Those who use relativistic mass and why

[pmb_phy]

I'm starting this thread since my response to Tom was too long for one post.

Misleading? Correct me if I'm wrong, but I think that the community of particle physicists is the majority of physicists who use relativity.

That's why I said is was misleading and not incorrect. You're giving the impression that there is an overwhelming number of physicists who use the concept you suggest.

Let me clarify by example: Suppose it were true that 60% of all relativistist use relativistic mass and 40% who didn’t. Then I'd say that a claim that the "majority of physicists don't use it" is incorrect. If the stats where 1% who use it and 99% who don't then I'd say that your statement was very accurate. If the stats were 40% who use it and 60% who don't then a statement that the majority do use it is misleading. It gives the wrong impression as far as how much its accepted. "The majority ..." makes one think that all but 10 or 20, who teach in community colleges, use it.

However I don't have the stats and I doubt that anyone does so its impossible to tell. One would actually have to poll all physicists who use relativity and ask them if they ever use it either in papers or in their thinking/motivation etc. That too is impossible.

Really? Every textbook I have teaches the concept of mass as the invariant norm of the 4-momentum, and they are written by relativists (Taylor and Wheeler, Ohanian and Ruffini, et al). What books do use it? And are there publications in the arxiv that use it?

If its not too much trouble, can you please list the relevant texts of which you speak?

Here is a list of the texts/books I'm speaking of --

Gravity from the ground up, Bernard F. Schutz, Cambridge Press, (2003)

Relativity: Special, General and Cosmological, Wolfgang Rindler, Oxford Univ., Press, (2001)

Cosmological Principles, John A. Peacock, Cambridge University Press, (1999)

Understanding Relativity: A Simplified Approach to Einstein's Theories, Leo Sartori, University of California Press, (1996)

Basic Relativity, Richard A. Mould, Springer Verlag, (1994)

Introducing Einstein's Relativity, Ray D'Inverno, Oxford Univ. Press, (1992)

Gravitation, Misner, Thorne and Wheeler (MTW)

Concepts of Mass in Contemporary Physics and Philosophy, Mass Jammer, Princeton University Press, (2000)

Classical Electromagnetic Theory, Vanderlinde, John Wiley & Sons, (1993)

A First Course in General Relativity, Schutz, Cambridge Univ. Press, (1990)

A Short Course in General Relativity, Foster & Nightingale, Springer Verlag, (1994)

Quantum Mechanics, Cohen-Tannoudji et al

The Cosmic Perspective, Bennet, Donahue, Schneider, Voit, Addison Wesley, (2001)

(dw uses a few of those, e.g. MTW and Rindler)

There are tricky little instances too. One such tricky thing can be found in Classical Electrodynamics - 2nd Ed., J.D. Jackson, page 617, [And, as I recall, there is something similar in Classical Mechanics 3rd Ed., Goldstein, Safko and Poole (2001)] problem 12.16. The student is supposed to find a a general relationship for the center of mass of an electromagnetic field. Now, as that term is used, in that problem, it can only be meaningful if the "mass" is relativistic mass. I worked out the solution here
http://www.geocities.com/physics_world/sr/momentum_conservation.htm

arxiv --

http://xxx.lanl.gov/abs/physics/0308039
http://xxx.lanl.gov/abs/physics/0103008
http://xxx.lanl.gov/abs/physics/0103051

You didn't ask about physics journal articles on this subject. For a list please see -- http://www.geocities.com/physics_world/mass_articles.htm as well as the other link and articles from AJP

There should be one more in the future when mine is proof read and all the typos and grammatical errors are out.

I have Alan Guth's lecture notes from his Early Universe course. He says in one place in his notes that he doesn't use it, yet in another place he actually uses it. I asked him about that and he said he didn't realize he was doing it.

Note: I don't hold that all relativists that use relativistic mass use it in all places at all times. That'd be silly for anyone to do. They use it where it is appropriate or useful to to use it.

All my undergraduate and graduate coursework.

See -- http://www.geocities.com/physics_world/relativistic_mass.htm

There are online class notes from universitys that use the concept as well as from particle accelerator labs such as CERN.

Thanks for the very direct response. It is greatly appreciated. Especially since you explained in a very professional tone. Thanks!

Question: Why do you refer to the magnitude of the 4-momentum as mass and not rest energy?

Factor in those solid state physicists who use relativisitc quantum mechanics or QED, and it's no contest.

I used Cohen-Tannoudji in grad school and they used the velocity dependence of mass.
As in particle physics these folks work with matter on a microscopic scale and they study the structure of matter. They don't really study the dynamics of matter. Consider this - Does the lifetime of a particle depend on the speed of the particle? Relativity says it does. Call the lifetime as measured in the particle's rest frame the proper lifetime. The proper lifetime is an inherent property of the particle and is one of the things particle physicists study. Do you think that when a particle physicist says "the lifetime of a free neutron is 15 minutes" that he didn't know that the lifetime depends on the speed of the particle? Do you think that particle physicist doesn't know that the particle's lifetime is different than the particle's lifetime? Sometimes people use the letter tau to represent proper lifetime and some use T. Quantites which appear in 4tensor equations are proper quantities, e.g. proper mass, proper time, proper distance, etc. Time does not appear in such equationsm, proper time does. Relativists know the difference right? The relativistic Lagrangian contains the proper mass of a particle, not the mass.

