Doctordick 42,287 Posted May 16, 2010 Author Report Share Posted May 16, 2010 First I am not exactly sure what do you mean by a probability wave representing an event. I was aligned to think of this in terms of probability waves representing the positions of defined objects.I would say that the "supposed position of a defined object" can be called "an event".Do you just see them as essentially the same thing, or do you refer to something like a probability wave representing something like a collision between virtual particle and a fermion?A collision between a virtual particle and a fermion can be seen as "an event", as can the simultaneous change in the form of the probability wave. Remember, the probability wave yields our expectations and, if a collision occurs, certainly our expectations change. In modern quantum mechanics that change is called "collapse of the wave function". I did not pick up though, why the Heisenberg uncertainty principle implies that the further apart the interaction fermions are, the less momentum transfer must be...If the exchanged particle is "virtual", it means, from a colloquial perspective, that it "really isn't there". From a quanta mechanical perspective, it means that it has insufficient energy to exist at that particular point in space. Since at this point we are talking about photons (which have no rest energy) the energy is given by its momentum and we can apply the uncertainty principal.[math]\Delta P \Delta x \geq \frac{\hbar}{2}.[/math] Essentially, so long as [imath]\Delta P \Delta x < \frac{\hbar}{2}[/imath] we can't detect it (to detect it would violate the uncertainty principal) and it qualifies as "virtual". Thus, so long as the exchange is virtual, the larger [imath]\Delta x [/imath] (the distance over which it is exchanged) the smaller [imath]\Delta P [/imath] (the momentum being exchanged) must be. So I suppose when you refer to the speed of an element, you must be referring to its speed in [imath]x,y,z,\tau[/imath]-space. I.e. it is because every element must move at the same fixed speed, that more wiggling means slower speed towards the average direction of the entire object.That is exactly correct.Right... When you say "total momentum", you are essentially referring to the number and density of the individual elements moving along [imath]\tau[/imath] (at fixed speed).Yes. It seems to me to be quite reasonable that the momentum transfer rate would be proportional to the density of the sources that provide that exchange: i.e., the greater the total momentum change can be over the same change in tau (the reading on a clock traveling with the object being influenced).I may be forgetting something, but where and when the massive boson exchange/nuclear forces were brought up...?The issue was brought up in the last six paragraphs of the "Anybody interested in Dirac's equation?" thread. I really do little with it beyond suggesting that such an interaction certainly appears to be required by my fundamental equation. By the way, the purpose of displaying these relationships to modern physics which I have uncovered was not to prove modern physics was a tautology. The purpose was to generate interest in the proof of my equation among people who had the education to subject that proof to hard analysis. I have been totally confounded by the fact that those who could analyze that proof absolutely refuse to look at it. Personally I have been quite astonished by the fact that so many of modern physics theories can be mapped into approximate solutions to my equation. That equation is a constraint and I certainly would not expect it to be the only constraint on modern physics as that in itself would imply physics is a tautology (the conclusion I have come to after many years). What I would expect (and have been unable to discover) is a solution to my equation which could not be found in modern physics: i.e., that some additional constraint would be required by physics. Such an event would give physics an independent status above and beyond a tautology, but I haven't found it and it seems that no one else is interested in looking.That is very interesting... Another very unexpected feature of modern physics turning out to be a very expected feature of the symmetry requirements. Should probably discuss that issue little bit at some point as well.Ah yes, parity is an interesting issue all by itself. Parity is often referred to as "mirror" symmetry. If you look in a mirror, you will find that the image is different from the image you would see if a duplicate "you" stood in front of you. That difference can be seen in three specific different ways. The most common way to see the difference is to note the fact that a duplicate of you would have to turn around to face you. When he did that, his right arm would be on your left and his left arm would be on your right: i.e., everything along the "right left" axis would appear to be inverted. All other relationships remain the same. But that is because we commonly think of people turning around by rotating around their vertical axis. If we are going to actually analyze the phenomena itself, we could just as well presume he could have turned around by rotating around that horizontal "right left" axis. In that case we would expect to see his head on the bottom and his feet on the top: i.e., in the image, everything along the vertical axis would be inverted. (Note that his right arm is on the same side as your right arm; wave at yourself and think about it.) The third way to see it is to see everything along the axis between the two of you inverted: i.e., you see the "front" of the image instead of the back you would expect to see if a duplicate "you" stood in front of you. In this case everything else is as expected. His head and your head are both on top and your right arm and his right arm are on the same side but his face is towards you and not away as it should be. What I am getting at is the fact that a mirror image can be seen as accomplished by the inversion of any one axis. Now, consider the inversion of all three axes. If you invert all three axes, you get a mirror image: i.e., three parity swaps are identical to one parity swap. That brings up an interesting conundrum. Why should the physics of such an inverted universe be different from the original: i.e., when you consider the fact that all the information about the universe is contained in "the information about the universe", why should plotting the three axes in a particular direction make a difference. How does one know "the correct direction". It seems entirely reasonable that mirror symmetry should be a symmetry of the universe. Things are a bit different in a four dimensional Euclidean universe. In that case, inversion of all the axes bring one back to the original configuration so inversion of all the axes have to be a symmetry of the universe. However, if we invert only the three axes of our perceived space (the tau axis not being inverted) then we have a parity inversion (a true inversion of all the axes would include inverting tau). This results in a rather strange circumstance when applied to massive boson exchange. If the velocity in the tau direction of an entity being exchanged is in the same direction as the source, one might very well expect the interaction to be subtly different from the case where they are traveling in opposite direction. If that is a second order effect, it might very well be on the same order as the differential effect which would suggest that the weak nuclear force might well display parity consequences. If you take a look at the fundamental equation, note that multiplying through by minus one changes the sign of all the terms. Changing the sign of all the momentum terms changes nothing as the total momentum is still zero and changing the sign of the energy term can be absorbed into the definition of "t" (which is unmeasurable anyway), thus the only real change is a change in the sign of the Dirac interaction term. That would change all attractive terms into repulsive terms and vice versa. (Kind of sets the direction of time in our explanations doesn't it?) Inversion of tau is essentially changing particles into antiparticles so the above suggests that the interaction differences showing parity effects should be related to which version of the exchange pair is involved. What is above is really speculation as I have never honestly worked out all the details of the thing and there are a couple of things about the deduction which bother me. That is another reason I would like some well educated people to look at it.Apart from this analysis, does any satisfying explanation exist as to why weak nuclear force violates parity symmetry?What one person finds satisfying may bother another. I will direct you to Erasmus as he is apparently one hundred percent satisfied with the modern physics version of such things.I can't work out what sort of interference one might expect between all the relevant bodies, but I'm just thinking if this explains the Allais effect as a gravitational effect, that would be remarkable.What you seem to be missing here is the realization that my fundamental equation is essentially unsolvable. The best one can do is to find approximate solutions. In order to do that one needs to know exactly what terms can be ignored and what approximations for the terms which can not be ignored are reasonable: i.e., in essence, you need to know the mathematical solution before you start making your approximations. In other words, my equation is not a means of deducing a solution but is rather a means of eliminating proposed solutions. That is exactly the mistake Bombadil insists on making. I read your reference and I have no idea as to exactly what the nature of the Allais effect is; that article does not give the details of the differences as compared to the expectations. It could be as simple as some kind of gravitational focusing effect due to the intermediate body or it may be some kind of local error not being taken into account. The Foucault pendulum is a very sensitive experiment and requires careful elimination of interfering forces. I was involved in setting one up in graduate school and we had some problems getting it to work properly. At any rate, my equation really doesn't provide any help on the thing that I can see. Sorry about that. Just as an aside, I appear to have upset modest so I wouldn't expect any help from him. Sorry about that too. Have fun -- Dick coldcreation 1 Quote Link to post Share on other sites

