi R^2 cos^2 (\phi) / t_0$$, since each atom is traveling along a circle of radius $$R cos(\phi)$$ and making a revolution in time $$t_0$$? 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).

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

 

[pmb_phy]

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?

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.

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?

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.

For example, if you imagine that the rotating sphere is made up of atoms of equal rest mass $$m$$ 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 $$m \pi R^2 cos^2 (\phi) / t_0$$, since each atom is traveling along a circle of radius $$R cos(\phi)$$ and making a revolution in time $$t_0$$? And wouldn't the total relativistic mass just be the sum of each atom's relativistic mass?

I don't believe so. What you're describing is listed here

http://www.geocities.com/physics_world/sr/rotating_cylinder.htm

which I believe is wrong/inaccurate (I need to either delete that page or correct it).

The relati

Pete

 

[Physicsguru]

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 $$dV = R^2 \, sin(\phi) \, dR \, d\phi \, d\theta$$. If the sphere makes one revolution in time $$t_0$$, then a volume element at coordinates $$(R, \phi, \theta)$$ should be moving at speed $$\pi R^2 cos^2 (\phi) / t_0$$, so its relativistic mass should be $$1 / \sqrt{1 - \pi^2 R^4 cos^4 (\phi) / {t_0}^2 c^2}$$ times its rest mass. So if the sphere has rest mass M and radius $$R_0$$, then I think the relativistic mass would be $$M \int_{0}^{2\pi} d\theta \, \int_{0}^{\pi} sin(\phi) \, d\phi \, \int_{0}^{R_0} R^2 / \sqrt{1 - \pi^2 R^4 cos^4 (\phi) / {t_0}^2 c^2} \, dR$$.

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)

 

[Physicsguru]

In this post I will keep track of my thoughts as they come.

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.

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.

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.

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 $$\omega$$ 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.

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.

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 let's continue.

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 $$dV = R^2 \, sin(\phi) \, dR \, d\phi \, d\theta$$.

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 doesn't 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

 

[JesseM]

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)

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.

 

[Garth]

Suppose that nothing in the universe existed, except for one solitary billiard
//////////////////////////
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?

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

Garth

 

[Physicsguru]

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

Garth

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

 

[Physicsguru]

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).

What do you mean by "a physical substance won't 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

 

[pervect]

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.

 

[pmb_phy]

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.

That's not quite a good idea since, in general, they are not proportional.

Pete

 

[Garth]

Guru

Its rotating in an inertial reference frame.

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

 

[pervect]

What do you mean by "a physical substance won't 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.

Summing the mass of all the atoms does not give the total system mass, because there is also energy involved in the binding energy between atoms and molecules.

The amount of this energy is very small, however, and it is usually neglected (with good reason - relativistic mass is simply another name for E/c^2, and the numerical value of E/c^2 can be seen to be very small).

If one takes a spring (this can be the usual coil sort of spring, or it can simply be a rod of metal which has a "stretchiness" given by its Young's modulus) and stretches it, one must do work on the spring (or rod) to perform the stretching operation. This stored energy is returned when the spring returns to its usual length. Where is this energy stored? The number of atoms in the spring remain constant as it is stretched. This stored energy is not in the form of thermal energy, so none of the atoms are moving any faster. The energy stored in a spring is mechanical energy, and it is stored not in any particular atom, or in the motion of any atoms, but in the bonds between atoms.

Clasically, this energy can be described by the non-relativistic stress-energy tensor. Relativistically, this mechanical energy is included in the relativistic stress energy-tensor.

Conceptually, it's easiest IMO to forget about the idea of "relativistic mass" and replace it by it's synonym, energy/c^2. Anything that does work on a system increases its energy. Anything that removes work from a system decreases its energy. The "relativistic mass" of a system is just the energy of the system divided by c^2.

In principle, it doesn't matter whether the energy in a system is stored as chemical bonds (chemical energy), nuclear bonds, thermal energy, mechanical strain, or in some other form. In practice, in everyday life, only nuclear binding energy has a large enough numerical value to significantly alter the mass of a system. Other sources of energy simply are not large enough to cause a measurable change in mass.

Exception: for technical reasons, the energy density T00 used in general relativity does NOT include any energy stored in "gravitational fields". The energy stored in gravitational fields is handled separately. As T00 is the source of gravitational fields (it's the right-hand side of Einstein's field equations), it would not be a good idea to include gravitational energy in its definition.

