The Final Piece Of The Puzzle!

[Doctordick]

So rather then use a fixed speed of light and see distance as scaled and try to find the shortest path by distance' date=' we see the distance as fixed and see light with a variable speed and then calculate the path that takes the shortest change in a fixed clock?[/quote'] Essentially, yes. It is certainly not the picture I use to examine the whys and wherefores standing behind my analysis but it is certainly a convenient model of the macroscopic outcome.

Will constructive interference between the elements that make up the object result in any related effects?

Constructive interference inherent in wave phenomena is the central theme behind analysis of “refraction”.

 

Take a look at the first image on this page. Each wave can be constructed by looking at the constructive interference of waves emanating from the previous wave.

I am assuming that it is the t coordinate you are using for this purpose.

”t” isn't a coordinate' date=' it is no more than a parameter expressing the evolution of path length in the Euclidean geometry used to display the information of interest. On a microscopic scale, the interactions produce “crooked paths” which, when looked at from a macroscopic scale of the “object” in an average perspective (as embedded in an object) appear to be slower: i.e., cover less distance than if the interactions weren't there.

If this is the case is it this path that we are looking for and not the actual scaling factor?

I do not understand your question.

I don’t like using the word ‘time’ here

Why not? It's nothing more than a parameter defining the distance an element moves in the Euclidean geometry being used.

But how do we have any idea of what the dependence of [imath]\Phi(\vec{x})[/imath] on r is.

We are simply talking about any interaction representable by particle exchange. The only important factor there is the one over r dependence which is purely a function of the geometry (the surface area of the exchange probability at a given distance). I gave an earlier argument as to the fact that the interaction had to be proportional to M (the mass of the source object) thus' date=' in the final analysis, kappa is nothing more than a parameter setting the strength.

Also I’m not sure how you conclude that the term that we are looking for will be the potential produced by the object in question.

I don't think you understand what I am doing. I am looking at the subtle consequences of exchange interactions between macroscopic objects. In order to understand that you have to understand the exchange interactions themselves.

 

I am sorry you are having so much trouble -- Dick

[Qfwfq]

That is a direct consequence of your failure to comprehend that “finite” can be a much larger number than you personally presume can be dealt with.

Do you mean, like, a googolplex raised to the power googolplex? Even larger maybe?

 

I don't know where you got that idea about me, just call it one of our many misunderstandings. :doh:

 

Mathematics is a tool which can extend logic far beyond what can be comprehended by a human mind. My proof is one of the best examples of that power.

Gosh there are some mightier examples, ones that mathematicians find no trouble in considering conclusive. Usually it's because the author's presentation isn't an impediment to the reader's comprehension.

 

The issue here is the fact that one of the primary axioms of my proof is: that “I know nothing of the true character of the information I am trying to explain” so, for you to accuse me of assuming I know how the universe works, does indeed smack of gall.

I think I already said, in the correct thread, that I hadn't accused you of that. You and Anssi both failed to see my point there.

 

It is very hard (if not impossible) to fight religious beliefs.

Which one are you implying that I hold?

[AnssiH]

In that same reference, you should look a equation 12.11 on page 52. Hooft comes up with

[math]\left(1-\frac{2 M}{r}\right)=E^2-\frac{J\left(r'\right)^2}{r^4}-J^2\left(1-\frac{2 M}{r}\right)\frac{1}{r^2}[/math]

 

which compares favorably with my

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

 

particularly if you look back to 11.41 on page 48. By the way he defines the potential, he is essentially using M to represent what I set as [imath]\frac{\kappa M}{c^2}[/imath]. His “E” corresponds to my [imath]c^2l^2[/imath] and his J corresponds to my h. Other than that, you have apparently had no problem doing the algebra.

 

Right okay... (Your quote of Hooft's equation is missing the squaring for the other J though)

 

Sorry about that. Fifty years ago the issue was obvious to me. Today I had to spend several hours trying to reconstruct the central aspects of the problem. Senility is a hard master and I am certainly up against that wall. Nevertheless, I think I have constructed a rational defense of the statement that, by choosing the appropriate constants one can compare Schwarzschild's solution to the classical result. The following is an attempt to make the issues clear.

 

First, you should take a look at the Kepler orbits. Check out equation #13 of this Wikipedia entry.

 

The equation given as the Kepler orbit in polar coordinates with the origin at the focal point is

[math]r=\frac{p}{1+e\cdot cos \theta}[/math].