Rade 37,414 Posted May 17, 2010 Report Share Posted May 17, 2010 Inversion of tau is essentially changing particles into antiparticles so the above suggests that the interaction differences showing parity effects should be related to which version of the exchange pair is involved. What is above is really speculation as I have never honestly worked out all the details of the thing and there are a couple of things about the deduction which bother me.Hi AnssiH. Given that DD not longer views my posts, I look to you to serve as middle person in possible dialog. I find this comment of DD to be of great interest. In his response to you, DD indicated that the purpose of his fundamental equation is to serve as "a means of eliminating proposed solutions". OK. Here is a "proposed solution" to the fundamental equation, and I would like to know the mathematics of how the fundamental equation can be used to "eliminate it". It involves what I can best explain as a combined real+virtual interaction of matter and antimatter along the tau dimension (that is, momentum in both the positive tau direction and what DD calls the inversion of tau direction). I would like to know how the fundamental equation would eliminate the possibility of predicting the "mass" for the proposed solution for the event of a matter object of 9 mass units (as quarks) as a coordinated fermion interacting with an antimatter object of 6 mass units (as antimatter quarks) as a coordinating boson along the tau dimension. It is my opinion that, if the fundamental equation CANNOT eliminate this possible event as a proposed solution to formation of mass along the tau dimension, then a new physics understanding of quark interactions within nucleons (protons, neutrons) will result. I hope this question will show DD I have a true interest in his fundamental equation. Quote Link to post Share on other sites

AnssiH 36,649 Posted May 23, 2010 Report Share Posted May 23, 2010 I would say that the "supposed position of a defined object" can be called "an event". ... A collision between a virtual particle and a fermion can be seen as "an event", as can the simultaneous change in the form of the probability wave. Remember, the probability wave yields our expectations and, if a collision occurs, certainly our expectations change. In modern quantum mechanics that change is called "collapse of the wave function". Right, of course. It represents the probability of actually detecting something. If the exchanged particle is "virtual", it means, from a colloquial perspective, that it "really isn't there". From a quanta mechanical perspective, it means that it has insufficient energy to exist at that particular point in space. Since at this point we are talking about photons (which have no rest energy) the energy is given by its momentum and we can apply the uncertainty principal.[math]\Delta P \Delta x \geq \frac{\hbar}{2}.[/math] Essentially, so long as [imath]\Delta P \Delta x < \frac{\hbar}{2}[/imath] we can't detect it (to detect it would violate the uncertainty principal) and it qualifies as "virtual". Thus, so long as the exchange is virtual, the larger [imath]\Delta x [/imath] (the distance over which it is exchanged) the smaller [imath]\Delta P [/imath] (the momentum being exchanged) must be. Hmm, okay, I don't really understand that... I suppose [imath]\Delta P[/imath] means uncertainty in momentum (/energy in this case), and [imath]\Delta x[/imath] means uncertainty in position. I understand that idea about "insufficient energy to exist at that particular point in space", but I don't understand why "less uncertainty in energy" means "less energy". (At least that's the implication I got from what you wrote) Likewise, I don't understand why "uncertainty in position" is related to the "actual distance over which the exchange occurs". I suppose this is traditional physics terminology, but it's a bit confusing/funny/odd to someone like me, that a virtual exchange, i.e. undetected exchange, is said to be the kind of exchange that has got "least uncertainty" to it; undetected exchange is considered to be the most certain kind of exchange :D Anyway, I do understand that as long as we are limited to staying below the minimal uncertainty limit, the larger uncertainty to position means smaller uncertainty to momentum, but again I do not understand why less "uncertainty to momentum" also means less "momentum". The issue was brought up in the last six paragraphs of the "Anybody interested in Dirac's equation?" thread. I really do little with it beyond suggesting that such an interaction certainly appears to be required by my fundamental equation. Okay, now I remember. Yeah that seems like valid speculation. [parity symmetry]...Inversion of tau is essentially changing particles into antiparticles so the above suggests that the interaction differences showing parity effects should be related to which version of the exchange pair is involved. What is above is really speculation as I have never honestly worked out all the details of the thing and there are a couple of things about the deduction which bother me. That is another reason I would like some well educated people to look at it. Hmmm, yes... That is interesting speculation too. What you seem to be missing here is the realization that my fundamental equation is essentially unsolvable. The best one can do is to find approximate solutions. In order to do that one needs to know exactly what terms can be ignored and what approximations for the terms which can not be ignored are reasonable: i.e., in essence, you need to know the mathematical solution before you start making your approximations. In other words, my equation is not a means of deducing a solution but is rather a means of eliminating proposed solutions. Actually the reason it popped up into my mind is because of how you arrive at "gravity" via virtual particle exchange. More about that below. I read your reference and I have no idea as to exactly what the nature of the Allais effect is; that article does not give the details of the differences as compared to the expectations. Yeah, I just picked up pretty much the first article that came up saying anything about the Allais effect, because I thought you were probably aware of it. Anyway, the original observation was, that the rate of change to the azimuthal angle changed abruptly during the solar eclipse. Before the solar eclipse, Allais had his swinging pendulum rotating at the steady rate of 0.19 degrees per minute, clockwise, as was expected at his latitude. At the beginning of the eclipse, the rotation changed direction to counter-clockwise and was changing an average of ~13 degrees per minute. In the middle of the eclipse the rotation changed direction again, and in the end of the eclipse the rate of rotation resumed to normal rate of 0.19 degrees per minute, clockwise. Corresponding anomalies to gravity during a solar eclipse have been observed since, with different kinds of pendulums, gravitometers and atomic clocks. Some experiments have come out negative though. There's a lot of information around the net, I just dug up couple that seem to contain little bit more information;http://arxiv.org/ftp/gr-qc/papers/0408/0408023.pdfReview on Gravitational Anomalies At any rate, there are two reason why that popped into my mind. One, it's one of the gravitation-related effects that current models of gravitation don't explain directly (in addition to the "pioneer anomaly" and the anomalous rotation speed of galaxies Galaxy rotation curve - Wikipedia, the free encyclopedia ). And second, since you are viewing gravity as an expected differential effect (inside our solution to reality), and since you are arriving at it via the idea of virtual photon exchange, it would not surprise me at all if that also necessitates one to expect disturbances to the total momentum exchange, due to the expected interference caused by the moon between the sun and the location of the pendulum on the surface of earth. The way the foucalt pendulum reacts in the very beginning and in the very end of solar eclipse just reminds me of interference patterns a lot. But as I said, I do not have the skills to work that out, and my knowledge of virtual photon exchange is far too limited to figure out if anything can be made out of it (I don't know what sort of effect a "middle body" puts to virtual photon exchange, if any...). But let it be said, that if someone can show that Allais effect is an expected effect of gravitation directly, in terms of your paradigm, I think it should generate some more interest towards your work. Just like the possibility of it explaining the pioneer anomaly and the rotation speed of large galaxies (and other things explained via dark matter), as you mention in the end of the OP. -Anssi Quote Link to post Share on other sites

