Photon Mass: Is Student Learning Misleading?

[bernhard.rothenstein]

Is the student not correctly taught if the intructor says:
the photon has no rest mass but it has a m=E/cc dynamic mass? Could such a statement have missleading consequences?

 

[robphy]

IMHO, to properly answer this, one needs the [precise] definition of "dynamic mass" [and "rest mass"] that is being given to the student. (To address the misleading issue, one should also consider the student's ability to reason with that precision.)

 

[Meir Achuz]

Is the student not correctly taught if the intructor says:
the photon has no rest mass but it has a m=E/cc dynamic mass? Could such a statement have missleading consequences?

Yes! Many of the questions and answers in this forum show that confusion.

 

[rbj]

Yes! Many of the questions and answers in this forum show that confusion.

to the contrary. the confusion (here in this forum) results when blanket statements are made that photons have no mass, without any qualification.

you end up having to explain why momentum [itex] p = mv [/itex] exists with photons, yet they have no mass. you end up having to explain why photons have energy [itex] E = h \nu = m c^2 [/itex], yet have no mass.

Photons have mass, but no rest mass.

 

[masudr]

I'd rather make the blanket statement that photons have no mass, and end up having to explain that [itex]p=m-0v[/itex] is no longer a helpful relation. And, in all fairness, [itex]E=m_0c^2[/itex] is not too helpful either. Of course, we all know the proper relation between energy, mass and momentum:

[tex]E^2 = m_0^2 c^4 + p^2 c^2[/tex]

Furthermore, in more advanced physics, "dynamic mass" or "relativistic mass" hardly ever enters into consideration, so it is more or less a useless concept. Ironically the concept of changing mass is a manifestation of the conservative human need to resist change from the classical relations of [itex]p = Mv[/itex] where [itex]M= \gamma m_0[/itex].

 

[pervect]

Is the student not correctly taught if the intructor says:
the photon has no rest mass but it has a m=E/cc dynamic mass? Could such a statement have missleading consequences?

I would avoid making up a new name for yet another sort of mass, if a perfectly servicable name already exists.

I'm not aware of the terms "dynamic mass" being used in textbooks or the literature.

It's confusing enough that we already have:

invariant mass, relativistic mass, Bondi mass, ADM mass, Komar mass, (and I think someone mentioned Dixon mass, which I want to learn more about someday).

There is no need at all to add another synonym for "relativistic mass" to this stew.

If you mean to ask "is the term relativistic mass outmoded", I would say, yes, it is. It's not incorrect, though - though it is frequently used incorrectly, one reason that it has become outmoded (students are too likely to misunderstand it).

 

[rbj]

I'd rather make the blanket statement that photons have no mass, and end up having to explain that [itex]p=m-0v[/itex] is no longer a helpful relation.

And, in all fairness, [itex]E=m_0c^2[/itex] is not too helpful either.

but, as long as we're being fair, i didn't say [itex]E=m_0c^2[/itex]. i said [itex]E=m c^2[/itex]

Of course, we all know the proper relation between energy, mass and momentum:

[tex]E^2 = m_0^2 c^4 + p^2 c^2[/tex]

and we can show that the above is true from

[tex] E = m c^2 [/tex]

[tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

and

[tex] p = m v [/tex]

Furthermore, in more advanced physics, "dynamic mass" or "relativistic mass" hardly ever enters into consideration, so it is more or less a useless concept. Ironically the concept of changing mass is a manifestation of the conservative human need to resist change from the classical relations of [itex]p = Mv[/itex] where [itex]M= \gamma m_0[/itex].

is that relation wrong ([itex]p = mv[/itex] where [itex]m = \gamma m_0[/itex])? it comes from my human need to have to commit to memory the fewest fundamental equations as possible. all i have to add to the list is [itex]m = \gamma m_0[/itex].

 

[Meir Achuz]

to the contrary. the confusion (here in this forum) results when blanket statements are made that photons have no mass, without any qualification.

you end up having to explain why momentum [itex] p = mv [/itex] exists with photons, yet they have no mass. you end up having to explain why photons have energy [itex] E = h \nu = m c^2 [/itex], yet have no mass.

Photons have mass, but no rest mass.

Q.E.D. I rest my case.

 

[rbj]

Q.E.D. I rest my case.

do you realize how anemic the "case" you're resting on is?

(non-existent is a more accurate term.)

 

[Meir Achuz]

Your confusion arises because you are trying to discuss a correct theory using equations and concepts from an incorrect theory.
Now, you can argue among yourselves.