A particle physicist, nor a solid state physicist, will never ask himself what the mass of a charged capacitor is or how to compute it. Its not as simple as it is for a particle. See
http://www.geocities.com/physics_world/sr/rd_paradox.htm

Nor will they compute the inertial mass of a gas. But it can be done and it has nothing to do with a magnitude of a 4-vector.

This stuff can be so confusing at times that even the best relativists can make serious mistakes when they don't fully think about what "mass" means. Even Schutz made a serious error in his new text and got a calculation wrong. But that's a topic for another thread.

As Scotty said How many times do I have to tell ye? The proper tool for the proper job!

Pete

 

[Garth]

Mass is to be measured and not just defined

 

[pmb_phy]

Mass is to be measured and not just defined

I agree. In fact I don't recall ever saying otherwise. However its impossible to measure something unless you first define what it is you're measuring.

Pete

 

[kurious]

When they've found a Higgs particle I'll believe that physicists
understand what mass is.Until then...

 

[pmb_phy]

When they've found a Higgs particle I'll believe that physicists
understand what mass is.Until then...

There are many mechanisms to inertia. One is the internal energy of a body. Another is the base rest mass (the mass a body has when it gives up all the energy it can besides the energy from the individual rest masses of the individual particles). Time dilation - that is the mechanism behind relativistic mass/inertial mass. It is why a moving body is harder to accelerate than the same body which is moving slower. The Higgs thingy is the mechanism behind the bare mass of a fundamental particle.

Pete

 

[quantumdude]

That's why I said is was misleading and not incorrect. You're giving the impression that there is an overwhelming number of physicists who use the concept you suggest.

*shrug*

I suppose everyone is free to interpret the word "majority" as they will. But the fact of the matter is that even if "only" 51% of physicists use the invariant mass convention, then it is neither misleading nor incorrect nor inaccurate to say that that is a majority. Your quibble is not with me, but with the dictionary.

However I don't have the stats and I doubt that anyone does so its impossible to tell. One would actually have to poll all physicists who use relativity and ask them if they ever use it either in papers or in their thinking/motivation etc. That too is impossible.

It's probably impossible to get an exact count of research papers that use it vs those that don't, but it doesn't seem that it would be that difficult to get a picture of how much research activity is being done in HEP vs. GR/QC. But I don't care enough about this to find out, so I'll concede the point.

If its not too much trouble, can you please list the relevant texts of which you speak?

I'll do it when I get home from work.

arxiv --

http://xxx.lanl.gov/abs/physics/0308039
http://xxx.lanl.gov/abs/physics/0103008
http://xxx.lanl.gov/abs/physics/0103051

Were any of those eventually publised anywhere?

You didn't ask about physics journal articles on this subject.

That's because I wanted to take a quick look, and the arxiv is easiest to access. I figure I can use the list of references to follow the paper trail into the journals.

For a list please see -- http://www.geocities.com/physics_world/mass_articles.htm as well as the other link and articles from AJP

OK, but AJP is a research journal of physics education. Does this concept appear in PRL or Phys Rev D? That's where working relativists (and not just teachers of relativity) publish.

Note: I don't hold that all relativists that use relativistic mass use it in all places at all times. That'd be silly for anyone to do. They use it where it is appropriate or useful to to use it.

Noted.

See -- http://www.geocities.com/physics_world/relativistic_mass.htm

There are online class notes from universitys that use the concept as well as from particle accelerator labs such as CERN.

Don't get me wrong: I did use the "noninvariant mass" concept as an undergrad in nuclear engineering. We used it to calculate the yield from fission reactions, and it works just fine. I was saying that my coursework led me to the conclusion that the noninvariant mass concept is not widely used in physics, not that it is wrong.

Question: Why do you refer to the magnitude of the 4-momentum as mass and not rest energy?

Because in natural units (hbar=c=1), the mass and the rest energy are identical. Being a particle physicist, I use natural units (as Feynman said, "Only stupid people carry c's and hbar's around" :tongue2: ).

I used Cohen-Tannoudji in grad school and they used the velocity dependence of mass.

The QM book? I took QM I out of volume I, and there is no relativity in it at all. Is there in volume II?

As in particle physics these folks work with matter on a microscopic scale and they study the structure of matter. They don't really study the dynamics of matter.

That's not true, QFT with interacting fields is our bread and butter, and it is the dynamic theory of matter par excellence.

Consider this - Does the lifetime of a particle depend on the speed of the particle? Relativity says it does.

I've snipped off the rest of the exposition on this and will just say, Yes, the particle physicist does understand that all mean lifetimes are proper lifetimes. I'll just note that when it comes to lifetimes (or lengths for that matter) we have no choice but to use speed-dependent lifetimes (and lengths) because, as DW correctly points out in the thread "Einstein's inconsistency", the Lorentz factor γ enters at the level of spatiotemporal intervals. But we do have a choice of convention when it comes to mass.

The relativistic Lagrangian contains the proper mass of a particle, not the mass.

It seems that you're making the same mistake as DW here, but in the opposite direction. The relativistic Lagrangian does indeed contain the mass of a particle if I adopt the convention that the norm of the 4-momentum is the mass. If one convention cannot be said to be wrong, then neither can the other.

 

[pmb_phy]

*shrug*

I suppose everyone is free to interpret the word "majority" as they will.

dw got carried away. But since I don't know you all that well yet I wanted to get to understand what you mean a bit more.

But the fact of the matter is that even if "only" 51% of physicists use the invariant mass convention, then it is neither misleading nor incorrect nor inaccurate to say that that is a majority. Your quibble is not with me, but with the dictionary.

