Photon momentum and energy

  • #1
bernhard.rothenstein
991
1
photon momentum and energy is a frequent topic on the forum. consider a tardyon (u<c) the momentum of which transforms as
p=gp'(1+V/u') (1)
E=gE'(1+Vu'/c^2) (2)
state that special relativity theory ensures a smooth transition from the properties of the tardyon to the properties of a photon and make in (1) and (2) u=u'=c in order to obtain in its case
p(c)=gp'(c)(1+V/c) (3)
E(c)=gE'(c)(1+V/c) (4)
Is there more to say?
sine ira et studio
 

Answers and Replies

  • #2
Meir Achuz
Science Advisor
Homework Helper
Gold Member
3,613
172
Why use u at all? Why not just p=g(p+VE) and E=g(E+vp)?
 
  • #4
bernhard.rothenstein
991
1
photon (energy and momentum)

Why use u at all? Why not just p=g(p+VE) and E=g(E+vp)?

Thanks. Because what I propose and what you propose represent one and the same thing (you consider c=1 as I see). Presenting as
p=p'g(1+V/u') and E=E'g(1+Vu'/cc) in the case of a tardyon I emphsize the part played by its speed in the I' inertial reference frame. I think the two equatioins are correct.
I state now (that is my right) that in special relativity we have a smooth transition from the equations which describe a tardyon to the equation which describe a photon i.e. if in the equations presented above i make u'=c I should recover the equations which account for the behaviour of a photon
p(c)=gp'(c)((1+V/c)/(1-V/c))^1/2
E(c)=gE'(c)((1+V/c)/((1-V/c(^1/2

showing that in the case of a photon the transformation is performed in both cases by the same Doppler factor.
Please let me know if there is a flow in my way of thinking.
sine ira et studio:rofl:
 
  • #5
bernhard.rothenstein
991
1
photon energy (mass??) momentum

Doesn't look right. Here are the correct ones:

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html

With all respect but I think your answers are stante pede)
If you look with more attention to the equations I propose and to the equations to which you send me you will se that they represent one and the same thing. My presentation presents the advantage that it underlines the part played by the speed of the tardyon in the two inertial reference frames between which we perform the transformation. Thank you for your participation at my thread and let me know if there is some flow in my presentation.:rofl:
 
  • #6
nakurusil
329
0
With all respect but I think your answers are stante pede)
If you look with more attention to the equations I propose and to the equations to which you send me you will se that they represent one and the same thing. My presentation presents the advantage that it underlines the part played by the speed of the tardyon in the two inertial reference frames between which we perform the transformation. Thank you for your participation at my thread and let me know if there is some flow in my presentation.:rofl:

Energy-momentum forms the coordinates of a 4-vector that transform together, i.e. each expresion contains both energy and momentum. Your equations transform momentum separate from the energy transformation. Can you prove that they mean the same thing as the standard transformations?
 
Last edited:
  • #7
bernhard.rothenstein
991
1
from the tardyon to the photon

Energy-momentum forms the coordinates of a 4-vector that transform together, i.e. each expresion contains both energy and momentum. Your equations transform momentum separate from the energy transformation. Can you prove that they mean the same thing as the standard transformations?

Thanks for your help. IMHO special relativity does not start and does not end with the four vectors. Yes I can. Start with (I have in mind the OX(O'X') components of the momentum. Start with (W.G.V. Rosser, Classical Electromagnetism via Relativity, (Butherworth 1967) p.14)
p=g(p'+VE'/cc)=gp'(1+V/u') (1)
E=g(E'+Vp')=gE'(1+Vu'/cc) (2)
(O.K.?)
What I have in mind is that the functions (1+V/u') and (1+Vu'/cc) have the same limit for u'=c and that is why I say that special relativity theory ensures a smooth transition from the equations that describe the behaviour of a tardyon to those that describe the behaviour of a photon.
I trhink that we can express the right hand side of (1) as a function of E' only and the right side of (2) as a function of p' only. Special relativity is a flexible theory.
Please let me know your oppinion.
sine ira et studio
 
  • #8
nakurusil
329
0
photon momentum and energy is a frequent topic on the forum. consider a tardyon (u<c) the momentum of which transforms as
p=gp'(1+V/u') (1)
E=gE'(1+Vu'/c^2) (2)

The above formulas are not correct. The RHS needs to be in expression in BOTH p' AND E'. You must be doing something like assuming E'=p'c but this is true for photons only (massless particles), and not for tardyons as you are claiming in your opening sentence above.
 
