The Final Piece Of The Puzzle!

[AnssiH]

The constant, [imath]\frac{2h^2}{r_0}[/imath], must be [imath]\frac{2\kappa M}{c^2}[/imath]

Hmmm, I don't understand where that came from...

Trivial beyond trivial! All I am doing is comparing my equation of an ellipse,

 

[math]1-\frac{2 h^2}{r_0r}=1-\frac{h^2}{r^2_0}(1-\epsilon^2)-\frac{h^2}{r^4}\left(r'\right)^2-\frac{h^2}{r^2}[/math]

 

to the Schwarzschild solution

[math]\left(1-\frac{2\kappa M}{c^2 r}\right)=c^2l^2 -\frac{h^2}{r^4}(r')^2-\frac{h^2}{r^2}\left(1-\frac{2\kappa M}{c^2r}\right)[/math]

 

There is only one term proportional to one over r in both expressions. The equation for the ellipse has the coefficient [imath]\frac{2h^2}{r_0}[/imath] and the Schwarzschild solution has the coefficient [imath]\frac{2\kappa M}{c^2}[/imath]. If these terms have exactly the same coefficients are then the terms are identical: i.e., that statement tells you exactly what r₀ has to be to make them the same.

 

 

Ahha, I had to spend some time refreshing my memory about all this and the [imath]c^2 l^2[/imath] was throwing me off for a while, but now I got it, they were defined in the following steps in the post that you were clarifying here (indeed I would have never remembered this had the math tags not started working now, I've been backtracking for the last hour or so :D)

 

The constant, [imath]\frac{2h^2}{r_0}[/imath], must be [imath]\frac{2\kappa M}{c^2}[/imath] which implies

[math]r_0=\frac{c^2h^2}{\kappa M}[/math]:

 

Substituting that for [imath]r_0[/imath] in the second constant, one gets:

[math]1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)[/math]

 

Setting this equal to [imath]cl[/imath] and solving for epsilon...

 

[math]cl = 1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)[/math]

 

Subtracting one from both sides I get

[math]cl-1= -\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)\quad\quad or \quad\quad (cl-1)\frac{c^4h^2}{\kappa^2 M^2}=-(1-\epsilon^2)[/math]

 

Adding one to each side of that expression and taking the square root, we have

[math]\epsilon =\left\{1+(cl-1)\frac{c^4h^2}{\kappa^2 M^2}\right\}^{\frac{1}{2}}[/math].

 

Yup. This explanation from post #75 overlaps with your explanation in post #64, and there the above equation still has the sign error in it. Thought I'd mention because it might confuse readers.

 

Substituting these results into the resulting differential equation yields exactly the representation I presented: i.e., we get,

[math]\left(1-\frac{2\kappa M}{c^2r}\right)=c^2l^2-\frac{h^2}{r^4} \left(r'\right)^2-\frac{h^2}{r^2}[/math].

 

There's something that troubles me with this... I'll re-summarize step by step because backtracking is becoming a chore:

 

[math]1-\frac{2 h^2}{r_0r}=1-\frac{h^2}{r^2_0}(1-\epsilon^2)-\frac{h^2}{r^4}\left(r'\right)^2-\frac{h^2}{r^2}[/math]

 

Since [imath]\frac{2h^2}{r_0} = \frac{2\kappa M}{c^2}[/imath]:

 

[math]

1-\frac{2 kM}{c^2 r}=1-\frac{h^2}{r^2_0}(1-\epsilon^2)-\frac{h^2}{r^4}\left(r'\right)^2-\frac{h^2}{r^2}

[/math]

 

Since [imath]r_0 = \frac{c^2h^2}{\kappa M}[/imath]:

 

[math]

1-\frac{2\kappa M}{c^2r}=1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)-\frac{h^2}{r^4}\left(r'\right)^2-\frac{h^2}{r^2}

[/math]

 

Then, we are essentially setting [imath]c^2 l^2[/imath] to equal [imath]1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)[/imath] to end up with:

[imath]\left(1-\frac{2\kappa M}{c^2r}\right)=c^2l^2-\frac{h^2}{r^4} \left(r'\right)^2-\frac{h^2}{r^2}[/imath].

