Re: Noether's Theorems (1918) in English translation

From: N:dlzc D:aol T:com \(dlzc\) (net_at_nospam.com)
Date: 03/14/05 

Date: Mon, 14 Mar 2005 07:17:43 -0700

Dear Ken S. Tucker:

"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1110788302.252439.279250@z14g2000cwz.googlegroups.com...
> N:dlzc D:aol T:com (dlzc) wrote:
>> Dear Ken S. Tucker:
>>
>> "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
>> news:1110740453.988903.85210@z14g2000cwz.googlegroups.com...
>> > Tom Roberts wrote:
>> >
>> >> But the planets in the solar system do NOT do
>> >> that (e.g. perihelion shift of mercury and other
>> >> objects).... And the binary pulsars certainly
>> >> do not do that (they are observed to be
>> >> spiraling in, in agreement with predictions
>> >> of GR)....
>> >
>> > Adding a "grain of salt", a well studied
>> > system is the Earth-moon, where we find
>> > the Moon receeding from Earth (~3cm/year)
>> > and Earths rotation decreasing by,
>> > .0018 sec/century.
>> > This deposits ~75 megatons (TNT) of energy
>> > into the earth daily,
>>
>> And you are sure you are clear on the *sign*
>> right? The loss of momentum is from the
>> Earth to the Moon. There is a net energy
>> *conversion* by the Earth, which radiates
>> excess energy (in general) to the depths
>> of space.
>
> Yes, that same energy input just killed 150,000
> people.

This was a change in potential -> kinetic -> frictional losses.
It was a *loss* of energy, not a "deposit" of energy. The crust
subducted.

>> > which is ultimately
>> > heating the Earth and is emitted as EM
>> > radiation (as heat).
>> > We find gravitational orbital change being
>> > converted to EM-radiant energy from a binary
>> > system consistent with Newton.
>>
>> The Earth should have a net decrease in energy, not a gain.

> I recommend your begin a new thread on this subject.
> I further recommend you aquaint yourelf with Earth-
> Moon dynamics, taking into account the conservation
> of angular momentum that renders the receeding moon
> and the slowing Earth rotation, those are facts.

Angular momentum transferred to the Moon, from the Earth, via
tides, is a loss of energy by the Earth. NOT a "deposit". A
withdrawal (in more ways than one).

> You should also calculate the rate of energy imparted
> to the Earth by the Moon, that is rather evident in
> the motion of water aka tides.

The motion of the water "deposits" energy & momentum to the Moon,
dude.

> I've posted that, but I don't want to spoil your
> fun by giving you an an answer.
>
> You deserve the pleasure of that solution.

I've already been there. Maybe you can finally get it.

David A. Smith

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