  • #9
nakurusil
329
0
Thanks for your help. IMHO special relativity does not start and does not end with the four vectors. Yes I can. Start with (I have in mind the OX(O'X') components of the momentum. Start with (W.G.V. Rosser, Classical Electromagnetism via Relativity, (Butherworth 1967) p.14)
p=g(p'+VE'/cc)=gp'(1+V/u') (1)
E=g(E'+Vp')=gE'(1+Vu'/cc) (2)
(O.K.?)

I am afriad that it is not ok.
You are starting with
p=g(V)*(p'+VE'/c^2) (1)
E=g(v)*(E'+Vp') (2)


This is ok (it is the standard 4-vector transformation). Then you are doing something strange to the above formulas that brings in u'. What is u'? Where is it coming from? What does all of this have to do with photons? How do you derive the second set of equalities?
 
  • #10
bernhard.rothenstein
991
1
tardyon to photon transition

I am afriad that it is not ok.
You are starting with
p=g(V)*(p'+VE'/c^2) (1)
E=g(v)*(E'+Vp') (2)


This is ok (it is the standard 4-vector transformation). Then you are doing something strange to the above formulas that brings in u'. What is u'? Where is it coming from? What does all of this have to do with photons? How do you derive the second set of equalities?[/QU:approve: QUOTE]

(1) and (2) hold in the case of a tardyon that moves with u relative to I and with u' relative to I'. I take into account that in I' E' and p' are related by
p'=Eu'/cc (3)
and I try to express the right hand side of (1) as a function of p' and u'
and the right hand side of (2) as a function E' and u'. I gain that way the functions (1+V/u') (4) and (1+Vu'/cc) (5) which have the property that they take the same algebrfaic structure for u'=c. The result is that making in (1) and (2) u'=c I obtain the transformation equations for the photon. Is that strange?
Is that the game of hazard or the result of the fact that special relativity theory ensures a smooth transition from the equations which describe the behaviour of a tardyon to those which describe the behaviour of the photon? I mention that the functions (4) and (5) are present in the addition law of relativistic velocities
u=u'(1+V/u')/(1+Vu'/cc) ensuring the fact that for u'=c we have u=c.
Thanks for giving me the opportunity to discuss a problem to which I hope others will also express theirs point of view.
 
  • #11
Meir Achuz
Science Advisor
Homework Helper
Gold Member
3,613
172
I think your fight could be resolved by realizing that u=p/E and u'=p'/E',
so BR is not wrong. I just think bringing in u, which has awkward transformation properties, leads to confusion (evidenced here) compared to just sticking with p and E, where it is trivial. You don't even have to consider any limit in treating photons. I long ago stopped using u (and c)because u is so close to 1 in HE physics that it is useless.
 
  • #12
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
6,650
2,016
In one sentence, can someone define u' physically?
Can u' be zero?
 
  • #13
nakurusil
329
0
(1) and (2) hold in the case of a tardyon that moves with u relative to I and with u' relative to I'. I take into account that in I' E' and p' are related by
p'=Eu'/cc (3)

Where is this one coming from? Especially, where did you get the c^2 factor?
 
  • #14
bernhard.rothenstein
991
1
Where is this one coming from? Especially, where did you get the c^2 factor?
So far as I know
p=mu=Eu/cc (relativistic mass exorcised?)
Thanks
 
  • #15
bernhard.rothenstein
991
1
I think your fight could be resolved by realizing that u=p/E and u'=p'/E',
so BR is not wrong. I just think bringing in u, which has awkward transformation properties, leads to confusion (evidenced here) compared to just sticking with p and E, where it is trivial. You don't even have to consider any limit in treating photons. I long ago stopped using u (and c)because u is so close to 1 in HE physics that it is useless.

Thanks! I think not everybody has the chance to work in HE (high energy I guess?) What are you using instead u and c? Have you published somewhere your point of view?
sine ira et studio
 
  • #16
bernhard.rothenstein
991
1
photon energy and momentum

In one sentence, can someone define u' physically?
Can u' be zero?

