Those who use relativistic mass and why [Archive]

pmb_phy

Jul25-04, 09:59 PM

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! :approve:

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 dependance 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. :smile:

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

Pete

Garth

Jul26-04, 01:55 AM

Mass is to be measured and not just defined

pmb_phy

Jul26-04, 05:24 AM

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

Jul26-04, 05:43 AM

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

pmb_phy

Jul26-04, 06:49 AM

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

Tom Mattson

Jul26-04, 09:43 AM

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 dependance 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

Jul26-04, 10:12 AM

*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 [i]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

Tom Mattson

Jul26-04, 10:29 AM

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 ([i]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? :biggrin:

pmb_phy

Jul26-04, 10:55 AM

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? :biggrin:

Ask the string people that, not me. :confused:

Pete

pmb_phy

Jul26-04, 11:13 AM

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 tendancy 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

Tom Mattson

Jul26-04, 11:20 AM

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. :confused:

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

Jul26-04, 11:32 AM

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 refering 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

Jul26-04, 11:35 AM

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

Tom Mattson

Jul26-04, 11:47 AM

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

Jul26-04, 11:58 AM

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

Jul26-04, 12:14 PM

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

Jul27-04, 02:42 AM

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

Jul27-04, 12:06 PM

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.

Jul27-04, 01:10 PM

pmb_phy

Jul27-04, 02:12 PM

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. :wink:

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

Jul27-04, 02:45 PM

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. :wink:

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 arguement 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

Jul27-04, 03:36 PM

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)

Tom Mattson

Jul27-04, 03:51 PM

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

Jul27-04, 04:12 PM

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 arguement 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

Tom Mattson

Jul27-04, 04:19 PM

I was providing an arguement 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

Jul27-04, 04:31 PM

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

Tom Mattson

Jul27-04, 04:33 PM

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

Jul27-04, 05:26 PM

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 arguement 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.

Jul27-04, 07:00 PM

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

Jul27-04, 10:52 PM

Pete,

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

Quart

pmb_phy

Jul28-04, 12:07 AM

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? :biggrin:

Smart move that you took quart. :approve:

Pete

Jul28-04, 12:14 AM

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.

Jul28-04, 12:19 AM

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

Jul28-04, 02:45 AM

pmb_phy

Jul28-04, 05:00 AM

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 in to 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

pmb_phy

Jul28-04, 05:40 AM

You ..

Me? Yeeesh! You sure have a lot to learn about the relativity community. Fine. Okay, me. But also Einstein, Wheeler, Thorne, Rindler, D'Inverno, Sartori, D'Inverno, Mould, Peacock, Guth, etc. etc. etc. etc. etc. etc. etc. etc.

...are wanting to replace the m in that form of Newton's second law with Planck's variable mass concept,...

In 1905 Einstein attempted to write the equations of a charged particle in an EM field in the form F = ma. That led to his use of transverse and longitudinal mass. In the year that followed, i.e. 1906, Planck showed that the Lorentz force could be written in the form

\bold F = \frac{d\bold p}{dt} = \frac{d(\gamma m_o \bold v)}{dt} = q(\bold E + \bold v \times \bold B)

Or substituting in relativistic mass m = gamma m_o

\bold F = \frac{d(m \bold v)}{dt} = q(\bold E + \bold v \times \bold B)

where m is the relativistic mass of the body. After that paper Planck never tried to prove that mass can in all cases be set equal to m = gamma m_o so he did not get the credit for showing that m = gamma m_o. Hence he does not deserved the credit. Three years later, in 1909, Tolman and Lewis argued that mechanics should be obtained from the conservation laws and the principle of relativity and without reference to electrodynamics. In their famous paper The Principle of Relativity and Non-newtonian mechanics they demonstrated the feasability of such a notion through the now famous collision thought experiment.Three years later in 1912 Tolman published a more general version in his famous paper Non-Newtonian Mechanics: The Mass of a Moving Body. All relativity texts (at least those which I know of) which derive the momentum equation p = gamma*m_o*v now use Tolman's method as described in that paper. Neither paper had anything to do directly with force. It was due to this work that, in part, was responsible for relativity papers to no longer be restrticted to EM journal references. Hence Tolman and Lewis are given the credit for being the ones to show that mass depends on velocity.

Jul28-04, 08:32 AM

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? :biggrin:

Smart move that you took quart. :approve:

Pete

Now you are back peddling on what you said. Originally you claimed position was a four-vector. Now you are trying to twist what you said to refer to a displacement. It is obvious you are not being honest particularly because you had considered the displacement OF that position in the numerator of your definition of four-vector velocity. You don't put a displacement of a displacement there! Just admit you were wrong and I corrected you and move on.

Jul28-04, 08:36 AM

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 in to the above equation gives

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

Pete

Corrections
The 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 = m c^2 \sqrt{1 - v^2/c^2} + q\Phi - \frac{q}{c}\bold v \bullet \bold A

where m is the particle's mass. However

E = \gamma mc^2

where E is the particle's energy. Solving for m and substituting in to the above equation gives

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

Jul28-04, 08:47 AM

In 1905 Einstein attempted to write the equations of a charged particle in an EM field in the form F = ma.
I am the one who has been telling you about this.
That led to his use of transverse and longitudinal mass.
No. He didn't use those. He introduced those concepts. The mass that he actually used in the paper was mass as invariant.

In the year that followed, i.e. 1906, Planck showed that the Lorentz force could be written in the form

\bold F = \frac{d\bold p}{dt} = \frac{d(\gamma m_o \bold v)}{dt} = q(\bold E + \bold v \times \bold B)

Or substituting in relativistic mass m = gamma m_o

\bold F = \frac{d(m \bold v)}{dt} = q(\bold E + \bold v \times \bold B)

where m is the relativistic mass of the body. ......

Which is why I keep referring to "relativistic mass" as Planck's mass concept which is a dead concept having no place in modern relativity.

