Duh ... you are so right ... and I know how to spell his name correctly too ... I will blame my fingers. Clicked "submit" without a proof-read:-(

Regards,

Gary Simpson

I must read in detail, but I understand that the quaternion algebra can be used like operator of translation and rotation (I read the use in the spacecraft control); it is completely new to me the use in relativity.

I don't understand - in this moment - if each Lorentz transformation can be a quaternion transformation, but I think can only a boost could be a quaternion transformation (the number of components of the Lorentz transformation are three only for a pure boost).

It is all clear, and the extension of the use of directional derivative to derive the quaternion is interesting.

    Thanks for giving it a read. You have the main idea I think.

    Regards,

    Gary Simpson

    Gary,

    I found your formulation of a quaternion derivative intriguing. I recall that with complex numbers the derivative of the conjugate is zero: for a complex number z, dz*/dz = 0. This can be deduced from the Cauchy-Riemann equations for a complex number. Taking the derivative of the square of the norm zz* then gives dzz*/dz = z*.

    It turns out things are less complex (if you will excuse a pun) for a quaternion q. I was curious about the derivative of its conjugate, expecting more complexity, but it turns out (thanks Wikipedia) that the conjugate of a quaternion can be expressed entirely using multiplication and addition, q* = -(q iqi jqj kqk) / 2, so that it is easy to see that dq*/dq = 1. Taking the derivative of the square of the quaternion norm qq* gives dqq*/dq = q q* or twice the real part of q.

    Thanks for your essay which helped advance my understanding of quaternions.

    Best wishes,

    Colin

      Colin,

      Many thanks for having a read. Yes, quaternions have some very nice algebraic behaviors. You can sum a conjugate pair to eliminate the vector. You can take the difference between a conjugate pair to produce a vector. You can multiply a conjugate pair to produce a scalar.

      It principle, d/dx(uv) = (du/dx)v u(dv/dx) should be applicable but I have not gone through it in detail and the order of multiplication shoyld matter. My next effort will be developing identities.

      Best Regards,

      Gary Simpson

      Colin,

      Many, many, thanks.

      Thinking about your post some more has made me realize something interesting. The product of a conjugate pair is a constant scalar value (it is the sum of the four squares). Therefore, the derivative of Y = (Q*)Q with respect to Q must be zero (dY/dQ = 0). But if I think of this as Y = AQ and take the derivative with respect to Q, the result is dY/dQ = A. So if A = Q* then something looks to be amiss. I've been wondering what to do with Equation 3 in the work and if there are any Eigen-value type problems that need to be identified and resolved. You have given me a big clue.

      Thanks again.

      Regards,

      Gary Simpson

      11 days later

      Dear Mr. Simpson,

      Thank you for thoughtfully warning me that I needed to have a "knowledge of quaternian and Linear Algebra" in order to understand your essay. Unfortunately, Newton was wrong about abstract objects having the option of being stationary or in motion, and Einstein was wrong for assuming that it was abstract light that was capable of obtaining constant speed. It is the real surface of all real objects that is in the same constant motion at the same constant speed, and as light does not have a surface, light is the only stationary substance in the real Universe.

      Regards,

      Joe Fisher

        Joe,

        Many thanks for making the effort. I hope the experience was not too frustrating.

        Linear Algebra is not too bad but quaternions were unknown to me prior to three years ago or so. It took a lot of effort on my part to stop the voice in my head from telling me that it makes no sense to add a scalar and a vector. What finally convinced me that it was ok was simply using them for what Hamilton intended ... namely, the ratio between non-collinear vectors.

        It is interesting that you believe that light is stationary. In the 2012 FQXI contest, I presented a scalar solution to the wave equation that was precisely that. Also, if you examine Equations 11.2 - 11.4 of this essay and set v = c for any of the velocity components and then apply the Lorentz Transform, the result is that the change in position is zero. Note that I did not say velocity but rather change in position. I'm still pondering the meaning of both of these things.

        I'm catching up with my reading and should be able to comment on your work soon.

        Best Regards,

        Gary Simpson

        5 days later

        Dear Gary,

        Your essay is superb! I gave it a 10 with only a regret that 11 or 12 are not options. It is worth pointing out that your matrix equation for the inverse of y = Qx has Q^{-1} that is the same form as the electromagnetic tensor. This is why Maxwell formulated electromagnetic theory with quaternions.

        If at all possible you might find my essay

        http://fqxi.org/community/forum/topic/2320

        of interest. I did not make explicit references to quaternions, but they are lurking in the background in the discussion on Bott periodicity.

        LC

          Lawrence,

          Many thanks for the kind words. I am very appreciative of your enthusiasm. If I can supply someone with a new tool to use when attempting to solve some of these difficult problems then I will count myself as fortunate.

