Miles Mathis

[StevenO]

Oh, as for "his numbers work", just try wrapping some string around a cylinder and see which value of the arc length is closer, 2πr or 8r. Miles tells us the value of π tells us nothing. It's "just a ghost". And yet, it's used quite practically in numerous everyday situations involving circular objects and circular motion. If the correct value for π was 4, everyone would have noticed, not just one revolutionary mathematician.

I think you just have a bad mood that you are ventilating against Miles without understanding his articles. Otherwise you would have known that using 2πr/t with v^2/r in itself yields the right proportions but not the correct numbers. For a geometric projection without time involved one would still use π=3.14.

[Trouserman]

I think you just have a bad mood that you are ventilating against Miles without understanding his articles. Otherwise you would have known that using 2πr/t with v^2/r in itself yields the right proportions but not the correct numbers. For a geometric projection without time involved one would still use π=3.14.

No, his actual derivation of π=4 makes no reference to time. It is purely geometric. It says this distance plus this distance is this distance, and it is wrong. But if one accepts that the geometric projection still uses π=3.14, try this on: A spool of radius r is placed on a spindle. You apparently accept that the length of one loop of thread about this spool is 2πr. Now, pull that length of thread off the spool (at a tangent) at a constant velocity. The end of the thread has moved in a straight line a distance 2πr at a speed of 2πr/t, and the spool has made one revolution at constant angular speed. Now, think about a small piece of thread still on the spool, and tell me which makes more sense to you:

(A) The small section of thread still on the spool has moved around a track with a geometric length of 2πr, so it has moved through a distance of 2πr. Its speed was 2πr/t, the same as a piece of thread moving in a straight line off of the spool.

(B) The small section of thread still on the spool has moved around a track with a geometric length of 2πr, but it has moved a distance of 8r because the path is curved. Its speed was therefore 8r/t.

If you answered (B), let me ask this: If we keep pulling the string out, what happens that causes the speed to decrease from 8r/t to 2πr/t when a piece of string reaches the spot where it comes off the spool?

In case there's some wild hand-wave to cover that, try this on: Go back to the diagram. There are four blue steps, and each one covers an arc of equal angle (11.25 degrees). Let the radius be 1, for simplicity. The first step touches the circle at (0,1) and (0.195,0.981), so it has a length of 0.195 + 0.019 = 0.214. The fourth step touches the circle at (0.556,0.831) and (0.707,0.707), so it has a length of 0.151 + 0.124 = 0.275. If the steps' lengths are equal to the arc lengths as Miles claims, how do you reconcile the fact that equal angles trace out arcs of different lengths depending on what part of the circle you're looking at?

[Trouserman]

Perhaps this way of looking at it will find more traction with Mathis's followers. Mathis says:

I am starting at point A in this drawing. In the drawing, the length of the vector stands for the velocity, and I must draw it at a tangent. Meaning, the angle at A must be 90 degrees. So, if you give me the same time to travel as I took to draw the radius, then I will end up at point B.

But if we want to draw a curve, we must keep the pencil on the circle, and here I am way off the circle. How do we solve this? Let us solve the way Newton did, by inserting a second motion. To create any real curve requires at least two simultaneous motions. As we know from classical orbital mechanics, this second motion must be a vector pointing at the center. This motion will take us from off the circle to back on the circle.

In other words, to draw an arc we need a tangent motion, plus a radial motion to get us back on the circle. When we divide an arc into a number of smaller pieces, should we not then do the same for each smaller arc? If we do this, we get not Mathis's stair-steps (blue, in the diagram below), but a series of tangents and small corrections (red).



For each small arc, we are doing exactly what Mathis tells us to do. Now, what is the outcome of this? If we break a 180 degree arc into 2 pieces in this way, we get Mathis's result, 4r. If we break it into 10 steps, we get a total length of 3.579605r. If we break it into 100 steps, we get 3.190420r. 1000 steps, 3.146522r. 1000000 steps, 3.141598r. The trend towards πr should be obvious. Now if you accept Mathis's result, ask yourself, why is the whole motion not equal to the sum of its parts?

[StevenO]

I think you should first read Miles' rederivation of the calculus to understand the differences between a circular orbit and a geometric circle.

[Trouserman]

The only challenge I've made to Mathis so far that you have responded to is the application of his kinematics to physical systems. Both you and Mathis apparently believe that his kinematics are equivalent to conventional kinematics, but expressed in a different mathematical formalism. One of two things is the case: his equations are equivalent to conventional kinematics (they describe the same motions using a different formalism), or they are not equivalent. If they are equivalent, then they cannot explain the Explorer anomalies, as he claims, or any other supposed failing of conventional kinematics. If they are not equivalent, then since he places his derivation on very generic kinematic grounds, they must differ from conventional description of circular motion in an enormous number of simple cases you can test at home, which are well described by conventional theory, and which therefore disprove Mathis's math.

You have not addressed the following questions: If the steps' lengths are equal to the arc lengths when the arc is broken down according to Mathis's "exhaustion" method, how can equal angles trace out arcs of different lengths depending on what part of the circle you're looking at? And, why is the whole motion not equal to the sum of its parts, when those parts are analyzed in precisely the way he proscribes? Either of these discrepancies shows a fatal flaw in his description of circular motion.