 

But when solving for epsilon in post #75 (quoted above) we set:

[math]

cl = 1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)

[/math]

 

So I was expecting to find that:

[math]c^2 l^2 = 1-\frac{\kappa^4 M^4}{c^8h^4}(1-\epsilon^4)[/math]

 

It looks to me like things have gone off-synch somehow at some point? I can't really figure out where. I don't even know why the epsilon was solved... :I But I guess there's a simple mistake when you set [imath]cl[/imath] in #75 and if so it would propagate itself to the solving for epsilon, but I'm not sure where else... (I'm starting to get confused, spent too much time backtracking and refreshing my memory :D)

 

-Anssi

[Doctordick]

Then, we are essentially setting [math]c^2 l^2[/math] to equal [math]1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)[/math] to end up with:

[math]\left(1-\frac{2\kappa M}{c^2r}\right)=c^2l^2-\frac{h^2}{r^4} \left(r'\right)^2-\frac{h^2}{r^2}[/math].

 

But when solving for epsilon in post #75 (quoted above) we set:

[math]

cl = 1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)

[/math]

 

So I was expecting to find that:

[math]c^2 l^2 = 1-\frac{\kappa^4 M^4}{c^8h^4}(1-\epsilon^4)[/math]

 

It looks to me like things have gone off-synch somehow at some point? I can't really figure out where. I don't even know why the epsilon was solved... :I But I guess there's a simple mistake when you set [math]cl[/math] in #75 and if so it would propagate itself to the solving for epsilon, but I'm not sure where else... (I'm starting to get confused, spent too much time backtracking and refreshing my memory :D)

Anssi, you simply have too much faith in the fact that my algebra is error free. The problem here is that when I got to the step where I had set r₀ in terms of kappa and M: i.e.,

[math]\left(1-\frac{2\kappa M}{c^2r}\right)=c^2l^2-\frac{h^2}{r^4} \left(r'\right)^2-\frac{h^2}{r^2}[/math].

 

And wanted to find the correct replacement for epsilon, I wrote down the equation to be solved as

[math]

cl = 1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)

[/math]

 

when I should have written down

[math]

c^2l^2 = 1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2).

[/math]

 

(See the first equation of the quote of you at the beginning of this post.) Clearly, if I solved that for the correct epsilon and substituted back in, I would have to get [math]c^2l^2[/math] back; so, presuming I had done the algebra correctly, I simply replaced

[math]

1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2) \quad with \quad c^2l^2.

[/math]

 

Unless I had made an error, that had to be the result. But the problem was, I had made an error. This is the kind of errors a person versed in the procedure makes; when one knows that no other answer is possible one tends to simply just write it down. I was doing the algebra for your sake, not for mine. Sorry about being sloppy.

 

I have fixed post #75 but not post #64. (I will take a look at that post later.) I note that, if you edit old posts which have the “math” tags, the system again strips out the backslashes. A pain in the butt by any measure.

 

I have taken a look at #64 and to fix it all I have to do is square that cl term in two equations; however, to accomplish that result, I would have to fix all the backslashes the system removes so I am going to just let it ride.

 

Have fun -- Dick

 

PS You should not have expected [math]cl = 1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)[/math] to yield [math]c^2 l^2 = 1-\frac{\kappa^4 M^4}{c^8h^4}(1-\epsilon^4)[/math]. You can't just square up all the terms in the result like that. You need to actually multiply the thing out. The correct procedure is as follows:

[math]

c^2l^2 = \left\{1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)\right\}^2

[/math]

 

[math]

= \left\{1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)\right\} \left\{1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)\right\}

[/math]

 

[math]

=1 \left\{1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)\right\}-\frac{\kappa^2M^2}{c^4h^2}\left\{1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)\right\}

[/math]

 

[math]

+\frac{\kappa^2 M^2}{c^4h^2}\epsilon^2\left\{1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)\right\}

[/math]

 

Which can be multiplied out further and then simplified a bit (but not much). You ought to be able to work it out the rest of the way if you want but it really isn't much fun.

[AnssiH]

when I should have written down

[math]

c^2l^2 = 1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2).

[/math]

 

Ah yeah okay... I suspected that's where the error was.

 

I have taken a look at #64 and to fix it all I have to do is square that cl term in two equations; however, to accomplish that result, I would have to fix all the backslashes the system removes so I am going to just let it ride.