(snipped some flamming)
Hence dw's (and his alter ego GRCQ} honored place on the ignore list.
Just because every knowledgable person disagrees with you does not make every knowledgable person me. I am not GRCQ and such a lie should warrent your dissmissal here. And how is it that you keep responding to someone you are ignoring? You never had me on any such list.

quartodeciman

Jul28-04, 12:16 PM

Pete,

You may have surmised by now that I'm right on the verge between (what I call) mass-first and energy-first points of view. That is why I posed my earlier question about validating SR momentum fundamentally. Then (I hope) one has (perhaps) a choice of directions for expanding SR dynamics.

SR kinematics --->SR momentum law--->exploitation of mass dynamics--->*
SR kinematics --->SR momentum law--->exploitation of energy dynamics--->*
*--->justification of 4-dimensional dynamic worldview

Quart

Jul28-04, 04:10 PM

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"?
Energy and mass are not equivalent. Mass is equivalent to "rest frame energy". Calling relativistic mass energy is NOT renaming it. Calling energy by relativistic mass IS renaming energy with a missnomer.Using Planck's concept of mass harminises nothing. Using mass as invariant is what is correct and as such haminises everything that encorporates it.

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.
Qualifying the word mass with proper is wrong because mass is invariant. The length of the momentum four-vector is the same value for ALL frames.

Garth

Jul29-04, 04:32 AM

pmb_phy

Jul29-04, 07:08 AM

Jul29-04, 09:46 AM

I recall somewhere someone quoting dw about his claim that the gamma factr is supposed to be associated with the velocity and not the mass. Unforunately for de that's an incorrect assumption. In the first place, any definition of mass should hold in all cases. dw's association of gamma with velocity is invalid for photons since in that case the gamma factor is infinite and the momentum is finite. The relativistic mass is still well defined. Also, as I've explained to dw on countless occassions, the complete description of mass requires a tensor. As an example, consider the case of pressureless dust. In the aforementioned tensor the momentum density, g for matter is given by

\bold g = \rho \bold v

where rho is the mass density given by

\rho = \gamma^2 \rho_0

In this case its not meaningful to associate gamma with v since one is still left with another gamma. And this relation is not valid in general and especially not for radiation.

Pete
My definition is not invalid for photons and you know it because you read it and are now misrepresenting it. You know very well that in general I define momentum in terms of a quantum frequency and wavelengths in short in terms of a wavelength k vector which applies for both massive and massless particles. I then define mass in terms of that vector and only afterward demonstrate the relation between momentum and four-vector velocity only as a second hand result for massive particles. How dare you knowingly missrepresent my position?

Jul29-04, 09:57 AM

DW - An atom has mass, which we would want to define as invariant, yet that mass may vary by the emission/absorption of a photon.
So we open up the atom and find atomic particles - nucleons and electrons and a total system energy. This energy is the source of that photon's energy. The particles have mass, which we would want to define as invariant, yet they enter into energetic interactions.
So we open up the nucleons and find quarks and strong/weak force energies which supply those interactions with energy. The quarks have mass, which we would want to define as invariant, yet they enter into energetic interactions.
So we open up the quarks??
Where does the process end?
It ends at the virtual particles comprising the field energy in the bindings, which are ultimatley what really account for why we can use the potential energy shortcut in the first place.
According to one popular theory it ends with strings, which have mass? or is it energy? At this level the strings are vibrations of energy, as the Schrodinger coordinate representation, with its wave packet functions, would have it all along in the first case, so where is mass at the most fundamental level?
Where it comes to strings, the mass of a string correspond to particular modes. Since I am not a string theorist and relativity was never intended to apply in the string domain, I don't see how bringing up those scales is relevent outside of being interesting if it does. As far as I have gone is to see how general relativity is applicable in electrodynamics all the way down to the subatomic particle quantum domain.

Garth

Jul29-04, 11:44 AM

DW - The question I am asking is whether mass is inevitably invariant or might it vary to include energies, especially potential energies?

I am not refuting any convention here to cause an argument, I am asking a serious question in order to seek the truth at the most fundamental levels.

I am not particularly happy with string theory either, as I don't like inventing things, like extra dimensions, which then conveniently roll themselves up so you can't see them. - Like the fairies at the bottom of my garden that are really there but you can never see them because they are so shy. Today's 'New Scientist' reports a new string theory that apparently does away with these extra dimensions so I shall be happier with that. We shall see.

However your phrase "all the way down " reminds me of the story of the Flat Earther who gave a lecture on how the Earth was a flat disc sitting on the back of four elephants, which stood on the back of a giant turtle. When he was asked what the turtle was standing on he replied, "Its turtles all the way down".

My point is that if it is not 'turtles all the way down' then we end up at a fundamental level which consists of objects (particles/strings or whatever) that are the final/ultimate repositories of mass. However, because their interactions will have to determine the interactions in the level above, they will have to enter into energetic interactions themselves. In this case as they use/release energy their mass will have to vary to accommodate that exchange of energy. If not there must be a deeper level consisting of particles with mass and an energy bank. And so on...

Perhaps it is "turtles all the way down."

Nereid

Jul29-04, 02:29 PM

However your phrase "all the way down " reminds me of the story of the Flat Earther who gave a lecture on how the Earth was a flat disc sitting on the back of four elephants, which stood on the back of a giant turtle. When he was asked what the turtle was standing on he replied, "Its turtles all the way down".OK, I confess; this is OT: Terry Pratchett wrote the Discworld series based on this idea, which itself has a very long history (x000 BC Hindu?). I'm not sure if the first use of the phrase is well determined :smile:

pmb_phy

Jul29-04, 06:15 PM

There are other situations when the mass = proper mass is inadequate. It's been shown that a rod is easier to accelerated when it is pulled rather than when it is pushed. This, of course, implies that a scalar cannot be associated with an object in all cases to describe the inertial properties of an object.

