Unifying Gravity and EM

sweetser · Apr 21, 2008

Covariant GEM field equations

Hello:

<Preamble>
This post is very similar to the last two. This time I have few of the set of the last two, but unify gravity and EM in the process. Nice.
</Preamble>

In this post I will derive gravity and the Maxweill field equations using quaternion operators in a manifestly covariant notation.

The Hamilton representation will be used for the Maxwell field equations. The even representation of quaternions - where all signs are the same, and the Eivenvalues ore excluded so the quaternion can be inverted - will be used in the first phase of this derivation. What the even representation does is flip the relative signs of the E and B signs, so now the e filed has two signs, and the b field has one, like so:

E = -\nabla_0 A ~-~ \nabla_u \phi
e = \nabla_0 A2 ~-~ \nabla_u \phi

B_w = \nabla_u A_v ~-~\nabla_v A_u = \nabla \times A
b_w = -\nabla_u A2_v ~-~\nabla_v A2_u = \nabla \Join A2 \quad eq ~0

The fifth play is the term that makes up the gauge field:

g = \nabla_0 \phi ~-~ \nabla \cdot A

None of these transform like tensors, but together they do from \nabla A.

Let's generate all 5 fields:

\nabla A = (\nabla_0 \phi ~-~ \nabla \cdot A2, \nabla_0 A2 ~+~ \nabla_u \phi ~+~ \nabla \times A) = (g, -E ~+~ B)
\nabla^* A2 = (\nabla_0 \phi ~-~ \nabla \cdot A2, -\nabla_0 A2 ~+~ \nabla_u \phi ~-~ \nabla \Join A2) = (g, -e ~+~ b)\quad eq ~1

The starting point for the derivation of the unified GEM equations is the Lagrangian which can be viewed as the difference between the scalars of E,b squared and B,e squared. This can be achieved by using the first set of quaternion operators used in the previous posts:

\frac{1}{8}((\nabla A)(A \nabla) ~-~(\nabla^* A2)(\nabla A2^*))

=\frac{1}{8}(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~+~ \nabla \times A)(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \times A)
-(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \Join A)(\nabla_0 \phi ~-~ \nabla \cdot A, -\nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \Join A)

=\frac{1}{2}(g, -E ~+~ B)(g, -E ~-~ B) ~-~ (g, e ~+~ b)(g, -e ~+~ b) = \frac{1}{2}(B^2 ~-~ E^2 ~-~ b^2 ~+~ e^2, 2 E \times B ~-~ b \Join b ~+~ e \Join e)\quad eq ~2

Any expression derived from this one will be invariant under a scalar gauge transformation which uses the terms \nabla_0 \phi, \nabla \cdot A because they have miraculously been cancelled!

Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling whose phase has spin 1 and spin 2 symmetry[/itex]:

\mathcal{L}_{EBeb} = -~(\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)
-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3

[sidebar: The assertion that the current coupling term has both spin 1 and spin 2 symmetry for all terms in the phase requires a small calculation. I decided to leave that as a problem. My guess is that until someone starts confirming these by hand, I'll be the only one to see that even these cool details work out.[/sidebar]

Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable:

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} \phi)}) = - \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho
= -\frac{1}{2}\nabla \cdot (E ~-~ e) - \rho = 0 \quad eq ~4

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_1)}) = -~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1
= \frac{1}{2}(\nabla_0 (E_1 ~+~ e_1) - (\nabla \times B)_1 - (\nabla \Join b)_1) + J_1 = 0 \quad eq ~5

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_2)}) = - \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2
= \frac{1}{2}(\nabla_0 (E_2 ~+~ e_2) - (\nabla \times B)_2 - (\nabla \Join b)_2) + J_2 = 0 \quad eq ~6

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_3)}) = - \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3
= \frac{1}{2}(\nabla_0 (E_3 ~+~ e_3) - (\nabla \times B)_3 - (\nabla \Join b)_3) + J_3 = 0 \quad eq ~7

This work can be summarized with the GEM gravity source equations:

\frac{1}{2} \nabla \cdot (E ~-~ e) = \rho \quad eq ~8

\frac{1}{2}(\nabla \times B ~+~ \nabla \Join b - \nabla_0 (E ~+~ e)) = J \quad eq ~9

This has been a manifestly covariant derivation. This derivation remains the same in flat or curved spacetime, in Cartesian or spherical coordinates.

I have spent the day trying get all the signs and factors right. If there are any questions, send me a private note, and I will recheck it. If needed, we can put up a new post.

Doug

Stills:
http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations

The talk:

(this derivation was not part of the talk)

sweetser · Apr 21, 2008

Covariant gravity field equations

Hello:

<Preamble>Looks like two of my posts were identical, sorry. The software here doesn't appear to like this much LaTeX.

This post is very similar to the last. I am starting out the draft by cutting and pasting the text, as all the same terms are the same. used in the same locations of the same equations. What changes are a few signs. The E field is made of two parts, -\nabla A_0 and -\nabla \phi. These happen to have the same sign. In this post, the case where these have opposite signs will be explored. A similar thing will be done for the magnetic field, where the two terms \nabla_u A_v and \nabla_v A_u have the same sign.

Those readers concerned about how these objects transform can rest easy. We know that \nabla A transforms like a rank 2 tensor - it was the justification behind developing the covariant tensor \nabla[/tex] in the first place. We also know that the EM field strength tensor, \frac{1}{2}(\nabla_u A_v - \nabla_v A_u), transforms like a tensor. The difference between these two tensors which is \frac{1}{2}(\nabla_u A_v + \nabla_v A_u), also transforms like a tensor since the difference of two tensors remains a tensor. </Preamble> In this post I will derive the gravity part of the GEM proposal using quaternion operators in a manifestly covariant notation. The even representation of quaternions - where all signs are the same, and the Eivenvalues ore excluded so the quaternion can be inverted - will be used in the first phase of this derivation. What the even representation does is flip the relative signs of the E and B signs, so now the e filed has two signs, and the b field has one, like so: E = -\nabla_0 A ~-~ \nabla_u \phi e = \nabla_0 A2 ~-~ \nabla_u \phi B_w = \nabla_u A_v ~-~\nabla_v A_u = \nabla \times A b_w = -\nabla_u A2_v ~-~\nabla_v A2_u = \nabla \Join A2 \quad eq ~0 None of these transform like tensors, but together they do from \nabla A. Notice that the quaternion differential operator acting on a 4-potential creates a scalar gauge field, and two fields the -e and b fields: \nabla^* A2 = (\nabla_0 \phi ~-~ \nabla \cdot A2, \nabla_0 A2 ~-~ \nabla_u \phi ~-~ \nabla \Join A2) = (g, e ~+~ b)\quad eq ~1 The starting point for the derivation of the GEM gravity equations is the Lagrangian which can be viewed as the difference between the scalars of b squared and e squared. This can be achieved by changing the order of the conjugation operator with respect to the 4-potential, which flips the sign of the time derivative of A and gradient of phi which make up e, but not the symmetric curl b. The scalar gauge field can be subtracted away: \frac{1}{8}(\nabla^* A2 ~-~ (\nabla^* A2)^*)(\nabla A2^* ~-~ (\nabla A2^*)^*) =\frac{1}{8}(0, \nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \Join A)(0, -\nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \Join A) =\frac{1}{2}(0, e ~+~ b)(0, e ~-~ b) = \frac{1}{2}(b^2 ~-~ e^2, b \Join b ~-~ e \Join e)\quad eq ~2 Any expression derived from this one will be invariant under a scalar gauge transformation which uses the terms \nabla_0 \phi, \nabla \cdot A because they have been explicitly subtracted away at this early stage. It is also of interest to think about the properties of the 3-vector, since in the EM case it was the Poynting vector. Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling, -\frac{1}{2 c}(J A2* + (J A2^*)^*): \mathcal{L}_{eb} = \frac{1}{2}((\nabla_1 \phi)^2 ~+~(\nabla_2 \phi)^2 ~+~(\nabla_3 \phi)^2 ~+~ (\nabla_0 A_1)^2 ~+~ (\nabla_0 A_2)^2 ~+~ (\nabla_0 A_3)^2 ~-~ (\nabla_3 A_2)^2 ~-~ (\nabla_2 A_3)^2 ~-~ (\nabla_1 A_3)^2 ~-~ (\nabla_3 A_1)^2 ~-~ (\nabla_2 A_1)^2 ~-~ (\nabla_1 A_2)^2) ~-~ (\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3) -\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3 Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable: \nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} \phi)}) = \nabla_1^2 \phi ~+~ \nabla_2^2 \phi ~+~ \nabla_3^2 \phi ~-~ \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho = -\nabla \cdot e - \rho = 0 \quad eq ~4 \nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} A_1)}) = \nabla_0^2 A_1 ~-~ \nabla_3^2 A_1 ~-~ \nabla_2^2 A_1 ~-~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1 = \nabla_0 e_1 - (\nabla \Join b)_1 + J_1 = 0 \quad eq ~5 \nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} A_2)}) = \nabla_0^2 A_2 ~-~ \nabla_3^2 A_2 ~-~ \nabla_1^2 A_2 ~-~ \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2 = \nabla_0 e_2 - (\nabla \Join b)_2 + J_2 = 0 \quad eq ~6 \nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} A_3)}) = \nabla_0^2 A_3 ~-~ \nabla_2^2 A_3 ~-~ \nabla_1^2 A_3 ~-~ \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3 = \nabla_0 e_3 - (\nabla \Join b)_3 + J_3 = 0 \quad eq ~7 This work can be summarized with the GEM gravity source equations: -\nabla \cdot e = \rho \quad eq ~8 \nabla \Join b - \nabla_0 e = J \quad eq ~9 This has been a manifestly covariant derivation. This derivation remains the same in flat or curved spacetime, in Cartesian or spherical coordinates. Once you get used to keeping track of all these terms, it is kind of fun, honest! Doug Stills: http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations The talk: <iframe allowfullscreen="" scrolling="no" allow="autoplay" style="background:linear-gradient(0deg,rgba(0,0,0,0) 0,rgba(0,0,0,0) 72%,rgba(0,0,0,.04) 79%,rgba(0,0,0,.1) 83%,rgba(0,0,0,.25) 88%,rgba(0,0,0,.67) 100%),url(https://i.ytimg.com/vi/Kd4nNb7nGOc/hqdefault.jpg) 50% 50% / cover" src="https://www.youtube.com/embed/Kd4nNb7nGOc" data-s9e-mediaembed-c2l="youtube" data-s9e-mediaembed-c2l-oembed-id="Kd4nNb7nGOc" data-s9e-mediaembed-c2l-src="https://www.youtube.com/embed/Kd4nNb7nGOc?autoplay=1"></iframe>

sweetser · Apr 26, 2008

Hatch's work on an exponential force equation

Hello Carl:

Ronald Hatch sent me his paper which has an exponential in the force equation. For me as a member of the ultra-conservative fringe, it was not a happy read. He did not define a Lagrangian, so the proposal felt ad hoc. The way to get the force equation is to vary the Lagrangian with respect to velocity.

He thinks there is an absolute ether which strikes me as wrong.

He wants to give up the equivalence principle. His logic for this struck me as muddled. The discussion of mass was particular confusing. It looked like he had the classic self-taught blindspot vis-a-vis E=mc², that the real relation is between the invariant m² c⁴ and the square of the covariant 4-momentum, E² - P² c². One of his equations looked like a trivial rearrangement of terms, a rearrangement he took seriously.

The precession of the perihelion of Mercury is a tough calculation. He wanted to add a second order effect in as if it was a first order effect.

By a strict application of the rules for the Independent Research area of Physics Forums, I don't think his work would be accepted. I forwarded my concerns to him directly, but he said I needed to think about things more carefully. The paper is published in Physics Essays.

Doug

sweetser · May 1, 2008

Seeing spin 1 and spin 2 symmetry

Hello:

I exchanged a dozen emails with Steve Carlip over the coupling current J^{\mu} A_{\mu}. The discussion started this way: "I have thought about Feynman's analysis of the spin of the vector current coupling." This post represents the back story, what I got from looking at chapter 3, pages 29-39 for about three to four weeks.

The first thing I did was work on my speed of going through the algebra in section 3.2, "Amplitudes and polarizations in electrodynamics, our typical field theory". You can only motor if you are confident about all the steps, just like in video games. The logic runs like this:

1. Start with the current-potential contraction.
2. Take the Fourier transformation of the 4-potential to get a current.
3. Simplify the current-current interaction along one axis, z.
4. Write out the contraction in terms of its components.
5. Use charge conservation to eliminate one term.
6. One term is the standard Coulomb interaction, the others are the relativistic corrections.