 

Phew, that took a while for me to dissect, I just had no idea what the parameters are supposed to be....

 

But after reading up around the web a bit, my understanding is that it is an equation describing an elliptical orbit of a planet, in terms of classical mechanics. So [imath]\theta[/imath] is the angle of the planet around the sun, and e is an eccentricity parameter between 0 and 1... [imath]cos \theta[/imath] yields 1, 0, -1, and 0 around the orbit, i.e. multiplies the eccentricity parameter differently at different points of the orbit; with 0 eccentricity it yields no changes to "r", and with eccentricity close to 1 it yields very elliptical orbits, as you would expect for any comet-like orbits with high eccentricity.

 

This was one of the helpful pages:

Kepler's laws of planetary motion - Wikipedia, the free encyclopedia

 

So am I correct to assume that "p" is simply defined as "the radius of the orbit when [imath]\theta[/imath] is at 90 degrees", i.e. at halfway between the perihelion and aphelion? They say "p is the semi-latus rectum", but I have no idea what that means. It just seems like some pulled a fancy name out of their... um... what's the word I'm looking for... yeah, their hat.

 

Anyway, I think I have an idea of what it is describing and how it works.

 

You should understand that, except for the actual definitions of the various constants, that is identical to my expression [imath]\frac{1}{r}=\frac{1}{r_0}(1-\epsilon \;cos (\theta))[/imath]: i.e., their “p” is my r₀, their “e” is my [imath]\epsilon[/imath] and their theta is 180 degrees greater (or less than) my theta.

 

i.e. in your definition, [imath]\theta[/imath] at 0 corresponds to aphelion while in the wikipedia definition it corresponds to perihelion...

 

Since these expressions are essentially the same, I will work with my representation.

 

The first step is to take the derivative of the equation with respect to theta: i.e., [imath]r'=\frac{dr}{d \theta}[/imath]. That will yield the following:

[math]-\frac{1}{r^2}r'=\frac{\epsilon}{r_0}sin (\theta)=\frac{\epsilon}{r_0}\sqrt{1-cos^2(\theta)}[/math]

 

I've been looking at that for a while but I can't figure out what you did there... :( Little help?

 

(I walked through the subsequent algebra little bit and looked valid so far)

 

-Anssi

[AnssiH]

Squaring both sides we have

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

 

The original equation can be solved for the value of [imath]cos(\theta)[/imath]:

[math]cos (\theta)=\frac{1}{\epsilon}\left\{1-\frac{r_0}{r}\right\}[/math].

 

Yup

 

Substituting that expression into our differential equation, we now have

[math]-\frac{1}{r^4}\left(r'\right)^2=\frac{\epsilon^2}{r^2_0} \left(1-\frac{1}{\epsilon^2}\left\{1-\frac{r_0}{r}\right\}^2\right)= \frac{\epsilon^2}{r^2_0}-\left\{\frac{1}{r_0} -\frac{1}{r}\right\}^2=\frac{\epsilon^2}{r^2_0}-\frac{1}{r^2_0}+ \frac{2}{r_0r}-\frac{1}{r^2}=-\frac{1}{r^2_0}(1-\epsilon^2)+\frac{2}{r_0r}-\frac{1}{r^2}[/math]

 

I don't actually know how to get to that third expression [imath]\frac{\epsilon^2}{r^2_0}-\left\{\frac{1}{r_0} -\frac{1}{r}\right\}^2[/imath], but I managed to walk my way to the expressions after that. Albeit I did get confused with the negatives many times and had to start over :D So right now, most of my confidence to my own result comes from the fact that we got the same result; it's always possible we have made the same mistake...

 

So now I do a little adjusting to make that look more like Schwarzschild's solution. If we multiply this through by [imath]h^2[/imath] and rearrange the terms (then, as a final step, add one to each side) the above equation can be rewritten as:

[math]1-\frac{2h^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].

 

Now here I got [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]

 

So it looks like there is an error somewhere, can you confirm?

 

-Anssi

[Qfwfq]

Yup

Nope. I find another expression for the cosine and I don't see how to get that one.

 

So it looks like there is an error somewhere, can you confirm?

I see a few dubious things there too but I don't have such a hi-tech miracle as a pencil on me ATM.

[AnssiH]

Nope. I find another expression for the cosine and I don't see how to get that one.

