Photon's mass is zero? [Archive]

yu_wing_sin

Sep4-05, 02:22 AM

It doesn't make sense to make "the extreme speed of cosmos" something other than the speed of light because then the speed of light wouldn't be a constant. Where did you read that a photon's mass isn't zero? My guess is that either it was a crackpot website or they were talking about relativistic mass (photons have zero invariant mass). Many at this forum are against the use of "relativistic mass", and I guess this is one more example of why it shouldn't be used.

But this link pointed out the photons have mass...
http://www.aip.org/pnu/2003/split/625-2.html

-------------------------------------------------
The content is:
Number 625 #2, February 19, 2003 by Phil Schewe, James Riordon, and Ben Stein
A New Limit on Photon Mass

A new limit on photon mass, less than 10-51 grams or 7 x 10-19 electron volts, has been established by an experiment in which light is aimed at a sensitive torsion balance; if light had mass, the rotating balance would suffer an additional tiny torque. This represents a 20-fold improvement over previous limits on photon mass.

Photon mass is expected to be zero by most physicists, but this is an assumption which must be checked experimentally. A nonzero mass would make trouble for special relativity, Maxwell's equations, and for Coulomb's inverse-square law for electrical attraction.

The work was carried out by Jun Luo and his colleagues at Huazhong University of Science and Technology in Wuhan, China (junluo@mail.hust.edu.cn, 86-27-8755-6653). They have also carried out a measurement of the universal gravitational constant G (Luo et al., Physical Review D, 15 February 1999) and are currently measuring the force of gravity at the sub-millimeter range (a departure from Newton's inverse-square law might suggest the existence of extra spatial dimensions) and are studying the Casimir force, a quantum effect in which nearby parallel plates are drawn together. (Luo et al., Physical Review Letters, 28 February 2003)
-------------------------------------------------------------

cefarix

Sep4-05, 04:38 AM

... A new limit on photon mass, less than 10-51 grams or 7 x 10-19 electron volts, has been established by an experiment in which light is aimed at a sensitive torsion balance; if light had mass, the rotating balance would suffer an additional tiny torque. This represents a 20-fold improvement over previous limits on photon mass.

Photon mass is expected to be zero by most physicists, but this is an assumption which must be checked experimentally. A nonzero mass would make trouble for special relativity, Maxwell's equations, and for Coulomb's inverse-square law for electrical attraction. ...

That does not mean the photon has mass. The experiment was conducted to confirm that the photon indeed has zero mass. The 10-51 grams figure cited above is the accuracy of the experiment, suggesting that IF the photon did have mass, it would have to be less than 10-51 grams, otherwise the experiment would have given different results.

DaTario

Sep4-05, 05:00 PM

IAN STINE

Sep4-05, 05:14 PM

Pair production. Positrons and electrons are often spoken of as having mass. So a photon having no mass becomes a produced pair having mass which turn into photons having no mass.........! I'm getting swim-headed!

Sep4-05, 06:14 PM

Energy is conserved, not mass. Mass is just a form of energy (E=mc^2).

yu_wing_sin

Sep5-05, 12:52 AM

That does not mean the photon has mass. The experiment was conducted to confirm that the photon indeed has zero mass. The 10-51 grams figure cited above is the accuracy of the experiment, suggesting that IF the photon did have mass, it would have to be less than 10-51 grams, otherwise the experiment would have given different results.

Quote:
"if light had mass, the rotating balance would suffer an additional tiny torque"

They set this hypothesis, but I don't know whether they had detected this, the abstract had not told. I think the limit of the experimental accuracy is not the main matter. So I am hard to judge your view on the matter photons are zero mass.

Doc Al

Sep5-05, 07:10 AM

Quote:
"if light had mass, the rotating balance would suffer an additional tiny torque"

They set this hypothesis, but I don't know whether they had detected this, the abstract had not told. I think the limit of the experimental accuracy is not the main matter. So I am hard to judge your view on the matter photons are zero mass.
Please read the abstract that you yourself had quoted. And reread cefarix's comments. This experiment puts a lower bound on the possible mass of light than ever before, thus confirming (so far) expectations of light having zero mass.

LURCH

Sep5-05, 08:08 AM

It may seem strange, Yu Wing Sun, but for experiments that attempt to measure the mass of light, the limit of experimental accuracy really is the main point. The reason for this can be explained historically. You see, all the best theories and models of physical law predict the mass of light to be zero. But how does one measure "zero"? All measurements so far have confirmed a result of Zero, but does that really prove anything? A sceptic could say, "the photon has mass, it's just such a small amount of mass that our current methods of measurement can't detect it".

So the problem facing the experimental researcher is that this is a valid argument, and one that is extremely hard to disprove. About the only thing they can do is to come up with more and more accurate ways of trying to measure the mass of the photon, with more and more sensitive equipment, and show that the result still comes up "Zero". Then they publish their results by saying, "I've got a new method that can detect a mass as little as x. I used it to try to measure the mass of a photon, and got no reading. Therefore, if a photon has mass, that mass must be less than x. The point that researchers are trying to verify is that we could put any number for the value of that "x", and the statement would still be true. The simple logic being that the statement:

"the mass of a photon is <x, no matter what x is"

can only be true if the mass of the photon is zero.

nmondal

Sep5-05, 10:32 AM

Photons has momentum!
Common misconception is that, as it has momentum, it must be having a mass.
That is untrue, as we know that any wave has a momentum!

pmb_phy

Sep5-05, 04:31 PM

Photons has momentum!
Common misconception is that, as it has momentum, it must be having a mass.
That is untrue, as we know that any wave has a momentum!Oy! Here we get yet once more! :cry:

It is not a "common misconception." What is a common misconception is that there are too many people who use the term "mass" to mean "proper mas."

A photon as a non-zero "relativistic mas" an zero "proper mass."

Pete

Chronos

Sep6-05, 01:51 AM

Momentum without mass seems to be a difficult concept. If photons are not massless [in the relativistic sense], theoretical physics is in serious trouble.

yu_wing_sin

Sep6-05, 03:51 AM

Logical view, every matter has mass, otherwise they are non-existence! Maybe in our now technologies are still unable to detect the photons' mass, but without doubts, photons were from the big bang (all mass were concentrated at it), since that, photons should be having mass.

Ich

Sep6-05, 04:45 AM

Juan R.

Sep6-05, 05:22 AM

Energy is conserved, not mass. Mass is just a form of energy (E=mc^2).

This is a common misconception.

If E=mc^2 and E is conserved (dE/dt = 0), then m is also conserved because c is just a constant => dm/dt = 0

The idea of that mass is just a form of energy "arises" from E=mc^2

but rewrite it like m = E/c^2, then energy is just a form of mass. Our current knowledge is about matter and the study of changes of energy on matter. Matter is more "fundamental" than energy.

As was admited by Feynman the more powerful idea of science is that universe is done of "atoms".

pmb_phy

Sep6-05, 05:37 AM

Photons exist and have no mass.
And don´t be so quick with "logic" - example:
A system of 2 similar parallel moving photons has no mass.
A system of 2 similar antiparallel moving photons has mass 2hf.
Maybe You should first read about the meaning of "mass" in modern physics before You draw any conclusions.Oh no!! Not another one!!. That's just a bunch of horse hockey! :biggrin: You learned the concept of mass wrong in "modern" physics. That's rest mass you're speakin of and not relativistic mass. When we speak of mass here some (most - hard to tell anymore) mean proper mass and some mean inertial mass, relativistic mass, etc. There's a reason you learned ir wrong but that's another story.

In you're case you're refering to the concept of the magnitude of free objects or an object which can be considered as a particle or a system which is isolated.

The more general case falls to hell if you want to define mass in the most general sense. See

http://www.geocities.com/physics_world/misc/relativistic_mass.htm

No selecctive reading please, i.e. don't skip the part which is from MTW's text that discusses it.

Notice how Argonne National Laboratory defines "mass."

http://www.neutron.anl.gov/hyper-physics/inertia.html

Massive Pete :rofl:

pmb_phy

Sep6-05, 05:50 AM

This is a common misconception.

If E=mc^2 and E is conserved (dE/dt = 0), then m is also conserved because c is just a constant => dm/dt = 0

The idea of that mass is just a form of energy "arises" from E=mc^2

but rewrite it like m = E/c^2, then energy is just a form of mass. Our current knowledge is about matter and the study of changes of energy on matter. Matter is more "fundamental" than energy.

As was admited by Feynman the more powerful idea of science is that universe is done of "atoms".
Writing mass as m = E/c2 is just bad juju! Einstein showed that this expression is not generally valid in On the inertia of energy required by the relativity principle. A. Einstein, Annalen der Physik 23 (1907): 371-384.

For a much much simpler derivation please see Energy vs. Mass located at - http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

and written by me. But this is standard stuff. It can be found in Rindler's 1982 Intro to SR text as well as Tolman's tex.

