| Chronology | Current Month | Current Thread | Current Date |
| [Year List] [Month List (current year)] | [Date Index] [Thread Index] | [Thread Prev] [Thread Next] | [Date Prev] [Date Next] |
-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu
[mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf
Of John Denker
Sent: Thursday, June 17, 2010 10:43 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] how to explain relativity
On 06/17/2010 05:50 AM, Jeffrey Schnick wrote:
John,
You seem to be solving a different problem.
I see no evidence of that.
There's is an inertial
reference frame O in which the two spaceships, one in front of the
other, are initially at rest.
Yes.
There is a rope stretched from the tail of the spaceship in front
(point A) to the nose of the other spaceship (point B).
Yes.
At time zero in that frame both spaceships start accelerating.
Yes.
The way I read the problem, the spacecraft are stipulated to
accelerate in such a manner that the projected (onto frame O)
separation of point A and point B never changes.
Yes.
You have themon the nose
accelerating such that an accelerometer on the tail of the lead
spaceship always has the same reading as an accelerometer
of the trailing spacecraft.
Yes. The proper acceleration profile a(τ) is the same.
That's a different problem.
No, it's not a different problem. It's an Ansatz.
That is to say, it is a hypothetical solution to the problem.
I wrote down the Ansatz and then proved that it does in fact
solve the problem -- the original, agreed-upon problem.
The core of the proof is little more than one line long. If
you blink you will miss it. The core is that all the
fundamental laws of physics commute with translation. So in
particular any specific acceleration profile a(τ) commutes
with translation.
If you want to get fussy there are additional details that
could be mentioned:
-- All this assumes flat spacetime; otherwise the problem
obviously has no solution. If I unwisely start rocket A high
above the north pole and rocket B high above the south pole,
both initially at rest, then when I let go the distance
between them changes, even if they don't fire their engines.
So, as Henny Youngman would say, don't do that.
-- The inertial of the rope itself is assumed negligible.
-- The rope does not interact with the rocket exhaust.
-- I proved that the distance was the same after one particular
time τ. I hoped it was obvious that τ was arbitrary, so in
fact the distance is the same all along the way.
-- Invariance with respect to translation is intimately
connected (via Noether's theorem) to conservation of momentum.
If this doesn't hold, then all of physics is in deep trouble.
People often complain that I tend to belabor the obvious. So
you tell me: should I have explained this in more detail?
To me, the whole analysis has involved nothing more than a
keen grasp of the obvious. I did read the original statement
of the problem, but even before I finished reading I knew
what the answer was going to be. At an impressionable age I
was taught how to do relativity: I can still hear Charlie
Peck saying "The goal here is not to teach you how to do
Lorentz trans- formations; the goal is to teach you how to
/avoid/ doing Lorentz transformations." By that he meant we
should think about spacetime, think about four-vectors, and
think about the invariances. Example:
http://www.av8n.com/physics/bevatron.htm
I also remember Kip Thorn saying "There are no paradoxes in
relativity. The only way you get paradoxes is by misstating
the laws of physics. The goal here is to teach you to
understand the laws so clearly that you cannot even utter a paradox."
We have the problem statement. We have a solution. Is it
really necessary to make it more complicated than that? Why?
In what way?
On 06/17/2010 07:20 AM, Jeffrey Schnick wrote:
By the way, the rope breaks.
No, it doesn't.
perpendicular to theConsider a taut horizontal rope segment of length L with the sun
directly overhead. It is casting a shadow of length L. A person
rotates the rope about a horizontal axis that is
rope while at the same time stretching the rope so that its shadow
remains at length L. The rope breaks.
That analogy is cute, but not apt.
If you want an analogy, try this:
One end of the rope is tied to the left hand of a hula dancer.
The other end is tied to the left hand of another hula dancer.
So long as the dancers are identical and perform the
identical moves, and ignoring possible interference and
tangling, then the rope does not break. More precisely, the
hand-to-hand distance remains the same. You can change the
angle of the sun, which will change the length of the rope's
shadow, but the shadow of a thing ought not be mistaken for
the thing itself.
http://faculty.washington.edu/smcohen/320/cave.htm
On the other side of the same coin, sometimes it is possible
to infer the properties of the real object from observations
of its shadow. This problem is particularly simple, in that
the rope moves always parallel to itself, so that with very
little effort you can set things up so that not only does the
rope maintain the same length, the shadow does too. This is
not automatic, but it is easy to set up.
On 06/17/2010 07:37 AM, chuck britton wrote:
distributed along the rope?How about having a bunch of identical rocket motors
Same answer. Same method of solution.
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l