In a previous thread I asked you if taking a limit is associated with a phyhsical procedure? My question remained without being answered. As far as I know there are at least two kinds of velocities:
1. u=proper length/coordinate time interval. The definition of the coordinate time interval involves clock synchronization, which involves the one way speed of light...Physicists also make a net difference between mean velocity (measurable?) and instantaneous velocity which as a result of a limit is not measurable. It is equal to zero when the measured proper length is zero (?)
2. U=proper length/proper time interval which does not involve clock synchronization requiring only initialization of the moving clock which measures the proper time interval with the clock of the stationary reference frame with which it is instantly at the same point in space imposing the condition that both read zero at that very moiment. Physicists establish a relationship between u and U and so the concept of instantaneous velocity can be avoided. By definition U is zero when u is zero.
I am convinced that I do not tell you anything new. I brought in discussion all that subject because I hope that I will receive competent points of view, concerning that fundamental problem and so the discussions could be usefull for the learners on the Forum as well.
Please consider that I come on the Forum with the handicap that English is not my first language which makes my sentences not very clear,:rofl: .
sine ira et studio
 
  • #17
nakurusil
329
0
So far as I know
p=mu=Eu/cc (relativistic mass exorcised?)
Thanks


Don't think so. The relationship between E and p for tardyons is:

[tex]E^2=(pc)^2+(mc^2)^2[/tex]
 
  • #18
bernhard.rothenstein
991
1
Don't think so. The relationship between E and p for tardyons is:

[tex]E^2=(pc)^2+(mc^2)^2[/tex]

But I do think so. IMHO the equation you propose m(rest mass) only proves the invariance of the left side of your equation. I find in my old and new books that the momentum of a tardyon is
p=g(u)mu
whereas its energy is
E=g(u)mcc.
Am I wrong?
 
  • #19
jtbell
Mentor
15,962
4,731
Yes, your equation [itex]p = Eu/c^2[/itex] (snipping out the [itex]mu[/itex] in the middle) is correct. I use it more often in the form

[tex]\frac{u}{c} = \frac{pc}{E}[/itex]

For example, an electron ([itex]mc^2[/itex] = 511 keV) with kinetic energy 500 keV has energy E = 511 + 500 = 1011 keV and [itex]pc = \sqrt{E^2 - (mc^2)^2} = \sqrt{1011^2 - 511^2}[/itex] = 872.35 keV, which gives u = (pc/E)c = (872.35/1011)c = 0.863c.
 
  • #20
nakurusil
329
0
But I do think so. IMHO the equation you propose m(rest mass) only proves the invariance of the left side of your equation. I find in my old and new books that the momentum of a tardyon is
p=g(u)mu
whereas its energy is
E=g(u)mcc.
Am I wrong?

Ok,

So you start with :

p=g(V)*(p'+VE'/c^2) (1)
E=g(V)*(E'+Vp') (2)

Then you use the fact that for TARDYONS ONLY (but NOT for photons) the following are true:
p=g(u)mu
E=g(u)mc^2 (I am using your notation).

This also gives you:

p=Eu/c^2

or:

E'=pc^2/u' (3)
p'=Eu'/c^2 (4)

Substitute (3) into (1) and (4) into (2) and you get:

p=g(V)p' (1+V/u') (5)
E=g(V)E' (1+Vu'/c^2) (6)

These are the relations you started with, now I understand how you got them. They are correct. Now what are you planning to do next? You cannot make u'=c as you claim in your initial post because (5) and (6) have been derived by using the tardyons equations. You have used :

p=g(u)mu
E=g(u)mc^2 remember? And they are applicable for massive particles only. You cannot turn around and apply a derivation that uses tardyon properties to photons.
In other words (3)(4) are valid for tardyons ONLY, therefore (5)(6) are valid for tardyons ONLY, so you CANNOT make u'=c because u' is the speed of a tardyon.
 
Last edited:
  • #21
bernhard.rothenstein
991
1
photon momentum energy

Ok,

So you start with :

p=g(V)*(p'+VE'/c^2) (1)
E=g(V)*(E'+Vp') (2)

Then you use the fact that for TARDYONS ONLY (but NOT for photons) the following are true:
p=g(u)mu
E=g(u)mc^2 (I am using your notation).

This also gives you:

p=Eu/c^2

or:

E'=pc^2/u' (3)
p'=Eu'/c^2 (4)

Substitute (3) into (1) and (4) into (2) and you get:

p=g(V)p' (1+V/u') (5)
E=g(V)E' (1+Vu'/c^2) (6)

These are the relations you started with, now I understand how you got them. They are correct. Now what are you planning to do next? You cannot make u'=c as you claim in your initial post because (5) and (6) have been derived by using the tardyons equations. You have used :

p=g(u)mu
E=g(u)mc^2 remember? And they are applicable for massive particles only. You cannot turn around and apply a derivation that uses tardyon properties to photons.
In other words (3)(4) are valid for tardyons ONLY, therefore (5)(6) are valid for tardyons ONLY, so you CANNOT make u'=c because u' is the speed of a tardyon.

