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

[John Denker <jsd@av8n.com>]

On 06/23/2009 07:27 AM, chuck britton wrote:

> Nomex undies have been donned!!

Shouldn't be needed.  The following question is a perfectly
reasonable well-posed question.  

Fortunately, the question has a good answer.

> Is it time to ditch the Lorentz equations?
> Some will say DEFINITELY.

Yes, we will, for good reason, as explained below.

> 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).

Fair enough.

But you are evidently more than halfway there.  If your standard
of length (the ruler) has a non-standard length, and your standard
of time (the clock) keeps non-standard time, that's awfully counter-
intuitive, and finding something "as intuitive" (or indeed much 
more intuitive) is not much of a challenge.

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

True.

That's all the more reason to make sure we interpret it correctly.

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

Quite a lot more can be said.  Quite a lot more has been said.
Alas, some of what has been said is untrue.

> 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?)

When using invariant mass, a factor of gamma is needed in some 
of the equations of motion, but not as many as you might think, 
and particularly not in the E=mc^2 equation.

Also note that hypothesizing a velocity-dependent mass does *not*
suffice to get rid of all the factors of gamma, unless you want
to have several different velocity-dependent masses!  See below.

> 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?)

The correct interpretation is that m is the rest mass and E is 
the rest energy, *not* the total energy.

> 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.)


It just works.  Try it.  Find the mass (i.e. rest mass) of the reactants
and the mass (i.e. rest mass) of the products.  Subtract.  Multiply by c^2.
That tells 'real people' how much mass is converted to non-mass forms of
energy.

  Historical note, FWIW:  I have never seen any credible evidence that 
  Einstein ever intended the E=mc^2 equation to apply in any frame other 
  than the rest frame of the particle.  If anybody has such evidence, 
  please speak up.

  Or don't bother.  I don't really care who thought what about velocity-
  dependent mass back in 1905;  I care what the right answer is.  Einstein 
  got some things wrong.  We are not required to do everything the way it 
  was done in 1905.  

To repeat: By interpreting m as the invariant mass, you do *not* lose any
ability to calculate mass/energy conversions.

Also note that the critical expression for momentum, when written in
terms of invariant mass, takes the simplest possible form, with no
factors of gamma:

      p = m d(x)/d(tau)                      [5]

Here p is the four-momentum, m is the mass (i.e. the invariant mass), x is 
the four-vector position, tau is the proper time, and of course d(x)/d(tau) 
is the four-velocity.   Equation [5] works at all speeds from zero on up.

By way of contrast, if you were silly enough to define velocity as d(x)/t(t)
using the coordinate time t rather than the proper time tau, then your
equation of motion would take the form

      p = m d(x)/d(t) d(t)/d(tau)             [6]

where the fudge factor d(t)/d(tau) is what we call gamma.  The gamma factor
is not noticeable at nonrelativistic speeds, so naive persons might hypothesize 
that p = m d(x)/d(t), but this hypothesis is untenable at relativistic speeds.

In equation 6, using t, which is not a Lorentz scalar but only one component
of a 4-vector, should set off all sorts of alarm bells.  You should know
without even a moment's thought that any expression involving d(x)/d(t) is 
likely to require ugly fudge factors (gamma factors).

So, as Henny Youngman would say, don't do that!  Use equation [5] instead.

The four-momentum p i.e. m d(x)/d(tau) is conserved.  The naive hypothetical
momentum-like quantity m d(x)/d(t) is not conserved.

Bottom line:  Using velocity-dependent definitions for mass is a mistake.

The right way to do things has been known since 1908, i.e. for 101 of the 104 
years that relativity has existed.










On 06/23/2009 07:53 AM, Moses Fayngold wrote:
 
> > The outcome qBR/v = qB/w turns out to depend on velocity, or, 
> > which is the same, the cyclotron frequency w depends on v. The latter
> > is the signature of velocity-dependent mass.
> >
> > That's it! 

That's it, indeed.  That's one of the many ways to show the absurd
inconsistency of the notion of velocity-dependent mass.

 -- For circular motion, as in the aforementioned cyclotron, the 
  3-acceleration (based on the coordinate time, t) differs from the 
  proper acceleration by two factors of gamma.  
 -- For straight-line motion, the 3-acceleration differs from the proper
  acceleration by three factors of gamma.
 -- Meanwhile the total energy (rest energy plus kinetic energy) differs 
  from the rest energy by one factor of gamma.

These are well-known results;  for a derivation see e.g.
  http://www.av8n.com/physics/spacetime-acceleration.htm

If you want to account for all three of these phenomena using velocity-dependent 
mass, it must be defined as rest mass times gamma, rest mass times gamma^2, 
and rest mass times gamma^3 all simultaneously.

Let us know how that works out for you.