Fixed Rotation?

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Re: Fixed Rotation?

Post by Sedna on 20th August 2009, 6:15 pm

Yes that's good now, Earth is spinning free. Mercury is still tidally locked, but its orbit is pseudo-synchronous (if I remember this term right). It shows that unities are very important: if one of them is not good, the whole formula is wrong.

Actually the formula with g/cm^3 gave a resonable result for young planetary system, such as 2M1207.

Bye

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no pseudo-synchronism

Post by efroimsk on 22nd January 2013, 4:32 pm

Lazarus wrote:Of course Gliese 581 d is on an eccentric orbit. Leads to pseudo-synchronisation not synchronisation. For e=0.38, pseudo-synchronous period is just over half the orbital period, at 34.6 days.

Dear Lazarus,

Browsing this Forum for topics related to tides, I saw your afore-quoted old post, and I cannot help from commenting on it. As you will see from http://iopscience.iop.org/0004-637X/764/1/27 , pseudosynchronous spin of terrestrial planets and moons is unstable and therefore cannot exist.

I am not sure how to extend this statement to objects with surface or internal oceans. However, I have strong arguments in favour of extension of this statement to purely viscous objects (stars, jupiters).

Best regards,

Michael

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Re: Fixed Rotation?

Post by Lazarus on 22nd January 2013, 4:35 pm

Indeed, as I stated later in the thread...
If I understand it correctly (and the tidal stuff is not my cup of tea!) a planet with a permanent asymmetry - which implies a solid planet can get captured into resonance, but fluid planets end up pseudosynchronised.

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Very likely, fluid planets never get pseudosynchronous either

Post by efroimsk on 23rd January 2013, 12:03 pm

Lazarus wrote:Indeed, as I stated later in the thread...
If I understand it correctly (and the tidal stuff is not my cup of tea!) a planet with a permanent asymmetry - which implies a solid planet can get captured into resonance, but fluid planets end up pseudosynchronised.

Very likely, the fluid planets will not get into pseudosynchronism either.

If you are interested, I can explain my point in more detail, but this will require some maths

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Re: Fixed Rotation?

Post by Lazarus on 23rd January 2013, 2:18 pm

If you feel you can contribute to a discussion then go ahead and contribute. You don't need to start asking people for permission!

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Re: Fixed Rotation?

Post by Sirius_Alpha on 23rd January 2013, 4:53 pm

efroimsk wrote:If you are interested, I can explain my point in more detail, but this will require some maths
No need to ask Smile Besides, this is a math-friendly board.

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Why pseudosynchronous rotation is an unlikely fate for many (perhaps, all) fluid bodies.

Post by efroimsk on 23rd January 2013, 6:38 pm

All right guys. You have opened the lid.


1. To warm up, recall the good old formula for the static Love number:

k_2 = (3/2)/[ 1 + (19/2) \mu/( g R \rho ) ] ,

where g, R, and \rho are the surface gravity, radius, and mean density of a near-spherical homogeneous body, while \mu is the relaxed (static) rigidity modulus. Similar (up to numerical factors) are the formulae for all other k_{\el} with \el > 2.

I prefer to rewrite the formula in terms of the relaxed (static) compliance J = 1/{\mu} :

k_2 = (3/2)/[ 1 + (19/2) 1/( J g R \rho ) ]

As we remember, J interconnects the strain tensor u with the tidal stress \sigma:

u = J \sigma ,

while k_2 interconnects the degree-2 additional tidal (static) potential U_2(r) with the tide-raising potential W_2(R) of a (stationary) perturber:

U_2(r) = (R/r)^3 k_2 W_2(R) .

Here r is some exterior point, while R is a surface point right beneath it.
For details, see http://i14.servimg.com/u/f14/18/07/03/56/figure14.png

As Peter Golrdeich pointed out back in mid-60s, in the denominator of the expression for k_{\el} we observe competition between self-gravitation (the `1' term) and rheology (the term with J).


2. With a lot of math, it can be proven that in many situations similar machinery works for time-dependent tides.

First, we have a similar formula for the Fourier components of the potentials:

\bar{U}_2(\omega) = (R/r)^3 \bar{k}_2(\omega) \bar{W}_2(\omega) .

Here I omitted the self-evident arguments r and R, but inserted the argument \omega, which is the Fourier tidal mode. I also put overbars, to remind that we are now talking about the Fourier components of U and W appropriate to the tidal mode \omega, and about the complex Love number \bar{k}_2(\omega) corresponding to this mode.

Second, and most important, the complex Love number \bar{k}_2(\omega) will be expressed through the complex compliance \bar{J}(\omega) in the same algebraic way as the static Love number was expressed via the static compliance:

\bar{k}_2(\omega) = (3/2)/[ 1 + (19/2) 1/( \bar{J}(\omega) g R \rho ) ] ,

where the complex compliance interconnects the Fourier components of the strain and stress tensors:

\bar{u}(\omega) = \bar{J}(\omega) \bar{\sigma}(\omega)

The old comment by Peter Goldreich remains in force. Were it not for the '1' in the denominator, the complex Love numbers would behave exactly as the complex compliances -- and there would be no difference between the frequency dependencies of materials and the frequency dependencies of celestial bodies made of those materials: \bar{k}_2 ~ \bar{J} = 1/{\bar{\mu}}. In reality, though, we have this term `1', wherefore self-gravitation comes into play, and the tidal response differs from the seismic response.


3. Suppose you tell me that a celestial body is composed of a material with rheology \bar{J}(\omega), and you ask me to tell you what its k_2/Q gonna be.