Thanks. We are on the right way. In what concerns your question Now what are you planning to do next start please with the addition law of parallel reletivistic velocities as
u=u'[(1+V/u')/(1+Vu'/cc) (a)
which hold obviously in the case of tardyons. In many textbooks devoted to SR I see that in order to show that (a) is in accordance with the invariance of c, make in the right hand side of (a) u'=c. Are they wrong?
I make the same thing in my equations (5) and (6). Am I wrong doing so?
Am I wrong stating that at the limit (u=u'=c) all the equations which account for the behaviour of a tardyon (as an example the Doppler Effect formula in an acoustic wave and the Doppler Effect in an electromagnetic wave popagating in empty space and I could offer some other examples).
I think that the way in which our discussion takes place is a good example for many participants on the Forum. Thank you for that.
 
  • #22
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
6,650
2,016
I am concerned that the expression
p=g(V)p' (1+V/u') (5)
will have problems [which would have to be explained] when u'=0...

The expression
p=g(V)*(p'+VE'/c^2) (1)
doesn't have that problem.

It seems to me that (5) and (6) tend to hide the "mixing" of spatial and temporal components... unless the goal is somehow naturally motivate (5) and (6), then obtain (1) and (2) which do show the mixing (as one has in a Euclidean rotation).

In any case, I still have the above concern about (5).
 
  • #23
nakurusil
329
0
Thanks. We are on the right way. In what concerns your question Now what are you planning to do next start please with the addition law of parallel reletivistic velocities as
u=u'[(1+V/u')/(1+Vu'/cc) (a)
which hold obviously in the case of tardyons. In many textbooks devoted to SR I see that in order to show that (a) is in accordance with the invariance of c, make in the right hand side of (a) u'=c. Are they wrong?

They are not wrong. The expression (a) is universal.

I make the same thing in my equations (5) and (6). Am I wrong doing so?

Yes, you are wrong. It is not the same thing, expressions (5)(6) are not universal, they do not apply to photons. To make matters worse, robphy has raised concerns for u'=0.
 
  • #24
nakurusil
329
0
I am concerned that the expression
p=g(V)p' (1+V/u') (5)
will have problems [which would have to be explained] when u'=0...

The expression
p=g(V)*(p'+VE'/c^2) (1)
doesn't have that problem.

It seems to me that (5) and (6) tend to hide the "mixing" of spatial and temporal components... unless the goal is somehow naturally motivate (5) and (6), then obtain (1) and (2) which do show the mixing (as one has in a Euclidean rotation).

In any case, I still have the above concern about (5).


Excellent point. The error is in the derivation. While :


p'=Eu'/c^2 (4) is valid for all u' (except u'=c)

the other expression needed in Bernhard derivation :

E'=pc^2/u' (3) is NOT valid for the trivial case u'=0.

Bernhard derivation has too many holes to be valid.
 
Last edited:
  • #25
bernhard.rothenstein
991
1
do not mix spatial and temporal components

I am concerned that the expression
p=g(V)p' (1+V/u') (5)
will have problems [which would have to be explained] when u'=0...

The expression
p=g(V)*(p'+VE'/c^2) (1)
doesn't have that problem.

It seems to me that (5) and (6) tend to hide the "mixing" of spatial and temporal components... unless the goal is somehow naturally motivate (5) and (6), then obtain (1) and (2) which do show the mixing (as one has in a Euclidean rotation).

In any case, I still have the above concern about (5).

Thank you very much for your concern with (5). I also had but as I mentioned in a previous thread I can present (5) as well as
p=g(V)p'(1+V/u')=E'c^-2[u'+V]:rofl:
and so there are no problems with u'=0. Your hint with the mixing is essential. Are there still problems with my derivation?
Could we state that the equations which account for effects generated by moving tardyon account for u=u'=c for events generated by photons?
 
  • #26
nakurusil
329
0
Thank you very much for your concern with (5). I also had but as I mentioned in a previous thread I can present (5) as well as
p=g(V)p'(1+V/u')=E'c^-2[u'+V]

Bernhard, your derivation is full of holes, why can't you accept that? Changing the above formula doesn't change anything.

and so there are no problems with u'=0. Your hint with the mixing is essential. Are there still problems with my derivation?