 

Okay, I hope it won't confuse people. If you need to fix something like that later, it's probably easiest via using some sort of search/replace with a text editor (you can probably build a macro so the messages can be typically fixed with one button if needed be)

 

PS You should not have expected [math]cl = 1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)[/math] to yield [math]c^2 l^2 = 1-\frac{\kappa^4 M^4}{c^8h^4}(1-\epsilon^4)[/math]. You can't just square up all the terms in the result like that. You need to actually multiply the thing out.

 

Doh, of course :I That's the same exact silly mistake that I did just few posts back, I probably would have remembered had we not had this pause here... :P

 

I'll get around to continue from here soon...

 

-Anssi

[Doctordick]

If you need to fix something like that later, it's probably easiest via using some sort of search/replace with a text editor (you can probably build a macro so the messages can be typically fixed with one button if needed be)

I use a text editor to fix the math/latex notation but then you have to fix the inserted latexematics entries. And to fix the stripped backslashes one would need a program. For example, corrections have to satisfy the following : corrections are only to be within the tags, the backslashes are inserted before specific entries (before right, left, frac, etc) and also before {, but only if it follows a right or left. That is not a trivial program and when you do it by hand, it is easy to miss some entries so you have to go back and read the result carefully. It's a pain in the butt! I think I will let Tormod fix it. By the way, I noticed he quoted the corrected cl - epsilon relationship in the news letter. :lol:

 

Have fun -- Dick

[Bombadil]

I have been a little slow to respond. Actually this is partly due to not getting to it before the forum was changed and now as you know the latex was in somewhat of a messy state to say the least, and there still seems to be problems that need worked out. Any how I thought that I would go ahead and reply.

 

What we are looking for is the unique path from P₁ to P₂ where any differential variation in that actual path yields no change in that integral. The actual path required to generate that result is what is desired. As I said earlier, differential equations define functions. What we are interested in is that differential equation which defines a solution which corresponds to that path which makes the differential variation vanish.

 

So in much the same way as one would find a stationary point of a function with differentiation you are looking at the stationary point of a path integral by looking at the differential equation satisfied by the path that minimizes the path integral.

 

[math]\left(1-\frac{2\kappa M}{c^2 r}\right)=c^2l^2 -\frac{h^2}{r^4}(r')^2-\frac{h^2}{r^2}\left(1-\frac{2\kappa M}{c^2r}\right)+\left[\frac{h^2}{r^4}(r')^2\left(\frac{2\kappa M}{c^2r}\right)\right],[/math]

I have certainly shown gravitational forces can be reduced to geodesics in my geometry by virtue of the fact that I have just done so. The fact that my result is not exactly the same as that obtained from Einstein's theory is not too troubling. It is possible that I have made a subtle deductive error in the above as none of my work has ever been checked by anyone competent to follow my reasoning; however, in the absence of an error, my result must be correct as it is deduced and not theorized as Einstein's solution is.

 

Don’t you still have to solve this equation and substitute it back into the original integral to find the minimizing function? That is, wouldn’t both the desired minimizing function and the maximizing function satisfy this equation, the question is what is the minimizing function. Or dose it really not matter since we are looking at the consequences of interference and there will be interference along both paths?

 

You really need to get yourself admitted to a good college or university. You would probably do quite well.

 

Any more I don’t expect such a thing to happening. At one time maybe I thought about it but not at the present time. But I’m not going to let that stop me from having fun.

[AnssiH]

Finally, substituting these constants into the above equation, one obtains the differential equation for a Newtonian elliptical orbit, with the gravitational attraction at one of the foci, which has taken the form:

 

[math]\left(1-\frac{2\kappa M}{c^2r}\right)=c^2l^2-\frac{h^2}{r^4}\left(r'\right)^2-\frac{h^2}{r^2}[/math]

 

That this is the Newtonian elliptical orbit directly analogous to the given Schwarzschild solution is defensible by two significant observations. First, it is indeed an equation for an elliptical orbit (as is shown above) and second, it contains all the terms of Schwarzschild's solution except one. What is important here is that the missing term is proportional to [imath]\frac{1}{r^3}[/imath]

 

Hmmm... oh, do you mean:

 

It may then be seen that the Schwarzschild solution amounts to a small adjustment to the energy contribution of the angular momentum term of Newton's solution (the most important consequence being the experimentally varified precession of the orbit of Mercury).