Garth

Jul30-04, 01:44 AM

OK, I confess; this is OT: Terry Pratchett wrote the Discworld series based on this idea, which itself has a very long history (x000 BC Hindu?). I'm not sure if the first use of the phrase is well determined :smile:
I had thought it was ancient Egypt.

selfAdjoint

Jul30-04, 09:31 AM

I had thought it was ancient Egypt.

India? With the elephants? I think Pratchett has only one level of elephants and one turtle, at least that's all the "disconauts" in one of his books saw.

Haelfix

Jul31-04, 09:45 PM

I just read this thread, wasting my time more or less.

I agree with Tom Matson btw, I can't see the argument as this is purely a matter of convention and notation. Feel free to rename popular concepts mass, energy, whatever all you want all that matters is the quantity an experiment measures and what a theory predicts is that *number*. I for one see your arguments as more or less equivalent.

All this is soo much easier if you just work in natural units hbar = c = 1. If you want set G = 1 too in the GR context.

Besides if you really want to quibble about semantics, all this stuff is moot game. Field theory and full general relativity is the language proffessional physicists talk in nowdays.

pmb_phy

Jul31-04, 10:25 PM

I just read this thread, wasting my time more or less.

I agree with Tom Matson btw, I can't see the argument as this is purely a matter of convention and notation. Feel free to rename popular concepts mass, energy, whatever all you want all that matters is the quantity an experiment measures and what a theory predicts is that *number*. I for one see your arguments as more or less equivalent.

Its not always wise to discard any future thought about something like this because you've decided its all a matter of semantics. It took a few years of studying this subject in detail before I realize that it was much more than that. Only then did I start asking myself more fruitful questions on this topic and it was then that it produced fruit.

Pete

Haelfix

Aug1-04, 12:36 AM

Lets be honest, mass is a completely nebulous concept either way.

Outside of classical mechanics, its somewhat arbitrarily defined depending on the theory.

Already in vanilla quantum mechanics, its hard to say exactly what *is* the mass. In special relativity there is your discussion thread. In GR, there exists metrics where no sensible global notion of what mass-energy is.

In field theory (particularly when talking about 1st order gravity), its just so painful to even think about such things, that no one has bothered muddling their head over what exactly *is* the physical meaning.

What we do have is a bunch of equations, that output a number for a specific situation and experimental setup, and thats that. I think nature has given us a pretty good hint that our intuitions are leading us down a blind alley in this case, and that we should just follow the tried and true equations that match experiment.

And based on those equations, I don't see any mathematical inconsistency between your choice of conventions and DW's. Now if you wish to debate that, please clearly outline the statement and show me that x is not equal to y in say an experiment.

Garth

Aug1-04, 02:37 AM

I agree with Tom Matson btw, I can't see the argument as this is purely a matter of convention and notation. Feel free to rename popular concepts mass, energy, whatever all you want all that matters is the quantity an experiment measures and what a theory predicts is that *number*. I for one see your arguments as more or less equivalent.

Thank you, a number of us have made the same observation.

All this is soo much easier if you just work in natural units hbar = c = 1. If you want set G = 1 too in the GR context.

So long as you are aware of the limitations that such a convention or 'language' places upon what you are able to say. G = 1 is fine in a strictly GR context but defining it as so would blind you to the possibility that G might vary as in the Brans Dicke theory, likewise defining mass to be invariant.

My intuition is that energy is fundamental and therefore we might take up Feynman's usage and call (rest) mass 'Rest Energy' and relativistic mass 'Total Energy'. This may be a more generalised way of conceiving of the world.

sal

Aug6-04, 11:35 AM

It's been shown that a rod is easier to accelerate when it is pulled rather than when it is pushed.

It has??

Can you explain that, or perhaps post a reference to a derivation? (BTW I'm happy with references to your website -- unlike DW I have no problem with it :wink: it's a good site, IMHO)

Or perhaps I should just ask where the acceleration is measured -- at the front end of the rod, the back end, or the "middle" (for some definition of "middle").

Garth

Aug6-04, 02:03 PM

The problem with the four-momentum equation P(mu) = mU(mu) is that U defines velocity with respect to proper time tau - dx(mu)/d(tau). But how do we measure proper time? What clocks keep proper time? It is only in the particle's rest frame that its proper time can be measured and therefore the fact that m is the constant rest mass is an observational tautology; it can only be measured in the rest frame. In any other frame of reference the frame dependent time t is measured and the relativistic mass which we may call M or m.gamma according to our convention as discussed above.

Aug6-04, 02:16 PM

The problem with the four-momentum equation P(mu) = mU(mu) is that U defines velocity with respect to proper time tau - dx(mu)/d(tau). But how do we measure proper time? What clocks keep proper time?

That is not a problem, but is not the definition of four-vector momentum anyway. It is only a result applicable to particles that don't happen to travel at the speed of light.

It is only in the particle's rest frame that its proper time can be measured and therefore the fact that m is the constant rest mass is an observational tautology; it can only be measured in the rest frame.
That simply isn't true. If it were we wouldn't know the mass of any particles, because we never measure it from their rest frames. The dynamics equation of relativity corresponding to Newton's second law is the four-vector equation F^{\lambda } = mA^{\lambda }. It is the m there that is measured in terms of that equation or an equivalent result from it. That m is the only real mass that there is and it is invariant.
In any other frame of reference the frame dependent time t is measured and the relativistic mass which we may call M or m.gamma according to our convention as discussed above.
And relativistic mass is a mistake anyway. If you mean relativistic energy then say relativistic energy because something other than that exact thing called by relativistic mass doesn't even exist in nature at all.

pmb_phy

Aug6-04, 02:39 PM

It has??

Can you explain that, or perhaps post a reference to a derivation? (BTW I'm happy with references to your website -- unlike DW I have no problem with it :wink: it's a good site, IMHO)

Or perhaps I should just ask where the acceleration is measured -- at the front end of the rod, the back end, or the "middle" (for some definition of "middle").

Hi sal

Nice to see you posting here.