Pete

Ich

Sep6-05, 09:37 AM

There's a reason you learned ir wrong but that's another story.
I don´t remember exactly where I got this from, but I´m quite sure it was Taylor/Wheeler "Spacetime Physics", 1992 ( I read the german "Physik der Raumzeit"). There they told me that one should say "mass" instead of "rest mass", because
a) that´s the only invariant quantity about mass and
b) the concept of "relativistic mass" is of little or no use.
They also declared this to be the view point of "modern physics".

In you're case you're refering to the concept of the magnitude of free objects or an object which can be considered as a particle or a system which is isolated.

Yes, I explicitly did. That was the most enlightening point of the whole book to me: m is invariant, but m(particle1+particle2) != m(particle1)+m(particle2). So eg thermal energy increases mass.

The more general case falls to hell if you want to define mass in the most general sense. See
http://www.geocities.com/physics_world/misc/relativistic_mass.htm
No selecctive reading please, i.e. don't skip the part which is from MTW's text that discusses it.

Yes, there he talks differently. I´m not too much into GR, so unfortunately I can´t judge what concept is of more use there.

Notice how Argonne National Laboratory defines "mass."
http://www.neutron.anl.gov/hyper-physics/inertia.html

I notice.
To clear things up, just replace every "mass" in my post by "rest mass". If this whole thing is about semantics, I have no problem with calling rest mass "mass", "Ruhemasse", or "not the mass pete means but the one Ich means".
If not, I´m looking forward to gain insight by understanding or even adopting Your point of view.

btw, in this thread yu_wing_sin quoted an article where they put an upper bound to the "mass" of the photon. This could hardly mean m=E/c². So my terminology is at least consistent.

Massive Pete :rofl:[/QUOTE]

russ_watters

Sep6-05, 11:27 AM

Logical view, every matter has mass, otherwise they are non-existence! That's true, but photons are not matter!

pmb_phy

Sep6-05, 11:33 AM

That's true, but photons are not matter!Who says so? It certainly wasn't Einstein that said that!

Pete

pervect

Sep6-05, 11:39 AM

Writing mass as m = E/c2 is just bad juju! Einstein showed that this expression is not generally valid in On the inertia of energy required by the relativity principle. A. Einstein, Annalen der Physik 23 (1907): 371-384.

For a much much simpler derivation please see Energy vs. Mass located at - http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

and written by me. But this is standard stuff. It can be found in Rindler's 1982 Intro to SR text as well as Tolman's tex.

Pete
A short comment:

m = E/c^2 can be fully and logically justified if p=0. (This is of course invariant mass, not relativistic mass). Before you get too ambitious in correcting other people's usage, remember that there are a lot of people who do NOT use "relativistic" mass and who prefer invariant mass.

TheAntiRelative

Sep6-05, 12:17 PM

Nice explanation by cefarix.

I, in turn, think that a physical entity which is able to transmit momentum (Einstein's gedanken experiment with the box) and also has energy, and also get attracted by gravitational fields deserves be interpreted as having mass. Our impossibility to measure mass in its own referential seems to be somewhat independent from the mass notion.

What is the current interpretation to the expression:

Planck's constant x frequency / c^2 ??

Photons certainly behave like they have mass. The compton effect is probably the best example. The path of a photon is actually deviated by collision with an electron if I remember correctly. It's a billiard-ball like effect.

Sep6-05, 12:27 PM

This is a common misconception.

If E=mc^2 and E is conserved (dE/dt = 0), then m is also conserved because c is just a constant => dm/dt = 0

What I wanted to say is that the total energy of a free particle comes from both it's kinetic motion and from it's mass. In E=mc^2 I'm of course talking about the invariant rest mass (I thought everyone by this time had abandoned the ugly and easlily confusing use of relativistic mass), and E is not ment to be the total energy, just the part coming from the (invariant rest) mass of the particle.

Juan R.

Sep6-05, 12:58 PM

Writing mass as m = E/c2 is just bad juju! Einstein showed that this expression is not generally valid

If E=mc^2 is valid (WHICH WAS A PREMISE OF MY POST) then m=E/c^2 is also! :rofl:

Juan R.

Sep6-05, 01:03 PM

What I wanted to say is that the total energy of a free particle comes from both it's kinetic motion and from it's mass. In E=mc^2 I'm of course talking about the invariant rest mass (I thought everyone by this time had abandoned the ugly and easlily confusing use of relativistic mass), and E is not ment to be the total energy, just the part coming from the (invariant rest) mass of the particle.

E = mc^2 in SR, where m =/= m0, the rest mass, in general.

If E is conserved, it is, then mass is also because the proportionality.

Mass is always conserved, In fact there is no violation of mass conservation in nuclear phenomena, for example, because mass is transformed to energy, that is also a form of matter via inverse Einstein relationship.

pervect

Sep6-05, 01:23 PM

Photons certainly behave like they have mass. The compton effect is probably the best example. The path of a photon is actually deviated by collision with an electron if I remember correctly. It's a billiard-ball like effect.

Photons have energy and momentum, but their invariant mass is zero, i.e. E^2 - (pc)^2 = 0, or E = pc.

This is usually shortened to say that photons have no mass, that they are massless particles (because their invariant mass is zero).

This is widely known, I have no idea why the question keeps popping up repeadetdly and endlessly.

Sep6-05, 04:24 PM

E = mc^2 in SR, where m =/= m0, the rest mass, in general.

If E is conserved, it is, then mass is also because the proportionality.

But the total energy of a free particle is not just mc^2 (where "m" is the rest mass, a convention I will stick to). That's unless you're in it's rest frame of course.
It is the total energy which is conserved, not mc^2.

Mass is always conserved, In fact there is no violation of mass conservation in nuclear phenomena, for example, because mass is transformed to energy, that is also a form of matter via inverse Einstein relationship.

Again, if we are talking about rest mass there are plenty of processes where mass is not conserved. Take electron-positron annihilation into photons for example.

Sep6-05, 04:27 PM

Who says so? It certainly wasn't Einstein that said that!

Pete

http://en.wikipedia.org/wiki/Matter

pmb_phy

Sep6-05, 06:29 PM

A short comment:

m = E/c^2 can be fully and logically justified if p=0.What is "p"? If its pressure/stress then that's correct. However you're beihng vauge about your terms. If you use E to mean inertial enegy then E = K + E0, not E = E0 unless you're making the unstated assumption that K is always zero.

If the object is extended object then its more commplicated.

Note Pressure contributes to inertia. E.g. see Shutz's new text Gravity from the ground up. Especially at the bottom of page 192 in the section entitled The extra inertia of pressure.

Before you get too ambitious in correcting other people's usage, ..Sorry lad but I didn't correct anybody in this thread. I stated my opinion that writing m = E/c^2 is bad juju.

... remember that there are a lot of people who do NOT use "relativistic" mass and who prefer invariant mass.There's no accounting for taste. Most peole learned SR wrong.

Pete

pmb_phy

Sep6-05, 06:32 PM

http://en.wikipedia.org/wiki/Matter

That's their opinion. And my opinion of them decreases evertime its referenced. Look at Einstein's 1916 GR paper

pervect

Sep6-05, 08:05 PM

If the object is extended object then its more commplicated.

Note Pressure contributes to inertia. E.g. see Shutz's new text Gravity from the ground up. Especially at the bottom of page 192 in the section entitled The extra inertia of pressure.

I don't have the text (though I hear it's quite good) - but you are right that pressure contributes to the stress energy tensor and hence the mass of a system.

Defining the mass of an extended system as the response to an external force is however an extremely bad idea, just because of the very fact that the mass of the system does depend on the distribution of the stresses. This means that different force distributions give different masses, an unsavory state of affairs. Fortunately, there are MUCH better ways of defining the mass of an extended system than considering the response of a system to an external force. More on this later.

Sorry lad but I didn't correct anybody in this thread. I stated my opinion that writing m = E/c^2 is bad juju.

There's no accounting for taste. Most peole learned SR wrong.

Pete

E = mc^2 is perfectly fine as long as p (momentum!) is zero, and m is invariant mass. (Apparently you did nor realize that p was momentum?????????????? Or were you pulling my leg? I think you were pulling my leg - though I doubt you'll admit it.)

Thus there's no reason for people to un-learn E=mc^2, all they have to do is realize that that equation only works when p=0, and that the real equation is more like E^2 - (pc)^2 = (mc^2)^2, (when p is not zero), which can be further simplfied by using geometric units to E^2 - p^2 = m^2. And they need to know that the m in this equation is the "invariant mass".

In addition, people eventually have to learn that the total energy E of a system is not the intergal of the energy density term of the stress-energy tensor (T_00), but also does includes contribution from the pressure terms as you state.

However, given that one has the necessary condtions - an isolated system and an asymptotically flat space-time will do, the resulting system has a well defined energy E in GR, and a well defined momentum p, as well. One confusing thing is that the energy of the extended system is not just the intergal of the energy density, but includes the intergal of some of the pressure terms as well. Note well that with these (standard!) defintions of energy and momentum, in geometric units E^2 - p^2 = m^2 works for both point particles AND extended systems.

