Effect of lunar/solar gravity on shape of earth

[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.

 

[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]

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.

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]

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?

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]

Well, causes are a philosophical thing.

I disagree, but perhaps that is the whole problem here...

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.

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]

I disagree, but perhaps that is the whole problem here...

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.

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.

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]

[..] 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.

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]

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?

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 \sqrt{2h/g}. 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, \sqrt{2h/g}. Just not as easily.

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

 

[D H]

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.

Better: We can say that centrifugal forces can be used to explain the tides. Saying that they cause the tides is going a bit too far IMO.

Consider a moon the same mass as ours that is in an orthogonal orbit about some planet the same size as the Earth. (Orthogonal orbit: An orbit with a zero angular momentum. When the moon hits the planet, the impact angle will be orthogonal to the ground.) At the instant that this hypothetical moon is one lunar distance from the planet, the tidal forces exerted by this moon on that planet will be exactly the same as the tidal forces exerted by our orbiting Moon on the Earth. Explaining that this is the case is a piece of cake if you use a gravity gradient based explanation. Now try from explaining using a centrifugal force based explanation.

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?

You most certainly could explain the tides from the perspective of an Earth-fixed frame. Explanation follows my response to harrylin's post.

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.'

That's because oceanographers are elided. (Elided because of the old mantra, If you can't say something nice don't say anything at all.)

The reason that centrifugal forces due to the Earth's rotation must be disregarded is because those forces have already been taken into account. The starting point for an Earth-fixed based explanation of the tides is the geoid. The geoid is an equipotential surface of the gravitational plus centrifugal forces. Accounting for those centrifugal force again would be a double booking.

 

[olivermsun]

That's because oceanographers are elided. (Elided because of the old mantra, If you can't say something nice don't say anything at all.)

Hmm...what does this have to do with anything?

 

[D H]

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.

What's wrong with it is that there is no math. Where people do use math in a centrifugal force-based explanation they inevitably do it wrong, and then hand-wave the mistakes away. I know that a correct explanation is possible, but I have yet to see one.

Compare that to a gravity gradient based approach. The desired outcome is the gravitational acceleration due to the Moon (or Sun) relative to the Earth as a whole. The Earth as a whole is accelerating toward the Moon by

{\boldsymbol a}_e = \frac {GM_m}{||{\boldsymbol R}||^3}\,{\boldsymbol R}

where R is the vector from the center of the Earth to the center of the Moon. The gravitational acceleration toward the Moon at some point r away from the center of the Earth is

{\boldsymbol a}_p = \frac {GM_m}{||{\boldsymbol R}-{\boldsymbol r}||^3}\,({\boldsymbol R}-{\boldsymbol r})

The relative acceleration is the difference between these two vectors:

{\boldsymbol a}_{\mbox{tidal}} = {\boldsymbol a}_p - {\boldsymbol a}_e

The vector R can always be expressed in terms of a unit vector \hat{\boldsymbol r} directed along \boldsymbol r and some other unit vector \hat{\boldsymbol{\theta}} normal to \hat{\boldsymbol r}:

{\boldsymbol R} = R(\cos\theta \,\hat{\boldsymbol r} + \sin\theta\,\hat{\boldsymbol{\theta}})

Using this notation and making a first order approximation for r<<R yields

{\boldsymbol a}_{\mbox{tidal}} \approx \frac {GM_m r}{R^3}<br /> \left((3\cos^2\theta-1)\,\hat{\boldsymbol r} + 3\cos\theta\sin\theta\,\hat{\boldsymbol {\theta}}\right)

Note that this covers the entire globe, not just the sublunar point. Try to do the same with a centrifugal-based approach. This is one of those cases where the inertial explanation is much easier, and much more general, than a rotating frame explanation.

 

[D H]

Hmm...what does this have to do with anything?

The places you are most likely to find very bad and/or mathematically incorrect explanations of the tides are textbooks and web sites written by oceanographers.