Plenty of problems, like the ones outlined in my post above.

Could we state that the equations which account for effects generated by moving tardyon account for u=u'=c for events generated by photons?

No.
 
  • #27
bernhard.rothenstein
991
1
from tardyon to photon

They are not wrong. The expression (a) is universal.



Yes, you are wrong. It is not the same thing, expressions (5)(6) are not universal, they do not apply to photons. To make matters worse, robphy has raised concerns for u'=0.

I appreciate the help of robphy. His crticism is constructive helping me out from the gap. My answer to its criticism is

Thank you very much for your concern with (5). I also had but as I mentioned in a previous thread I can present (5) as well as
p=g(V)p'(1+V/u')=E'c^-2[u'+V]
and so there are no problems with u'=0. Your hint with the mixing is essential. Are there still problems with my derivation?
Could we state that the equations which account for effects generated by moving tardyon account for u=u'=c for events generated by photons?

I think that the answer holds in the case of your thread as well!
You did not answer my question:
Could we state that all equations that account for relativistic effects generated by a moving tardyon become for u=u'=c the equations which account for relativistic effects generated by photons?
 
  • #28
nakurusil
329
0
I appreciate the help of robphy. His crticism is constructive helping me out from the gap. My answer to its criticism is

Thank you very much for your concern with (5). I also had but as I mentioned in a previous thread I can present (5) as well as
p=g(V)p'(1+V/u')=E'c^-2[u'+V]
and so there are no problems with u'=0. Your hint with the mixing is essential. Are there still problems with my derivation?

Yes, you are using tardyon formulas that do not apply to photons. How many times do you need to be told?

Could we state that the equations which account for effects generated by moving tardyon account for u=u'=c for events generated by photons?

No.


I think that the answer holds in the case of your thread as well!
You did not answer my question:
Could we state that all equations that account for relativistic effects generated by a moving tardyon become for u=u'=c the equations which account for relativistic effects generated by photons?

No, a tardyon can never move at u'=c.Surely you knew that. Try making u'=c into (5) or (6).
 
  • #29
nakurusil
329
0
p=E'c^-2[u'+V]

The stuff above is not even a coherent formula.You probably mistyped.Either way, the formula does not address robphy's criticism.
 
Last edited:
  • #30
bernhard.rothenstein
991
1
photon tardyon

The stuff above is not even a coherent formula.You probably mistyped.Either way, the formula does not address robphy's criticism.
I missed g(V) a fact that is evident if you compare with the previouequations. I wonder that you do not see that the formula is coherent
momentum=momentum!
Let Robphy and others to decide. That is all. Thanks.
 
  • #31
nakurusil
329
0
I missed g(V) a fact that is evident if you compare with the previouequations. I wonder that you do not see that the formula is coherent
momentum=momentum!
Let Robphy and others to decide. That is all. Thanks.

Never mind. The formula that both robphy and I had problems with is this one:

E'=pc^2/u' (3)

which you are attempting to use in your derivation of eq (5). Obviously, it is not valid for u'=0.
So the error occurs BEFORE p=g(V)p' (1+V/u') (5).

As an aside, you are conveniently not addressing the more severe criticisms of blindly trying to apply tardyon formulas to the photon when it is well-known that they do not apply.
 
Last edited:
  • #32
nakurusil
329
0
I am concerned that the expression
p=g(V)p' (1+V/u') (5)
will have problems [which would have to be explained] when u'=0...

The expression
p=g(V)*(p'+VE'/c^2) (1)
doesn't have that problem.

Correct. All Bernhard would have to do is what was suggested to him early on, to use the obvious (and correct) [tex]E'=p'c[/tex], substitute it in (1) and obtain:

[tex]p=\gamma(V)p'(1+V/c)[/tex]

Meir told him this, I told him this, you told him this but for an unexplainable reason, Bernhard clings to his [tex]u'[/tex]
 
  • #33
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
6,650
2,016
Mathematically, I didn't see anything wrong or inconsistent with p=g(V)p'(1+V/u'). Of course, while you can't SET u'=0 in the expression, you can take the limit of the whole expression as u' --> 0. That is what "would have to be explained".

