Effect of lunar/solar gravity on shape of earth

[russ_watters]

For sake of example, take center at Earth-Moon CoM, rotating with the same period as the system, so both objects are roughly stationary in that coordinate system. Clearly, there is centrifugal tide in this setup.

I didn't ask about the moon - I already knew what you chose for that case. I asked about the example I gave, where there is no obvious choice of rotation rate. What would you choose in that case?

 

[K^2]

Are we looking at tidal force on the dumbbell? Ok. How about this. Ceter - Earth's center at t0, traveling at 30,000km/s in direction tangential to the line connecting Earth and dumbbell. Rotation - omega=1 with axis perpendicular to the direction of CS center's travel and the line connecting Earth and dumbbell. Dumbbell is stationary at t0 in this coordinate system, yet clearly experiences centrifugal tide.

Does that help?

 

[olivermsun]

He doesn't specifically, he just talks about centrifugal forces. It's the meat of the explanation - the center third of the post.

The explanation invokes angular momentum around the orbital focus, by Kepler's convention.

 

[A.T.]

if the Earth and moon were not orbiting each other, ... the tidal force would be the same.

The tidal force due to gravity gradient would be the same. But would the bulges be the same without rotation?

 

[russ_watters]

The tidal force due to gravity gradient would be the same. But would the bulges be the same without rotation?

What bulges are you referring to? The Earth has a doughnut shaped bulge of 23 km due to its rotation and two additional bulges due to the tidal force.

 

[russ_watters]

Hurkyl, what I think I'm seeing for the near side doesn't sound like any of those:

The tidal force creates a near-side bulge and flattens the far side.

Then #4 would explain the far side bulge.

It just so happens someone posted a discussion of exactly what I'm referring to in another thread (the second diagram shows it):

One of the few books that clearly defines "tide" at the outset is The Planetary System by Morrison and Owen [1966]: "A tide is a distortion in the shape of one body induced by the gravitational pull of another nearby object." This is definition (2) above. It clearly says that tides are the result of gravitation, without any mention of rotation effects...

http://www.lhup.edu/~dsimanek/scenario/tides.htm

Now here's where it gets really interesting:

In this representation [the centrifugal force misconception] we can treat this system as if it were an inertial system, but only at the expense of introducing the concept of centrifugal force. It turns out that when this is done, the centrifugal force on a mass anywhere on or within the Earth is of constant size, and is therefore equal to the size of the gravitational force the moon exerts on the same amount of mass at the center of the Earth... ...

We are now focusing on the effects due only to the Earth-moon system. The motion of the Earth about the Earth-moon center of mass, causes every point on or within the Earth to move in an arc of the same radius. This is a geometric result most books totally ignore, or fail to illustrate properly. Therefore every point on or within the Earth experiences the same size centrifugal force. A force of constant size throughout a volume cannot give rise to tidal forces (as we explained above). The size of the centrifugal force is the same as the force the moon exerts at the Earth-moon center of mass (the barycenter), where these two forces are in equilibrium. [This barycenter is 3000 miles from the Earth center—within the Earth's volume.]

That's a twist I actually wasn't aware of, but of course it makes sense: The barycenter is not a fixed point that the Earth rotates around, but a virtual point that moves as the moon moves around the earth.

 

[russ_watters]

And just to be as clear as possible, here's the equation for tidal force:

F=2G \frac{mM}{r^3}dr

http://burro.cwru.edu/Academics/Astr221/Gravity/tides.html

Does anyone have an equation for tidal force that includes a centrifugal force?

 

[russ_watters]

Are we looking at tidal force on the dumbbell? Ok. How about this. Ceter - Earth's center at t0, traveling at 30,000km/s in direction tangential to the line connecting Earth and dumbbell. Rotation - omega=1 with axis perpendicular to the direction of CS center's travel and the line connecting Earth and dumbbell. Dumbbell is stationary at t0 in this coordinate system, yet clearly experiences centrifugal tide.

Does that help?

Yes, thanks. So:

1. How did you come up with this coordinate system?
2. Can you show me how you would compute the tidal force using your coordinate system? By that I mean, show me the equations you would plug that information into to compute the force.

 

[K^2]

1. I didn't want to deal with Coriolis effect, so I picked one where dumbbell system is stationary. I also like to use heavy body as center for CM for simpler symmetry. All other numbers are arbitrary.