 

[olivermsun]

What's wrong with it is that there is no math. Where people do use math in a centrifugal force-based explanation they inevitably do it wrong, and then hand-wave the mistakes away. I know that a correct explanation is possible, but I have yet to see one.

Compare that to a gravity gradient based approach. The desired outcome is the gravitational acceleration due to the Moon (or Sun) relative to the Earth as a whole…

Note that this covers the entire globe, not just the sublunar point. Try to do the same with a centrifugal-based approach. This is one of those cases where the inertial explanation is much easier, and much more general, than a rotating frame explanation.

What is easy to explain in a few words as a "reason" for tides and what is easiest to compute for the surface of the Earth are not necessarily the same thing. That doesn't necessarily make the explanation with "no math" less right, nor does it make the most expedient mathematical approach more right.

Personally, I found it kind of an interesting exercise to convince myself (mathematically) that several of the explanations offered earlier in the thread were indeed equivalent. As a side bonus, it tends to help you understand where people arguing from the various viewpoints are (mutually) confused.

The places you are most likely to find very bad and/or mathematically incorrect explanations of the tides are textbooks and web sites written by oceanographers.

Oh boy. I don't suppose you'd care to back this up with any math, or it it just going to be hand-waving?

 

[russ_watters]

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.

Yes. So if we ignore the effect of the centrifugal force, we'll get a different result for an measured acceleration than if we include it -- a wrong prediction about reality. So:

1. Because including a centrifugal force component has a real/non-zero effect on the acceleration of a dropped object, we can say that both gravity and rotation play a role in causing the measured acceleration. One causes positive acceleration, one causes negative acceleration, but both together form the cause of the observed experimental result.

The corollary of this is what I am trying to convey in this thread:

2. Because including a centrifugal force component has no effect on the calculated tidal forces, we can say that the centrifugal force does not play any role in the cause the tidal force.

Better: We can say that centrifugal forces can be used to explain the tides. Saying that they cause the tides is going a bit too far IMO.

Well if I was unclear before, that difference is the entire point I'm trying to make. But I would be more affirmative with it: Because the centrifugal force does not cause the tides, it should not be used in the explanation of the tides because using it in the explanation will fool people into thinking that that means it is the cause. Again, a corollary: explaining the far side bulge using a centrifugal force causes people to wrongly believe that static gravitational force cannot explain the far side bulge. The reason websites exist that are dedicated to dispelling that misconception is because explaining the tides using centrifugal force causes that misconception.

 

[D H]

What's wrong with it?

What's wrong with it? Everything.

Starting with the first sentence.

The tide-raising forces at the Earth's surface thus result from a combination of basic forces: (1) the force of gravitation exerted by the moon (and sun) upon the earth; and (2) centrifugal forces produced by the revolutions of the Earth and moon (and Earth and sun) around their common center-of-gravity (mass) or barycenter.​

Right of the bat, this is just wrong. First off, what centrifugal force? To have a centrifugal force you have to have a rotating frame. There is no centrifugal force in an inertial frame in which the moon orbits the earth. There is a centrifugal force in a frame that rotates with the earth-moon system, but in that frame the Earth and the moon are stationary. They are not orbiting one another in that frame.

A centrifugal force explanation of orbits can be useful at times. A geosynchronous satellite sits in a fixed position in the sky from our earthbound perspective. From our perspective, such a satellite indeed does have gravitational and centrifugal forces that are in balance, and per Newton's first law (extended to non-inertial frames), the satellite remains stationary. A rotating frame in which the Earth and moon do not move can be useful at times. Such a frame is very useful for describing the Lagrange points, for example, and also is very nice for depicting the trajectory of a satellite in transit from the Earth to the moon.

Continuing with the next paragraph,

With respect to the center of mass of the Earth or the center of mass of the moon, the above two forces always remain in balance (i.e., equal and opposite). In consequence, the moon revolves in a closed orbit around the earth, without either escaping from, or falling into the Earth - and the Earth likewise does not collide with the moon.​

That is the silly centrifugal force description of an orbit. And it is wrong. Look at it from the perspective of Newton's first law. If the two forces truly are in balance (i.e., equal and opposite), there is no closed orbit. There is only straight line motion or no motion at all.