I do not need to read Mathis's "rederivation" of calculus to see that his understanding of conventional calculus and vector arithmetic are deeply flawed. That much is apparent from his criticism in his first article on a=v2/r of the standard calculus derivation of that relationship. Briefly: He says the orbital velocity and tangential velocity are not the same, but in conventional mathematics they are. Because the path length is equal by the limit of tangential displacements, of course the orbital velocity is equal to the tangential velocity! He says Δv is not equal to v-v0, which is absurd because that is the definition being used. If Mathis wants to define the words "change in velocity" to mean something other than the difference between the velocity and the beginning and end of the interval for his own work, it's his prerogative to be silly in that way, but it is not a legitimate criticism of conventional mathematics. "The numerical value of v is the same as the numerical value of v0, so Δv could only be zero here. The difference between v and v0 is only an angle here." Here he shows his misunderstanding of vector arithmetic. Subtracting one vector from another yields a perfectly well defined vector result. No one even remotely qualified to attempt a rederivation of vector calculus could possibly be confused about this.

I have no desire to read any more of Mathis's garbage, and it saddens me to see non-mathematicians buying into it.

[johnmartin]

Hello Trouserman

I’ve recently read some of Mathis’s papers and found them interesting. I have done some thought on what he says and I believe he has some interesting points worthy of discussion. After reading his paper on the trouble with tides I think he has come up with a problem with the standard model. I have seen your comments and will make my own comments below.

There are no centrifugal forces on the Earth directly caused by the Moon, since there is no angular velocity around the Moon. http://milesmathis.com/tide.html

You responded as follows –

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.

OK. Point taken, but if the earth-moon system is to be analysed, we need to account for the centrifugal forces caused by the earth orbiting the earth-moon barycentre. The barycentre is located inside the earth at 4671 km from the center of the Earth and this causes the inner side of the earth facing the moon to have a negative radius and the outsides earth face, a positive radius.

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, Ftotal=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 = mω2r

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 = mω2R + mω2x

The first term, mω2R, is a uniform contribution at all points on Earth, and is therefore tidally neutral. The second term, mω2x, 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.

Please demonstrate “The second term, mω2x, is axisymmetric about the Earth's center”.

(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 the crux of your argument. I’d request you explain how the earth-moon rotating system has a uniform centrifugal force around the earth surface. Claiming the centrifugal acceleration varies in an axisymmetric way, is only a claim and not a demonstration.

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.)

So far I agree, but your demonstration should be interesting.

JM

[Trouserman]

Hello, John. Thanks for your reply.

OK. Point taken, but if the earth-moon system is to be analysed, we need to account for the centrifugal forces caused by the earth orbiting the earth-moon barycentre. The barycentre is located inside the earth at 4671 km from the center of the Earth and this causes the inner side of the earth facing the moon to have a negative radius and the outsides earth face, a positive radius.

We don't have to account for them if we analyze the motion in a non-rotating reference frame. Any effects they would have in a rotating reference frame will show up in a non-rotating frame as accelerations. I understand that it can be difficult for people to get their heads around centrifugal force, when it's appropriate to apply, and the correspondence between rotating and non-rotating frames, though.

Please demonstrate “The second term, mω2x, is axisymmetric about the Earth's center”.

The x in that relationship was defined as the vector displacement (within the plane of rotation) of our little test mass from the center of the Earth. Its magnitude is the distance from the center of the Earth, which doesn't depend on the direction, and its direction is radially outward from the center of the earth. mω2 is just a constant multiplier; it doesn't depend on the distance or direction to the little test mass. The whole thing, then, mω2x has a magnitude that depends on the distance from the center of the Earth and does not depend on the direction, and it is directed radially outward. That is axial symmetry. (It is axial, not radially symmetric in three dimensions, because if we leave the plane of rotation, the perpendicular component of the displacement does not affect the centrifugal force.)

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.)

So far I agree, but your demonstration should be interesting.

Are you asking for a fuller explanation of why the variation in the gravitational field is not axisymmetric, or is seeing why the centrifugal force is axisymmetric sufficient? I'll be happy to do the demonstration for the gravitational field, but it could get a bit hairy.

[jjohnson]

Miles has recently gotten into more calculus (bashing; what else?) with interesting results for mathematicians. I am referring to his two papers, on the derivations of a^x and e^x, and on variable acceleration. (He wrote some stuff on variable acceleration a while back and I responded to him with a discussion on what you call a varying rate of acceleration. He uses "cition" and I am more familiar with the term "jerk", which he had not heard of since, I presume, it comes out of engineering rather than math theory. Interestingly he picked up the term jerk for variable acceleration in his recent paper.) I was also drawn to read some of his earlier papers on Newton's work with calculus and subsequent mathematicians' manipulations of things, which is another intellectual romp. My wife can't believe that I do this for fun, but hey...

I see this collection of papers investigating what has happened and what goes wrong with the calculus as an interesting analogue to his looking at various historical aspects of physics theories, and showing where original authors may have erred, or their original works have been modified uncritically as time progressed by others who contributed their own mistakes, resulting in an edifice which has become The Current Standard. That it is accepted by nearly everybody in both math and physics goes without saying, except that it IS being said by Miles, and by physicists in EU and PC who are trying to get better alternates to be considered in how we tackle our knowledge, mathematics and engineering in the context of a real universe.

I wouldn't expect Miles to tackle EU's particular mathematical issues except as theoretical challenges. He is not into that subject particularly, and it is up to us to think through the things he is saying and adapt that which is useful and see if it works for us. Thinking that we may have been taught our math the wrong way since day one is unsettling, to say the least, so I've got to sit down with the calculator or Excel and play with this stuff. This is useful, if true, since, as Twain wrote, "It's not what you don't know that gits you into trouble; it's what you know for sure that just ain't so." I don't know if he meant math and physics by this, but if the shoe fits...

-and there are sure a lot of people who believe that they know for sure...