Perhaps it was a poor choice of wording on my part.

re - "Were any of those eventually publised anywhere?"

This one was -- http://xxx.lanl.gov/abs/physics/0103051
It was publsihed in Physics - Uspekhi, 43 (12), 1267 (2000). The one I wrote was reviewed by AJP. They said that while it was new and it was correct, it wasn't spectacular enough for them to print. Ah well!

OK, but AJP is a research journal of physics education.

I'm sorry Tom but I don't see how that makes the physics any less meaningful or valid. Working relativists publish papers there and not soley for pedagogical reasons per se. Articles which appear there are often of theoretical interest in relativity.

Does this concept appear in PRL or Phys Rev D? That's where working relativists (and not just teachers of relativity) publish.

That is where physicists who use relativity publish articles whose subject fit under the general title of Particles, Fields, Gravitation and Cosmology. Its not where theoretical relativity is published.

The QM book? I took QM I out of volume I, and there is no relativity in it at all. Is there in volume II?

Yes. If you ever pick up Vol. II then turn to page 1214 and read section b. Interpretation of the various terms of the fine structure Hamiltonian Subsection alpha Variation of the mass with the velocity (Wsub]mv[/sub] term). The authors write

This term represents the first energy correction, due to the relativistic variation with the velocity.

I can scan and e-mail you that section if you'd like. But its also online at
http://minty.caltech.edu/Ph195/wednesday1c.pdf

That's not true, QFT with interacting fields is our bread and butter, and it is the dynamic theory of matter par excellence.

That's quantum dynamics, not classical dynamics. In QFT velocity does not have a meaning outside of a statistical sense.

The relativistic Lagrangian does indeed contain the mass of a particle if I adopt the convention that the norm of the 4-momentum is the mass. If one convention cannot be said to be wrong, then neither can the other.

I wasn't going in that direction. I was saying that since the Lagrangian is an explicity function of m₀ and not m(v) (call them what you will, I was just making this point).

Thanks Tom

Pete

 

[quantumdude]

I'm sorry Tom but I don't see how that makes the physics any less meaningful or valid. Working relativists publish papers there and not soley for pedagogical reasons per se. Articles which appear there are often of theoretical interest in relativity.

AJP articles aren't any less meaningful or valid. It's just that they aren't at the forefront of research in either gravitation or cosmology.

That is where physicists who use relativity publish articles whose subject fit under the general title of Particles, Fields, Gravitation and Cosmology. Its not where theoretical relativity is published.

Plenty of theoretical and experimental relativity is published there. The "Gravitation and Cosmology" part includes classical GR, and a quick perusal of the table of contents in recent issues reveals current research in gravitational waves.

Yes. If you ever pick up Vol. II then turn to page 1214 and read section b. Interpretation of the various terms of the fine structure Hamiltonian Subsection alpha Variation of the mass with the velocity (Wsub]mv[/sub] term). The authors write

I can scan and e-mail you that section if you'd like. But its also online at
http://minty.caltech.edu/Ph195/wednesday1c.pdf

No need, I know people who have a copy of volume II.

That's quantum dynamics, not classical dynamics. In QFT velocity does not have a meaning outside of a statistical sense.

Quantum Dynamics: Is there any other kind?

 

[pmb_phy]

AJP articles aren't any less meaningful or valid. It's just that they aren't at the forefront of research in either gravitation or cosmology.

How did "forefront of research" get into this conversation?

Plenty of theoretical and experimental relativity is published there.

You're saying that some of it contains physics which is not related to Particles, Fields, Gravitation and Cosmology?

Quantum Dynamics: Is there any other kind?

Ask the string people that, not me.

Pete

 

[pmb_phy]

AJP articles aren't any less meaningful or valid. It's just that they aren't at the forefront of research in either gravitation or cosmology.

I don't agree. People have a tendency to think that things like cosmology are just more "sexy" than things like a homopolar generator etc. There are papers in it which report experimental results. For example

Measurement of the relativistic potential difference across a rotating magnetic dielectric cylinder, J. B. Hertzberg, S. R. Bickman, M. T. Hummon, D. Krause, Jr., S. K. Peck, and L. R. Hunter, Am. J. Phys. 69, 648 (2001)

According to the Special Theory of Relativity, a rotating magnetic dielectric cylinder in an axial magnetic field should exhibit a contribution to the radial electric potential that is associated with the motion of the material's magnetic dipoles. In 1913 Wilson and Wilson reported a measurement of the potential difference across a magnetic dielectric constructed from wax and steel balls. Their measurement has long been regarded as a verification of this prediction. In 1995 Pelligrini and Swift questioned the theoretical basis of the experiment. In particular, they pointed out that it is not obvious that a rotating medium may be treated as if each point in the medium is locally inertial. They calculated the effect in the rotating frame and predicted a potential different from both the Wilsons' theory and experiment. Subsequent analysis of the experiment suggests that the Wilsons' experiment does not distinguish between the two predictions due to the fact that their composite steel–wax cylinder is conductive in the regions of magnetization. We report measurements of the radial voltage difference across various rotating dielectric cylinders, including a homogeneous magnetic dielectric material (YIG), to unambiguously test the competing calculations. Our results are compatible with the traditional treatment of the effect using a co-moving locally inertial reference frame, and are incompatible with predictions based on the model of Pelligrini and Swift.