[math]-\frac{h^2}{r^2}\quad\Rightarrow\quad -\frac{h^2}{r^2}\left(1-\frac{2\kappa M}{c^2r} \right).[/math]

 

Since;

[math]

-\frac{h^2}{r^2}\left(1-\frac{2\kappa M}{c^2r} \right)

\equiv

-\frac{h^2}{r^2} + \frac{2\kappa M h^2}{c^2r^3}

[/math]

 

The only difference between Newtonian form and Schwarzschild form is the additional [imath]\frac{2\kappa M h^2}{c^2r^3}[/imath]

 

And that corresponds to the famous precession of Mercury's orbit... I think I get it.

 

In the same vein, my "error" (if indeed it is an error) amounts to an equally small adjustment to the energy contribution of the radial momentum term. It should be noted that both Einstein's change to Newton's solution and my change to Einstein's solution are not only small but they also both contain an additional factor of one over r relative to their respective base terms. Whereas Einsteins theory makes no changes whatsoever in the radial term of Newton's equation,

[math]-\frac{h^2}{r^2}(r')^2\quad\Rightarrow\quad -\frac{h^2}{r^2}(r')^2,[/math]

 

my solution makes a very small change to that term:

[math]-\frac{h^2}{r^2}(r')^2\quad\Rightarrow\quad -\frac{h^2}{r^2}(r')^2\left(1-\frac{2\kappa M}{c^2r} \right).[/math]

 

I guess that should be [imath]r^4[/imath] in all of those.

 

I'll continue from here later...

 

-Anssi

[Doctordick]

I guess that should be [imath]r^4[/imath] in all of those.

As usual, you are absolutely correct. I have edited the OP to reflect your corrections. You will never know how much I appreciate your efforts in proof reading this stuff.

 

For the moment, Bombadil needs a little help.

 

See you later -- Dick

[Doctordick]

So in much the same way as one would find a stationary point of a function with differentiation you are looking at the stationary point of a path integral by looking at the differential equation satisfied by the path that minimizes the path integral.

Essentially yes; however the problem is much more difficult than the finding the stationary point of a function. That is what the Euler-Lagrange thing is all about.

 

Don’t you still have to solve this equation and substitute it back into the original integral to find the minimizing function?

No, whole issue of the Euler-Lagrange relationship is that the function which minimizes the path length must be a solution to the Euler-Lagrange equation.

 

That is, wouldn’t both the desired minimizing function and the maximizing function satisfy this equation, the question is what is the minimizing function.

Think about it for a moment. There is no single minimizing function (or maximizing function) for that matter. We are talking about the length of a path between two points where the instantaneous path length at any given point is a function of both that position and the velocity along that element at that position. The minimization is essentially equivalent to defining a straight line in a Euclidean geometry as the shortest distance between two points.

 

In this case, we are using a four dimensional Euclidean geometry and altering the measurement of distance into a change in the associated velocity. (The two are equivalent ways of looking at the same phenomena.) The result is what is normally called free-fall orbits and there are an infinite number of such orbits through every point; each with a different apparent (three dimensional) velocity. All such orbits must satisfy the Euler-Lagrange equation.

 

Now, is the solution a minimum or a maximum? Certainly you should see that, just as a straight line is a singular phenomena of an old fashion Euclidean geometry this solution has to be a minimum. In the geometry, even my somewhat distorted Euclidean geometry, no matter what path between two points you care to look at, I can plot a path which is longer (just think of a twisted involved path) by adding something between any two points on the path you provided. It follows that there is no “maximum length” path.

 

Or dose it really not matter since we are looking at the consequences of interference and there will be interference along both paths?

Here you are confusing two different issues. The solution to the problem has nothing to do with interference; it has to do with the actual mathematical description of differential path length and I have shown that altering the velocity is totally equivalent to altering the actual geometry. My arguments concerning the consequences of interference goes to the defense of the fact that average path lengths have to be a function of interaction density; a separate issue.

 

Any more I don’t expect such a thing to happening. At one time maybe I thought about it but not at the present time. But I’m not going to let that stop me from having fun.

People go back to college all the time. We just had some woman here who, after she retired from her job, got admitted to college and just graduated last spring. The sooner you get at it the better. God, I used to think I was old when I was fifty.

 

Have fun -- Dick

[AnssiH]

As usual, you are absolutely correct. I have edited the OP to reflect your corrections.