2. Schwarzschild metric in the rotating frame:

ds^2 = \left(r^2\omega^2 sin^2\theta + \frac{2M}{r} -1\right) dt^2 + \left(1 - \frac{2M}{r}\right)^{-1} dr^2 + r^2 d\theta^2 + r^2 sin\theta d\phi^2 - 2 r \omega dt d\phi

I'm sure you know how to compute free-fall Reiman tensor.

 

[A.T.]

What bulges are you referring to?

The bulges that cause tides. In your scenario without rotation, would the tides will still be the same, as in the case of Moon and Earth orbiting each other?

And just to be as clear as possible, here's the equation for tidal force:

F=2G \frac{mM}{r^3}dr

http://burro.cwru.edu/Academics/Astr221/Gravity/tides.html

Does anyone have an equation for tidal force that includes a centrifugal force?

That is the tidal force due to gravity gradient alone, so it obviously doesn't include the effects of rotation. My question is if the effective tidal force for a rotating system would be different, and cause different tides than in a non rotating system.

For simplicity let's replace the Earth with two masses connected via a string :

O-------O to the moon --->

Will the force in the string will be the same, regardless if the system rotates or not, or will it be different?

 

[harrylin]

Agree or disagree:

1. Tidal force from the moon causes a bulge on the near and far sides of the Earth, and flattens the places in-between.
2. Tidal forces from the sun cause a bulge on the sides of the Earth near and far from the Sun, and flattens the places in-between.
3. The Earth's rotation causes a bulge evenly distributed around the equator, and flattens the poles. [..]

The last one quoted (no.3) is misleading: due to the centrifugal action the land mass is bulged the same as the oceans, in a steady state (this is verified with atomic clocks: at mean sea level they have the same rate everywhere). Thus the Earth's rotation cannot explain the tides.

The common (and probably correct) explanation is that the acceleration of the water due to the moon's "pull" is higher than that of the Earth on the side of the moon, and lower than that of the Earth on the other side.

 

[russ_watters]

The bulges that cause tides. In your scenario without rotation, would the tides will still be the same, as in the case of Moon and Earth orbiting each other?

The tides are the same size whether there is an orbit or not and whether there is a rotation or not.

That is the tidal force due to gravity gradient alone, so it obviously doesn't include the effects of rotation. My question is if the effective tidal force for a rotating system would be different, and cause different tides than in a non rotating system.

I'm sorry, but that's nonsense. By definition that's all the tidal force is.

For simplicity let's replace the Earth with two masses connected via a string :

O-------O to the moon --->

Will the force in the string will be the same, regardless if the system rotates or not, or will it be different?

If the system rotates, the force on the string will be greater, but that isn't a tidal force. What you are describing sounds to me like the constant height equatorial bulge, not the tidal force.

 

[russ_watters]

2. Schwarzschild metric in the rotating frame:

ds^2 = \left(r^2\omega^2 sin^2\theta + \frac{2M}{r} -1\right) dt^2 + \left(1 - \frac{2M}{r}\right)^{-1} dr^2 + r^2 d\theta^2 + r^2 sin\theta d\phi^2 - 2 r \omega dt d\phi

I'm sure you know how to compute free-fall Reiman tensor.

You're claiming you can calculate a tidal force from that equation? Doesn't look like it to me, but I'd love to see it. I'd also like to see a reference for that being an equation for tidal force.

 

[A.T.]

I'm sorry, but that's nonsense. By definition that's all the tidal force is.

I'm not asking about the definition of the term "tidal force". I'm asking if the tides on Earth are solely an effect of the "tidal force" due to gravity gradient, or if the effect is modified by the gradient in the centripetal force required for the orbit of the Earth's COM around the common COM of Earth & Moon. (In the co rotating frame it would be the gradient of the centrifugal force)

If the system rotates, the force on the string will be greater, but that isn't a tidal force.

I don't care how it's called. If the force is greater in a rotating system then it means the Earth's is stretched more, than it would be due to gravity gradient alone. That would mean that the rotation of the system does affect the magnitude of the tides on Earth.

What you are describing sounds to me like the constant height equatorial bulge, not the tidal force.

No. The equatorial bulge is due to the spin of the Earth around it's own axis in 24h periods. I'm talking about the effects of the rotation of the Earth's COM around the common COM of Earth & Moon in 27d periods.

 

[russ_watters]

I'm not asking about the definition of the term "tidal force". I'm asking if the tides on Earth are solely an effect of the "tidal force" due to gravity gradient...

Yes. The tides are soley due to the tidal force.

 

[daniel6874]

The primary effect of the gravity of an object is to pull other objects towards it.