Now let's look at the image that follows. You will see a similar image in most descriptions that invoke centrifugal force.

[URL]http://co-ops.nos.noaa.gov/images/restfig1a.gif[/URL]

The moon and Earth are clearly shown as orbiting one another here. There is no centrifugal force in this diagram.

Let's go on to the next diagram.

[URL]http://co-ops.nos.noaa.gov/images/restfig2.gif[/URL]

And this is flat out wrong. What's wrong is that this site (and almost every text that invokes centrifugal force) has the centrifugal force identical at the center of the earth, at the sublunar point, and at its antipode. The centrifugal force at a point removed from the axis of rotation by a distance r is rΩ². The sublunar point is only 1068 miles from the barycenter; the center of the earth, 2895 miles, and the antipodal point, 6858 miles. The centrifugal force is not the same at these three points. Do the math right and you do not get the tidal forces as shown in the diagram from the perspective of this rotating frame.

That's okay, though. Force is a frame dependent quantity when one allows fictitious forces to enter the picture. We don't want the tidal forces as observed from the perspective of this frame. We want the tidal forces as observed on the earth. To get a correct explanation we need to move the origin to the center of the Earth -- and we need to get rid of that monthly rotation. The centrifugal forces that we added in need to be subtracted out.

What we're left with is a non-rotating with origin at the center of the Earth. This frame is the earth-centered inertial frame. Despite its name, this isn't really an inertial frame; it is accelerating toward the moon and the sun. This acceleration of the origin results in a fictitious force, and it is a uniform fictitious force. This fictitious force is the inertial force, not the centrifugal force.

There is little need to invoke the earth-moon barycenter here and there is absolutely no need to invoke the centrifugal force.

 

[harrylin]

[..] The reason that centrifugal forces due to the Earth's rotation must be disregarded is because those forces have already been taken into account. The starting point for an Earth-fixed based explanation of the tides is the geoid. The geoid is an equipotential surface of the gravitational plus centrifugal forces. Accounting for those centrifugal force again would be a double booking.

Yes, I already mentioned that in post #41.

 

[olivermsun]

What's wrong with it? Everything.

You seem to start with the conclusion that the explanation is wrong (or more accurately, not to your tastes), and then go on to interpret everything in the "most wrong" way possible. Since you do indeed get the same numerical answers for the tide-producing "forces" in either an inertial frame or a moving frame with "centrifugal" forces, let's try and see why this could happen despite "everything" being wrong.

Starting with the first sentence.

The tide-raising forces at the Earth's surface thus result from a combination of basic forces: (1) the force of gravitation exerted by the moon (and sun) upon the earth; and (2) centrifugal forces produced by the revolutions of the Earth and moon (and Earth and sun) around their common center-of-gravity (mass) or barycenter.​

Right of the bat, this is just wrong. First off, what centrifugal force? To have a centrifugal force you have to have a rotating frame. There is no centrifugal force in an inertial frame in which the moon orbits the earth.

At this point, no frame has been specified, so I would tend to assume that the usage of centrifugal force is colloquial, or else that a rotating frame is about to be used.

There is a centrifugal force in a frame that rotates with the earth-moon system, but in that frame the Earth and the moon are stationary. They are not orbiting one another in that frame.

I think it's fair to assume that a reader will understand that the centrifugal forces "observed" in the rotating frame are the result of orbital motion.

With respect to the center of mass of the Earth or the center of mass of the moon, the above two forces always remain in balance (i.e., equal and opposite). In consequence, the moon revolves in a closed orbit around the earth, without either escaping from, or falling into the Earth - and the Earth likewise does not collide with the moon.​

That is the silly centrifugal force description of an orbit. And it is wrong. Look at it from the perspective of Newton's first law. If the two forces truly are in balance (i.e., equal and opposite), there is no closed orbit. There is only straight line motion or no motion at all.