 

Looks like you made the correction in a bit of a haste:

 

Whereas Einsteins theory makes no changes whatsoever in the radial term of Newton's equation,

[math]-\frac{h^2}{r^2}(r')^4\quad\Rightarrow\quad -\frac{h^2}{r^4}(r')^2,[/math]

 

:D

 

Since the energy attributable to the radial momentum of Mercury is very small compared to the energy attributable to its orbital angular momentum, the perhelic shift of Mercury is certainly not a test of the validity of Einstein's theory as, within the accuracy of experiment, I obtain exactly the same result. The only test which could possibly remain is the deflection of star light. In that regard, please notice that in the above deduction, I selected my constants l and h so as to reproduce the exact form of Schwarzschild's solution. By doing so I forced my two constants of first integration to be related.

 

Under normal circumstances, one would simply write the first integrals as equal to an arbitrary constant: i.e., one would ordinarily write,

[math]\frac{\partial f}{\partial \dot{\tau}}= \left(1-\frac{2\kappa M}{c^2r}\right)^{-\frac{1}{2}}\frac{\dot{\tau}}{c}=\frac{A_1}{c} \quad and \quad \frac{\partial f}{\partial \dot{\theta}}= \left(1-\frac{2\kappa M}{c^2r}\right)^{-\frac{1}{2}}r^2\frac{\dot{\theta}}{c}=\frac{A_2}{c}[/math]

 

In this case, one would obtain a differential equation for the geodesic of the form

[math]A^2_1\left(1-\frac{2\kappa M}{c^2 r}\right)=c^2 -\frac{A^2_2}{r^4}(r')^2-\frac{A^2_2}{r^2}\left(1-\frac{2\kappa M}{c^2r}\right)+\left[\frac{A^2_2}{r^4}(r')^2\left(\frac{2\kappa M}{c^2r}\right)\right].[/math]

 

Walked through that over again now that you have got the division by [imath]c[/imath] in the definition of the constants, and got the exact same result.

 

...will continue from here later...

 

-Anssi

[Doctordick]

Looks like you made the correction in a bit of a haste:

I guess you got that right! I have fixed it. Thanks again.

 

Walked through that over again now that you have got the division by [imath]c[/imath] in the definition of the constants, and got the exact same result.

I guess I finally got something right. I have always been quite confident as to the outcome of my algebra but seemed to be somewhat sloppy when it came to the details. I think I get ahead of my self when working these things out.

 

Thanks for your help -- Dick

[AnssiH]

Under normal circumstances, one would simply write the first integrals as equal to an arbitrary constant: i.e., one would ordinarily write,

[math]\frac{\partial f}{\partial \dot{\tau}}= \left(1-\frac{2\kappa M}{c^2r}\right)^{-\frac{1}{2}}\frac{\dot{\tau}}{c}=\frac{A_1}{c} \quad and \quad \frac{\partial f}{\partial \dot{\theta}}= \left(1-\frac{2\kappa M}{c^2r}\right)^{-\frac{1}{2}}r^2\frac{\dot{\theta}}{c}=\frac{A_2}{c}[/math]

 

In this case, one would obtain a differential equation for the geodesic of the form

[math]A^2_1\left(1-\frac{2\kappa M}{c^2 r}\right)=c^2 -\frac{A^2_2}{r^4}(r')^2-\frac{A^2_2}{r^2}\left(1-\frac{2\kappa M}{c^2r}\right)+\left[\frac{A^2_2}{r^4}(r')^2\left(\frac{2\kappa M}{c^2r}\right)\right].[/math]

 

In this representation, the geodesic for a photon is immediately obtained by setting A₁ equal to zero (which corresponds to no initial motion in the tau direction

 

Right I see, because if you set it to zero, then according to [imath]\left(1-\frac{2\kappa M}{c^2r}\right)^{-\frac{1}{2}}\frac{\dot{\tau}}{c}=\frac{A_1}{c}[/imath], it means [imath]\dot{\tau}[/imath] must also be zero.

 

It is very interesting and somewhat expected; yet another demonstration of how definitions are defined by each others and the original symmetry constraints.

 

Again, except for the term in square brackets, the result is exactly the same result Schwarzschild obtains by setting his invariant interval to zero and resolving the problem.

 

Hmmm, I don't know how to perform that move to Schwarzschild solution, so I think I'm going to need to take that on faith, unless you want to point it out to me.

 

Here also, the term in square brackets can be seen as an energy adjustment related to that part of the photons energy which can be seen as due to it's radial motion. Clearly the deflection of star light by the sun does not test the existence of my additional term as, in grazing the sun, the impact of this term is negligible as the radial motion of the photon is only comparable to the angular motion for large r.