Tidal force is the difference in gravitational attraction. If you have an elastic object and you tug (along a line) on the opposite sides with different force, then you stretch the object out along that line. This is the secondary effect of gravity -- to stretch objects. To deform spheres into things like seen at this wiki page.

Your reply is given without reference to other posts. To suggest that a prolate spheroid results from the moon's gravitational pull on the Earth begs the question. The question is, why would a pull in the direction of the moon cause the waters on the far side of the Earth to move in the *opposite* direction? At worst we would expect a bulge at the far side that did not exceed the "equatorial bulge."

You suggest that a *difference* in attraction might result in a rebellion of waters at the far side of the earth, but this is contrary to the idea that the moon's g-field is continuous. It may fall off quickly as the distance from the moon, but it will not change directions.

While I agree with Russ, who only acknowledges me in the third person, that resort to tensor calculus is unhelpful, and I can be persuaded that the argument he cites (Siminak, who directly contradicts Gamow) regarding centrifugal forces is correct, I do not see in the cited page the calculation of the "gradient of the moon's gravitational force" resulting in "tidal forces" whose sense is precisely opposite to that of the force creating them--the moon's pull. It's counterintuitive and I would like a cite for the calculation.

 

[Hurkyl]

The question is, why would a pull in the direction of the moon cause the waters on the far side of the Earth to move in the *opposite* direction?

The waters on the far side don't move away from the moon.

The waters on the far side move away from the center of the Earth for exactly the same reason that the center of the Earth moves away from the waters on the near side.

 

[daniel6874]

The waters on the far side don't move away from the moon.

The waters on the far side move away from the center of the Earth for exactly the same reason that the center of the Earth moves away from the waters on the near side.

If you are agreeing with Russ, can you explain Simanek's "gradient"? The moon's gravitational field is a vector field--he speaks of finding the "gradient of the moon's gravitational force upon...[the volume of the earth]." Well, then it would be tensorial.

You are proposing that the Earth arranges itself along the axis of lunar pull, and that if we suddenly placed the moon on the opposite side, in a while the (tidal) shape of the Earth would be essentially unchanged.

 

[A.T.]

Yes. The tides are soley due to the tidal force.

Okay, thanks. I found this website:
http://www.vialattea.net/maree/eng/index.htm

It comes to the same conclusion:
The orbiting of the Earth's COM around the common COM of Moon & Earth has no effect on the tides.

I have not checked the math in every detail. It goes into great detail explaining common errors, and it has the calculation in all possible reference frames.

 

[xts]

"Tensorial"? "Simanek's gradient"? "Schwarzschild's metric"? Guys, stick to common sense!
Tide is a phenomenon that on a full/new moon you may go to the beach in the afternoon to pick crabs, moules, and other tasty living beings, while on the quartermoon you could drown in the place you used to pick crabs a week before, rather than being able to pick something for a dinner.

No tensorial analysis is able to explain why on Bretonian beach tides are 2 metres high, while on Lofoten they may be 8 meters high. And even Simanek won't help you hunting crabs.

 

[daniel6874]

Okay, thanks. I found this website:
http://www.vialattea.net/maree/eng/index.htm

It comes to the same conclusion:
The orbiting of the Earth's COM around the common COM of Moon & Earth has no effect on the tides.

I have not checked the math in every detail. It goes into great detail explaining common errors, and it has the calculation in all possible reference frames.

This appears helpful. I will be interested to see if the effect is not slightly attenuated on the far-side of the Earth w/r to the moon. And perhaps I can get a refund from Dover for Gamow's book?

 

[K^2]

You're claiming you can calculate a tidal force from that equation? Doesn't look like it to me, but I'd love to see it. I'd also like to see a reference for that being an equation for tidal force.

Yeah, except I messed up transform, and the metric I gave you isn't Ricci flat. I'll fix it and get back to you.

Edit: Ok, here we go. Corrected form of the metric, first.

ds^2 = \left(r^2\omega^2 sin^2\theta + \frac{2M}{r} -1\right) dt^2 + \left(1 - \frac{2M}{r}\right)^{-1} dr^2 + r^2 d\theta^2 + r^2 sin\theta d\phi^2 - 2 r^2 \omega sin^2\theta dt d\phi

Given that metric, all you have to do is compute the Rieman tensor.