AnssiH 36,649 Posted May 23, 2010 Report Share Posted May 23, 2010 Hi AnssiH. Given that DD not longer views my posts, I look to you to serve as middle person in possible dialog. I find this comment of DD to be of great interest. In his response to you, DD indicated that the purpose of his fundamental equation is to serve as "a means of eliminating proposed solutions". OK. Here is a "proposed solution" to the fundamental equation, and I would like to know the mathematics of how the fundamental equation can be used to "eliminate it". It involves what I can best explain as a combined real+virtual interaction of matter and antimatter along the tau dimension (that is, momentum in both the positive tau direction and what DD calls the inversion of tau direction). I would like to know how the fundamental equation would eliminate the possibility of predicting the "mass" for the proposed solution for the event of a matter object of 9 mass units (as quarks) as a coordinated fermion interacting with an antimatter object of 6 mass units (as antimatter quarks) as a coordinating boson along the tau dimension. It is my opinion that, if the fundamental equation CANNOT eliminate this possible event as a proposed solution to formation of mass along the tau dimension, then a new physics understanding of quark interactions within nucleons (protons, neutrons) will result. I hope this question will show DD I have a true interest in his fundamental equation. I'm sorry but I'm not entirely sure what you are asking about, and I suspect even if I did understand, I would not have the necessary know-how to actually analyze it properly... :coffee_n_pc: -Anssi Quote Link to post Share on other sites

Doctordick 42,287 Posted May 27, 2010 Author Report Share Posted May 27, 2010 Hmm, okay, I don't really understand that...Well the real problem here is that you cannot solve the relevant equations. Without mathematics the issue becomes rather meaningless. That being the case, I will try to explain it from a pseudo classical perspective. Let's start with a real photon and two entities interacting with that photon. If one of the entities emits a photon and the other absorbs that emitted photon what do you expect? The photon starts in one place and ends up in another. That would be taken to imply that the photon had momentum in the direction it traveled. Conservation of momentum implies the emitter gains a momentum in the opposite direction and the absorber ends up gaining the photons momentum. The end result is that the two entities gain momentum in the opposite directions: i.e., their change in momentum is always away from one another. This case implies a repulsive force due to the exchange. Now, let's look at a “virtual” photon exchange: i.e., it doesn't really exist but does nonetheless exchange momentum. In this case, which is emitting and which is absorbing is unknown. Not only that but the time of these events us unknown. In fact, from a quantum mechanical perspective (and that would be a solution of the dynamic equations of quantum mechanics) it could even be absorbed before it was emitted (has to do with the uncertainty in time). Its momentum could be in the opposite direction of its motion (remember it isn't really there so we can't check it, this is a sort of “let's look at this strange possibility from a classical perspective”). If that were the case (the momentum of the photon could be opposite to its travel direction), the two entities could then gain momentum towards one another: i.e., one can obtain an attractive force. (That is the real value of "virtual exchange forces".) This type of thing can only occur if the uncertainty principal yields [imath]\Delta P \Delta x \geq \frac{\hbar}{2}[/imath]. From the classical perspective this thing just can't happen. If the actual real distance between the interacting entities is known to be x, the uncertainty in position of the photon must be at least x (it has to be able to get from one entity to the other somehow). Divide through the uncertainty relationship above by the factor [imath]x=\Delta x[/imath] and you obtain [imath]\Delta P \geq \frac{\hbar}{2x}[/imath]. If follows that, if [imath]\Delta P \geq \frac{\hbar}{2x}[/imath], it's a classical exchange and the momentum of the photon is a measurable thing. In order to keep it in the realm of “unknowable”, the transferred momentum must fall off by more than one over x. I don't understand why "less uncertainty in energy" means "less energy". (At least that's the implication I got from what you wrote)Because the thing can only violate conservation of energy if it does it quickly. If it took long enough to observe (i.e., obeyed the uncertainty principal) it would be detectable from the classical perspective and it wouldn't be “virtual” and wouldn't violate conservation of energy.Likewise, I don't understand why "uncertainty in position" is related to the "actual distance over which the exchange occurs".Again if it doesn't violate the uncertainty principal, it is detectable as a classical entity: i.e., it's not virtual.I suppose this is traditional physics terminology, but it's a bit confusing/funny/odd to someone like me, that a virtual exchange, i.e. undetected exchange, is said to be the kind of exchange that has got "least uncertainty" to it; undetected exchange is considered to be the most certain kind of exchange :DIt isn't “undetected” which is important here, it's rather “undetectable”. Photon exchange, real or virtual, is essentially undetectable by classical means except by detecting the energy and/or momentum exchange. I read a few of the references you gave for the Allias effect and got the impression that the effect has not really been verified yet and there are people who believe it's bogus. As I said earlier, my equation doesn't produce solutions, it only points out flaws in solutions where you know what can and can not be ignored. If someone hands me a solution to the Allias effect then, knowing what kinds of terms they are ignoring, I can determine if or if not that solution does indeed obey my equation (it often takes some deep thought but it can generally be done). But no one has offered a solution that I can understand.But as I said, I do not have the skills to work that out, and my knowledge of virtual photon exchange is far too limited to figure out if anything can be made out of it (I don't know what sort of effect a "middle body" puts to virtual photon exchange, if any...).The problem here is that gravity is such a small effect that only the grossest exchange is detectable (two body exchange). If one adds in a third body the equation becomes quite difficult to solve and is probably solvable only for very special cases.But let it be said, that if someone can show that Allais effect is an expected effect of gravitation directly, in terms of your paradigm, I think it should generate some more interest towards your work.I doubt that very much. All it would do is add further support to the “crackpot” designation, a title I have already well earned. :lol: :lol: :lol: Have fun -- Dick Quote Link to post Share on other sites