 

Sorry for my unclear quote, I was just quoting post #64 from where I left off. If you look at #64, you can see the whole thing; by "original equation" DD is referring to [imath]\frac{1}{r}=\frac{1}{r_0}(1-\epsilon \;cos (\theta))[/imath], which is not visible in my quote... I should try to be clear with my quotes, but it's nice to see someone is watching because I am certainly not the most qualified person to find all the errors! :D

 

I see a few dubious things there too but I don't have such a hi-tech miracle as a pencil on me ATM.

 

Oh one of those things that work in zero gravity? :D

 

-Anssi

[Doctordick]

Right okay... (Your quote of Hooft's equation is missing the squaring for the other J though)

Anssi, you are a god send. It just goes to show one that proof reading their own stuff is a bear because your mind just doesn't see those little things. From the preface of my first attempt at publishing:

 

It should be clear, even to the uninitiated, that errors in the logical deductions only occur when an attack is newborn and are quickly eliminated by careful examination of those deductions. Errors in deduction are the easiest to eliminate and, in fact,seldom persist long enough to pervade the field. Certainly, if any idea survives long enough to be part of the body of knowledge passed from one generation to another, one can expect to find few if any errors in the deductions; too many people will have been led through those deductions to allow anything but extremely subtle errors to stand for long.

My work has never endured any such examination and it really is a necessary step. The first person (a professional physicist by the way) that I got to read my stuff found an error and utterly refused to read any further even after I provided the correction. He said, “you can alibi errors till the sun goes down and I don't have time to pay any attention to trash!”. Anssi, I thank you again.

 

I have fixed the “J” squared thing and added two steps in the displayed algebra going from

[math]-\frac{1}{r^4}\left(r'\right)^2=\frac{\epsilon^2}{r^2_0} \left(1-\frac{1}{\epsilon^2}\left\{1-\frac{r_0}{r}\right\}^2\right)\quad\quad to\quad\quad=-\frac{1}{r^2_0}(1-\epsilon^2)+\frac{2}{r_0r}-\frac{1}{r^2}[/math]

 

Note the sign error on the first term! Squarinng a negative yields a positive!

 

[math]\frac{1}{r^4}\left(r'\right)^2=\frac{\epsilon^2}{r^2_0} \left(1-\frac{1}{\epsilon^2}\left\{1-\frac{r_0}{r}\right\}^2\right)= \frac{\epsilon^2}{r^2_0}-\frac{\epsilon^2}{r^2_0} \frac{1}{\epsilon^2}\left\{1-\frac{r_0}{r}\right\}^2= \frac{\epsilon^2}{r^2_0}-\frac{1}{r^2_0}\left\{1-\frac{r_0}{r}\right\}^2[/math]

 

[math]=\frac{\epsilon^2}{r^2_0}-\left\{\frac{1}{r_0} -\frac{1}{r}\right\}^2=\frac{\epsilon^2}{r^2_0}-\frac{1}{r^2_0}+ \frac{2}{r_0r}-\frac{1}{r^2}=-\frac{1}{r^2_0}(1-\epsilon^2)+\frac{2}{r_0r}-\frac{1}{r^2}[/math]

Originally, when I squared the left side of the differentiated equation (the step just prior to the above algebra) I didn't get rid of the minus sign on the r' term. I later discovered that error and had corrected the term immediately above the algebra (the algebra I just elucidated here) and the final result but didn't correct that complex algebraic expression at the time. (I have now fixed the posts; I left it in the line before my quote so you could see where the problem came from.) You also caught the fact that I dropped the minus sign on the first term of that result when I wrote down the next step. (See your last observation.)

 

I've been looking at that for a while but I can't figure out what you did there... :( Little help?

We want to differentiate both sides of the equation

[math]\frac{1}{r}=\frac{1}{r_0}(1-\epsilon cos (\theta))[/math]

 

You need to know the following information (common rules of differential calculus). The derivative of [math]Ax^n[/math] is given by [math]Anx^{n-1}x'[/math] where x' is the derivative of x. This implies that the derivative of [math]\frac{1}{x}\equiv (x)^{-1}[/math] is given by [math]-(x)^{-2}x'[/math]. It also implies that the derivative of a constant is zero ( since the x dependence can be written as [math]x^0[/math]). And finally you need to know that the derivative of the sine function is the cosine function and the derivative of the cosine function is minus the sine function. In this case, the derivative of [math]cos (x)\;\; is\;\;-sin(x)x'[/math].