          Having said that, I should mention that it would be better not to indicate how you might have rated an essay. The administrators at FQXI might construe that to be vote trading and that could be cause for disqualification.

          One of my objectives is to get back to Maxwell in quaternion form. I need to spend some time working on identities and simple kinematics prior to that. I also think that it should be possible to formulate and solve a quaternion style wave equation. That should closely resemble Dirac's solution without the need for factorization using Clifford Algebras. Something that puzzles me regarding that work by Dirac is the equation ab = -ba. I understand non-commutation etc ... What seems odd to me is that the first thing that I think of is simply the cross product of two vectors. If a and b are both vectors then (a cross b) = - (b cross a) for any arbitrary vectors. Obviously, I need to study the subject some more.

          I have read your essay once. I will need to read it one or two more times before being able to make any meaningful comments. I see from your e-mail address that you are no stranger to quaternions.

          Best Regards and Good Luck,

          Gary Simpson

          The Dirac equation can be looked at as the multiplication of quaternions. The Dirac operator is a quaternion valued set of differential operators and the spinor field is also quaternion valued.

          What you might be pondering is the role of forming differential forms from quaternions that are antisymmetric and quantum commutators. One can think of the 1, i, j, k as one-forms that give wedge products that are a generalization of the cross product. I think this is some question with the relationship between the Heisenberg group and quaternions. I think this relationship involves the AdS_5 spacetime.

          My email address golden field quaternions refers to the 120 quaternions in the icosian group, that is half of the E8 lattice. That gets into octonions.

          Cheers LC

          Dear Gary D. Simpson

          Your essay is an example of how learning enables identifying or finding new problems and persistence leads to proposed solutions. Moreover, the presentation of your ideas was logical and accessible, which is considerate of the reader, likely reflecting a desire that reader can share in the benefit of your hard work.

          Does the following question in relation to your essay make sense? Can a space of points described by quaternions each of which isotropically scales by the same increasing (scalar) scale factor model the isotropic cosmological expansion of space?

          Regards.

          Bob Shour

            Dear Bob,

            Thank you much for your generous comments. You are most gracious. I try to write as simply as possible with short sentences in a linear, logical sequence with clear endpoints. Sometimes I have to leave an idea dangling so that I can merge it with something else.

            Regarding your question, I think that I understand your question and I think that the answer is yes. I'll expand on this a little. If you look at Equation 3 in my essay and you make the vectors collinear, the cross product term becomes zero. Now, you only have to worry about the scalar term. I have done the calculation using 13.8 billion light-years as the radius of the universe. This allows me to predict an expansion rate equal to 70.75 km/sec per Mpc. The observed expansion rate is 67.80 km/sec per Mpc.

            So it looks like I can get to within roughly 5% of the accepted value by using a fairly simple analysis. But I think there may be a problem. The way the cross product term is defined, it looks to me like linear velocity and angular velocity are linked (ie, they are not independent of each other). The implication is that if something is far away and moving linearly very fast then the space associated with it must be rotating fast.

            Best Regards,

            Gary Simpson

            5 days later

            Dear Gary,

            You forewarned that knowledge of quaternions and Linear Algebra would be required to fully grasp your essay. I don't have much knowledge of either but was able to decipher that your essay was trying to bring out something important from the depths using calculus. I am happy that your essay suggests a way of reconciling irreconciliables like Special relativity and the Aether.

            My reservation about calculus is how the limit or the infinitesimal can be equated with zero and in the same breadth not zero. This looks like a mathematical trick, even if a useful one. Under a magnifying lens what appears infinitesimal or tending towards zero can be seen to be physically non-zero.

            What is your take on these statements about the infinitesimal, dx

            dx ~ 0 (~ = indistinguishable from)

            neither dx = 0 nor dx тЙа 0

            dx2 = 0

            dx тЖ' 0 (тЖ' = becomes vanishingly small)

            The statement dx = 0 and dx тЙа 0 cannot both be logically true thus raising suspicion. While not denying the usefulness, suggest to equate dx with the Planck length, 10-35m . You may read my efforts and comment when you have the time.

            Regards,

            Akinbo

            Akinbo,

            Thank you for reading my essay. In truth, I think that all of the FQXI readership is easily able to follow my mathematics but I do feel obliged to give a small warning regarding Linear Algebra and quaternions. Having said that, all that is really needed is to know regarding LA is how to multiply matrices and to understand the meaning of an inverse matrix.

            I have read your essay and will be commenting shortly.

            Regarding infinitesimals in general, I am perfectly comfortable with them as mathematical entities. Regarding dx^2, I can only say that it goes to zero faster than dx.

