Milli360, you seem to have convinced Richard J. Hanak about this revolution/rotation thing – he’s even amended his web site accordingly – but I’ve been waiting for some resolution to the relevance of centrifugal force in raising the tides.

Your response to his amendment was:

and there followed by a minor exchange about centrifugal force before we all sailed off up (or is it down?) the Mississippi – but it didn’t seem to resolve anything.

I believe your central assertion can be found here:

As I understand it, this conclusion follows from your observation that, when all other rotations are removed, there is no net centrifugal tidal force due to earth’s rotation about the (earth/moon) barycentre – since all points on (and indeed, within) the earth exhibit have the same rotation.

(I hope this is an accurate statement of you position – I am not contending anything here).

Given, then, that there is some rotation (albeit the same at all points) what, in your view, would be the effect of that rotation. What effect does the centrifugal force, arising from this rotation, have on the oceans or the surface generally. Presumably we can’t have a force that does nothing at all and if it doesn’t raise a tide what does it do?

Note. For the sake of argument I’m going along with this ‘centrifugal force’ business – but I won’t be able to keep it up for long!

I have switched usernames to A Thousand Pardons. Mea culpa.
It seems accurate to me.
Centrifugal force is sometimes called a fictional force, because it depends upon the reference frame. I prefer to treat it as a real force (especially since Einstein showed that gravity itself can depend upon ones reference frame), but with the understanding that we need careful treatment within that reference frame.

But since the centrifugal force is constant in magnitude and direction, there is a reference frame in which that force disappears. The great thing about that particular reference frame is that the gravity field is constant and unmoving in that reference frame--which makes things easier to compute.
Imagine a body falling in a gravity field. Gravity acts on all parts of the body, so that the body experiences no net force--it's in free fall. It "feels" as if there is no force at all. Of course, in a real gravity field there will be a small tidal force, but that is exactly what we are talking about. If the gravity field will truly constant (as the centrifugal force is), there wouldn't even be a tidal force.

So, that's why I say the centrifugal force contributes nothing to the tidal force.

The most common mistake in these analyses, that I've seen, is that someone will compute the centrifugal force on the nearside and compare it to the farside, and conclude that those are what is responsible for the tide. But if they were to make a similar computation for the sideside (eh), they would find it to be the same--and all of them are directed away from the center of the body, making it clear that they are not really tidal forces, but equatorial bulge forces arising from the rotation of the body.

And I tried (unsuccessfully) to get him to actually try the conventional explanation and see if it works. What I got instead was something to the effect of:

'The conventional doesn't work because the conventional explanation is being applied wrong (by every scientist, ever). So if the conventional explanation is applied correctly, it doesn't work and if it is applied incorrectly, it works. Therefore it is wrong.'

uh huh... :roll:

For me, the issue is easily solved by seeing if the conventional explanation actually works. And we have ample evidence that it does. If the calculated predictions compared with reality aren't good enough, comet SL9 provided a pretty good demonstration absent of centrifugal forces.

A Thousand Pardons wrote:What is the reference frame in which the centrifugal force of orbiting the barycenter disappears? Does something about that reference frame disable Newton’s law of gravitation as well, so that the moon and earth no longer attract each other?

P.S You were right again! I have corrected my model for the solar centrifugal force and will have revision 2 up soon. Thanks again.

How is it you all would explain the fact of high tides on the opposite sides of the earth simultaneously? Does the moon not seem a little inadequate for this purpose? To say it could pull one side's water is one thing, but how does it pull the opposite water out in the opposite direction?

Thanks,

Sincerely,

Gary Shelton

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Gary Shelton

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Ooo! Ooo! Can I say it?

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Yes, I do understand your position on what the (monthly) centrifugal force does NOT do.

My question is what do you think it DOES do?

You see, this is a cardinal force in Richard Hanak’s model. If you are to make a credible challenge to it you need to show (i) that it does not exist or (ii) that it acts in some way contrary to that claimed by Richard.

You’ve never claimed that it does not exist so the question remains: what DOES it do?

"Disable"? No, of course not.

The reference frame is that attached and fixed to the body. Since the body is not moving in that reference frame, there is no centrifugal force. If there had been differences in the centrifugal force across the body, the differences would also show up in that reference frame.
I think I answered that before. It is a constant force, applied with equal direction and magnitude across the entire body--hence, no tidal force.
I've done that, ii. Richard has thanked me for those corrections, graciously.
No, in some reference frames, it does not even exist at all, as I point out above. But if one is to use a reference frame where it does exist, one has to be very careful. It turns out that any and all computations involving centrifugal force (in the idealized form that we are using) do not contribute a tidal effect.

A Thousand Pardons et al:

Having been led to the light, I have seen that the barycentric centrifugal forces on earth are constant in magnitude and always directed away from the moon. Therefore, they cannot contribute differences giving rise to tides. HOWEVER, the resultant of the lunar gravitational force and the barycentric centrifugal force changes direction only because on the far half of the earth the lunar gravitational force directed toward the moon is weaker than the barycentric centrifugal force directed away from the moon. Including barycentric centrifugal force eliminates the paradox of gravitational attraction of the moon causing a force directed away from the moon.

When the gravitational force at the center of the earth is taken as a reference force and is then subtracted from any lunar gravitational force (making the force at the center zero), what is really taking place is a shell game with words. That sleight of words hopes that you will not notice that changing the sign of that central lunar gravitational force (by subtraction) converts it to the barycentric centrifugal force with which subsequent resultant forces will be formed. If not an intentional shell game, it would seem to be self-delusion or maybe only plain forgetfulness. I won’t cast stones about the last two.

