Photon Mass: Is Student Learning Misleading?

[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 E = h \nu. 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:

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

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

$$m_0 = m \sqrt{1 - \frac{v^2}{c^2}}$$

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

The physical meaning of c is given by:

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

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 \mathbf{E} of one equation into the \mathbf{E} of another, and same for \mathbf{B} 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 c = 1/\sqrt{\epsilon_0 \mu_0}.

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 c, but the speed of the light is less than c. 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 \mathbf{E} of one equation into the \mathbf{E} of another, and same for \mathbf{B} 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 c = 1/\sqrt{\epsilon_0 \mu_0}.

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 c, but the speed of the light is less than c. 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:

$$p = \frac{h \nu}{c}$$

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

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

$$E^2 = \left( m_0 c^2 \right)^2 + (p c)^2$$

without first using E = m c^2, p = m v and

$$m = m_0 \left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}}$$.

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 E = m c^2 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 $$m_0 \left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}}$$ 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 $$m$$ 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 $$m_0 \left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}}$$ 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 $$m$$ and call it "relativistic mass" and run with it.

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

$$p = m_0 v \left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}}$$ ?

let's assume we know about the invariancy of c, 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 p = m v, 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.

 

[robphy]

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

$$p = m_0 v \left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}}$$ ?

let's assume we know about the invariancy of c, 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 p = m v, how do we get the more accurate expression of momentum above?

Maybe I misspoke or am misunderstanding something. Let me be clear (at the expense of possibly being redundant).
I'm advocating not to use the relativistic mass $$m$$, however, one could use the above expression... and one could keep the grouping $$m_0\left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}}$$ together if it helps you calculate.

How would I obtain your formula for the "relativistic momentum", that is, to say, the "spatial-component of the spacetime momentum vector" (or simply, the "spatial-component of the 4-momentum")?

For particles with nonero rest-mass, we write the 4-vector expression
$$\tilde P=m_0\tilde v$$, (quite a natural generalization)
where $$\tilde v$$ is the 4-velocity [the unit tangent-vector to the worldline...] of the particle, and
the norm of the 4-momentum is (up to conventional factors of c) the rest mass $$m_0$$. This is product of an observer-independent scalar and an observer-independent vector.

The spatial component of this 4-vector is $$(norm)\sinh(rapidity)$$, and the temporal component is $$(norm)\cosh(rapidity)$$, where the rapidity $$\theta$$ is the spacetime-angle from your observer's worldline to the worldline of the object.

In this language, the fractional relative velocity ($$\beta$$) is $$\frac{v}{c}=\tanh\theta$$;
the combination $$\left(1-\left(\frac{v}{c}\right)^2\right)^{-1/2} =\cosh\theta$$, aka $$\gamma$$;
the combination $$\left(\frac{v}{c}\right)\left(1-\left(\frac{v}{c}\right)^2\right)^{-1/2} =\sinh\theta$$, aka $$\beta\gamma$$.
So, the spatial component of the 4-momentum is
$$\tilde P_{spatial}=m_0 \sinh\theta=m_0 v \left(1-\left(\frac{v}{c}\right)^2\right)^{-1/2}$$ (after restoring the c's)
and the temporal component is
$$\tilde P_{temporal}=m_0 \cosh\theta=m_0 \left(1-\left(\frac{v}{c}\right)^2\right)^{-1/2}$$ (after restoring the c's)
You'll recognize this as the "relativistic momentum" and the "relativistic energy"... up to appropriate factors of c.

For a photon, the 4-momentum has a tangent-vector along the light cone... so it has zero norm. Further, the magnitudes of its temporal and spatial components are equal. (Rapidity cannot be used here because it is inifinite.) In spite of this, its spatial and temporal components of the 4-momentum are still interpreted as relativistic momentum and relativistic energy, and are proportional to the frequency of the photon.

By the way... to introduce force, one can write the 4-vector equation
$$\tilde F=m_0\tilde a = \frac{d}{d\tau}\tilde p$$, (quite a natural generalization).
As you may be aware, there are problems writing down an expression like
"$$\vec F=m\vec a$$", where $$\vec F$$ and $$\vec a$$ are [spatial] relativistic vectors.

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

Not quite a whole textbook... but a syllabus for teaching introductory kinematics and dynamics using [Galilean] spacetime concepts... to facilitate the future transition to Special Relativity.

 

[masudr]

how do we get to:

$$p = m_0 v \left( 1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}}$$ ?

It's a good question, and it has a good answer. The first thing to say is that the special thing about momentum is that it is conserved (in isolated systems; this caveat will apply from here on, so I will omit it). But using the p=m_0v we know for particles isn't conserved, especially when considering relativistic motion. So we say p=\gamma m_0v is conserved, and gives the p=m_0v in the non-relativistic limit. Deriving E=\gamma m_0 c^2 requires working out the work done on a particle.

I think one of the reasons I like to think of invariant mass as the mass is because I feel that mass is something that should be fundamental to a particle, not dependent upon how I'm observing it. That's what I've always known classically, and maybe, just maybe, I'm too stupid to accept that mass should depend on the observer. But I like to think it doesn't. And with the concept of invariant mass, I can still happily hold on to this "childish" concept if you will. And I get to things like four-momentum easily: p^\mu=m_0u^\mu where u^\mu = dx^\mu/d\tau.

As for getting to E^2-(pc)^2=(m_0c^2)^2 from p=\gamma m_0 v, E=\gamma m_0 c^2:

$$E=\gamma m_0 c^2$$

$$E^2=\frac{m_0^2 c^4}{1-v^2/c^2}=\frac{m_0^2c^6}{c^2-v^2} \ \ \ (1)$$

$$pc=\gamma m_0 v c$$

$$(pc)^2=\frac{m_0^2 v^2 c^2}{1-v^2/c^2}=\frac{m_0^2 v^2 c^4}{c^2-v^2}\ \ \ (2)$$

Subtracting (2) from (1):

$$E^2-(pc)^2=\frac{m_0^2c^6-m_0^2v^2c^4}{c^2-v^2}=m_0^2c^4 \frac{c^2-v^2}{c^2-v^2}=(m_0c^2)^2$$

$$\mbox{Q.E.D.}$$

I think the above is certainly pedagogically simpler, to borrow a phrase, than rbj's derivation of photon momentum. It is also more general.

Assuming we used rbj's approach, and used m=\gamma m_0 without ever writing any \gammas or m_0s, how would rbj expect beginning undergraduates to understand more advanced treatments where we need to speak of four-vectors? Please could rbj show how to get to four-momentum without invoking invariant mass.

EDIT: robphy's approach to getting relativistic momentum is perfectly acceptable, but I was trying to demonstrate an ideal approach to be given to a beginner in the subject, as it were.

 

[pervect]

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.

I think that how one learns has a big influence, one of the goals is to figure out what works well so we can encourage people to learn that first - so that they get comfortable with what works.

One of the biggest points I want to make is that people (at least ones who are serious about physics) need to be able to understand what is meant by statements like "photons have no mass".

It seems to many people don't understand, or perhaps just get argumentative when they see that statement. (The fact that they get argumentative suggests that they don't understand fully, too).

It has a definite meaning, one that people need to be able to learn, and I think to some extent that "relativistic mass" idea, learned first, is interfering with people's ability to understand what physicists mean when they say "photons have no mass".

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].

Some good points, that I'll have to think over a bit. It seems to me that the traditional formulation of Maxwell's equations (rather than the tensor form) sticks around because it is useful, and because it does take some steps towards coordinate invariance (it's not all the way there, as you point out). One can still do Maxwell's equations in cartesian or polar coordinates, for instance, so they offer a freedom to tailor the coordinate system to the problem at hand.

 

[robphy]

EDIT: robphy's approach to getting relativistic momentum is perfectly acceptable, but I was trying to demonstrate an ideal approach to be given to a beginner in the subject, as it were.

Yes, my approach does require a little preparation... but not that much if it's done right. I like the structure of the approach... it answers why those are the combinations of factors. I feel the traditional approach doesn't emphasize this enough, resulting in an exercise in algebra to use these equations to get that equation.

I do like the idea that momentum conservation can be used to motivate the appropriateness of the expression $$\gamma m_0 v$$... but I haven't found a demonstration that I like which derives the explicit form of $$\gamma$$... that I can use in class, that is.

 

[masudr]

I do like the idea that momentum conservation can be used to motivate the appropriateness of the expression $$\gamma m_0 v$$... but I haven't found a demonstration that I like which derives the explicit form of $$\gamma$$... that I can use in class, that is.

Well the point is that momentum is nothing really, but a useful quantity. Of course, the reason that is the case is because it is the spatial component(s) of the four-momentum which is a proper geometric object living in Minkowski space. But that aside, momentum is something that we define because it makes writing out our equations easier. So if the \gamma is in the definition, then so be it. Time will show if it's a useful quantity or not.

Remember, we defined momentum classically, so we can define it relativistically as well, however we want (and to be consistent, they should agree in the appropriate limit, otherwise we would have to re-write old textbooks which is just plain inconvenient).

 

[robphy]

Well the point is that momentum is nothing really, but a useful quantity. Of course, the reason that is the case is because it is the spatial component(s) of the four-momentum which is a proper geometric object living in Minkowski space. But that aside, momentum is something that we define because it makes writing out our equations easier. So if the \gamma is in the definition, then so be it. Time will show if it's a useful quantity or not.

Remember, we defined momentum classically, so we can define it relativistically as well, however we want (and to be consistent, they should agree in the appropriate limit, otherwise we would have to re-write old textbooks which is just plain inconvenient).

In my opinion, it's nice when we are [mathematically and physically] forced to a definition... even nicer when we can construct it from first principles. Otherwise, it appears rather magically [http://www.thegiftedchildlearning.com/Images/miraclelx.jpg" ]... (If you choose "this" for \gamma, this definition for momentum works for momentum conservation. Would another choice work? Is it really a choice?)

 

[Ich]

At school, I got a little conufed about relativisic mass. Now I think the reason was the following:
We were introduced to the concept by calculating how a moving body with rest mass m0 would react on a force F in the rest frame. The conclusion was that it would react more inert, like having a greater mass.
My problem: the whole effect is in the transformations. If you use relativistic mass, you are no longer allowed to use the transformations. I mean, m0 transforms correctly and everything is ok until you claim that mass is increasing. Then you have to forget about the transformations (which you actually used to derive all this). It´s like first proving that everything comes from the usual time and space effects and then claiming that these don´t exist and all effects come from relativistic mass.

 

[rbj]

It's a good question, and it has a good answer. The first thing to say is that the special thing about momentum is that it is conserved (in isolated systems; this caveat will apply from here on, so I will omit it). But using the p=m_0v we know for particles isn't conserved, especially when considering relativistic motion. So we say p=\gamma m_0v is conserved,

but i'll bet the way you get that is precisely the way relativistic mass is derived.

and gives the p=m_0v in the non-relativistic limit. Deriving E=\gamma m_0 c^2 requires working out the work done on a particle.

which is how kinetic energy T = \gamma m_0 c^2 - m_0 c^2 is derived.

As for getting to E^2-(pc)^2=(m_0c^2)^2 from p=\gamma m_0 v, E=\gamma m_0 c^2:

i know how that is done. i was wondering if there was an independent method and relativistic momentum gets derived from that.

I think the above is certainly pedagogically simpler, to borrow a phrase, than rbj's derivation of photon momentum. It is also more general.

it may be more general (or better for advanced study of physics), but it is not pedagogically simpler.

$$E = m c^2$$

$$E = h \nu$$

$$m c^2 = h \nu \ \ \Longrightarrow \ \ m = \frac{h \nu}{c^2}$$

$$p = m v = m c = \frac{h \nu}{c}$$

this is more complicated than teaching someone who knows neither about this nor 4-vectors and such what they need to know about 4-vectors to do it your way?

