effect of lunar/solar gravity on shape of earth

K^2

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.

[tex]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[/tex]

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

[tex]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}[/tex]

Where the Christoffel symbols are defined as follows.

[tex]\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)[/tex]

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

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

The tidal acceleration from curvature is easy.

[tex]a^\alpha = - R^\alpha_{\beta r \delta}u^\beta u^\delta L[/tex]

Since object starts out at rest, u is trivial.

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

(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.

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

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

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

daniel6874

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

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

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.

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

daniel6874

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.. 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:
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/top...erns_hires.tif
Movie: http://svs.gsfc.nasa.gov/vis/a000000...01332/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.

russ_watters

No matter how many times I say I understand that the math works out, people keep harping on it, so I must not be making myself clear. Let me try a completely different approach:

I dropped my pen and it fell to the ground. The centrifugal force due to the earth's rotation caused the pen to fall to the ground. True or false? And why or why not?

olivermsun

What's wrong with it? It's pretty much what Gamow and Feynman were talking about.

If you think about it carefully, rotating coordinates aren't so counter-intuitive for explaining tide-producing forces. The tidal "bulges" (in all the simplified examples here) are stationary relative to the earth-moon line, not to relative to fixed points on a non rotating earth in an inertial frame.

olivermsun

Well, causes are a philosophical thing.

But physics-wise, I'd like to hear why anybody would explain a pen falling toward the ground as being due to the centrifugal force due to the earth's rotation.

russ_watters

I disagree, but perhaps that is the whole problem here...
Since as people harped on me previously, we can construct rotating or non-rotating frames from which to analyze the question and all will give the same result, right? So then any effect captured by the calculations can said to be the cause.

olivermsun

Well, not to point at you specifically, but there does seem to be much ado about nothing in some recent threads. There are a few obviously wrong explanations being tossed about, and then there are the persistent disagreements which are based on people's opinions about what is the "right" frame of reference for physics.

As I said, if you think it makes any sense to explain a falling pen that way, then I'd like to hear that construction.

russ_watters

You're missing my point: I don't think it works that way, but that is what I think I am seeing from others.

What I'm seeing from others in this thread is this:

1. Since we can construct a mathematical model of the tides that incorporates a centrifugal force component, we can say that the centrifugal force causes the tides.

So why can't we then say this:

2. Since we can construct a mathematical model of the motion of a falling pen that incorporates a centrifugal force component, we can say that the centrifugal force causes the tides.

Why is one correct and the other not?

harrylin

At first sight they seem to give exactly the same explanation:

'the effect of the external gravitational force produced by another astronomical body may be different at different positions on the earth because the magnitude of the gravitational force exerted varies with the distance of the attracting body.' (but why "may be"?!)

Also (and happily!) they do point out that 'It is important to note that the centrifugal force produced by the daily rotation of the earth on it axis must be completely disregarded in tidal theory.'

PS: note that I fully agree with your criticism on the way they phrase things.

D H

False, of course. Centrifugal force is directed outwards. It explains why your pen takes longer to fall to the ground than it would on a non-rotating planet with the exact same acceleration due to gravity.

Let g' be the local gravitational acceleration vector due to the Earth's mass only. From the perspective of a frame rotating with the Earth, your pen accelerates Earthward at g' due to the Earth's mass and accelerates outward at r×Ω² due to the Earth's rotation. Here r is the vector from the Earth's rotation axis to the pen and Ω is the Earth's rotation vector, 1 revolution per sidereal day. The pen's net acceleration is g=g'-r×Ω². Assuming that gravity and centrifugal acceleration change very little over the short fall, the time the pen takes to fall is given by [itex]\sqrt{2h/g}[/itex]. This calculation becomes even easier if you just use g in the first place rather than g'. For example, the standard value for Earth gravity, g₀=9.80665 m/s² includes both the gravitational and centrifugal forces at sea level and at a latitude of 45.5 degrees.

From the perspective of an inertial observer, the only force acting on the pen is gravity. The pen has an initial horizontal velocity due to the Earth's rotation that puts it on a curved trajectory. That initial horizontal velocity coupled with the Earth's curvature means the pen will fall for a distance slightly greater than the initial height h above the surface of the Earth. Account for that increase in the vertical distance traveled by the pen and make a simple second order assumption (the first order correction is zero) and you will get the exact same answer as above, [itex]\sqrt{2h/g}[/itex]. Just not as easily.

This is one of those occasions where working in a rotating frame is easier than an inertial one.