 

Hmmm, well this gets slightly complicated I think, especially due to my lack of familiarity with traditional physics. If I think of the situation in terms of a single star and a single photon, then I suppose I could view the photon's almost straight path as a small portion of a very large orbit, in which case its radial velocity would be small at the perihelion, but would become very large very quickly. I'm not sure if its radial momentum is taken as "large" at that point (since it doesn't have rest mass), but nevertheless I have no idea how such expectations could be measured, and whether it's valid to just ignore the rest of the universe there.

 

At any rate, considering the actual observations that exist about the deflection of star light, I certainly cannot judge whether your extra term would produce a measurable difference to the expectations produced by Schwarzschild solution. And yes I can see that the term [imath]\left[\frac{A^2_2}{r^4}(r')^2\left(\frac{2\kappa M}{c^2r}\right)\right][/imath] should bring fairly small adjustment to the rest of the equation since either [imath]r'[/imath] is small or [imath]r[/imath] is large at all times.

 

With this final result, I have clearly demonstrated that Einstein's assertion of Minkowski space-time is in no way necessary to explain the validity of either the Lorentz transformations or the so called tests of general relativity. I think it is very possible that Einstein would have given me some serious thought had he lived to see what I have done.

 

Well I would think so too. More than once, he was pushing ideas that went against the grain of common intuition... Relativity, photons, and supporting de Broglie hypothesis while most of physics community didn't think it made sense. I think his attitude was usually that the truth can be seen in the relationships between valid definitions, and "doesn't make sense" doesn't really come to play in that game. Even if he did walk into that exact pitfall himself few times in his career too.

 

I find it interesting that Einstein himself had doubts concerning field theory. In particular with regard to Qfwfq's insistence that the correct way of doing quantum mechanics was to quantize the fields: i.e., he considered field theory to be more fundamental than quantum mechanics whereas my position was quite the reverse. (Qfwfq never responded to that post.)

 

I also don't think he quite picked up that you were talking about the nature of knowledge, not ontological beliefs.

 

In addition to the above, there might be some evidence that Einstein's theory is actually wrong. Note that, if my work is correct (and I certainly admit I could be wrong) the difference between my solution and Einstein's solution would be that simple factor multiplying the [math](r')^2[/math] term

[math]\sqrt{1-\frac{2\kappa M}{c^2r}}[/math].

 

.

.

.

 

I'll work that last bit out a bit later, when I can focus a bit better... (but yes the prospect that this could explain few things that modern physics currently doesn't, is very interesting)

 

...I'll continue from here soon...

 

-Anssi

[Doctordick]

I'll work that last bit out a bit later, when I can focus a bit better... (but yes the prospect that this could explain few things that modern physics currently doesn't, is very interesting)

I just noticed a rather confusing fact about that last comment of mine you quote. The term I was referring to is the right term except for the fact of the square root. In the reference I refer to the term as being related to the square of r prime: i.e., the correct comparison would be to r prime itself. The term multiplying r prime squared should be without the square root; it has to be squared. Sorry about that, I have edited the OP.

 

Thanks again -- Dick

[AnssiH]

I just noticed a rather confusing fact about that last comment of mine you quote. The term I was referring to is the right term except for the fact of the square root. In the reference I refer to the term as being related to the square of r prime: i.e., the correct comparison would be to r prime itself. The term multiplying r prime squared should be without the square root; it has to be squared. Sorry about that, I have edited the OP.

 

Oh okay, that's nice to hear because that is exactly what I could not immediately figure out, and thought I'm probably just missing something. But I was feeling too cloudy to try and think it through, I'm sure I would have realized it's just a simple error after a while :)

 

-Anssi

[AnssiH]

In addition to the above, there might be some evidence that Einstein's theory is actually wrong. Note that, if my work is correct (and I certainly admit I could be wrong) the difference between my solution and Einstein's solution would be that simple factor multiplying the [math](r')^2[/math] term

[math]\left(1-\frac{2\kappa M}{c^2r}\right)[/math].

