Re: kinematics, traditional or not

["John S. Denker" <jsd@MONMOUTH.COM>]

Tim Folkerts wrote:
> If I say an electron has "a speed of 0.999c", I think I mean that the
> electron will move a distance of 299,792,458*0.999 meters (by my
> meterstick) in one second (by my watch).

That's allowed.

> I don't see why I need to redefine v.

"Need" is too strong a word.
But you might want to;  see below.

> It seems I am still using v==dx/dt.  Sure, someone in a
> different frame will get a different answer, but that is true even in
> Galilean transformations.

Presumably everybody gets the _same_ final answer to any
physically-meaningful question.  Only the intermediate
steps look different.  I'm not sure why this point came up;
I doubt it is the main point.

Probably the more-important points have to do with
 -- convenience, and
 -- communication.

If you know for sure that you'll always be working in
the lab frame, then yes, it might be convenient to
calculate (d/d t)x to the exclusion of (d/d tau)x.
But sooner or later most people need to calculate
something in the center-of-mass frame, or some other
non-lab frame, whereupon it becomes more convenient
to deal with (d/d tau)x, which is a well-behaved
four-vector.

Another convenience issue is that the laws of physics,
notably conservation of momentum, are most simply
stated in terms of (d/d tau)x.  That quantity, times
the invariant mass, is the momentum that is conserved.

> Certainly, F <> ma in the Newtonian sense.

OK.

> I would tend to look to F = dp/dt.

That's your private force.  If you use it privately
and consistently, you'll be fine.
But when communicating with others, be prepared for the
possibility that they will define F = d(p)/d(tau)

> Then I would say that applying a force (say from a uniform E) to an
> electron that becomes highly relativistic means:
>    Force is constant
>    speed approaches c
>    acceleration approaches 0 (since the speed mostly stops changing)
>    a = F/gamma*(rest mass)
>    momentum continues to increase linearly:  dp = qE dt

OK, but note that it's linear as a function of frame-time,
not linearly as a function of proper time.

> Is this not a reasonable way to view things?

It's not completely unreasonable.  It just lacks something
in elegance.  In particular, we understand the importance of
writing the laws of physics in a rotationally-invariant
form:  It is better to write
        Force dot distance
rather than writing some more-complicated expression that
assumes that the force is always in a particular direction.

By the same token, we should sieze every opportunity to
write the laws of physics in a way that is Lorentz invariant.
We think p ought to be Lorentz invariant, so writing
        F = dp/dt
wastes an opportunity to keep things invariant.

> I would keep the gamma explicit, rather than bury it
> in either m or v or t.

In a Lorentz-invariant calculation, you'll never need to
put in a gamma of any kind.  You might see things like
(d/d tau)t ... but only if you go looking for them.

For instance, the increase-of-momentum equation above
would come out
        dp = v_0 qE d(tau)
and the reason the timelike piece of the four-velocity
appears is that E is essentially the timelike piece of
the Lorentz force equation;  the spacelike pieces of
the four-velocity interact with the magnetic field,
which is absent in this example.  But by definition,
        v = (d/d tau)x
so in particular
        v_0 = (d/d tau)t
and plugging into the previous equation we find to
nobody's surprise that
        dp = qE d(t)
So you see that in the invariant approach, the formalism
takes care of itself.  It automagically produces factors
of gamma (i.e. v_0) in just the right places.