 

[rbj]

Your confusion arises because you are trying to discuss a correct theory using equations and concepts from an incorrect theory.
Now, you can argue among yourselves.

your argument remains vapid. you make no defense of it and expect it to be taken for granted. do you also get paid for doing no meaningful work?

must be nice, if you do.

 

[bernhard.rothenstein]

your argument remains vapid. you make no defense of it and expect it to be taken for granted. do you also get paid for doing no meaningful work?

must be nice, if you do.

as I see the discussion is far from what I have asked. Consider an atom located on a sensitive balance and equlibrated by a mass located on the other pan. the atom emits two photons of the same frequency in horizontal and opposite directions. if the balance goes out from the state of equilibrium then you explain the fact using which mass of the photon?

 

[rbj]

Consider an atom located on a sensitive balance and equlibrated by a mass located on the other pan. the atom emits two photons of the same frequency in horizontal and opposite directions. if the balance goes out from the state of equilibrium then you explain the fact using which mass of the photon?

someday they will redefine the kilogram in such a way to fix Planck's constant to a defined value (sorta like the meter is defined today to fix the speed of light to a defined value 299792458 m/s):

The kilogram is the mass of a body at rest whose equivalent energy is equal to that of a collection of photons with frequencies that sum to exactly (299792458)²/6626069311 × 10⁴³ Hz

my hypothetical (and rhetorical) question is, if you had a box of negligible mass with perfectly mirrored internal surfaces with such a collection of photons in it and put it on a scale, how much would it weigh? at least 1 kg?

 

[bernhard.rothenstein]

someday they will redefine the kilogram in such a way to fix Planck's constant to a defined value (sorta like the meter is defined today to fix the speed of light to a defined value 299792458 m/s):



my hypothetical (and rhetorical) question is, if you had a box of negligible mass with perfectly mirrored internal surfaces with such a collection of photons in it and put it on a scale, how much would it weigh? at least 1 kg?

whom are you quoting?

 

[ZapperZ]

I would avoid making up a new name for yet another sort of mass, if a perfectly servicable name already exists.

I'm not aware of the terms "dynamic mass" being used in textbooks or the literature.

It's confusing enough that we already have:

invariant mass, relativistic mass, Bondi mass, ADM mass, Komar mass, (and I think someone mentioned Dixon mass, which I want to learn more about someday).

There is no need at all to add another synonym for "relativistic mass" to this stew.

If you mean to ask "is the term relativistic mass outmoded", I would say, yes, it is. It's not incorrect, though - though it is frequently used incorrectly, one reason that it has become outmoded (students are too likely to misunderstand it).

I would also add "effective mass" to that list you have there. And I think you have made a terrific point.

This whole thread is going WAY beyond what is called for. If someone who doesn't undertand physics that much would ask for the mass of something, I would seriously doubt that he/she is asking for "effective mass", "relativistic mass", etc.. etc. These are concepts that he/she does NOT understand, or maybe not even aware of. So if we start dealing out all of these things, we do nothing but add to the confusion.

Photons have no mass. Done! We all can safely assume that answers the question in the simplest manner. We can THEN, address the issue of "momentum". I have zero problem in addressing that, because even in classical E&M where the concept of photons does not exist, you can still find radiation "pressure" of an EM wave. We had no need to invoke any exotic explanation for such a thing. There is no issue whatsoever in explananing a wave having a momentum (no mass concept is involved here either last time I checked Jackson's text). And there is also no mass in defining the crystal momentum in solid state physics. So defining a momentum of something with no mass concept is no big deal. It cannot be used to argue for the usage of a "relativistic mass".

If people are so uptight about making sure we define and distinguish between "rest mass" and "relativistic mass" each time a simple question like this comes up, then I would insist that you also distinguish between "bare mass" and "effective mass" of whatever particle you are using. I can easily make the case that MOST of the masses that you measure are "effective masses" and not the particle's bare mass, and that such effective masses can CHANGE! So I can play this annoying game too till the cows come home.

Zz.

 

[bernhard.rothenstein]

photon

I would also add "effective mass" to that list you have there. And I think you have made a terrific point.

This whole thread is going WAY beyond what is called for. If someone who doesn't undertand physics that much would ask for the mass of something, I would seriously doubt that he/she is asking for "effective mass", "relativistic mass", etc.. etc. These are concepts that he/she does NOT understand, or maybe not even aware of. So if we start dealing out all of these things, we do nothing but add to the confusion.

