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Re: [Phys-l] velocity-dependent mass (or not)



Nomex undies have been donned!!


On Jun 23, 2009, at Jun 23(Tue) 9:57 , Moses Fayngold wrote:

It is true that relativistic mass is essentially relativistic energy.
But it is also true that it can be independently determined as MASS
(measure of inertness - ratio of external force perpendicular to acceleration when the force is perpendicular to velocity) by an executable experimental procedure (e.g., from known charge, velocity, and radius of the circle traced out by a particle in a known magnetic field perpendicular to the plane of the circle).

Moses Fayngold,
NJIT

The results of this 'executable experimental procedure' can also be explicated in terms of momentum change, without calling on an inertial mass term as such??

I grew-up with the Lorentz equations (space-time, and mass) and only remember seeing the Minkowsky treatment in very off-hand remarks of 'another way of looking at things'.

Is it time to ditch the Lorentz equations?
Some will say DEFINITELY.
I'll reserve judgement until I become as 'intuitive' with Minkowsky diagrams as I am with shrinking meter sticks and slowing clocks (both of which are required by the underlying fact of Simultaneity Lose).

E=mc^2 is an equation that won't be going away anytime soon.

Time magazine declared it to be the 'Equation of the Century'. What more can be said ;-)

To make it acceptable to those who say that mass is ONLY to be considered the 'Rest Mass', a factor of gamma would have to be added. (right?)

Since Time magazine doesn't do 'gamma', I took to saying (to classes) that it would be better to say that

delta E = delta m c^2.

But since m must be REST mess - it's hard to see how it can 'delta'. (right?)


Bottom line - I'd like for 'real people' to be able to use this 'Equation of the Century' to calculate how much mass is converted into energy in a nuclear reactor without needing to get down and dirty with the specific nuclear equations (which are WAY too many to consider anyway).

so how can we use E = m c^2 in this regime of non-variable mass??


(maybe just by declaring all mass in the reactor to be 'at rest' - but this seems to be a bit of a cop-out.)