Thus, the total mass of a system is not equal to the spatial integral of T00 when a system has a significant amount of gravitational self-energy.

There are actually at least two sorts of system mass in General relativity that do include gravitational self energy. These are known as the Bondi mass and the ADM mass. Roughly speaking, the ADM mass includes all sources of gravitational self-energy, including gravitational waves. The Bondi mass does not include the energy stored in gravitational waves. A useful popular discussion of energy in General Relativity is contained in the sci.physics.faq

Is energy conserved in General Relativity

And another question.

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

No. Charge is a conserved quantity.

 

[pmb_phy]

Summing the mass of all the atoms does not give the total system mass, because there is also energy involved in the binding energy between atoms and molecules.

Note: If the system is not closed then when external forces act on the system one cannot add 4-momenta and get a unique 4-vector.

The amount of this energy is very small, however, and it is usually neglected (with good reason - relativistic mass is simply another name for E/c^2, and the numerical value of E/c^2 can be seen to be very small).

That is incorrect in general. There are cases when your assertion is incorrect. E.g. suppose that in the inertial frame S' there is a rod lying on the x-axis. Let S be in standard configuration with S'. Now subject the rod to stress such that the rod doesn't accelerate, and thus remains at rest in S'. The momentum p of the object will not be related to the energy E by p/v = Ec².

Pete

 

[pervect]

Note: If the system is not closed then when external forces act on the system one cannot add 4-momenta and get a unique 4-vector.

I assume you are objection to my definition of the stress-energy tensor?
For a large cube of matter, different observers will have diffrent notions of simultaneity, so the notion of the momentum and energy of the cube at a given time will be different for different observers. However, the stress-energy tensor appies to a very small cube of matter. When one takes the limit as the size of the cube edge goes to zero, the simultaneity problem becomes unimportant.

I assume that you agree for a point particle that there is no problem with representing the energy and momentum of the particle as a 4-vector, so I don't see why you would have a problem with representing the momentum and energy of an infintesimal volume with a 4-vector.

I write

The amount of this energy is very small, however, and it is usually neglected (with good reason - relativistic mass is simply another name for E/c^2, and the numerical value of E/c^2 can be seen to be very small).

you respond

E.g. suppose that in the inertial frame S' there is a rod lying on the x-axis. Let S be in standard configuration with S'. Now subject the rod to stress such that the rod doesn't accelerate, and thus remains at rest in S'. The momentum p of the object will not be related to the energy E by p/v = Ec².

Pete

I'm not positive, but I think you are objecting to my statemnet that relativistic mass is just another name for E/c^2? As you know, I am not a big fan of the term "relativistic mass".

However, I thought you just got through arguing that a distributed system subject to external forces would not in general have a momentum, or that it's momentum would not be represented by a 4-vector, or something like that, so I don't see how you justify giving it a momentum, or a relativistic mass.

There's no problem with the standard approach of giving a distributed system a stress-energy tensor - it seems to me you are creating your own problems here.

 

[pmb_phy]

I assume you are objection to my definition of the stress-energy tensor?

Nope.

For a large cube of matter, different observers will have diffrent notions of simultaneity, so the notion of the momentum and energy of the cube at a given time will be different for different observers. However, the stress-energy tensor appies to a very small cube of matter. When one takes the limit as the size of the cube edge goes to zero, the simultaneity problem becomes unimportant.

Not all bodies are infinitely small and the energy is still not proportional to the mass for such a body and thus E/c^2 does not equal p/v in all cases. But what you assert here is still incorrect since even an infinitely small body can have stresses. What you're referring to is a point object which has no internal structure and thus cannot have internal stresses.

I assume that you agree for a point particle that there is no problem with representing the energy and momentum of the particle as a 4-vector, so I don't see why you would have a problem with representing the momentum and energy of an infintesimal volume with a 4-vector.

Who said I was speaking about an infintesimal volume?

I'm not positive, but I think you are objecting to my statemnet that relativistic mass is just another name for E/c^2?

Yes.

However, I thought you just got through arguing that a distributed system subject to external forces would not in general have a momentum, or that it's momentum would not be represented by a 4-vector, or something like that, so I don't see how you justify giving it a momentum, or a relativistic mass.

You are asserting that relativistic mass = E/c². However the definition of "relativistic mass" is m = p/v. I think you're assuming that E/c². As such you're changing the definition of relativistic mass.