Then Feynman wants to know what that correction term is. This is were it gets a little odd. He talks about plane polarized light, and how looking at that you can see the angular momentum projections. I admit, I never quite saw those. What I did instead was try to strip away all the physics-speak, and just find the kernel of math underlying the operation. Looking back, that is what took the time: reducing the physics to a simple math expression. The Rosetta stone was a line on page 39:

Feynman said:

...we know that (x ~\pm ~ i y)(x ~\pm~ i y) are evidently of spin 2 and projection \pm 2; these products are (xx ~-~ yy ~\pm~ 2 i xy), which have the same structure as our terms (3.4.1)

This is the pure math way to spot a system with spin 2: start with the product of two complex numbers, and check that the imaginary part has a 2ixy. This will require a change in x of pi radians to get back to the start point since there is an multiplier of 2. To speak like Feynman, I should talk up the projection operators, but I like to keep the math kernel free of that jargon.

Looking back on an earlier calculation, I was able to transcribe an earlier bit of algebra into a similar set of complex numbers:

(x ~\pm ~ i y)(x' ~\pm~ i y')^* = (xx' ~+~ yy' ~+~ (yx' ~-~ xy') i \quad eq~1

This is a system which has spin 1 projection operators, to use physics-speak. In math terms, the xy' does not help out the yx', there are no factors of 2 or 1/2, so this would take 2 pi radians to get the imaginary part of this back to where it was.

In terms of the math, the difference between a spin 1 system (eq 1) and a spin 2 system (see the quote) is no more complicated than looking at the imaginary part of a complex product.

Doug

sweetser · May 3, 2008

Feynman, Steve Carlip, and I

Hello:

I have the exchange of emails between Prof. Steve Carlip and myself up on the screen. It is not easy for me to read. Steve is a professional, I am not. On occasion, I babble. Babbling is a form of exploring, a process used to learn how to speak a language like a native. Studies have shown that deaf children do so learning sign language, and that baby birds do so before they can sing exactly like adults do. Recognizing this process, I have great patience for others that babble physics. Steve probably is that way in the right context, but in this email exchange, I got the book tossed at my head.

The discussion began with a http://picasaweb.google.com/dougsweetser/SpinAndPhaseOfCurrentCoupling/photo#5196141258835741330 . I went by the book, section 3.2 of Feynman's lectures, for three of four steps:

1. Start with the coupling term, J^{\mu} A_{\mu}.

2. Take the Fourier transformation, J^{\mu} A_{\mu} = -\frac{1}{K^2} J^{\mu} J'_{\mu} \quad eq~3.2.2

3. Write out 2 in terms of the components:
J^{\mu} A_{\mu} = -\frac{1}{K^2}(\rho \rho' - J_1 J'_1 - J_2 J'_2 - J_3 J'_3 ) \quad eq~3.2.5

[Note on imprecise notation: in the slide I use x, y, and z which imply a coordinate choice. I should have used numbers for subscripts. I also didn't toss in the minus sign as Feynman does for the Fourier transformation step.]

Up to this point, I have exactly walked down the path Feynman wrote about. When it came to writing up the slides, I initially put in a further step Feynman used: he imagines picking a coordinate system such that all the current goes along one direction (the 3 axis). Everyone is accustom to this step. Yet it bothered me. I would have to rewrite the derivation if someone chose the 2 axis instead. A more general position would involve choosing no axis, yet spotting the symmetries of the spin in the phase anyway. I decided to work with that as a goal.

4. Multiply out the two currents as quaternions:

-J J' = (-rho rho' + J₁ J'₁ + J₂ J'₂ + J₃ J'₃,

-rho J'₁ - J₁ rho' - J₂ J'₃ + J₃ J'₂,

-rho J'₂ - J₂ rho' - J₃ J'₁ + J₁ J'₃,

-rho J'₃ - J₃ rho' - J₁ J'₂ + J₂ J'₁)The terms in italics are equation 3.2.5, the underlined terms are in an expression about spin 2 symmetry on page 39, and the terms in bold are in 3.2.10 in a discussion of spin 1 symmetry.

Steve had no idea what I was doing, none. Communication was broken by step 4. I was at work, trying to do my job, or appear to be doing my job, and quickly come up with a response to someone with far more intellectual precision. It did not work out so well. The worse thing I did was about equation 3.2.10 concerning circularly polarized light. Feynman writes out the currents for two circularly polarized light whose imaginary parts cancel. I got a sign wrong, so they didn't cancel, and I got to look stupid.

Both Steve and Feynman talked about projection operators, the relevant machinery from quantum field theory. I did not talk about projection operators at all. I am not going to do so now since I would probably just babble about them. I understand why the well-trained would say that if I don't discuss projection operators intelligently, then this has nothing to do with the spin of particles. Nothing. For me, projection operators are a patina on the underlying algebra (patina def.: the sheen on the surface of an old object, caused by age and much handling).

Steve left the discussion convinced I didn't even understand the basics of spin in physics. He won the word game, but the algebra in step 4 still stands consistent with what Feynman wrote. At the end of the day, algebra trumps words.

Doug

sweetser · May 5, 2008

Three new bits of math

Hello:

Preparing for my talk in Brazil, I had an interesting insight. The thesis behind "Doing Physics with Quaternions" at quaternions.com is that physics describes patterns of events in spacetime using quaternions up to an isomorphism. Most of physics works great without quaternions because the quaternion expression would not provide new information.

My epiphany was this: new quaternion math equals new physics. Here in this thread, the new math is the Even representation for quaternions (a reinvention of Clyde Daven's hypercomplex numbers). The current coupling J2 A2 has the spin 2 symmetry in the phase, and the field strength tensor \nabla A2 contains the symmetric curl needed for the symmetric field b.

I have also mentioned here the work in animating quaternions which lead to an understanding why the groups U(1), SU(2), SU(3) and Diff(M) must be all that makes up the symmetry forces of Nature. With the visual perspective, the forces are more tightly linked algebraically. The standard model is written as U(1)xSU(2)xSU(3), which would have 1+3+8=12 generators for its Lie algebra. One mystery of the standard approach is why the electro and weak forces should team up to form the electroweak force. Another mystery is why should confinement exist for the strong force SU(3)? These phenomena are suggesting something more like [U(1)xSU(2)]xSU(3) than three equal players. That is what happens with the quaternion representation of the symmetry of forces:

A^* B = (\frac{A}{|A|} exp (A - A^*))^* (\frac{B}{|B|} exp (B - B^*))\quad eq~1

which is (U(1) SU(2))^*(U(1) SU(2)). The Lie algebra only has 8 generators. This smaller model has a chance to provide a cause for the confinement of quarks.

The third quaternion math innovation I do not talk about much because I have yet to see how it impacts a calculation, although it helps with a big riddle, a "why" question, in physics. Many who work with quaternion derivatives accept the idea of a left handed versus right handed derivative. This comes from the limit definition, putting the differential on the left or right. This definition is ineffective since one cannot show that a function as simple as f(q)=q² is analytic in q. For me that indicates the definition has no utility. What I did was steal a move from L'Hospital's rule and use a dual limit process. Let the pesky 3-vector with its three imaginary basis vectors go to zero first, leaving only the real number which commutes with all. Effectively this is a directional derivative along the real axis. Things work out great for proofs using this definition (if one is good at doing proofs, which I am not). All events are ordered by the real scalar. If this definition is applied to events in spacetime, the scalar is time, and thus all the events are order in time like a movie.

What happens when the limit processes are reversed, and the pesky 3-vector goes to zero after the scalar gets frozen? All one can do in this case is to determine the norm of the derivative. Although not well known, there is a branch of math that studies norms of derivatives. I think this is the domain of quantum mechanics. We cannot order things in time, but we can tell on average how much change is going to happen after making many measurements.

Now to make the slides...
Doug

mhill · May 5, 2008

but when doing the product of quaternions if Q is a quaternion then where you put Q.Q=Q^{2} it should read Q.Q^{*} , for example for Minkowsky metric

dQ=dt-idx-jdy-kdz then ds^{2}=(dQ).(dQ^{*})=dt^{2}-dx^{2}- dy^{2}-dz^{2}

and as i pointed in other part of the forum, you have the problem of non-commutativity so

ij(dx.dy)=-ji (dy.dx),

anyway the idea seems very interesting

sweetser · May 5, 2008

The Even representation

Hello mhill:

In the standard Hamilton representation, for a quaternion dQ = (c dt, dx, dy, dz) we have the product:

dQ dQ = (c^2 dt^2 ~-~ dx^2 ~-~ dy^2 ~-~ dz^2, 2 c ~dt~ dx, 2 c~ dt~ dy, 2 c ~dt~ dz)\quad eq~1

When one uses tensors, the Einstein summation convention ignores the three other terms. That is a mistake in my opinion. One thing I have noticed about the GEM proposal: the 3-vector (2 dt dx, 2 c dt dy, 2 c dt dz) is an invariant in the presence of a gravitational source. That is really cool because special relativity is about the invariant scalar, and for GEM, gravity is about the invariance of the 3-vector.

If we look at the product of two quaternions that are nearby each other, a dQ and dQ', we get a similar result with a cross product:

dQ dQ' = (c^2~ dt~ dt' ~-~ dx~ dx' ~-~ dy~ dy' ~-~ dz ~dz',
c~ dt ~dx' ~+~ c ~dx~ dt' ~+~ dy ~dz' ~-~ dz~ dy',
c ~dt ~dy' ~+~ c ~dy ~dt' ~+~ dz ~dx' ~-~ dx ~dz',
c ~dt ~dz' ~+~ c ~dz ~dt' ~+~ dx ~dy' ~-~ dy ~dx')\quad eq~2

Hamilton could have done that one, so this is not new math. For the Even representation of quaternions, the rules are easy to remember: everything is sunny and positive in California, and so it goes for the Even representation:

i^2 = j^2 = k^2 = ijk = 1\quad eq~3

ij = ji = k \quad eq~4

ik = ki = j \quad eq~5

jk = kj = i \quad eq~6

There is not a minus sign to write. This will be a division algebra if and only if the eigenvalues of the real 4x4 matrix representation are excluded from the set of quaternions.

There will be math wonks who insist that the name "quaternions" is reserved for non-commutative 4D division algebras. Historically, that has been the use. Doing new math can cause conflict, so be it. The non-commutative aspect is a result of an arbitrary choice in how to represent a 4D division algebra. I have chosen a different representation. It still is a division algebra, but one where multiplication commutes. Clyde Daven did this first and called them hypercomplex numbers. That makes them sound like a separate animal, and I don't think it is. This is representation theory in practice, applied to quaternions.

To indicate the Even representation is being used, I toss in a "2" every now and then. Repeat eq. 1 with the even representation:

dQ2^* dQ2 = (c^2 ~dt^2 ~-~ dx^2 ~-~ dy^2 ~-~ dz^2, - 2~ dy ~dz, - 2 ~dx ~dz, - 2 ~dx ~dy)\quad eq~7

The scalar is the same, but the 3-vector has changed. Instead of rescaling the (dx, dy, dz) by a 2 dt factor, this time it is all about the symmetric curl. The dt factor does not make it into the 3-vector.

Repeat the dQ2 dQ2' in the Even representation:

dQ2^* dQ2' = (c^2 ~dt ~dt' ~-~ dx ~dx' ~-~ dy ~dy' ~-~ dz ~dz',
c ~dt ~dx' ~-~ c ~dx ~dt' ~-~ dy ~dz' ~-~ dz ~dy',
c ~dt ~dy' ~-~ c ~dy ~dt' ~-~ dz ~dx' ~-~ dx ~dz',
c ~dt ~dz' ~-~ c ~dz ~dt' ~-~ dx ~dy' ~-~ dy ~dx')\quad eq~8

If dQ2' where to get acted on by the conjugate operator instead of dQ2, then the only terms to flip signs involve dt dR.

What makes the Even representation interesting is it gives a more complete way of viewing a quaternion product. In the Hamilton representation, the rescaling (dt dR) was unaltered by changing the order of multiplication, while the cross product flipped signs. In the Even representation, the symmetric curl is unaltered by changing the order of conjugation, while the rescaling flips signs. The scalar remains steady no matter what. That sounds like a more complete survey of the product of two sets of four numbers.

