## Unifying Gravity and EM

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

 Recognitions: Gold Member 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
 Recognitions: Gold Member 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
 Recognitions: Gold Member 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
 Recognitions: Gold Member 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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 Quote by sweetser 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.
Hi Doug,

I don't reconise the gentleman's definitions. Isn't force dP/dt or dE/dx ?

Four days seems rather short for the average professional to learn GR.

Regards,
Lut

 Recognitions: Gold Member Hello Lut: I have used 3 definitions for force. One is the derivative of the 4-momentum with respect to the interval tau. I saw that in a discussion of the Lorentz force of EM in Jackson, the end of chapter 11. This is the dP/dt and dE/dx one. The second definition is based on varying a Lagrange density with respect to velocity. Again that was used to derive the Lorentz force, in Landau and Lifshitz "Classical Theory of Fields". The third definition is the integral definition discussed in post #585, p37. All three give the same answer for force, love that consistency. The class was a survey of issues in gravity and cosmology, not a professional introduction to GR. We got a bit deeper than Scientific American, using a few equations. This thread is more "equation dense" than the course was. A fun aspect was being able to spend the days thinking of physics. I miss that spirit, and have been adrift since then. Doug
 Blog Entries: 3 Recognitions: Gold Member Of course, any expression with the right dimensions can be interpreted as force. My view was a bit telescopic (?). I found gravity as a coordinate acceleration independent of mass, by letting inertial mass depend on a field. It doesn't quite work but it was fun doing it. Lut

 Quote by sweetser Hello: This would eliminate the need for dark energy.
Doug:

Feng and Gallo have been working on the luninocity density relation for a solution to the dark matter issue. You may find their paper useful.

"Galactic Rotation Described with Thin-Disk Gravitational Model"
http://arxiv.org/abs/0803.0556

DT

 Hey Sweets, I was reading this and noticed something which seemed familiar from how I was considering the problem. Would you object to saying your interpretation of Gravity is more akin to a "stretching" of Space in the direction of the body in question, and EM could be considered a "stretch and a twist" in the direction of the charge orientation? I didn't come into physics from the mathematics side, but from the thinking about explanations for processes which can be converted into mathematics side. It could just be that I'm reading what you're describing mathematically wrong, but when I model it in my head, it looks like the explanation I've been developing from the other "direction", if you will.
 Recognitions: Gold Member Hello: Preparing the talk for Midwest Relativity Meeting was quite productive. It forced an interesting shift: I actually need to use quaternions for the action, tensors will not suffice. It is ironic that it has taken so long for me to insist on using quaternion on technical grounds since I have owned the domain quaternions.com since 1997. Tensor have a few great properties. For example, one does not choose a coordinate system for a tensor expression. The same tensor expression applies if one uses Cartesian, spherical, cylindrical, or some sort of oddball coordinate system that applies to a problem at hand. This property is shared by quaternions. Tensors can be added to other tensors, or multiplied by another scalar. This property is shared by quaternions. Tensors can be in arbitrary dimensions. Quaternions are limited to 4D. Folks working with strings will find that utterly inadequate. I have no interest in making folks who get the units of spacetime wrong happy. Every measurement of events in spacetime, every measurement of energy and momentum, ever made in physics has been 4D. That covers an absurd volume of data, so I will stick with it. One cannot take two tensors and form a product. It is this limitation which has forced me to write my Lagrangian only in terms of quaternions. If one wants to make a Lagrangian that has connections to EM, the weak, and the strong forces, then one needs to build a bridge to group theory. Group theory involves having an identity, multiplication, and inverses. That is an option for quaternions, but not so for tensors. Sure, one can tack on a group symmetry operator, but that feels phoney. I recall seeing the symmetry stapled onto terms of the action. Yukko. Better for group theory is to start with a division algebra to see if various ways to write it are equivalent to the group. In the case of U(1), the canonical representation is a complex number normalized to itself, forming a unit circle in the complex plane, $\frac{z}{|z|}$. Take either n roots, or n powers, whichever you prefer, and a unit circle in the complex plane can be formed. The exact same approach can be used for quaternions, $\frac{q}{|q|}$. The group U(1) is Abelian, members commute with each other. Although quaternions do not commute in general, they do commute if all the quaternions point in the same direction, which would be the case for n roots or powers of a normalized quaternion. Nice. The weak force has the symmetry SU(2), also known as the unitary quaternions. Sure enough, there is a way to write quaternions as unit quaternions. The Lie algebra has 3 generators. A common approach is to take the exponential of a quaternion where the scalar has been subtracted away, $exp(q-q*)$. It is natural to combine the U(1) and SU(2) together so as to capture the four degrees of freedom in a quaternion with $\frac{q}{|q|} exp(q-q^*)$. So wherever there was the rank 1 tensor $A^{\mu}$, drop in its place the quaternion $\frac{q}{|q|} exp(q-q^*)$ so the resulting expression has electroweak symmetry. There is only one more symmetry in the standard model to get, that for the strong force, SU(3). Its Lie algebra has 8 parts, which is what is found in two quaternions. Multiplying two U(1)xSU(2) symmetries just creates another member of the group. That is a basic property of groups. To make the multiplication table different, one needs to take the conjugate of one quaternion, then multiply it by the other, $(\frac{q}{|q|} exp(q-q^*))^*\frac{p}{|p|} exp(p-p^*)$. Drop this one in for $A^{\mu}$ and you have the symmetries required by three of the four known forces of Nature. This connection to the known forces makes it worthwhile to insist on quaternions. My sense is that those formally trained will reject moving to quaternions out of hand. I have seen conversations stop because I use the q word, so that has been the source of my reservations. That is too bad, because a quaternion is in fact a tensor whose diminsions happened to be constrained to 4 but has additional abilities, namely multiplication and division, useful to reach out to group theory. Doug