Re: Error in Weisstein's redshift


Henry Haapalainen wrote:
"John C. Polasek" <jpolasek@xxxxxxxxxx> kirjoitti
viestissä:qr5sf2hu4he5vcqid031p17tkunhc7a356@xxxxxxxxxx
On Fri, 17 Feb 2006 16:19:40 GMT, John C. Polasek
<jpolasek@xxxxxxxxxx> wrote:

On 13 Feb 2006 01:48:08 -0800, "harry" <harald.vanlintel@xxxxxxx>
wrote:


John C. Polasek wrote:
On Thu, 2 Feb 2006 09:59:49 +0100, "Harry" <harald.vanlintel@xxxxxxx>
wrote:


"John C. Polasek" <jpolasek@xxxxxxxxxx> wrote in message
news:l9npt11hv0r2eqk2ceei8jrcp6brpurrji@xxxxxxxxxx
On 28 Jan 2006 22:47:51 -0800, xxein@xxxxxxxxxxxxx wrote:


John C. Polasek wrote:
On 28 Jan 2006 18:39:20 -0800, xxein@xxxxxxxxxxxxx wrote:


John C. Polasek wrote:
On 28 Jan 2006 16:44:42 -0800, xxein@xxxxxxxxxxxxx wrote:


John C. Polasek wrote:
Take a look at this site:

http://scienceworld.wolfram.com/physics/GravitationalRedshift.html
I believe they have the wrong formula for change in L
(lambda) vs
gravity. They show
L/L0 = sqrt(1 - 2MG/r^2c),
where L0 is the "rest" WL and L is the "shifted" WL. It
shows the
WL
would be shifted lower, a blue shift, when gravity makes
red
shift.
It would be OK with f/f0, but they derived it using Newton,
at
which
point they got it upside down, or so it seems to me.

John Polasek
http://www.dualspace.net

xxein: L/L0 = sqrt(1 - 2MG/r^2c) is not the same as sqrt(1 -
2MG/rc^2).

It should not seem upside down with this correction to your
copy.
I mis-typed and should have put rc^2. But it still looks
wrong. The
"shifted" WL L can't be lower than the "rest" L.
John Polasek

xxein: Why would you expect L0 to be greater than L?

What is your primary logic here?
The frequency of an oscillator is reduced by gravity. Then its WL
in
gravity should be greater; we are talking about the "shifted" L
clock.

The clock removed from gravity will run faster; its WL is
shorter;
that's the L0 customer with the short WL.

Their equation says L = L0*(1-eps), making L lower, but it should
be
higher.
John Polasek

xxein: I knew you couldn't be that dumb. It was my mistake in
reading
what L and L0 was representing. I was thinking 'frequency'
recieved.
My sincere apologies.

Yes. It would appear that the originator of the text was either
confused or lacking logical threads of physics. Wasn't there an
editor
or proofreader? I've always known that I needed one (or more).

Even though you pointed out f/f0, I missed it.
Not only that, they have no business telling about lambda in a
gravity
well. It's frequency and energy of the oscillator in the well, that
are depressed there. The idea that lambda is increased there depends
on c being constant.

But c isn't constant. It is similarly depressed (in my theory), so
lambda = c/f is unchanged down there.

Relativity has to make some accommodation if they want to finally
admit (and now flaunt) the Shapiro effect. And this is just the
place
to admit that c is reduced.

I'm puzzled by that remark! c is locally the same everywhere, but that
light
speed is reduced was already described in 1911 and 1916 by Einstein.

Harald
H:
It's not all that clear.
Look at the Pound & Rebka experiment in 1960. They compute
df/f = -gh/c^2 = -dE/E on the way up. They regard c as constant and
they think the frequency reduces on the way up to produce a red shift.
If in addition, c increases as you say on the way out of the well you
get a double red shift in WL
WL = c(1+e)/f(1-e) = WL(1+2e).
Please clarify.
John P

Oops, I only see this one on Google, sorry for the delay.

I don't remember the interpretation of Pound and Rebka; but there is
only one interpretation that works, exactly for the reason you give:
One can compute the effect on clock rate due to gravitational
potential, which yields exactly the measured redshift.

Light behaves as a wave and its frequency in empty space can't change
in stationary conditions, that would result in creation or
disappearence cycles -- instead its wavelength changes. The time of
flight is constant over time. This was already pointed out by Einstein
in 1911...
You can understand it by throwing leaves in a river, one per second.
When they enter a rapid, their speed and distance will increase but the
frequency with which they pass by will not change -- one can't increase
the number of leaves by increasing their speed!

Cheers,
Harald
The redshift detected outside is the normal WL stretching as c
resumes
its proper rate out of the well of course with the same frequency.
(i.e. WL is unchanged by gravity)

Isn't Wikipedia one of those blog things? I didn't notice any
invitation to make a comment.

John Polasek
Harry: I have several times pointed out to Weisstein/Wolfram of their
error in their "Email us" utility, but never received any response.
This is what I sent them this morning;

"I would at least appreciate a reply to my messages pointing out that
your expression for Lam/Lam0 is inverted. If you stick with your first
3 terms, then the 4th one is inverted, as it shows E0 to be afflicted
by Mmg/r, when this 2-piece term should be in the denominator.
(Aside from this I can show that WL is not affected at all, only
frequency and c together are reduced in gravity.)"

John Polasek

This error of inverting L and L0 has still not been corrected since I
wrote them in January.
If they had used f and f0 it would have the right sense. A clock's
frequency f0 is reduced to some lesser value f in a gravity well.

But in fact the wavelength does not change, because c is reduced by
the same factor as f, and lambda = kc/kf.

The site is
http://scienceworld.wolfram.com/physics/GravitationalRedshift.html

I sent them another reminder.
John Polasek

It feels good to read a comment from someone who understand physics.

Henry Haapalainen

xxein: So, would you say that some are finally weaning themselves off
of Einstein's mothermilk?

.

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