Personally I find things like the homopolar generator more sexy that black holes and dark energy etc.

Pete

 

[quantumdude]

How did "forefront of research" get into this conversation?

It got there by me inquiring about it.

You're saying that some of it contains physics which is not related to Particles, Fields, Gravitation and Cosmology?

No. "Gravitation and Cosmology" includes GR, and research in that field is published there.

Ask the string people that, not me.

What I mean is this:

When you raise the objection, "But that's quantum dynamics...", my response is, "That's the only kind of dynamics that takes place in reality."

 

[pmb_phy]

It got there by me inquiring about it.

You inquired about the forefront of research? Sorry. I didn't see that inquiry.

What I mean is this:

When you raise the objection, "But that's quantum dynamics...", my response is, "That's the only kind of dynamics that takes place in reality."

I had a feeling you were going to say that. But since this thread is on a classical concept based on a classical thing like velocity. There are many relativistic topics which are meaningless in QFT but are quite meaningful otherwise. Relativistic mass is one such quantity. To this end I'm referring to the notion that QFT people don't use relativsitc mass as if that's something meaningful. They don't seek of things like velocity in relativistic quantum mechanics but that doesn't mean that people who do that work never use the notion of velocity.

Pete

 

[quartodeciman]

My burning question is whether one can get by with a fundamentally given relativistic concept of 3-momentum (mvγ) and just avoid the mass issue by always talking in momentum terms in any high speed mass scenario. I am not particularly happy about elementary derivations of this 3-momentum usually offered, but that might just be on account of my prejudices.

In short, what is the sense in which speed-dependent mass is an essential rather than a derived (specified) concept?

Thank you,
Quart

 

[quantumdude]

You inquired about the forefront of research? Sorry. I didn't see that inquiry.

I inquired about it by asking about articles from PRL and Phys Rev D.

 

[Garth]

I agree. In fact I don't recall ever saying otherwise. However its impossible to measure something unless you first define what it is you're measuring.

Sorry I blinked and the thread shot off way ahead of me!

The measurement problem has two components: what is to be used as a standard unit, and how is that standard to be transported around the universe for the comparison to be made?
In GR, based on the EEP, energy-momentum is conserved therefore, in that theory, the standard is an atom, and because mass is defined to be invariant by the EEP it is assumed that masses, lengths and times on the far side of the universe can be measured by that standard.
In the Jordan frame of the conformal gravity theory of self-creation it is energy and not energy-momentum that is conserved. (Incidentally this allows a form of continuous creation.) The standard becomes a “standard photon”, carefully defined – cosmologically it is a photon sampled at the peak intensity of the CMB – its energy, which is invariant in this frame, yields a measure of mass, its frequency time and hence length. (c is constant in the theory)

 

[pmb_phy]

My burning question is whether one can get by with a fundamentally given relativistic concept of 3-momentum (mvγ) and just avoid the mass issue by always talking in momentum terms in any high speed mass scenario.

Depends on what you want to know.

In short, what is the sense in which speed-dependent mass is an essential rather than a derived (specified) concept?

I'm sorry but I don't know what that means or what you're asking. E.g. Regarless of what anyone defines things the physics is the same and the equations as as well. The only thing that changes is the symbols and the names.

It should be stated at this point that, unless the subject matter contains potential energy, or the subject matter is energy, the concept of energy in SR is not required. I don't think its required in GR either. One can replace energy by mass in all such cases.

Pete

 

[Garth]

Back to the original question. "Those who use relativistic mass and why"

One reason for the use of relativistic mass is simply "the faster it gets the harder it is to push".

Of course this experimental fact can be interpreted in two ways.
1. The first is to say the mass of a body moving relative to an observer increases when measured by the observer - relativistic mass.
2. The second is to say that Newton's law of inertia, F = ma, has to be modified by the relativistic gamma factor and the mass of the body does not change.

Both are equally correct, one is equivalent to the other, it is simply a matter of convention as to which is the more convenient. As we have seen above this is a matter of opinion, however, the No. 2 formulation is generally adopted because it is also consistent with Einstein's Equivalence Principle (the EEP). Mass is invariant, and the word refers to what otherwise might be called "rest mass", that is the mass as measured by a co-moving observer in whose frame of reference the body is at rest and equal to the body's four-momentum.

However to the simple mind the idea of relativistic mass may be more in keeping with E = mc^2, that is to say energy can actually turn into mass (and vice versa) and not just have a mass equivalent value.

In the No. 2, standard convention you are stuck with the mass in those particles that you have, apart from particle/antiparticle creation/annihilation, all the energy released in an nuclear reaction for example, is that "system energy" formely bound with the nuclear particles. Particle mass has not been converted into pure energy at all, as is popularly thought. Furthermore, if the mass of a particle actually does change, due to absorption of energy,
this convention would be incapable of recognising the fact, it is tautological in definition and blind to any variation that might be taking place.

Yet at a fundamental level a particle is not some indestructible mass but seems to be 'just' energy, in the form of energetic vibrating strings, or whatever, which has acquired inertia; so although the No. 2 convention is adopted because it is convenient for a lot of applications it may be the No.1 convention that is more fundamentally 'true' and favoured by the 'String people'.
Our experiments may yield the data about the universe out there, but we have to interpret that data and that requires an act of faith, in which we choose one particular interpretation over another. The choice is yours!