Jim

[Trouserman]

I take it back. Curiosity got the better of me, and I read the paper on redefining the derivative. Miles Mathis's work does have a certain pull to it. It's like the mathematical equivalent of watching a bad horror movie. If it was just me, I would wince and chuckle and move on, but I find it troubling that some people here are taking his work seriously. I think a lot of the ideas in EU theory seem worthy of further analysis, and I'd hate to see it sidetracked and diluted by bad math. If you're a follower of Mathis's work, this is for you. Make no mistake, it is bad.

Let's start with a philosophical double standard from the preamble to "A Redefinition of the Derivative":

So, you have picked a point. I am not even going to be rigorous and make you worry about whether that point is truly dimensionless or indivisible, since, again, that is just quibbling as far as this paper is concerned. Let us say you have picked the corner of a table as your point. The only thing I am going to disallow you to do is to think of that point in relation to an origin.1

When Mathis disallows thinking about a point in relation to an origin, he disallows thinking about the relationships between points in general. If you think about the relationships between points, you can always pick one point arbitrarily as the origin from which to describe those relationships. Without relationships between points, one cannot begin to consider any aspect of geometry, kinematics, or dynamics. This applies for Mathis as much as anyone else. Without an origin to measure from, he cannot specify so much as a single Δx from the first line of his table. When he says, in effect, "You're not allowed to do this, but I am," he is imposing a ridiculously illogical double standard.

If I tell my friend to meet me at 3 blocks, 2 hours, it's nonsensical. If I tell my friend to meet me 3 blocks east of the park, 2 hours from now, we'll have no problem finding the same spot. I might just as well specify the exact same point in space and time using another reference, such as 2 blocks west of my house, at 4:30 (assuming, of course, that I live 5 blocks east of the park, and the time now is 2:30). 3 blocks and 2 hours are intervals, not points, but in relation to a given origin, they specify a point; not just a mathematical point, but a physical point at which my friend and I can meet for a debate about whether numbers can specify points in space and time. It will be a short debate, because having arrived at the same point and time as specified by numbers exchanged over the phone, we will be in agreement that there is no problem doing so.

Firstly, the curve is defined by the graph. When we discover a curve equation by our measurement of the curve, the equation will depend entirely on the graph. That is, the graph generates the equation. Secondly, if we use a Cartesian graph, with two perpendicular axes, then we have two and only two variables. Which means that we have two and only two dimensions. Thirdly, every point on the graph will likewise have two dimensions.1

The curve is not defined by the graph, it is defined by the relationships of its parts. The equation is defined by the choice of coordinate system (origin and axes), in addition to the relationships of the parts of the curve. The coordinate system is indeed a feature of the graph and not of the relationships. However, the equation still describes the relationships, which are what we are really interested in. When Miles talks about the dimensions of a point here, he is talking about the dimensions used to characterize a point; that is, the dimensions of the relationship that point has with the chosen origin. When he goes on to rant about how this contradicts the definition of a point as having no dimensions, he is missing the fact that this definition refers to the extent of the point itself, not to its relationships with other points. It's irrelevant rambling, triggered by his apparent conflation of these two related, but different, uses of the word "dimension".

There's a lot more to say about why Mathis's objections to infinitesimal calculus are invalid, but I'd like to restrict myself mostly to a few places where Mathis's own work is poorly defined, inconsistent with his own definitions, or just wrong. There is a lot to say about that, as well.

We use the calculus to tell us what the rate of change is for any given variable and exponent. Given an x, we seek a y’, where y’ is the rate of change of x. And that is what we just found. Currently, calculus calls y’ the derivative, but that is just fancy terminology that does not really mean anything. It just confuses people for no reason. The fact is, y’ is a rate of change, and it is better to remember that at all times.2

This is almost right, although Mathis is not being rigorous in his definitions and use of language. Given a relationship between y and x (y as a function of x), we seek y', which is the rate of change of y with respect to x. When Mathis presents the equation y’=nxn-1, without specifying the relationship y=xn, he is neglecting a critical portion of the problem's definition. But yes, y' is a rate of change. What is a rate of change?

To measure a length you don’t need a watch. To measure velocity, you do. Velocity has a “t” in the denominator, which makes it a rate of change. A rate is just a ratio, and a ratio is just one number over another number, with a slash in between.2

I must take exception to Mathis's characterization of a ratio as "just one number over another number, with a slash in between". The notation of placing a number over another with a slash between them stands for something. It has a mathematical meaning: the operation of division. If I go 120 miles in 2 hours at a constant velocity, that velocity is 120miles/2hours, which is the same as 60miles/hour. Since 1hour=60minutes, this is also the same as 60miles/60minutes, or 1mile/minute. Different units, same velocity. This is important because, by ignoring the denominator and treating differences as rates of change, Miles occasionally slips up.

It is also worth pointing out that velocity is a specific kind of rate of change: the rate of change of position with respect to time. This can be generalized to variations of anything with respect to any parameter. So if y is a function of x, giving (say) the elevation at a given horizontal distance along a trail, y' is the rate of change of elevation with respect to horizontal distance, Δy/Δx. In other words, the slope of the trail. If the term "rate of change" has too much time-related baggage, then perhaps the technical term "derivative" is not so useless after all.

Another important point is that this is the definition of velocity: the rate of change of position with respect to time. Miles sometimes confuses this with "flattening the curve", which is not the same thing at all. If the velocity is changing unevenly from one interval to the next, or within an interval, a graph of velocity versus time will not produce a straight line. A vector is straight, but a graph of the vectors that apply at different times (or over different intervals) need not be.