For me, it's always helpful to think geometrically.
In terms of rapidities, the equations translate to
(1) p=gp'(1+V/u')
[tex] p=\cosh\theta p' (1 + \frac{c \tanh\theta}{c\tanh \phi'})[/tex]
(2) E=gE'(1+Vu'/c^2)
[tex] E=\cosh\theta E' (1 + \frac{c\tanh\theta\ c\tanh \phi'}{c^2})[/tex]
which don't look very recognizable [in terms of an analogy to Euclidean geometry] like (upon minor rearrangement)
[tex] E=\cosh\theta E' + \sinh\theta(cp')[/tex]
[tex] p=\sinh\theta(E'/c) + \cosh\theta p'[/tex]
which, of course, follows Minkowski's viewpoint and interpretation as a spacetime geometry. I would think that composing boosts would be rather tedious in the proposed equations (1)-(2).

The " setting of u' = c " corresponds to the " setting of [itex]\tanh\phi' = 1[/tex] ".

The " setting of u' = 0 " corresponds to the " setting of [itex]\tanh\phi' = 0[/tex] ", which has a problem in (1) which needs to be addressed by taking a limit involving p'/u' .


Here are some thoughts on the tardyon to photon "limit".
  • For a timelike 4-momentum-vector (representing a tardyon, a massive particle) (E/c) > |p|.
    For a lightlike 4-momentum-vector (representing a photon, a massless particle) (E/c) = |p|.
  • There is no finite set of boosts that will make a tardyon into a photon.
    Geometrically, you can't boost a timelike-vector (whose tip is on a hyperbola asymptotic to the light cone) into a [null] lightlike-one (whose tip is on that light cone)... The rapidity (the Minkowski angle) would tend to inifinty, without ever reaching it.
  • In the spatial-velocity representation where V=c tanh(theta), the V=c limit appears as a mere finite endpoint, which [by our Galilean intuition] naively suggests that getting from-0.98c-to-0.99c is probably about as easy from getting from-0.99c-1.00c.
  • It seems the two limiting procedures are different. In one case, we boost, taking the momentum 4-vector up the hyperbola asymptotically to the light cone. (This involves V and theta.) In the other case, we slide the tip of the momentum 4-vector onto the light cone in such a way which continuously but asymmetrically changes its temporal and spatial components while discontinuously changing its invariant norm. (This involves u'.)

Pedagogically, I think that the necessary explanation of the u'-->0 limit and the breaking of the functional symmetry between E and p are too high a price to emphasize the interpretation sought. So, IMHO, while it might be an interesting interpretation [i.e. side comment], I can't see it forming the foundations of a pedagogical presentation of relativity.
 
Last edited:
  • #34
nakurusil
329
0
Pedagogically, I think that the necessary explanation of the u'-->0 limit and the breaking of the functional symmetry between E and p are too high a price to emphasize the interpretation sought. So, IMHO, while it might be an interesting interpretation [i.e. side comment], I can't see it forming the foundations of a pedagogical presentation of relativity.

Absolutely. Especially in the context of being able to get the answer by the much simpler (and rigurous) applycation of E'=p'c.
 
  • #35
bernhard.rothenstein
991
1
photon tardyon

Absolutely. Especially in the context of being able to get the answer by the much simpler (and rigurous) application of E'=p'c.

What nakurusil does not understand is that my intention was to show that
p=gp'(1+V/u') (1)
E=gE'(1+Vu'/cc) (2)
in the case of a tardyon
and
p=gp'(1+V/c) (3)
E=gE'(1+V/c) (4)

in the case of a photon could be an illustrations of the fact that the formula which accounts for an effect generated by a photon can be obtained from the corresponding formula that accounts for a simillar effect generated by a tardyon by simply replacing u=u'=c. On the basis of that I asked if the statement "Special theory of relativity ensures a smooth transition from the properties of a tardyon to the properties of a photon i.e there is not a discontinuity in that transition (0<u'=c)" is correct or it could receive a better formulation.
My humble approach is not intended to produce radical changes in the teaching of special relativity, considering that it is a good exercise. I did not find in the literature an explicit mentioning of the fact that the functions (1+V/u') and (1+Vu'/cc) have the same limit for u' going to c.
I propose to all participants on the Forum to restrain from personal addressing the discussion partner as long as they act behind a pseudonym
 

Suggested for: Photon momentum and energy

Replies
19
Views
688
Replies
2
Views
411
Replies
3
Views
138
Replies
2
Views
418
Replies
4
Views
1K
Replies
8
Views
358
Replies
36
Views
2K
  • Last Post
Replies
6
Views
490
  • Last Post
Replies
2
Views
791
Replies
21
Views
812
Top