R^\alpha_{\beta\gamma\delta} = \frac{\partial \Gamma^\alpha_{\beta\delta}}{\partial x^\gamma} - \frac{\partial \Gamma^\alpha_{\beta\gamma}}{\partial x^\delta} + \Gamma^\alpha_{\gamma\rho}\Gamma^\rho_{\beta\delta} - \Gamma^\alpha_{\delta\sigma}\Gamma^\sigma_{\beta \gamma}

Where the Christoffel symbols are defined as follows.

\Gamma^\alpha_{\beta\gamma} = \frac{1}{2}g^{\alpha\delta}\left(\frac{\partial g_{\beta\delta}}{\partial x^\gamma} + \frac{\partial g_{\gamma\delta}}{\partial x^\beta} - \frac{\partial g_{\beta\gamma}}{\partial x^\delta}\right)

And using the line element I posted earlier, the non-zero elements of the metric tensor are following.

g_{tt} = r^2\omega^2 sin^2\theta + \frac{2M}{r} - 1
g_{rr} = \left(1-\frac{2M}{r}\right)^{-1}
g_{\theta\theta} = r^2
g_{\phi\phi} = r^2 sin\theta
g_{t\phi} = -r^2\omega sin^2\theta

The tidal acceleration from curvature is easy.

a^\alpha = - R^\alpha_{\beta r \delta}u^\beta u^\delta L

Since object starts out at rest, u is trivial.

u^\alpha = (\left(1-\frac{2M}{r}\right)^{-1/2}, 0, 0, -\omega \left(1-\frac{2M}{r}\right)^{-1/2})

(Feel free to verify that u^αu_α=-1)

So all you really need are 3 vectors: R^α_trt, R^α_φrφ, and R^α_trφ.

Because I'm obviously not going to do all that work, I threw it into a Mathematica package and crossed my fingers.

a^\alpha=\left(0, \frac{2M}{r^3}L, 0, 0\right)

Which is exactly the same result as I should get without rotation, so that's a good sign. If I substitute in the numbers for M, L, and r, I get tidal acceleration of 0.00015m/s². At the mass of 1000kg, that's 0.15N of tidal force. You should get exactly the same from classical method. (Especially since the classical formula gives you 2GM/r³.)

 

[olivermsun]

...I can get a refund from Dover for Gamow's book?

Perhaps, but from what I could see from the OP, there was nothing wrong with Gamow's explanation.

 

[daniel6874]

Which is exactly the same result as I should get without rotation, so that's a good sign. If I substitute in the numbers for M, L, and r, I get tidal acceleration of 0.00015m/s². At the mass of 1000kg, that's 0.15N of tidal force. You should get exactly the same from classical method. (Especially since the classical formula gives you 2GM/r³.)

Since you have gone to the trouble, perhaps you could explain why this might imply a bulge on the far side of the earth? The page Russ cited says the author found the gradient of the force-field to arrive at his conclusions about the movement of water. The gradient field in his picture, if correct, answers the question. I have access to Mathematica, so if you outline the calculation, that would be sufficient. It is clear we must take into account the shape of the earth.

 

[daniel6874]

Perhaps, but from what I could see from the OP, there was nothing wrong with Gamow's explanation.

The problem would be that if what Russ said about centrifugal force is true, the explanation cannot be correct. I am suspending judgment until I understand the argument better.

 

[K^2]

The computation I showed gives you a differential in forces acting on the front and far sides of the planet or other object. All it says is that the object will be stretched out. Id est, there will be some sort of bulging. Whether that bulging is symmetric or not is a matter of higher order effects which we have not yet discussed.

If you want to use Mathematica to graph what the Earth's surface would look like if it was a perfect fluid, taking into account Earth's gravity, Earth's rotation, and Moon's gravity, I can help you out with this. But maybe you should send me a PM, so that we don't clutter this topic.

Edit: Unless there are other people who want to see it. I suppose, I can just set it up in Mathematica and upload the notebook.

 

[olivermsun]

The problem would be that if what Russ said about centrifugal force is true, the explanation cannot be correct. I am suspending judgment until I understand the argument better.

I'm not sure the posts in the thread have much if anything to do with Gamow's explanation (which appears to be equivalent to Feynman's description, quoted in the other, similar discussion), so you won't understand the argument any better by reading what Russ said.

 

[A.T.]

"Tensorial"? "Simanek's gradient"? "Schwarzschild's metric"? Guys, stick to common sense!
Tide is a phenomenon that on a full/new moon you may go to the beach in the afternoon to pick crabs, moules, and other tasty living beings, while on the quartermoon you could drown in the place you used to pick crabs a week before, rather than being able to pick something for a dinner.