 

[quartodeciman]

Granted the SR ordinary momentum defined by mvγ, I can readily derive the energy-momentum-restmass relationship. If I opt for a relativistic mass M defined by mγ, then the momentum is a cinch: it is just Mv, just as Newton said "quantity of motion" should be quantified. But the energy-first school thinks that is improper, reifying something as substantial without warrant. So can I render the relativistic 3-D momentum as a primary dynamic quantity and avoid the conflict? If so, M would just be a convenient calculus substitution variable, enough to pull out the derivation.

 

[DW]

Back to the original question. "Those who use relativistic mass and why"

One reason for the use of relativistic mass is simply "the faster it gets the harder it is to push".

Of course this experimental fact can be interpreted in two ways.
1. The first is to say the mass of a body moving relative to an observer increases when measured by the observer - relativistic mass.
2. The second is to say that Newton's law of inertia, F = ma, has to be modified by the relativistic gamma factor and the mass of the body does not change.

We have already been through all this here already. The law of motion for special and general relativity is the four-vector law:
$$F^{\lambda } = mA^{\lambda }$$.
The mass m in that equation does NOT change with speed! How many times do I have to say that here? I really wish those who are still using the mistake called "relativistic mass" would actually read this:
http://www.geocities.com/zcphysicsms/chap3.htm

You are wanting to replace the m in that form of Newton's second law with Planck's variable mass concept, but in doing so would wind up with an equation of motion that is just plain wrong according to relativity, even accoding to Planck, Tolman, and Lewis version of special relativistic dynamics.

 

[pmb_phy]

One reason for the use of relativistic mass is simply "the faster it gets the harder it is to push".

I agree that this is why many people think in terms of relativistic mass.

Of course this experimental fact can be interpreted in two ways.
1. The first is to say the mass of a body moving relative to an observer increases when measured by the observer - relativistic mass.
2. The second is to say that Newton's law of inertia, F = ma, has to be modified by the relativistic gamma factor and the mass of the body does not change.

Let

$$\bold a_{\|}$$ = Component of acceleration parallel to the particle's velocity

$$\bold a_{\bot}$$ = Component of acceleration perpendicular to the particle's velocity

The force, F, on a particle whose proper mass is m₀ is related to those components of acceleration through the relation

$$\bold F = m_{\|}\bold a_{\|} + m_{\bit}\bold a_{\bot}$$

where

$$m_{\|} = \gamma m$$ = Longitudinal mass

$$m{\bot} = \gamma^3 m$$ = Transverse mass

For proof please see - http://www.geocities.com/physics_world/sr/long_trans_mass.htm

The relationship F = ma not valid. It is only valid when the mass, m, is not a function of time. In SR, when a particle is accelerating the mass is a function of time. The correct relation between mass and force is F = dp/dt where p = mv.

To measure the mass of a charged particle one can place the particle in a uniform magnetic field and observer the particle's trajectory, measure its velocity and then, once you've made the appropriate calculations, you can be said to have "measured" its mass. E.g. let the charge be q. Let the magnetic field be parallel to the z-axis and have a magnitude B. Let the velocity vector be parallel to the xy-plane. The trajectory will be a cirlce (for the most part). Measure the radiius of that circle and call it r. Then the mass is found to be

$$m = \frac{qBr}{v}$$

For a derivation of this please see
http://www.geocities.com/physics_world/sr/cyclotron.htm

...however, the No. 2 formulation is generally adopted because it is also consistent with Einstein's Equivalence Principle (the EEP). Mass is invariant, and the word refers to what otherwise might be called "rest mass", that is the mass as measured by a co-moving observer in whose frame of reference the body is at rest and equal to the body's four-momentum.

Do you think that the qualifier proper should be left off of the term proper time and simply call d(tau) "time"? If not please explain why.

However to the simple mind the idea of relativistic mass may be more in keeping with E = mc^2, that is to say energy can actually turn into mass (and vice versa) and not just have a mass equivalent value.

So long as one does not confuse simplicity with stupidity.

Furthermore, if the mass of a particle actually does change, due to absorption of energy, this convention would be incapable of recognising the fact, it is tautological in definition and blind to any variation that might be taking place.

I disagree. The proper mass of a particle can change. You can still legitimately call that "invariant mass" since the term "invariant", as aplied to mass means "unchanged by a change in coordinates." It does not mean "unchanged with time."

Pete

 

[DW]

I agree that this is why many people think in terms of relativistic mass.

Let

$$\bold a_{\|}$$ = Component of acceleration parallel to the particle's velocity

$$\bold a_{\bot}$$ = Component of acceleration perpendicular to the particle's velocity

The force, F, on a particle whose proper mass is m₀ is related to those components of acceleration through the relation

$$\bold F = m_{\|}\bold a_{\|} + m_{\bit}\bold a_{\bot}$$

where

$$m_{\|} = \gamma m$$ = Longitudinal mass

$$m{\bot} = \gamma^3 m$$ = Transverse mass

For proof please see - http://www.geocities.com/physics_world/sr/long_trans_mass.htm

The relationship F = ma not valid. It is only valid when the mass, m, is not a function of time. In SR, when a particle is accelerating the mass is a function of time. The correct relation between mass and force is F = dp/dt where p = mv.