AnssiH 36,649 Posted June 14, 2010 Report Share Posted June 14, 2010 Well the real problem here is that you cannot solve the relevant equations. Without mathematics the issue becomes rather meaningless. That being the case, I will try to explain it from a pseudo classical perspective.......If the actual real distance between the interacting entities is known to be x, the uncertainty in position of the photon must be at least x (it has to be able to get from one entity to the other somehow). Ah, okay, now that's clear. I was thinking of the emission and absorption as two different positions, both having their own uncertainty. I don't understand why "less uncertainty in energy" means "less energy". (At least that's the implication I got from what you wrote)Because the thing can only violate conservation of energy if it does it quickly. If it took long enough to observe (i.e., obeyed the uncertainty principal) it would be detectable from the classical perspective and it wouldn't be “virtual” and wouldn't violate conservation of energy. Hmm, I'm still confused... I now understand how larged distance between the bodies implies larger uncertainty in position (of the exchange particle), and also I understand why that implies smaller uncertainty in total momentum/energy being exchanged. I still don't understand why small uncertainty to the total energy transfer means the amount of energy transfer is also small. From your response, I'm guessing virtual energy exchange is always seen as something violating conservation of energy, by definition(?) And I gather that the time it takes for the energy to transfer (i.e. the time between emission and absorption events?), is somehow related to the amount of energy transfer? But I still have no idea why that is... I read a few of the references you gave for the Allias effect and got the impression that the effect has not really been verified yet and there are people who believe it's bogus. Yeah, it's not easy to setup experiments because you have to have a solar eclipse for them. Still it's generated enough interest that people have performed different types of experiments, and actually appears to be generally accepted as a real phenomena. At least real enough for mainstream press to write articles about its existence without getting frowned upon. It may be that most physicists believe that a conventional explanation will be found. Okay, back to the OP. In it, I just spotted something that looks like an editing error of some kind: Of issue is the fact that, if the influence of the distant object is ignored, the influenced element will inexplicitly appear to be proceeding at a slightly slower velocity than it would if the distant object didn't exist. I don't think you meant to say that if the influence of the distant object is ignored, then the influenced element will appear to proceed slower. But let us get back to gravitational effects. It is my position that gravity is a direct consequence of the fact that the presence of an object with momentum in the tau direction yields a secondary differential effect which causes the speed of elements through my four dimensional Euclidean space to slow: i.e., refraction of the wave solution to my fundamental equation occurs. In order to check my proposition, I need to be able to work out the geodesic paths implied by such a notion. To begin with, my assertion is that the four dimensional velocity of all elements can be seen as slowed in the presence of a gravitational potential. Thus it is that the “index of refraction” of a gravitational field is given by the instantaneous value of [imath]n=\frac{c}{c'}[/imath] evaluated a specific point in that field (where c is the velocity of light far removed from any massive influence). Geodesic paths can be obtained by minimizing nds where “ds” is the distance differential along that geodesic path in my four dimensional Euclidean geometry. Right I think I understand that. If there was no gravitational field in the situation, the "n" would be taken to be "1" everywhere, and in that case minimizing the nds would just yield a straight line in the coordinate system. In the case that there is a gravitational field, which yields larger "n" as you move closer to the "source" of that field, minimizing the nds means the path must curve away from the source little bit. In order to accomplish that result, we need to use what is called “the calculus of variations”. We want the variation of the path integral along the geodesic to vanish: i.e., [math]\delta \int^{P_2} _{P_1}nds = 0\quad\quad where \quad\quad n=\frac{1}{\sqrt{1+\frac{2}{c^2}\Phi(\vec{x})}}[/math]. Thankfully this is a problem already solved by the physics community. The vanishing of the path integral occurs when the function being integrated satisfies the Euler-Lagrange equation. Okay, I started up by reading your references, and wikipedia article about the calculus of variations. The wikipedia article is much more useful for me because it contains more extensive explanations. I didn't yet read it all too carefully, I think I'll have to spend little bit more time on this. I did not yet understand what it means that "the variation of the path integral along the geodesic vanishes". I can sort of guess what it's about, but I would just like to be sure I understand what "variation of an integral" refers to exactly. Overall, my understanding of this is very shady right now. I'm just picking up that the calculus of variations is a method for finding the function which corresponds to some minimum (or other extreme) of some integral of a given set of functions. Like the functions corresponding to the geodesic path in the case there are some constraints that yield non-linear paths, just like our problem with the the refraction index. I'll continue from here later... -Anssi Quote Link to post Share on other sites

Doctordick 42,287 Posted June 16, 2010 Author Report Share Posted June 16, 2010 I was thinking of the emission and absorption as two different positions, both having their own uncertainty.Generally speaking, the uncertainty in the positions of the two entities being influenced by virtual exchange is not a significant factor. In many cases those entities (and their motion) can be expressed via classical mechanics. I don't think you meant to say that if the influence of the distant object is ignored, then the influenced element will appear to proceed slower.If the object being observed lies between the observer and the distant object being ignored then, yes, the influenced object will appear to proceed slower than expected as it lies in a lower gravitational potential than the observer. If the distant object is being ignored, one’s calculations would essentially be presuming the two were at the same gravitational potential. I can sort of guess what it's about, but I would just like to be sure I understand what "variation of an integral" refers to exactly.In this case we are talking about the integral of a function defined along a specific path through a space. The variation being spoken of is a variation in that defined path. We are looking for the minimum in the length of that path: i.e., if the path is varied in any way, the integral will increase. This is analogous to “a straight line is the shortest distance between two points” used in Euclidean geometry. Have fun -- Dick Quote Link to post Share on other sites

AnssiH 36,649 Posted June 16, 2010 Report Share Posted June 16, 2010 If the object being observed lies between the observer and the distant object being ignored then, yes, the influenced object will appear to proceed slower than expected as it lies in a lower gravitational potential than the observer. If the distant object is being ignored, one’s calculations would essentially be presuming the two were at the same gravitational potential. Oh that's how you meant it, actually existing in the situation but being ignored.... I thought you meant ignored as in "taken out of the situation being considered"... I think someone else might make the same mis-interpretation... In this case we are talking about the integral of a function defined along a specific path through a space. The variation being spoken of is a variation in that defined path. We are looking for the minimum in the length of that path: i.e., if the path is varied in any way, the integral will increase. This is analogous to “a straight line is the shortest distance between two points” used in Euclidean geometry. Ah, okay I see. I'll try to get around to continue from here soon... -Anssi Quote Link to post Share on other sites

AnssiH 36,649 Posted June 17, 2010 Report Share Posted June 17, 2010 In order to accomplish that result, we need to use what is called “the calculus of variations”. We want the variation of the path integral along the geodesic to vanish: i.e., [math]\delta \int^{P_2} _{P_1}nds = 0\quad\quad where \quad\quad n=\frac{1}{\sqrt{1+\frac{2}{c^2}\Phi(\vec{x})}}[/math]. Thankfully this is a problem already solved by the physics community. The vanishing of the path integral occurs when the function being integrated satisfies the Euler-Lagrange equation. Just in order to avoid making unfortunate false assumptions, that delta sign before the integration sign is simply to say "variation of..." right? And the equation for "n", I see you got it simply from [imath]n=\frac{c}{c'}[/imath] and [imath]c'=c\sqrt{1+\frac{2\Phi}{c^2}}[/imath] established earlier in the OP. ...if [imath]J[/imath] is defined by an integral of the form [math]J =\int f(t,y,\dot{y})dt[/math] where [math]\dot{y} \equiv \frac{dy}{dt}[/math] then [imath]J[/imath] has a stationary value if the Euler-Lagrange differential equation [math]\frac{\partial f}{\partial y} - \frac{d}{dt}\left ( \frac{\partial f}{\partial \dot{y}} \right ) = 0[/math] is satisfied. I take it the dot on [imath]\dot{y}[/imath] is just to denote a derivative? So... how I'm reading that is, that [imath]\int f(t,y,\dot{y})dt[/imath] states an integration of some function over [imath]t[/imath], and it's a function which takes the arguments [imath]t[/imath], [imath]y[/imath] (which is a function of [imath]t[/imath]), and the derivative of [imath]y[/imath] with respect to [imath]t[/imath]. (I reckon [imath]y[/imath] is required to be a function of [imath]t[/imath] because it states [imath]\dot{y}[/imath] means derivative of [imath]y[/imath] with respect to [imath]t[/imath]...) Then [math]\frac{\partial f}{\partial y} - \frac{d}{dt}\left ( \frac{\partial f}{\partial \dot{y}} \right ) = 0[/math] looks to me like requiring the derivative of [imath]f[/imath] with respect to [imath]y[/imath], to exactly negate the derivative of "the derivative of [imath]f[/imath] with respect to [imath]\dot{y}[/imath]" with respect to [imath]t[/imath] (sic) Like if they are seen as two different curves, they exactly negate each others. Sooo, if that's at all right, I think I'm starting to get some sort of grasp of this idea... Not very crystal clear yet though. In order to simplify the problem, I would like to turn to a spherically symmetric case. Essentially a point source with a very large momentum in the tau direction and negligible momentum in the other three dimensions. This is in order to look at geodesic paths in the vicinity of a single very massive object where the other properties of the object can be ignored. In such a case, I can write [imath]\Phi(\vec{x})[/imath] in a very simple form consistent with standard notation:[math]\Phi(\vec{x})=-\frac{\kappa M}{r}[/math] where [imath]\kappa[/imath] is the proper proportionality constant to yield the correct potential generated by the mass M. Um, right, so that is just to say that the gravitational potential is proportional to the mass, and inversely proportional to the distance... Hmmm, shouldn't it be the square of the distance? This expression can be further simplified by changing to cylindrical coordinates (omit z as by symmetry we need only consider motion in the x-y plane). Yup. Thus it is that the differential of the path along which our metric is to be minimized is given by [imath]ds=\sqrt{(d\tau)^2+(dr)^2+r^2(d\theta)^2}[/imath]. Struggling here a bit... I get the squared [imath]d\tau[/imath] and [imath]dr[/imath], but I can't figure out what that last term means exactly and where you got it... I'll have a rest here... -Anssi Quote Link to post Share on other sites