I read an article about it in the American Journal of Physics (AJP) several years back. I don't recall the exact reasons but I think it was related to gravitation time dilation. There was a similar article in another journal which I'm trying to get my hands on.

If you'd like I can scan that AJP article in an e-mail it to you?

Pete

sal

Aug6-04, 02:52 PM

I read an article about it in the American Journal of Physics (AJP) several years back....

If you'd like I can scan that AJP article in an e-mail it to you?

Sure, thanks, I'd like to see it -- off hand I can't imagine how it could work out that way.

This seems like a really pleasant forum. And if I could only figure out how to get threaded message display enabled, I'd be reasonably happy with the user interface, too...

Garth

Aug6-04, 04:04 PM

That is not a problem, but is not the definition of four-vector momentum anyway. It is only a result applicable to particles that don't happen to travel at the speed of light.

That is obvious of course, what we are talking about in this thread is the definition of mass, i.e. a particle's mass.

That simply isn't true. If it were we wouldn't know the mass of any particles, because we never measure it from their rest frames.
I wasn't talking about the measurement of mass here but proper time, how would you measure it? The measurement of mass is then derived from that, so that is the problem.

And relativistic mass is a mistake anyway. If you mean relativistic energy then say relativistic energy because something other than that exact thing called by relativistic mass doesn't even exist in nature at all.
The phrase "relativistic mass" is used by so many authoritative people that I do not think you can dismiss it that easily, what we are trying to do is ask whether it is a useful concept of not. Of course, as I have said before, we could also use the phrase "Total energy" and use "Rest energy" for "Rest Mass" - or in your convention - mass. If we use the term "Rest energy" we also leave open the question as to whether it is invariant or not under translations and boosts, say within a gravitational field. A postulate that is open to experimental verification/falsification.

Aug6-04, 07:59 PM

That is obvious of course, what we are talking about in this thread is the definition of mass, i.e. a particle's mass.
Right, and it is not p^{\mu } = mU^{\mu }. That is a result, not a definition.
I wasn't talking about the measurement of mass here but proper time, how would you measure it? The measurement of mass is then derived from that, so that is the problem. No its not so it is not a problem. You have it backwards. The definition of mass involves no proper time derivatives. Expressions that contain them are are derived from it. Step by step here is how I am currently defining things so as to avoid circularity:
1. Take wavelength to be a primative concept.
2. Define a quantum frequency in terms of that wavelength.
3. Define 3 component momentum in terms of wavelengths and energy in terms of frequency.
4. Define the momentum four-vector in terms of the 3 component momentum and energy.
5. Define particle mass as the length of the momentum four-vector.
(In no way is the definition of mass in terms of proper frame coordinates. Everything so far is in terms of your coordinate frame and applicable to both massive and massless particles)
6. Define four-vector velocity.
7. Derive the relationship between massive particle's four-vector momentum and four-vector velocity
p^{\mu } = mU^{\mu }
8. Define four-vector force in terms of four-vector momentum
F^{\lambda } = \frac{Dp^{\lambda }}{d\tau }
9. Define four-vector acceleration
A^{\lambda } = \frac{DU^{\lambda }}{d\tau }
10. Derive the relativistic version of Newton's second law:
F^{\lambda } = mA^{\lambda }
11. Derive the relativistic power equation
g_{\mu }_{\nu }F^{\mu }U^{\nu } = 0
12. Derive for special relativity the work energy relation
\Delta E_{K} = \int(\frac{d\vec{p}}{dt})\cdot d\vec{r}
The familiarity of this equation is the ONLY motivation for the next step:
13. Define ordinary force which is not a four-vector as
f^{\lambda } = \frac{dp^{\lambda }}{dt}
(The zeroth element is zero)
The phrase "relativistic mass" is used by so many authoritative people that I do not think you can dismiss it that easily, what we are trying to do is ask whether it is a useful concept of not.
Yes I can, I just did, and no it is not.
Of course, as I have said before, we could also use the phrase "Total energy" and use "Rest energy" for "Rest Mass" - or in your convention - mass.
Calling is "rest" mass would is wrong because the definition of mass by step 5 is frame invariant. The mass is not just the length of the momentum four-vector according to the proper frame. In fact to that point no reference to the proper frame had been made. The mass is the length according to any frame arbitrarily.
If we use the term "Rest energy" we also leave open the question as to whether it is invariant or not under translations and boosts, say within a gravitational field.
No it doesn't. Calling mass rest energy in fact infers that it is invariant because every frame must agree on what the rest frame energy is.
A postulate that is open to experimental verification/falsification. What postulate?
The ONLY thing I took axiomaticaly was the idea that the four-element combination of wavelength and frequency into a four-element momentum constituted a four-vector in step 4.

Garth

Aug7-04, 01:48 AM

Now we are talking instead of just contradicting each other! Thank you for a considered reply.

1. Take wavelength to be a primative concept.
2. Define a quantum frequency in terms of that wavelength.

Why do we take a wavelength as a primitive concept? What do we mean by a wavelength and in which frame of reference is that length measured or defined? To accept step 1. we have to adopt a preferred foliation of space-time, to use Butterfield and Isham’s expression: cf. Butterfield, J. & Isham, C. J.: 2001, Physics meets Philosophy at the Planck Scale, ed. by C. Callender and N. Huggett. Cambridge University Press.

To go from step 1. to step 2. we have to introduce the postulate or definition that c is invariant, which I am happy to accept but I recognise others who do not.

It is a physical system X that is under observation, the observation event consists of some object X and the observer O with her apparatus. The observation of a real observable A is normally made by the exchange of photons between the two at some stage, especially if they are separated across cosmological distances.

It would be an orthodox approach to consider the X system's state vector Psi to be the primitive concept. To obtain the observable A the state vector has to be solved, either using the Heisenberg's representation in which A is time dependent or the Schrodinger representation in which Psi state vector is time dependent. But time in which frame of reference? X is in one frame and A is observed in another.

pmb_phy

Aug7-04, 08:58 AM

Why do we take a wavelength as a primitive concept? What do we mean by a wavelength and in which frame of reference is that length measured or defined?

