Photon Mass: Debunking the Myth of Zero Mass in SR Equation

[pmb_phy]

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 got to 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. 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

 

[EL]

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]

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

 

[EL]

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!

 

[Sam Owen]

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

 

[Ich]

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]

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]

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

 

[pmb_phy]

Maybe we should put up a poll about this!

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

 

[EL]

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.]

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]

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]

I am very doubtful about relativistic formulas anyway...

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

 

[EL]

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]

Does this thread need to keep banging on?

 

[jtbell]

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]

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 got to 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.]

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.]

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]

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.

Moving particle with every second C speed?
Traveling with the speed of light for mass object - does it means having infinite mass --speeding with no time along thread of possible paths in expanding spacetime will be possible practically measure it?

 

[janusz]

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 I am 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]

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.

 

[EL]

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... ).

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]

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.]

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.]

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

 

[Juan R.]

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... ).

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]

For the pair e⁺ e^- m = 2 m_electron =/= 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.]

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

$$E^2 = p c^2$$

 

[EL]

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]

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.

 

[Sojourner01]

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]

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.

Pete

 

[Juan R.]

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.]

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'.

Well, initially i began with that convention and remarked that m =/= m₀ but it appears that is not standard now and anyone write m for anything.