            Infinitesimals are not really a difficulty in mathematics since they are only used to integrate or differentiate. When used for integration, you multiply by them and take a sum. So you are summing the infinitesimals to produce an area. When used to differentiate, they cancel each other out in part of the calculation and simply go to zero in another part ... consider the following:

            let y = mx

            dy/dx = (m(x delta) - mx)/((x delta) - x)

            this simplifies to

            dy/dx = m*delta/delta = m

            So when you take the limit as delta goes to zero, there is no delta in the equation. SO it is not really a problem at all. When higher order polynomials are differentiated, a similar behavior is observed except that there will be one or more delta terms. When the limit is taken, they become zero and only the first term (where there is no delta) remains.

            This brings out a more general problems of zero divided by zero, infinity divided by infinity, and zero multiplied by infinity. These are all considered to be indeterminate and L'Hopital's Rule is used to evaluate them. Essentially, the question is asked ... "which function goes to zero the fastest or which function goes to infinity the fastest". L'Hopital evaluates this by looking at the derivatives.

            So you see, dx is not really a trick.

            It is interesting that you should mention the Aether. I will confess to being a believer therein. The problem of course is how to measure something if it is not understood what that something is. It was thinking about that problem that ultimately produced the thinking that went into my essay.

            Best Regards,

            Gary Simpson

              Dear Gary,

              Can dx have a smallest possible size?

              As you pointed out, infinitesimals may not be a difficulty for mathematics, but what of for physics?

              Calculus and Cauchy's solution are used to solve Zeno's paradox, what do you think about the last step, which though indeterminate must be taken?

              Give these some thought.

              Regards,

              Akinbo

              Akinbo,

              To me, the infinitesimal can be zero. It is simply a concept. Whether or not it has physical meaning does not alter its usefulness to me as a concept.

              For Physics, it does not make any sense to me to think about distances that are smaller than the size of a proton or a neutron since those are the smallest stable particles that have a physical dimension. I'll discuss the electron further in your forum since that is where you pose the question. It is true that both the proton and neutron can be destroyed by high-energy collisions. However, it is not true that either can simply be divided into parts. To do so would simply destroy them.

              Regarding Zeno's Paradox for motion ... to me, it is not a paradox. He makes note that you must go 1/2 of the remaining distance each step but he does not mention that at constant velocity it will only take 1/2 of the time of the previous step. To me, Zeno is an example of a missed opportunity. He knows that motion is possible yet his logic gives him doubts. What he needed was a new concept (sum of infinite series) that would connect the two. But he was not able to make the mental step and humans had to wait for 2000 years for Isaac Newton to come along with Calculus. I do not give Zeno a high score ... of course, everything is obvious after someone else shows you the truth.

              The discussion between Dr. Klingman and Dr. Maudlin reminds me of Zeno. There is a deeper truth to be had there. I just hope we don't have to wait 2000 years to get a solution.

              Best Regards,

              Gary Simpson

                Gary,

                Thanks for your response over at my forum topic. I admit Calculus has achieved a lot but I don't want to let you off so easily :).

                You say, to you the infinitesimal can be zero. This runs against the common notion that the infinitesimal is a quantity almost indistinguishable from zero, but not zero, at least as is discussed HERE.

                Anyway, the reason for my post is that I have been giving Calculus more thought following your responses. It appears clear that for calculus dx can have no lowest finite value and that being the case, a line would be continuous, having an infinite number of points. Because you have also once considered motion as the creation and destruction of space, and motion as the destruction of the moving object and its creation and reappearance in the next adjacent space, it means all scenarios have at least once featured on your table.

                What I want to now know is how the continuous line in space can be cut since between any two points there is always a third by definition. Where on this line can you cut, since a point is uncuttable by definition?

                Regards,

                Akinbo

                Akinbo,

                Thanks for the continued dialog.

                I will answer your question concerning making a cut for both a physical structure such as a piece of string and for a mathematical concept such as a continuous line.

                Let us say that a piece of string is cut by a knife. The knife is harder than the string. Its atoms are bound to each other more strongly than the atoms in the string. The knife is forced into the space of the string and the bond between some of the atoms in the string is broken. The cut occurs in the space between atoms.

                Now let us consider an abstraction such as a continuous geometric line. A cut is to be made between two points but there are an infinite number of such points between any two points ... so where is the cut to be made? Let us be smarter than Zeno and introduce a new concept ....

                Let an interval near point x be defined as follows:

                (x - delta) < x < (x delta) where x and delta are both reals

                We can now make a cut at point x by taking the limit as delta approaches zero. Essentially, the line is cut into segments by removing point x. Something to remember about real numbers is that they have an infinite number of digits. So, the real number one is 1.000000000 ... ad infinitum. This is equivalent to infinite precision or to taking the limit of the above interval as delta goes to zero.

                The more interesting question is what happens to point x? Both line segments approach it but neither segment includes it. I suppose that it could be reconnected to one of the segments but not to both.

                Regards,

                Gary Simpson