Given that a system has an energy E and a momentum P, and that these quantities transform as a 4-vectors in the appropriate asymptotically flat space-time (which they do), it's trivial to compute the invariant mass 'm' of the system, usually just called the mass. If one needs to be especially precise, the mass of a system can be sub-categorized into at least two types of mass - the ADM mass, and the Bondi mass. The ADM mass is mathematically based on the behavior of the system at space-like infinity, and includes gravitational radiation terms. The Bondi mass is mathematically based on the behavior of the system at null infinity, and does not include gravitational radiation contributions.

Usually such detail is not needed or even wanted at the introductory level. The best source for information on mass in GR is probably Wald's "General Relativity", which is not an introductory book.

pmb_phy

Sep6-05, 10:02 PM

Defining the mass of an extended system as the response to an external force is however an extremely bad idea, just because of the very fact that the mass of the system does depend on the distribution of the stresses.So what? This is based on your wishing to define mass as an intrinsic property of an object where there is no need to do so. In fact in general its impossible to do so. Mass is that which defines momentum an
This means that different force distributions give different masses,...Absolutely correct.
.. an unsavory state of affairs.The elite say that beluga caviar is quite savory ubut I find that al caviar are quite unsavory.

Fortunately, there are MUCH better ways of defining the mass of an extended system than considering the response of a system to an external force.This I gotta hear. Please keep in mind that those who have defined mass in such a way have done so as he definer of momtum. The momentum of an object is dependant of the environment its in and the details of the object.

E = mc^2 is perfectly fine as long as p (momentum!) is zero, and m is
invariant mass.

At first I did but upon reflection I realize that the statement is incorrect as it stands but correct if p = pressure. Upon further reflection I see that is wrong too. Since you're not one to make such a glaring error I assumed it was pressure. But once you read Schutz you'll see why that is incorrect. He gives an example where he calculates kinetic energy as being the definer of mass. He ends up, with (for v << c) {Note: Schutz uses E}

K = \frav{1}{2}(\rho + \frac{p}{c^2})v^2 v

This gives an inertial mass densit of \rho + p/c^2

re - " (Apparently you did nor realize that p was momentum?????????????? Or were you pulling my leg? I think you were pulling my leg - though I doubt you'll admit it.)" - Shame on you. :frown: Whenever I note an error of mine I make it my highest priority to make sure I correct myself. My response is sxplained above.
Thus there's no reason for people to un-learn E=mc^2,....I don't see how you got from my response that this was something to be unlearned. Its place should never be forgotten since all currect appliccations I'm aware of that is 100% correct. But current applictions is not 100% of its possible concievable uses.

A hile bal I wrote an article on this entire subject. The pupose being so I woudn't have to keep getting into the same ole disucssions. Here's the article fof the topic at hand - http://www.geocities.com/physics_world/mass_paper.pdf

It'd be incorrect to dissmiss this debnate as mere semantic since the concepts and words we use detemine our thoughts and thoutht paterns.

Pete

Sep7-05, 03:24 AM

That's their opinion. And my opinion of them decreases evertime its referenced. Look at Einstein's 1916 GR paper

Well, it's just a definition. You are of course free to define "matter" however you want, but I think it is convenient to define it in the way most people do nowadays.

pmb_phy

Sep7-05, 03:53 AM

Well, it's just a definition. You are of course free to define "matter" however you want, but I think it is convenient to define it in the way most people do nowadays.I never chose to define it myself. It always fails at a definition. E.g. if you have Taylor and Wheeler's text read the section on mass.

Pete

Sep7-05, 04:24 AM

I never chose to define it myself. It always fails at a definition. E.g. if you have Taylor and Wheeler's text read the section on mass.

Pete

Maybe we should put up a poll about this! :smile:

Sam Owen

Sep7-05, 08:09 AM

just a thought and no doubt many have had it before but...

...can you isolate light from Einsteins equation to show that everything is made from it by making breaking c down to distance over time ?

if not why not and if so would the answer be something like 42

:smile:

Ich

Sep7-05, 09:30 AM

after knowing the answer, please try to remember the question. "...can you isolate light from Einsteins equation to show that everything is made from it by making breaking c down to distance over time ?" certainly doesn´t make sense.

Sam Owen

Sep7-05, 09:57 AM

sure it does...

...the only difference is you aren't me

frames and terms of reference aren't my strong points as you may have guessed...

...I'm more of a visual type of guy, pictures in my head, that sort of thing

Can you really not get the gist of what i'm asking ?

Ich

Sep7-05, 10:34 AM

sorry, no.
Maybe later, I´m gonna drink some beer :)

pmb_phy

Sep7-05, 10:56 AM

Maybe we should put up a poll about this! :smile:
Why? It'd be a huge waste of time. Stop worrying who calls things what. Concentrate only what is measurable

Sep7-05, 11:33 AM

Why? It'd be a huge waste of time. Stop worrying who calls things what. Concentrate only what is measurable

Que? I'm worried?
I have to remind you that you were in fact the one who started worrying about the definition of "matter".
Relax for god's sake. Did you see the smiley at the end of my post?

Juan R.

Sep8-05, 06:52 AM

But the total energy of a free particle is not just mc^2 (where "m" is the rest mass, a convention I will stick to). That's unless you're in it's rest frame of course.
It is the total energy which is conserved, not mc^2.

Again, if we are talking about rest mass there are plenty of processes where mass is not conserved. Take electron-positron annihilation into photons for example.

From special relativity

E^2 = (pc)^2 + (m0c^2)^2

and

E = mc^2

The total energy E is conserved <=> the total mass m is also conserved because c is a constant.

As said m =/= m0. You may be wrong about what mass is entering on general formula E = mc^2. (Your "m" is my m0)

The formula E = m0c^2 that you claim in some of your posts is valid only when p is zero as pervect already said.

In a pair annihilation process, precisely rest mass is transformed on energy (which is another form of mass). Total mass -or total energy- continues being constant. Precisely photons have mass (equal to total mass of initial pair annihilated), this is the reason of light bending around Sun computed from Newtonian mechanics. Since photons have mass they feel gravitation. Of course, rest mass m0 of photons is (at least today) zero but m =/= 0 because photons travel to light velocity.

Redfox

Sep8-05, 07:15 AM

Photon's mass is zero?
-----------------------------------

According to the SR's equation : m=m0/√(1-v2/c2), if the photon's velocity is the speed of light, then the photon's mass should be zero in value, but recently scientists have measured the photon having mass, but I forget the very long value of the photon mass.

you never know. photons are supposed to have only impuls mass (pressure of light) however lightwaves still change direction in a strong gravitational field.
I am very doubtful about relativistic formulas anyway...

inha

Sep8-05, 08:41 AM

I am very doubtful about relativistic formulas anyway...

You mean the formulae that give results validated by thousands of experiments?

Sep8-05, 10:02 AM

As said m =/= m0.

As I said before I am only using the invariant mass (you call it m0, I call it m). You are unfortunately stuck in that you believe I mean relativistic mass by m.
The reason why I only use the invariant mass, and never speak about any relativistic mass is because all (ok at least all) modern texts adopt this convention. The concept of relativstic mass is an easily confusing one, and should in my opinion (and many others) not be used.
I stress it again: I am always talking about the invariant (rest) mass, and I call it just m.

By this convention photons are massless, and mass is not (in general) conserved in interactions.

Sojourner01

Sep8-05, 10:56 AM

Does this thread need to keep banging on?

jtbell

Sep8-05, 12:32 PM

By this convention photons are massless, and mass is not (in general) conserved in interactions.

To be more precise, the mass of an isolated system is conserved in interactions, but mass is not an additive property (the mass of the system does not necessarily equal the sum of the masses of its component particles or subsystems).

In this convention, the mass of a system is defined via (mc^2)^2 = E_{total}^2 - ({\vec p}_{total} c)^2. Both {\vec p}_{total} and E_{total} are conserved in an isolated system, so m must be also. That is, m of the system is the same before and after the interaction.

If you do external work on the system (e.g. in separating the proton and electron in a hydrogen atom), then the total energy of the system changes, and so does its mass (the mass of a separated proton and electron is greater than the mass of a bound proton and electron). In this case the system is no longer isolated.

Again, this is under the convention "mass = invariant mass". By the way, I prefer not to use the term "rest mass" because "rest" doesn't apply to massless particles. The term "rest mass" can also confuse people when applied to a system of particles which are individually in motion, but the total momentum of the system is zero. I prefer "invariant mass."

pervect

Sep8-05, 04:07 PM

So what? This is based on your wishing to define mass as an intrinsic property of an object where there is no need to do so. In fact in general its impossible to do so.