 

Term by term, the derivative with respect to theta of [math]\frac{1}{r}[/math] is [math]-\frac{1}{r^2}r'[/math], the derivative of [math]\frac{1}{r_0}[/math] is zero and the derivative of [math]-\frac{1}{r_0}\epsilon cos (\theta) [/math] is [math]+\frac{1}{r_0}\epsilon sin (\theta)[/math] since the derivative with respect to theta of theta is “one”. Putting those terms together, one has, as the result of differentiating that equation, the following

[math]-\frac{1}{r^2}r'=\frac{\epsilon}{r_0}\sin (\theta)\equiv\frac{\epsilon}{r_0}\sqrt{1-cos^2 (\theta)}[/math]

 

Yup

Nope. I find another expression for the cosine and I don't see how to get that one.

I showed the start equation in the original post to my wife and she was able to solve it for [math]cos (\theta)[/math] in her head (no pencil) so I think your “Yup” was quite appropriate. Oh, she got the same thing we did.

 

So am I correct to assume that "p" is simply defined as "the radius of the orbit when [math]\theta[/math] is at 90 degrees", i.e. at halfway between the perihelion and aphelion? They say "p is the semi-latus rectum", but I have no idea what that means. It just seems like some pulled a fancy name out of their... um... what's the word I'm looking for... yeah, their hat.

Now that is really funny! I believe the word “rectum” in there has to do with the fact that they are referring to the “latus” orthogonal to the one through the foci (the same concept as “rectilinear”). Sometimes using Latin can be dangerous. :lol: Anyway, as far as I can tell, you got the whole thing right. (Note, Qfwfq is probably correct in his comment concerning the meaning of “semi-latus rectum”.)

 

I don't actually know how to get to that third expression [math]\frac{\epsilon^2}{r^2_0}-\left\{\frac{1}{r_0} -\frac{1}{r}\right\}^2[/math], but I managed to walk my way to the expressions after that.

I think my edited fix above should make the issue clear.

 

Albeit I did get confused with the negatives many times and had to start over :D So right now, most of my confidence to my own result comes from the fact that we got the same result; it's always possible we have made the same mistake...

Yeah, sign errors are easy to make. Actually, together I think we are doing a pretty good job of finding and fixing them. The real problem here is overlooking an error I have made and you have been doing an excellent job of catching them; including the two errors in

[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]

 

The first error (the one you caught) is a simple failure to include that negative sign on resultant first term on the right side of the equation (the error I mentioned above). The second error is the fact that you carried through my squaring error (also mentioned above). Thus the correct expression should be

[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]

 

(Neither of us managed to get the right result; but I think this has fixed the thing.) This result still has a form identical to the Schwarzschild solution except for two embedded constants so the form is preserved; the only change is that the solution for epsilon is incorrect (on redoing it I discovered my original result also included an algebraic error; I'll just plead senility) . Plus another error I am in the process of fixing with this edit [math]cl[/math] should be [math]c^2l^2[/math] in calculations of [math]\epsilon[/math].

 

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

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

 

which is exactly what I had before.

 

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

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

 

(I didn't square the kappa or M in the original post.) Anyway, setting this equal to [math]c^2l^2[/math] and solving for epsilon I get, via the following algebra,

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

 

Subtracting one from both sides I get

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

 

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].

 

Now, once again, I don't guarantee I have made no errors there; however, the issue that the equation is identical for appropriate values for those two constants is pretty fundamental. All I have done is solve for those constants. If you find an error, please let me know.

 

Oh, after I worked out the algebra above, which took several hours (I am slow, senile and plus, I had some other things to do today), I discovered your post to Qfwfq. It didn't dawn on me that he failed to look up what you were referring to, but that is pretty typical of him. He tends to scan things and then presume he knows what you are talking about. Typical of all trained physicists I have known (Erasmus included), they are 100% convinced that the rules of modern physics require the presumptions they are taught and any presentation which doesn't make those assumptions cannot possibly be correct. After all, Occam's razor says, “don't make any presumptions which are not required” and modern physics certainly has never done anything as dumb as that! :rolleyes:

 

I am afraid I have made an enemy of Qfwfq as he never once has explained the details of of any of his complaints; all he ever tells me is that what I am saying cannot be true. Notice that he never gave any information on his algebra in this case either. The only conclusion I can come to is that he is just trying to do his best to cast dispersions on my efforts. I am sorry but I get tired of giving people the benefit of the doubt when they don't return the favor. “You're a better man than I am, Anssi Hyytiäinen!” :shrug:

 

Have fun -- Dick

Edited September 12, 2010 by Doctordick
two more errors found by Anssi

[Qfwfq]

Anssi you're a naughty little boy! :naughty:

 

Do you know that anus can also be an elderly woman or Granny?

bibit soror, bibit frater

bibit anus, bibit mater

Apart from this anus means ring and rectum is the neutral for straight or right; if it ain't, you can always rectify it!