To measure the mass of a charged particle one can place the particle in a uniform magnetic field and observer the particle's trajectory, measure its velocity and then, once you've made the appropriate calculations, you can be said to have "measured" its mass. E.g. let the charge be q. Let the magnetic field be parallel to the z-axis and have a magnitude B. Let the velocity vector be parallel to the xy-plane. The trajectory will be a cirlce (for the most part). Measure the radiius of that circle and call it r. Then the mass is found to be

$$m = \frac{qBr}{v}$$

For a derivation of this please see
http://www.geocities.com/physics_world/sr/cyclotron.htm


Do you think that the qualifier proper should be left off of the term proper time and simply call d(tau) "time"? If not please explain why.


So long as one does not confuse simplicity with stupidity.


I disagree. The proper mass of a particle can change. You can still legitimately call that "invariant mass" since the term "invariant", as aplied to mass means "unchanged by a change in coordinates." It does not mean "unchanged with time."

Pete

Your own site is not a reference. Look how your argument goes against the very first postulate of relativity. Look what a mess it makes of the physics. Look how there is no transition from it to general relativity. You are blatantly wrong. Now look at what the physics is. For a massive particle the momentum four-vector is related to the velocity four-vector by
$$p^{\mu } = mU^{\mu }$$.
The force four-vector
$$F^{\lambda } = \frac{Dp^{\lambda }}{d\tau }$$
is related to the four-vector acceleration
$$A^{\lambda } = \frac{DU^{\lambda }}{d\tau }$$
by
$$F^{\lambda } = mA^{\lambda }$$
where m is the mass and is completely invariant not depending at all on speed or position within the gravitational field etc. This in a manifestly frame invariant relationship. Contrary to your claim, it works relativistically for every possible choice of frame. This is a simple relation. This is general relativistically correct. That is what the physics is whether you like it or not.

 

[pmb_phy]

Look how there is no transition from it to general relativity.

Yet another invalid comment posted without any attempt to back it up. I've wrote and posted proofs of all your claims. You've chosen to ignore them. Most likely because you are unable to back your claims up.

You are blatantly wrong.

I only expected people who understood physics to understand. Both I, and now Tom, have explained your errors to you. You have been unable to succeed in supporting your comments because, as always, you refuse to define the quantity which whose definiton your trying to discuss. E.g. in this case you don't have, or refuse to post, the definition of mass in a well defined way. All you're able to do now and in the past is to assume a particular definition, i.e. the m₀ such that m₀ U is a conserved quantity. You then try to prove that m₀ does not equal m. Well DUH.

For a massive particle the momentum four-vector is related to the velocity four-vector by
$$p^{\mu } = mU^{\mu }$$.

That relation is incorrect. The correct relation is (Note: I use the convention that capital letters are 4-vectors and small letters are 3-vectors)

$$\bold P = \mu \bold U = (cm, \bold p)$$

where

$$m = \gamma \mu$$

$$\bold p = m \bold v = \gamma \mu \bold v$$

I use the symbol mu in this post for the proper mass of the particle. davy - Do you claim that its impossible in physics to represent proper mass by the Greek letter mu? Or don't you understand what the term "proper mass" means? The greek letter for m is used because if one is going to use Greek letters for proper quantities (such as proper time) then one should be consistent about it.

4-force is defined as

$$\bold F = d\bold P/d\tau$$

Tell us all something - Why do you use tau for proper time? Time is always represented by the letter "t" in ever single relativity artiucle and text which has ever been written with few exceptions (Mould being one of them). Hmnmmm ... Hold on a sec! davy - Do you mean to tell us that tau is PROPER TIME?? Hmmm. Then if tau is proper time and since "t" (coordinate time) does not appear in 4-vector equations, then why don't you use "t" for proper time The answer is you don't understand the difference between (relativistic) mass and proper mass.

(snip)

Sorry but until you have something logical to post then its back to oblivion for you once more.

Plonk (i.e. back to the ignore list for you)

 

[quantumdude]

DW: $$p^{\mu } = mU^{\mu }$$

PMB: That relation is incorrect. The correct relation is (Note: I use the convention that capital letters are 4-vectors and small letters are 3-vectors)

Hold up a second. Why is DW's relation wrong? The only differences between his and yours seems to be that you associate γ with m, and he doesn't, and what he calls m, you call μ.

I've been following this thing between you and DW for a while now (even though I haven't joined in), and for the life of me I can't see how the difference between your respective points of view is anything other than a difference in choice of convention. But each of you is so convinced that the other is wrong, that I'm wondering if the difference is deeper than that.

 

[pmb_phy]

Hold up a second. Why is DW's relation wrong? The only differences between his and yours seems to be that you associate γ with m, and he doesn't, and what he calls m, you call μ.

I was providing an argument identical to dw's in hopes that he'd finally get the point that al he's commented on so far is the semantics of a symbol. I was hoping for that 1 in a million chance that he'd understand that all he's complaining about is what a letter means.

But each of you is so convinced that the other is wrong, that I'm wondering if the difference is deeper than that.

Have you ever see me say or imply that the geometric view of relativity was wrong? No. That's because I don't believe its wrong. I explained all that in that paper I wrote in fact. I believe that the geometric view is simply another view of relativity. There is the 3+1 view (what you see me discuss in this thread) and the geometric view (the onlyu think waite was taught in school) and they are two views of identically the same thing. I understand that. dw is unable to understand that.

What I disagree with waite on is what quantity in relativity is most deserving of the word "mass" and I have very good reasons for that which waite has failed to argue against successfully.

Pete

 

[quantumdude]

I was providing an argument identical to dw's in hopes that he'd finally get the point that al he's commented on so far is the semantics of a symbol. I was hoping for that 1 in a million chance that he'd understand that all he's complaining about is what a letter means.