My goals are actually to provide a path from "mass" to ADM mass and Bondi mass, the two sorts of mass that are commonly used in General Relativity. Also, I wish to defend the correct statements made by other posters that with the proper defintions, m = E/c^2 when momentum = 0.

At first I did but upon reflection I realize that the statement is incorrect as it stands but correct if p = pressure. Upon further reflection I see that is wrong too. Since you're not one to make such a glaring error I assumed it was pressure. But once you read Schutz you'll see why that is incorrect. He gives an example where he calculates kinetic energy as being the definer of mass. He ends up, with (for v << c) {Note: Schutz uses E}

K = \frav{1}{2}(\rho + \frac{p}{c^2})v^2 v

This gives an inertial mass densit of \rho + p/c^2

I'm not sure I follow this. I would expect that with isotropic pressure in nearly flat space-time, the mass of an extended system should be the volume intergal of \rho + 3P/c^2, i.e in orthonormal cartesian coordinates the intergal of T^{00}+T^{11}+T^{22}+T^{33} -- here T^{00} = \rho, and T^{11} = T^{22} = T^{33} = P

\rho is density, and P is pressure in the above paragraph.

[add]
Upon reflection, I'm guessing that perhaps the pressure in Schutz's example is NOT isotropic, but oriented in some particular direction only (along the length?), hence explaining the factor of 3 difference.
[end]

I'm moving this last section to the back, as it will probably scare people :-)

Pete[/QUOTE]

This I gotta hear. Please keep in mind that those who have defined mass in such a way have done so as he definer of momtum. The momentum of an object is dependant of the environment its in and the details of the object.

This gets rather technical - but you asked for it. See for instance Wald's chapeter on energy in "General Relativity" for more information if you want more.

Momentum conservation comes from space translation invariance and Noether's theorem, (given that GR satisfies an action principle, which it does).

One might be lucky enough to have a static space-time metric which gives a direct space translation invariance - which gives a direct and natural definition of mass and energy in static space-times.

In general, one is not so lucky as to have a static space-time. But if one has a non-static space-time that is still asymptotically flat, one can still define an asymptotic space translation invariance, even though an exact space translation invariance isn't available. It turns out that the formal group structure at null infinity for asymptotically flat space-times is an infinite dimensional group (bummer!) called the BMS group. The good news is that while the BMS group is infinite dimensional, it has a preferred and unique 4 parameter normal subgroup. The space-time symmetries associated with this unique 4-parameter sub-group give one the defintions of momentum and energy at null infinity - the Bondi energy and momenta - in a very natural way. You can think of this 4-parameter group as definining preferred asymptotic time translations, and asymptotic space translations (hence the 4 parameters). The asymptotic time translations give a conserved energy, and the asymptotic space translations give a conserved momentum.

This requires an asymptotically flat space-time, but just about all the definitions of mass in GR require that (with the exception of static space-times, which already have a defintion of momentum and energy which arise from exact translational invariances).

Juan R.

Sep9-05, 03:55 AM

As I said before I am only using the invariant mass (you call it m0, I call it m). You are unfortunately stuck in that you believe I mean relativistic mass by m.
The reason why I only use the invariant mass, and never speak about any relativistic mass is because all (ok at least all) modern texts adopt this convention. The concept of relativstic mass is an easily confusing one, and should in my opinion (and many others) not be used.
I stress it again: I am always talking about the invariant (rest) mass, and I call it just m.

By this convention photons are massless, and mass is not (in general) conserved in interactions.

Ok! it was a misconception. I wrote that i was talking of total mass even wrote that m =/= m0 and your reply was again that "mass" was not conserved using the same m that i said that was not rest mass.

I do not see why total is more confusing that rest mass. About "rest" mass it was replied to you. Note that photons rest mass is zero, but note also that photons are newer at rest -move at c, except in the cone reference- and, therefore, its effective (total) mass is newer zero. This is the reason of the bending of light around sun, computed from Newtonian mechanics.

Moreover, the idea of a variable mass is again recovered in QFT, where mass of an electron is dependant of interaction, due to the cloud of virtual photons. I see no reason for emphasis on a constant "rest" mass like the real mass.

Juan R.

Sep9-05, 04:05 AM

To be more precise, the mass of an isolated system is conserved in interactions, but mass is not an additive property (the mass of the system does not necessarily equal the sum of the masses of its component particles or subsystems).

In this convention, the mass of a system is defined via (mc^2)^2 = E_{total}^2 - ({\vec p}_{total} c)^2. Both {\vec p}_{total} and E_{total} are conserved in an isolated system, so m must be also. That is, m of the system is the same before and after the interaction.

If you do external work on the system (e.g. in separating the proton and electron in a hydrogen atom), then the total energy of the system changes, and so does its mass (the mass of a separated proton and electron is greater than the mass of a bound proton and electron). In this case the system is no longer isolated.

Again, this is under the convention "mass = invariant mass". By the way, I prefer not to use the term "rest mass" because "rest" doesn't apply to massless particles. The term "rest mass" can also confuse people when applied to a system of particles which are individually in motion, but the total momentum of the system is zero. I prefer "invariant mass."

Are you claiming that the rest mass of pair e- e+ is the same that rest mass of two photons?

I think that you are a bit confused about conservation laws in a pair anihilation m (my previously denoted m0) ---> 0

but total mass and total energy is conserved.

janusz

Sep9-05, 05:41 AM

janusz

Sep9-05, 06:03 AM

you never know. photons are supposed to have only impuls mass (pressure of light) however lightwaves still change direction in a strong gravitational field.
I am very doubtful about relativistic formulas anyway...
I think that relativity is the greatest achievement of human kind and Im very enthusiastic about grasping the idea. Photons do not change direction under influence of gravity but the gravity itself change the spacetime -the path of the light. I wonder if anyone think about straight line of traveling light in the expanding Universe .The line is it curved by swelling space? Playing darts where the middle point is not accessible by players because every second is born new central point?

Ich

Sep9-05, 07:03 AM

Are you claiming that the rest mass of pair e- e+ is the same that rest mass of two photons?

I think that you are a bit confused about conservation laws in a pair anihilation m (my previously denoted m0) ---> 0

but total mass and total energy is conserved.
It´s definitely not jtbell who is confused. He is claiming that those two photons (seen as a system) have mass (¡ notpetesmassbutmine !), and that´s just how it is defined.

Sep9-05, 07:39 AM

Ok! it was a misconception. I wrote that i was talking of total mass even wrote that m =/= m0 and your reply was again that "mass" was not conserved using the same m that i said that was not rest mass.
Ok,at lest we have cleared this out now. I had problems to understand what =/= ment at first, but now I get you...

I do not see why total is more confusing that rest mass. About "rest" mass it was replied to you. Note that photons rest mass is zero, but note also that photons are newer at rest -move at c, except in the cone reference- and, therefore, its effective (total) mass is newer zero. This is the reason of the bending of light around sun, computed from Newtonian mechanics.
Well, I still say it's more convenient nowadays to just say that the photons have mass zero, and that the reason they bend around the sun is because the space-time is curved (also note that the object determining the curvature is called "energy-momentum" tensor). However you are of course free to see it the way you like, it doesn't matter for the result. The reason why I think it's confusing to define mass as relativistic mass is because present text hardly ever do so. Otherwise it could be quite hard to understand each other (as this thread is a good example of... :wink: ).

Moreover, the idea of a variable mass is again recovered in QFT, where mass of an electron is dependant of interaction, due to the cloud of virtual photons. I see no reason for emphasis on a constant "rest" mass like the real mass.
This is not the same thing.
The physical mass of an electron does not vary.

jtbell

Sep9-05, 09:24 AM

Are you claiming that the rest mass of pair e- e+ is the same that rest mass of two photons?

I am claiming that the invariant mass of the system, as defined by the equation that I gave, is the same for the e- e+ pair and for the two photons that they annihilate into.

If you're claiming that the rest mass of the e- e+ pair is not the same as the rest mass of the two photons, then your definition of the rest mass of the system (for one or the other or both) must be different from my definition of the invariant mass of the system.

Juan R.

Sep9-05, 09:50 AM

It´s definitely not jtbell who is confused. He is claiming that those two photons (seen as a system) have mass (¡ notpetesmassbutmine !), and that´s just how it is defined.

either Ich or jtbell (or Juan R.) is confused.

jtbell exactly said that (rest) mass m is conserved after and before the interaction. Then i ask if this is also true for pair anihilation.

Now you say that jtbell said that those two photons have mass, but what mass is m?

It is not "my" m0. which is the mass that enter in above jtbell definition. The m defined by jtbell above (i call m0) is zero for photons still total mass is conserved via E=mc^2 where m =/= m0

Juan R.

Sep9-05, 09:56 AM

I am claiming that the invariant mass of the system, as defined by the equation that I gave, is the same for the e- e+ pair and for the two photons that they annihilate into.

