Re: [Phys-l] Teaching Special Relativity

[John Denker <jsd@av8n.com>]

On 07/03/2009 08:29 AM, Richard Tarara wrote:

> OK--then help me out here, for my gen-ed students and in fact myself. 

OK.

> Low 
> velocity momentum is mass x velocity.  

The 4-momentum is *always* mass × 4-velocity.

Here of course mass means invariant mass.

> We know the velocity of a an object 
> is limited by 'c'. 

I assume that means to say that the magnitude of the 3-velocity
is limited by c.  That is, |d(x)/d(t)| < c.  Equivalently we
can say that d(x)/d(t) • d(x)/d(t) < c^2.

The corresponding expression for the 4-velocity is 
  d(x)/d(τ) • d(x)/d(τ) = -c^2 = constant.

meaning that like any other 4-vector, it is a Lorentz invariant.
A boost in the X-direction just rotates the 4-velocity in the 
XT plane.

Beware that the definition of 3-velocity is slightly squirrely:
  -- The 3-momentum is the spatial part of the 4-momentum
  -- The 3-vector position is the spatial part of the 4-vector position.
  -- The 3-velocity is d(x)/d(t) which is *not* the spatial part of
   d(x)/d(τ).  There is an extra factor of d(t)/d(τ) in there.
   This is a notorious trap for the unwary.  There's no physics in
   it;  it is just another example of misleading terminology.

Note that d(t)/d(τ) is conventionally called gamma (γ).

>  The momentum of a fast object increases faster than the 
> velocity.  

Again:  The 4-momentum is *always* mass × 4-velocity.

> Is there really any 'conceptual' way to view this other than that 
> the mass has effectively increased.  

Yes indeed.  See above.

> Saying just that the momentum increases 
> non-linearly may be the most 'correct' thing to say, but not very 
> satisfying.  

Again:  The 4-momentum is *always* mass × 4-velocity.  

There's nothing nonlinear about it.  Do you want a simpler 
expression?  Is it even possible to imagine a simpler 
expression than this???

> I know JD will say use time-space diagrams 

Not even.  For the problem of continued steady acceleration, 
the problem is so easy that we don't need to draw the diagram,
although we could if we wished.  (The world line would be a
hyperbola.)

All we really need is to take seriously the idea that time is 
the fourth dimension ... and a boost in the X direction is just
a rotation in the XT plane.

For continued steady rotation of a vector in the XY plane, i.e. 
rotation in a purely spatial direction, you would without 
hesitation formulate the problem in terms of the angle.  There 
would be a steady (linear!) rate of change of the angle.  To 
find the _slope_ given the angle, we use simple trigonometry.  
The slope is a very nonlinear function of angle, but not beyond
the reach of ordinary high-school trigonometry: 
     slope = tan(angle)

For continued steady proper acceleration, i.e. steady rotation
in the XT plane, the same logic applies.  The angle in this
situation is conventionally called the _rapidity_ (ρ) although 
the terminology is not very important;  you can just call it 
the angle if you wish.  For continued steady acceleration, the
rapidity increases at a steady (linear!) rate.  The 3-velocity 
is a nonlinear function of rapidity, but not beyond the reach of 
hyperbolic trigonometry:
     3-velocity / c = tanh(rapidity)

On any scientific calculator, students can calculate tanh(...)
just as easily as they can calculate tan(...).

Nothing happens to the mass.  The essential nonlinearity here is 
in the 3-velocity.  And that nonlinearity is entirely in the 
the projective geometry of the situation.  If you rotate a ruler, 
the length of the ruler does not change, but the length of its 
shadow (projection) does change, in a nonlinear way.  Also the 
slope changes.  It's just trigonometry.

Similar statements apply to the lifetime of the proverbial fast-
moving muon.  Nothing happens to the "clocks" onboard the muon,
or any other physics onboard the muon.  The only thing that
changes is the _projection_ of the muon's time-axis onto the
laboratory time-axis.  This is how you know that relativity
affects all particles and all clocks the same:  it's because
it has got nothing to do with the particles and nothing to do
with the clocks!  It's just the geometry of projection from one 
frame to another.

Special relativity is not weird or paradoxical; it is just the 
geometry and trigonometry of spacetime.






    Von Stund′an sollen Raum für sich und Zeit für sich 
    völlig zu Schatten herabsinken 
    und nur noch eine Art Union der beiden 
    soll Selbständigkeit bewahren.

                             Hermann Minkowski (1908)