I recalll dw attempting to define mass in such a manner but its not a very meaningful, or practicle, way to do so.

Pete

Aug7-04, 10:02 AM

Now we are talking instead of just contradicting each other! Thank you for a considered reply.

Why do we take a wavelength as a primitive concept?
Because I thought you and anyone at this level of physics should already know what wavelength means. That is what is meant by a primative concept, something that is commonly understood so that it doesn't need to be defined in terms of other more "primative" words.

What do we mean by a wavelength and in which frame of reference is that length measured or defined?I think you know what wavelength means and any observer's inertial frame is appropriate for special relativity.

To go from step 1. to step 2. we have to introduce the postulate or definition that c is invariant, which I am happy to accept but I recognise others who do not.
No you don't but its fine by me if you want to.

It is a physical system X that is under observation, the observation event consists of some object X and the observer O with her apparatus. The observation of a real observable A is normally made by the exchange of photons between the two at some stage, especially if they are separated across cosmological distances.
For the most part I was defining particle mass, but fine if you want to define system mass so an "object" can be considered then it should be defined with the object property that is most consistent with the property known as particle mass and for a system that property would be center of momentum frame energy.

It would be an orthodox approach to consider the X system's state vector Psi to be the primitive concept. To obtain the observable A the state vector has to be solved, either using the Heisenberg's representation in which A is time dependent or the Schrodinger representation in which Psi state vector is time dependent. But time in which frame of reference? X is in one frame and A is observed in another.
I don't take that to be primative because hardly anyone knows what a state vector is, though most of us have some understanding of what wavelength is. As for solving for the wave equation, both the Klein-Gordon equation and the Dirac equation come from the definition of mass as I have proposed it.
From the definition of particle mass
m^{2}c^{2} = g_{\mu }_{\nu }p^{\mu }p^{\nu }
Consider the introduction of a four-vector potential adding potential energy to relativistic energy and vector potential elements to momentum terms so that a "second kind" of four-vector momentum can be defined:
P^{\mu } = p^{\mu } + (q/c)\phi ^{\mu }
In terms of the momentum four-vector of the second kind, the mass definition becomes:
m^{2}c^{2} = g_{\mu }_{\nu }[P^{\mu } - (q/c)\phi ^{\mu }][P^{\nu } - (q/c)\phi ^{\nu }]
To get the Klein-Gordon equation simply replace the elements of the momentum four-vector of the second kind with the energy and momentum opperators of quantum mechanics then opperate what you get on the wave equation. Getting the Dirac equation from this is a little trickier, but comes directly from this mass definition as well. See problem 3.1.8 on page 26 at
http://www.geocities.com/zcphysicsms/chap3.htm#BM26

Garth

Aug9-04, 02:15 PM

I think you know what wavelength means and any observer's inertial frame is appropriate for special relativity.

I was genuinely confused by your answer, I wasn't asking, "What is wavelength?" but rather, "The wavelength of what?"
In order to make a measurement we have to make a comparison with a standard, so what is the fundamental standard either of your wavelength, or mass, and which observer's frame is it defined in?

Aug9-04, 02:21 PM

I was genuinely confused by your answer, I wasn't asking, "What is wavelength?" but rather, "The wavelength of what?"
In order to make a measurement we have to make a comparison with a standard, so what is the fundamental standard either of your wavelength,...
There are plenty of length standards in common place. Take your pick.
..., and which observer's frame is it defined in?
I already answered that one.

pmb_phy

Aug9-04, 03:25 PM

I was genuinely confused by your answer, I wasn't asking, "What is wavelength?" but rather, "The wavelength of what?"
In order to make a measurement we have to make a comparison with a standard, so what is the fundamental standard either of your wavelength, or mass, and which observer's frame is it defined in?

dw is speaking of the DeBroglie wavelength. However that only has a statistical meaning and cannot be measured for a single particle. Hence its not useful, or even meaningful, for a single particle. Especially if the object in question is large. E.g. how does one measure the DeBroglie wavelength of an asteroid?

Pete

Aug9-04, 06:12 PM

Garth

Aug10-04, 06:58 AM

dw is speaking of the DeBroglie wavelength. However that only has a statistical meaning and cannot be measured for a single particle. Hence its not useful, or even meaningful, for a single particle. Especially if the object in question is large. E.g. how does one measure the DeBroglie wavelength of an asteroid?

Pete
Thank you for that; in which case not only is the wavelength statistical in nature (as with the rest of q-m) but also the argument is circular; the wavelength is defined in terms of the particle's mass and velocity and the mass is defined in terms of the wavelength; both wavelength and velocity are frame dependent. Do not quantum mechanical definitions require a preferred foliation of space-time, a preferred frame of reference - normally that of the observer? The subject is all about predicting and observing observables and an apparatus that experimentally is most often in that same frame of reference, I don't recall any double slit experiments being observed on a passing asteroid, for example.

The question is, "What standard do we use to measure mass?" "How do we know that in a mass-field theory such as Hoyle's, or the Jordan frame of self creation that masses do not secularly increase with some cosmic field?" "How would we measure such a variation if our definitions blind us to any such change?"

Tom Mattson

Aug10-04, 11:04 AM

Do not quantum mechanical definitions require a preferred foliation of space-time, a preferred frame of reference - normally that of the observer?

No, it doesn't. The quantization procedure is consistent with relativity, though neither is implied by the other. The Klein-Gordon equation that was referred to before is the relativistic version of QM for spinless particles, and it sits with SR just fine.

