# Fixed Rotation?

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## Re: Fixed Rotation?

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

Sedna

Sedna
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## no pseudo-synchronism

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

efroimsk
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## Re: Fixed Rotation?

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.

Lazarus
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## Very likely, fluid planets never get pseudosynchronous either

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

efroimsk
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## Re: Fixed Rotation?

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

Lazarus
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## Re: Fixed Rotation?

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

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Sirius_Alpha

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

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

efroimsk
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## Re: Fixed Rotation?

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?

Stardust
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## Re: Fixed Rotation?

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 )

efroimsk
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