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*From*: Moses Fayngold <moshfarlan@yahoo.com>*Date*: Wed, 26 Nov 2014 15:45:37 +0000 (UTC)

On Thursday, November 20, 2014 7:57 PM, John Denker <jsd@av8n.com> wrote:

>" ... for any particle with nonzero mass you can define

> the 4-velocity u = dr / dτ.

>...Then you can define the 4-momentum p = m u (for some "m").

>... massless particles have undefined 4-velocity > and undefined proper time, but they have a perfectly good 4-momentum"

Wouldn't it look strange if such a fundamental relativistic characteristic as 4-velocity were undefined for the most fundamental of relativistic particles - a photon? Actually, the norm of a free photon's 4-velocity is perfectly defined and equal to c, as well as for any massive particle. It is true that its temporal and at least one of its spatial components are infinite, but we do not usually consider this as undefined, just as we do not call "undefined" the mass or charge density of a point particle. At least these infinite components produce the finite norm since they are of the same power and enter the expression for norm with opposite signs. I think, the revelation that all objects of Nature flow through space-time with the same invariant 4-speed u=c is one of the most beautiful discoveries of the theory of relativity. While all things such as a bullet or even a stationary rock enjoy, according to this discovery, a privilege to have invariant 4-speed u=c, it would be really surprising if such a privilege would be "undefined" for a photon.

>" Suppose we pick out the timelike component of the > 4-momentum in some chosen frame, and choose to call it

> the "energy". Then by the basic definition of norm of

> a 4-vector, we have

E^2 - p_xyz^2 = m^2 [1]

> where E is the timelike part of the momentum and p_xyz

> is the spatial part ... "

True

The classical energy is not defined uniquely; you can

shift it by a gauge transformation. Therefore you can

always pick a gauge so as to /make/ the classical energy

equal to mc^2 -- which an awful lot of people have done --

and it would even be consistent with the correspondence

principle. However, you could equally well make it equal

to (mc^2 / 2) or any other thing that suits your fancy."

The last two quotations do not look consistent with each other and with [1] (with c taken as c=1). First, it is universally acknowledged that E for a single object is defined under condition that the value of 4-potential of any external field at infinity is taken as a reference point. This fixes the specific gauge. Second, E=mc^2 follows directly from [1] at p_xyz =0, as the equivalence between the rest mass m and rest energy E. But generalized equivalence E(v)=m(v)c^2 also follows directly from [1] if you express there p in terms of m and v, denote m gamma(v) =m(v) and call it relativistic mass (which turns out to be exactly equal to transverse mass - nothing undefined here). It is totally irrelevant whether one likes these terms and notations or not - this is just mathematical corollary of [1]. And more important, [1] is, in turn, corollary of E(v)=m(v)c^2. In other words, these two equations are mathematically equivalent. Therefore, if we accept [1], we cannot discard E(v)=m(v)c^2 as generalized mass-energy equivalence.

Moses Fayngold,NJIT

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**References**:**Re: [Phys-L] How Einstein discovered E = mc2***From:*John Denker <jsd@av8n.com>

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