In another thread, it was recommended to read Miles Mathis's essay,
"The Trouble with Tides". I commented on his apparent misunderstandings and incorrect conclusions. This post is to elaborate on those issues.
Miles Mathis wrote:We know the ocean tides are caused by the Moon, since they follow lunar cycles. But are they caused by the Moon’s gravity? Let’s look at some numbers. Let’s compare the Sun’s field to the Moon’s field, at the Earth.
Miles's initial intuition appears to be that tidal forcing ought to be proportional to gravitational field strength. As MoscaicDave pointed out, this is simply wrong. Perhaps this is not actually his belief, but only a rhetorical device to guide readers who do have this intuition. I cannot tell.
aS = force on the Earth by the Sun
aS = GMS/r2
Actually, that's the gravitational field (an acceleration), not a force. This kind of sloppy use of language can only confuse the subject. Similarly, when we get to applying the 1/r
3 law for tidal forcing:
FS = GmMS/r3
= 2.4 x 1011 N
The units of this expression are N/m, not N. The tidal force on an object is proportional also to the distance of the object from the center of the Earth. ("Tidal force" here refers to the difference between the actual gravitational force on the object and what it would be at the Earth's center. Also, this only applies to the linearized approximation along the line between the centers of Earth and Sun.) Miles has, furthermore, used the mass of the entire Earth for m, when in fact m refers to the mass of an object (or a piece of the Earth) at a particular location. If Earth was more massive, it would not directly increase the tidal effects other bodies have on the Earth.
So the gravitational attraction of the Sun is 178 times greater than that of the gravitational attraction of the Moon. But how can this be? We all know the moon is more effective in producing tides than the Sun. There is a simple explanation for this, and it is not that we have been lied to! It is only the proportion of the gravitational force not balanced by centripetal acceleration in the Earth’s orbital motion that produces the tides.2
Two major problems here. One, the gravitational force causes the centripetal acceleration. There can be no lack of balance.
The centripetal acceleration referred to is that of the Earth as a whole. If the Earth were completely rigid and not rotating, it would be the same for all points on Earth. (The details of Earth's rotation are separable and tidally neutral.) The gravitational force, however, is not the same at all points on Earth, so the equation F
gravity=ma
centripetal does not balance for each part of Earth. Instead, we see tidal effects in the form of tidal stresses (in addition to F
gravity) and dynamic responses (in addition to a
centripetal).
As for the gravitational and centrifugal forces, although they are caused separately, they cannot cancel, since they both tend to create tides. In fact, most physics books and websites use a summation of centrifugal effects and gravitational effects to create tides on the Moon, as I will show below, since both are tidally positive.
Miles does not cite sources here, so I cannot determine whether the "most physics books and websites" in question are giving incorrect explanations (which is, sadly, possible), or Miles has simply misunderstood them. However, centrifugal forces
cannot contribute to tidal forcing, as I'll show in a moment.
There are no centrifugal forces on the Earth directly caused by the Moon, since there is no angular velocity around the Moon.
This shows that Miles does not understand that centrifugal forces are not real forces. They are pseudoforces one must introduce to balance F=ma in a rotating coordinate system. They are not caused by physical objects, but are artifacts of non-inertial reference frames. They cannot have physical effects that are not accounted for by analysis in an inertial frame. Any mention of centrifugal forces without a well defined rotating reference frame is sloppy, and should be avoided because it only leads to confusion.
So, let's define a suitable rotating reference frame. Let the barycenter lie at the origin, and let the frame rotate with angular speed ω, such that the Earth and Moon (or Sun, as appropriate) are at rest in the frame. That is, they have an acceleration of 0. This is why one says the centrifugal and gravitational forces are in (imperfect) balance, since with a=0, F
total=0. Now, let us look as a piece (of mass m) of Earth in the plane of rotation, at a (vector) position
r relative to the barycenter. (I'll use underscores for vectors here, since I'm finding bold isn't showing clearly enough.) The centrifugal force acting on this piece is
Fc = mω
2r
Let the position of the center of the earth relative to the barycenter be
R (constant, in this reference frame), and let
x be the position of the piece relative to the center of the Earth. Then
r =
R +
x
If we substitute this simple vector relationship into the expression for centrifugal force, we get
Fc = mω
2R + mω
2x
The first term, mω
2R, is a uniform contribution at all points on Earth, and is therefore tidally neutral. The second term, mω
2x, is axisymmetric about the Earth's center, and is therefore also tidally neutral, just like the effect of the Earth's rotation about its axis. (In fact, it is exactly equivalent to the rotation one would observe in an inertial reference frame if the Earth were to appear not to rotate in our rotating reference frame.) When Miles later calculates the centrifugal force at the near and far edges of Earth, and notes their difference, he is missing the fact that it varies in an axisymmetric way, not preferentially along the line between Earth and Sun/Moon. This is not the case for the linearized gravitational variations. Although the derivation often leaves it our for simplicity, it can be shown that there is no linear term in the variation of the gravitational field in the perpendicular direction. (Higher order variations are much smaller than the linear terms when the gravitational source is distant compared to the size of Earth.)
