Re: TomTom's stupidity (re: was always TomTom's stupidity)
- From: stevendaryl3016@xxxxxxxxx (Daryl McCullough)
- Date: 5 Oct 2005 06:41:02 -0700
schoenfeld1@xxxxxxxxx says...
>Daryl McCullough wrote:
>> >> Nobody said otherwise. You are confusing two different things:
>> >> (1) Is it possible to use SR with noninertial coordinates? Yes,
>> >> it certainly is possible.
>> >
>> >You cannot do this without additional postulate(s).
>>
>> Maybe you can't, but I certainly can.
>
>If that is the case, why haven't you shown how the postulates of SR
>imply that proper distances are preserved in noninertial frames?
Who said that they imply that? All I'm saying is that if
you have a differential equation describing the dynamics
of particles in one coordinate system, for instance, the
law for force-free motion:
(d/dTau)^2 x^u = 0
then you can describe the same physics in any other coordinate
system, inertial or not: To change from coordinates x^u to
noninertial coordinates y^v (letting @/@y^v mean the partial
derivative with respect to y^v):
(d/dTau)^2 x^u
= (d/dTau) (d/dTau) x^u
= (d/dTau) ((d/dTau) y^v) @x^u/@y^v
= (d/dTau)^2 y^v @x^u/@y^v + ((d/dTau) y^v) ((d/dTau) @x^u/@y^v))
= (d/dTau)^2 y^v @x^u/@y^v
+ ((d/dTau) y^v) ((d/dTau) y^w) (@/@y^w) (@/@y^v) x^u)
Now, Let A^u_v be the matrix @x^u/@y^v, and let A*^v_u be the
inverse @y^v/@x^u. Let U^v be d/dTau y^v. Then the above equation
for y^v can be rewritten as: (multiplying through by A*)
d/dTau U^v + U^u U^w G^v_uw = 0
where G^v_uw = A*^v_q @/@y^u A^r_w.
You don't need any new *physical* assumptions to know how things
behave in terms of noninertial coordinates y^v. Just use calculus
and the chain rule.
But looking at the free-particle equation in terms of y^v, you
can see that, in addition to the usual d/dTau U^v, there is also
a velocity-dependent "force" term U^u U^w G^v_uw.
>> >What justification have you, from the SR postulates,
>> >to assume that accelerating frames preserve their proper
>> >lengths?
>>
>> Nothing physical depends on a choice of coordinates. So if
>> you have a description of physics that works in one set of
>> coordinates, then you can figure out the description in
>> any other set of coordinates. It's just calculus.
>
>Well, why don't you show the calculus then.
I gave you one example above. For any equation written
using inertial coordinates, there is a corresponding
equation (often with additional terms) using noninertial
coordinates.
>Problem:
>A ruler is stationary relative to stationary frame A.
>The ruler instantaneously accelerates to velocity 0.866c.
>What is the length of the ruler in A's frame?
>What is the length of the ruler in the rulers frame? (the proper
>length)
>
>Solve that from the SR postulates alone.
The problem is ill-posed as it stands. You can't actually
answer such a problem (either in inertial coordinates, or
noninertial coordinates) without an assumption about the
physical nature of the ruler. For example, if you accelerate
one end of a *spring*, then the other end will accelerate
with a delay, after a compression wave travels from one end
of the spring to the other. After the acceleration, the spring
will go through an oscillation period: stretching, contracting,
stretching contracting, etc.
In contrast, if you push a stick made of putty, then it will
just compress, and never spring back.
>> Do you think that any additional physical assumptions are
>> needed to use spherical coordinates to do Newtonian physics?
>> It's just calculus.
>
>But we aren't talking about inertial motion through spherical
>coordinates.
But there is no difference, in principle between using spherical
coordinates and using noninertial coordinates. Spherical coordinates
mix x, y, and z in a nonlinear way to produce r, theta, and phi.
Inertial coordinates mix x, y, z, and t in a nonlinear way to
produce new coordinates X, Y, Z, and T. The same issues come up
either way.
>We are talking about non-inertial motion through
>rectangular coordinates. You raise so many non-issues
>and red-herrings it's impossible to stay on-topic.
It's the *same* topic. A noninertial coordinate system is
just a nonlinear tranformation of an inertial coordinate
system, in exactly the same way that a curvilinear coordinate
system is a nonlinear transformation of rectangular coordinates.
The fact that you think of these as essentially different just
shows that you aren't considering time on an equal footing with
the spatial coordinates.
What, in principle, is the difference between a coordinate
change such as:
r = square-root(x^2 + y^2)
theta = arctan(y/x)
and a coordinate change such as:
X = square-root(x^2 - c^2 t^2)
T = arctanh(ct/x)
They are both just nonlinear changes of coordinates.
>> >Consider a 1 unit ruler initially at rest in frame A.
>> >At time t = 0 the ruler is accelerated instanteously
>> >to velocity v = 0.866c.
>>
>> That's ambiguous. Do you mean that, as measured in frame
>> A, both ends accelerate at the same time? (Perhaps by attaching
>> rockets to both ends, and setting them off at a designated time)
[stuff deleted]
>> In the first case, the proper length will increase,
>
>Well then you should re-learn your noninertial SR because the proper
>length of the ruler does not change in this model.
Well, then you should re-learn your inertial SR because the proper
length of the ruler *does* change in this model. You're making a
common mistake in understanding accelerating rods. If you have
a rod oriented left-to-right, and you accelerate it towards the
right, then:
1. If the proper length of the rod remains constant, then
the right end of the rod must accelerate *less* than the
left end of the rod.
2. If the right end of the rod and the left end of the rod
accelerate identically, then the proper length of the rod
will *increase*.
The intuitive way to see number 1 is to use length contraction:
Initially, the left end of the rod is at x=0, and the right
end of the rod is at x=L. After the acceleration is over,
the left end of the rod will be at some new position x_final
(and moving at speed v towards the right) and the right end
of the rod will be at some new position x_final + L/gamma
(and also moving at speed v towards the right). The total
distance travelled by the left end is x_final. The total
distance travelled by the right end is x_final + L/gamma - L,
which is *smaller* than x_final. Clearly, the right end
travels a smaller distance than the left end, and so they
do not accelerate identically.
>The proper length of the ruler remains constant in this model.
No. If you accelerate both ends in the same way, then the
length in the initial frame remains constant, but the length
in the instantaneous rest frame of the ruler *increases*.
--
Daryl McCullough
Ithaca, NY
.
- References:
- Re: TomTom's stupidity (re: was always TomTom's stupidity)
- From: schoenfeld1
- Re: TomTom's stupidity (re: was always TomTom's stupidity)
- Prev by Date: Re: The true crackpots
- Next by Date: Re: Sal the crackpot
- Previous by thread: Re: TomTom's stupidity (re: was always TomTom's stupidity)
- Next by thread: Re: TomTom's stupidity (re: was always TomTom's stupidity)
- Index(es):
Relevant Pages
|