Doug

sweetser · May 13, 2008

Pitching the program

Hello:

I spend time and effort pitching this research project to both the upper elite and the technical masses. I took off a Wednesday from work to see a talk by Nobel Laureate Sheldon Glashow, who got one in the 70s for the electroweak theory. I told him I had an animation for U(1)xSU(2) on my MacBook Pro. He was too busy to look right then, but I could send him an email, which I did.

Later that evening, Michio Kaku was doing book promo work. He agreed to sign his latest book, "Physics of the Impossible", which is doing well on the New York Times Best Seller list. I got in line late, and the book was sold out by the time I got there. I bought a different book, and had him sign "Maxwell is the best!" I also dropped off a version of my proposal where I derive the Maxwell equations first - to established I am much better than your average crank - and with a variation get equations for a metric approach to gravity. He thanked me as he went on to sign another book.

Max Tegmark gave a talk on the Physics of Super Heroes, along with a screening of Superman. He is a big survey astronomer by day, so it was fun to cruise through the known Universe with his software. He knew me from a previous outreach program he did, and had traded a few stories back then about quaternions. I told him of the quaternion animation project, how it could be found on YouTube with a search for "Quaternions Standard Model". I gave him a business card with the search instructions after my 20 second pitch.

Low odds on getting a reply from these busy folks, but I need to try, so I do.

There are many more folks who read a high traffic site such as slashdot.com They had an article on Lectures On the Frontiers of Physics Online. Buch of big names in physics have videos up there:

Neil Turok's 'What Banged?,' John Ellis with 'The Large Hadron Collider,' Nima Arkani-Hamed with 'Fundamental Physics in 2010,' Paul Steinhardt with 'Impossible Crystals,' Edward Witten with 'The Quest for Supersymmetry,' Seth Lloyd with 'Programming the Universe,' Anton Zeilinger with 'From Einstein to Quantum Information,' Raymond Laflamme with 'Harnessing the Quantum World,'

You might be able to see the site in a day or two here, http://perimeterinstitute.ca/index.html

So I posted a note there which reflects my current thoughts:

Title: Maxwell Trumps General Relativity

General Relativity rocks. It is elegant in its minimialism. All efforts to add a little extra have failed, usually by allowing a dipole gravity wave mode of emission which has been ruled out by binary pulsar data.

The only field theory that is manifestly better than GR is the Maxwell field equations. Every time we have added to it in the name of symmetry, the theory has done more. James did it himself by tacking on the Ampere current. Einstein looked to get rid of a duplicate law, and so special relativity was born. With the huge supply of new particles coming out of atom smashers, the gauge symmetry in EM (U(1)) was expanded to SU(2) for the weak force, and SU(3) for the strong.

None of those smart cats listed in the initial post will be talking about the Maxwell equations. Too bad, the history of physics is clear: expand Maxwell, you win.

Max depends on the field strength tensor d_u A_v - d_v A_u. There is a subtraction in there, a great thing (called an exterior derivative). But in the name of symmetry, we need to work with the rest of it, d_u A_v + d_v A_u. Do that right, and you get a unified field theory that Einstein failed to find by looking for workable extensions of GR. Extend Max, not GR.

If anyone here wants to see the nuts and bolts of deriving the Maxwell equations using the Euler-Lagrange equations, search for "GEM action" on YouTube. A small variation - two minus signs - on the Maxwell equations leads to equations for gravity. Yes, I show that there is a metric solution (the Rosen metric if you are up on your GR jargon, a bunch of exponentials if not). Yes I know there is an issue of spin 1 and spin 2 which can be addressed if you get what the phase of current coupling really is.

YouTube can survive being slashdotted.

Doug

sweetser · May 24, 2008

Brazil Talk Video

Hello:

I am about to power down the laptop and get it ready for a trip to a foreign land. I will be giving this Thursday, May 29, in Campinas, Brazil at 5:30 pm. Stop by if you are in town. Since I figure about zero of you can do that, I just uploaded a video of the talk to YouTube. The video is 41 minutes long, but I only have a 25 minute slot, so I will have to be more efficient. As always, critiques are appreciated.

I really like the central thesis: that only by doing new math can I do new physics. Will report back on if I can get others excited by this work.

Doug

Mentz114 · May 24, 2008

Bon voyage, Doug.

Lut

sunsphere · May 24, 2008

I have a brand new theory on the mechanics of gravity I believe the whole science world should review for possible validity.

sweetser · May 25, 2008

Hello Sunsphere:

The rules for starting your own thread are https://www.physicsforums.com/showthread.php?t=82301. The rules may appear harsh, but there are many people who make grand claims and the rules are there to filter out ideas that are not precise enough to have a discussion about.

Doug
Brazil is tropical, what can I say?

sweetser · Jun 1, 2008

Report from ICCA8 in Brazil

Hello:

I am back in the US after the trip to the 8th International Conference on Clifford Algebras and the Applications to Mathematical Physics in Campinas Brazil. The exotic location was irrelevant. If the meeting were held in Indiana, it would have made little difference. I went from the plane to the hotel to one building on the campus, and back to the hotel. Sure the vegetation was odd, but when technical talks go from 8:30 AM to 6:30 PM, there was not time or structure to get any exposure to Brazilian culture. Campinas means "grass fields" in Portuguese, a place not far outside San Paulo. thirty years ago it had 100,000 people. Now there are over a million, attracted by high tech companies. When a farm town blows up quickly, the effort is focused on infastructure and not on art and public works. In the opening remarks, one of the organizers said one of the benefits of the meeting was that there was nothing to see in Campinas so we could focus on work.

There were about 70 participants. This was an international conference, with accents in French, Italian, German, British, Polish, and South American (few people from the Far East). Each day had a dozen talks. The morning and early afternoon plenary talks were an hour long. The last afternoon session had half hour talks for the "not invited" who wanted to present their work. This was a PC crowd (I may have had the only Keynote presentation on a Mac). Much of the meeting was video taped, but how they hope to make that available is not known. My own session was not recorded since I was shuffled to the alternate room of a parallel session.

There were many lessons I learned. One of the basic ones is that geometric algebra and Clifford algebra are one and the same (the similarities were clear, but not the formal identity). The story is that David Hestenes has made a big push to bring Clifford algebras onto the center stage of physics. He did not think that the name "Clifford" had information content, preferring to use the name Clifford himself used, geometric algebra, to discuss this area of work.

David attended the meeting. It was clear he was the "star", but he was a decent and approachable guy. It is also clear that despite his efforts, geometric algebra has remained a minor contributor to physics. I could relate to his position (excluding the star part). Quaternions are an even smaller footnote in physics. David said he discovered things using geometric algebra that he needed to translate into tensor lingo to publish. I also find myself translating quaternions insights back to tensors to communicate with the broader physics community.

During a coffee break, I asked him what he thought about quaternions. He said they are just a part of the bigger Clifford algebra (specifically CL(0, 2)). I expected that one, the typical male thing - mine is bigger. I will write a separate post about my critique about such a perspective. He invited me, like I saw him do with others, to grow up and embrace the real math tool of Nature, geometric algebra.

David's talk was great. He brought up an idea proposed by Louis de Broglie, that each atom has an internal clock with a frequency of B = m c^2/h. The math is simple but has been largely ignored by the physics community. The frequency is in the zeptoseconds range (if they skipped this in your schooling, it is because it is new, representing a factor of 10^{-21}). David proposed a way to detect this absurdly fast clock as a resonance in a crystal. The message is darn deep: TIME IS MASS (an ironic aside: I recall Archimedes Plutonium, an Internet crackpot who has passed away, that would always write this IN CAPS).

David did not give a mechanism for the clock. My own work might provide that. In GEM, one has a relativistic 4D wave equation. That has a 1/R² potential solution, leading to a 1/R³ force law. Dipoles have this sort of potential and force law. Since charged massive particles will both attract and repel their brothers, they are born as dipoles. I have been wondering what those dipoles do. Now I have a candidate: it could be the basis of the de Broglie internal clock and connected to mass. Neat.

Over the course of the next few weeks, I may discuss other talks that have already influenced my research. As for my own talk, I was scheduled to be the next to last talk on the next to last day. I had pitched the talk to a number of people. Most of the people I gave the personal pitch showed up, about 20 folks in the room. This may have been the biggest crowd with good technical credentials I have presented to (Hestenes was in the group).

The thesis was that quaternions, as invented by Hamilton and Rodrigues, cannot make new contributions to physics, but new math with quaternions can lead to new physics. I went through three examples.

First, I have a new definition of a quaternion derivative that splits into a directional derivative along the real line for classical physics, and a normed derivative for quantum mechanics. Getting the derivative of a quaternion derivative on a spacetime manifold might resolve the Bohr/Einstein debate as to why quantum mechanics is different from classical physics.

Second, the analytic animations represents a logical extension of analytic geometry. We can be precise, exact, with the treatment of time's relationship to space. For this crowd, I developed some new animations about cross products which I will discuss in another post.

Third, the Maxwell equations have a better track record at being extendable and connected to quantum mechanics than general relativity. I have figured out how to write the Maxwell action using quaternions. When I wanted to write a symmetric tensor contraction using quaternions, I faced a wall. The way to tunnel through the wall was to find a 4D division algebra where elements commute so long as the Eigen values are excluded. That leads to the GEM action, a way to unify gravity and EM. I showed the solution to the field equations that was physically relevant. There was a slide for the spin of the current coupling term.

The crowd reaction was standard: no one asked a question (other than one clarification). Most other talks had two or three questions. No one came up to ask about the content of the talks during later coffee breaks or over lunch. People were interested in the fact that this was a hobby of mine.

This trip was well worth the time and money, I am glad I went.

Doug

Mentz114 · Jun 2, 2008

Hi Doug,
thanks for ther report. It's very interesting. The thing is that if we're geometrizing something, then to handle curved spaces you need tensors It's very handy that all the observables are scalar contractions of tensors.

I will have to bone up on geometric algebra, which may be has a different way of doing curved space.

Lut

sweetser · Jun 2, 2008

Curved spacetime quaternions

Hello Lut:

It would take less effort to "bone up" on how to handle curved spacetime with quaternions. In flat spacetime, an event at:

event = (t e_0, x_1 e_1, x_2 e_2, x_3 e_3) \quad eq~1

where:

|e_0| = |e_1| = |e_2| = |e_3| = 1 \quad eq~2

e_0^2 = +1
e_1^2 = -1
e_2^2 = -1
e_3^2 = -1\quad eq~3-6

In spacetime curved by a static, spherically symmetric mass charge, equation 1 remains the same, while the other ones are altered:

|e_0| = \frac{1}{|e_1|} = \frac{1}{|e_2|} = \frac{1}{|e_3|} = exp(-G M/c^2 R) \quad eq~7

e_0^2 = +exp(-2 G M/c^2 R)
e_1^2 = -exp(+2 G M/c^2 R)
e_2^2 = -exp(+2 G M/c^2 R)
e_3^2 = -exp(+2 G M/c^2 R)\quad eq~8-11

This might not be too hard to implement, now that I have written it down clearly (the key step to programming). q_metric -mass 10e-6 1 2 3 4 would return the event (1, 2, 3, 4) in a space curved by a 10e-6 mass.

The problem with my software as written is handling this very tiny numbers.

Doug

Mentz114 · Jun 2, 2008

Hi Doug,
yes, one way of losing tensors is to use differential geometry ( if I haven't got my nomenclatures wrong). I think that's what you did above, by introducing basis vectors in your your quaternion space. Makes sense.

What technology are you using for computing ? I have floating point down to 10^-300.

Lut

sweetser · Jun 2, 2008

Precision in C

Hello Lut:

I am using C. I print to the command line which by default prints to 8 digits. I bet I can adjust printf to do more. I am not certain how to set the internal precision accurately of a standard C program.

I have always had the basis vectors, but had not put them to use in a calculation.
Doug

Mentz114 · Jun 2, 2008

Hi Doug,

sounds like you need a better C compiler.

You seem to have all the apparatus for space-time in your algebra. Letting gravity appear in the basis and not the coordinates is a good practice.

When you conjugate this

event = (t e_0, x_1 e_1, x_2 e_2, x_3 e_3) \quad eq~1

do you change sign and invert the basis vectors ? I'm thinking about a conjugate space, and what scalars you can make by multiplying a vector times a conjugate ?