In the rotating frame, which as you pointed out is the only frame in which the centrifugal "force" is observed, the forces are balanced. Hence in the rotating frame, we do not observe the moon falling into or away from the earth. The reader probably realizes that such motion (or lack thereof) observed in the rotating frame corresponds to a closed orbit in the inertial frame.

Is centrifugal force really this confusing?

Now let's look at the image that follows. You will see a similar image in most descriptions that invoke centrifugal force.
...
The moon and Earth are clearly shown as orbiting one another here. There is no centrifugal force in this diagram.

The orbiting satellite problem is commonly illustrated in physics textbooks using a schematic such as this. The orbital motion is shown in the inertial frame so that we understand the "big picture," but the solution includes defining (and sketching in) the proper moving frame axes. Here no particular coordinates have been defined (yet?).

Let's go on to the next diagram.

[URL]http://co-ops.nos.noaa.gov/images/restfig2.gif[/URL]

And this is flat out wrong. What's wrong is that this site (and almost every text that invokes centrifugal force) has the centrifugal force identical at the center of the earth, at the sublunar point, and at its antipode. The centrifugal force at a point removed from the axis of rotation by a distance r is rΩ². The sublunar point is only 1068 miles from the barycenter; the center of the earth, 2895 miles, and the antipodal point, 6858 miles. The centrifugal force is not the same at these three points. Do the math right and you do not get the tidal forces as shown in the diagram from the perspective of this rotating frame.

I'll grant you, this figure is not really clear or rigorous regarding what quantities are really being represented by what "vectors." Amusingly, however, the centrifugal forces actually do become equal at points A and B when the math is "done right," and you do in fact get the correct tide-producing forces.

That's okay, though. Force is a frame dependent quantity when one allows fictitious forces to enter the picture. We don't want the tidal forces as observed from the perspective of this frame. We want the tidal forces as observed on the earth. To get a correct explanation we need to move the origin to the center of the Earth -- and we need to get rid of that monthly rotation. The centrifugal forces that we added in need to be subtracted out.

We didn't add in or take away anything at all. The Earth is observed to be rotate even in the "inertial" frame, and we simply choose to decompose its total rotation into parts which are either convenient or intuitively useful. For the rotating frame, some may choose to keep the part of the rotation which locks the Earth and moon faces together and then deal with the consequences later.

What we're left with is a non-rotating with origin at the center of the Earth. This frame is the earth-centered inertial frame. Despite its name, this isn't really an inertial frame; it is accelerating toward the moon and the sun. This acceleration of the origin results in a fictitious force, and it is a uniform fictitious force. This fictitious force is the inertial force, not the centrifugal force

There is little need to invoke the earth-moon barycenter here and there is absolutely no need to invoke the centrifugal force.

This just sounds like frame (or word) games to me.

And finally, to respond to this oft repeated point:

The reason that centrifugal forces due to the Earth's rotation must be disregarded is because those forces have already been taken into account. The starting point for an Earth-fixed based explanation of the tides is the geoid. The geoid is an equipotential surface of the gravitational plus centrifugal forces. Accounting for those centrifugal force again would be a double booking.

Which is why, at the end of the rotating frame computation, you subtract the "symmetric" component of centrifugal force which is due to rotation around the Earth's center. Then, you get a nice final answer which is numerically correct, leaves the symmetric part in the geoid where it's already done for you, and shows you some interesting equivalences between frames rotating around different axes to boot.

 

[A.T.]

Because including a centrifugal force component has no effect on the calculated tidal forces, we can say that the centrifugal force does not play any role in the cause the tidal force.

I think in general that cause/effect arguments tend to get too philosophical. But I think the point you trying to make is:

- The gravitational gradient is the same in every reference frame
- The tidal effect computed in every reference frame must be the the same

Therefore:

- The sum of all inertial forces (which might arise from the choice of the reference frame) must not affect the tidal effect.