Garth

Aug10-04, 02:12 PM

The Klein-Gordon equation that was referred to before is the relativistic version of QM for spinless particles, and it sits with SR just fine.
Apart from the problem of defining the mass by the Klein-Gordon equation of spin 1/2 particles such as an electron, the question is whether QM sits with GR just fine. I think you will find that the problem in developing a quantum-gravitational theory is that QM requires the preferred foliation of space-time referred to above. There may be confusion here; all along I have not been questioning SR but GR and its problems with defining time and mass. If I may repeat the first of my "questions" see "Questions of the equivalence principle" http://physicsforums.com/showthread.php?t=32285.
"1. In the presence of gravitational fields the Einstein
Equivalence Principle (EEP) is a necessary and sufficient condition
for the Principle of Relativity, (PR). Here I summarise PR as the
doctrine of no preferred frames of reference. In the absence of such
fields the EEP becomes meaningless, although then the PR does come
into its own and is appropriate in Special Relativity (SR), which was
formulated for such an idealised case. However, if we now re-
introduce gravitational fields, i.e. gravitating masses, do we not
then find that the PR collapses? For in that case is it not possible
to identify preferred frames of reference? Such frames being those of
the Centre of Mass (CoM) of the system in question and the universe
as a whole, (that in which the Cosmic Microwave Background is
globally isotropic.) The CoM is preferred in the sense that only in
that frame of reference, that is the centroid measured in the frame
co-moving with the massive system, is energy conserved as well as
energy-momentum. But if the PR is not valid in the presence of
gravitational masses then surely the EEP cannot be either? "

Tom Mattson

Aug10-04, 02:24 PM

Apart from the problem of defining the mass by the Klein-Gordon equation of spin 1/2 particles such as an electron, the question is whether QM sits with GR just fine.

First, there is no conflict between the definition of mass as an invariant and the KG equation. The KG equation is just the quantized version of the relation E2-p2=m2.

Second, the KG equation does apply to spin-1/2 particles. It just applies to components of the Dirac wavefunction.

Third, the Dirac equation, which is the equation that describes the quantum mechanics of spin-1/2 particles, is also Lorentz covariant, and it also sits just fine with SR.

And finally, there is no conflict between QM and GR. There is nothing at all stopping one from doing QM in curved spacetime. The problem comes into play when one tries to come up with a quantum theory of gravity, but that's not QM.

I think you will find that the problem in developing a quantum-gravitational theory is that QM requires the preferred foliation of space-time referred to above.

But QM doesn't require a preferred frame of reference. Relativistic QM is just that: QM that conforms to SR.

Aug10-04, 03:24 PM

Thank you for that; in which case not only is the wavelength statistical in nature (as with the rest of q-m) but also the argument is circular;
No it isn't.
the wavelength is defined in terms of the particle's mass
No it isn't.
Do not quantum mechanical definitions require a preferred foliation of space-time, a preferred frame of reference - normally that of the observer?
No.

pervect

Aug10-04, 07:54 PM

ISuch frames being those of
the Centre of Mass (CoM) of the system in question and the universe
as a whole, (that in which the Cosmic Microwave Background is
globally isotropic.) The CoM is preferred in the sense that only in
that frame of reference, that is the centroid measured in the frame
co-moving with the massive system, is energy conserved as well as
energy-momentum. But if the PR is not valid in the presence of
gravitational masses then surely the EEP cannot be either? "

There isn't any frame of reference in which the CMB is globally isotropic. If you pick a point, there's a local frame in which the CMB is isotropic. But it's a different frame at each point, i.e. the frame in which the CMB is isotropic at point A is moving with respect to the frame in which the CMB is isotropic at point B.

I don't understand why you think the CoM frame is special for the conservation of energy-momentum. AFAIK, to define a conserved energy, you need either a local timelike symmetry (a timelike Killing vector), or an asymptotically flat space-time. I've never read about any requirement to be in the center of mass "frame", though it's certainly convenient to calculate in.

I don't follow your argument about the EEP either, but that's probably my own failing. Though I do note that you seem to reject the EEP for the same reasons you require it, which makes me doubt your argument.

Garth

Aug11-04, 03:36 AM

Tom - Thank you for your comments, I am approaching the QM/GR interface from the GR side and I value your constructive criticisms in order to deepen my understanding.

I am not suggesting there is a conflict between QM and SR, indeed QM might be seen to be derived from SR through the resolution of the "de' Broglie paradox".

I was interested in your remark that the KG equation applied to spin-1/2 particles; my understanding was that as it involves the second time derivative of the Psi state vector the probability density associated with its solutions is not positive definite and therefore it could not represent such particles.

Also your affirmation that “there is no conflict between QM and GR” I find debateable. For example, the statement "No prediction of spacetime, therefore no meaning for spacetime is the verdict of the Quantum Principle. That object which is central to all of Classical General Relativity, the four dimensional spacetime geometry, simply does not exist, except in a classical approximation." (Misner, Thorne and Wheeler, Gravitation, p. 1183) would suggest otherwise.

I appreciate it depends on whether you approach the subject from the QM or the GR side but any problems of either approach should be mirrored in the other I would have thought. I would suggest another such problem to be the energy density and hence curvature associated with the quantum vacuum.

Pervect – Thank you too for your comments.

I appreciate the CMB is globally isotropic in a different frame at each point. I was taking that as understood.

I don’t think the CoM frame is special for the conservation of energy-momentum, which is conserved in all inertial frames as a consequence of the EEP, it is the conservation of energy that is the crucial point.

Finally it is GR that requires the EEP.

Tom Mattson

Aug11-04, 10:08 AM

I am not suggesting there is a conflict between QM and SR, indeed QM might be seen to be derived from SR through the resolution of the "de' Broglie paradox".

I don't know what the deBroglie paradox is, but I do know that you cannot derive QM from SR. Schrodinger's QM contradicts SR, inasmuch as the expectation values of observables do not satisfy the classical relativistic Hamiltonian.

I was interested in your remark that the KG equation applied to spin-1/2 particles; my understanding was that as it involves the second time derivative of the Psi state vector the probability density associated with its solutions is not positive definite and therefore it could not represent such particles.