Lut

sweetser · Jun 2, 2008

Conjugates

Hello Lut:

That is a great question! My initial reaction was the signs just flip. The thought is that conjugation is about the work of mirrors. In flat spacetime, three sign flips is all one needs to do a mirror reflection. If the mirror is curved, I am not sure if flipping the sign is enough. I'll have to think about it some more.

I use the gcc compiler. I am sure there is a way to make the precision high, but at this time I don't know how to do it. What is your programming trick/tools to get such high precision?

Doug

Mentz114 · Jun 2, 2008

Hi Doug,
I was free thinking there. I'll have to ponder it too.

I use Delphi Win32 and I've got the following floating type amongst others

Extended 3.6 x 10^-4951 .. 1.1 x 10^4932 digits 19-20 uses bytes 10

10^-4951 !

Lut

sweetser · Jun 2, 2008

Precision in gcc

Hello Lut:

I figured out how to crank up the precision. The default single precision float has about 7 significant digits. At double precision, there are 16 significant digits. All I have to do is set the -DFmode flag and set all programs to print 16 digits. There is the third level, -XFmode which will do 24 digits, but that may just be wasteful of resources, using 12 bytes. There is a Tetra Floating number for those that need 16 bytes.

Thanks for getting me to look this up.
Doug

sweetser · Jun 6, 2008

Quaternions versus Clifford algebras

Hello:

Quaternions as events in spacetime physics are not the Clifford algebra CL(0, 2). Let me first explain what that bit of jargon means.

Clifford algebras were an attempt to generalize the math started by Hamilton, passing through Grassman. These algebras are independent of coordinates, and can be written in arbitrary dimensions. As algebras, it is not important that they are invertible, although that can happen.

The Clifford algebra CL(0, 2) is a multivector, meaning it has two parts. First there is the scalar, the 0 part, a pure number. The bivector is the second part. They use a wedge product, \wedge, a generalized cross product to form a 3-term bivector.

q(t, x_1, x_2, x_3) = t ~+~ x_1 e_2 \wedge e_3 ~+~ x_2 e_3 \wedge e_1 ~+~ x_3 e_1 \wedge e_2 \quad eq~1

Why use a wedge product? The reason is that wedge products are axial vectors, those that switch handedness in a mirror. The other sort is a polar vector. If one looks at an event in a mirror, this makes physical sense.

It is not worth the time to argue with math types who think they can define whatever they want, however they see fit. I have the perspective of a mathematical physicists, where any math definition can be cow-roped by physical meaning (I am a crude mathematical physics who uses rodeo analogies).

The unspoken assumption behind CL(0, 2) is that one should use a mirror on this event. That is like taking (t, x₁, x₂, x₃) to (t, -x₁, -x₂, -x₃). That sort of transformation does have the handedness needed for an axial vector. Yet it is easy enough to think of counter examples. What about a time reflection, where an event (t, x₁, x₂, x₃) goes to (-t, x₁, x₂, x₃)? This transformation does not change handedness, it is represented by a polar vector. Compare the two functions

mirror reflection: q -> q' = q^*
time reflection: q -> q' = -q^*

Since these are so close to each other as functions, I don't think one should take precedence over the other, as asserting one must use a bivector for the spatial part of a quaternion does.

The fundamental currency of the Universe is a bare event which stands alone in a vacuum, not in front of a mirror waving a left hand. General relativity has a message about basis vectors. Should we have a toy Universe with two events, the interval between these two events is found by taking the dot product of the difference between the two events:

(c dt, dx_1, dx_2, dx_3).(c dt, dx_1, dx_2, dx_3) = c^2 dt^2 ~+~ dx_1^2 (e_2 \wedge e_3).(e_2 \wedge e_3) ~+~ dx_2^2 (e_3 \wedge e_1).(e_3 \wedge e_1) ~+~ dx_3^2 (e_1 \wedge e_2).(e_1 \wedge e_2)

= c^2 dt^2 ~-~ dx_2^2 ~-~ dx_2^2 ~-~ dx_2^2 \quad eq~2

So far, so good. Now repeat the measurement in a toy Universe where nothing has been altered for the generators of the two events, but one has added a mass. A metric theory of gravity does not alter events, but does change the measure of events. In this setup, the differences are the same. The only thing that could change are the sizes of the basis vectors themselves. The problem here is that there is no basis vector associated with the dt² term.

The deep message of special relativity is to treat time as we do space. In the geometric algebra approach, time is a scalar, whereas the space parts use two basis vectors as a bivector. Contrast this with the way I define a quaternion:

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

This is not a Clifford algebra because the basis vector e_0 commutes with the others. In the definition of a Clifford algebra, all the basis elements anti-commute. Physicists would call this a 4-vector, because one can add them or multiply it by a scalar, and it transforms like a 4-vector. It is this last phrase, on how it transforms under a Lorentz boost, that holds the magic. The goal of a great definition is to remove the magic and let the math speak for itself. Square the difference between two events using the quaternion definition as written in equation 3:

(c dt, dx_1, dx_2, dx_3)^2 = (c^2 dt^2~ e_0^2 ~+~ dx_1^2 ~e_1^2 ~+~ dx_2^2 ~e_2^2 ~+~ dx_3^2 ~e_3^2, 2 c ~dt ~dx_1~ e_0 ~e_1, 2 c ~dt ~dx_2 ~e_0 ~e_2, 2 c ~dt ~dx_3 ~e_0 ~e_3) \quad eq~4

To be consistent with special relativity, we make the following map:

e_0^2 = +1
e_1^2 = -1
e_2^2 = -1
e_3^2 = -1 \quad eq~5-8

To be consistent with general relativity to first order tests of weak gravity fields for non-rotating, spherically symmetric sources, all we need is:

e_0 = exp(-G M/c^2 R)
1/e_0 = e_1 = e_2 = e_3 \quad eq~9-10

so according to the GEM proposal,

e_0^2 = exp(-2 G M/c^2 R)
e_1^2 = -exp(2 G M/c^2 R)
e_2^2 = -exp(2 G M/c^2 R)
e_3^2 = -exp(2 G M/c^2 R) \quad eq~11-15

I forget who said it, but one can only hope to find an important law in physics if one also finds an invariance principle. When there is a gravitational source, the terms in the 3-vector part of the square (2 dt dx e_0 e_1, 2 dt dy e_0 e_2, 2 dt dz e_0 e_3) remain invariant under the presence of such a gravitational field.

In summary, I see two flaws to the assertion that quaternions are the Clifford algebra CL(0, 2). First, while some quaternions are axial vectors, others are polar vectors. Second, the scalar needs to have a basis vector associated with it to work with metric theories of gravity. As far as I can tell, everyone working with geometric algebra in physics works with the assumption that the quaternions are faithfully represented by CL(0, 2).

Doug

sweetser · Jun 10, 2008

The geometric product

Hello:

Today I will discuss the geometric product, a tool used extensively for people applying Clifford algebras to problems in physics and robotics. Here is the definition:

ab = a.b ~+~ a \wedge b \quad eq~1

There are a number of great features to this definition. It will work for whatever dimension you have in a Clifford algebra. The product is conformal, meaning angles do not change. The dot product is the magnitude of a time b times the cosine of the angle between a and b, while the wedge is the magnitude of a time b times the sine of the angle between a an b. How clean!

The road to Clifford's geometric product is convoluted. It started with Hamilton writing out the 4D product one day, carving it into a bridge, and the very next day working to get rid of one dimension with "pure quaternions" that had a zero scalar. The scalar that looked completely alien to Hamilton for the square of a quaternion, t^2 ~-~ x^2 ~-~ y^2 ~-~ z^2, looks familiar to us as the interval of special relativity. By setting t=0, one gets something the Egyptians would have recognized instead of Einstein, so long as one ignores a pesky minus sign.

Yet that minus sign did bother people. The nice thing about a.b that appears in the geometric product is that all the signs are the same and positive (fears of minus signs have wasted more time historically than almost any other math issue). It was Grassman that focused on the wedge product, the new math to come out of the quaternions. He built up an entire system of algebra using basis vectors and wedge products. Clifford put the dot product together with the wedge product to make the geometric product. All is good...

...unless you are a stickler for technical details. Hamilton developed quaternions that are defined for one scalar dimension, and three spatial vectors. One starts with 4 numbers. Look for the product of those four numbers with another set of 4 numbers, and the complete story necessarily has 16 parts. The geometric algebra only has 10. Oops. What it is missing is a scalar times a 3 vector, done twice. This is the math omission needed to make the product conformal - there is no vector stretching.

The geometric product, written as a quaternion operation, looks like so:

ab = (a^* b ~+~ b^* a)/2 ~+~ (a b ~-~ b a)/2 \quad eq~2

Recall the history. Quaternions were used first to form a product that always had an inverse. Grassman isolates on the wedge, then Clifford glues things back together. The result to my eye does not look pretty. The geometric product makes sense if you fear the unknown Lorentz invariant interval, but that is not where we are today. The conformal nature of the geometric product looks to me like a rigged outcome because the 6 problem terms are dropped.

I understand why people have put a huge investment in the geometric product, yet it does not look like a safe bet to me.

Doug

Mentz114 · Jun 10, 2008

Hi Doug,

most illuminating. What exactly does this

The geometric product makes sense if you fear the unknown Lorentz invariant interval,

say ?

Lut

sweetser · Jun 10, 2008

Hello Lut:

I was projecting to the minds of three dead men, Hamilton, Grassmann, and Clifford, whose lives where lived within the 1800's, well before the insights of 1905.

Doug

CarlB · Jun 11, 2008

Doug, glad to see you're looking healthy, happy, and in the middle of things, at least as of Brazil:
http://www.ime.unicamp.br/~icca8/images/GRUPOWEB.jpg

sweetser · Jun 24, 2008

Screws and Quaternions

Hello:

How would you described the motion of a screw? It is a combination of linear and angular motion, with the angular motion perpendicular to the linear. At the meeting in Brazil, I learned the way most people handle this problem: they use dual quaternions. I had never heard about dual quaternions, so let me give you a brief introduction.

As the name suggests, the dual quaternions have two quaternions. One of these also has an epsilon that when squared equals zero. Linear motion goes in the first quaternion, rotation in the dual. They don't mix with each other, staying nice and perpendicular. Dual quaternions are vital for many calculations in robotics.

At a dinner at the meeting, one fellow said that it would not be possible to use quaternions to do the work of dual quaternions. This made no sense to me. After all, the motion of a screw is a collection of events in spacetime. It might take some work to see the connection between quaternions and dual quaternions, but I had confidence it could be done. Another person thought it would be big news to find such a trick. He sent me a reference to a short paper on the topic. I read the paper, and with a bit of effort, was able to spot how to make things work without the epsilon.

Here is the idea. Screw motion is most easily characterized by two quaternion: one for linear motion, the other for angular motion about the axis of the linear motion. These two quaternions can point in arbitrary directions to each other. For a screw, we want to ensure that the angular motion is at a right angle to the linear motion, good old 90 degrees. Consider two arbitrary quaternions. The norm of each is zero if and only if the quaternion is zero, otherwise it is positive definite:

a^* a = (a_0^2 ~+~ a_1^2 ~+~ a_2^2 ~+~ a_3^2, 0, 0, 0)
b^* b = (b_0^2 ~+~ b_1^2 ~+~ b_2^2 ~+~ b_3^2, 0, 0, 0) \quad eq ~1,2

What happens when we form the same product with a and b? Here we could get a zero if a and b between them have 4 zeros, like so:

a = (0, 3, 0, 4)
b = (1, 0, 2, 0)

scalar(a^* b) = 0 \quad eq ~ 3-5

A more general case happens if the two 3-vectors are at 90 degrees to each other, and one of the two scalars happens to be zero:

Let A = (0, a1, a2, a3)
A = (0, b1, b2, b3)
a \angle b = 90, so \quad a.b = |a||b| cos(90) = 0

scalar(a^* b) = 0 \quad eq ~ 6-10

A pair of quaternions where one of the scalars happens to be zero has a conformal Euclidean product (my name for a^*). The angle does not change because it must remain fixed at 90 degrees.

