[gem (OP)]

Hi all,



BC my bold

What's the mechanism by which the Earth pushes back on the Sun in order to conserve angular momentum for the (isolated) system as a whole??

So at the risk of repeating myself the Earth is not an isolated system and you cannot ignore the fact that the action/interaction of the surface and atmosphere has the SI unit of  joule-second, for momentum.

Also these transfer happen by way of friction and friction is a non-conservative force.

BC

Please show us where anyone here said that the solid Earth and its atmosphere don't exchange angular momentum?

I have been stating on numerous posts, that the transfers between the solid surface and the atmosphere are partially inelastic, So the exchange of angular momentum will not be able to return the same value of momentum back to the rotating solid earth, to maintain the LOD.

The rotation of the Earth is actually its "terminal angular velocity"

IE a balance between frictional couplings and torque/accelerating forces.

Similar to sky divers who can alter their terminal velocity. 

[Bored chemist]

Also these transfer happen by way of friction and friction is a non-conservative force.

At the atomic level, all the interactions are conservative.
So, in every interaction momentum and energy are conserved.
When you add them all up the total is also conserved.

Momentum is (unlike energy) conserved in non-elastic collisions where heat is formed.

So at the risk of repeating myself the Earth is not an isolated system

The problem is not the repletion; the problem is repeating an error.

I already explained this to you in a previous thread.
Quote from: Bored chemist on 10/09/2020 08:47:48
It is, as you keep stressing the importance of the fact, an isolated system.
The atmosphere can't push the world for the same reason that the man on the truck bed can't push the truck.

You seem to be trying to elevate the terms isolated and closed to the status of holy writ.

It's not magical.
Classically:
(1) A system to which you can't add or remove stuff will not change mass
(2) A system to which you can't add or  remove energy will  change energy
(3)A system to which you can not apply a force will not change momentum.
(4) A system to which you can't apply a torque  will not change angular momentum.
Those are pretty much tautologically true.
Since Einstein's day the first pair of those are a bit more flexible, but only in a very clearly defined way.

Historically, in thermodynamics it was important to lump together some of those statements of the obvious.
If you have a system to which you can add or subtract matter then you can't sensibly define what it will do- because it depends on the matter you might add.The same goes for applying forces to it. Those forces just complicate the issue.
So, they invented the "closed system" where tautologies 1 and 3 apply.
 

Similarly, for some calculations, you want to prevent energy entering or leaving the system.
And that's why they invented the "isolated system"- as  a shorthand for a system where tautology 2 also applies.

If thermodynamics dealt with angular momentum they would have invented another term for systems where tautology 4 also applied. They didn't.

But, if they had done, the Earth would be on the list.

[Bored chemist]

I have been stating on numerous posts, that the transfers between the solid surface and the atmosphere are partially inelastic, So the exchange of angular momentum will not be able to return the same value of momentum back to the rotating solid earth, to maintain the LOD.

Momentum is conserved during inelastic collisions.
Energy is conserved in  inelastic collisions.

Only kinetic energy is not conserved- some is converted to thermal energy.

So the elasticity of a collision is immaterial to a discussion on momentum.

[gem (OP)]

Hi all

For the elasticity of the collisions to be immaterial, then it requires explanations of phenomena such as the transfer of momentum via sound waves the majority that also end up as heat, retrieved to the angular momentum.

For your position to be credible.

[Bored chemist]

then it requires explanations of phenomena such as the transfer of momentum via sound waves t

The explanation isn't very complex.
Molecule A hits molecule B.
Momentum and energy are conserved in this process.
So they are also conserved in any process which is a succession of repetitions of that step.

[gem (OP)]

Hi all
ok

then it requires explanations of phenomena such as the transfer of momentum via sound waves t

The explanation isn't very complex.
Molecule A hits molecule B.
Momentum and energy are conserved in this process.
So they are also conserved in any process which is a succession of repetitions of that step.

Ok so the collision occurs transferring momentum away from direction of the angular velocity of the solid Earth to the atmosphere, and its a partially inelastic collision, in which direction do these subsequent conserved momentum of sound waves of molecule A and molecule B  colliding go ? that  enable the 100% transfer back in the correct direction later, actually go ?
Given the Inverse square law of sound waves how does  that  enable the 100% transfer of momentum back in the correct direction later.
 
http://hyperphysics.phy-astr.gsu.edu/hbase/Acoustic/invsqs.html

Also given the time aspect between transfers how do you exclude thermal radiation in your calculations spontaneously transfering  away from that side of the equation, and away from the isolated system the joules in your original calculation ? which are the SI values of the action under consideration ?
https://en.wikipedia.org/wiki/Thermal_radiation

[Bored chemist]

Ok so the collision occurs transferring momentum away from direction of the angular velocity

You are muddling a whole bunch of things there.
Please clarify your question in terms of angular momentum;
linear momentum and energy, all of which are strictly conserved.