Photons have no mass. Done! We all can safely assume that answers the question in the simplest manner. We can THEN, address the issue of "momentum". I have zero problem in addressing that, because even in classical E&M where the concept of photons does not exist, you can still find radiation "pressure" of an EM wave. We had no need to invoke any exotic explanation for such a thing. There is no issue whatsoever in explananing a wave having a momentum (no mass concept is involved here either last time I checked Jackson's text). And there is also no mass in defining the crystal momentum in solid state physics. So defining a momentum of something with no mass concept is no big deal. It cannot be used to argue for the usage of a "relativistic mass".

If people are so uptight about making sure we define and distinguish between "rest mass" and "relativistic mass" each time a simple question like this comes up, then I would insist that you also distinguish between "bare mass" and "effective mass" of whatever particle you are using. I can easily make the case that MOST of the masses that you measure are "effective masses" and not the particle's bare mass, and that such effective masses can CHANGE! So I can play this annoying game too till the cows come home.

Zz.

would you say all that in front of students you teach?

 

[rbj]

The kilogram is the mass of a body at rest whose equivalent energy is equal to that of a collection of photons with frequencies that sum to exactly (299792458)²/6626069311 × 10⁴³ Hz

whom are you quoting?

it was originally from a paper by Peter Mohr and Barry Taylor that i thought was the one that can be found at

http://ejde.math.unt.edu/conf-proc/04/m1/mohr.pdf

but, it appears (from a slight rewording and change of numbers) that i got it from somewhere else (but from Mohr and Taylor, if you Google "kilogram" and "definition", you get a lot of hits with their names on it - some of the papers are not free) and i cannot find the exact quote at present. but the paper above has an equivalent quote.

Mohr and Taylor are physicists at NIST ( http://physics.nist.gov/cuu/ ) and when i hear someone at NIST saying "the kilogram should be redefined to this:", i tend to think that someday soon (within years) it will be.

if they did this (which would have the effect of defining Planck's constant to h = 6.626069311 × 10^-34 J/Hz) and, if someday, we could get a really good measure of the universal gravitational constant, G (we likely will never measure G to 9 digits), then they could also redefine the second so that G is also a defined value, and then the meter, kilogram, and second would effectively be defined as constant multiples of the Planck length, Planck mass, and Planck time. for experimental purposes, it's probably best to leave the definition of the second to what it is. these Cesium clocks are getting cheaper and cheaper to duplicate for reference use in modern physical experiments.

 

[ZapperZ]

would you say all that in front of students you teach?

Yes I would, because they won't come back with a retort "But is that rest mass or relativistic mass?"

If I'm teaching a higher level physics course, then I would tell you that such a question doesn't come up. Having taken so many advanced classes throughout my academic years, I have never once heard a student asking such a question. And trust me, we were known to ask the instructors a lot of questions.

Zz.

 

[rbj]

Yes I would, because they won't come back with a retort "But is that rest mass or relativistic mass?"

but if i was your student, Z, there would be other questions that would bounce back at you. (i have identified them previously.)

 

[ZapperZ]

but if i was your student, Z, there would be other questions that would bounce back at you. (i have identified them previously.)

And I have answered them, especially on your reason for using the "relativistic mass" term JUST to avoid explaning the existence of momentum without a mass.

And guess what, if you were in an intro physics class where you have zero clue on what special relativity is, why would you even know the existence of a relativistic mass, much less understand what it is? When asked for the mass of an electron, you do also go on in a lengthy treatise on it's bare mass and effective mass without caring if you are adding to the confusion BEYOND what that person can comprehend?

This has NOTHING to do with the physics. It has everything to do with finding the appropriate answer to right audience. I have to deal with such a thing regularly since I am actively involved in many outreach programs - I just finished participating in a week-long activity for high school girls during the Women in Science week here at the lab. I have been asked this VERY question many times. Do you think this type of audience would know the difference between such a thing? Do you think when they think about the concept of "mass" that they have a clear idea of the significance of "relativistic mass", "effective mass", "inertial mass", mass renormalization, etc, etc? Do you go into a lengthy spiel on this regardless of who asked?

Zz.

 

[pervect]

as I see the discussion is far from what I have asked. Consider an atom located on a sensitive balance and equlibrated by a mass located on the other pan. the atom emits two photons of the same frequency in horizontal and opposite directions. if the balance goes out from the state of equilibrium then you explain the fact using which mass of the photon?

It's very easy. The mass (invariant mass!) of a system is not the sum of the masses of its parts. That's all there is to it. This rather trivial observation makes the whole thing a total non-issue.

The energy of an isolated system is the sum of the energies of its parts.
The momentum of an isolated system is the sum of the momenta of its parts.

The mass of an isolated system is given by the relation E^2 - p^2 = m^2 (in geometric units, add factors of 'c' if you desire.

This is all there is to it. The only remaining complexity is for non-isolated systems, which students rarely have to deal with.

 

[Doc Al]

Latex testing...

[tex]E = m c^2[/tex]

(Seems to work fine here too.)

 

[krab]

A lucid explanation of the concept of mass is here:
http://arxiv.org/abs/hep-ph/0602037

 

[rbj]

A lucid explanation of the concept of mass is here:
http://arxiv.org/abs/hep-ph/0602037

i like Lev Okun. I've had a few email conversations with him (along with Michael Duff and Gabriele Veneziano) about their "Trialogue on Fundamental Constants" paper (i think Duff might call me a "4-constants" partisan, even though i completely agree with Duff's main thesis).

but i just do not see his point regarding "The pedagogical virus of 'relativistic mass' ". it seems to me, pedagogically, that it's precisely the other way around:

1. after time-dilation, length-contraction, and Lorentz transformation is derived, then, given the same axioms, relativistic mass is derived. it is equivalent to saying that the momentum of an inertial body as viewed by an observer it is moving past is

[tex] p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

and we all agree with that statement. even the relativistic mass deniers. it's just that i continue to recognize that as


[tex] p =m v [/tex]

where

[tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]


2. then it might be asked, what is the kinetic energy, [itex]T[/itex], of a body with mass at rest [itex]m_0[/itex] and velocity [itex]v[/itex]? what is the amount of work required to accelerate such a body to velocity [itex]v[/itex]? and we might hope that, when velocity [itex]v \ll c[/itex] that [itex]T = \frac{1}{2} m_0 v^2[/itex].

when the work (in one dimension) is

[tex] T = \int F dx [/tex]

[tex] v = \frac{dx}{dt} [/tex]

and force is

[tex] F = \frac{dp}{dt} [/tex]

and

[tex] p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

and after you do a little fanagling (substitution of variable and integration by parts), you get this result.

[tex] T = m_0 c^2 \left( \left(1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}} - 1 \right) [/tex]

which is the same as

[tex] T = m c^2 - m_0 c^2 [/tex]

or, as interpretation:

[tex] T = E - E_0 [/tex]

where the total energy of the body is:

[tex] E = m c^2 [/tex]

and the energy the body has when it isn't even moving is

[tex] E_0 = m_0 c^2 [/tex].

If commonly agree symbols are used (i would ask you to use [itex]m_0[/itex] for "invariant mass"), we all agree with this last equation. but, given the expression for "relativistic mass" (that you guys want to dispute the existence of):

[tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

then even

[tex] E = m c^2 [/tex]

is also true and is perfectly consistent with

[tex] E^2 = E_0^2 + (p c)^2 = \left( m_0 c^2 \right)^2 + (p c)^2 [/tex]

and we all agree with that equation.

3. The pedagogical problem that i just can't understand from the deniers or relativistic mass is that

[tex] E = m c^2 = E_0 + T [/tex]

compactly (and accurately) relates total energy of a body as the sum of "rest energy" and "kinetic energy" and the kinetic energy becomes the classical expression

[tex]T = \frac{1}{2} m_0 v^2[/tex]

in the limit where the velocity of the body is at nonrelativistic speeds.

How is it that anyone disputes this?

4. Then, for photons, we start with the result of the photoelectric effect:

[tex] E = h \nu [/tex]

which we interpret as total energy (since the photon was exchanged for added kinetic energy for an electron after paying for the "work function" of the surface):

[tex] E = m c^2 [/tex]

so we equate the two and get an inherent (relativistic) mass for the photon as

[tex] m = \frac{h \nu}{c^2} [/tex]

which you guys don't seem to like, but it's the simplest pedagogical path to get the photon momentum of

[tex] p = m v = m c = \frac{h \nu}{c} [/tex]

since the velocity of the photon is, by definition, the speed of light.

And the fact that the rest mass of the photon is zero comes out simply from

[tex] m_0 = m \sqrt{1 - \frac{v^2}{c^2}} [/tex]

when [itex] v = c [/itex].

Now, none of you guys disagree with this momentum expression:

[tex] p = \frac{h \nu}{c} [/tex]

What is your pedagogically simpler means to get to that expression? why should the momentum of a photon be that expression (from pedagogically fundamental principles)? if, in your derivation, you use

[tex] E^2 = E_0^2 + (p c)^2 = \left( m_0 c^2 \right)^2 + (p c)^2 [/tex]

then you need to pedagogically justify that.

 

[robphy]

If commonly agree symbols are used (i would ask you to use [itex]m_0[/itex] for "invariant mass"), we all agree with this last equation. but, given the expression for "relativistic mass" (that you guys want to dispute the existence of):

[tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

Who disputes its existence? You've clearly written it down, you have used it in some computations, and it can be given a physical interpretation.
The complaint is that it is probably not pedagogically good as the invariant rest-mass.

 

[pervect]

The main point IMO in favor of invariant mass is that the concept is independent of coordinates, as I have said before and will say again.

It is generally good to delay introducing coordinates into a problem for as long as possible. By sticking with invariant mass, this goal can be accomplished.

Invariant mass, like 4-velocities and the energy-momentum 4-vector, stands alone, independent of any choice of coordinate system. These coordinate independent entities "package" information so that all the information needed to change views (coordinates) is included in the package.

Relativistic mass demands that one chose a specific coordinate system and stick with it. Every new coordinate system needs a completely new set of variables.

To make an analogy, coordinate indepenent physics is like modular programming. The information that one needs is collected into complete, self-sufficient modules. These modules have the capability to be represented by different "views" (coordinate systems). The process of converting the viewpoint (coordinate system) is conceptually packaged as part of the object itself - it does not have to be re-written on a case-by-case basis, or re-learned. It can be written (and learned) only once.

 

[Hans de Vries]

given the expression for "relativistic mass" (that you guys want to dispute the existence of
[tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

If you want to be consistent:

According to NIST, the rest-mass of a photon is less than 6 10^-17 eV
That makes about 10^-52 kg. The rest-mass should be positive according
to the formula above. A negative rest-mass becomes negative energy.
A mathematically zero rest-mass would give an undefined energy (?!)


A photon with a wavelenght of 700 nm trails the speed of light by a
maximum of about 70 nm / "age-of-the-universe" if it has the maximum
NIST rest-mass.


Regards, Hans

 

[robphy]

It is generally good to delay introducing coordinates into a problem for as long as possible.

[stream of consciousness]
Certainly...

Some related questions...

Do we ask students to do the algebra first, then plug in the numerical givens last? (this reveals the physics, independent of many of the specific givens)
Or do we plug in the numbers first, then solve the problem?

Do we ask students to draw a free-body diagram with vector-forces? (which involves the physics)
Or do we immediately write down components? (which involves the physics, a choice of measuring instruments, and the mathematics problem of determining the components with respect to our choice of [measuring] axes... hopefully, we chose a good set of axes... or else we'll have more math (but not any more physics) to do).

If we do the latter, maybe we should present Maxwell's Equations as a system of scalar equations written in our favorite choice of axes.

Hmmm...
Maybe that's the key to the issue... it's a matter of comfort level... and how one learned [or learned to like] the subject.

We write Maxwell's Equations vectorially because we have [developed] an intuition about vectors and vector fields. If we're not advanced enough, we write more-familiar-looking, special case versions... in terms of more-digestible components [possibly tied to instruments associated with the choice of axes]. Maybe someday, if one so desires, one may then discover certain curious combinations of terms that yield observer-independent "invariants". [This is consistent with regarding a tensor as "a quantity with components that transforms as such and such..."]

Of course, one could be more advanced and write Maxwell's Equations tensorially. Invariants are then much easier to find... often one can count them. From that level, discussions of [observer-dependent] "electric fields" and "magnetic fields" [rather than the observer-independent field tensor] are akin to discussions of [observer-dependent] "relativistic mass" [rather than the observer-independent rest-mass scalar].

[/stream of consciousness]

 

[rbj]

According to NIST, the rest-mass of a photon is less than 6 10^-17 eV
That makes about 10^-52 kg. The rest-mass should be positive according to the formula above.

i have never, ever heard of a non-zero rest mass for photons. could you give a reference to that NIST statement attributing a non-zero rest mass to a photon?

A mathematically zero rest-mass would give an undefined energy (?!)

no, it is a 0/0 problem, but there is a defined energy from [itex] E = h \nu [/itex]. the zero rest mass comes from the velocity of the photon being c.

A photon with a wavelenght of 700 nm trails the speed of light by a maximum of about 70 nm / "age-of-the-universe" if it has the maximum NIST rest-mass.

i don't know if you're pulling my leg. is not the velocity of a photon, by definition, the speed of light?