 

Note though that the semi latus rectum is not an orbital radius at all. The gravitating body is at one of the two foci, not half-way between, so when the orbiting object is at the end of the semi latus rectum the radius vector has a length greater than it and is not at a right angle to the major axis.

 

Sorry for my unclear quote, I was just quoting post #64 from where I left off. If you look at #64, you can see the whole thing; by "original equation" DD is referring to [imath]\frac{1}{r}=\frac{1}{r_0}(1-\epsilon \;cos (\theta))[/imath], which is not visible in my quote...

Now that's a relief! Yeah now with the corrected sign Dick already mentioned it matches up and, yeah, I can do that too, even without a gadget used by any космона́вт.

[Qfwfq]

I believe the word “rectum” in there has to do with the fact that they are referring to the “latus” orthogonal to the one through the foci (the same concept as “rectilinear”).

:P

 

I discovered just now that semilatus rectum is not the semiminor axis at all, I was wondering about this name and pondering what they might call the semimajor axis and, looking it up, I find it's a trickier thing.

Semilatus Rectum -- from Wolfram MathWorld

[AnssiH]

I have fixed the “J” squared thing and added two steps in the displayed algebra going from

[math]-\frac{1}{r^4}\left(r'\right)^2=\frac{\epsilon^2}{r^2_0} \left(1-\frac{1}{\epsilon^2}\left\{1-\frac{r_0}{r}\right\}^2\right)\quad\quad to\quad\quad=-\frac{1}{r^2_0}(1-\epsilon^2)+\frac{2}{r_0r}-\frac{1}{r^2}[/math]

 

Note the sign error on the first term! Squarinng a negative yields a positive!

 

Ahha, I was totally overlooking that one myself too :P

 

Substituting that expression into our differential equation, we now have

[math]\frac{1}{r^4}\left(r'\right)^2=\frac{\epsilon^2}{r^2_0} \left(1-\frac{1}{\epsilon^2}\left\{1-\frac{r_0}{r}\right\}^2\right)= \frac{\epsilon^2}{r^2_0}-\frac{\epsilon^2}{r^2_0} \frac{1}{\epsilon^2}\left\{1-\frac{r_0}{r}\right\}^2= \frac{\epsilon^2}{r^2_0}-\frac{1}{r^2_0}\left\{1-\frac{r_0}{r}\right\}^2[/math]

 

Thus far everything looks right to me...

 

[math]=\frac{\epsilon^2}{r^2_0}-\left\{\frac{1}{r_0} -\frac{1}{r}\right\}^2=\frac{\epsilon^2}{r^2_0}-\frac{1}{r^2_0}+ \frac{2}{r_0r}-\frac{1}{r^2}=-\frac{1}{r^2_0}(1-\epsilon^2)+\frac{2}{r_0r}-\frac{1}{r^2}[/math]

 

Now here I get different results starting from the first expression. I get [math]=

\frac{\epsilon^2}{r^2_0}-

\left\{\frac{1}{r^2_0} - \frac{1}{r_0 r} \right\}^2[/math]

 

So unless I'm missing something, it looks like you have accidentally dropped the square from that [imath]r^2_0[/imath], which was still there in the previous expressin. And that the error has propagated its way further. I think I shouldn't quote that post more until you have confirmed this...

 

Hmm, I am not really sure how I ended up with the same end result still though... It's hard to check now, as my russian space apparatus produced a lot of scribblings on many papers, because I got confused so many times along the way... Like I said, my confidence to my own result was pretty low because of that. It was a lot of scribblings also because I'm so unfamiliar still with algebra that I can't remember even some of the basic rules from the top of my head, and I keep trying to figure them out on the fly, which is really time consuming. But I suppose that is what allows me to spot mistakes, I have to look at every detail carefully since I don't yet have any expectations as to what should/would be there.

 

We want to differentiate both sides of the equation

[math]\frac{1}{r}=\frac{1}{r_0}(1-\epsilon cos (\theta))[/math]

 

You need to know the following information (common rules of differential calculus). The derivative of [imath]Ax^n[/imath] is given by [imath]Anx^{n-1}x'[/imath] where x' is the derivative of x.