 

[A.T.]

Amusingly, however, the centrifugal forces actually do become equal at points A and B when the math is "done right,"

1) In which rotating frame would the centrifugal force vector be the same for A & B (as shown in the picture)?

2) If the centrifugal force vector is the same for A & B & C (as shown in the picture) then it cannot cause/affect the tides, so why show/mention it in an explanation of tides?

 

[olivermsun]

1) In which rotating frame would the centrifugal force vector be the same for A & B (as shown in the picture)?

The residual vectors really are equal, pointing in the same direction and everything, once the Earth rotation at 1/lunar orbit is taken out. I suppose that's part of what's amusing about it. The vectors make sense since every point on the Earth (if you take out all the rotation around its axis) translates in a circle with the same radius as the distance from the center of the Earth to the barycenter.

2) If the centrifugal force vector is the same for A & B & C (as shown in the picture) then it cannot cause/affect the tides, so why show/mention it in an explanation of tides?

It provides the constant "offset" that gives you a tide-producing force toward the moon on the near side, away on the far side, and with the whole system static in the moving frame (the moon and Earth maintain their relationship).

 

[A.T.]

The residual vectors really are equal, meaning pointing the same direction, once the Earth rotation at 1/lunar orbit is taken out.

I have not idea what you mean by "residual vectors" and "taking out the Earth rotation". The inertial centrifugal force in rotating frames points away from the frame's axis of rotation, and has the magnitude m \omega² r, where \omega is the angular velocity of the frame, and r is the distance to the frame's axis of rotation. So again:

In which rotating frame would the centrifugal force vector be the same for A & B (as shown in the picture)?

It provides the constant offset which allows the tide-producing force to be pointing toward the moon on the near side and away on the far side while

A uniform inertial force field causes uniform acceleration, so it cannot cause or contribute to any deformation (tides).

 

[olivermsun]

A uniform inertial force field causes uniform acceleration, so it cannot cause or contribute to any deformation (tides).

The moon's gravitational force is nonzero and pointed toward the moon (even if there is a gradient), yet the Earth center stays still in the rotating frame, and the "tide-producing force" on the antipodal side of the Earth even has opposite sign. Hence the centrifugal force is "necessary" to describe what is observed in the rotating frame.

I have not idea what you mean by "residual vectors" and "taking out the Earth rotation"...So again:
In which rotating frame would the centrifugal force vector be the same for A & B (as shown in the picture)?

Let r be the radius of the earth, x the distance from the barycenter to the center of the earth, and \omega the orbital frequency. In the rotating frame, the "centrifugal acceleration" at the center of the Earth is \omega^2 x directed to the right, at A is -\omega^2 (r - x), directed toward the left, and at B is \omega^2 (r + x) toward the right again. Since people like to remove the component of centrifugal force which is due to rotation around the Earth's axis and put that into the geoid, take out outward accelerations -\omega^2 r, +\omega^2 r from the accelerations at A and B, respectively. The remaining centrifugal bits are what I called the "residual" vectors. What are they?

 

[A.T.]

The remaining centrifugal bits are what I called the "residual" vectors.

So the Fc vectors shown in the picture are actually not the centrifugal force, but the uniform components of the centrifugal force, left after the radial component in respect to the Earth's center was removed.

What are they?

They form a uniform field, which cannot create tidal deformation, so they are not a good "reason" for the tides. But I see how in that particular frame they can be used to "explain" the opposite bulge on a static Earth:

<- Moon's gravity is pulling to the left with a r^-2 fall off.
-> Uniform centrifugal component pulls to the right

- In the Earth's center they cancel
- In A Moon's gravity wins and creates a bulge to the left
- In B Uniform centrifugal component wins and creates a bulge to the right

 

[D H]

et r be the radius of the earth, x the distance from the barycenter to the center of the earth, and \omega the orbital frequency. In the rotating frame, the "centrifugal acceleration" at the center of the Earth is \omega^2 x directed to the right, at A is -\omega^2 (r - x), directed toward the left, and at B is \omega^2 (r + x) toward the right again. Since people like to remove the component of centrifugal force which is due to rotation around the Earth's axis and put that into the geoid, take out outward accelerations -\omega^2 r, +\omega^2 r from the accelerations at A and B, respectively. The remaining centrifugal bits are what I called the "residual" vectors. What are they?