So you were kidding then. I didn't pick up on your true meaning.

Have you ever see me say or imply that the geometric view of relativity was wrong? No.

Well, it looked like you were doing it here.

 

[pmb_phy]

So you were kidding then. I didn't pick up on your true meaning.

Yes. I was kidding. But it was done to make a point.

Well, it looked like you were doing it here.

Please point out what it was that I said that gave that impression so that I may avoid giving that impression in the future. Thanks

Pete

 

[quantumdude]

Please point out what it was that I said that gave that impression so that I may avoid giving that impression in the future. Thanks

It was the part where you were kidding, and I didn't get it.

 

[pmb_phy]

It was the part where you were kidding, and I didn't get it.

Ah! Okay. So it was just that one post. I undestand. Thanks for clarifying.

For the longest time I've been trying to teach dw the difference between proper quantities and, for lack of a better term at the moment, coordinate quantities. In this case I'm teaching him to use his (ahem) "knowledge" of proper time being different, but related to, (coordinate) time to understand the relationship between proper mass and (relativistic) mass.

Let me elaborate. In what follows, unless otherwise stated, all capital letters represent 4-vectors and lower case letters represent represent 3-vectors.

Define R as

$$\bold R \equiv (ct, x, y, z) = (ct, \bold r)$$

Obvioulsy in this expression there is a difference between R and r. Each has a physical meaning in relativity. The former quantity is a 4-vector and the later is a 3-vector. Now consider the differential of R

$$d\bold R = (cdt, dx, dy, dz) = (cdt, d\bold r)$$

That is, of course, the spacetime displacement. Define $$d\tau$$ through

$$c^2 d\tau^2 = \bold g(d\bold R, d\bold R) = d\bold R \bullet d\bold R$$

as you know, $$d\tau$$ is related to dt through

$$dt = \gamma d\tau$$

Define 4-velocity as

$$\bold U = \frac{d\bold R}{d\tau} = (c\gamma, \gamma \bold u)$$

Similar to above, in this expression there is a difference between U and u where u = 3-velocity. As you know, 4-momentum is defined as (once more using mu for proper mass)

$$\bold P = \mu \bold U = (c\gamma\mu, \bold p)$$

Once again, in this expression there is a difference between P and p where p = 3-velocity.

Call dt the "time component of dR." The exact relation is dt = dR⁰/c. With this as motivation defined the time component of all 4-vectors A as A⁰/c. What's the time component of P? It's

$$m \equiv \gamma \mu = P^0/c$$

From this we obtain

$$P^0 = cm$$

We can now write P as

$$\bold P = (cm, \bold p)$$

Also as you know

$$c^2 \mu^2 = \bold g(d\bold P, d\bold P) = \bold P \bullet \bold P$$

This is one of the ways to motivate the definition of m since now we have a relation similar to that above, i.e,

$$m = \gamma \mu$$

Summary:

$$time = \gamma (proper time)$$
$$mass = \gamma (proper mass)$$

The time component of the two 4-vectors (dR and P) above have the classical name and letter. The magnitude of the 4-vectors has the "proper" qualifier appended to the classical name (Its a shame that dw doesn't understand relativity/physics/logic well enough to grasp all of this).

See where I'm going here? Nothing above can be read to mean that I dislike the geometric view. That would be like saying that me explaining dt = gammma*d(tau) is an argument against the geometric view. Both claims are silly.

Pete

ps - Yes yes yes dw. We all know. You'll post a few insults and call us stupid. We've heard all that nonsense before so don't bother repeating your insults and illogical/ill-thought out arguements again.

 

[DW]

Let me elaborate. In what follows, unless otherwise stated, all capital letters represent 4-vectors and lower case letters represent represent 3-vectors.

Define R as

$$\bold R \equiv (ct, x, y, z) = (ct, \bold r)$$

Position is not a four-vector. Displacement is. In the following corrections I use $$[p^{\mu }]$$ in the place of P because it is standard to use one capitalisation of p for the four-vector momentum of the first kind and the other capatilitation to include potentials so that its elements correspond to energy and momentum opperators in relativistic quantum mechanics, though which capitalisation is used for which is not universal.

The former quantity is a 4-vector and the later is a 3-vector.

No. Displacement is a vector, but position is not.

$$d\bold R = (cdt, dx, dy, dz) = (cdt, d\bold r)$$

That is, of course, the spacetime displacement.

That is first vector you've presented here.

As you know, 4-momentum is defined as (once more using mu for proper mass)

$$\bold P = \mu \bold U = (c\gamma\mu, \bold p)$$

Correction:
As you know, 4-momentum for the case of a particle with mass m is
$$[p^{\mu }] = m\bold U = (c\gamma m, \bold p)$$
there is no term proper as mass is invariant.

where p = 3-velocity

Correction
p is the spatial components of the momentum four-vector.

What's the time component of P? It's

$$m \equiv \gamma \mu = P^0/c$$

Correction, it is

$$\frac{E}{c^{2}} \equiv \gamma m = p^0/c$$

From this we obtain

$$P^0 = cm$$

Correction
$$p^0 = \gamma mc$$

We can now write P as

$$\bold P = (cm, \bold p)$$

Correction
We can now write $$[p^{\mu }]$$ as
$$[p^{\mu }] = (cm\gamma , \bold p)$$

Also as you know

$$c^2 \mu^2 = \bold g(d\bold P, d\bold P) = \bold P \bullet \bold P$$

Correction
$$c^2 m^2 = \bold g([p^{\mu }],[p^{\mu }]) = [p^{\mu }] \bullet [p^{\mu }]$$

$$m = \gamma \mu$$

Correction
$$E = \gamma mc^2$$

 

[quartodeciman]

Pete,

That was a good tip: put 'DW' in my ignore list. The topic looks much better now.