[Bored chemist]

Given the Inverse square law of sound waves

Are you aware that sound does not follow the inverse square law?

[Bored chemist]

Also given the time aspect between transfers how do you exclude thermal radiation in your calculations spontaneously transfering  away from that side of the equation

I don't need to.
All transfers, including those carried by photons, conserve momenta and energy

[gem (OP)]

HI all,

OK BC

You are muddling a whole bunch of things there.
Please clarify your question in terms of angular momentum;
linear momentum and energy, all of which are strictly conserved

All of which are strictly related in wave form, for example,

momentum = energy/phase velocity.

wave momentum is related to wave energy in
a simple and seemingly universal way. there are many different
kinds of wave motion that links them, for example, electromagnetic waves, sound waves, water waves, and certain kinds of traveling waves such as one's on strings under tension.

BC

Are you aware that sound does not follow the inverse square law?

Sorry BC, I don't agree, and I did provide a link in the previous post that states:

The sound intensity from a point source of sound will obey the inverse square law

also given the sound intensity is defined as the sound power per unit area.

http://hyperphysics.phy-astr.gsu.edu/hbase/Acoustic/invsqs.html


gem

Also given the time aspect between transfers how do you exclude thermal radiation in your calculations spontaneously transferring  away from that side of the equation

BC

I don't need to. All transfers, including those carried by photons, conserve momenta and energy

The problem with that is,
the momentum and energy of sound  is carried off in all directions.

which brings us back to, how is the vector direction total maintained for later transfer back to the angular momentum of the solid earth, due to the partially inelastic collisions sending momentum away from the transfer point, in all directions ?

indeed as you say photons conserve momenta and energy but the system does not.
     

[Bored chemist]

Sorry BC, I don't agree,

It is unfortunate that you do not agree with reality, but it is no surprise to me.

"Stokes's law of sound attenuation is a formula for the attenuation of sound in a Newtonian fluid, such as water or air, due to the fluid's viscosity. It states that the amplitude of a plane wave decreases exponentially with distance travelled,"

https://en.wikipedia.org/wiki/Stokes%27s_law_of_sound_attenuation

[Bored chemist]

The problem with that is,
the momentum and energy of sound  is carried off in all directions.

Which means that, on average, it is zero.
So, as it fades, there's no problem.

which brings us back to, how is the vector direction total maintained

It is zero.
As you say, it points equally in all directions and the only way to do that is to have all the vectors "cancel out".

[Bored chemist]

indeed as you say photons conserve momenta and energy but the system does not.

Does it worry you that you are the only one who believes this?

[gem (OP)]

Hi all

OK BC

"Stokes's law of sound attenuation is a formula for the attenuation of sound in a Newtonian fluid, such as water or air, due to the fluid's viscosity. It states that the amplitude of a plane wave decreases exponentially with distance travelled,"

SO Acoustic attenuation is a measure of the energy loss of sound propagation in media. Most media have viscosity, and are therefore not ideal media. When sound propagates in such media, there is always thermal conversion of wave energy caused by viscosity.

Which just confirms what I posted earlier

gem

For the elasticity of the collisions to be immaterial, then it requires explanations of phenomena such as the transfer of momentum via sound waves the majority that also end up as heat, retrieved to the angular momentum.

Attenuation does not include the decrease in intensity due to inverse-square law geometric spreading. Therefore, calculation of the total change in intensity involves the inverse-square law and an estimation of attenuation over the path due to the viscosity of the medium.

gem

how is the vector direction total maintained for later transfer back to the angular momentum of the solid earth, due to the partially inelastic collisions sending momentum away from the transfer point, in all directions ?

BC

It is zero.
As you say, it points equally in all directions and the only way to do that is to have all the vectors "cancel out".

precisely the solid earth transferred angular momentum to the atmosphere reducing the solid earths momentum total.

Which as you say the momentum carried away by sound to be available to be transferred back later is ZERO   

[Bored chemist]

Therefore, calculation of the total change in intensity involves the inverse-square law and an estimation of attenuation over the path due to the viscosity of the medium.

