# A question for relativists

1. Nov 21, 2007

### bernhard.rothenstein

Consider please that you face the following question:
The kinetic energy of a tardyon is given by
T=mcc[g(u)-1]
Somebody could ask: What does it become in the case of a photon?
Thanks

2. Nov 21, 2007

### Phred101.2

Can you please define all terms in less than a paragraph each?

3. Nov 21, 2007

### Ich

In the case of a photon, the formula becomes irretrievably inapplicable.
In the limit, you can argue that for the energy to remain finite, mass has to approach zero.
At v=c, the formula is meaningless and you can choose energy as you like.

4. Nov 21, 2007

### bernhard.rothenstein

question

T kinetic energy E(0)rest energy u speed

5. Nov 21, 2007

### DrGreg

The only conclusion you can reach is that anything that travels at the speed of light must have zero mass. (I.e. "invariant mass" = "rest mass".)

The equation tells you nothing about the energy of a photon. "Tending to the speed of light" doesn't really help, because it makes no physical sense -- something is either definitely slower than or definitely equal to the speed of light; something either has mass, or it hasn't. An individual particle's mass can never "tend to zero".

$$T = \sqrt{p^2c^2 + m^2c^4} - mc^2$$

which works for both tardyons (particles that move slower than light) and photons.

6. Nov 21, 2007

### neutrino

I'm not sure what you mean by "what does it become"? Do you mean to say, what happens when v=c, or what form does the expression for kinetic energy take in the case of a photon?

If it is the former, then I would ask the questioner to read the statement again: It clearly states that formula gives the KE for a tardyon, which is a particle that travels at a speed less than that of light. Therefore, the question is meaningless.

7. Nov 21, 2007

### robphy

Some of this post echoes comments by DrGreg and others.

It seems to me the focus on this discussion should be on the 4-momentum vector and its components relative to an observer (the norm of that 4-momentum does not tell the whole story).

Geometrically, the rest-mass of a particle determines a "hyperboloid in momentum space" ("the mass shell"). The 4-momentum vector $$p^a$$ of that particle has its tip on that hyperboloid, which can be decomposed by an observer $$\hat u^a$$ into two components: the "relativistic energy" (the temporal-component $$p^a \hat u_a$$) and the "relativistic momentum" (the remaining spatial-component).
The relativistic kinetic energy (the non-rest part of the relativistic energy) of that particle determined by that observer is the difference $$T=p^a \hat u_a - p^a \hat p_a=m(\cosh\theta-1)$$, which could be visualized as the "remaining length" of the temporal-component less the "radius" m of the hyperboloid.

For a photon, its "mass-shell" hyperboloid degenerates onto the nullcone... and thus has zero radius. In addition, its 4-momentum lies on the nullcone. Thus, the analogous "remaining length" is the entire size of the temporal-component [although the above expression in terms of m can't be used since $$m=0$$ and $$\cosh\theta \rightarrow \infty$$]. Physically, this means that all of the photon's energy is kinetic. [And since the 4-momentum of a real photon is finite, its energy, and thus its kinetic energy is finite.]

To take the limit from the massive to the massless case, it seems one needs to hold fixed the energy of the particle (the temporal component), then push the 4-momentum onto nullcone. So, one should use $$T=p^a \hat u_a - p^a \hat p_a$$ as the starting point, keeping $$p^a \hat u_a$$ [and $$\hat u^a$$] fixed while $$p^a$$ is varied.

(EDIT: I should mention that to attempt the limiting process while keeping the mass m constant amounts to pushing the 4-momentum vector up along the hyperboloid, with the 4-momentum asymptotically approaching but never reaching a null direction. In this process, the relativistic energy [and the relativistic kinetic energy] increases without bound to infinity.)

(EDIT: It's also worth mentioning that the relativistic kinetic energy doesn't appear to have any special geometrical significance beyond the construction described above. So, I suspect that its transformation properties, though calculable, is probably of limited value on its own. It's probably better to work with more geometrically significant quantities first, then derive this when needed.)

Last edited: Nov 21, 2007
8. Nov 21, 2007

### Phred101.2

robphy: the statement that "mass can't be used", in the case of a photon (a "massless" particle), is a remnant (that we all seem to cling to) of the physical sciences discovering that mass has momentum (so we leave "mass" in the equation, even when it's zero), and we seem unable (or extremely reluctant), to discard this notion, and substitute the concept of a field (as Einstein wanted) for the concept of mass. But mass must be meaningless to photons -their momentum is a 'result' of the energy transferred by an electron changing its (quantum energy) state. We are kind of catching up with the need to 'ignore' the older physical concepts...
I've had extended 'discussions' on other forums about photons and momentum and mass(lessness), and lots of people who claim to have an understanding of the science seem to be completely thrown by this one.