The most general case happens when the contribution from the 3-vector dot product happens to cancel with the scalars, when:

scalar(a^* b) = (a0 b0 ~-~ a1 b1 ~-~ a2 b2 ~-~ a3 b3) = 0 \quad eq ~ 11

I have been spending my time writing programs to make visualizations of these screw motion equations. Nothing obscene, honest. It has been easiest to set up the case where there are four zeroes between the two quaternions. Then you let the numbers travel linear along the z axis, while doing sin t along x, cos t along y, and t along t. A parameterized quaternion:

q(t) = (t, sin(t), cos(t), k t) \quad eq ~ 12

That was so easy to do, I started tossing in different coefficients:

q2(t) = (t, sin(t), cos(t), k t) \quad eq ~ 13
q3(t) = (t, sin(t), cos(2 t), k t) \quad eq ~ 14
q4(t) = (t, sin(t), cos(.5 t), k t) \quad eq ~ 15

What do these look like? The answer is http://picasaweb.google.com/dougsweetser/Screws . This evening's fun was to take equation 12 and put the sin(t) in the t position like so:

q(t) = (sin(t), t, t, cos(t)) \quad eq ~ 16

What does that look like? Nothing is linear because time is doing the trig function thing. The tx and ty complex planes look like standard trig functions since it is just a plot of t versus sin(t). The tz plane ends up being a circle. And the animation? Well, you will have to go to picasaweb.

Odd simple math. Neat.

Doug

Note: reached the 50k views of thread milestone. At least I keep doing new things with quaternions.

sweetser · Jun 25, 2008

Picturing quantum field theory

Hello:

There are two groups that understand the odd ways of relativistic quantum field theory.

The first group are the elite people who hang around the physics departments of major research universities. They don't bother to teach quantum field theory to undergraduates. I got my exposure at Harvard University while actually working as a molecular biologist over at the School of Public Health (and probably the only person to ever make such a commute). I recall going to Mitch Goldstein's office hours, where he would emphasize as he did in class that the only way to understand quantum field theory was to recognize that it involved the behavior of many particles. There was all that stuff about creation and annihilation operators, details which I could follow at the time but would require some refresher work. Few people in a few cities on the planet know how to work with quantum field theory.

The second group is every collection of particles anywhere in the universe. That is a far larger group, ridiculously large. Each and every member of this group is stupid beyond conception. They all know how to do quantum field theory.

The riddle is how all this stupid stuff knows how to do things the smart boys barely can grasp.

Then I took another http://picasaweb.google.com/dougsweetser/Screws/photo#5215627885126508242 created by equation 16 of the previous post

q(t) = (2 sin(3 t), t, t, 2 cos(3t)) \quad implementation ~ of ~eq ~ 16

[note: on a slow, wireless connection, picasaweb shows a still, so you might have to download it, then open it up in a browser to see the animation]. The images starts out blank. Then at 3 separate locations, a pair of events are created. The events move apart, then pairs annihilate each other. It can help to watch the shadows, since that represents a 2D simplifications of the motion in 3D. I had no intention of doing quantum field theory. Yet I know that sines and cosines are solutions to 4D relativistic wave equations. Toss the trig function into the time slot, and you have 3 pairs of events creating and then annihilating each other. This may be how quantum field theory looks.

Doug

shmengie · Jul 4, 2008

Watched a norwegian documentary downloaded as a torrent. The plasma theory of the universe would seem to give credence to Mr. Sweetser's theory, IMHO. Please accept my apologies if this post is unwelcome.

http://thepiratebay.org/tor/3501592/No_Big_Bang________Universe_-_The_Cosmology_Quest

This is not the exact same one I watched. The one I watched didn't have subs. It was mostly English but there was some spoken Norwegian (I think), which this one may have translated subs.

sweetser · Jul 7, 2008

Cosmology

Hello Joe:

I have not been able to download the 2.8G torrent yet, so I will present some of my thoughts about cosmology independent of that video.

Cosmology has major problems, and they know it. I try and be careful in discussing these issues, to call it by the problem with the data and not by the hypothesis that has gathered the most attention. Dark matter is a hypothesis. There is a problem with using Newtonian gravity - as a good approximation to GR in these low mass density, non-relativistic situations - to create a mathematically stable solution for the motion of a thin disk galaxy where the outermost stars have the same velocity. One might be tempted to call this the velocity profile problem. My issue with that is velocity is not a conserved quantity. We should be focusing on momentum and changes in momentum. I call this the galaxy rotation problem, where we do not have a force equation that yields a stable, constant velocity solution. I have posted in this thread a sketch of an idea that uses the product rule which stands a chance of providing a new classic stable constant velocity solution involving gravity. That equation has not been applied to the data at hand, so remains an area of study.

The standard big bang is known to have two problems. They have the labels of the horizon and flatness problems. We have measured the cosmic microwave background radiation, and it is absurdly consistent, the same everywhere you look to one part in a hundred thousand. If we roll this Universe back to within 100,000 years after its creation in the big bang model, we realize that there is no way for different parts of the Universe to be in causal contact with each other to agree on what speed all the particles should be traveling. That is a major problem which cosmologist are aware of and discuss openly.

The flatness problem comes about by looking at the stability of the standard big bang theory. There is a math solution for how a pencil can be on a table and balanced on the point. The solution is unstable, and pencils fall over if you try to confirm this experimentally. The standard big bang model has the same problem. A bunch of matter either wants to collapse quickly or expand so fast nothing else in the neighborhood. Instead, the Universe went through a constant velocity stage, and keep going apart at a reasonable clip.

I do not like the labels for these two problems because they don't reflect the math issue clearly. The early cosmology of the Universe needs a new classic stable constant velocity solution involving gravity (editor's note: I cut and pasted that from the galaxy rotation profile above). A different problem, but I hope I can apply the same math solution because the problem when put in math terms is the same: stable constant velocity gravity solutions. Since momentum is the product of mass and velocity, changes in momentum caused by a force are the sum of a constant mass times changes in velocity (acceleration, mA) plus a constant velocity times a change in mass distribution. I call this the relativistic rocket science term, V c \frac{d m}{d R} \hat{V} which teaches gravity to work in a new direction, along V instead of R. I have even less idea how to apply thi product rule term to cosmology data than I do for a spiral galaxy. A few times I looked into doing the spiral galaxy problem numerically, and got too scared by the complexity to proceed.

The best known hypothesis to explain the stable constant velocity problem of the big bang is known as inflation. The idea is that at the start of the Universe, there was an epoch were the Universe grew at an extraordinary rate. The equations that govern this process do produce a mathematical stable solution that drives all the participating matter to the same speed. The hypothesis predicted that there would be a particular type of power series for the wee bit of variation seen the cosmic microwave background radiation spectrum. One should show respect for inflation because it was predictive of data gathered from surveys of the cosmic background radiation.

Scientists are aware of the significant problem with the inflation hypothesis: there is no cause. One could speculate that there is a type of matter that could lead to the inflation epoch, yet it would have properties that are not credible. The problem of cause was there in the 1980's when the hypothesis was first proposed, and remains to day. It is why I am skeptical about the proposal, one I feel comfortable making since I have a specific alternative.

As I said earlier, I have not see the video. A common complaint at the fridge of physics is the there an alternative way to explain the cosmic microwave background radiation and redshifts in spectrums. I will give you a cosmological experiment you can do in your own home that demonstrates the Universe is expanding.

One must completely embrace the idea of spacetime - avoid thinking of space separately from time! In my dinning room hangs "The Speed of Light According to René Magritte". At the bottom of the work, it reads: "Il n'y a pas d'espace sans temps, pas de temps sans expace" (translation: "There is not time without space, no space without time"). If the Universe uses quaternions as the fundamental system of accounting, there may be situations where the time or space time could be zero, but never a situation where the quaternion was missing a time or space part (zero being a value).

Take a look at your watch. If the hand or digits move, that is proof that spacetime is expanding. The watch is gathering information about spacetime. It so happens that the watch is not moving relative to the person wearing the watch, so all the change in the Universe appears as a change in time. This is a universal property: if the relative velocity between two parts of the Universe is zero, any change will appear as a change in time. The corillary is that if the relative velocity is large, the expansion of spacetime appears as an expansion in space.

In the entire beyond-comprehension-huge Universe, the only matter that travels through spacetime at the same speed as yourself is a thin crust of the Earth. Everything else has a relative velocity to you: we spin here once a day looking at the rest of the Universe. The Universe is a story of motion: once a year we go around the Sun, once every 250,000 our solar system goes around the Milky Way, our Milky Way travels at nearly the speed of light relative to high redshift galaxies.

I view relative velocity as a numerical measure of shared history. The Earth, the only object we can ever know that has zero relative velocity to us, has been together for a least four billion years. The atoms that make up you, me, and the planet made it through a supernova explosion. The high redshift galaxies, the ones traveling at nearly the speed of light, while we may have shared the big bang moment, that is all we share except for a few photons an hour delivered over space. We have no visitation rights for such galaxies only ephimeral signals that ultrasensitve telescopes can see (the human eye is blind to sources below a magnitude of 16).

We have much to learn about cosmology, and cosmologist are collecting the needed data as we speak. I hope I can provide a critical equation and apply it to real data someday.

Doug

shmengie · Jul 7, 2008

Perhaps I should have posted a the link to the torrent I originally downloaded:

http://thepiratebay.org/tor/4223510/Universe_The_Cosmology_Quest_-_Big_Bang_Big_Crap_[Documentary]

It came in faster with a slightly smaller foot print. I'm inclined to repost the torrent with an even smaller foot print, but have yet to do such a thing.

re: your comments on cosmology are somewhat in line with what is said in the documentary. Basicaly, "The Universe, The Cosmology Quest" series debunks the big bang theory. Halton Arp's research which does not fit with the "big bang" mainstream. I found Arp's papers intriguing. Arp has, from what I gather, concludes redshift changes with the age of galaxies, and younger galaxies are being ejected from older ones. He makes a strong argument.

http://bigbangneverhappened.org/index1.htm, which is why I posted here postulates that electric charge in the original plasma formed the super structures of galaxies into filaments. Which sounds to me like EM & GR combined.

shmengie · Jul 7, 2008

http://bigbangneverhappened.org/p17.htm

sweetser · Jul 9, 2008

Force

Hello:

I am taking the General Relativity course 8.06s at MIT's Professional Institute. It is great fun that I get to think and chat about physics for four days. It has been work the 4 vacation days and $2k for the class.

At lunch, one fellow said his biggest concern was with force. He pressed us to define it. I decided to just play it quiet, so he said it was either energy integrated over space or momentum integrated over time. Most discussion emphasize then energy integrated over time, while spending less effort on the impulse force. There is no reason for why one should be more important than the other except for accidents in the history of teaching physics.

This gentleman wanted to know why those two made up force. I looked at the problem with my quaternion hat stapled onto my head, realizing that force involves two quaternions: momenergy (E, Pc) and differential spacetime (dt/c, dR). Multiply and integrate these quaternions:

F = \int{(dt/c, dR)(E, Pc)} = \int{(E dt/c ~-~ c ~P.dR, E dR ~+~ P dt ~+~ c ~ dR \times P)}

What the gentleman did not mention - but was aware of - was the angular momentum. That is a great property of quaternions: the calculation makes it impossible to forget. An additional bonus is this equation calculates the 4-force at little extra cost. If a force is conservative, the scalar is zero. It is unclear to me how the c's effect the meaning.

Doug

Ghost_of_PL · Jul 9, 2008

Hello Doug,

The algebra you describe below is not the quaternions, that is to say it is not isomorphic to the quaternions. What you have is not a division algebra. For example: what is 1/(1+e_0) ?
1+e_0 has no inverse.

sweetser said:

Hello:

Quaternions as events in spacetime physics are not the Clifford algebra CL(0, 2). Let me first explain what that bit of jargon means.

Clifford algebras were an attempt to generalize the math started by Hamilton, passing through Grassman. These algebras are independent of coordinates, and can be written in arbitrary dimensions. As algebras, it is not important that they are invertible, although that can happen.

The Clifford algebra CL(0, 2) is a multivector, meaning it has two parts. First there is the scalar, the 0 part, a pure number. The bivector is the second part. They use a wedge product, \wedge, a generalized cross product to form a 3-term bivector.

q(t, x_1, x_2, x_3) = t ~+~ x_1 e_2 \wedge e_3 ~+~ x_2 e_3 \wedge e_1 ~+~ x_3 e_1 \wedge e_2 \quad eq~1

Why use a wedge product? The reason is that wedge products are axial vectors, those that switch handedness in a mirror. The other sort is a polar vector. If one looks at an event in a mirror, this makes physical sense.

It is not worth the time to argue with math types who think they can define whatever they want, however they see fit. I have the perspective of a mathematical physicists, where any math definition can be cow-roped by physical meaning (I am a crude mathematical physics who uses rodeo analogies).