If you're claiming that the rest mass of the e- e+ pair is not the same as the rest mass of the two photons, then your definition of the rest mass of the system (for one or the other or both) must be different from my definition of the invariant mass of the system.

Now i am completely perplexed. From SR

E^2 = (mc^2)^2 + (pc)^2

but the m is rest mass. For the pair e+ e- m = 2 melectron =/= 0 but for the two fotons m = 0, because photonds are massless.

:bugeye:

Juan R.

Sep9-05, 10:21 AM

Well, I still say it's more convenient nowadays to just say that the photons have mass zero, and that the reason they bend around the sun is because the space-time is curved (also note that the object determining the curvature is called "energy-momentum" tensor). However you are of course free to see it the way you like, it doesn't matter for the result. The reason why I think it's confusing to define mass as relativistic mass is because present text hardly ever do so. Otherwise it could be quite hard to understand each other (as this thread is a good example of... :wink: ).

Well, the curvature of spacetime has been not measured. Therefore i do not accept it. In fact, in torsion theories gravity is due to torsion of a flat spacetime, in force theories, like Feynman FRG, is a force, etc.

GR says that the bend is due to spacetime curvature but is not correct. Take the limit c--> infinite. One obtains a flat metric but still there is a residual Newtonian force. Bending of light is predicted by Newtonian mechanics because photons have nonzero mass (total mass) and interact with gravitational field.

The problem with Newtonian mechanics is that bending is half the experimental value, but modified newtonian-like theories can obtain the exp value. As said, curvature GR interpretation is unsustainable because a half of bending is due to Newtonian force which cannot be explained by just curvature.

To call to matter just energy is a matter of taste. From Wald GR textbook pag 77

In the general relativistic viewpoint, the mass-energy of the Sun produces a curvature of the spacetime

However when sumarizes the entire content of GR, Wald talks only of mass.

This is not the same thing.
The physical mass of an electron does not vary.

That is not true. One begins with a Lagrangian with the rest electron mass but due to interactions, one is obligated to substitute initial electron mass by a renormalized mass. This is the principal flaw of QFT, initial mass and charges are unphysical when there are interactions. Take a book on the topic.

jtbell

Sep9-05, 12:19 PM

For the pair e+ e- m = 2 melectron =/= 0

Correct, assuming the electron and positron are at rest.

but for the two fotons m = 0, because photonds are massless.

In this situation he two photons are traveling in opposite directions, each with energy E = m_{electron}c^2, and magnitude of momentum pc = E. The total energy of the system of two photons is E_{total} = 2m_{electron}c^2. The total momentum of the system is {\vec p}_{total} = 0. Therefore the invariant mass of the system is

mc^2 = \sqrt{E_{total}^2 - ({\vec p}_{total} c)^2} = \sqrt{(2m_{electron}c^2)^2 - 0^2} = 2m_{electron}c^2

The invariant mass of a system does not necessarily equal the sum of the invariant masses of its component particles or subsystems.

Juan R.

Sep9-05, 01:10 PM

Correct, assuming the electron and positron are at rest.

Incorrect, if both are not at rest m entering in above SR formula -post #61- is the same.

In this situation he two photons are traveling in opposite directions, each with energy E = m_{electron}c^2, and magnitude of momentum pc = E. The total energy of the system of two photons is E_{total} = 2m_{electron}c^2. The total momentum of the system is {\vec p}_{total} = 0. Therefore the invariant mass of the system is

mc^2 = \sqrt{E_{total}^2 - ({\vec p}_{total} c)^2} = \sqrt{(2m_{electron}c^2)^2 - 0^2} = 2m_{electron}c^2

The invariant mass of a system does not necessarily equal the sum of the invariant masses of its component particles or subsystems.

The invariant mass for the photons is zero because their energy is purely due to total momentum p already said that m is zero for photons

E^2 = p c^2

Sep9-05, 04:15 PM

One begins with a Lagrangian with the rest electron mass but due to interactions, one is obligated to substitute initial electron mass by a renormalized mass. This is the principal flaw of QFT, initial mass and charges are unphysical when there are interactions. Take a book on the topic.

No, you begin with the BARE electron mass. What do you mean by "take a book on the topic"?

jtbell

Sep10-05, 01:56 AM

Sojourner01

Sep10-05, 05:50 AM

It would definitely help if the board as a whole adopted a common convention.

I'm no expert, but I'd say that the general trend is to subscript the algebraic term when you're referring to a specific case. That is, for invariant mass i.e. mass at zero relative velocity, one should be representing such a quantity as m_0 i.e. following the convention of l_0 for 'proper length' and t_0 for 'proper time'.

pmb_phy

Sep10-05, 06:28 AM

The definition of "invariant mass" of a system of particles via

(mc^2)^2 = E_{total}^2 - ({\vec p}_{total} c)^2

is the one that is universally used among particle physicists, as far as I know.

You've been totally englufed in one and only one are of relativity - relativity of particles. And then only in particle physics. There are other areas such as relatiuvity itself which is barely touched by particle physicists and then there is general relativity and cosmology.

There is no question that that the Q in Q^2 = E^2 - P^2 is called the invariant mass of a particle. But invariant mass has its limitations. What one labels it is the choice made by the author for whatever reasons. I too would probably do that if I were a particle physicists since otherwise there'd be too many subscripts. I tried that once (using subscripts) and it was a nightmare so chose to toss it out in those kinds of things.

I will not argue this point further, since it is purely a matter of semantics.
No offence my good sir but this is a terrible view since one is dismissing "semantics" as meaningless. It makes a huge difference of what our terminology means. Words effect thought more that we'd like to admit.

If you want a good list then see
http://www.geocities.com/physics_world/misc/relativistic_mass.htm

Notice the Gravitation, Misner, Thorne and Wheeler example. :biggrin:

Pete

Juan R.

Sep10-05, 09:56 AM

No, you begin with the BARE electron mass. What do you mean by "take a book on the topic"?

Ok! What is the Dirac lagrangian for an electron please?

Juan R.

Sep10-05, 10:03 AM

Juan R.

Sep10-05, 10:21 AM

The definition of "invariant mass" of a system of particles via

(mc^2)^2 = E_{total}^2 - ({\vec p}_{total} c)^2

is the one that is universally used among particle physicists, as far as I know. This is based on my own experience in experimental particle physics (in which I got my Ph.D.), and on some Google searches which turned up pages such as

www.hep.lu.se/staff/eerola/relativitet.pdf (see bottom of page 3)

www.yorku.ca/marko/ComPhys/RelDynamics/RelDynamics.html
(see section 8, "Invariant mass for systems of particles")

http://www.hep.man.ac.uk/u/tamsin/dzeroweb/page10.html

http://faculty.cua.edu/sober/635/relkin.pdf
(see eq. 11 on the first page)

I have yet to find any explicit mathematical definition of "invariant mass" that differs from this one.

I will not argue this point further, since it is purely a matter of semantics. I will simply try to remember in the future that when Juan refers to the invariant mass of a system, he means the sum of the invariant masses of its component particles, and will interpret his remarks accordingly.

I talked of rest mass. In fact the SR expression i used is STANDARD and m is the rest mass. I initially used the notation m0 for rest mass but i abandoned due confusion of people, perhaps now would i recover it again?

You are talking of "invariant mass" which is not rest mass, which is also invariant. At least, the name invariant mass is misleading because rest mass is also invariant. Since rest mass is both invariant and constant and "invariant mass" is just invariant but is not constant. The better name would be non-constant mass or similar.

On http://www.hep.man.ac.uk/u/tamsin/dzeroweb/page10.html

the notation W is used for "invariant mass" which is different from m i used. The web you provided says

they use invariant mass, a physical quantity involving some tricky concepts.

Perhaps the most tricky is that the vectorial character of the impulse is used in the definition, which is not the SR definition.

What is the classical relativistic Lagrangian that you use for a free particle?

In your other link

http://www.yorku.ca/marko/ComPhys/RelDynamics/RelDynamics.html

it is said in "(8) Invariant mass for systems of particles" that

The dynamic state of a relativistic system of particles can be described by the following expression for the invariant total rest mass in terms of energy and momentum

and definition used next is the same i used! :yuck:

And after says in "9) Massless particles: photons,...?"

The only particles known to be definitely massless are photons.

that is i said! :yuck:

Your link

www.hep.lu.se/staff/eerola/relativitet.pdf (see bottom of page 3)

is a bit misleading. Define invariant mass to be mass invariant in any frame but before 1.2 explicitely says that (E^2 - p^2), is also an invariant. Are two different concepts of invariant mass.

Seeing the other of your links

http://faculty.cua.edu/sober/635/relkin.pdf

I do not understand because you was confounded. eq (5) is the same equation i used including notation, whereas equation 11 is the definition i use but notation is again W.

I think that you did not note that my m is also an invariant and therefore is properly the invariant mass.