There's no problem with the standard approach of giving a distributed system a stress-energy tensor - it seems to me you are creating your own problems here.

Nope. There is no problem with the usual definition of the energy momentum tensor. In fact what I've been stating is directly related to the that tensor. The expression E/c² = p/v only holds for an entire closed system. A cube that has stresses exerted upon it is not a closed system. I don't think you got what I've been saying - I'm saying that when you add 4-momenta of of particles which belong to a non-closed system then the 4-momentum of such a system is not well defined. I'm also saying that, for such a non-closed system, relativistic mass is always well defined and has the value p/v. For the same exact system the energy is always well defined and has the usual value. Momentum of such a system is always well defined and is given by the momentum density times volume. But the relation E/c² = p/v is not always valid.

Another example (besides the one I already gave) would be the mass of the field between the plates of a charged capacitor. This mass does not transform according to the usual transformation for a particle. The energy does not transform as $E = \gamma E_0$ and thus the mass does not have the value $m = \gamma m_0$ so the relation E/c^2 = p/v is invalid.

Pete

 

[Physicsguru]

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

The quick answer is "center of mass of the universe." But, that is too quick of an answer.

There are many inertial reference frames, not just one, and this can be shown.

Suppose that the temperature of the billiard ball is at absolute zero. Thus, there are reference frames in which nothing interior to the ball is moving relative to anything else interior to the ball. Now, pick a point exterior to a billiard ball to be the origin of a coordinate system in which the motion of the system(a billiard ball at absolute zero) is going to be analyzed.

Consider coordinate systems in which the center of mass of the billiard ball is accelerating. Any such coordinate system will be a non-inertial reference frame because there will be an object which is accelerating in absence of an externally applied force, contrary to Newton's third law.

Thus, if we insist on finding all the inertial reference frames in which to analyze the motion of the billiard ball (to figure out whether or not there is just one, or more than one), we simply cannot choose a coordinate system in which the center of mass is accelerating. This puts a constraint on how to choose the origin of an inertial reference frame for the ball to be moving in.

So here is a coordinate system constraint for a coordinate system to be an inertial reference frame the billiard ball can be observed in:

$$\vec a = 0 = \frac{d\vec v}{dt}$$

where

$$\vec v = \frac{d\vec r}{dt}$$

Where $$\vec r$$ is the position vector of the center of mass of the billiard ball in the coordinate system.

(As for how a time measurement is to be made, the clock making the measurement would have to have its center of mass be at rest in the coordinate system, and have it's absolute temperature be above 0)

Now consider a coordinate system in which the billiard ball isn't spinning, and the center of mass is at rest (there are many such coordinate systems depending where you choose the origin). Therefore there is more than one inertial reference frame.

QED

Regards,

Guru

PS

Any reference frame in which the center of mass of the billiard ball obeys the constraint above:

$$\vec a = 0$$

Will have Newton's third law be true in it, which is a necessary but not sufficient criterion for a reference frame to be an inertial reference frame.

Also, any frame of reference in which the center of mass of the billiard ball is moving at a constant velocity will satisfy the constraint above, so consider only such frames. So long as all the parts of the interior of the billiard ball obey Newton's laws, these frames will be inertial too.

At this point it shouldn't be hard to see that there is more than one inertial reference frame/inertial coordinate system.

Let F1 be an inertial reference frame in which the center of mass of the ball is at rest, and all the parts interior to the ball are at rest. We know there is at least one since the temperature of the ball is zero.

Now, let F2 be a frame in which the origin of F1 is moving in a straight line at a constant speed, and all the parts inside the ball are moving in a straight line at a constant speed. F2 will be another inertial reference frame, since all three of Newton's laws will be true in F2.

 

[Physicsguru]

Note: If the system is not closed then when external forces act on the system one cannot add 4-momenta and get a unique 4-vector.
That is incorrect in general. There are cases when your assertion is incorrect. E.g. suppose that in the inertial frame S' there is a rod lying on the x-axis. Let S be in standard configuration with S'. Now subject the rod to stress such that the rod doesn't accelerate, and thus remains at rest in S'. The momentum p of the object will not be related to the energy E by p/v = Ec².

Pete

What do you mean by standard configuration?

 

[pmb_phy]

What do you mean by standard configuration?