The unspoken assumption behind CL(0, 2) is that one should use a mirror on this event. That is like taking (t, x₁, x₂, x₃) to (t, -x₁, -x₂, -x₃). That sort of transformation does have the handedness needed for an axial vector. Yet it is easy enough to think of counter examples. What about a time reflection, where an event (t, x₁, x₂, x₃) goes to (-t, x₁, x₂, x₃)? This transformation does not change handedness, it is represented by a polar vector. Compare the two functions

mirror reflection: q -> q' = q^*
time reflection: q -> q' = -q^*

Since these are so close to each other as functions, I don't think one should take precedence over the other, as asserting one must use a bivector for the spatial part of a quaternion does.

The fundamental currency of the Universe is a bare event which stands alone in a vacuum, not in front of a mirror waving a left hand. General relativity has a message about basis vectors. Should we have a toy Universe with two events, the interval between these two events is found by taking the dot product of the difference between the two events:

(c dt, dx_1, dx_2, dx_3).(c dt, dx_1, dx_2, dx_3) = c^2 dt^2 ~+~ dx_1^2 (e_2 \wedge e_3).(e_2 \wedge e_3) ~+~ dx_2^2 (e_3 \wedge e_1).(e_3 \wedge e_1) ~+~ dx_3^2 (e_1 \wedge e_2).(e_1 \wedge e_2)

= c^2 dt^2 ~-~ dx_2^2 ~-~ dx_2^2 ~-~ dx_2^2 \quad eq~2

So far, so good. Now repeat the measurement in a toy Universe where nothing has been altered for the generators of the two events, but one has added a mass. A metric theory of gravity does not alter events, but does change the measure of events. In this setup, the differences are the same. The only thing that could change are the sizes of the basis vectors themselves. The problem here is that there is no basis vector associated with the dt² term.

The deep message of special relativity is to treat time as we do space. In the geometric algebra approach, time is a scalar, whereas the space parts use two basis vectors as a bivector. Contrast this with the way I define a quaternion:

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

This is not a Clifford algebra because the basis vector e_0 commutes with the others. In the definition of a Clifford algebra, all the basis elements anti-commute. Physicists would call this a 4-vector, because one can add them or multiply it by a scalar, and it transforms like a 4-vector. It is this last phrase, on how it transforms under a Lorentz boost, that holds the magic. The goal of a great definition is to remove the magic and let the math speak for itself. Square the difference between two events using the quaternion definition as written in equation 3:

(c dt, dx_1, dx_2, dx_3)^2 = (c^2 dt^2~ e_0^2 ~+~ dx_1^2 ~e_1^2 ~+~ dx_2^2 ~e_2^2 ~+~ dx_3^2 ~e_3^2, 2 c ~dt ~dx_1~ e_0 ~e_1, 2 c ~dt ~dx_2 ~e_0 ~e_2, 2 c ~dt ~dx_3 ~e_0 ~e_3) \quad eq~4

To be consistent with special relativity, we make the following map:

e_0^2 = +1
e_1^2 = -1
e_2^2 = -1
e_3^2 = -1 \quad eq~5-8

To be consistent with general relativity to first order tests of weak gravity fields for non-rotating, spherically symmetric sources, all we need is:

e_0 = exp(-G M/c^2 R)
1/e_0 = e_1 = e_2 = e_3 \quad eq~9-10

so according to the GEM proposal,

e_0^2 = exp(-2 G M/c^2 R)
e_1^2 = -exp(2 G M/c^2 R)
e_2^2 = -exp(2 G M/c^2 R)
e_3^2 = -exp(2 G M/c^2 R) \quad eq~11-15

I forget who said it, but one can only hope to find an important law in physics if one also finds an invariance principle. When there is a gravitational source, the terms in the 3-vector part of the square (2 dt dx e_0 e_1, 2 dt dy e_0 e_2, 2 dt dz e_0 e_3) remain invariant under the presence of such a gravitational field.

In summary, I see two flaws to the assertion that quaternions are the Clifford algebra CL(0, 2). First, while some quaternions are axial vectors, others are polar vectors. Second, the scalar needs to have a basis vector associated with it to work with metric theories of gravity. As far as I can tell, everyone working with geometric algebra in physics works with the assumption that the quaternions are faithfully represented by CL(0, 2).

Doug

sweetser · Jul 10, 2008

Inverse of curved quaternions

Hello Ghost_of_PL:

Great question! I had not looked at all the details, and your post got me to think about them.

There is hope that the algebra represented in equations 11-15 will be a division algebra since as M approaches zero, the algebra approaches the standard quaternion division algebra of equations 5-8.

To solve this problem, I fired up Mathematica and defined the quaternion using a 4x4 real matrix. This definition is identical to the standard one except for factors of Exp(G M/c² (x² + y² + z²)^1/2 (a minus factor on the t term).

In the limit of M->0, the inverse matrix is exactly the same as a Hamilton's quaternions. Those are defined for all possible values of t, x, y, and z because the divisor is the norm, t + x² + y² + z², which is positive definite. The inverse is only undefined if t=x=y=0.

In the general case, the inverse looks like so:

\frac{1}{q(t,x,y,z)} = (t exp(\frac{G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -x ~ exp(\frac{G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -y ~exp(\frac{3 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -z ~exp(\frac{3 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm) \quad eq ~1

where: ~norm = t^2 ~+~ (x^2 ~+~ y^2 ~+~ z^2) exp(\frac{4 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}) \quad eq ~1

Again the norm is positive definite. With Hamilton's quaternions, the point set that must be excluded is zero for all four numbers. With the curved by mass quaternions, one must also exclude where x=y=z=0, which is the entire real line. This calls for a refinement of the definition of the algebra in equation 11-15, that should R=0, then the algebra becomes the standard Hamilton case, equations 5-8.

To answer your question specifically, the inverse in spacetime curved by mass of (1, 1, 0, 0) is:

\frac{1}{(1, 1, 0, 0)} = (\frac{e^{\frac{G M}{c^2}}}{1 ~+~ e^{\frac{4 G M}{c^2}}},-\frac{e^{\frac{3 G M}{c^2}}}{1~+~ e^{\frac{4 G M}{c^2}}}, 0, 0) \quad eq ~ 2

Let's do a sanity check at least for the first term. If we hope to get a 1 out of multiplying (1, 1, 0, 0) by its inverse, we have to get out a 1~+~ e^{\frac{4 G M}{c^2}}, which shows up in the denominator of the inverse. For the time terms, there will be a positive and a negative exponent that cancel each other out, yielding a 1. From the x terms, we get a positive exponent plus an exponent times 3 which combine for the exponent to the fourth power. Looks good. If anyone is interested, the notebook is here.

By refining the definition to cover the case where R=0, I hope I have shown the curved basis quaternions are a division algebra isomorphic to the flat spacetime quaternions defined by Hamilton.

Thanks for the question,
Doug

Ghost_of_PL · Jul 10, 2008

Hi Doug,
Mathemaica is a wonderful tool but I sometimes think too much of it can be the crutch that cripples so I’m going to use good old fashioned algebra. You have stated (e_0)^2=1, correct? Well then (1+ e_0)(1- e_0)=0 since we have 1^2 – e_0 + e_0 – (e_0)^2 =0. That is to say 1+e_0 and 1-e_0 are zero divisors and have no inverse. Taking the limit as M->0 won’t get us to the quaternion’s so there’s an inconsistency somewhere.

sweetser said:

Hello Ghost_of_PL:

Great question! I had not looked at all the details, and your post got me to think about them.

There is hope that the algebra represented in equations 11-15 will be a division algebra since as M approaches zero, the algebra approaches the standard quaternion division algebra of equations 5-8.

To solve this problem, I fired up Mathematica and defined the quaternion using a 4x4 real matrix. This definition is identical to the standard one except for factors of Exp(G M/c² (x² + y² + z²)^1/2 (a minus factor on the t term).

In the limit of M->0, the inverse matrix is exactly the same as a Hamilton's quaternions. Those are defined for all possible values of t, x, y, and z because the divisor is the norm, t + x² + y² + z², which is positive definite. The inverse is only undefined if t=x=y=0.

In the general case, the inverse looks like so:

\frac{1}{q(t,x,y,z)} = (t exp(\frac{G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -x ~ exp(\frac{G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -y ~exp(\frac{3 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -z ~exp(\frac{3 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm) \quad eq ~1

where: ~norm = t^2 ~+~ (x^2 ~+~ y^2 ~+~ z^2) exp(\frac{4 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}) \quad eq ~1

Again the norm is positive definite. With Hamilton's quaternions, the point set that must be excluded is zero for all four numbers. With the curved by mass quaternions, one must also exclude where x=y=z=0, which is the entire real line. This calls for a refinement of the definition of the algebra in equation 11-15, that should R=0, then the algebra becomes the standard Hamilton case, equations 5-8.

To answer your question specifically, the inverse in spacetime curved by mass of (1, 1, 0, 0) is:

\frac{1}{(1, 1, 0, 0)} = (\frac{e^{\frac{G M}{c^2}}}{1 ~+~ e^{\frac{4 G M}{c^2}}},-\frac{e^{\frac{3 G M}{c^2}}}{1~+~ e^{\frac{4 G M}{c^2}}}, 0, 0) \quad eq ~ 2

Let's do a sanity check at least for the first term. If we hope to get a 1 out of multiplying (1, 1, 0, 0) by its inverse, we have to get out a 1~+~ e^{\frac{4 G M}{c^2}}, which shows up in the denominator of the inverse. For the time terms, there will be a positive and a negative exponent that cancel each other out, yielding a 1. From the x terms, we get a positive exponent plus an exponent times 3 which combine for the exponent to the fourth power. Looks good. If anyone is interested, the notebook is here.

By refining the definition to cover the case where R=0, I hope I have shown the curved basis quaternions are a division algebra isomorphic to the flat spacetime quaternions defined by Hamilton.

Thanks for the question,
Doug

sweetser · Jul 10, 2008

Basis vectors and quaternions

Hello Ghost_of_PL:

Good old fashioned algebra rocks! The only way I really learn something is with paper and pencil (so I can erase a lot). I did not understand what your issue was, but now I have a better grasp of it. I did define what I meant by a quaternion in this expression:

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

Notices that every term has a basis vector, no exceptions. With equation 3, there is never a pure number like the 1 in 1 + e_0. One can work with 1 e_0 and 1 e_0 + 1 e_0, both of which have inverses. So your question looks poorly defined, a way of saying it does not work nicely with equation 3. If I were to allow a term such as 1 + e_0, then there would be 5 numbers that could be put into a quaternion. Oops.

The reason I needed Mathematica was to answer a different problem because I misunderstood your question, namely finding the inverse of a quaternion with the basis vectors in equations 5-8. That was a fun calculation anyway :-)

Doug

Ghost_of_PL · Jul 11, 2008

Hello Doug,
The problem is equation 3 can not mitigate the fact that you introduced basis elements that square to plus or minus “1” . (e_0)^2= 1 So there you have it, “one”, in all its naked glory. Either the number one is a legitimate element, or our “quaternions” are not closed under the operation of multiplication and therefore not an algebra. Maybe what you really want is e_0 as an idempotent, something along the lines of: (e_0)^2= e_0 , (e_0)(e_1)= e_1, (e_0)(e_2)= e_2 , (e_1)(e_2)= e_3 and so on…

sweetser said:

Hello Ghost_of_PL:

Good old fashioned algebra rocks! The only way I really learn something is with paper and pencil (so I can erase a lot). I did not understand what your issue was, but now I have a better grasp of it. I did define what I meant by a quaternion in this expression:

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

Notices that every term has a basis vector, no exceptions. With equation 3, there is never a pure number like the 1 in 1 + e_0. One can work with 1 e_0 and 1 e_0 + 1 e_0, both of which have inverses. So your question looks poorly defined, a way of saying it does not work nicely with equation 3. If I were to allow a term such as 1 + e_0, then there would be 5 numbers that could be put into a quaternion. Oops.

