# Unifying Gravity and EM

Gold Member
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

Gold Member
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 [Broken]. 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.

Last edited by a moderator:
Gold Member
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 [Broken] 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

Last edited by a moderator:
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 [Broken]

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.

Last edited by a moderator:
Gold Member
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

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.

The Plasma Theory of the Universe, 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.

Last edited:
Gold Member
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

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.

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

Gold Member
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

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.

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

Gold Member
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

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…

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

Gold Member
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

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

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:
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.

Gold Member
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

Gold Member
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 R1/4. 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 [Broken] for the work I hope to accomplish over the next few months:
1. Draw it
2. Label Everything
3. Write out all the equations
4. Write out all the kg, m, s units and conversions
5. 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
6. 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

Last edited by a moderator:
Gold Member
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

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

Gold Member
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

Gold Member
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

Gold Member
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

Gold Member
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 travelling uniformily. This would eliminate the need for dark energy.

If I could only quantify such a claim...

Doug

Gold Member
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.