Doctordick 42,287 Posted June 21, 2010 Author Report Share Posted June 21, 2010 Just in order to avoid making unfortunate false assumptions, that delta sign before the integration sign is simply to say "variation of..." right? And the equation for "n", I see you got it simply from [imath]n=\frac{c}{c'}[/imath] and [imath]c'=c\sqrt{1+\frac{2\Phi}{c^2}}[/imath] established earlier in the OP.Right!I take it the dot on [imath]\dot{y}[/imath] is just to denote a derivative?The symbol, “[imath]\equiv[/imath]”, is commonly read as “equivalent” which essentially means “identical” so, yes, that is exactly what it means.Like if they are seen as two different curves, they exactly negate each others. Sooo, if that's at all right, I think I'm starting to get some sort of grasp of this idea... Not very crystal clear yet though.This is a rather deep and complex mathematical relationship and you really shouldn’t expect see it as an obvious idea. That is the essence of the power of mathematics: once one has proved a mathematical relationship one knows it is true and seeing it as something other than an expressible mathematical relationship is unnecessary. In fact, trying to see it in terms of a mental picture can be dangerous because it is always possible that your mental picture may either add aspects not contained in the mathematical expression or omit aspects contained in the mathematical expression. I am not saying that the mental model you have in mind is erroneous; but rather that, when it comes to advanced mathematics, one should get in the habit of seeing such things as “required mathematical relations”. As I have said in that post about rational thought, mathematics is a powerful tool capable of extending logical thought far beyond what we can comprehend in our minds eye. Mathematicians are a very dependable as a group; once they accept that something has been proved, you can pretty well accept it as fact. To prove all the stuff yourself would take a lifetime. What is important is that you understand the various symbols used and you seem to be picking up on that just fine.Um, right, so that is just to say that the gravitational potential is proportional to the mass, and inversely proportional to the distance... Hmmm, shouldn't it be the square of the distance?You are thinking of the gravitational force field (almost all “fields” referred to in physics are “vector force fields”). The idea of a potential field arises from the fact that most force fields (remember, force is a vector) can be expressed via the expression [math] \vec{F}=\vec{\nabla}G(x,y,z)\equiv \frac{\partial G}{\partial x}\hat{x}+\frac{\partial G}{\partial y}\hat{y}+\frac{\partial G}{\partial z}\hat{z}[/math] Or, if the function G has only a radial dependence, [math] \vec{F}=\vec{\nabla}G®\equiv \frac{\partial G}{\partial r}\hat{r}.[/math] In other words, the force is proportional to the derivative of the potential and the derivative of [imath] \frac{1}{r}[/imath] is [imath]-\frac{1}{r^2}[/imath]: i.e., if the force is proportional to [imath]\frac{1}{r^2}[/imath], the potential must be proportional to [imath]-\frac{1}{r}[/imath].Struggling here a bit... I get the squared [imath]d\tau[/imath] and [imath]dr[/imath], but I can't figure out what that last term means exactly and where you got it... This is nothing more than dl expressed in radial coordinates (radial coordinates for the x and y directions). See Polar Coordinates -- from Wolfram MathWorld The differentials dr and d theta are always orthogonal to one another so the distance dl can easily be expressed in either set. The only difficulty to be handled is the fact that the actual distance moved when theta is changed is proportional to “r”. I think you can figure the thing out. If you still have problems, let me know and I will clear it up in detail. Have fun -- Dick Quote Link to post Share on other sites

Bombadil 14,583 Posted June 26, 2010 Report Share Posted June 26, 2010 More importantly, the above suggests an attack towards determining the geometry which will yield gravity as a pseudo force in our four dimensional Euclidean geometry. I have already shown how static structures appear as three dimensional objects in this geometry so let us examine what is commonly called “a gravitational well”. The gravitational well consists of a vertical hole where there is a gravitational field in the vertical direction. If an experimenter in a gravitational well sets up a clock via a light pulse traveling back and forth between two horizontally displaced mirrors, since we can establish horizontal measure (simple vertical lines carry those measures to different heights in the hole) and his clock must run slow, we must see the apparent velocity of light to be [math]c'=c\sqrt{1+\frac{2\Phi}{c^2}}[/math] I’m somewhat puzzled by how we can measure the distance that the light travels and conclude that the light in the hole is traveling slower then the light outside of the hole because I don’t see how we can establish the horizontal measures and know that the distance at the top will be the same as that in the hole. Unless we can say that any scaling can only have an effect in the vertical direction, And then don’t we also have to consider the clock to be small enough that the scaling over the hole clock is in the same direction. For what we seem to be interested in it seems that this effect can be easily ignored. Also won’t the person in the gravity well see a clock outside of the gravity well run fast? As a result they will see the speed of light outside of the hole to be faster? Again I don’t see how we know that the distance between the mirrors is not being scaled unless we say that the result of gravity is the same as an object that is accelerating and an accelerating object can only have a vertical scaling and can‘t have a scaling along the horizontal direction due to it being independent of the vertical direction. Secondly, the faces of the mirrors cannot be perpendicular to the floor of the elevator; it they were, the light would miss the original mirror on the return (the component of the elevator vertical velocity has changed, the vertical component of the light has not). It must now have an additional component of its velocity in the direction the elevator is accelerating. As a result, both mirrors must lean slightly towards the outside such as to assure that the reflected light always picks up that required change in velocity to keep up with the elevator. A little thought about the situation should be enough to prove to the reader that, at the instant immediately prior to reflection (as seen by the rest observer), the path of the light must be parallel to the floor of the elevator and immediately after reflection the path must be towards the position the other mirror will be in when the light gets to the other side. But, is the angling of the mirrors a consequence of the acceleration or is it a consequence of the design of the clock. While a clock could be designed that had this built into it, do we need to build it into it? If it is a consequence of it accelerating then wouldn’t this have to be taken into effect when we try to measure the distance between the two mirrors. If this is not an effect of the acceleration then are you in fact saying that the physics are different for an accelerating object and so the original clock must be redesigned? Thus it is that the clock we have proposed will run even slower than what was predicted a few paragraphs ago. The error is clearly a function of the physical size of the clock. This effect can be eliminated by defining our clock's size to be small enough to make the error negligible. I point this out so that we can discuss further ramifications of those instantaneous relativistic effects. We know that these are different since one is a consequence of the size that we build the clock while the other one is a differential consequence of the Lorenz transformation and being in a gravity well. Both seem to be a consequence of the laws of physics not being the same in a gravity well. So are we in fact looking for a transformation that will result in the objects being correctly scaled so that the fundamental equation will be unchanged by being in a gravity well? But won’t this scaling result in the faces of the mirror being parallel in this frame and so it would make sense that the clock is the same as one not in the gravity well? Quote Link to post Share on other sites