Quart

 

[pmb_phy]

Pete,

That was a good tip: put 'DW' in my ignore list. The topic looks much better now.

Quart

Hi quart

Don't get me wrong. I never put people on my ignore list except for people who spam threads with the same comments they've posted a thousand times before without end. It is always a good idea to hear somone elses views (in fact you're almost always a better person for it) dw has badly abused that idea by repeating himself to the same person, the same comments a times while ignoring proof under all occasions. Its irritating after the first 100 times.

E.g. to show you what I mean I took a gander at his last one for purposes of illustration. This is the 1,000 th time that he's claimed that the position 4-vector is not 4-vector. I explained to dw why his claim is wrong 1,000 times. He ignores it 1000 times and then stgarts repeating himself

In this case R = (ct, x, y, z) is a Lorentz 4-vector. Its defined as the displacement displacement from a chosen event which is defined as the "origin of coordinates". This is standard stuff found everywhere and in nearly all relativity/em texts (e.g. Ohanian, J.D. Jackson, Thorne and Blanchard etc). Yet dw can't seem to learn it. (sigh)

To be precise, define

$$\bold X^P \equiv (ct_P, x_P, y_P, z_P)$$ = Event P

$$\bold X^Q \equiv = (ct_Q, x_Q, y_Q, z_Q)$$ = Event Q

$$\Delta \bold X \equiv \bold X^P - \bold X^Q = (ct_P, x_P, y_P, z_P) - (ct_Q, x_Q, y_Q, z_Q) = (c\Delta t, \Delta x, \Delta y, \Delta z)$$

$$x \equiv x_P - x_Q = \Delta x$$

$$y \equiv y_P - y_Q = \Delta y$$

$$z \equiv z_P - z_Q = \Delta z$$

Now define event Q as the "Origin" of the coordinate system. This means, for example, that x is the x-component of a displacement R from something called the "origin" and is written as

$$\bold R \equiv \Delta \bold X = (ct, x, y, z)$$

That is the template of all Lorentz 4-vectors.

A previous example was when he claimed that what I was using couldn't be readily used to translate to GR. Thus he took my explanation of what is equivalent of defining and describing the components of 4-vectors and has ignored the numerous times where I've used it in equations in GR wiuth 4-vectors. Here is a perfect example

http://www.geocities.com/physics_world/gr/grav_force.htm

In that derivation you can see how the relativistic mass falls out of a derivation which starts with all 4-vectors. See Eq. (8a) in above link. I assume you'll understand why I'll ignore dw's claims on its correctness when he tries to respond to this right?


Smart move that you took quart.

Pete

 

[DW]

Pete,

That was a good tip: put 'DW' in my ignore list. The topic looks much better now.

Quart

I guess fake physics looks better to you.

 

[DW]

While It doesn't hurt to hear somone elses views, in fact you're a better person for it, dw abuses that notion by repeating his same bogus claims a thousand times while ignoring proof.

You are the one ignoring the proof I and others have given that you are wrong.

This is the 1,000 th time that he's claimed that the position 4-vector is not 4-vector.

See, as I said you never stopped reading my posts. And no, position is not a four-vector.

I explained to dw 1,000 times that R = (ct, x, y, z) is a Lorentz 4-vector.

No it is not.

Its defined as the displacement ...

No it is not. The four-element coordinate position is not a displacement and is not a vector. Displacement is a vector. Position is not.

 

[Garth]

All that has gone before seems to illustrate the point in my last post.
Is it the case that one way to harmonise the two points of view would be to accept that energy and mass are equivalent concepts and 'relativistic mass' can simply be renamed "Total energy"? Of course particles have other properties too, inertia and other charges, in which case the defining characteristic of 'proper mass' would seem to be its inertia.
Finally, are there any thoughts about my point that the 'relativistic mass' concept, although not in favour with DW and the standard convention in particle physics, may actually be the more fundamental point of view?

 

[pmb_phy]

The relativistic Lagrangian does indeed contain the mass of a particle if I adopt the convention that the norm of the 4-momentum is the mass.

That is only true for the covariant Lagrangian. It is not true for the relativistic, non-covariant Lagranmgian. The (relativistic, non-covariant) Lagrangian for a charged particle in an EM field is given by (Reference: Classical Electrodynamics - 2nd Ed., J.D. Jackson, page 574, Eq. (12.9)}

$$L = \mu c^2 \sqrt{1 - v^2/c^2} + q\Phi - \frac{q}{c}\bold v \bullet \bold A$$

where mu is the particle's proper mass. However

$$m = \gamma \mu$$

where m is the particle's inertial mass. Solving for mu and substituting into the above equation gives

$$L = \frac{mc^2}{1 - v^2/c^2} + q\Phi - \frac{q}{c}\bold v \bullet \bold A$$

The Lagrangian can therefore be expressed in terms of either the proper mass or the relativistic mass and there is no significant difference inherent in the Lagrangian which prefers one over the other. The only diference is that the later Lagrangian is not defined for v = c. That's a direct result of m not being defined for v = c.

Pete