Current notation has one less delta at each point than I do. Current notation assumes that curve-equation variables are naked variables: x, t. I assume they are delta variables, Δx, Δt. But I agree with current theory that velocity is a change of these variables. Therefore velocity must be ΔΔx/ΔΔt.1

The "naked variables" x and t are always expressed as numbers relative to some origin, so it is reasonable to view them as delta variables in this way. However, the extra delta makes for a lot of notational clutter, and is unnecessary. Furthermore, although expressing them as numbers in an equation assumes a particular origin, the origin itself is of little interest. The relationships between points of interest do not depend on the arbitrary choice of origin, and it is thus useful to distinguish them notationally with deltas. But, that's just a notational choice. The thing to notice here is that Mathis has identified both y' and ΔΔx/ΔΔt as the velocity. Differential calculus says simply that y' is equal to whatever ΔΔx/ΔΔt approaches in the limit of small intervals ΔΔt. That is the definition of "instantaneous velocity", whether you think that term is misleading or not. Call it the "asymptotic limit of the average velocity" over small intervals about x=x0, if you prefer that to the "instantaneous velocity" at x0. That's what it means, by definition.

For finite intervals, ΔΔx/ΔΔt is simply the average velocity during that interval. Don't take me as saying that the average velocity is wrong or useless. It can be quite useful. But it does not have the same properties as the instantaneous velocity (which is defined in a particular way, whether you like the term or not), because the velocity is not the same during the entire interval.

All this can also be applied to velocity and acceleration. The magic equation can be applied to velocity and acceleration, too. If x is a velocity, then y’ is an acceleration. This is because acceleration is the rate of change of the velocity. Acceleration is v/t.2

Again, a slight technical correction to the language: If y is a function expressing the velocity as a function of time, then y' is the acceleration. This is the definition of acceleration. Mathis sometimes confuses this with a curved path, or a curved plot of position as a function of time, saying that the curve is an acceleration. But a curved path is not the same as an acceleration. Path and acceleration are both properties of motion, and a curved path indicates a non-zero acceleration, but the path is not the same thing as the acceleration.

In summary:

This is important, because it means that mathematically, a velocity is not a distance, and an acceleration is not a velocity. They have to be kept separate. A calculus equation takes you from one sort of variable to another sort.2

Unfortunately, Mathis isn't consistent about this, because he sometimes treats differences (Δx) as rates of change (Δx/Δt), and other peculiarities. Take this:

Acceleration is traditionally Δv/Δt. By current notation, that is (ΔΔx /Δt)/ Δt. By my notation of extra deltas, that would be [Δ(ΔΔx)/ΔΔt] / ΔΔt . My variables have been upside down this whole paper, meaning I have been finding slope and velocity as t/x instead of x/t. So flip that last equation
[Δ(ΔΔt)/ΔΔx] / ΔΔx1

This is certainly not the definition of acceleration above, on which Mathis appeared to agree with everyone else. What he should say here is that he's derived a formula for the derivative of a function of a variable Δx, and that derivation is independent of what Δx represents. If Δx represents time, then these are time derivatives, and if the functions are velocities, then you have the time derivative of velocity, which is the acceleration.

Here's another example:

Notice that the derivative of a curve is a tangent, which is a straight line. But (we are told) the derivative of a velocity with respect to time is an acceleration. In one instance, we get a straight line from a curve; in the other, we get a curve from a straight line.3

Here, Mathis is confused about the difference between a velocity vector at a particular time (or average velocity over an interval) being straight, and a possibly curved plot of the velocity as a function of time. Remember, Mathis agreed earlier that the acceleration is by definition the time rate of change of velocity, and the velocity is by definition the time rate of change of position.

Now, getting back to the "magic equation" y’ = nxn-1, Mathis's derivation is dead wrong, because the lines he pulls out of his table do not correspond to the equation at all. Remember, both y' and ΔΔx/ΔΔt represent the rate of change. Miles builds a table of values based on Δz, specifies that ΔΔz is 1, and notes:

2ΔΔz = ΔΔΔz2
3ΔΔΔz2 = ΔΔΔΔz3
4ΔΔΔΔz3 = ΔΔΔΔΔz4
5ΔΔΔΔΔz4 = ΔΔΔΔΔΔz5
6ΔΔΔΔΔΔz5 = ΔΔΔΔΔΔΔz6

All these equations are equivalent to the magic equation, y’ = nxn-1.2

But that's not what these relationships say at all! Never mind the confusion of different y, x, z, and t variables. If y=Δz4, then y', which is by definition the rate of change with respect to Δz, ΔΔy/ΔΔz, or (by choosing ΔΔz=1) simply ΔΔy, the "magic equation" says this:
ΔΔz4 = 4Δz3

That is not the same equation as
4ΔΔΔΔz3 = ΔΔΔΔΔz4

Since in those last equations we have z on both sides, we can cancel a lot of those deltas and get down to this:
2z = Δz2
3z2 = Δz3
4z3 = Δz4
5z4 = Δz5
6z5 = Δz6
Now, if we reverse it, we can read that first equation as, “the rate of change of z squared is two times z.” That is information that we just got from a table, and that table just listed numbers.2

You cannot simply cancel deltas. They are not multiplicative factors, they are difference operations. If two sequences are equal, their differences will be equal, but the reverse is not necessarily the case. If you take the "magic equation" at face value, and apply the definitions, and consult the table for ΔΔz4 and 4Δz3, you will find that they are not equal.

4 Δz3 1, 8, 27, 64, 125, 216, 343
4a 4Δz3 4, 32, 108, 256, 500, 864, 1372
11 ΔΔz4 1, 15, 65, 175, 369, 671

The reason for this is that y' in the magic equation, the derivative of y with respect to Δz (or whatever you want to name your variables), is by definition not the average rate of change ΔΔy/ΔΔz over a finite interval, but the limiting value for small ΔΔz. ΔΔy/ΔΔz takes on different values over different sub-intervals of the original interval, thus leading to the discrepancy between the average and "instantaneous" rate of change.