AnssiH 36,649 Posted June 30, 2010 Report Share Posted June 30, 2010 Right!You are thinking of the gravitational force field (almost all “fields” referred to in physics are “vector force fields”). The idea of a potential field arises from the fact that most force fields (remember, force is a vector) can be expressed via the expression [math] \vec{F}=\vec{\nabla}G(x,y,z)\equiv \frac{\partial G}{\partial x}\hat{x}+\frac{\partial G}{\partial y}\hat{y}+\frac{\partial G}{\partial z}\hat{z}[/math] Or, if the function G has only a radial dependence, [math] \vec{F}=\vec{\nabla}G®\equiv \frac{\partial G}{\partial r}\hat{r}.[/math] In other words, the force is proportional to the derivative of the potential and the derivative of [imath] \frac{1}{r}[/imath] is [imath]-\frac{1}{r^2}[/imath]: i.e., if the force is proportional to [imath]\frac{1}{r^2}[/imath], the potential must be proportional to [imath]-\frac{1}{r}[/imath]. Ah, right... Okay, that's clear now. This is nothing more than dl expressed in radial coordinates (radial coordinates for the x and y directions). See Polar Coordinates -- from Wolfram MathWorld The differentials dr and d theta are always orthogonal to one another so the distance dl can easily be expressed in either set. The only difficulty to be handled is the fact that the actual distance moved when theta is changed is proportional to “r”. I think you can figure the thing out. If you still have problems, let me know and I will clear it up in detail. Hmmm, ahha, okay, while I don't understand that completely, I can take it on faith, being that that Wolfram page about polar coordinates states that a line element is given by [imath]ds^2 = dr^2 + r^2d\theta^2[/imath]. (If you want to expand on it, feel free to, but I'm pretty convinced it is valid anyway) It follows that the metric to be minimized to yield gravity consistent with a pseudo force is [imath]dl=nds[/imath] which can be written [math]dl={\left[1-\frac{2\kappa M}{c^2r}\right]^{-\frac{1}{2}}}\left[(d\tau)^2+(dr)^2+r^2(d\theta)^2\right]^{\frac{1}{2}}[/math] Hmm, right, looks like [imath]\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}[/imath] so yes I understand how you got that expression. Using the standard Euler-Lagrange notation, the path integral over which the variation is to vanish (“J” in equation #1) is given by[math] J =\int f(t,y,\dot{y})dt=\int^{P_2}_{p_1}\frac{1}{\sqrt{1-\frac{2\kappa M}{c^2r}}}\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}dt[/math] Since “t” in my reference frame is actually a path length measure (essentially a reference as to where one is on the referenced path) I have changed the “ds” into “dt” thus converting the original expression given for ds, [imath]\left[(d\tau)^2+(dr)^2+r^2(d\theta)^2\right]^{\frac{1}{2}}[/imath], into [imath]\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}dt[/imath]. Hmm, I'm struggling here a bit... So I take it that [imath]d\tau \equiv \dot{\tau}[/imath]; that that's just a matter of notation. Is "t" actually referring to the evolution parameter of your universal notation? What does it mean that it's a path length measure? I don't really understand why it appears in the euler lagrange equation; does it always have something to do with time evolution of something? Needless to say, I'm a bit confused... :I -Anssi Quote Link to post Share on other sites

Doctordick 42,287 Posted July 2, 2010 Author Report Share Posted July 2, 2010 I’m somewhat puzzled by how we can measure the distance that the light travels and conclude that the light in the hole is traveling slower then the light outside of the hole because I don’t see how we can establish the horizontal measures and know that the distance at the top will be the same as that in the hole.The thesis here is that the observed force is due entirely to a distortion of the coordinate system thus the consequences of that acceleration is exactly what would be expected of any other acceleration. Acceleration is a change in velocity. That change in velocity results in Lorentz-Fitzgerald contraction in the direction of that velocity change but no change in measures orthogonal to that acceleration. Thus, so long as the acceleration is uniformly in the same direction (that is, the source of the gravitational field is sufficiently far away as to make the sides of the gravitational well essentially parallel), the distance across the well is the same at the top as it is at the bottom. The clock consists of two mirrors on the opposite sides of the well and thus a slow clock requires light to move slower in the well as the distance moved in a cycle is the same at both heights.While a clock could be designed that had this built into it, do we need to build it into it? ... If this is not an effect of the acceleration then are you in fact saying that the physics are different for an accelerating object and so the original clock must be redesigned?Of course this kind of clock has to be redesigned; I am talk about that very fact when I bring up the issue of the size of the clock (the clock would be any clock dependent upon electromagnetic harmonic resonance).The error is clearly a function of the physical size of the clock. This effect can be eliminated by defining our clock's size to be small enough to make the error negligible. I point this out so that we can discuss further ramifications of those instantaneous relativistic effects.The issue is that the path of a photon is influenced by gravity.But won’t this scaling result in the faces of the mirror being parallel in this frame and so it would make sense that the clock is the same as one not in the gravity well?All this says is that, if we make our clock so large that we have to take into account the gravitational refraction of the photon path, we have to include that secondary problem. Think about that! Essentially it says that if the gravitational field is more than the Sun’s, your clock should be somewhat smaller than the sun (that is the circumstance which made this refraction measurable)! If your clock is less than a foot across, the acceleration necessary to make this a problem would be utterly beyond your interest; human beings simply couldn’t survive in such an environment. Hmm, I'm struggling here a bit... So I take it that [imath]d\tau\equiv \dot{\tau}[/imath]; that that's just a matter of notation.Not quite. The "dot" signifies the derivative with respect to "t" so [imath]\dot{\tau}[/imath] is equivalent to [imath]\frac{d \tau}{dt}[/imath]; slightly different from what you have stated. In my geometry, [imath]\frac{d \tau}{dt}[/imath] is essentially the velocity of an object in the tau direction just as [imath]\frac{dx}{dt}[/imath] would be the velocity in the x direction.Is "t" actually referring to the evolution parameter of your universal notation? What does it mean that it's a path length measure? I don't really understand why it appears in the euler lagrange equation; does it always have something to do with time evolution of something? Needless to say, I'm a bit confused... :I In this case, we are looking at how physics looks inside a gravitational field when observed from outside that gravitational field (or perhaps from a different place in that field). The physical process being observed is in freefall: i.e., there are actually no forces on it other than the pseudo force due to the supposed geometric distortion caused by the gravitational potential. A clock at that point in space will run exactly as defined by that potential in its own reference frame. Of course, as seen by the outside observer, it must also be corrected by velocity effects (special relativistic effects); but I have already shown that all clocks actually measure changes in tau along their path in my four dimensional space. Thus dt (as measured by a clock following a geodesic path) and [imath]d\tau[/imath], path length in my geometry, are the same thing (the fact that everything is momentum quantized in the tau direction is immaterial except for projecting out that tau direction when it comes to actual observations). Think of it this way, [imath]d\tau[/imath] in Einstein's picture is what is called the “invariant interval”: i.e., everyone gets the same value when they calculate that term. But dx=dy=dz=0 in the rest frame of a moving clock thus the clock measures exactly the same value everyone else gets for [imath]d\tau[/imath]: i.e., the clock measures exactly that invariant path length. In the Euler-Lagrange equation, all the mathematical terms may be regarded as abstract concepts: i.e., they are talking about abstract solutions to abstract equations. If one can interpret one’s equation in such a way that it conforms to the Euler-Lagrange equation then the solutions to that equation are solutions to his equation. It is as simple as that. That is the great power of mathematics. Have fun -- Dick Quote Link to post Share on other sites

