sci.physics.research Google Group

Re: Questions About Nuclear Radii

2010-03-18T16:22:04Z

This months issue of Physics Today has a "Physics Update" on page 22.
It summarizes a paper (K. Tanaka et al., Phys. Rev. Lett. 104, 062701,
2010) about carbon-22 with 16 neutrons and a nuclear radius of twice
the more common C-12 isotope. FYI
Rich L.

Re: Definition of "momentum operator" in quantum theory

2010-03-18T09:21:11Z

Thanks. My first reply was to your second post. This second reply is
to your first post. Such are the vagaries of the SPR moderation, as
Jonathan just pointed out.
That is just a question of ordering of presentation, but I agree that
(5) logically precedes (3).
A typo.
Good.
I guess based on my experience with Lie groups (such as 2x2 = 3+1 for

Re: Motion Molehill

2010-03-18T09:21:06Z

[[description of conceptual confusion & muddles in trying to teach
the concept of "work" to 3rd-grade students]]
I have a couple of suggestions of useful resources (these are mostly
aimed at more general problems in introductory teaching, and not
specifically focused on the original poster's topic):

Re: Time dilation and constant velocity vs. acceleration

2010-03-18T08:38:58Z

Yes, for one method of determining this (see below).
But not for this method:
This is wrong for watching through a telescope.
For that method of determining the aging of the distant twin, and for a
very quick turn-around, the traveling twin observes the stay-at-home
twin to age very little during turn-around. Using a telescope, the

Re: Time dilation and constant velocity vs. acceleration

2010-03-18T08:38:58Z

Fran,
When viewed through a telescope, the moving twin is seeing his past
light cone. I.e. what he is seeing at home is not "now" but is some
time in the past. This is still true during turn-around. Nothing
special happens at that time. During the return trip the telescope is
still seeing the past light cone, but now the moving twin is getting

Re: Second reply to question about momentum operators

2010-03-18T08:38:58Z

If anybody cares, I did, in fact, drop a factor of 1/2 in (4), which
should be [a,a*] = 1/2 as Igor says. I haven't gone through the rest to
see how it should be changed, and I don't intend to. There are more
interesting and important things to think about.
Stephen Parrott

moderator's note: how moderated newsgroups work

2010-03-17T22:46:21Z

[[Mod. note --
In a recent posting in this newsgroup, someone whose name isn't relevant
here wrote:
It's useful to understand how this & other moderated newsgroups work:
1. author writes an article and submits it
(either via a dedicated "newsreader" program or via a web gateway
like groups.google.com)

Re: Time dilation and constant velocity vs. acceleration

2010-03-17T22:19:54Z

[note to moderator: I submitted this posting about three hours earlier
than a posting of mine that just showed up, and the earlier one HASN'T
shown up, so I'm re-submitting it.]
[[Mod. note -- See a moderator's note which I'll post in a few minutes.
-- jt]]
No...during an abrupt speed change, the traveler would see no change in

Re: Definition of "momentum operator" in quantum theory

2010-03-17T22:18:30Z

Not all of your questions are clear to me, but one seems clear enough to
answer, and maybe the answer will help with the rest. If it doesn't, no
harm is done.
You're getting ahead of yourself.
To justify this, you have to specify the commutation relations [a,a*]=
*something* between a and a*, as you did in (5) below. Also, the right

Re: Time dilation and constant velocity vs. acceleration

2010-03-17T22:17:29Z

Mike_Fontenot says...
I just wanted to point out that it is a little misleading to
talk about rapid aging of the stay-at-home twin. To see the
problem, let's make it a triplet paradox:
Assume that there are three earthlike planets in a straight
line: Earth1, Earth2, and Earth3. Earth2 is midway between

Cocycles in definition of "momentum operator"

2010-03-17T22:16:57Z

Previous posts have been concerned with the definition of "momentum
operator" P in quantum mechanics. Some people think P has to be
defined as (Pf)(x) := -i df/dx, while others think it might be more
general, Pf(x) := -idf/dx + A(x)f(x), where A is a given function.
This note indicates where the more general form comes from.

Re: Time dilation and constant velocity vs. acceleration

2010-03-17T22:15:46Z

This is really a question about SR, not GR. That's because it does not involve
gravitation, and can be analyzed in Minkowski spacetime. SR can handle
accelerations perfectly well, it just cannot handle spacetime curvature (i.e.
gravitation).
This is not really "time dilation", it is an example of the elapsed proper time

Re: R. Feynman left a puzzle.

2010-03-17T22:15:20Z

Khrapko, I think what you might have thought about the average
power of the photon beam.
If so then the heat produced is the average power (Pave)*time/4.
Where Pave = N*h/nu where N is the number of photons per second
or any time division you want.

Re: Time dilation and constant velocity vs. acceleration

2010-03-17T19:26:52Z

That concept has no UNIVERSAL or INVARIANT meaning. But it certainly
DOES have meaning for a PARTICULAR observer. It's easiest to see in the
case of an inertial (unaccelerated) observer...there, the "current age
of a distant object" (CADO) can be determined from the observer's simple
and elementary measurements. Basically, the inertial observer CAN'T

Re: Time dilation and constant velocity vs. acceleration

2010-03-17T19:26:51Z

[NOTE TO MODERATOR: This has not appeared in 2 days, so I'm re-sending it.]
This is really a question about SR, not GR. That's because it does not involve
gravitation, and can be analyzed in Minkowski spacetime. SR can handle
accelerations perfectly well, it just cannot handle spacetime curvature (i.e.