Perhaps by this reason equation 11 defines "invariant mass" instead of invariant mass. The use of " is interesting.

pmb_phy

Sep10-05, 05:56 PM

Now i am completely perplexed. From SR

E^2 = (mc^2)^2 + (pc)^2

but the m is rest mass. For the pair e+ e- m = 2 melectron =/= 0 but for the two fotons m = 0, because photonds are massless.

:bugeye:You two have a concept in your minds which differ to a confussion of terms. I believe that each of you are refering to the "Total rest mass" of the system. "Total" can mean one of two things. It can be "of the whole" or it can mean "sum of its parts." jtbell is using the former while Juan R. is using the later. This notion was addressed in Max Jammer's most recent book on mass.

Well, the curvature of spacetime has been not measured.
Woa there! That's a clear falsehood. All spacetime curvature is is tidal force. Tidal forces are easy to measure and are done so all the time.

The physical mass of an electron does not vary.Clearly not a truth but that which follows jtbell's definition/usauage of the term "physical." Nothing in physics is ever measured besides things like distance and time. So what is physical is what you define it to be.

For a very clear discussion on this matter I prepared a web page for it. See
http://www.geocities.com/physics_world/sr/invariant_mass.htm

I'm pretty certain at this point someone will once again claim that this is all semantics etc. But let me remind you errors that appear in the modern literature on relativity due to the incorrect ideas about mass. E.g. take Special Relativity: A Modern Introduction, Hans C. Ohanian, Physics Curriculum & Instruction, page 149. Question 14 reads
The strongest magnetic fields produced in laboratories, by explosive compression of magnetic field lines, is 103 T. In such a magnetic field, what is the energy density and what is the mass density?If you try the definitions given by the "mass = rest mass" definition.

Pete

jtbell

Sep11-05, 12:38 AM

"Total" can mean one of two things. It can be "of the whole" or it can mean "sum of its parts." jtbell is using the former while Juan R. is using the later.

Precisely. I referred to the "invariant mass of the system", implicitly meaning the whole system. In fact, I did not even use the word "total" in connection with mass (only with energy and momentum), precisely because I wanted to avoid a connotation of "sum of parts".

The physical mass of an electron does not vary.

I didn't write that, EL did. I haven't gotten involved in that aspect of this thread.

pmb_phy

Sep11-05, 06:52 AM

Precisely. I referred to the "invariant mass of the system", implicitly meaning the whole system. In fact, I did not even use the word "total" in connection with mass (only with energy and momentum), precisely because I wanted to avoid a connotation of "sum of parts".
You may not have noticed it yet but different modern texts on physics will give different answers to the question Is mass conserved even though the authors use the exact same definition of mass. Bizzare!

I didn't write that, EL did. I haven't gotten involved in that aspect of this thread.I wish I could avoid these threads on this subject but they're the threads where I know exactly what I'm talking about. Back in 1997 I decided to look into this mass/rest thingy and try to decide which one I thought worked and which didn't. I had no predisposition to either so as to remain as neutral as possible (and that, of course, is very hard to do) but eventually I could only come to one conclusion - Mass = relativistic mass since the otherjust can't work in all situations and still make sense. The question in Ohanian is a perfect example. This will give you an idea of why Ohananian made a mistake.

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

Pete

Juan R.

Sep11-05, 09:35 AM

jtbell,

I am interesed by definition of W, the "invariant mass"

please what is the lagrangian for a free particle?

Sep11-05, 09:43 AM

Ok! What is the Dirac lagrangian for an electron please?

Don't you know that, or is this some kind of examination?

Juan R.

Sep11-05, 09:47 AM

You two have a concept in your minds which differ to a confussion of terms. I believe that each of you are refering to the "Total rest mass" of the system. "Total" can mean one of two things. It can be "of the whole" or it can mean "sum of its parts." jtbell is using the former while Juan R. is using the later. This notion was addressed in Max Jammer's most recent book on mass.

Woa there! That's a clear falsehood. All spacetime curvature is is tidal force. Tidal forces are easy to measure and are done so all the time.

I am not very happy with your evaluation. I am not using total like the "sum of its parts" for the pair. "My" definition of total energy is the definition for the whole system because the asymptotic character of both initial and final states of the scattering process. E is the energy for the whole system electron positron and E is the whole energy for the two photons system.

Moreover "my" definition is standard one.

S. Wilson. Relativity and Quantum mechanics. Chapter 14 in Handbook of molecular physics and quantum chemistry. Volume 1 (Fundamentals). Whiley (2003).

Regarding spacetime curvature, you are supporting my point.

"Tidal forces are easy to measure"

In the torsion version of general relativity the curvature of spacetime is zero. and forces are due to torsion.

Spacetime curvature is not measured.

Moreover in the strict Newtonian limit c--> infinite, the curvature is exactly zero, but the gravitational force is not, doing the curvature interpretation of standard GR incorrect. If curvature is the cause of gravity, Earth cannot suffer gravity with zero curvature.

Juan R.

Sep11-05, 10:05 AM

Don't you know that, or is this some kind of examination?

i think that i knew!

But i said that one begins with the rest mass in the lagrangians and you said that in #65

"NO, you begin with the BARE mass".

Please write down the Lagrangian.

Sep11-05, 10:26 AM

i think that i knew!

But i said that one begins with the rest mass in the lagrangians and you said that in #65

"NO, you begin with the BARE mass".

Please write down the Lagrangian.

You can find the Lagrangian in any quantum field theory book, or just by googling a little...

What I ment with "NO, you begin with the BARE mass", was that I think you are mixing up two different things. I would say that both the bare mass and the physical mass are rest masses.
I don't really see what the relation between bare mass and physical mass has to do with the concept of rest mass? I don't get your point, could you explain more?

pmb_phy

Sep11-05, 04:52 PM

Spacetime curvature is not measured.Where in the world did you get that crazy idea from????

For a description on how to calculate the Riemann tensor from tidal force measurements simply see Ohanian's text.

Pete

pervect

Sep11-05, 06:18 PM

The definition of "invariant mass" of a system of particles via

(mc^2)^2 = E_{total}^2 - ({\vec p}_{total} c)^2

is the one that is universally used among particle physicists, as far as I know. This is based on my own experience in experimental particle physics (in which I got my Ph.D.), and on some Google searches which turned up pages such as

www.hep.lu.se/staff/eerola/relativitet.pdf (see bottom of page 3)

www.yorku.ca/marko/ComPhys/RelDynamics/RelDynamics.html
(see section 8, "Invariant mass for systems of particles")

http://www.hep.man.ac.uk/u/tamsin/dzeroweb/page10.html

http://faculty.cua.edu/sober/635/relkin.pdf
(see eq. 11 on the first page)

I have yet to find any explicit mathematical definition of "invariant mass" that differs from this one.

I will not argue this point further, since it is purely a matter of semantics. I will simply try to remember in the future that when Juan refers to the invariant mass of a system, he means the sum of the invariant masses of its component particles, and will interpret his remarks accordingly.

I'm not sure how much more this point can be profitably argued either, but the difference is more than one of semantics. Your defintions are correct, Juan's are not.

Note that when you have a system of particles that have significant interaction energies, the formulas for computing the mass of the system get to be more complex than the formulas you have presented. For the purposes of particle physics, however, the formulas you quote are sufficient.

The most general case requires that one includes fields to model the interactions between particles. In addition, the gravitational interaction between particles is NOT modelled as a field, but nonetheless must be taken into account when one desires to arrive at a conserved energy for a system.

In spite of the remarks of some posters, it is possible to assign an energy-momentum 4-vector to an isolated, extended system in GR, including the effects of both strong non-gravitational binding energies between particles and even strong gravitational interactions, when the proper conditions are met. (Proper conditions, such as an asymptotically flat space-time or a static space-time, must be present, however).

For a specific formula on how to do this, see Wald's "General Relativity", pg 289. I've also posted some of the relevant formulas here on this board, I'll dig up the references if anyone is really interested.

When energy and momentum can be defined, the usual equation

E^2 - p^2 = m^2 (using geometric units where c=1) works just fine. To quote Wald (pg 290)

We turn, now, to the definition of energy-momentum in the general, non-stationary, asymptotically flat case. Since there now is no notion of "holding a test mass in place" and thus no notion of "gravitational force,", it is not obvious that a well-defined notion of mass or energy-momentum exists. Furthermore, it is far from clear to what vector space an "energy-momentum vector" of the spacetime should belong. However, we shall see that the properties of asymptotically flat spactimes give just enough structure to define the energy-momentum vector P in terms of an asymptotic limt as one "goes to infinity", and that the properties of infiinty provide an appropriate vector space structure for P.

For this defintion to lead to a meaningful notion of energy-momentum at spatial infinity, E and p should depend only on the aymptotic behavior of \Sigma, should be conserved (i.e. be unchanged if \Sigma undergoes a "time translation" near infinity), and should transform as a 4-vector if \Sigma undergoes a "Lorentz boost" near infinity.