AnssiH 36,649 Posted July 5, 2010 Report Share Posted July 5, 2010 Not quite. The "dot" signifies the derivative with respect to "t" so [imath]\dot{\tau}[/imath] is equivalent to [imath]\frac{d \tau}{dt}[/imath]; slightly different from what you have stated. In my geometry, [imath]\frac{d \tau}{dt}[/imath] is essentially the velocity of an object in the tau direction just as [imath]\frac{dx}{dt}[/imath] would be the velocity in the x direction. Ah, okay. ...but I have already shown that all clocks actually measure changes in tau along their path in my four dimensional space. Thus dt (as measured by a clock following a geodesic path) and [imath]d\tau[/imath], path length in my geometry, are the same thing (the fact that everything is momentum quantized in the tau direction is immaterial except for projecting out that tau direction when it comes to actual observations). Right okay, and I suppose then that... Since “t” in my reference frame is actually a path length measure (essentially a reference as to where one is on the referenced path) I have changed the “ds” into “dt” thus converting the original expression given for ds, [imath]\left[(d\tau)^2+(dr)^2+r^2(d\theta)^2\right]^{\frac{1}{2}}[/imath], into [imath]\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}dt[/imath]. ...that move is based on the fact that every element is moving at constant velocity, so any change in "t" also entails a similar change in [imath]\sqrt{\tau^2 + r^2 + r^2\theta^2}[/imath] Following the standard Euler-Lagrange attack, if the expression[math]f(t,y,\dot{y})=\frac{1}{\sqrt{1-\frac{2\kappa M}{c^2r}}}\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}[/math] satisfies the Euler-Lagrange equation, the variation of the integral over the implied path will vanish. In this case we have three independent variables to deal with (note that “y”, in this case, stands for the variable of interest) which leads to three independent differential equations. Just to be absolutely sure, by "three independent variables" you refer to [imath]\tau[/imath], [imath]r[/imath] and [imath]\theta[/imath]... (We are not concerned of [imath]M[/imath]) However, we have one additional constraint which simplifies the problem considerably: [imath](\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2=c^2[/imath]. Yup. The partials of f with respect to tau and theta vanish Hmmm, I should probably understand by now why that is, but I'm a bit lost apparently... You are saying that changing only [imath]\tau[/imath] or [imath]\theta[/imath] does not change the output of the function [imath]f[/imath]? But I don't quite understand how do we know that... :/ ...little help...? -Anssi Quote Link to post Share on other sites

Doctordick 42,287 Posted July 6, 2010 Author Report Share Posted July 6, 2010 Hi Anssi, I want you to know that I really appreciate your attempts to understand each and every step in my presentation. If some of these other people would make that kind of examination, this whole thing would have been settled years ago. Again, mathematics is not an easy subject because sometimes one overlooks what would be utterly obvious if one were used to the manipulations being performed. It's a learning thing. I know you will kick yourself when you see what you overlooked. ...that move is based on the fact that every element is moving at constant velocity, so any change in "t" also entails a similar change in [imath]\sqrt{\tau^2 + r^2 + r^2\theta^2}[/imath]First, that statement is not exactly correct. The actual fact is that a change in “t” entails a change in tau, r and theta (these are the polar coordinates of the position). The problem is that [imath]\hat{r}[/imath] and [imath]\hat{\theta}[/imath] although orthogonal to one another point in varying directions during a change. If we were in rectilinear coordinates, x,y, tau (z being immaterial by the fact that the motion must be in the x,y plane), [imath]\sqrt{\tau^2 + x^2 + y^2}[/imath] would be the distance from the origin by the Pythagorean theorem. That theorem depends very much on the fact that [imath]\hat{\tau}[/imath], [imath]\hat{x}[/imath] and [imath]\hat{y}[/imath] continue to point in the same directions as the defined position changes. Opposed to this, [imath]\sqrt{\tau^2 + r^2 + r^2\theta^2}[/imath], is not an easily definable measure. On the other hand, [imath]\sqrt{\dot{\tau}^2 + \dot{r}^2 + r^2\dot{\theta}^2}[/imath] and [imath]\sqrt{\dot{\tau}^2 + \dot{x}^2 + \dot{y}^2}[/imath] are essentially the same thing: i.e., they are the magnitude of the instantaneous velocity. What you should have said is that “a change in t also entails a similar change in position”; however, [imath]\sqrt{\tau^2 + r^2 + r^2\theta^2}[/imath] is not the position. Just to be absolutely sure, by "three independent variables" you refer to [imath]\tau[/imath], [imath]r[/imath] and [imath]\theta[/imath]... (We are not concerned of [imath]M[/imath])That is correct! But let's get back to your real problem.Hmmm, I should probably understand by now why that is, but I'm a bit lost apparently... You are saying that changing only [imath]\tau[/imath] or [imath]\theta[/imath] does not change the output of the function [imath]f[/imath]? But I don't quite understand how do we know that... :/ ...little help...?First, look at the line immediately prior to the one you quote: i.e., just beforeHowever, we have one additional constraint which simplifies the problem considerably: [imath]\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2=c^2.[/imath]That line isFollowing the standard Euler-Lagrange attack, if the expression[math]f(t,y,\dot{y})=\frac{1}{\sqrt{1-\frac{2\kappa M}{c^2r}}}\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}[/math] satisfies the Euler-Lagrange equation, the variation of the integral over the implied path will vanish. In this case we have three independent variables to deal with (note that “y”, in this case, stands for the variable of interest) which leads to three independent differential equations. If [imath]\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2=c^2[/imath], then f must be [math]f(t,y,\dot{y})=\frac{c}{\sqrt{1-\frac{2\kappa M}{c^2r}}}:[/math] i.e., the only argument in “f” is “r” ; there is no dependence upon either tau or theta. (Actually, it is not necessary to make that substitution as all it really does is make f look simpler; however, it does hide an "r" dependence but that has no consequence as we don't solve the third equation.) The Euler-Lagrange equations of interest in this case (see equation #3) are,[math]\frac{\partial f}{\partial \tau}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)=0 [/math] [math]\frac{\partial f}{\partial \theta}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)=0 [/math] and [math]\frac{\partial f}{\partial r}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{r}}\right)=0[/math] The first two are trivial; if the differential with respect to t is zero one can integrate the function and obtain a constant which is exactly what I do (that constant is exactly identical to what occurs in the parenthesis between [imath]\frac{d}{dt}[/imath] and the dt being integrated over). [math] \int \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}\right\}\right)dt[/math]The third equation is quite complex but I don't need to solve it because I have the fact that [imath]\sqrt{\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2}=c[/imath]. I can use that relationship to obtain [imath]\dot{r}[/imath] from the solutions to the first two. Let me know if you have any trouble following the conversion from [imath]\dot{r}[/imath] and [imath]\dot{\theta}[/imath] to [imath]\frac{dr}{d \theta}[/imath]. I know you are enjoying this! You are aren't you? ;) :lol::lol: Have fun -- Dick Quote Link to post Share on other sites

