Relativistic Mass of an Orbiting Billiard Ball

[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

 

[Physicsguru]

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.

Could an external magnetic field change it?

Regards,

Guru

 

[quantumdude]

Tom: What are you going to change the number of vibrations with?

PG: Could an external magnetic field change it?

Not according to any observational evidence.

 

[pmb_phy]

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?

Its easily shown by a simple example which I've provided somewhere here.

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.

Such a body will be Lorentz contracted when you change frames or the body's speed is increased.

Now, an external agent exerts two equal forces in opposite directions which "stretch the iron bar."

Such an operation will change the energy of the system.

At any rate, ok so someone stretches this bar.

You've added a new complication to the problem, i.e. stretching. That is different than Lorentz contraction which is at the heart of the derivation.

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

No.

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

The maintainer of that page doesn't want to make changes to it even when the change is a correction. When I did make recommendation that is what he told me. But I'm not referring to online sources (since I don't have any faith in them). I'm referring to the physice/relativity literature.

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.

Why would I want to do that? I rarely seen a relativity text which uses relativistic mass which has the definition wrong. I always seem them get it right. As I stated above the person who maintains that page doesn't want to change it. The change I recommended was not signifigant enough for him to change it. He said other things too so I didn't bother anymore. I know I'm right so there is no reason for me to make more suggestions on this matter.

I don't personally use or like the term "relativistic mass", and since the textbooks I own (Wald & MTW)

That is incorrect. MTW does use relativistic mass in certain places. I think Wald does too in at least one place. But I know of at least two significant places where MTW uses rel-mass. Peacock uses it too.

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".

"Fan" is an not an accurate description. I post on the matter only when people make an error on this subject.

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?

Sorry. I made an error there. Invariant mass (magnitude of 4-momentum) is well defined for a closed system but not well-defined for a non-closed system.

If all the individual momenta are definied, why isn't the sum defined?

I didn't say that the sum is not defined. I said it is not "well" defined.

I would say the sum of momenta of a group of particles acted on by external forces is a "function of time", not "undefined".

When you add the momenta of such particles for a sysem then to do this one normally adds the space and time components at a given time in the observers frame of reference. But "at the same time" is not a Lorentz invariant concept. This is too hard to explain here. However I did make an entire page for it. See

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

See the section labled "Invariant Mass of a System of Particles - Non-Closed System "

I recommend that you give what I said a try. I.e. take a box in S' and let the edges coincide with the x'y'z' axes. Let there be two forces acting on the box. One on the +x face acting in the -x direction. Let the second force act on the -x face acting in the +x direction. Find the total energy of the box. Find the total momentum of the box. Compare them and you'll see that p/v = E/c^2 does not hold in this case.

Pete

 

[pervect]

That is incorrect. MTW does use relativistic mass in certain places. I think Wald does too in at least one place. But I know of at least two significant places where MTW uses rel-mass. Peacock uses it too.

I've never seen the actual phrase "relativistic mass" used in MTW. If you can point out a section in MTW that favors your defintion of relativistic mass over the one in the FAQ I'd be very interested to see it.

"Fan" is an not an accurate description.

If you say so.

I didn't say that the sum is not defined. I said it is not "well" defined.
When you add the momenta of such particles for a sysem then to do this one normally adds the space and time components at a given time in the observers frame of reference. But "at the same time" is not a Lorentz invariant concept. This is too hard to explain here. However I did make an entire page for it. See

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

OK, that webpage was a lot clearer (due to the extra length and the diagrams and examples), but it boils down to what I thought you were saying earlier.

The way I view the situation is slightly different from yours. I regard the momentum and energy of a distributed non-closed system as being a well-defined function of time once the system is exactly defined. Part of the definition of the "system" is a defintion of simultaneity.

Two systems with different concepts of simultaneity are not the same, physical system - the events that make up the system are different.

I also think I disagree with your concluding statement

"the invariant mass is not a physically meaningful quantity"

but I have to think about it some more to get the details right. For starters, though, while there is a magnetic field and no electric field in the frame of the diagram of figure 2, in a moving frame there should be both an electric and a magnetic field. If the magnetic field is taken to be in the x direction, and the motion of the system in the z direction, the electric field should be in the y direction.

The electric field in the moving frame will of course be doing work on the charges.

 

[pmb_phy]

I've never seen the actual phrase "relativistic mass" used in MTW. If you can point out a section in MTW that favors your defintion of relativistic mass over the one in the FAQ I'd be very interested to see it.

I didn't mean to imply that they used that term. I meant to state that in some places they use the term "mass" to referr to "relativistic mass."

The way I view the situation is slightly different from yours. I regard the momentum and energy of a distributed non-closed system as being a well-defined function of time once the system is exactly defined. Part of the definition of the "system" is a defintion of simultaneity.

Suppose you're in the inertial frame S where there is a box at rest which is mirrored inside so that photons can bounce back and forth. Let two photons of the same energy be at the center of the box and traveling in opposite directions. Then in S they will hit opposite walls at the same time as measured in S. If you evaluate the "invariant mass" of the two photon system then it will be constant as measured in S. Now move to S' which is moving relative to S. In S' the "invariant mass" will have two different values. That is to say that the "invariant mass" depends on time. As such I've just shown you two different ways of evaluating the "invariant mass" of the system. Yet since it is time dependant it can't be invariant. Hence the term "invariant mass" must be used with caution.