Now, let's consider the claim Mathis makes that his results are exact, and those of infinitesimal calculus are poor estimates, in a particular concrete example. Direct your attention to the calculation of the slope of a tangent to an exponential curve, found in http://milesmathis.com/expon.html.

In closing, let us look again at the current derivative for exponential functions, y' = axln(a). If a=2, then y'=.693(2x). That is still a curve, so the derivative is still a curve. How can the tangent be a curve? I will be told that there is a tangent only at a particular point or solution. For instance, at x=3, the slope of the tangent is 5.545. Fair enough, if it were true. I will be taken to the Cartesian graph for x,y and shown a long proof that it is true. And on the graph it IS true (almost). I don't deny it. The modern calculus has found a way to estimate (poorly) the slope of that line drawn on the x,y graph. Unfortunately, that line is not the rate of change of our function, since I have just shown you the rate of change of the function from a table of differentials.3

Excellent. Let's see who can compute the best tangent to y=2x at x=3. The tangent should touch the curve at (3,8) without crossing. Differential calculus tells us its slope should be m=y'=2xln2=8ln2, or approximately 5.545. The equation for a line with slope m, passing through a point (x0,y0) can be written:

y = y0 + m(x-x0)

(Notice that when x=x0, y=y0, so it passes through the point. And it can be rearranged as y = mx + b, where b=y0-mx0, so its slope is m.) We'll plug some values into this shortly.

In sum, the modern calculus must make up its mind. Does it want the derivative as the rate of change at x or does it want the derivative as the slope of the tangent drawn at x? With the exponential function, it cannot have both, since the two numbers are different. And the difference is not small. The difference between 5.545 and 8 is not small.3

Where does he get 8 here? It is neither the derivative as specified by differential calculus nor the slope of the tangent. Differential calculus says that they are both the same, and equal to 23ln2 (approximately 5.545). I suspect that the 8 Mathis refers to is his derivative, the average rate of change over the interval from x=3 to x=4, which is not the definition of the "instantaneous" rate of change at x=3 (again, whether you like that terminology or not). It is true, the difference between 5.545 and 8 is not small, and that is because the instantaneous rate of change varies considerably over that interval, and it is the lowest at the point under consideration, x=3.

Amazingly, we can use my new derivative here to find the slope of the drawn tangent, without using either the current equation or any “long and tedious process.” Notice that my value for y' @x=3 was 8. Now, y' by itself is not the slope of anything, since slope is and always has been defined as y/x. But if we find y/x at the point (x,y'), we find 8/3 = 2.667. Doubling that gives us 5.333, which is tantalizingly close to 5.545. Have I stumbled upon something important here? Let us try that for other values of a and x. If a=2 and x=4, my derivative is 16. Dividing by 4 is 4. And if we multiply by (x – 1) we get 12. The current slope is 11.1. Another set of values: a=4 and x=2. My slope is 24. The current slope is 22.

The general equation for this is

m = (x – 1)y'/x
or m = (x – 1)(a – 1)ax/x 3

Look what Mathis has done. He has treated his y' as merely the amount by which y changes over the interval from x=3 to x=4, missing the fact that it is properly the average rate of change (or slope) over that interval, a conceptual error resulting from setting Δx=1 and forgetting about the ratio in the definition y'=Δy/Δx. He then picks out y/x as a starting slope. This is the slope of the line between the origin and (x,y). It has no bearing on the slope of the tangent line, as should be obvious from the fact that the origin can be chosen arbitrarily, without changing relationships like the slope of the tangent. He then arbitrarily doubles this. The result is sort of close to the accepted value, which he thinks is wrong, so he takes the fantastically absurd leap of logic to conclude that he must be on to something. He tries a couple more examples, finds 3 points where the result is within 10% of the accepted value, and concludes that it is entirely general. Wow. I wouldn't be surprised to find out he's wrong. Would you?

Well, let's take a look at the first one, a=2 and x=3. Mathis's proposed slope of the tangent at (3,8) is
m=(8/3)*2=16/3
or approximately 5.333. The equation for the tangent line would thus be:
y=8+(x-3)*16/3

Here's a plot of it, along with the original curve, y=2x:



It looks pretty good, but if you have a keen eye you might spot that it looks a little different for x<3 than for x>3. Let's zoom in.



Wait a minute! It looks like that "tangent" is crossing the curve in two places. It's not a tangent at all! You can check this yourself. Pick out a few values of x, near 3, and compute the following values at each:

• y=2x, the original curve
  yM=8+(x-3)*16/3, the "tangent" line with the slope Mathis proposed
  yT=8+(x-3)*8ln2, the tangent one finds by the application of standard differential calculus

Code: Select all

x       2^x           8+(x-3)*16/3  8+(x-3)*8ln2
2.80    6.96440451    6.93333333    6.89096451
2.85    7.21000370    7.20000000    7.16822338
2.90    7.46426393    7.46666667    7.44548226
2.95    7.72749063    7.73333333    7.72274113
3.00    8.00000000    8.00000000    8.00000000
3.05    8.28211939    8.26666667    8.27725887
3.10    8.57418770    8.53333333    8.55451774

Notice that Mathis's line starts below the curve, crosses it somewhere between x=2.85 and x=2.90, then crosses again at exactly x=3. By way of contrast, notice that the standard tangent touches at exactly x=3, but is lower than the original curve everywhere else. Zoom in further, if you like. You won't find a discrepancy until you get too close for the precision of the values you put in the calculation.