Well, this a turn-of-the-crank situation. I take your \bar{J}(\omega), plug it into the above formula for \bar{k}_2(\omega) -- and into all the similar formulae for \bar{k}_{\el}(\omega) with \el > 2 -- and I obtain the complex Love numbers.

Then I calculate the negative imaginary parts of of my complex Love numbers, i.e., k_2 \sin\epsilon_2 , where k_2 = k_2(\omega) is the absolute value of the complex Love number, while \epsilon = \epsilon(\omega) is the negative phase angle (negative argument) of the complex Love number.

The quantity k_2(\omega) = | \bar{k}_2 (\omega) | is what they often call the dynamical Love number, while \sin\epsilon_2(\omega) is often misdenoted with 1/Q. In fact, one should use 1/Q_2 (and, likewise, 1/Q_{\el} for \el > 2). Recall that the tidal Q is not the same as the seismic Q ! See the image http://i14.servimg.com/u/f14/18/07/03/56/figure16.png


4. What remains is to grab the expansion of the tidal potential (or force, or torque) into the Fourier series over the tidal modes \omega_{lmpq} , and to insert the right k_l(\omega_{lmpq}) \sin\epsilon_l(\omega_{lmpq}) into each {lmpq} term of the series. This is now rheology will enter the tidal theory in a consistent manner. See the image
http://i14.servimg.com/u/f14/18/07/03/56/figure15.png

Now let us limit ourselves, for simplicity, to the degree-2 Love number and to the principal (semidiurnal) tidal mode {lmpq} = {2200}. Instead of \omega{2200}, I shall simply write \omega.

Whatever realistic rheology \bar{J}(\omega) you take, its insertion into the expression for the complex Love number will render you the k_2(\omega) \sin\epsilon_2(\omega) -- or, as many will say, k_2/Q -- shaped as a kink http://i14.servimg.com/u/f14/18/07/03/56/figure13.png

The kink goes continuously through nil when the tidal mode goes through zero -- so we cross spin-orbit resonances smoothly.


5. Now, the final point. Let us take a viscous body lacking rigidity completely. Its complex compliance will read as

\bar{J}(\omega) = - i/(\chi \eta) ,

where \eta is the viscosity, while \chi = |\omega| is the physical forcing frequency. (It is positive definite, while the tidal mode \omega can be of either sign.)

Plug this into the formula (**) given at http://i14.servimg.com/u/f14/18/07/03/56/figure16.png , then take its negative imaginary part (which is k_2/Q) -- and get a kink!


6. The bottomline is that a purely viscous rheology is incompatible with the linear tidal law k_2/Q ~ \omega. The latter rotten law stems from the assumption that \Delta t is the same for all frequencies. We now see that this law and therefore its underlying assumption are mighty unphysical.

At the same time, it is this CTL (`constant time lag') law which underlines the calculation leading to pseudosynchronism. (Mind a mistake in the book by Murray & Dermott who derived pseudosynchronism from a `constant geometric lag' assumption http://iopscience.iop.org/0004-637X/764/1/26 )

I would say that all depends upon the viscosity of the body. The peak frequency is about G \rho^2 R^2/\eta. If it so happens that the peak frequency much exceeds n e^2 , then perhaps pseudosynchronism still has a chance. But not otherwise.

Important disclaimer:

In realistic stars and jupiters, complex hydrodynamical processes are going on. Among other things, these entail deviation from the afore-described wonderful similarity between the static and dynamical formalisms. Therefore the above back-of-the-envelope estimate cannot be taken as a serious calculation of the k_2/Q. Nonetheless, even this, simplified treatment is sufficient to demonstrate how far the resulting k_2/Q is from being linear in frequency. In reality, its frequency dependence will be even more complex.



7. Although you did not ask me about that, I cannot help from mentioning that employment of the correct kink-shaped k_2/Q (or, better to say, k_2/Q_2) renders dramatic consequencies for the theory of entrapment into spin-orbit resonances. For example, Mercury had absolutely no chance in the world to avoid the 3:2 trap, as was demonstarted by Valeri Makarov http://iopscience.iop.org/0004-637X/752/1/73



Thank you for urging me to write all this discourse. I now have to go and shape it into a short Letter to the Editor, and probably send it to the ApJ Letters


Last edited by efroimsk on 24th January 2013, 2:27 pm; edited 1 time in total

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Re: Fixed Rotation?

Post by Stardust on 23rd January 2013, 8:59 pm

I would like to know if these formulae and the situation they describe could also explain the slow rotation of Venus... or Venus is realy a diferent situation envolving colisions with other solar system bodies? A Venus-like planet in a Venus-like orbit always will have that slow rotation or could it have a rotation period more earth-like?

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Re: Fixed Rotation?

Post by efroimsk on 23rd January 2013, 9:12 pm

Stardust wrote:I would like to know if these formulae and the situation they describe could also explain the slow rotation of Venus... or Venus is realy a different situation envolving colisions with other solar system bodies? A Venus-like planet in a Venus-like orbit always will have that slow rotation or could it have a rotation period more earth-like?

Not sure. Have to cogitate on this.

I am now looking at the paper by Correia and Laskar in Icarus 163, 24 (2003). There they did a lot of good work, but unfortunately their tidal model was not very advanced: k_2/Q linear in the tidal frequency at low frequencies, and a constant k_2/Q at higher frequencies.

It would be surprising if their results stand the test of the consistent tidal model.

(Compare their results on Mercury with the computation by Makarov http://iopscience.iop.org/0004-637X/752/1/73 )

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