The definitions used satisfy do in fact satisfy all of these conditions - thus, an isolated extended system does have an energy-momentum 4-vector, and an invariant mass, when it is in an asymptotically flat space-time or a static space-time.

jtbell

Sep11-05, 07:34 PM

(definition of invariant mass)

(mc^2)^2 = E_{total}^2 - ({\vec p}_{total} c)^2

Note that when you have a system of particles that have significant interaction energies, the formulas for computing the mass of the system get to be more complex than the formulas you have presented.

Assuming we're in a domain where GR effects are negligible, is it more complex than simply including interaction energies, or energy and momentum carried by e-m fields and the like, in E_{total} and {\vec p}_{total}?

pmb_phy

Sep11-05, 08:58 PM

Assuming we're in a domain where GR effects are negligible, is it more complex than simply including interaction energies, or energy and momentum carried by e-m fields and the like, in E_{total} and {\vec p}_{total}?
Ouch. This is troublesome if you're talking about an actual "point" particle since if thats true the energy of the field would be infinite. If this is not an actual particle but has structure (e.g. a little ball) then you need to use the stress-energy-momentum tensor to account for Poincare stresses.

:zzz: Sorry. I'm so totally exhausted lately I don't have the energy to continue here with you gentelmen.

Pete

pervect

Sep11-05, 10:06 PM

Assuming we're in a domain where GR effects are negligible, is it more complex than simply including interaction energies, or energy and momentum carried by e-m fields and the like, in E_{total} and {\vec p}_{total}?

I can tell you for sure that in nearly Minkowskian space-time (i.e. a flat metric) with orthogonal cartesian coordinates (x,y,z,t) one can find the mass of an isolated system from the stress-energy tensor in the center of momentum frame by taking the intergal

\int_V \left(T^{00}+T^{11}+T^{22}+T^{33}\right) \,dx dy dz

While I could be mistaken on the following point, I do not believe that this expression is equivalent to the intergal of the energy density of particles and fields, which I would expect to be simply \int T^{00} dV, without the pressure terms T^{11},T^{22},T^{33}.

In many cases, the difference does not matter. For instance, if you have a *small* spherical pressure vessel containing pressurized gas, (small meaning that the metric coefficients are still close to unity everywhere including the center of the pressure vessel) there is no net contribution to the mass of the system from the pressure terms, for the positive pressure in the interior of the sphere is exactly balanced out by the negative pressure (tension) in the exterior of the sphere. To give one a general idea of it's applicability, this approximation is pretty good even for a sphere of gas as large as our sun (if one considers errors measured in units of parts per million as being "pretty good").

One thing that makes it difficult to construct an example is that classically, systems of positve and negative charges do not posses tensile strength, making it hard to construct a purely electromagnetic example.

Unfortunately, I do not see any simple way to motivate the inclusion of the pressure terms in the formula for mass, though it is clear from various sources (such as Wald) or Carlip's paper on kinetic energy in GR

http://arxiv.org/abs/gr-qc/9909014

that these pressure terms are needed. I have a suspicion that Einstien's full field equations are needed to motivate these terms (that the equivalence principle alone is not sufficient). Someday I hope to understand at a more fundamental level why these terms are needed - at the moment, I simply have to point to the sources that say they are needed.

Meanwhile, though the presence of the pressure terms may be somewhat puzzling, I also have to point out that the 4-vector approach still does work to find the invariant mass of extended systems via the formula E^2-p^2 (c == 1) when one has necessary pre-conditions (asymptotically flat and/or static space-times) needed to define energy and momentum.

So is the energy of a extended system the sum of the energies of its parts? Probably not, definitely not if it's a large system (i.e. one where the metric coefficients become non-Minkowskian). Is the invariant mass of an extended system still given by E^2 - p^2 when c=1? Yes.

Juan R.

Sep12-05, 07:28 AM

You can find the Lagrangian in any quantum field theory book, or just by googling a little...

What I ment with "NO, you begin with the BARE mass", was that I think you are mixing up two different things. I would say that both the bare mass and the physical mass are rest masses.
I don't really see what the relation between bare mass and physical mass has to do with the concept of rest mass? I don't get your point, could you explain more?

Ok I find. It begins with the rest mass m and AFTER in the renormalization procedure one introduces the renormalization constant Z defining the bare mass

m ---> Z m.

But one begins with m which IS rest mass. I do not know any book where one begins with a Lagrangian in terms of the bare mass.

My initial motivation for talking on QFT was

Moreover, the idea of a variable mass is again recovered in QFT, where mass of an electron is dependant of interaction, due to the cloud of virtual photons. I see no reason for emphasis on a constant "rest" mass like the real mass.

Juan R.

Sep12-05, 08:00 AM

Where in the world did you get that crazy idea from????

For a description on how to calculate the Riemann tensor from tidal force measurements simply see Ohanian's text.

Pete

Again you are supporting my point.

"how to calculate the Riemann tensor from tidal force measurements" is not the same that measuring spacetime curvature.

Crazy idea? :rofl:

As said and repeat again: in torsion gravity curvature of spacetime is ZERO. Read literature please.

Nobody has measured curvature of spacetime and, as already said, in the nonrelativistic limit the usual curvature GR interpretation is just wrong because force is non zero and curvature is zero.

It is really interesting to compare your statement "crazy idea" with words of some "specialists" in gravitation. See the post #76 on

http://www.physicsforums.com/showthread.php?t=75197&page=6&pp=15

my point

2) A theory for gravity on a flat spacetime. Unless one can measure curved spacetime, all our experimental evidence is for flat space and time.

was replied

Point 2: Ok, fine. In some sense, "curved spacetime" is just semantics. It's a very nice mental picture, though, and I don't know why you want to spoil it.

Why would we spoil it?

i) Elimination of spacetime curvature could permit to us to quantize gravity directly.

ii) There is no experimental support.

iii) Flat spacetime permits to unificate gravity with others three forces.

iv) Curved spacetime interpretation is unsustainalbe in the non relativistic limit, where curvature is zero but force is not. A basic epistmological principle says that if A is cause of B then elimination of A mays eliminate B. If there is gravity with zero curvature then curvature is not the real cause of gravity.

For me, it is very clear that spacetime curvature interpretation of GR is just wrong. Precisely, this is my research program.

From Thorne, the same guy of classical textbook on gravitation by Thorne, Misner, and Wheeler wrote in his popular book "Black Holes and Time Warps"

Is spacetime REALLY curved? Isn't it conceivable that spacetime is actually flat, but the clocks and rulers with which we measure it... are actually rubbery? Wouldn't... distortions of our clocks and rulers make truly flat spacetime appear to be curved? Yes.

:biggrin:

Juan R.

Sep12-05, 08:09 AM

I'm not sure how much more this point can be profitably argued either, but the difference is more than one of semantics. Your defintions are correct, Juan's are not.

:eek:

My definition is not only correct one, it is also standard. See the Handbook that cited above. Do you know that difference between an Handbook and a textbook?

Moreover, i already showed that definition i use is in the links that jtbell provided.

Expression used in above link

http://www.yorku.ca/marko/ComPhys/R...elDynamics.html

is same i used.

Equation (5) of above link

http://faculty.cua.edu/sober/635/relkin.pdf

is the same equation i used including notation.

Etc.

Sep12-05, 09:30 AM

Ok I find. It begins with the rest mass m and AFTER in the renormalization procedure one introduces the renormalization constant Z defining the bare mass

m ---> Z m.

But one begins with m which IS rest mass. I do not know any book where one begins with a Lagrangian in terms of the bare mass.

No, the parameter "m" in the QED-Lagrangian is the bare mass. If it was taken as the physical mass it would lead to infinities. Of course most books starts out by just saying that "m" is the rest mass of the electron (i.e. the finite "physical" mass) just to find out that when including corrections of higher orders a finite electron mass would lead to infinities. These infinities are then removed by introducing the concepts of bare and physical masses.
However, note that both the bare and the physical mass are invariant (rest) masses (particle physicists hardly ever talks about relativistic masses), and hence I can still not see what this has to do with relativistic/rest mass.

Juan R.

Sep12-05, 10:06 AM

No, the parameter "m" in the QED-Lagrangian is the bare mass. If it was taken as the physical mass it would lead to infinities. Of course most books starts out by just saying that "m" is the rest mass of the electron (i.e. the finite "physical" mass) just to find out that when including corrections of higher orders a finite electron mass would lead to infinities. These infinities are then removed by introducing the concepts of bare and physical masses.
However, note that both the bare and the physical mass are invariant (rest) masses (particle physicists hardly ever talks about relativistic masses), and hence I can still not see what this has to do with relativistic/rest mass.

Still i said is correct. One begins with rest mass m. E.g. equation 12.1.1 of Michel le Bellac. Quantum and statistical field theory. Oxford university Press, 1991.