Assuming we used rbj's approach, and used m=\gamma m_0 without ever writing any \gammas or m_0s, how would rbj expect beginning undergraduates to understand more advanced treatments where we need to speak of four-vectors? Please could rbj show how to get to four-momentum without invoking invariant mass.

the fact is, i don't know. i am a physics pedestrian myself. I'm an electrical engineer, not trained anymore in physics than was necessary for my degree (so Modern Physics and Solid-State Physics was what was required). i never took a course in GR or QM (which didn't help for the astrophysics course i did take). we had introductory SR and QM (up to the H atom) in the Modern Physics introductory course. in the EE department, we had a reasonably rigorous classical E&M course. had some applied math in diff eq., complex analysis, probability theory, matrix theory & linear algebra, numerical analysis, approximation theory, functional analysis. never really used concepts like Hamiltonian, etc. nor tensors or 4-vectors.

but for some reason if you are saying that that concepts like 4-vectors, 4-momentum, etc. is simpler than that of relativistic mass for explaining something (like what the mass of a photon is) to a neophyte or pedestrian, that seems dubious to me.

r b-j

 

[masudr]

but for some reason if you are saying that that concepts like 4-vectors, 4-momentum, etc. is simpler than that of relativistic mass for explaining something (like what the mass of a photon is) to a neophyte or pedestrian, that seems dubious to me.

I didn't use 4-vectors in my derivation of the more general result. Nor did I invoke relativistic mass. Whereas the approach shown by rbj took the same number of lines of working as my approach, I arrived at the result applicable to all particles, whereas rbj's only works for photons. Again, I remind you, no 4-vectors were brought up.

If one is expected to study a serious course on physics, then one should be prepared to put up with squares and fractions, which is all I used.

My point was when the "neophyte", having mastered relativistic mass, comes to the more general approach of tensors on manifolds, we have to re-teach them the importance of invariant mass. When both concepts are easily understood, I feel that this is a waste of time. I don't think I have any more to say on the matter, unless I am misquoted or misinterpreted.

 

[pmb_phy]

My point was when the "neophyte", having mastered relativistic mass, comes to the more general approach of tensors on manifolds, we have to re-teach them the importance of invariant mass. When both concepts are easily understood, I feel that this is a waste of time. I don't think I have any more to say on the matter, unless I am misquoted or misinterpreted.

Why reteach? It seems quite obvious how important they are to begin with. Rel-mass is simply the time component of the 4-momentum 4-vector (not energy as most people claim). And 4-vectors and invariants are widely used in SR (i.e. inertial frames). Take the sum of the 4-vector of non-interacting particles (or those who act on contact and elastically) then form the magnitude. This quantity will be invariant and is used in particle physics during analysis of particle interactions.

Pete

 

[bernhard.rothenstein]

could somebody sumarize the discussions. could that be: the photon has no rest mass, but has momentum, energy and if you want it has mass as well?

 

[Meir Achuz]

and if you want it has mass as well?

If you want, you can call apples oranges.

 

[bernhard.rothenstein]

If you want, you can call apples oranges.

yes! if they are related by
apple=c^2oranges

 

[jtbell]

If you want, you can call apples oranges.

`When _I_ use a word,' Humpty Dumpty said in rather a scornful tone, `it means just what I choose it to mean -- neither more nor less.'

`The question is,' said Alice, `whether you CAN make words mean so many different things.'

`The question is,' said Humpty Dumpty, `which is to be master - - that's all.'

(hmmm, I had to add something to get this to post... yadda yadda yadda)

 

[jtbell]

Apropos "photon mass" discussions, I can't resist adding one more Humpty Dumpty-ism:

` [...] Impenetrability! That's what _I_ say!'

`Would you tell me, please,' said Alice `what that means?`

`Now you talk like a reasonable child,' said Humpty Dumpty, looking very much pleased. `I meant by "impenetrability" that we've had enough of that subject, and it would be just as well if you'd mention what you mean to do next, as I suppose you don't mean to stop here all the rest of your life.'