"Only"? No, it's only because of the particular reference frame that one uses that it changes direction.
There is no paradox, really.
No, it's simple mathematics.
O good. But you have missed one other possibility in your list. It is not an intentional shell game, and it is not self-delusional, and it is not forgetfulness. Sometimes people just don't have enough mathematical sophistication. I know I don't. Please don't hold it against me.

That's what everyone thinks the first time they consider the question. Its normal. But some people never get past it. Please read the explanation UT linked so you don't get hung up on that. Nothing (measurable) at all. That should be obvious - if the moon's gravity causes all of our measured tidal variation, the centrifugal force can't contribute anything....

Wait, isn't that circular? If I say (using ATP's words with one small change): my chair contributes nothing to the tidal force, are you going to follow up with well, what DOES your chair do? I sure hope not. Yes, I see now: its much more likely that every physicist, ever is wrong (either delusional or part of a conspiracy) and you alone are right. Why didn't I see that before! :roll:

He has indeed thanked you, frequently and most graciously – but he’s not quite on board yet is he?

His model still requires a contribution from centrifugal force but you say centrifugal force contributes nothing. This is what I’m trying to resolve. I thought the easiest way would be for you to state clearly what the centrifugal force actually does. Unfortunately I can’t get you to say – and I won’t be asking for a third time.

I suggest you reconsider this Richard – your conclusion is incorrect.

Lunar gravity (acting on earth) also has the same direction – in this case towards the moon – but this does not preclude it being a tide raising force. In fact uniform direction is precisely what you DO need for a tide raising force.

Notice that a uniform force cannot cause oblateness, this requires a force that is directed radialy. Earths oblateness is caused by (in your terms) centrifugal force acting radialy and arising from axial rotation about the poles.

The fact that the force is constant in magnitude (for the case under consideration) is not important, this only affects the magnitude of the subsequent tide. What makes it a tide raising force is its uniform direction.

If you are still unclear about this why don’t you add, say, a white dot to your Figure 3, opposite to the black dot. Then draw in the direction of your centrifugal forces for both dots. If these forces are the same then when one is pointing outwards - tide raising - the other must be pointing inwards – tide depressing. I do urge you to draw it – don’t try to imagine it. You might as well draw in a lateral pair of dots (side to side) as “A thousand Pardons” suggests – you’ll find both of these cause a tan force at the surface, exactly as required.

Tidal forces are uni-directional.

Oblateness forces are radial.

I could not agree with you more Russ. Your logic is quite unassailable. Unfortunately it’s also incomplete. It contains an ‘if’ and a ‘then’, for completeness it needs an ‘else’ because that’s what we are talking about aren’t we?

Well actually Russ, yes I would.

If Richard Hanak had made a credible case that your chair contributes to the tidal force, and if you wished to dispute it, then I most certainly would expect your challenge to explain what your chair actually DOES do. Do you not think this reasonable?

(Else ‘do nothing’ !!).

I must confess I was a bit thrown by your reference to Adenosine Triphosphate.

A Thousand Pardons wrote:The body under consideration is the earth. In that reference frame the earth does not rotate and is not subject to centrifugal force. What force, in that reference frame, causes the earth’s equatorial bulge? If the reference frame were the sun, what would cause the spectrographic shifts we observe at the limbs of the sun? If the reference frame were a star exhibiting spectrographic line widening, what would cause the spectrographic line widening we observe for that star?

RichardMB: I will respond to your post later.

Gravity, as I've maintained all along.
Any reference frame can be employed, and I am comfortable using one in which there are centrifugal forces. But they must be computed correctly, if we are to draw appropriate conclusions from their magnitudes.

It appears that RichardMB would have us compute the gravity vectors and then add the centrifugal force vectors and arrive at the tidal vectors. I do not have a problem with that approach at all. At the center of the Earth, the centrifugal force is equal and opposite to the gravity force--so adding in the centrifugal force is the same thing, mathematically, as subtracting that gravity force. The BA explanation takes advantage of the latter.

The main problem I have seen with the centrifugal force attempts at explaining the tides is that confusion often arises about how to compute the quantity--rotational effects cause equatorial bulges which are not tidal bulges, and they should not be included in the computations. Those are the corrections I have made to your work.

The other problem I have with such explanations--and this has nothing to do with the science or math of it--is that often the explainers fail to understand the other approaches, and criticize them. In your case, you've attacked them as misleading, or self-delusioned. That's very unfair, and ironic.

Richard, I am concerned that you might be trying to incorporate my observation into your September revision. This won’t work, you need to revert to your original, June, model. The September revision is flawed because it is based on a misunderstanding about rotation:

This is not correct. The moon rotates once per month. That’s it. No ‘revolves’.

Likewise the earth also rotates once per month and this what you must base your analysis on – as indeed you did initially.

But for goodness sake don’t take my word for it. Look it up in any maths or physics book.

I was referring to the monthly orbit of the moon about the Earth.
??

Generally, it's said that the Earth rotates once per day, approximately. What are you referring to?
Or even a good dictionary, although there are some small discrepancies. Even most sources on the internet are adequate.

Is it not clear from the context? The topic is the tides – so I am referring to rotation about the barycentre. The moon and the earth rotate once a (sidereal) month around the earth/moon barycentre.

Of the numerous rotations characterising earths passage through space, its daily rotation (about the polar axis) is the only one that does not contribute to the tides. As previous posts clarify, the daily rotation is responsible for earth’s oblateness.

Yes, most sources on the net are adequate but, as Richard Hanak has discovered, not all. Given his recent experience I think he is better advised to avoid the net, which is precisely why I suggested books. Rotation is well defined in the literature so it should not give him a problem.