 

If you solve the Schwarzschild solution to Einstein's field equations for “dr” and then divide by “dt” you will obtain

[math]\frac{dr}{dt}= \frac{r^2}{h}\left[c^2l^2-\left(1-\frac{2\kappa M}{c^2 r}\right)\left(1+\frac{h^2}{r^2}\right)\right]^{\frac{1}{2}}\frac {d\theta}{dt}[/math]

 

Hmm, looks like I ran into some trouble immediately. I tried to find my way into that result, but the closest I got is:

 

[math]

\frac{dr}{dt}

=

\frac{r^2}{h}

\left [c^2l^2

-

\left(1-\frac{2\kappa M}{c^2 r}\right)

-

\frac{h^2}{r^2}\left(1-\frac{2\kappa M}{c^2r}\right)

\right ]^{\frac{1}{2}}

\frac{d\theta}{dt}

[/math]

 

Can't really find my way through the rest of the way, and started to suspect if there's an error, or if I'm just missing something... Either way, little help...? :)

 

-Anssi

[Doctordick]

Sorry Anssi but your algebraic skills just aren't up to snuff. The missing step is so easy that I know you will be kicking yourself when you see it.

 

Hmm, looks like I ran into some trouble immediately. I tried to find my way into that result, but the closest I got is:

[math]

\frac{dr}{dt}=\frac{r^2}{h}\left [c^2l^2-\left(1-\frac{2\kappa M}{c^2 r}\right)-\frac{h^2}{r^2}\left(1-\frac{2\kappa M}{c^2r}\right)\right ]^{\frac{1}{2}}\frac{d\theta}{dt}

[/math]

 

Can't really find my way through the rest of the way, and started to suspect if there's an error, or if I'm just missing something... Either way, little help...? :)

Take a careful look at the last two terms inside the square root in your result. They are exactly -1 times the following two terms:

[math]

\left(1-\frac{2\kappa M}{c^2 r}\right)+\frac{h^2}{r^2}\left(1-\frac{2\kappa M}{c^2r}\right)

[/math]

 

The first term is exactly "one" times the factor [math]\left(1-\frac{2\kappa M}{c^2 r}\right)[/math]. And the second is exactly [math]\frac{h^2}{r^2}[/math] times the same factor. That means that the two terms added together are exactly [math]\left(1+\frac{h^2}{r^2}\right)[/math] times that same factor. That makes your result exactly the same as my result.

 

Another way to see the same thing is to realize the [math]\left(1-\frac{2\kappa M}{c^2r}\right)[/math] can be factored out of those last two terms. In some ways it is nice that such things don't appear obvious to you; it helps you spot my errors. Sorry about not making an error this time.

 

Thanks -- Dick

[AnssiH]

Ahha, okay that's about clear then....

 

Add to this the fact that conservation of angular momentum requires [math]v_\theta = r\frac{d\theta}{dt}[/math] = a constant or [math]\frac{d\theta}{dt}=\frac{C}{r}[/math] and we have,

[math]\frac{dr}{dt}= \frac{r}{h}\left[c^2l^2-\left(1-\frac{2\kappa M}{c^2 r}\right)\left(1+\frac{h^2}{r^2}\right)\right]^{\frac{1}{2}}C=F®.[/math]

 

Yup, got the same result.

 

If you use exactly the same procedure to solve my equation for [math]\frac{dr}{dt}\sqrt{1-\frac{2\kappa M}{c^2 r}}[/math] you will obtain exactly the same result: i.e.,

[math]\frac{dr}{dt}\sqrt{1-\frac{2\kappa M}{c^2 r}}=F®,[/math]

 

where F( r ) is exactly the same function obtained from Schwarzschild's solution.

 

Indeed, so I do! :D

 

This implies that the radial velocity obtained from my solution multiplied by [math]\sqrt{1-\frac{2\kappa M}{c^2 r}}[/math] must be exactly Schwarzschild's solution. That, in turn, implies that the radial velocity for a given radius will be greater in my case than it was in Schwarzchild's solution (or Newton's solution for that matter); however, if the object is far far away, the accurate measurement here is the radial velocity itself, not the exact value of r. The radial velocity is determined via the Doppler effect and, with an accurate clock on board the object, its velocity can be measured quite accurately.

 

Right, um, you are referring to measuring how fast something is receding from us via measuring a doppler effect of some light source.

 

The radius on the other hand would be estimated by solving for the orbital motion of the object (using either Einstein's field theory or Newtonian gravitational theory).

 

So, the radius is measured with the assumption that Schwarzschild's solution is correct, and if it isn't, we also get a corresponding error to our results.