For an even bigger laugh, look at x=1, where Mathis tells us the slope of the tangent line should be:
m = (x – 1)(a – 1)ax/x = 0

In other words, the tangent should be a horizontal line! You don't even have to plot it to see that it will simply slice across the curve without looking like a tangent at all. That's how absurd Mathis's generalization from a few sort-of-okay points is.

I'll close with a point on which I agree with Mathis 100%:

If the math in chapter 1 and 2 is false, you know the math in chapter 30 is false.4

[Lloyd]

* If we laugh at Mathis's math, we'd have to laugh at our own even more, because he seems to understand it much better than most of us do. So, if he's screwed up, we're worse. I was pretty good in math and even took some calculus in college, but I thought Mathis sounded like he knew what he was talking about. But, now, you sound like you do too. And I expect that you're more likely right about calculus and geometry than is Mathis.
* But being wrong about several things doesn't mean necessarily being wrong about all things. I liked Mathis's reasoning on gravity, because it suggests that normal orbits of planets etc include an electrical component, which would seem plausible to me. Then I heard from Thornhill, who looked at Mathis's analysis of Newton's formula for gravity, and he says Mathis is partly right, but usually for the wrong reasons. And he sent me a link to his article from April on his site, http://holoscience.com. And it covers Newton's formula. He discussed there the idea that plasma tails move charged mass from an inner planet to an outer one and that the mass changes the planetary orbits, shrinking the orbit of the one that loses mass, and expanding the one that gains mass.
* Had I kept up with the articles on his site, I might not have been so impressed with Mathis's view on the subject. Thanks for taking the time to explain Mathis's errors, assuming you're right. It might be helpful if you could explain his gravity errors, too, if you find any. His math explanation of the plasmaspheres of the planets and about the tilts of the planets, as well as tides, also sounded reasonable to me. Out of the many subjects that he at least has the courage to tackle publicly, maybe a few of them will turn out to be correct. His ideas about photons seem fairly plausible, too. And I agree with him that "attraction at a distance" seems implausible, i.e. attractive forces, like gravity. To me it seems more likely that there is a pushing force from outside that gives the appearance of attraction.
* I'll try to remember to email Mathis to link to your post.

[johnmartin]

johnmartin wrote:OK. Point taken, but if the earth-moon system is to be analysed, we need to account for the centrifugal forces caused by the earth orbiting the earth-moon barycentre. The barycentre is located inside the earth at 4671 km from the center of the Earth and this causes the inner side of the earth facing the moon to have a negative radius and the outsides earth face, a positive radius.


We don't have to account for them if we analyze the motion in a non-rotating reference frame. Any effects they would have in a rotating reference frame will show up in a non-rotating frame as accelerations. I understand that it can be difficult for people to get their heads around centrifugal force, when it's appropriate to apply, and the correspondence between rotating and non-rotating frames, though.

If the earth is taken in a non rotating reference frame, I understand you mean by this the earth is understood to be stationary relative to the moon when doing the analysis. Yet even so, if we believe the earth and moon orbit the earth-moon (e-m) barycentre (BC), then we need to include the fictional centrifugal force of the earth orbiting the BC, when doing our analysis.

Please demonstrate “The second term, mω2x, is axisymmetric about the Earth's center”.

The x in that relationship was defined as the vector displacement (within the plane of rotation) of our little test mass from the center of the Earth. Its magnitude is the distance from the center of the Earth, which doesn't depend on the direction, and its direction is radially outward from the center of the earth. mω2 is just a constant multiplier; it doesn't depend on the distance or direction to the little test mass. The whole thing, then, mω2x has a magnitude that depends on the distance from the center of the Earth and does not depend on the direction, and it is directed radially outward. That is axial symmetry. (It is axial, not radially symmetric in three dimensions, because if we leave the plane of rotation, the perpendicular component of the displacement does not affect the centrifugal force.)

Ok. According to Mathis the barycentre of the e-m system is located at 4671 km from the center of the Earth and from his equations below, the earth has a radius of 6371km. Below he calculates the accelerations on the inside, centre and outside of the earth caused by the earth orbiting the e-m barycentre once per 27.32 days.

Vo = 2π*11042 / 2360448 = 29.39m/s
Vi = 2π*(-1707) / 2360448 = -4.54m/s

Using a = V 2 /R

a = 29.39 2 / 11042 = 7.82 x 10-5 m/s2

ai = -4.54 2 / (-1707) = -1.2 x 10-5 m/s2

ao = 12.43 2 / 4671 = 3.31 x 10-5 m/s2

Δao = 7.82 x 10-5 - 3.31 x 10-5 = 4.51 x 10-5 m/s2

Δai = 3.31 x 10-5 - -1.2 x 10-5= 4.51 x 10-5 m/s2

The last term is different to what Mathis gets = 5.51 x 10-5 m/s2 so I assume it’s a typo.

R = 4671 km
v = 2πR/t
t = 27.32d = 2360448s
v = 12.43 m/s
R + r = 11042 km
R – r = -1707 m
vo = outer velocity = 29.39 m/s
vi = inner velocity = -4.54 m/s
a = 3.31 x 10-5 m/s2
ao = 7.82 x 10-5 m/s2
ai = -1.2 x 10-5 m/s2

Δao = 4.51 x 10-5 m/s2

Δai = 5.51 x 10-5 m/s2

From this we conclude that the centrifugal accelerations are far greater than the gravitational accelerations from the moon and sun and therefore the standard tidal model fails.

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.)


So far I agree, but your demonstration should be interesting.


Are you asking for a fuller explanation of why the variation in the gravitational field is not axisymmetric, or is seeing why the centrifugal force is axisymmetric sufficient? I'll be happy to do the demonstration for the gravitational field, but it could get a bit hairy.