AnssiH 36,649 Posted July 7, 2010 Report Share Posted July 7, 2010 Hi Anssi, I want you to know that I really appreciate your attempts to understand each and every step in my presentation. Thank you, that is always nice to hear :) First, that statement is not exactly correct......What you should have said is that “a change in t also entails a similar change in position”; however, [imath]\sqrt{\tau^2 + r^2 + r^2\theta^2}[/imath] is not the position. Ah right! Of course... But yes, what I tried to say was indeed "change in position" But let's get back to your real problem.First, look at the line immediately prior to the one you quote: i.e., just beforeThat line isIf [imath]\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2=c^2[/imath], then f must be [math]f(t,y,\dot{y})=\frac{c}{\sqrt{1-\frac{2\kappa M}{c^2r}}}:[/math] i.e., the only argument in “f” is “r” ; there is no dependence upon either tau or theta. Ah right that's what you meant, okay. By the way, all those parentheses in [imath]\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2=c^2[/imath], I guess they are there because we are taking the square of the derivative i.e. [imath]\left (\frac{dr}{dt} \right )^2[/imath], while writing [imath]\dot{r}^2[/imath] would be taken as [imath]\frac{dr^2}{dt}[/imath]? (the derivate of the squared r?) I'm not sure if that even makes any sense, perhaps it would be just ambiguous notation without the parentheses or something? (Actually, it is not necessary to make that substitution as all it really does is make f look simpler; however, it does hide an "r" dependence but that has no consequence as we don't solve the third equation.) The Euler-Lagrange equations of interest in this case (see equation #3) are,[math]\frac{\partial f}{\partial \tau}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)=0 [/math] [math]\frac{\partial f}{\partial \theta}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)=0 [/math] and [math]\frac{\partial f}{\partial r}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{r}}\right)=0[/math] Ahha, right I see. The first two are trivial; if the differential with respect to t is zero one can integrate the function and obtain a constant which is exactly what I do (that constant is exactly identical to what occurs in the parenthesis between [imath]\frac{d}{dt}[/imath] and the dt being integrated over).[math] \int \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{-\frac{1}{2}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}\right\}\right)dt[/math] Um, hmmm... I don't understand integration enough to know anything about how or why you can "integrate the function and obtain a constant if the differential with respect to t is zero"... :I I was looking at those equations in the OP and couldn't really understand the moves you do there, I suspect simply because I don't understand what you mean by that comment exactly... The third equation is quite complex but I don't need to solve it because I have the fact that [imath]\sqrt{\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2}=c[/imath]. I can use that relationship to obtain [imath]\dot{r}[/imath] from the solutions to the first two. Let me know if you have any trouble following the conversion from [imath]\dot{r}[/imath] and [imath]\dot{\theta}[/imath] to [imath]\frac{dr}{d \theta}[/imath]. Well, one step at a time... I know you are enjoying this! You are aren't you? ;) :lol::lol: Yes! :D -Anssi Quote Link to post Share on other sites

Doctordick 42,287 Posted July 7, 2010 Author Report Share Posted July 7, 2010 I'm not sure if that even makes any sense, perhaps it would be just ambiguous notation without the parentheses or something?Actually no; there is no good reason for those parentheses other than to emphasize the existence of the “dot” on top. I am reminded of a math professor I once had who got his Ph.D. in math long before the existence of air conditioners. The “dot” notation was invented by Newton and the [imath]\frac{d}{dt}[/imath] notation by Leibniz. My professor referred to them as “dot-ism” and “deism”. His position was that “deism” was preferable, as with “dot-ism” one often had to contend with flies performing unwanted differentiations. :lol: :lol: Your idea that [imath]\dot{r}^2[/imath] could be interpreted as the differential of [imath]r^2[/imath] is quite new to me. I suppose it could. Anyway, it does appear that “dot-ism” is fraught with possible erroneousness interpretations. I only used it because the web page on the Euler-Lagrange equation did.Um, hmmm... I don't understand integration enough to know anything about how or why you can "integrate the function and obtain a constant if the differential with respect to t is zero"... :ISorry about that. You do so well at interpreting what I say that I forget you don't have a formal education in integral calculus. You need to take a quick look at the definition of an indefinite integral. Expression #1 is what is commonly called an “indefinite integral”; indefinite in that its value is not known. As is commented there, it is also called an “antiderivative”; which would be the opposite of a derivative. Essentially, what that means is that if [imath]\int f(x)dx = F(x)[/imath] then [imath] \frac{d}{dx}F(x)=f(x)[/imath] by definition: i.e., one operation is the inverse of the other. The definite integral is when the integral sign has specific limits: i.e., [imath]\int^b_a f(x)dx [/imath]. In that case the (now “definite”) answer is F(b)-F(a). You might enjoy reading the Wikipedia entry for calculus terminology, both for its “history of calculus” entry and its excellent “Introduction” to calculating integrals. I know I did. But let's get back to your question. Changing the variable “x” to “t”, if [imath]\int f(t)dt = F(t)[/imath] then it must be true that [imath]\int \frac{d}{dt}\left\{F(t)\right\}dt = F(t)[/imath]; correct? But what happens to that expression if we look at the differential representation [imath] \frac{d}{dt}F(t)=f(t)[/imath] and, instead of looking at the differential of F(t), we look at G(t)=F(t)+C where “C” is any constant. Since the differential of a constant is zero, we have the fact that [imath] \frac{d}{dt}G(t)=f(t)+0\equiv f(t)[/imath]. This implies an interesting fact. If F(t)=0, then the “indefinite integral” is indefinite to the extent of some arbitrary constant: i.e., [imath]\int \frac{d}{dt}\left\{F(t)\right\}dt = 0 + C[/imath]. That is exactly the reason behind that adjective “indefinite” attached to indefinite integrals. Think about it a little and I think you will understand exactly what is going on there. Have fun -- Dick Quote Link to post Share on other sites

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