Not at all. While it's true that the problem of non-positive definiteness of the KG probability density r originally was thought to be fatal to the theory, a reinterpretation (by Pauli? can't remember whom) of r as a charge density qr solved the problem. Charge densities are not required to be positive definite.

Also your affirmation that “there is no conflict between QM and GR” I find debateable.

It's not debatable, general relativistic quantum mechanics exists. It's just not found in textbooks in a standard graduate curriculum because it's so specialized. All you have to do is replace the SR metric with the GR metric of your choice, and start calculating. The theory is just as well-defined as KG or Dirac.

Tom Mattson

Aug11-04, 10:15 AM

I neglected to comment on the following important point.

That object which is central to all of Classical General Relativity, the four dimensional spacetime geometry, simply does not exist, except in a classical approximation." (Misner, Thorne and Wheeler, Gravitation, p. 1183) would suggest otherwise.

You seem to be consistently mixing up the ideas of QM, QFT, and Quantum Gravity. When people say that "GR and quantum theory are not compatible", they mean that a theory of quantum gravity doesn't exist. They do not mean that GR is inconsistent with QM or QFT. Indeed, QM and QFT can be done in a curved spacetime.

When people talk about "quantization", they refer to the quantization of 3 things: dynamical variables, fields, and spacetime (the metric). The quote from MTW is referring to the latter type. But the metric in QM and QFT is perfectly classical. It's the metric in Quantum Gravity that is not.

Here's a summary:

Quantum Mechanics
Dynamical variables: quantized
Fields: classical
Spacetime: classical

Quantum Field Theory
Dynamical variables: quantized
Fields: quantized
Spacetime: classical

Quantum Gravity
Dynamical variables: quantized
Fields: quantized
Spacetime: quantized

Garth

Aug11-04, 10:26 AM

Tom, thank you again.
I don't know what the deBroglie paradox is, but I do know that you cannot derive QM from SR. Schrodinger's QM contradicts SR, inasmuch as the expectation values of observables do not satisfy the classical relativistic Hamiltonian.
I understand that in 1923 de Broglie's paradox was one of Schrodinger's starting points in his formulation of QM. The paradox is that between on the one hand the de Broglie frequency, which is derived from the de Broglie wavelength and which is proportional to total energy and therefore increases with relative velocity and on the other hand the observed frequency of a moving clock which decreases with relative velocity. I believe it was the recognition that these were two totally different frequencies that led Schrodinger to develop the idea of the de Broglie's 'wave' into a wave function. However both the de Broglie's 'wave' and the wave function are not observables themselves but may be used to predict such.

Not at all. While it's true that the problem of non-positive definiteness of the KG probability density r originally was thought to be fatal to the theory, a reinterpretation (by Pauli? can't remember whom) of r as a charge density qr solved the problem. Charge densities are not required to be positive definite.
Thank you I live and learn!

It's not debatable, general relativistic quantum mechanics exists. It's just not found in textbooks in a standard graduate curriculum because it's so specialized. All you have to do is replace the SR metric with the GR metric of your choice, and start calculating. The theory is just as well-defined as KG or Dirac.
"the metric of your choice" in which frame may I ask, a preferred one perhaps? Or does general relativistic quantum mechanics work for any inertial frame in general?

pervect

Aug11-04, 03:56 PM

I don’t think the CoM frame is special for the conservation of energy-momentum, which is conserved in all inertial frames as a consequence of the EEP, it is the conservation of energy that is the crucial point.

But the conservation of energy-momentum implies the conservation of energy. For a closed system, the conservation of energy-momentum means that one has four quantites which are conserved by the system as it evolves in time - these quantities are E, a scalar, and the three components of P.

You are apparently using some different defintion. You seem to be demanding that energy by conserved over some sort of transformation of the system (a Lorentz Boost?). This isn't the right definition.

I think this has been pointed out before, even in classical mechanics, or special relativity, there is no guarnatee that different observers will compute the same value of energy for a closed system, there is only the guarantee that each observer will find that the energy of the closed system is constant as (their) time evolves.

Anyway, to be able to define the conserved energy of a closed system in GR, one needs either a local timelike symmetry, in which case the conservation of energy is obvious by Noether's theorem, or one needs an asymptotically flat space time.

Garth

Aug11-04, 05:02 PM

You seem to be consistently mixing up the ideas of QM, QFT, and Quantum Gravity.
Tom - Indeed I have been talking about Quantum Gravity, sorry about the confusion and thank you for the clarification.

Pervect - It was Noether's second theorem that demonstrated that GR was a type of "improper energy theorem" and that in general it did not, and would not be expected to, conserve energy.
I am thinking about a single observer in an inertial frame freely falling towards a central gravitational mass, the Earth for example. That observer would conclude that the Earth was accelerating towards him even though both he and the Earth were in free fall with no (first order) gravitational forces acting, just the convergence of their geodesics over curved spacetime. In the observer's frame of reference the components of the metric would be time dependent and hence the time component of the Earth's energy momentum vector, its total energy, would not be conserved.

Tom Mattson

Aug11-04, 07:41 PM

I understand that in 1923 de Broglie's paradox was one of Schrodinger's starting points in his formulation of QM. The paradox is that between on the one hand the de Broglie frequency, which is derived from the de Broglie wavelength and which is proportional to total energy and therefore increases with relative velocity and on the other hand the observed frequency of a moving clock which decreases with relative velocity. I believe it was the recognition that these were two totally different frequencies that led Schrodinger to develop the idea of the de Broglie's 'wave' into a wave function. However both the de Broglie's 'wave' and the wave function are not observables themselves but may be used to predict such.

I still don't see the paradox. The frequency of timekeeping the moving clock is not in any way related to the deBroglie frequency. But the deBroglie frequency of the moving clock (the frequency of the matter wave associated with the moving clock) varies with momentum just as QM says it should.

"the metric of your choice" in which frame may I ask, a preferred one perhaps? Or does general relativistic quantum mechanics work for any inertial frame in general?