 

[rbj]

Who disputes its existence? You've clearly written it down, you have used it in some computations, and it can be given a physical interpretation.
The complaint is that it is probably not pedagogically good as the invariant rest-mass.

i don't think that anyone disputes that rest-mass of a particle or body is invariant. the dispute comes from saying that there is no other concept of mass (or there should be no other concept of mass) than invariant rest mass. with that we get to blanket statements that "photons are massless particles" which is, IMO, wrong when made without qualification. it is, at least, misleading.

i still would like to see the pedagogically simpler way of getting to:

[tex] p = \frac{h \nu}{c} [/tex]

without first using [itex] E = h \nu = m c^2 [/itex] and [itex] p = m v [/itex] and [itex] v = c [/itex].

i would also like to see a pedagogically simple way to get to:

[tex] E^2 = \left( m_0 c^2 \right)^2 + (p c)^2 [/tex]

without first using [itex] E = m c^2 [/itex], [itex]p = m v[/itex] and

[tex] m = m_0 \left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}} [/tex].

I don't see how 4-velocities and the energy-momentum 4-vector is a pedagogically simpler way (than differentiating "mass" from "rest mass") to get there for a novice. since the issue is: is it pedagogically better (whatever that is) to tell a neophyte that has some concept of physics that "photons are massless particles" or to tell that person "photons have mass, but no rest mass" (and to have to explain the difference between "mass" and "rest mass")? i would like to see the former explained. to introduce the concept, how do you get there without relativistic mass and [itex]E = m c^2 [/itex] being total energy, not just rest energy?

 

[Hans de Vries]

i have never, ever heard of a non-zero rest mass for photons. could you give a reference to that NIST statement attributing a non-zero rest mass to a photon?

It comes from the Particle Data Group actually:

http://pdg.lbl.gov/2005/listings/contents_listings.html

http://pdg.lbl.gov/2005/tables/gxxx.pdf

It's of course an upper bound for the photon rest-mass. It doesn't mean
that they have measured a rest-mass at all.

no, it is a 0/0 problem, but there is a defined energy from [itex] E = h \nu [/itex]. the zero rest mass comes from the velocity of the photon being c.

I don't know if we can be completely sure about this. Testing if a value
is mathematically zero seems to be in principle impossible. The best one
can do is determine upper bounds. For the general formula m/m_o to be
valid at all times one needs a photon rest-mass which can be infinitely
small but must stay positive.

i don't know if you're pulling my leg. is not the velocity of a photon, by definition, the speed of light?

The physical meaning of c is given by:

[tex]
\left(\frac{\partial^2 }{\partial t^2} - c^2 \frac{\partial^2
}{\partial x_2} - c^2 \frac{\partial^2 }{\partial y^2} - c^2
\frac{\partial^2 }{\partial z^2}\ \right) \phi\ =\ m_o^2 \phi
[/tex]

Where phi is a scalar, a spinor or a vector in case of spin 0, spin 1/2 or
spin 1 particles. For a massless spin 1 particle this leads to Maxwell's
equations. The lefthand side operator represents a deconvolution with
the light-cone. This means that c is the speed of information propagation
which is independent of the physical speed of the particle. Regards, Hans.

 

[rbj]

It comes from the Particle Data Group actually:

http://pdg.lbl.gov/2005/listings/contents_listings.html

http://pdg.lbl.gov/2005/tables/gxxx.pdf

It's of course an upper bound for the photon rest-mass. It doesn't mean
that they have measured a rest-mass at all.

I don't know if we can be completely sure about this. Testing if a value is mathematically zero seems to be in principle impossible. The best one can do is determine upper bounds. For the general formula m/m_o to be valid at all times one needs a photon rest-mass which can be infinitely small but must stay positive.

but the reciprocal of that formula

[tex] m_0 = m \sqrt{1 - \frac{v^2}{c^2}} [/tex]

makes perfect sense with a zero [itex]m_0[/itex], even if [itex]m = E/c^2 > 0[/itex]. it just requires that the particle speed be [itex]c[/itex].

The physical meaning of c is given by:

[tex]
\left(\frac{\partial^2 }{\partial t^2} - c^2 \frac{\partial^2} {\partial x_2} - c^2 \frac{\partial^2 }{\partial y^2} - c^2 \frac{\partial^2} {\partial z^2}\ \right) \phi\ =\ m_0^2 \phi
[/tex]

Where phi is a scalar, a spinor or a vector in case of spin 0, spin 1/2 or
spin 1 particles. For a massless spin 1 particle this leads to Maxwell's
equations.

i thought it was the other way around. i thought that Maxwell's Equations were a unification of observations of electromagnetic behavior that was quantified, that is Gauss's Law (which come from the static inverse-square law), Faraday's Law, Ampere's Law, and the fact that magnetic monopoles are (or at least were) believed to not exist. set charge density to zero (which sets current density to zero) which is what you get in free space, plug in the [itex] \mathbf{E} [/itex] of one equation into the [itex] \mathbf{E} [/itex] of another, and same for [itex] \mathbf{B} [/itex] and you get the wave equation above. Maxwell's Equations come first, and the wave equation results and the propagation speed of the wave is [itex] c = 1/\sqrt{\epsilon_0 \mu_0}[/itex].