Secondly, the math above is dishonest. If we look at the Sun/Earth system, then the center of gravity of the two bodies is so close to the center of the Sun that it makes no difference. The Earth has almost no effect on the Sun. Therefore, the distance L is just the radius R, and the equation is the same as
Ft = GmMR/R3
Miles is responding here to the equation F
t = GmML/R
3 found in a brief FAQ answer which defines L as the distance from "the center of gravity of the system". I believe this is meant to be the linearized result given later as F
t = 2GmMr/R
3. That is, L is supposed to be the distance from the center of the Earth (not the barycenter of the Earth/Moon or Earth/Sun system), and they have neglected a factor of 2. A derivation or even a diagram would help make the intended meaning clear, but there is none. It is a disappointingly sloppy and misleading answer. However, everything Miles concludes based on treating L as equal to R is the result of a simple error due to sloppy language, not dishonest mathematics.
"Linearizing" means differentiating the equation with respect to R, so that this new equation represents a change in the field, rather than the strength of the field.
Linearizing doesn't actually
mean differentiating. It involves differentiating, but it
means replacing a nonlinear relationship with a linear approximation. This is done when a linear approximation is easier to work with mathematically, and is close enough to the original relationship that the differences are irrelevant within the region of interest.
It is clear that the differentiating proves that there would be an inverse cube effect in the tide-producing differentials, supposing that the postulates of this theory are true.
At least we have some agreement here. But the way he phrases this, and statements like "They tell us this equation is approximately equal to..." suggest that Miles does not understand how linearization works. It is a result of calculus. If you have a well behaved function F(x), then for x near some particular R, F(x) is approximately a constant, plus a linear term, plus a quadratic term, plus a cubic term, and so on, given by
F(R+r) ~= F(R) + rF'(R) + 1/2 r
2 F''(R) + ...
where F'(R) is the derivative of F at R, F''(R) is the second derivative of F at R, and so on. In the analysis of tidal forces, F(R) gives a uniform effect at all points which accounts for the acceleration of the Earth as a whole. rF'(R) gives the inverse cube term GM2r/R
3 responsible for the bulk of tidal forcing. 1/2 r
2 F''(R) gives a correction of 3GMr
2/R
4, which is smaller by a factor on the order of r/R. Since the size of the Earth is much less than the distance to the Moon or Sun, this correction is negligible. Further corrections are progressively smaller still.
I don't know that I would call it an inverse cube "law", since it does not apply to the field itself. It applies to the differential field.
Miles apparently interprets "law" in the scientific sense of a fundamental natural law. It is used, rather, in the mathematical sense of a power law relationship.
But the Earth is not in simple freefall around the Sun. It is in orbit.
Orbit is a special case of free fall.
We must therefore add a centrifugal effect to the static effect of the field.
Wrong, as I've demonstrated already.
The Standard Model, as expressed in Wikipedia and elsewhere, adds the centrifugal effect using this equation:
Δa = ω2mr
Without seeing the actual derivation in question, I cannot comment on whether the centrifugal force was used correctly or incorrectly. If incorrectly, that's a shame, but it's a good thing people were around to correct the error.
Note added August 2007: Confronted with parts of this paper in late 2005, Wikipedia deleted all its tidal theory math, its tidal theory page, and ordered a rewrite with lots of new illustrations. It appears they are perfecting their propanganda rather than admitting that their math and theory doesn't work.
Correcting an error is wildly different from perfecting propaganda.
Another major problem with tidal theory concerns its use and misuse of the barycenter. The barycenter is the center of gravity of the Earth/Moon system, which both bodies are said to orbit. Feynman was one of the most famous to suggest that the Earth has a non-negligible tide created by orbiting this barycenter.
A citation here would be interesting. Even very smart people make mistakes. It would be interesting to see just how far he took this idea.
Is this true? Let's do the full math.
What follows is some scattered mathematical expressions without formal definitions, in which he makes an arithmetic error in the computation of Δa
i = a - a
i. (This definition is inferred, and has the opposite sign convention of Δa
o, which is strange given that Miles is otherwise consistent with his signs here.)
The effect doesn’t just raise a tide on the far side of the Earth from the Moon. In fact, it raises an even larger tide toward the Moon.
As I've already discussed, his conclusion that there is a huge tidal forcing from this ignores the fact that it is actually an axisymmetric variation. His conclusion that it's a larger effect on the side facing the moon is due to his arithmetic error compounding the problem.
The barycenter falsifies the entire standard analysis, since it would swamp all effects from the Sun and Moon. You cannot include effects from the barycenter, since they cannot be made to fit the given data. And you cannot fail to include effects from the barycenter, since current gravity theory demands a barycenter. This is called a failed theory.
You cannot simply look at the difference in centrifugal force on the near and far edges of the Earth without taking the symmetry into account. Doing so is much like looking at the difference between the effect of gravity from the Earth itself on the near and far edges.
a
i = 9.8m/s
2
a
o = -9.8m/s
2
This is 216,000 times as strong as the barycentric effect Miles tells us should swamp the tidal forcing of the Moon's gravity, and in the opposite direction. Applying the same error of not taking the symmetry into account might lead us to conclude that there should be
enormous tides perpendicular to the direction towards the Moon.
It is at this point, roughly half way through the article, that I gave up completely. Whether his failings are in the physical concepts or the mathematics, it is clear that Miles Mathis is not equipped to criticize the standard model of tidal forcing.