Metrics don't vary from frame to frame. Metrics are what tell you how to transform from frame to frame. When I say "metric of your choice", I mean equivalently "energy momentum tensor of your choice". That is, near a stationary black hole, you will have one metric. Near a spinning black hole, you'll have another. Near a charged spinning black hole, you'll have still another.

But for any given metric, it is possible to formulate a generally covariant quantum mechanics in that spacetime.

Haelfix

Aug12-04, 02:55 AM

Just to specify, the KG equation works for spin 1/2 particles, but doesn't encapsulate the full theory. The Dirac equation is more general in that context as it contains the equations of motion of antiparticles in addition.

And then qft generalizes that one step further with full Grassmann mechanics for fermions and the polarization of the vacuum.

The problem with Quantum field theory in curved spacetime is fivefold IMO.

1) Its a nonrenormalizable theory (which is fine in the modern context, it just means that its not the full story)
2) Its hard to find meaningfull local observables in the theory (which is troubling), indeed perhaps only the S Matrix makes sense.
3) Its conceptually hard to take into account back reactions from the geometry, particularly so in very strong curvature regimes.
4) Half of QFT becomes intrinsically problematic, unless you believe in the Euclidean path integral being applicable. You really need to work in the Hamiltonian context, but then you have to figure out clever ways to foliate spacetime so your configuration space is meaningfull. Its best to work in the algebraic framework in this context.
5) Weird things show up, like conformal anomalies and the measurement problem is further complicated. There are solutions to this however, but the technical details are still debated about in the specialist circles.

But the theory seems to work fairly well albeit being a difficult subject with lots of unpleasant results and some unbeautiful brute force methodology. Its just been a little dated since String theory and the other quantum gravity theories have surfaced, which look at the problem more from the bottom-up, and seem to generalize things several steps further (and presumably remove all ambiguities in that limit).

Garth

Aug12-04, 03:13 AM

I still don't see the paradox.
That's the point, Schrodinger resolved it by formulating his representation of QM. However prior to QM it was a paradox, the study of which inspired/informed Schrodinger to make his conceptual leap.

Metrics don't vary from frame to frame.
Again my apologies I sometimes make mistakes reading from the screen and I read "frame" when you had written "metric". I am interested in the problem of time in canonical quantum gravity.
Haelfix - Thank you, being a relativist I am a stranger straying into another discipline in QFT, seeking understanding. I was aware there were problems on the interface between the two disciplines but were unsure of what they were exactly.

"Why keep your mouth closed so as not to appear a fool when you can open it and prove you are!"

pmb_phy

Aug12-04, 08:02 AM

Isn't this getting a bit off topic? Perhaps you can start a new thread in the appropriate forum so that those interested in the current suibject (i.e. QM) can participate or follow along. If you keep it here then there is little reason to assume that QM people would read this thread.

Just a suggestion mind you.

Pete

Garth

Aug12-04, 06:38 PM

Opening my mouth again (!), the present discussion may be on topic if it is (as it started out) a consideration of how mass may be defined in the various theories and conventions. At a fundamental level, if mass is the energy of string vibrations, de Broglie waves, or whatever, then the distinction between mass and energy maintained by the "mass is invariant" convention breaks down; it is all a sea of energies and virtual particles transmitting forces. If so at the most fundamental level, then perhaps for consistency sake, might it be thought of as such at higher levels?

pmb_phy

Aug13-04, 05:38 PM

Opening my mouth again (!), the present discussion may be on topic if it is (as it started out) a consideration of how mass may be defined in the various theories and conventions.
That is not the topic of this thread. The topic is Those who use relativistic mass and why

Pete

pmb_phy

Aug16-04, 11:41 AM

Aug16-04, 11:55 AM

I never realized it before today but Wald does use the term "mass" in at least one place to refer to "relativistic mass" in his text General Relativity.

If you have his text see page 62 right below Eq. (4.2.15). He speaks of conservation of mass. What he's refering to is the conservation of inertial energy (i.e. T00) which is proportional to relativistic mass.

Pete

Not only is T00 NOT mass, but it is NOT energy either. It is an energy density.

pervect

Aug19-04, 02:12 AM

I never realized it before today but Wald does use the term "mass" in at least one place to refer to "relativistic mass" in his text General Relativity.

If you have his text see page 62 right below Eq. (4.2.15). He speaks of conservation of mass. What he's refering to is the conservation of inertial energy which is proportional to (relativistic aka inertial) mass.

Pete

I read this differently. I think that Wald was saying that if we consider a perfect fluid, and take the equations of motion \partial^{a}T_{ab}=0 , one of the unsurprising results in the non-relativsitic limit is that the mass of the perfect fluid is conserved.

pmb_phy

Aug19-04, 03:58 AM

I read this differently. I think that Wald was saying that if we consider a perfect fluid, and take the equations of motion \partial^{a}T_{ab}=0 , one of the unsurprising results in the non-relativsitic limit is that the mass of the perfect fluid is conserved.
Thanks. That is what Wald says. My mistake. Thanks for pointing that out. I was flipping through Wald and saw that and it seemed to agree with what I posted here in Eq. (1) - http://www.geocities.com/physics_world/sr/mass_tensor.htm

Note: There are other authors who do use the term "mass" to mean "relativistic mass". See example listed above, i.e. Cosmological Physics, John A. Peacock, page 18

The only ingredient now missing from a classical theory of relativistic gravitation is a field equation: the presence of mass must determine the gravitational field. [...] Now, if this equation is to be covariant, T^uv must be a tensor and is known as the energy-momentum tensor (or sometimes as the stress-energy tensor). The meanings of its components in words are T^00 = c^2x(mass density) = energy density, T^12 = x-component of current of y-momentum etc. From these definitionsl the tensor is readily seen to be symmetric. Both momentum density and energy flux density are the product of a mass density and a net velocity, so T^0m = T^m0.

By "mass density" he means what you'd call "relativistic mass density" and what Wald(page 60)/MTW call "mass-energy density."

Pete