For starters, though, while there is a magnetic field and no electric field in the frame of the diagram of figure 2, in a moving frame there should be both an electric and a magnetic field.

Yep. I agree.

If the magnetic field is taken to be in the x direction, and the motion of the system in the z direction, the electric field should be in the y direction.

The electric field in the moving frame will of course be doing work on the charges.

Of course. But nobody said that the mass was conserved or not. In the rest frame of the source if the B-field there will be no electric field and thus no work can be done on a charged particle. When you transform to a new frame moving relative to the first then in that frame work will be done. You'll recall that whether work is done or not will depend on the frame of reference. One wouldn't expect it to be since in general there are external forces which do work and the amount of work will depend on the frame of reference. The energy of the system is conserved in S but not in S'.

Pete

 

[pervect]

To help convince myself of what was going on, I generated some plots of the transverse component of position (the position at right angles to the boost),of a pair of particles moving in a circular path, as per Pete's example #2. If things go well, there will be three jpeg files, showing a plot of the transverse position in the unboosted frame, in the boosted frame, and a plot of the center of mass in the boosted frame.

 

[pmb_phy]

To help convince myself of what was going on, I generated some plots of the transverse component of position (the position at right angles to the boost),of a pair of particles moving in a circular path, as per Pete's example #2. If things go well, there will be three jpeg files, showing a plot of the transverse position in the unboosted frame, in the boosted frame, and a plot of the center of mass in the boosted frame.

The figure in the page I referred you to was done by a program, i.e. it drew a sinewave. I drew them displaced because the figure would have looked to weird, i.e. too much gunk in too small of a space.

What do you mean by "plot of the center of mass"? How do you define "center of mass"? Do you mean as it is defined here?
http://www.geocities.com/physics_world/sr/center_of_mass.htm

Pete

 

[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.

The centre of mass must be the 8-ball itself, there is nothing to 'hang your origin on' otherwise. But how do you define a non-rotating frame other than by the 8-ball itself? In other words can a solitary spherically symmetric ball be seen to rotate in an otherwise empty universe?

Garth

 

[pervect]

The figure in the page I referred you to was done by a program, i.e. it drew a sinewave. I drew them displaced because the figure would have looked to weird, i.e. too much gunk in too small of a space.

What do you mean by "plot of the center of mass"? How do you define "center of mass"? Do you mean as it is defined here?
http://www.geocities.com/physics_world/sr/center_of_mass.htm

Pete

My plots were also computer generated.


What I did exactly is took

x1 := a*cos(w*lambda1);
y1 := a*sin(w*lambda1);
t1 := lambda1;

x2 := -a*cos(w*lambda2);
y2 := -a*sin(w*lambda2);
t2 := lambda2;

This is sufficient to generate the first plot, which is just a plot of a simple sine function, y1 and y2.

Of course I needed specific variable values to do the plot, they were
a=1;w=1;c=1;v=.9;

Obviously y1=-y2, and x1=-x2, and the center of mass is at the origin (x1+x2)/2 = 0, (y1+y2)/2 = 0.

Now comes the interesting part.
I do a boost in the 'x' direction to get

xx1 := a*cos(w*lambda1)-v*lambda1;
yy1 := y1;
tt1 := lambda1-v/c^2*a*cos(w*lambda1);


xx2 := -a cos(w lambda2) - v lambda2;
yy2 := y2;
tt2 := lambda2+v/c^2*a*cos(w*lambda2);

Then I solve tt1=t for lambda1 as a function of time, and tt2=t for lambda2 as a function of time. There was no closed form solution, the computer output was of the "placeholder" form:

lambda1 := (v*a*cos(RootOf(-_Z*c^2+w*v*a*cos(_Z)+w*t*c^2))+t*c^2)/c^2;

lambda2 := -(v*a*cos(RootOf(_Z*c^2+w*v*a*cos(_Z)-w*t*c^2))-t*c^2)/c^2;

With the same values for constants, i.e.
a=1;w=1;c=1;v=.9;

the second plot is a plot yy1 and yy2 as a function of the boosted time t=tt1=tt2, and represents the y-component of the same motion as seen from the frame boosted in the x direction.

The center of mass is just (yy1+yy2)/2. It was stationary in the unboosted frame, but because of the change of the defintion of simultaneity, it moves (quite a bit!) in the boosted frame.

After variable substititions, the appropriate equations for the second plot are fairly simple, they are just

lambda1 := RootOf(-10*_Z+9*cos(_Z)+10*t);
lambda2 := RootOf(10*_Z+9*cos(_Z)-10*t);

yy1 := sin(lambda1);
yy2 := -sin(lambda2);

 

[pervect]

As I proceeded to phase 2, finding the actual momentum and energy, I relaized that I made an unfortunate choice of constants. Sepcifically, I need to change the radius a or the angular velocity w so that a*w < 1 to make the solution physical. The revised position plots (attached) of yyy1 and yyy2 with a=.5 aren't that much different though.

The center of mass of the system in the boosted frame moves, which is a direct consequence of the fact that it isn't an isolated system, external forces are both accelerating it (changing its momentum) and doing work on it (changing its energy).

In the unboosted frame, the external forces at any given instant of time cancel - in the boosted frame, they do not, because of the relativity of simultaneity.