No tensorial analysis is able to explain why on Bretonian beach tides are 2 metres high, while on Lofoten they may be 8 meters high. And even Simanek won't help you hunting crabs.

Or in the words of Bill O'Reilly: You can't explain that!

 

[daniel6874]

I'm not sure the posts in the thread have much if anything to do with Gamow's explanation (which appears to be equivalent to Feynman's description, quoted in the other, similar discussion), so you won't understand the argument any better by reading what Russ said.

I think that Russ and K^2 did address the OP. A quick summary of the positions so far might include:

1. Gamow as per the OP.

2. Feynman, and [extrinsic to this post] [ crackpot link deleted ] daniel6874, the link you cited is not a good link for this site.[/color]. And that F. himself saw problems with his idea.

3. A quaint older text suggesting that the pull of the moon is stronger on the Earth than the water, so the Earth shifts like an egg-yolk toward the moon, leaving a bulge on the far side.

4. A reasonable note by xts that water for high tides is shifted from low-tide areas, and glosses from others [outside this post] on the effect of resonant land formations and separation of oceans to explain tidal height discrepancies.

5. An series of notes by K^2 that suggest the explanation of forces depends on choice of coordinate system.

6. Cited by Russ, http://www.lhup.edu/~dsimanek/scenario/tides.htm, an author who calculates the "tide-generating force" as the gradient of the moon's gravitational potential [I mistakenly said "force"], with a picture that seems to explain tidal bulges in terms of gravitation alone. The author then proceeds to outline what appears to be K^2's approach--the use of a rotating coordinate system with barycenter at the origin, noting that the mathematical origin of the "fictitious" forces is often omitted in this approach.

7. A gloss by harrylin on essentially the egg-yolk theory of 3 above, but updated I think to account for discussion in the post.

8. A.T. posts http://www.vialattea.net/maree/eng/index.htm, which seems to agree with K^2 but like Russ emphasizes the non-use of fictitious forces.

I am tempted to conclude that Russ, K^2, Gamow, and Feynman are all correct, taking into account simplifications and assumptions. After reading the author's home-page in 2 above I no longer think it bears pursuing. Thanks to all who helped with this apparently not-so-easy question.

 

[D H]

I'll start off by noting that I am not a fan of the centrifugal force explanation of the tides.

That said, one cannot say that a centrifugal force explanation is incorrect. K^2 was spot-on correct way back on page 2:

It. Is. Arbitrary. You can chose ANY coordinate system. In any coordinate system, the results will be the same. In some systems, part of the tidal bulging is due to centrifugal force. In other systems, there is no centrifugal force.

You can choose ANY system you like and do the computations. You'll get the SAME answer.

General relativity says that all frames of reference are equally valid. Do the math right and you will come up with the same correct answer regardless of perspective. How can you say that one explanation is right and another is wrong when both explanations yield the same end results? You can't. What you can say is that one explanation is clean and simple in comparison to another that is twisted and ugly. The extra terms that arise in a rotating frame explanation vanish from the perspective of a non-rotating frame. While those extra terms are not essential to explaining the tides, they are not wrong.

The explanations that result from a non-rotating perspective and a barycentric frame rotating with the Moon's orbital rate are valid with respect to the tidal forcing functions. Both are correct with regard to the shape of the Earth itself in response to these forcing functions. The Earth itself undergoes tidal deformations. See http://en.wikipedia.org/wiki/Earth_tide, for example.

Neither explanation is correct with regard to oceanic tides. There is no tidal bulge. There are two huge land masses, the Americas and Eurasia+Africa, that prevent this from arising. To get a correct picture of the oceans' responses to the tidal forces you need to go to yet another perspective, one based on the rotating Earth. There you'll get a picture of amphidromic points, points with virtually no tides, about which the tides appear to rotate.

Picture: http://svs.gsfc.nasa.gov/stories/topex/images/TidalPatterns_hires.tif
Movie: http://svs.gsfc.nasa.gov/vis/a000000/a001300/a001332/full.mov

Since all frames are equally valid, I suppose one could come up with an explanation of the ocean tides from the perspective of a non-rotating frame or a frame that rotates with the orbit of the Earth and Moon about one another. Nobody does that because such an explanation would be hideously complex.

Addendum
An example of a hideously bad explanation based on centrifugal force, courtesy of our government: http://co-ops.nos.noaa.gov/restles3.html. This is so convoluted that I can't see if they've done the math right. There's just too much hand waving and there is no math to check. If they did do the math right they would of course have come up with the same answer as the much, much simpler gravity gradient based explanation.