The reason I needed Mathematica was to answer a different problem because I misunderstood your question, namely finding the inverse of a quaternion with the basis vectors in equations 5-8. That was a fun calculation anyway :-)

Doug

sweetser · Jul 11, 2008

Square of a quaternion

Hello Ghost_of_PL:

There are only 4 slots in a quaternion, something that is obvious from the 4x4 real matrix representation used in the Mathematica notebook. Looking back, I not only used a variation on a common definition (which would normally have i, j, k for e_1, e_2, e_3, no e_0):

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

Sometimes people have complained about the meaning of the plus sign since this is not like the sum of real numbers. I also wrote out explicitly what is meant by squaring a quaternion:

(c dt, dx_1, dx_2, dx_3)^2 = (c^2 dt^2~ e_0^2 ~+~ dx_1^2 ~e_1^2 ~+~ dx_2^2 ~e_2^2 ~+~ dx_3^2 ~e_3^2, 2 c ~dt ~dx_1~ e_0 ~e_1, 2 c ~dt ~dx_2 ~e_0 ~e_2, 2 c ~dt ~dx_3 ~e_0 ~e_3)\quad eq~4

I did this so I could work with basis vectors whose length was not equal to unity, which is why I am not working with idempotent such that (e_0)^2= e_0. Equations 5-8 (unitary basis vectors) and 11-15 (non-unitary basis vectors) are applied in the context of equation 4. In curved spacetime, e_0^2 = exp(-2 G M/c^2 R) and it goes in the first of the four slots of a quaternion. A quaternion sum of a^2 ~+~ b \ne 0 would always have an inverse in flat or curved spacetime.

There certainly is a chance that my way of defining an algebra in flat spacetime with equations 3-8, and in curved spacetime with 3, 4, 11-15 does not work with people more formally trained define an algebra. If that happens to be the case, then please use the 4x4 real matrix representation which carries exactly the same information:

 \[ \left[ \begin{array}{cccc} e_0 t & -e_1 x & -e_2 y & -e_3 z\\\ e_1 x & e_0 t & -e_3 z & e_2 y \\\ e_2 y & e_3 z & e_0 t & -e_1 x \\\ e_3 z & -e_2 y & e_1 x & e_0 t \end{array} \right]\]

In flat spacetime, all the e's have a magnitude equal to one. If the additive identity (zero) is excluded, this matrix always has an inverse.

Doug

Ghost_of_PL · Jul 12, 2008

Hi Doug,
most likely this is my last post on this subject:

sweetser said:

Hello Ghost_of_PL:

There are only 4 slots in a quaternion, something that is obvious from the 4x4 real matrix representation used in the Mathematica notebook. Looking back, I not only used a variation on a common definition (which would normally have i, j, k for e_1, e_2, e_3, no e_0):

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

You have defined at least 5 slots since you defined e_0^2 =1

sweetser said:

Sometimes people have complained about the meaning of the plus sign since this is not like the sum of real numbers. I also wrote out explicitly what is meant by squaring a quaternion:

(c dt, dx_1, dx_2, dx_3)^2 = (c^2 dt^2~ e_0^2 ~+~ dx_1^2 ~e_1^2 ~+~ dx_2^2 ~e_2^2 ~+~ dx_3^2 ~e_3^2, 2 c ~dt ~dx_1~ e_0 ~e_1, 2 c ~dt ~dx_2 ~e_0 ~e_2, 2 c ~dt ~dx_3 ~e_0 ~e_3)\quad eq~4

Do you not notice eq. 4 violates:

sweetser said:

Notices that every term has a basis vector, no exceptions. With equation 3, there is never a pure number like the 1 in 1 + e_0. One can work with 1 e_0 and 1 e_0 + 1 e_0, both of which have inverses. So your question looks poorly defined, a way of saying it does not work nicely with equation 3. If I were to allow a term such as 1 + e_0, then there would be 5 numbers that could be put into a quaternion. Oops.

The first term of eq.4 is a pure number without a factor of e_0 and violates "that every term has a basis vector, no exceptions" since c^2dt^2e_0^2 +dx1^2e_1^2 +dx2^2e_2^2 +dx3^2 = c^2dt^2- dx1^2- dx2^2 -dx3^2 . The right hand side of this last equation has no factor of e_0. Your matrix representation can fare no better since it also admits stand alone real numbers.

sweetser · Jul 13, 2008

Hello Ghost_of_PL:

> most likely this is my last post on this subject

This discussion is becoming repetitive, so that is good news for general readers of this thread.

Your weakest statement is this assertion:

> Your matrix representation can fare no better since it also admits stand alone real numbers.

What behaves no different from the real numbers is this diagonal matrix:

\[ \left[ \begin{array}{cccc} e_0 t & 0 & 0 & 0\\\ 0 & e_0 t & 0 & 0 \\\ 0 & 0 & e_0 t & 0 \\\ 0 & 0 & 0 & e_0 t \end{array} \right]\] \quad eq ~1

All real numbers are formally a subgroup of quaternions, so there is no way a real number can stand alone from the mathematical field of quaternions.

What behaves no different from anything that can ever be done with complex numbers or complex analysis are the following three matrices:

\[ \left[ \begin{array}{cccc} e_0 t & -e_1 x & 0 & 0\\\ e_1 x & e_0 t & 0 & 0 \\\ 0 & 0 & e_0 t & -e_1 x \\\ 0 & 0 & e_1 x & e_0 t \end{array} \right]\]

\[ \left[ \begin{array}{cccc} e_0 t & 0 & -e_2 y & 0\\\ 0 & e_0 t & 0 & e_2 y \\\ e_2 y & 0 & e_0 t & 0 \\\ 0 & -e_2 y & 0 & e_0 t \end{array} \right]\]

\[ \left[ \begin{array}{cccc} e_0 t & 0 & 0 & -e_3 z\\\ 0 & e_0 t & -e_3 z & 0 \\\ 0 & e_3 z & e_0 t & 0 \\\ e_3 z & 0 & 0 & e_0 t \end{array} \right]\] \quad eq ~2-4

Quaternions have as subgroups equations 1-4, and can be viewed as the union of three complex numbers that share the same real value.

I remain open to the idea that as an exercise in algebra, my notation could be wrong. The most likely thing would be I have to consider keeping all the basis vectors, so in flat spacetime e₀² = +1 e₀², whereas in curved spacetime e₀² = +exp(-2 GM/c² R) e₀². Then one would have to figure out an efficient way to say you can add an e₀² to an e₁² to an e₀. The way I tried to avoid this complexity was to say the numbers mapped back to equations 3 and 4 of the original post (#574). Instead, equations 5-8 were removed from their context to create the stand-alone issue. The real numbers are a subgroup of quaternions, so are of the form (a, 0, 0, 0). A real number cannot stand alone from quaternions, ever. The relation of real numbers to quaternions may not be understood or used widely in practice, but ignoring rules is not a defense.

Perhaps I can say it this way: the equations that no longer have an e₀² explicitly have it implicitly up to a cofactor of +1 in flat spacetime, or +exp(-2 GM/c² R) in curved spacetime. It would take a well-trained math wonk to know how to say these things correctly.

Doug

sweetser · Aug 3, 2008

Disk galaxy momentum profile - the plan

Hello:

Personal stuff:
I spent the month of July drifting. I was bummed that neither the trip to Brazil nor the MIT GR course created lasting dialogs with the participants. Although I am familiar with not connecting, both events involved interactions over a week long period. I also spent much time thinking about cell phones. I almost got an iPhone, but that would require sending too many dollars to AT&T over time, so bought a Nokia N82. Those Nokia folks sure know how to make a cell phone.

Physics stuff:
For more than 5 years, I have wanted to do a calculation, but never got the courage up to work all the way through it. The calculation involves figuring out the momentum profile of a thin disk galaxy. We know three things about disk galaxies: the distribution of the visible mass, the velocity of such mass, and that Newton's law is not consistent with these two observations. General relativity is not necessary in such a system since the speeds and mass densities are low. The angular momentum must balance the force. To explain this problem, the most popular hypothesis is dark matter. I call that hypothesis "stuffing the matter box". We know an absurd amount of things about particles, yet do not have a reasonable candidate for dark matter. Given that the dark matter needed to fix problems with gravity exceeds all known matter, this is a huge missing gap in our knowledge, or an indication we should seek an alternative, testable hypothesis.

I have such a hypothesis. What a force does is change momentum. By the product rule, that means the effect of a force is the sum of a constant mass times a change in velocity plus a constant velocity times a change in mass. The former term is the famous Newtonian mA and is the only term used by folks modeling the momentum profiles of galaxies. The latter term is used in rocket science, because as the rocket moves in time, there is less mass in the rocket itself. The rocket term accounts for this change in mass with respect to time.

Galaxies do not change rapidly in time, so it is reasonable to ignore the change in mass with respect to time. The mass of a galaxy does change with respect to space. The mass density of thin disk galaxies has two parts. One is spherical, falling off approximately like R^1/4. The disk has a mass that often falls off from the center as an exponential decay. The velocity of matter of the disk stays constant, while the mass drops off exponentially with increasing distance from the center. As a word description, that says to me we are dealing with a term of the form V c dm/dR, what I call the relativistic rocket effect.

The relativistic rocket effect, as a way to express a change in momentum, has to my knowledge never been discussed, let alone seen. My guess is that the factor of c needed to get the units right means that it is only seen for systems where the mass is spread over areas of space as big as a galaxy or larger. For such grand systems, most of the force of gravity goes into dictating where mass should be on that grand scale, and not into making mass move faster.

One of the nice or scary things about the relativistic rocket hypothesis is that it has no free parameters. The equation either matches the data for the visible mass distribution profile with the velocity profile, or the hypothesis is wrong. I have no parameters to "tweak", the way good science proceeds.

The standard approach to figuring out the rotation profile of a thin disk galaxy uses elliptical integrals. That is a math topic I do not understand. I do not have peers who could walk me through that bit of mathematics.

I must do something simpler, more direct. I could take a galaxy, chop it up into lots of little bits, figure out the effect of gravity on two little bits - with and without the relativistic rocket effect - and add it all up. If I were to do the calculation correctly, I would see the problem with Newton's law alone, and then how the velocity profile goes dead flat if I take into account the relativistic rocket effect. If the relativistic rocket effect does not change the velocity profile much from Newton's law alone, then the hypothesis is wrong.

I work from a position of zero self-confidence, but do embrace the process of science. It is most important to ask a specific question clearly. I will be looking at the momentum profile of one galaxy, NGC 3198, because astronomers have figured out an approximate exponential function for the visible mass distribution profile.

Here is the http://picasaweb.google.com/dougsweetser/GalaxyDiskMomemtumProfile/photo#5229379345512696162 for the work I hope to accomplish over the next few months:
http://picasaweb.google.com/dougsweetser/GalaxyDiskMomemtumProfile/photo#5229379345512696162

Draw it
Label Everything
Write out all the equations
Write out all the kg, m, s units and conversions
mo_profile - a program in Perl to do the calculation
- List options, input, and output
- List modules and variables
- Write test programs to check area, mass calculations
Experiment: Given the mass distribution profile for NGC 3198
- Does the pure Newtonian term get the correct maximum velocity, but have the velocity fall with increasing distance from the center?
- Does the Newtonian term plus the relativistic velocity term get the correct maximum velocity and a flat velocity curve with increasing radius?

As I make steps along this master plan, I will post the results here. The plan was drawn up on Thursday, and my N82 was able to upload it to the web. I spent Saturday and Sunday drawing pictures of the galaxies and labeling all the parts, so by mid-week I may be able to clear the next hurdle. At least the goals are clear.

Doug

sweetser · Aug 4, 2008

The plan

Hello:

Here is a picture of the plan I have for calculating the rotation profile of a galaxy:

I had included a higher resolution image in the previous post, but apparently it was too large to be accepted here (600x450 pixels, 57k works).

Doug

Varnick · Aug 4, 2008

Doug,

Your grasp of physics is, frankly, much beyond mine, but I would disagree on one point in you previous "Wordy" post. It is of no real consequence to your calculation, but SUSY particles are considered excellent candidates for dark matter, as far as I am informed. Would you disagree, or was it just something you missed? Either way, this calculation appears both impressive and immersive, I look forward to seeing it finished.

V

sweetser · Aug 5, 2008

SUSY and dark matter

Hello Vanick:

Supersymmetric (SUSY) proposals have been with us for over 30 years. During that time, every time a new more powerful collider has come on line, part of the justification was the search for the SUSY particles. The same is claimed for the Large Hadron Collider: it might find the first evidence for supersymmetric. Based on the previous decades of effort, I will not bet on it, although there is 8 billion dollars on the line.

As a hypothesis, supersymmetry has two failings in my eyes. The first is that one of the motivating factors behind the proposal is to explain aspects of the standard model. One big mystery is why there should be three groups, U(1), SU(2) and SU(3) - and why those groups in particular. In my reading of this article on grand unified theories, using a group like SO(10) does not answer that question. Oops.