What they are is invalid. That is exactly the hand-waving stuff I was talking earlier. It is invalid.

Here's a website where an oceanographer does it right: http://oceanworld.tamu.edu/resources/ocng_textbook/chapter17/chapter17_04.htm. Emphasis mine:

Note that many oceanographic books state that the tide is produced by two processes: i) the centripetal acceleration at Earth's surface as Earth and moon circle around a common center of mass, and ii) the gravitational attraction of mass on Earth and moon. However, the derivation of the tidal potential does not involve centripatal acceleration, and the concept is not used by the astronomical or geodetic communities.​

This is a very nice description of the tides. Stewart starts off in terms of potential, develops the horizontal component of the tidal force, and then proceeds to dive deep, eventually reaching the concept of Doodson numbers.


Let's see what that constant r\omega^2 term of yours really is. The angular velocity of a pair of objects orbiting one another circularly due to gravity is given by

\omega^2= \frac{G(m_1+m_2)}{R^3}
where ω is the angular velocity, G is the universal gravitational constant, m₁ and m₂ are the masses of the two bodies, and R is the distance between the two bodies. (This generalizes very nicely to elliptical orbits; it is Kepler's third law.)

Multiply both sides of the above by r₁, the distance between the center of mass and m₁, and we have the desired rω² term on the left hand side:

r_1\omega^2 = \frac{G(m_1+m_2)r_1}{R^3}
The relationship between r₁ and R is given by m₁r₁=m₂r₂=m₂(R-r₁) or (m₁+m₂)r₁=m₂R. Substituting this into the right hand side of the equation for r₁ω² yields

r_1\omega^2 = \frac{Gm_2}{R^2}
Recognizing that our generic body 1 is the Earth and that body 2 is the Moon, we have

r\omega^2 = \frac{Gm_{\text{M}}}{R^2}
The right hand side is the translational acceleration of the Earth toward the Moon. This is what you really want because the answer you really want is in a frame with origin at the center of the Earth and rotating with the Earth.

You can use a centrifugal force concept to develop those forces in that frame with origin at the barycenter and rotating at the Earth-Moon orbital rate. Do it right and you will get some goofy answers. That's okay though because those answers are to a question other than the question we are trying to answer. To answer the question that was asked, you are going to have to transfer those forces from that frame to the Earth-centered, Earth-fixed frame. All of those centrifugal terms you so painstakingly added in have to be painstakingly subtracted out. A barycentric POV misses one fictitious force, that due to the translational acceleration of the Earth toward the Moon. This missed term also needs to be taken into account when transforming those forces from the rotating barycentric frame to an Earth-centered frame. Do the math right and you will get the same answer as you would have gotten by starting in an Earth-centered frame in the first place. So why not start in an Earth-centered frame in the first place? For one thing, the math is a lot easier, and for another, it avoids all this going around your elbow to get to your thumb business.

Just because the fictitious centrifugal force in the rotating barycentric frame at the center of the Earth has the same value as the fictitious inertial force in an Earth-centered at the center of the Earth does not mean these are the same fictitious force. They are very different fictitious forces.

 

[A.T.]

A barycentric POV misses one fictitious force, that due to the translational acceleration of the Earth toward the Moon.

I don't think it misses that inertial force. I think the uniform component of the inertial centrifugal force in the barycentric frame (that olivermsun called "residual vectors", and you called "invalid", and is called Fc in the picture) is exactly that inertial force.