The lefthand side operator represents a deconvolution with
the light-cone. This means that c is the speed of information propagation
which is independent of the physical speed of the particle.

that's pretty hard to accept. we say that E&M radiation has wave-like properties (for some phenomena) and particle-like properties (for other observed phenomena) and i don't know that any physicists know precisely what light is. sometimes it acts like a wave and other times it acts like a bunch or particles with energy. both are true. light is waves and light is photons. anyway, to say that the wavefront of light arrives before the counterpart photons makes absolutely no sense. it is like saying the speed of light is [itex] c [/itex], but the speed of the light is less than [itex] c [/itex]. that is self-contradictory.

anyway, to get back to pedagogy and how to talk (to newbies and pedestrians) about light, particles, energy, and mass, i can't understand why anyone would claim that concepts like "a spinor or a vector in case of spin 0, spin 1/2 or spin 1 particles" or "4-velocities and the energy-momentum 4-vector" so that you can say that photons are massless particles is pedagogically simpler than saying "photons have mass, but no rest mass" and having to explain what (relativistic) mass is as opposed to rest mass (invariant mass). that pedagogical claim, made by the deniers of photon mass, has yet to be defended. at least here.

 

[Hans de Vries]

i thought it was the other way around. i thought that Maxwell's Equations were a unification of observations of electromagnetic behavior that was quantified, that is Gauss's Law (which come from the static inverse-square law), Faraday's Law, Ampere's Law, and the fact that magnetic monopoles are (or at least were) believed to not exist. set charge density to zero (which sets current density to zero) which is what you get in free space, plug in the [itex] \mathbf{E} [/itex] of one equation into the [itex] \mathbf{E} [/itex] of another, and same for [itex] \mathbf{B} [/itex] and you get the wave equation above. Maxwell's Equations come first, and the wave equation results and the propagation speed of the wave is [itex] c = 1/\sqrt{\epsilon_0 \mu_0}[/itex].

It's just to show that Maxwell's equations fit in a broader picture of
wave functions governed by similar equations.

that's pretty hard to accept. we say that E&M radiation has wave-like properties (for some phenomena) and particle-like properties (for other observed phenomena) and i don't know that any physicists know precisely what light is. sometimes it acts like a wave and other times it acts like a bunch or particles with energy. both are true. light is waves and light is photons. anyway, to say that the wavefront of light arrives before the counterpart photons makes absolutely no sense. it is like saying the speed of light is [itex] c [/itex], but the speed of the light is less than [itex] c [/itex]. that is self-contradictory.

These last statements are incorrect indeed, don't attribute them to me.
The speed c governs much more then just "the speed of light" It covers
the speed of all interactions. The fact that c is a basic ingredient of the
equations of all massive particles does not mean that these particles
have to move at the speed of light.

It means that the value of the wave function at each point is propagated
to all directions with speed c. Even so, this can lead to stable solutions
which don't move at c but at any speed below c. For example the hydrogen
solutions.

If the photon would have a rest-mass equal to the currently established
upper bound then light of 700 nm (waves and photons) would be delayed
by 70 nm after traveling for 13 billion years or so.

anyway, to get back to pedagogy and how to talk (to newbies and pedestrians) about light, particles, energy, and mass, i can't understand why anyone would claim that concepts like "a spinor or a vector in case of spin 0, spin 1/2 or spin 1 particles" or "4-velocities and the energy-momentum 4-vector" so that you can say that photons are massless particles is pedagogically simpler than saying "photons have mass, but no rest mass" and having to explain what (relativistic) mass is as opposed to rest mass (invariant mass). that pedagogical claim, made by the deniers of photon mass, has yet to be defended. at least here.

Talking about mass as either rest-mass or relativistic mass is a matter
of taste. I don't really care as long as it is well defined and understood.
It just that in general "mass" is reserved for rest-mass.


Regards, Hans.

 

[robphy]

Who disputes its existence? You've clearly written it down, you have used it in some computations, and it can be given a physical interpretation.
The complaint is that it is probably not pedagogically good as the invariant rest-mass.

i don't think that anyone disputes that rest-mass of a particle or body is invariant.