 

Okay I'm trying to be a bit careful here because it's easy to get confused... When you say "solving for orbital motion", do you mean the standard procedure here is to measure the velocity and direction of the object (I have no idea how... Can it be done exclusively with doppler measurement?), and from that data figure out the orbit using Schwarzschild solution. Your solution would yield different results, in that in your plotted orbit, you would expect the radius to increase faster while the object is moving towards the aphelion.

 

Since the velocity is slowing as an object leaves the solar system, if my equation is the correct calculation, Schwarzschild would presume the object was somewhat further away than I would (I would have obtained that same value at a somewhat closer radius). If the object were actually slightly closer to the sun than they thought, its deceleration would be slightly greater.

 

Now I'm confused, I think I misunderstood something because in my mind, if the future of the object was predicted by figuring out its orbit, and if your result generated an orbit where the radius was increasing more rapidly, then I would have said Schwarzschild would presume to object to be closer than you would. But with my lacking knowledge, there are just few too many unknowns in what you said, so maybe you could clarify a bit...

 

That is apparently what the measurements on that recent satellite leaving the solar system have yielded.

 

Yeah the so-called Pioneer anomaly. I'm wondering if the anomalous orbital speeds of large galaxies could be affected by the same issue:

http://en.wikipedia.org/wiki/Galaxy_rotation_curve

 

I really don't know how the associated parameters are measured so it's a bit difficult to judge.

 

It appears (at least to me at the moment) that the effect of that extra term in my solution is to make the gravitational field appear to be slightly stronger than estimated via Einstein's field theory. If that conclusion is correct, then it could also explain the “dark matter” problem.

 

Yeah okay so you are basically referring to the galaxy rotation curve anomaly here as well I guess. Anyway, I really don't understand what you are saying properly, but it's very interesting, so let's discuss this a bit more.

 

It would also be interesting to hear if Qfwfq has got any comments to this little detail.

 

-Anssi

[Bombadil]

No, whole issue of the Euler-Lagrange relationship is that the function which minimizes the path length must be a solution to the Euler-Lagrange equation.

 

I’m assuming that the converse is also true in this case, that is any solution to the Euler-Lagrange equation is a minimizing path and there are no stationary paths. Actually the idea of a stationary path doesn’t make much sense as it would imply the existence of a path that any infinitesimal change in would result in the path having the same length, at least in the limiting case.

 

Now, is the solution a minimum or a maximum? Certainly you should see that, just as a straight line is a singular phenomena of an old fashion Euclidean geometry this solution has to be a minimum. In the geometry, even my somewhat distorted Euclidean geometry, no matter what path between two points you care to look at, I can plot a path which is longer (just think of a twisted involved path) by adding something between any two points on the path you provided. It follows that there is no “maximum length” path.

 

So after we have defined the initial conditions, is the shortest path and hence the solution to the equation that you have derived define a unique path in the geometry that you have defined. Or is there some further requirements needed to define a unique path for an object to follow? That is, will the differential equation you have derived on its own define a unique equation or is some further constraint needed to insure uniqueness.

 

Also I am wondering if we can expect this equation to be solvable and if it is are we more likely to learn more from studying the equation itself then we would from actually solving it?

 

If you use exactly the same procedure to solve my equation for [math]\frac{dr}{dt}\sqrt{1-\frac{2\kappa M}{c^2 r}}[/math] you will obtain exactly the same result: i.e.,

[math]\frac{dr}{dt}\sqrt{1-\frac{2\kappa M}{c^2 r}}=F®,[/math]

 

where F( r ) is exactly the same function obtained from Schwarzschild's solution. This implies that the radial velocity obtained from my solution multiplied by [math]\sqrt{1-\frac{2\kappa M}{c^2 r}}[/math] must be exactly Schwarzschild's solution. That, in turn, implies that the radial velocity for a given radius will be greater in my case than it was in Schwarzchild's solution (or Newton's solution for that matter); however, if the object is far far away, the accurate measurement here is the radial velocity itself, not the exact value of r. The radial velocity is determined via the Doppler effect and, with an accurate clock on board the object, its velocity can be measured quite accurately.

 

Doesn’t this imply that there is a limit to the mass per radius that is allowed otherwise the term in the square root could equal zero for large masses which would seem to imply an infinite radial velocity limit? Or a negative under the square root which seems no less problematic.

 

Also is this the only place that a difference between your equation and the Schwarzschild solution exists or is there other differences that exist?