Well done.
You have caught up with what I said a while ago.

[Bored chemist]

precisely the solid earth transferred angular momentum to the atmosphere reducing the solid earths momentum total.

The solid earth (over a reasonable period of time) transfers ZERO momentum to the atmosphere.
Because the momentum carried by the sound sums to zero.
So the loss by the solid earth is zero.


If the Earth lost angular momentum to the air then the air would have to rotate WRT the earth.
Friction would slow it down again until the angular momentum was transferred back.

In the short term, there can be transfers (though they are very near zero on average across the Earth).
Averaged over time they are exactly zero. |(Give or take atmospheric tidal drag).

[gem (OP)]

Hi all,

OK

precisely the solid earth transferred angular momentum to the atmosphere reducing the solid earths momentum total.
Which as you say the momentum carried away by sound to be available to be transferred back later is ZERO 

The solid earth (over a reasonable period of time) transfers ZERO momentum to the atmosphere.
Because the momentum carried by the sound sums to zero.
So the loss by the solid earth is zero.


If the Earth lost angular momentum to the air then the air would have to rotate WRT the earth.
Friction would slow it down again until the angular momentum was transferred back.

In the short term, there can be transfers (though they are very near zero on average across the Earth).
Averaged over time they are exactly zero. |(Give or take atmospheric tidal drag).

In the short term the transfers may seem small when we are considering fractions of a milli second daily variations in the
LOD "however they are not"
If we consider the approximate 0.3 of a milli second variation for one day given in the dates earlier, corresponds to a change in rotational kinetic energy of the solid earth in the region of 1.6 x 10^21 joules for the day in question.
These daily fluctuations are values that are not remotely near to zero.

The fact that the LOD appears consistent over longer time periods can also be explained
by a balance of frictional forces and accelerating forces ( terminal angular rotation) rather than the "born that way" explanation.   

   

[Bored chemist]

If we consider the approximate 0.3 of a milli second variation for one day given in the dates earlier, corresponds to a change in rotational kinetic energy of the solid earth in the region of 1.6 x 10^21 joules for the day in question.

I realise it's terribly "English" to always talk about the weather but...

We are talking about the momentum transfers which definitely sum to zero over the long term.
You are talking about the transfer of heat from the earth to the air.
And that's broadly what drives the weather.
So it is irrelevant.

The hint is in the units.

[evan_au]

The annual component of the change of the length of day of approx  0.34 ms...
If we consider the approximate 0.3 of a milli second variation for one day

I don't know where the variation of 0.3ms in one day came from. This quotes an annual change, not a 1 day change.

The current long-term variation in Earth's rate of rotation is quoted as"+1.7 ms/d/cy", ie every century, it changes by 1.7 ms/day.
- So in a year, you could expect it to change by about 170μs/day (plus short-term variations which might take it up to 300μs/day over the course of some years).
See: https://en.wikipedia.org/wiki/%CE%94T_(timekeeping)#Earth's_rate_of_rotation

[gem (OP)]

Hi all,

If we consider the approximate 0.3 of a milli second variation for one day given in the dates earlier, corresponds to a change in rotational kinetic energy of the solid earth in the region of 1.6 x 10^21 joules for the day in question.

I realise it's terribly "English" to always talk about the weather but...

We are talking about the momentum transfers which definitely sum to zero over the long term.
You are talking about the transfer of heat from the earth to the air.
And that's broadly what drives the weather.
So it is irrelevant.

The hint is in the units.

MMMMM, I believe we already covered the universal link (momentum = energy/phase velocity.)
and if you look at the original question, the units for energy are correct and relevant, to put a scale to what you tried to dismiss as close to zero.
However if you prefer a 0.3 milli sec daily variation in LOD should give a corresponding  ±Δ angular momentum

of aprox = 2.019 x 10^25  Kg m^2  for the day in question

Evan_au

I don't know where the variation of 0.3ms in one day came from. This quotes an annual change, not a 1 day change.

You probably need to look back to post 60 and a few on from there.

But to help, the LOD tends to skip about quite a bit more than you believe, also it has an annual speeding up in the northern hemisphere's summer and slowing down in its winter to a well established pattern of ± 1.5 milli sec as per the link below.

https://en.wikipedia.org/wiki/Day_length_fluctuations#/media/File:Deviation_of_day_length_from_SI_day.svg


And bringing that data into closer detail

https://datacenter.iers.org/singlePlot.php?plotname=FinalsDailyIAU1980-LOD-BULA&id=12