The decomposition of the centrifugal force in the barycentric rest frame is explained here in Fig 12
http://www.vialattea.net/maree/eng/index.htm

To answer the question that was asked, you are going to have to transfer those forces from that frame to the Earth-centered, Earth-fixed frame.

I agree. And in the Earth-centered frame you will always have that uniform inertial force field, pointing away from the Moon. This is even true without any rotation, just linear fall as described in Case 1 of:
http://www.vialattea.net/maree/eng/index.htm

This uniform inertial force field turns this:

into this:

So I think it is OK to use that uniform inertial force field to explain why the water on the opposite side is accelerated away from the fixed Earth's center.

What is misleading, is calling that uniform inertial force field "centrifugal force".

 

[D H]

I don't think it misses that inertial force. I think the uniform component of the inertial centrifugal force in the barycentric frame (that olivermsun called "residual vectors", and you called "invalid", and is called Fc in the picture) is exactly that inertial force.

There is no such fictitious force in the barycentric frame from a Newtonian perspective. A non-rotating barycentric frame is, ignoring the solar tides, an inertial frame. There are no fictitious forces here. None. You get fictitious forces in a barycentric frame only if it is rotating. Fictitious forces that result from rotation are not everywhere uniform.

There is one, and only one, fictitious force in Newtonian mechanics that is everywhere uniform: The fictitious force that results from choosing an origin that is accelerating with respect to an inertial frame.

So I think it is OK to use that uniform inertial force field to explain why the water on the opposite side is accelerated away from the fixed Earth's center.

No dispute.

The simple answer, that the tidal acceleration is the difference between the gravitational acceleration of the Earth as a whole toward the Moon from the gravitational acceleration at some point of interest toward the Moon, looks just as hand-wavy as getting rid of those "residual vectors". What justifies this subtraction? What justifies it is that the question implicitly asks us to look at things from the perspective of an accelerating reference frame (accelerating from a Newtonian perspective, that is). Such a perspective results in a d'Alembert force, the inertial force resulting from the acceleration of the origin of the reference frame.

What is misleading, is calling that uniform inertial force field "centrifugal force".

Exactly.


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You referenced this website, http://www.vialattea.net/maree/eng/index.htm, in your post. The author opens with

At many websites, but also in some textbooks, one finds plainly wrong explanations of the lunar and solar tides. In particular much confusion arises when authors try to explain the existence of the second tidal bulge, the one opposite to the Moon. Often they invoke the centrifugal force as an "explanation". But centrifugal force is a fictitious force, and we can't justify a real effect with a fictitious force, can we?​

That is perhaps a bit too dismissive. Moreover, a physicist who comes from a GR background would most likely say that it is wrong. Let me parody Sirtoli's opening paragraph to demonstrate why (these are not Sirtoli's words):

At many websites, but also in some textbooks, one finds plainly wrong explanations of the lunar and solar tides. In particular much confusion arises when authors try to explain the existence of two tidal bulges. Often they invoke the gravitational force as an "explanation". But gravitational force is a fictitious force, and we can't justify a real effect with a fictitious force, can we?​

 

[A.T.]

A non-rotating barycentric frame is, ignoring the solar tides, an inertial frame.

Okay, sorry. I was talking about a frame rotating around the common center of mass, the one that Sirtoli used in case 3.

What justifies this subtraction? What justifies it is that the question implicitly asks us to look at things from the perspective of an accelerating reference frame (accelerating from a Newtonian perspective, that is).

Yes, exactly. The explanation of the opposite bulge in the inertial frame that invokes the difference in accelerations, between the Earth's center and point B, is in fact implicitly doing a transformation into the non-inertial rest frame of the Earth's center, where an inertial force creates that bulge. Mathematically these two approaches are equivalent.

But centrifugal force is a fictitious force, and we can't justify a real effect with a fictitious force, can we?

That is perhaps a bit too dismissive.

I agree that this is not a matter of real vs. inertial forces. Treating gravity as an inertial force doesn't change the tides.