 

Hmmm, I'm a bit confused here too... x' is the derivative of x, with respect to what? Itself? I.e. it's just "1"? That would make sense in that I can't see it included in wikipedia or mathworld explanation about this same issue derivatives, but then it doesn't make sense in that I can't figure out why you have it included... :I *scratching head*

 

ON THE SECOND THOUGHT, while trying to undestand the next paragraph it dawned on me that you probably mean "the derivative of x with respect to whatever variable we are interested of", i.e. if it happened to be "x" itself, it would be "1", and that is exactly why it's not even mentioned at Wikipedia et al... Right?

 

Term by term, the derivative with respect to theta of [imath]\frac{1}{r}[/imath] is [imath]-\frac{1}{r^2}r'[/imath],

 

where r' is the derivative of r with respect to theta...

 

the derivative of [imath]\frac{1}{r_0}[/imath] is zero

 

...since [imath]r_0[/imath] is constant...

 

and the derivative of [imath]-\frac{1}{r_0}\epsilon cos (\theta) [/imath] is [imath]+\frac{1}{r_0}\epsilon sin (\theta)[/imath] since the derivative with respect to theta of theta is “one”.

 

Hmm, okay so the fact that there's also [imath]-\frac{1}{r_0}\epsilon[/imath] in that term doesn't have any consequences? I don't really understand that in a way that would satisfy myself but I can take it on faith for now...

 

Putting those terms together, one has, as the result of differentiating that equation, the following

[math]-\frac{1}{r^2}r'=\frac{\epsilon}{r_0}sin (\theta)\equiv\frac{\epsilon}{r_0}\sqrt{1-cos^2 (\theta)}[/math]

 

Hmmm, I can't figure out how to get that last expression... :( [imath]sin (\theta)[/imath] doesn't seem to correspond to [imath]\sqrt{1-cos^2 (\theta)}[/imath] at all... What am I missing?

 

Phew, I'm starting to feel cloudy, I'll have a pause here... I guess I should wait for your response to the above questions anyway...

 

Oh, after I worked out the algebra above, which took several hours (I am slow, senile and plus, I had some other things to do today), I discovered your post to Qfwfq. It didn't dawn on me that he failed to look up what you were referring to, but that is pretty typical of him. He tends to scan things and then presume he knows what you are talking about.

 

Well Qfwfq doesn't strike me as someone deliberately trying to confuse matters... I remember he said once something to the effect of not really wanting to work out through the whole thing, and I guess people feel that the next best thing then is to keep a casual eye on things and voice up when they see individual assertions that seem wrong, like that quote of mine. Generally speaking, people don't backtrack; sometimes they'd have to backtrack to the very root of things through multiple threads, just to find out what the associated definitions are for whatever issue they are commenting on... So instead people they just comment on what they can immediately see with whatever 20 minutes they have at their hands. (dear lord it would be messy to explain relativity like this, how many people could pick up something like the relationship of simultaneity and lengthts, without having the original definitions understood :D)

 

And yeah, I'm sure Qfwfq feels frustrated too for not being able to speak the same language, for instance, I remember the issue with shift symmetry in probabilities, and the implications to the associated symmetries in [imath]\psi[/imath] via [imath]P=\vec{\psi}^\dagger \cdot \vec{\psi}[/imath]... Something about shift symmetries to [imath]\psi[/imath] itself and purely mathematical relationships... I had too many other things on my plate to pay good attention so I don't know what such circumstance might mean, but I should revisit that issue (among few others) at some point.

 

I am sorry but I get tired of giving people the benefit of the doubt when they don't return the favor. “You're a better man than I am, Anssi Hyytiäinen!” :shrug:

 

Heh... Well, Qfwfq, not that I think this is especially characteristic to you, but I think what especially rubs DD the wrong way is when people just blatantly point out something they see as "obviously wrong" without taking the trouble of at least asking why something is stated in so seemingly odd manner. I think that is exactly when he doesn't want to (once again) put in the effort to try and understand what the other party is thinking (the odds are, they have probably made up their mind already anyway :) ). He's had this analysis in his mind for decades in one form or another, and most of the issues are quite carefully thought out; I think it would be quite unlikely to have so obvious errors in there that someone could actually spot in matter of minutes. (That's not to say there aren't errors, trivial ones too)

 

Anssi you're a naughty little boy! :naughty:

 

Do you know that anus can also be an elderly woman or Granny?