Last edited: Nov 21, 2007
9. Nov 21, 2007

### bernhard.rothenstein

Please treat with patience the following way of reasoning:
Consider a tardyon that moves with speed u relative to I and with speed u' relative to I'. I' moves with speed V relative to I all the speeds showing in the same direction. Accept the following notations: b(V)=V/c, b(u)=u/c,
b(u')=u'/c; g(u')=1/sqrt(1-u'u'/cc), E(0) rest energy of the tardyon, T and T' the kinetic energy of the same tardyon detected from I and I' respectively.
Express T as a function of physical quantities measured in I' via the addition law of parallel speeds. The result is

T=E(0)g(u'){[1+b(u')b(V)]g(V)}-E(0) (1)
Try to apply (1) in the case of a photon. Put E(0)=0, b(u')=1 in order to obtain
T=(0/0)sqrt[(1+b(V))/(1-b(V))]. (2)

From a mathematical point of view there is an indeterminacy. A physicist could say that 0/0 is the signature of a photon a particle which has zero rest energy and moves with c relative to I and I' as well.
Under such conditions T should account for the energy of the photon in I and 0/0 should account for the energy of the same photon measured in I' i.e. among other possible presentations
hniu=hniu'sqrt[(1+b(V))/(1-b(v))] (3)
One more question: Are there equations in special relativity that hold only for tardyons others holding for tardyons and photons as well?
Thanks to all the participants on my thread. I respect their oppinions.
Use please soft words and hard arguments.

10. Nov 22, 2007

### Ich

True, a reformulation of (1) without referring to E(0) gives
E'=E*(1+-uv)/(1-v²).
That's a general formula for both tardyons and photons. You can derive it also by transforming the energy-momentum vector of said particle.
This is not the first time you derived interesting relationships from invalid equations by interpreting indeterminacies. The common pattern behind this is that instead of simply transforming nullvectors, you derive their transformation properties as the limit of transformation formulas that rely heavily on the norm of the vector, like mass.
An example: m*(4-velocity)=(energy-momentum). Energy-momentum is ok for every particle, 4-velocity only for tardyons. There is a completely unnecessary gamma*m on the left side, which cancels anyways but becomes indeterminate for v=c, where you have to interpret it.
Never use formulas that depend on the norm of a vector being nonzero, whenever there are generally valid expressions.

11. Nov 22, 2007

### Staff: Mentor

I agree, it makes no sense to go from the general formula, derive a special case valid only under certain conditions, and then try to re-generalize in the limit when those conditions no longer hold. It is a lot of work with no result other than confusion.

On the other hand, within their limits special-case formulas are usually easier to compute or understand and therefore useful. I wouldn't typically dismiss them, I would just warn about ever trying to generalize from them.

Last edited: Nov 22, 2007
12. Nov 22, 2007

### bernhard.rothenstein

Thanks again. But I started by finding out a transformation equation for the kinetic energy which I think has nothing to do with four vectors. So I was obliged to start with E(0)g(u).

13. Nov 22, 2007

### Staff: Mentor

I would highly recommend that you learn the four-vector approach. It is really not very difficult, and once you learn it many concepts become clear, easy to calculate, and easy to understand.

14. Nov 22, 2007

### Ich

There's a problem, because the very nature of the Lorentz Tranformation is a rotation in spacetime. They don't work with time only (Energy), or space only (Momentum), because all they do is to mix these components.
When you derive a transformation equation for energy (or, worse, kinetic energy), you implicitely use the whole four vector. In this case, it has two relevant components. Instead of using both components explicitly, you use only the first one (energy) and include the other one indirectly by using the norm of the vector (mass). This causes unnecessary problems if the norm vanishes.
If you're concerned with the didactics of SR, I'd recommend you use the geometric approach (as robphy did earlier in this thread) rather than cementing from the beginning the time only & space only view that the students must overcome anyway.

15. Nov 22, 2007

### bernhard.rothenstein

Thanks. Please tell me why do you consider that "worse, kinetic energy". I consider that in special relativity theory there are physical quantities which transform and physical quantities which do not transform. Consider
that you are a teacher and a student of yours ask you how does kinetic energy transform? Thinking about the problem and learning from what the participants on my thread told me I have reconsidered the problem avoiding the involvment of the rest energy. I start with the definition of the kinetic energy in I (T) and in I'(T') divide the two exprtession in order to eliminate the rest energy and express the right side of the quotient as a function of u' via the addition law of parallel speeds. The result is
T=T'{(g(V)(1+b(u')b(V))-1/g(u')}/(1-1/g(u')). (1)
As you can see (1) works in the case of the tardyon but also in the case of a photon leading for b(u')=1 to
T=T'sqrt[(1+b(V))/(1-b(V))]
I am not convinced that we should mix four vectors in all the problems we have to solve. A teacher of mine told me: Do not use a machine gun when you hunt rabbits.
With respect

16. Nov 22, 2007

### robphy

One doesn't have to always explicitly invoke 4-vectors into a problem (as you may have seen from numerous posts of mine).... however, the methods should be certainly consistent with them. Ideally, it would be nice if the methods were simplified versions of them.

There is some value for a coherent and mathematically-consistent strategy [which 4-vectors and spacetime geometry provides, either explicitly or implicitly]... with emphasis on ideas which can solve current problems, as well as new ones [e.g. more advanced special relativity, general relativity, and quantum field theory]. Otherwise, one is likely leading students along a path that may have to be abandoned.

