Photon Mass: Is Student Learning Misleading?

[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 $\gamma$s or $m_0$s, 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 $\gamma$s or $m_0$s, 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.'