Lets concentrate on the centrifugal forces calculated above. According to Mathis, the centrifugal accelerations are dependent upon the distance away from the e-m barycentre. As the Earth far side from the moon distance is R+r and the earth close side to the moon distance is R-r, then we must expect the centrifugal accelerations on the far and close sides of the earth to be different due to the different distances from the e-m barycentre. When we do the calcs, this is what we see. These accelerations are then included in the Δao and Δai calcs and are far greater than the gravitational accelerations from the sun and moon.

What’s your response to the conclusion that the centrifugal forces swamp the gravity forces and invalidate the standard tidal model? I don’t see how the standard tidal model can be defended. I believe Mathis' observation that the radius from the BC to the earth inside face is negative, is correct.

JM

[Trouserman]

We don't have to account for them if we analyze the motion in a non-rotating reference frame. Any effects they would have in a rotating reference frame will show up in a non-rotating frame as accelerations. I understand that it can be difficult for people to get their heads around centrifugal force, when it's appropriate to apply, and the correspondence between rotating and non-rotating frames, though.

If the earth is taken in a non rotating reference frame, I understand you mean by this the earth is understood to be stationary relative to the moon when doing the analysis. Yet even so, if we believe the earth and moon orbit the earth-moon (e-m) barycentre (BC), then we need to include the fictional centrifugal force of the earth orbiting the BC, when doing our analysis.

Actually, by "a non-rotating reference frame", I mean one in which the directions of the axes are fixed, which means the Earth and Moon would both be observed to be moving about the barycenter. Centrifugal forces do not enter into the analysis of motion in such a reference frame, as they are a formal mathematical device for analyzing motion in a rotating reference frame, such as the one you mention in which the Earth and Moon are understood to be stationary.

Δao = 7.82 x 10-5 - 3.31 x 10-5 = 4.51 x 10-5 m/s2

Δai = 3.31 x 10-5 - -1.2 x 10-5= 4.51 x 10-5 m/s2

The last term is different to what Mathis gets = 5.51 x 10-5 m/s2 so I assume it’s a typo.

Whether it's a typo or arithmetic error makes little difference. Either way, it's not a big deal. Do note, however, that Mathis's conclusion that the effect is stronger on the inside than the outside, is a consequence of that mistake.

Lets concentrate on the centrifugal forces calculated above. According to Mathis, the centrifugal accelerations are dependent upon the distance away from the e-m barycentre. As the Earth far side from the moon distance is R+r and the earth close side to the moon distance is R-r, then we must expect the centrifugal accelerations on the far and close sides of the earth to be different due to the different distances from the e-m barycentre. When we do the calcs, this is what we see. These accelerations are then included in the Δao and Δai calcs and are far greater than the gravitational accelerations from the sun and moon.

What’s your response to the conclusion that the centrifugal forces swamp the gravity forces and invalidate the standard tidal model? I don’t see how the standard tidal model can be defended. I believe Mathis' observation that the radius from the BC to the earth inside face is negative, is correct.

My response to that is exactly what I've been saying this whole time: The variations in the centrifugal force are axisymmetric, and thus cannot raise tides. Since showing this with a general vector equation didn't make it clear, let's look at a particular point off the line between the Earth and Moon:



The analysis at the "inner" and "outer" points i and o is as you give. I'm calling the new point p for "perpendicular" (due to the right angle made at the Earth's center c). The centrifugal force at p due to rotation about the barycenter b must be directed away from b, and will therefore have both x and y components, complicating the analysis slightly compared with i and o, where everything was along the x axis. It's a little more natural in this situation to calculate the acceleration associated with the centrifugal force as ω2r, but I'll do it in terms of v2/r to match the analysis for i and o.

The x component of the displacement from b to p is x = 4671km, and the y component is y = 6371km. The distance to the barycenter is thus r = sqrt(x2+y2) = 7900km.

vp = 2π*7900km / 2360448s = 0.02102km/s = 21.02m/s
ap = vp2/r = (21.02m/s)2 / (7900000m) = 5.597 x 10-5 m/s2

In order to take the vector difference Δap = ap - a, it is necessary to break ap into x and y components.

apx = ap cos(angle cbp) = ap x/r = 3.31 x 10-5 m/s2
apy = ap sin(angle cbp) = ap y/r = 4.51 x 10-5 m/s2

Now find the x and y components of the difference Δap:

Δapx = apx - ax = 3.31 x 10-5- 3.31 x 10-5 = 0
Δapy = apy - ay = 4.51 x 10-5 - 0 = 4.51 x 10-5 m/s2

So, Δap has a magnitude of 4.51 x 10-5 m/s2, and is directed in the +y direction, away from the center of the earth. Comparing that to Δai and Δao, you can see that the centrifugal force doesn't have a stretching effect along the direction between the Earth and Moon. It pulls the Earth outward equally about an axis through its center. (It is in fact precisely the effect of the rotation implicitly assumed by taking the Earth to appear stationary and non-rotating in the rotating reference frame.)

Hopefully that makes my point about symmetry clear.

[johnmartin]

After Mathis has completed his calcs he says -

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.

So according to your analysis, Mathis has correctly calculated the acceleration differences at the near and far sides of 4.51x10^-5m/s^2, but has made a false conclusion, because the acceleration differences are the same all around the earth. So in effect the acceleration differences from the barycentre analysis can be ignored in calculating the tides. Is this what you are saying?

JM

[Trouserman]

So according to your analysis, Mathis has correctly calculated the acceleration differences at the near and far sides of 4.51x10^-5m/s^2, but has made a false conclusion, because the acceleration differences are the same all around the earth. So in effect the acceleration differences from the barycentre analysis can be ignored in calculating the tides. Is this what you are saying?