Animating physics with quaternions does give a visual reason why Nature is constrained to the symmetries of U(1), SU(2), SU(3) and Diff(M) for EM, the weak force, the strong force, and gravity respectively. If you go though the math of an expanding and contracting unit sphere with quaternions, these are the groups that come into play. If an observer is at the origin and sees an event, that event must be a member of these groups. It is less that SUSY is wrong than I have a concrete counter proposal. I too have limitations in my math skills, and have no idea how to pitch a model that is smaller than the standard model to particle physicists. The smaller nature of the quaternion proposal might have an advantage: it could justify containment, the observation that there are no free gluons anywhere in the Universe. Formalizing that line of logic is beyond my reach. A search on YouTube for "standard model" should have my blue & yellow colored animation.

That is the backdrop. Now to your specific question about SUSY and dark matter. It is not enough for a light SUSY particle to only interact with gravity, to have the properties of a dark matter particle. That particle would have to be distributed in space around disk galaxies with a distribution that would lead to a momentum profile where the velocity was constant, but the visible mass falls off exponentially. How Nature using only gravity could get the distribution of dark matter such that it is consistent with what we see is a core issue that is unexplained. The distribution of dark matter must be different for clusters of galaxies. Getting non-interacting matter except for gravity into the right place looks like a tough problem.

Doug

sweetser · Aug 5, 2008

Disk galaxy momentum profile - the drawing

Hello:

To use Newton's law of gravity, masses and distances must be known. For a disk, one cannot treat it as an effective point mass. Instead, the galaxy needs to be chopped up into little bits, and add up each contribution. My plan is to do this discretely, but it may be possible to do things continuously. I will have to see.

Here is a picture of how I plan to slice up a galaxy:

There are quite a few R's:

Rmax - the maximum radial distance
Rpi - the radius to the passive mass, i steps from the origin
Raj - the radius to the active mass, j steps from the origin
Rij - the distance between the active and passive masses.
Rr - the portion of Rij in the direction of the radius
Rv - the portion of Rij perpendicular to the radius, parallel to V

The passive mass is the one that appears on both sides of Newton's force law, and always gets cancelled. With the relativistic rocket effect, that cancelation might not happen (since the mass distribution is an exponential, and we are taking the derivative of an exponential which returns the exponential times the derivative of the exponent, it might drop out in this special case). There are three forces that point in these directions: Fij, Fr, and Fv. For the pure Newtonian calculation, only Fr will be needed. The rocket effect uses Fv.

I have partially revealed how I will be slicing up the galaxy: in i steps with the passive gravitational mass and in j steps for the active gravitational mass. The active mass also needs to revolve around the origin, which will be done in k steps. The differential active mass is dMa, and the differential passive mass is dmp.

The force term will be the sum of these differential terms:

F_g = \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n (dFr + dFv) \quad eq~1

For the Newtonian case, the Fv is not included. Can all these terms be written in terms of i, j, k, and Rmax? It is not simple, but a weekend of doodling created this picture of the terms involved:

The first three terms are about the progression along i, j, and k:

Rpi = \frac{i}{n} ~ Rmax

Raj = \frac{j}{n} ~ Rmax

B = 2 \pi \frac{k}{n} \quad eq ~ 2-4

Although we don't use it everyday, Rij is calculated using the law of cosines, a variation on the Pythagorean theorem with a cosine to account for a non-right angle triangle:

Rij = \frac{Rmax}{n} \sqrt{i^2 + j^2 - 2 i j ~Cos(2 \pi \frac{k}{n})} \quad eq ~5

Break down Rij into that pointing along Rv - a simple application of the definition of a sine, and Rr, which is the venerable Pythagorean theorem at work:

Rv = Rij ~Sin (2 \pi \frac{k}{n})

Rr = Rij \sqrt{1 - \frac{Raj^2}{Rij^2} ~Sin^2 (2 \pi \frac{k}{n})}\quad eq ~6, ~7

There are corresponding equations for the forces that are proportional to these:

dFv = dFij ~\frac{Rai}{Rij} ~Sin (2 \pi \frac{k}{n})

dFr = dFij \sqrt{1 - \frac{Raj^2}{Rij^2} ~Sin^2 (2 \pi \frac{k}{n})} \quad eq ~8,~9

With this much algebra going on, it is good to think of quality controls. The distance Rij should only equal 2 Rmax for one set of values of i, j, and k. The differential forces dFij, dFr, and dFv should satisfy the Pythagorean theorem.

The mass for disk galaxies is often given in terms of mass per unit area. As such, determine what a differential area is. The area of the complete galaxy is pi Rmax^2. This is being sliced into n pieces, so the area of a sector is pi Rmax^2/n. As we step from i-1 to i, how much area is there?

dAi = \pi ~\frac{Rmax}{n} ~ (\frac{i^2}{n^2} ~-~ \frac{(i ~-~ 1)^2}{n^2}) = \pi ~ \frac{Rmax}{n^3} ~ (2 i ~-~ 1) \quad eq ~10

One problem with this approach is that it samples the galaxy near the core more densely than the outer regions, where i is greater. That may be acceptable since the mass distribution is exponential, so most of the mass comes from the center.

I have skimmed from a paper that the number of solar masses/parsec² is 37 exp (R'/2.23'). Combine the mass/area with the differential area to get the differential masses:

dmp = 37 \pi ~\frac{Rmax}{n^3} ~ (2 i ~-~ 1) ~ exp (Rpi'/2.23')

dMa = 37 \pi ~\frac{Rmax}{n^3} ~ (2 j ~-~ 1) ~ exp (Raj'/2.23')\quad eq ~11, ~12

These are the players needed to calculate the force: the distance Rij and the two masses, dmp and dMa. Yoda has said, simple it is not, the way of relativistic rocket astrophysics.

Doug

sweetser · Aug 7, 2008

Symmetry and torque forces

Hello:

Symmetry is useful to think about because it can pinpoint what can be ignored. Feynman emphasized looking for things that add up, knowing we can ignore things that cancel. This is what I drew on my board this morning:

The two Fv's point in opposite directions. Oops, those will never add up and amount to anything. Therefore, none of my calculations should involve Fv, which appeared in the previous post #598, eq 1 and 8.

So is this galaxy momentum profile calculation over? No, because there is a change in momentum that is being omitted from the standard calculation. Recall that energy is force times distance. One only does work in the direction of motion, or the cosine of the angle between two 3-vectors. In the drawing, that would be Fr dotted to Rr. That is what goes into a classical calculation. Fr dotted with Rv is zero. My objection is that our analysis is not complete. Something which has the same units as work, but is not a scalar like energy, is the sine of a force and a distance vector, or Fij ~ Rv ~Sin(\theta). That will not be zero. It is the 3-momentum times c. This is the torque force term that pairs with the relativistic rocket effect.

Doug

sweetser · Aug 7, 2008

Dark energy and the relativistic rocket effect

Hello:

All forces must have a cause and an effect. For a disk galaxy, there is the Newtonian force, Fij Rr Cos(\theta), whose effect is the centripital motion. At the current time, this is all that is used to calculate the momentum profiles of disk galaxies. This is not enough, something is missing. The leading hypothesis is dark matter. I am proposing that the omission is the gravitational force in a different direction, Fij Rr Sin(\theta), causes the new relativistic rocket effect, which helps determine where masses are located in space. This would eliminate the need for dark matter. It is important that I get through all the details of this calculation to test the hypothesis.

The proposal at this early stage has consequences for cosmology. As one moves to larger scales, the effect of gravity becomes more about where mass is distributed than about making things move faster or slower. At the big bang - the farthest distance we can go in spacetime, the cosmic background radiation data indicates that gravity was exclusively about the relativistic rocket effect, where the velocity is constant, at least to five significant digits. No matter was changing its velocity relative to other matter at the time of recombination. As the Univserse has aged, it may be that the balance has shifted a bit towards the mA term from the relativistic rocket effect. If so, then the net effect of gravity would be the same, but the Universe would appear to be accelerating more instead of all traveling uniformily. This would eliminate the need for dark energy.

If I could only quantify such a claim...

Doug

sweetser · Aug 21, 2008

Bicycle wheels and disc galaxies

Hello:

In this post we will think about momentum for a disk bicycle wheel and a disk galaxy. The similarities between the two systems may give support to the relativistic rocket effect discussed here.

A disk has all of its mass in a plane. Both of the disks in question spin. The bicycle wheel is a rigid body, while the galaxy is not. All parts of the wheel rotate at the same rotational velocity, but travel at a different tangential velocity which depends on how far out from the center a point is. For disk galaxies, the stars at the center of a galaxy travel slowly, yet quickly reach a maximum velocity outside the core. From there on, the stars, and then the helium gas, rotate at the same tangential velocity.

If one applies a force to the axis of a spinning bicycle wheel, the wheel will move a distance. This is an easy problem. It is just like applying a force to a brick, which moves, so the energy is the distance times the force.

Now apply exactly the same amount of force to the spinning bicycle disk, but somewhere along the rim. The problem gets much messier! The axle might move a bit - so you can do the force times distance calculation for that energy - but there will also be a wobble. I don't know how to deal mathematically with the energy that goes into wobbling, sorry. I doubt many people are confident about the subject since it involves torques and all that jazz. My big picture view is that if I apply the same amount of energy to moving the bicycle wheel along the axis as to somewhere on the disk, then the disk must distribute that energy between moving the center of mass and the energy of wobbling. Energy must be conserved.

Now think about a disk galaxy. The mass for disk galaxies has two components, a spherical one, and the disk, whose visible mass drops off exponentially with increasing distance from the center. That means the vast majority of mass can be viewed as being central: due to spherical symmetry and the exponential decay of the disk.

Newton's law of gravity has a cause, the inverse square attraction of gravity, and an effect, the centripetal force that keeps things moving in toward the center:

-\frac{G M m}{R^2} ~\hat{R} = m~\frac{V \cdot V}{R} ~\hat{R} \quad eq ~ 1

What I would like to point out is both cause and effect point in exactly the same direction, as they must, along the radius, \hat{R}. This is why modeling of galaxies by Newton's law works near the core, getting the right maximal terminal velocity. For a bicycle wheel, the closer to the axis you apply the force, the less you need to worry about all the messy torque stuff.

On physical grounds, I cannot accept it as reasonable that gravity does not exert a torque force. Gravity would somehow have to favor doing work along the radial direction. One might argue that perhaps all the torques cancel out nicely. Please recall the bicycle wheel. That problem is hard, there is no cancellation going on for the rigid body. There is no reason to expect the disk galaxy to be so clean. Yet in all my readings, I have never seen anyone discuss gravity working in any direction other than \hat{R}. I have a paper from 1963 where Alar Toomre did the calculation of the momentum profile for highly flattened galaxies the right way (using Bessel integrals, too technical for me, but one of the papers that started this area of study). It is clear he was only dealing with force in the radial direction.

Gravity from a system with its mass distributed in a disc must have torque forces that need to be taken into account. That would make the cause term point in a new direction, perpendicular to the radius, or \hat{V}. The effect of gravity must point in this same direction. Dimensional analysis leads to this proposal:

-\frac{G M m}{R^2} ~\hat{V} = c~ V~ \frac{d m}{d R} ~\hat{V} \quad eq ~ 2

Since the torque force points in the V direction, a V can only appear once, and naturally points along \hat{V}. As discussed in earlier posts, this term is a direct result of viewing a force as a change in momentum, and using the product rule, which has a constant V and a changing m.

I call myself a member of the "ultra-conservative fringe". Folks as bright as Alar Toomre frighten me. I don't think it is reasonable to say someone with such skills is wrong. On rare occasions, it may be OK to document something they did not account for, the error by omission. All the complexity of a torque force for gravity of a disc galaxy have been omitted. Based on energy conservation, I think that omission must be corrected.

It is reasonable to ignore this proposal since I have yet to go numerical. It might be the effect is trivial. Yet this problem works in the trivial end of the force spectrum. The accelerations are on the order of 10^-10 m/s²! They are measuring how helium II gas moves way out from the core. One needs to be complete. I cannot defend the models used today.

Doug

Note added in proof: Notice in equations 1 and 2, the cause term has in the denominator R², which is equal to the dot product of R with itself, and thus has no direction. Newton tacked on the radial direction part, as has every one of his students since. There is no "natural" directionality to the universal law of gravitation. The terms on the effect side both get there directions from single vectors, the dividing by R of the directionless V dotted to V for the centripetal force, and the V in the relativistic rocket effect. Nice.

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