It seems to me that many of these discussions are akin to developing Euclidean geometry from the numerous ways that one can prove or obtain the Pythagorean theorem. Certainly, main of them are cute and may have some (possibly limited) pedagogical advantage... but many of them take advantage of some special property that is not likely found in more general cases that might be considered in more advanced topics.

My \$0.02, as a relativist.

Last edited: Nov 22, 2007
17. Nov 22, 2007

### Staff: Mentor

Kinetic energy transforms only as part of the timelike component of the 4-momentum.
If I were a teacher and a student asked me about kinetic energy transforming I would teach him about 4-vectors, as everyone on this thread is trying to do with you. It is an extremely easy concept to learn. If the student refused to learn 4-vectors I would have to give the student a failing grade.
Are you so certain you are hunting rabbits? A half-dozen people who understand energy and relativity all suggest that you use 4-vectors.

Trying to understand relativity without 4-vectors is a little like trying to understand Newtonian mechanics without calculus. Yes, you can memorize a bunch of separate formulas and apply them in their various domains, but you cannot see the underlying and unifying concepts and you cannot really understand how all of the separate formulas are derived.

18. Nov 22, 2007

### bernhard.rothenstein

Start from the begining:I respect your opinion! I don't know at which level do you teach phyhsics and I do not know how many time have you at your disposal to do it. Inspecting introductory textbooks I could quote (and I think that I am in a good company):
1. A.P. French, Special Relativity . Four vectors are mentioned on page 214 telling us: "We shall not go any further with it here, however, since it is not essential and really adds nothing to the basic physics of relativity."
2. David Bohm The Special Theory of Relativity Four Vectors are not mentioned at all. Even LT for energy and momentum are derived without using them using instead a very productive relativistic identity convincing the learner about the kinematic roots of relativistic dynamics.
3. Hans C. Ohanian, Special Relativity, A modern introduction in which four vectors are not mentioned at all.
4. Thomas A. Moore, A traveler's guide to space time. The four momentum vector is mentioned on page 164
I stop here. Could you mention a textbook which starts with four vectors?
Of course each teacher acts in accordance with his convictions.
De gustibus....
The problem is not that a student refuses to learn four vectors but when should he learn them.
If you continue the discussion please use soft words and hard arguments.
Kind regards

19. Nov 22, 2007

### Ich

Because kinetic energy doesn't even transform as a component of a vector. There's no easy geometric counterpart.
I tell her the same I would tell him (political correctness ) in Newtonian mechanics: it transforms awfully, because there simply isn't enough information in it. You need momentum, too.
This should equal
,unless I made an error. Remember that I got this equation by merely boosting the energy-momentum vector - which is a quite simple and intelligible operation.
Well four vectors are rather like a swiss army knive: Once you got it, it's a knife, screwdriver, botlle-opener, corkscrewer, saw, tooth pick ant whatnot.
Here is one of the rare occasions where relativity is conceptually simpler than Newtonian mechanics: you need the connection between energy and momentum anyway, but relativity brings them into context and shows that both are mere aspects of the same thing. And that a rotation (read: "boost") of that thing of course doesn't change the thing, but only our point of view. And that there are many such things, like spacetime distance or current density, which naturally behave exactly the same way.
Given the well-know geometric interpretation of relativity, I ask you (as an analogy):
You hold a stick perpendicular to your view. If you rotate it, how would you explain to your students the fact that it appears shorter? By shrinking its z-dimension and invoking calculations that connect the z-dimension with the x-dimension? Or by telling them simply that the stick is still tha same, did not shrink, but that we see only its projection?

20. Nov 22, 2007

### robphy

4-vectors are admittedly advanced, just as 3-vectors are advanced in an introductory textbook. However, what I am advocating is a use of SPACETIME concepts [akin to the distance-vs-time graphs in the introductory texts (often before the use of vectors)] instead of merely mystical "formulas of transformation". I made reference to 4-vectors in the earlier post as if I were speaking to a teacher (rather than a student) of relativity... to help clarify the concept that the teacher is pursuing in his or her line of thought. So, it is appropriate to be precise to that audience.

And kinetic energy is mentioned on pp. 171-172 in Moore.

Although the use of 4-vectors is delayed or absent in the above, the Minkowski spacetime diagram is featured prominently and many of the calculations are arranged in component-form (as one does in introductory physics) that is one step from grouping them into a vectorial quantity.

Let us recall that the original question asked about the kinetic energy of a photon.
While the final result desired might not make use of spacetime diagrams and 4-vectors, I see no harm in using them to clarify the underlying nature of the problem TO A RESEARCHER.... then, if so desired, find a non-vectorial non-spacetime approach for whatever reason or whatever target audience. In my mind, the spacetime analysis is cleaner and more rigorous than a loosely phrased question of some not-so-clearly spelled-out limiting procedure.

If the above is not what you desire, it might be better to phrase your question with something akin to "without using spacetime concepts how would you .... for an introductory audience".

...what kind of answer were you expecting to get a from a [modern] relativist, the target audience of your question?

Last edited: Nov 22, 2007