It means to use the two coordinate systems that people use most often. I.e. let S be the system where the observer is at rest and S' be the system where the particle of interest is at rest. Then the transform

x' = gamma(x - vt)
y' = y
z' = z
t' = gamma(t - vx/c^c)

transforms the events as measured in S to the coordinates of events as measured in S'. The xyz axes are parallel to the x'y'z' axes and +x ray is in the direction of +x' ray etc. When the above is true then S and S' are said to be in standard configuration.

Pete

 

[Physicsguru]

It means to use the two coordinate systems that people use most often. I.e. let S be the system where the observer is at rest and S' be the system where the particle of interest is at rest. Then the transform

x' = gamma(x - vt)
y' = y
z' = z
t' = gamma(t - vx/c^c)

transforms the events as measured in S to the coordinates of events as measured in S'. The xyz axes are parallel to the x'y'z' axes and +x ray is in the direction of +x' ray etc. When the above is true then S and S' are said to be in standard configuration.

Pete

I thought that was what you meant(just checking), about the +x ray and the +x` ray having the same direction, the +y ray and the +y` ray having the same direction, and the +z ray and the +z` ray having the same direction, at some moment in time t in S, simultaneous to moment in time t` in S`. That's a rather quick way to say it.

I also notice that you wrote the lorentz transformations, rather than the Galilean transformations, you will now encounter problems with simultaneity.

Regards,

Guru

PS You have $$c^c$$ instead of $$c^2$$

 

[Physicsguru]

No. Charge is a conserved quantity.

How do you know?

Just for the sake of argument, what if electrons don't exist, and instead they are just nodes in a wave which is 'standing' in some particular frame, meaning that the nodes aren't moving relative to one another.

Think of a uniform line charge of N electrons, as being a standing wave with N nodes. As you change the vibration frequency, you change the number of nodes in the standing wave. Then wouldn't the amount of 'electric charge' have changed?

Regards,

Guru

 

[quantumdude]

How do you know?

He knows because charge conservation has never been observed to be violated.

Just for the sake of argument, what if electrons don't exist, and instead they are just nodes in a wave which is 'standing' in some particular frame, meaning that the nodes aren't moving relative to one another.

This sort of argument already exists. It's called "quantum field theory", which says that all electrons are just excitations of the one single electron field.

Think of a uniform line charge of N electrons, as being a standing wave with N nodes. As you change the vibration frequency, you change the number of nodes in the standing wave. Then wouldn't the amount of 'electric charge' have changed?

What are you going to change the number of vibrations with? Experimentally, the only way to create new electrons is to create an electron-positrion pair. Naturally, that process conserves charge as the produced pair has zero net charge.

 

[Physicsguru]

Note: If the system is not closed then when external forces act on the system one cannot add 4-momenta and get a unique 4-vector.
That is incorrect in general. There are cases when your assertion is incorrect. E.g. suppose that in the inertial frame S' there is a rod lying on the x-axis. Let S be in standard configuration with S'. Now subject the rod to stress such that the rod doesn't accelerate, and thus remains at rest in S'. The momentum p of the object will not be related to the energy E by p/v = Ec².

Pete

I am curious, exactly why do you tell him that the momentum p of the object will not be related to the energy E by p/v= E/c²?

You have a steel bar at rest in some inertial reference frame S. The steel bar is at absolute zero degrees kelvin, so no radiation is being emitted. Furthermore nothing inside is moving relative to anything else inside. Thus, the system is certainly closed, nothing is entering it or exiting it.

Now, an external agent exerts two equal forces in opposite directions which "stretch the iron bar." Obviously there will be some internal motion, some radiation even. (One time I found a steel bar in my basement, really thin it was part of a baby crib. It was fairly strong, but kind of thin so that I could bend it. I bent the bar very well using my foot, so that this steel bar was now in the shape of a U. Much to my surprise, when I put my hand on the "crease" it was extremely hot, and I got burned, but points further away on this bar were still cold. The bar was about 4 feet long i think.)

At any rate, ok so someone stretches this bar. Its center of mass remains at rest during the process. The initial momentum of the bar in this frame is its inertial mass M, times its speed V in this frame. Its speed in this frame is zero, so that its total momentum in this frame is zero, initially. Then after the two forces act simultaneously on the bar, the speed of the bar is still zero in this inertial frame. In fact, its center of mass never moved throughout the stretching. Its initial momentum is equal to the final momentum, so that we have conservation of momentum, as expected.

As for the system energy that has increased. Energy was put into the system. Is this what you mean?