Yes, that is exactly what I am saying. It is equivalent to the effect of the Earth rotating about its own axis.

[johnmartin]

So when the tide model uses acceleration differences, we then move to calculating the tide heights from the following formula (Mathematical explanation of tides, http://mb-soft.com/public/tides.html) -

The mass of the Earth and G cancel out and we have the proportion of the accelerations equal to Rhigher2 / Rlower2.
alower Rhigher2
_____ = _________________
ahigher Rlower2
http://mb-soft.com/public/tides.html

Using this equation the author calculates the tide heights at the near and far sides of the earth to the moon – Near side –

In our case, we know the proportion of the accelerations, different by 1.129 * 10-6 m/s2 /9.8 m/s2 so the left fraction is 1.0000001152. The radius R therefore has to be in the proportion of 1.0000000576 so that its square would be the acceleration proportion. If we multiply this proportion (difference) by the radius of the Earth, 6.378 * 106 meters, we get 0.367 meter, which is the required difference is Earth radius to account for the difference in the acceleration.
http://mb-soft.com/public/tides.html

And far side –

When you plug in the numbers, you get a slightly smaller differential acceleration, 1.074 * 10-6 m/s2 (instead of the slightly larger 1.129 * 10-6 m/s2 that we got for the front side water.) Using the same calculation as above, we get a Moon-caused tidal change in radius of 0.349 meter (instead of 0.367 meter as before) or 13.8 inches. That is only 0.018 meter or around 3/4 of an inch difference in the heights of the front and back (Moon-caused) tides. The rear side tide IS therefore slightly smaller, but not by very much! The same reasoning and calculation applies to the Sun-caused tidal bulges, so the total Spring Tide difference is actually around one inch.
http://mb-soft.com/public/tides.html

This author uses the gravity equation to derive an equation showing the relationship between the acceleration at the earth surface and the top of the tide surface, which is related to the inner and outer radius from the centre of the earth to the earth surface and top of tide surface respectively.

He calculates the acceleration differences caused by the gravity force of the moon. The differences are taken by deducting the accelerations at the near side from the acceleration at the center of mass and then the second calculation by deducting the far side acceleration from the acceleration at the centre of mass. From this he arrives at two acceleration differences at the near and far sides of the earth.

Now from the past exchange we have established that the centrifugal forces are eliminated from the tidal calculations because the acceleration differences are uniform around the earths surface. From this we can say we are using a “law” of deduction of equal acceleration differences.

So stated simply we have Δa=4.51x10-5m/s2 all around the earth. So when we take any point on the earth, we know there is the same acceleration difference, adjacent to that point and also on the opposite side of the earth. So if we only look at the centrifugal accelerations as causing tides, we would have the same tidal acceleration all around the earth simultaneously, which is to have the same tidal force and therefore the same tidal height all around the earth surface. As this is not observed, we say the same acceleration difference due to centrifugal forces “cancels out”, because we have the same acceleration difference all over the earth surface.

So this is the same as saying we cancel the centrifugal force because the there is no difference between the accelerations differences around the earth. Which is the same as saying – when the acceleration differences are the same around the earth, we take the difference of the acceleration differences and find the difference of the differences is zero, therefore the tidal effect of the centrifugal force is zero and therefore we can neglect the centrifugal forces when calculating the tides.

Moving onto the gravitational accelerations – The tides are calculated using a similar procedure whereby the acceleration differences are calculated by taking the differences of the gravity acceleration at the earths surface from the gravity acceleration on the earths surface. This is the method used here

Here is the mathematical explanation of the tides. Call the distance from the Earth's center to the Moon's center, D. Call the radius of the Earth R. Anything on the Earth's surface where the Moon is directly overhead is then at (D - R) distance from the Moon's center. Using Newton's gravitational equation F = G * M1 * M2 / R2 we know what the FORCE on it is. But Newton also told us that F = M * a. So the Moon attracts it (upward, there) with an ACCELERATION of F / M1 = atide = G * Mmoon / (D - R)2. THAT describes the acceleration that is caused in the water (or you!) when the Moon is exactly overhead. However, the entire Earth itself is ALSO being attracted, accelerated by the Moon at F / Mearth = aearth = G * Mmoon / (D)2. The effective tidal acceleration on our object is then just the difference of these accelerations, or (G * Mmoon) * ((1 / (D - R)2) - (1 / D2)). Simplifying, this becomes G * Mmoon * (2 * D * R - R2) / (D2 (D - R)2).

The gravity acceleration differences at the near earth side = 1.129 * 10-6 m/s2 and the far earth side of 1.074 * 10-6 m/s2.

So this is where there seems to be a consistency problem when we determine what acceleration differences to include and exclude. According to the procedure described above, we have a difference of the acceleration differences for the centrifugal accelerations, that cancel and are then said to be eliminated from the tidal calculations. Yet this author is inconsistent when he calculates the tide heights, without first taking the differences of the acceleration differences for the gravity accelerations. To be consistent it seems to me, he should have taken the difference from say the opposite side of the earth first and then proceeded. Accordingly it seems he has not done this –

1.129 * 10-6 m/s2 - 1.074 * 10-6 m/s2 = 0.065 * 10-6 m/s2

And then used this acceleration instead of the 1.129 for the near side and 1.074 for the far side. However if this is done the tide heights are wrong and the model is invalidated. If however the acceleration difference inconsistency is not resolved, then even if the tidal heights come close to what we observe, it is still invalidated.

It’s currently just a thought I’ve had about the tidal calcs, that at least on the surface shows an inconsistency in dealing with acceleration differences. I’m fully prepared to be corrected on this and invite your comment.


JM