Photon Momentum and Energy: A Smooth Transition to a Tardyon's Properties?

[nakurusil]

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.

 

[nakurusil]

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) E'=p'c, substitute it in (1) and obtain:

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

Meir told him this, I told him this, you told him this but for an unexplainable reason, Bernhard clings to his u'

 

[robphy]

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')
p=\cosh\theta p' (1 + \frac{c \tanh\theta}{c\tanh \phi'})
(2) E=gE'(1+Vu'/c^2)
E=\cosh\theta E' (1 + \frac{c\tanh\theta\ c\tanh \phi'}{c^2})
which don't look very recognizable [in terms of an analogy to Euclidean geometry] like (upon minor rearrangement)
E=\cosh\theta E' + \sinh\theta(cp')
p=\sinh\theta(E'/c) + \cosh\theta p'
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 \tanh\phi&#039; = 1[/tex] &quot;.<br /> <br /> The &quot; setting of u&#039; = 0 &quot; corresponds to the &quot; setting of \tanh\phi&amp;#039; = 0[/tex] &amp;quot;, which has a problem in (1) which needs to be addressed by taking a limit involving p&amp;#039;/u&amp;#039; .Here are some thoughts on the tardyon to photon &amp;quot;limit&amp;quot;.&lt;br /&gt; &lt;ul&gt; &lt;li data-xf-list-type=&quot;ul&quot;&gt; For a timelike 4-momentum-vector (representing a tardyon, a massive particle) (E/c) &amp;gt; |p|.&lt;br /&gt; For a lightlike 4-momentum-vector (representing a photon, a massless particle) (E/c) = |p|.&lt;/li&gt; &lt;li data-xf-list-type=&quot;ul&quot;&gt;There is no finite set of boosts that will make a tardyon into a photon.&lt;br /&gt; Geometrically, you can&amp;#039;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.&lt;/li&gt; &lt;li data-xf-list-type=&quot;ul&quot;&gt;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.&lt;/li&gt; &lt;li data-xf-list-type=&quot;ul&quot;&gt;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&amp;#039;.)&lt;/li&gt; &lt;/ul&gt;&lt;br /&gt; Pedagogically, I think that the necessary explanation of the u&amp;#039;--&amp;gt;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&amp;#039;t see it forming the foundations of a pedagogical presentation of relativity.

 

[nakurusil]

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.

 

[bernhard.rothenstein]

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

 

[nakurusil]

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.

...and everybody that answered to you said that you shouldn't attempt that since it is incorrect. We also tried to correct you by showing how you can get the right derivation starting from:

p=\gamma(p&#039;+\beta E&#039;/c)

by using E=p&#039;c

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.

...and this is for good reason, no self-respecting book would show such an incorrect approach, two or three of us tried to explain this to you.

 

[bernhard.rothenstein]

photon tardyon

...and everybody that answered to you said that you shouldn't attempt that since it is incorrect. We also tried to correct you by showing how you can get the right dericvation startig from:

p=\gamma(p&#039;+\beta E&#039;/c)

by using E=p&#039;c




...and this is for good reason, two of us tried to explain this to you.

As I see you use the plural (we) showing that you do not conspect all the answers: see the competent answer of robphy who considers that the equations are at least mathematically correct. Who are the two?
You did not explain why the statement concerning the transition from the properties of a tardyon to the properties of a photon simply stating no. Could you extend your oppinion?