Regards,

Guru

 

[pervect]

You are asserting that relativistic mass = E/c². However the definition of "relativistic mass" is m = p/v. I think you're assuming that E/c². As such you're changing the definition of relativistic mass.

Ah, I see. Your definition of the term relativisitc mass appears to be in conflict with the definition used in the sci.physics.faq

http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

There is sometimes confusion surrounding the subject of mass in relativity. This is because there are two separate uses of the term. Sometimes people say "mass" when they mean "relativistic mass", mr but at other times they say "mass" when they mean "invariant mass", m0. These two meanings are not the same. The invariant mass of a particle is independent of its velocity v, whereas relativistic mass increases with velocity and tends to infinity as the velocity approaches the speed of light c. They can be defined as follows:

mr = E/c2
m0 = sqrt(E2/c4 - p2/c2)

You might want to write to the maintainer of the sci.physics.faq and give them both your defintion of the term "relativistic mass" along with appropriate references to the literature for why your definition is right.

Unfortunately, there isn't any standard online dictionary to resolve such questions of defintion. At the moment I've got your word, and the sci.physics.faq's word, which unfortunately conflict, and no other sources of information about the proper usage of the term "relativistic mass". Since I don't personally use or like the term "relativistic mass", and since the textbooks I own (Wald & MTW) don't use the term, either, it's rather difficult for me to find examples of proper usage. Since you're the biggest fan I know of the term "relativistic mass", I'll at least accept that p/v is what you mean when you say "relativistic mass".

Nope. There is no problem with the usual definition of the energy momentum
tensor. In fact what I've been stating is directly related to the that tensor. The expression E/c² = p/v only holds for an entire closed system. A cube that has stresses exerted upon it is not a closed system. I don't think you got what I've been saying -

I'm not sure I've been getting what you've been trying to say either

I'm saying that when you add 4-momenta of of particles which belong to a non-closed system then the 4-momentum of such a system is not well defined. I'm also saying that, for such a non-closed system, relativistic mass is always well defined and has the value p/v.

OK, I'm definitely not getting what you are saying. You are claiming that the sum of the momenta of a group of particles in a non-closed system isn't well defined, but you are claiming that the relativistic mass, the ratio of this undefined momentum to a velocity v, is well definied?

If all the individual momenta are definied, why isn't the sum defined? If I define n numbers, their sum is also defined. I think perhaps you meant something else other than "is not defined", but I'm afraid I can't guess what that something might be. I would say the sum of momenta of a group of particles acted on by external forces is a "function of time", not "undefined".

A possibly related question - are you assuming that all the particles have the same velocity 'v', or are you using the velocity of the center of mass for 'v'?

 

[selfAdjoint]

Unfortunately, there isn't any standard online dictionary to resolve such questions of defintion

That's because there isn't any resolution. Both definitions continue in use. The formalism accepts either and adjusts momentum and energy formulas accordingly. Particle physics like the invariant mass, because it means they can plug in the "mass of the particle" where they need it and not worry about dilation. I believe those who do relativistic hydrodynamics prefer the other, but I am not sure of this.

 

[JesseM]

Although it's true that $$m_r = E / c^2$$, it's also true that $$m_r = p/v$$, as long as you make it clear that you're talking about the relativistic momentum. After all, if $$m_0$$ is the rest mass, relativistic momentum would be $$p = m_0 v / \sqrt{1 - v^2/c^2}$$, so $$p/v = m_0 / \sqrt{1 - v^2/c^2} = m_r$$. If you plug $$m_0 / \sqrt{1 - v^2/c^2}$$ into the equation $$E = m_r c^2$$, you get back the equation $$E^2 = {m_0}^2 c^4 + p^2 c^2$$.

 

[Physicsguru]

That's because there isn't any resolution. Both definitions continue in use. The formalism accepts either and adjusts momentum and energy formulas accordingly. Particle physics like the invariant mass, because it means they can plug in the "mass of the particle" where they need it and not worry about dilation. I believe those who do relativistic hydrodynamics prefer the other, but I am not sure of this.

There doesn't need to be confusion.

If $$M = \frac{m0}{\sqrt{1-v^2/c^2}}$$

then $$E = Mc^2$$

Where E=m0c^2+T

M is relativistic mass
m0 is rest mass, also called invariant mass

From the above you can derive the result that:

$$E^2 = (Pc)^2 + (m_0c^2)^2$$

Where

$$P = Mv$$