Then after in the renormalization procedure -section 12.4.3- one MAY substitute rest mass by bare mass, introducing, ad hoc, renormalization constant Z. Then the renormalized lagrangian -equation 12.4.28- contains the NEW mass concept -denoted by m0 in that book-, which is not the original mass concept m I used. In fact, the fact one may change mass and charge is one of recognized flaws of QFT.

From J. Sanchez Guillen and M. A. Braun. Física cuántica. Alianza Editorial S.A., 1993. pag 362.

Con ello las constantes m and e que caracterizan al electrón 'desnudo' y entran en el hamiltoniano en realidad están mál definidas [incorrectly defined], lo que constituye una cierta debilidad teórica [teorical flaw] de la física cuantica relativista que no se ha sabido superar todavía [nobody has corrected still this flaw of relativistic QM].

Estrictly speaking the mass of the electron is its rest mass, that one can find in a table of universal constants. The observed mass -often called 'physical mass'- is not the physical mass of the electron alone. It is the mass of the pair electron more cloud of virtual photons. The same about the charge.

I already said that emphasis on constant masses is unnecesary since in QFT mass is variable. The bare mass is variable. The only constant mass is rest mass. This was my motivation for talking of QFT post ago...

Sep12-05, 11:36 AM

E.g. equation 12.1.1 of Michel le Bellac. Quantum and statistical field theory. Oxford university Press, 1991.

Then after in the renormalization procedure -section 12.4.3- one MAY substitute rest mass by bare mass, introducing, ad hoc, renormalization constant Z. Then the renormalized lagrangian -equation 12.4.28- contains the NEW mass concept -denoted by m0 in that book-, which is not the original mass concept m I used. In fact, the fact one may change mass and charge is one of recognized flaws of QFT.

Sorry I don't have le Bellac.
How does the renormalized Lagrangian differ from the unrenormalized in that book? Is it just that m has been substituted by m0?

pmb_phy

Sep12-05, 01:32 PM

again: in torsion gravity curvature of spacetime is ZERO. Read literature please. :grumpy: I know GR thanks. It is you who does't know what you're talking about regarding GR. It sounds like you have no idea what torsion is. There is no torsion in GR by assumotion/postulate. If you have your own theory then you're posting messages in the wrong forum. This is not the forum for posting personal theories. When they are found those threads are moved to the correct forum. And I'm not interested in anyone's personal theories at this time in my life.

Pete

pervect

Sep13-05, 05:07 AM

My definition is not only correct one, it is also standard. See the Handbook that cited above. Do you know that difference between an Handbook and a textbook?

Moreover, i already showed that definition i use is in the links that jtbell provided.

While the first link didn't work, the second link clearly supported J.T.Bell's correct argument (in equations 9 and 10), and did not support your point of view.

I'm not quite sure what sort of intellectual blinders you are wearing not to see this, but I guess I don't really want to know.

Juan R.

Sep13-05, 05:24 AM

Sorry I don't have le Bellac.
How does the renormalized Lagrangian differ from the unrenormalized in that book? Is it just that m has been substituted by m0?

It is not exclusive of le Bellac. The book is standard. See also Fisica quantica what i cited. Also celebrated Weinberg manual on QFT use the same m i used in the QED lagrangian (before the chapter in renormalization).

See equation 8.6.1 of Weinberg. Also Feynmann book on QED, etc

One always begins with real physical mass m. After one may apply renormalization for eliminate infinites, but real electron mass is m, what is the mass that appears in tables of universal constants.

1) The physical renormalized mass is a mathematicla trick due to theoretical flaw of standard QFT. 2) That mass is not the electron mass, it is the observed mass due to virtual cloud.

Only m? I think that i replied to this in #89. Even if you cannot read spanish, you can see symbol "e".

On any case i see no signifcant error in my initial claim one begins with mass m before renormalization. You said "NO, one begins with bare mass", but in at least three standard books one begins with the same m i used in my definition.

You can verify that m used by Weinberg for the QED lagrangian verifies expresion i used and appears in the Handbook.

Am i wrong?

pmb_phy

Sep13-05, 05:44 AM

My definition is not only correct one, it is also standard. See the Handbook that cited above. Do you know that difference between an Handbook and a textbook?There is no universal definition since there was no term used = As I said, the terms were in your minds and not on the board. This happened between both posters but each was unable to nail quite rightly what the meat of the subject was. This happens throughout the physics literature for this very same reason.

Pete

Juan R.

Sep13-05, 05:45 AM

:grumpy: I know GR thanks. It is you who does't know what you're talking about regarding GR. It sounds like you have no idea what torsion is. There is no torsion in GR by assumotion/postulate. If you have your own theory then you're posting messages in the wrong forum. This is not the forum for posting personal theories. When they are found those threads are moved to the correct forum. And I'm not interested in anyone's personal theories at this time in my life.

Pete

If you know GR then you know that GR cannot -rigorously- explain the Newtonian limit in the linear regime.

I know both standard GR which is based in the asumption of spacetime curvature and torsion gravity which is based -i say again- in ZERO spacetime curvature.

Personal theory? There are many people that does not follow curvature description. Since it appears that you cannot search literature by yourself i will do.

arXiv:gr-qc/0403074

You claimed that spacetime curvature had been measured which is, of course, completely false. One measure forces.

Are those forces the result of

a) Spacetime curvature

b) spacetime torsion

c) A force in Weinberg sense?

You said that idea of no curvature was crazy. Let me cite again to Thorne, the same guy of classical textbook on gravitation by Thorne, Misner, and Wheeler.

Is spacetime REALLY curved? Isn't it conceivable that spacetime is actually flat, but the clocks and rulers with which we measure it... are actually rubbery? Wouldn't... distortions of our clocks and rulers make truly flat spacetime appear to be curved? Yes.

Juan R.

Sep13-05, 06:01 AM

While the first link didn't work, the second link clearly supported J.T.Bell's correct argument (in equations 9 and 10), and did not support your point of view.

I'm not quite sure what sort of intellectual blinders you are wearing not to see this, but I guess I don't really want to know.

Please, click in the second link of #81

"The second link clearly supported jtbell arguments"

Are you sure of that?

Perhaps i am reading a different pdf document that you, but definition of mass m used in equation (5) of above PDF link is exactly i used.

Equations (9) and (10) you cite are used in equation (11) for the definition of

W

or more exactly the definition of

W2

but the definition of

m

continues to be that of equation (5) which is the same i used several post ago...

For systems of particles one may introduce the potential energy BUT in the study of scattering asymptotic states like that of pair anhilitation process, the potential energy is zero by cluster decomposition principle and then

E^2 = p^2 + m^2

Juan R.

Sep13-05, 06:06 AM

There is no universal definition since there was no term used = As I said, the terms were in your minds and not on the board. This happened between both posters but each was unable to nail quite rightly what the meat of the subject was. This happens throughout the physics literature for this very same reason.

Pete

sorry but Handbook definition is standard one.

pmb_phy

Sep13-05, 12:49 PM

sorry but Handbook definition is standard one.That's total nonsense. That certainly can't be taken to mean that all "handbooks" agree nor does it mean that the terms in a "handbook" are adhered to by all professionals - In fact they usually disagree. Go look for yourself. Find 5 different "handbooks" and see if each and every one defines the term "mass" identically. Then go to several GR and SR texts and see if they agree with the handbooks.

If I were you I'd worry less about who defines what as what but concern yourself more on the various definitions used and how they are used. Any handbook that defines "mass" as the magnitude of a 4-vector is wrong. That is a quite limited use. A genral definition is what is desired, i.e. one that works all the time. Then you can choose your own way in your own work and simply define your terms. This is how its usually done.

Pete

pmb_phy

Sep13-05, 12:52 PM

If you know GR then you know that GR cannot -rigorously- explain the Newtonian limit in the linear regime. wrong.

I know both standard GR which is based in the asumption of spacetime curvature and torsion gravity which is based -i say again- in ZERO spacetime curvature.

Wrong.

Personal theory?
Yes. Are you going by Einstein's general theory of relativity or Juan's general theory of relativity?

There are many people that does not follow curvature description. Since it appears that you cannot search literature by yourself i will do.

This is where the discussion ends. When you start being a wiseguy/flaming etc.

No more interaction between us until you act more like a professional or more like an adult.

(snip)

Pete

Juan R.

Sep13-05, 01:13 PM

That's total nonsense. That certainly can't be taken to mean that all "handbooks" agree nor does it mean that the terms in a "handbook" are adhered to by all professionals - In fact they usually disagree. Go look for yourself. Find 5 different "handbooks" and see if each and every one defines the term "mass" identically. Then go to several GR and SR texts and see if they agree with the handbooks.

If I were you I'd worry less about who defines what as what but concern yourself more on the various definitions used and how they are used. Any handbook that defines "mass" as the magnitude of a 4-vector is wrong. That is a quite limited use. A genral definition is what is desired, i.e. one that works all the time. Then you can choose your own way in your own work and simply define your terms. This is how its usually done.

Pete

curiously they agree :approve: