Re: [Phys-l] Teaching Special Relativity

[John Denker <jsd@av8n.com>]

On 07/04/2009 05:35 AM, Bob Sciamanda wrote:

> The driving essence of the principle of (special) relativity is that
> physics is the same in all (inertial) reference frames. 

That is consistent with -- and indeed deeply intertwined with --
the assertion I've been making, namely that special relativity
is the geometry and trigonometry of spacetime.

  By way of background, consider ordinary high-school plane
  geometry.  For this to make sense, we must require rotational
  invariance.  Geometry must recognize a triangle and a rotated
  version of that triangle as being the same.

  So it is with spacetime.  We cannot even talk about its geometry 
  unless we require invariance with respect to rotations in the XT
  plane i.e. boosts i.e. changes in velocity.

>  This means
> that measured quantities are related by the same quantitative
> equations.  This is achieved by (Lorentz) transforming the
> measurables in the invariant equations.

That's true, but it requires some sophistication to understand
what we mean by "invariant equations".  Not all correct equations
are invariant.  Suppose we have an equation written in terms of
vectors, in such a way that it is manifestly invariant with respect
to rotations.  So far so good.  If we decompose that equation in 
terms of components, relative to some particular basis, the new
version will be correct in that basis but the component-values 
will not be invariant.  The vector equation is an apt model of 
the underlying physics, while the component equations are not so 
apt.

> This is the same relativity which describes the E and B components of
> an electromagnetic field when viewed in different frames.  There is
> no universal "proper" frame in which E and B are measured as their
> "real" values. (Is it that we want to resurrect the "aether frame"?)

See previous discussion of components.  Expressing E and B as
vectors is not the most apt model of the electromagnetic field.  
If you want the field to be manifestly invariant with respect 
to boosts, you should represent it as a _bivector_ (F).  In a 
particular frame you can decompose F into an E-like component
and a B-like component, but the component values (E and B) are
strongly frame-dependent.


> The measurable quantities appearing in the physics equation are the
> physicists "reality";

If "measurable" means "directly measurable", then I must disagree.
Appearances can be deceiving, and conversely reality may be only
indirectly measurable.  Consider the folks in Plato's cave.  They
can directly measure the shadows, but they cannot directly measure
the real objects.  I have to go with Plato on this one;  the objects, 
not the measurable shadows, are the physicist's "reality".

As the icon representing this idea, I use the image of saturn
  http://www.av8n.com/physics/img48/saturn.jpg

Before the days of interplanetary spacecraft, every observation of 
saturn's rings resulted in an elliptical image.  Any direct measurement
of the image confirms that the image is an ellipse, with a quite
significant eccentricity.  Meanwhile the "physicist's reality" is
that the rings are in reality circular (to an exceedingly good 
approximation).  The apparent ellipticity is well explained by the
projective geometry of the situation.

The analogy to special relativity is direct and profound:  When folks
think they are measuring a Fitzgerald-Lorentz contracted ruler, or a
dilated clock, or a velocity-dependent mass, they are not measuring
physical reality;  they are only measuring a shadow projected onto
some particular reference frame.



On 07/04/2009 07:49 AM, Moses Fayngold wrote in part:

> > ... And how does one actually measure the invariant (rest) mass of a photon?

1) We can infer the mass in the usual way:  We can directly determine 
the energy and momentum of the photon.  We can then construct the 
(energy, momentum) 4-vector.  The gorm of this 4-vector, i.e. the 
dot product of the 4-vector with itself, is a Lorentz invariant and 
is (as always) equal to -m^2.  Theory and experiment show that m is 
zero for photons.

1a) The theory is unequivocal.

1b) If you want high accuracy, the experiment is tricky, because it 
involves a small difference between large numbers, but there are
ways to alleviate this problem.

2) Another independent way of reaching the same conclusion is to plug
a hypothetical mass into the Yukawa potential exp(-m r)/r for the 
electromagnetic interaction.  By all accounts, the electromagnetic 
interaction has infinite range, which corresponds to zero mass.  
According to some guy named John David Jackson, this has been checked 
experimentally, and any correction to the 1/r potential must be less 
than one part in 10^12.  That looks like zero to me.

> > Based on this, should we discard the notion of the invariant mass
> > for photons and use only their relativistic mass?

No.