I agree... but that's not what I was referring to.
Explicitly, I don't think anyone disputes that the notion of "relativistic mass" exists.

the dispute comes from saying that there is no other concept of mass (or there should be no other concept of mass) than invariant rest mass. with that we get to blanket statements that "photons are massless particles" which is, IMO, wrong when made without qualification. it is, at least, misleading.

Agreed... as it is also wrong (or at least misleading) to make blanket statements that "mass depends on velocity" or "photons have mass". (In my opinion, it is important to emphasize that photons are different from particles with rest-mass.) As we all know and easily see, SR has forced us to distinguish concepts which were once blurred by Galilean goggles into one concept of mass.

i still would like to see the pedagogically simpler way of getting to:

[tex] p = \frac{h \nu}{c} [/tex]

without first using [itex] E = h \nu = m c^2 [/itex] and [itex] p = m v [/itex] and [itex] v = c [/itex].

i would also like to see a pedagogically simple way to get to:

[tex] E^2 = \left( m_0 c^2 \right)^2 + (p c)^2 [/tex]

without first using [itex] E = m c^2 [/itex], [itex]p = m v[/itex] and

[tex] m = m_0 \left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}} [/tex].

I don't see how 4-velocities and the energy-momentum 4-vector is a pedagogically simpler way (than differentiating "mass" from "rest mass") to get there for a novice. since the issue is: is it pedagogically better (whatever that is) to tell a neophyte that has some concept of physics that "photons are massless particles" or to tell that person "photons have mass, but no rest mass" (and to have to explain the difference between "mass" and "rest mass")? i would like to see the former explained. to introduce the concept, how do you get there without relativistic mass and [itex]E = m c^2 [/itex] being total energy, not just rest energy?

Well... if the only goal is obtain these formulas, then your method may be fine, which can be also be accomplished by using the combination [tex]m_0 \left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}}[/tex] without giving it a name or providing more than a passing statement of a possible interpretation. You can of course go ahead and call it [tex]m[/tex] and call it "relativistic mass" and run with it.

If the goal is appreciate to relativity as a whole and its methods and interpretations, then I would choose another route.

The route you provide above is probably one seen in standard introductory textbooks.. and it works to some extent... and has probably been refined and woven into our common knowledge [of how to think about relativity from a standard intro physics textbook viewpoint]. Of course, most intro textbooks follow a rough historical storyline when discussing Modern Physics... so part of the pedagogical appeal comes from following the historical struggles with these ideas.

So, any other route I might offer, say, using spacetime-vectors, as advocated by Minkowski, would probably meet with resistance in the standard textbook community... especially if it doesn't get woven into the story... or have a story woven around it. [I'm actually working on a storyline.]

 

[rbj]

... as it is also wrong (or at least misleading) to make blanket statements that "mass depends on velocity" or "photons have mass". (In my opinion, it is important to emphasize that photons are different from particles with rest-mass.)



Well... if the only goal is obtain these formulas, then your method may be fine, which can be also be accomplished by using the combination [tex]m_0 \left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}}[/tex] without giving it a name or providing more than a passing statement of a possible interpretation. You can of course go ahead and call it [tex]m[/tex] and call it "relativistic mass" and run with it.

if no concept of name "relativistic mass" and [itex]m[/itex] is created, what is the means of getting to the true momentum of a body of mass [itex]m_0[/itex] as observed by someone as it is flying by at velocity [itex]v[/itex]? how do we get to:

[tex]p = m_0 v \left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}}[/tex] ?

let's assume we know about the invariancy of [itex]c[/itex], time dilation, length contraction, Lorentz transformation of coordinates, and velocity addition. without an idea of a different "relativistic mass" and plugging that into what we previously defined momentum to be [itex] p = m v [/itex], how do we get the more accurate expression of momentum above?

If the goal is appreciate to relativity as a whole and its methods and interpretations, then I would choose another route.

The route you provide above is probably one seen in standard introductory textbooks.. and it works to some extent... and has probably been refined and woven into our common knowledge [of how to think about relativity from a standard intro physics textbook viewpoint]. Of course, most intro textbooks follow a rough historical storyline when discussing Modern Physics... so part of the pedagogical appeal comes from following the historical struggles with these ideas.

So, any other route I might offer, say, using spacetime-vectors, as advocated by Minkowski, would probably meet with resistance in the standard textbook community... especially if it doesn't get woven into the story... or have a story woven around it. [I'm actually working on a storyline.]

are you working on an intro textbook? however it is, this sounds very cool.