 

Yeah I know few people like that.

 

bibit soror, bibit frater

bibit anus, bibit mater

Apart from this anus means ring and rectum is the neutral for straight or right; if it ain't, you can always rectify it!

 

Right... Hmm, but can anyone explain to me, why is it that the word "right" can mean either the direction "right", or "correct" in so many languages? And likewise, why "left" never seems to also mean "wrong"?

 

Note though that the semi latus rectum is not an orbital radius at all. The gravitating body is at one of the two foci, not half-way between, so when the orbiting object is at the end of the semi latus rectum the radius vector has a length greater than it and is not at a right angle to the major axis.

 

Yup, this was the picture that I found helpful:

File:Ellipse latus rectum.PNG - Wikipedia, the free encyclopedia

 

I.e. when the [imath]\theta[/imath] is at 90 degrees... Yeah, in terms of distance, that's not mid-way between perihelion and aphelion. But that means it is orthogonal to the major axis isn't it? It's just not the same thing as semi-minor axis, but it is parallel to it...

 

-Anssi

[Doctordick]

Now here I get different results starting from the first expression. I get [math]=

\frac{\epsilon^2}{r^2_0}-

\left\{\frac{1}{r^2_0} - \frac{1}{r_0 r} \right\}^2[/math]

You are missing a simple mathematical relationship known by anyone who works with algebra long enough to use it regularly; you have to be careful here. Notice that, in your expression, you can factor out one of those [imath]\frac{1}{r_0}[/imath] factors which, when you take it outside the curly brackets, becomes [imath]\frac{1}{r^2_0}[/imath]. You need to work this out in detail. I think you agree with expression immediately after I cancel out the two epsilon squared terms: i.e.,

[math]\frac{\epsilon^2}{r^2_0}-\frac{1}{r^2_0}\left\{1-\frac{r_0}{r}\right\}^2[/math].

 

In this case, it is convenient to see that product between squared factors in an alternate way:

[math]\frac{1}{r^2_0}\left\{1-\frac{r_0}{r}\right\}^2\equiv \frac{1}{r_0}\frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}\left\{1-\frac{r_0}{r}\right\}\equiv \frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}\frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}[/math].

 

[math]\equiv\left(\frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}\right)\left( \frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}\right)\equiv\left(\frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}\right)^2\equiv\left\{\frac{1}{r_0}-\frac{1}{r_0}\frac{r_0}{r}\right\}^2[/math]

 

Clearly the two [imath]r_0[/imath] in the second term cancel out and you get exactly what I got.

I'm so unfamiliar still with algebra that I can't remember even some of the basic rules from the top of my head, and I keep trying to figure them out on the fly, which is really time consuming. But I suppose that is what allows me to spot mistakes, I have to look at every detail carefully since I don't yet have any expectations as to what should/would be there.

Yes it is an advantage when it comes to proof reading. When it comes to doing algebra, when you get confused, the best bet is to do everything one step at a time as I did above. When you get competent, you can jump across steps easily; and easily make lots of errors. :wave:

Hmmm, I'm a bit confused here too... x' is the derivative of x, with respect to what?

That is what makes the “prime” notation so convenient; you don't have to specify what the derivative is respect to. It is nothing more than a convenient way of asserting the differentiation rules; that is why I used “x” instead of r and theta. The rule has nothing to do anything except what kind of function is being differentiated.

ON THE SECOND THOUGHT, while trying to undestand the next paragraph it dawned on me that you probably mean "the derivative of x with respect to whatever variable we are interested of", i.e. if it happened to be "x" itself, it would be "1", and that is exactly why it's not even mentioned at Wikipedia et al... Right?

You are right on the money!

where r' is the derivative of r with respect to theta...

That is absolutely correct. Just like any language, mathematics has expressions who's meanings must be picked up from context. One I like to point out to people is that F(x) can mean a function of x or it can also mean multiply x by F; this can get difficult if x=k+m and we get the expression F(k+m) so one must always be aware of the context. That seems to many to be a contradiction to the exactness of mathematics; but it really isn't as one of the rules of the language is that the context cannot allow more than one interpretation (a rule not seen in other languages).

Hmm, okay so the fact that there's also [imath]-\frac{1}{r_0}\epsilon[/imath] in that term doesn't have any consequences? I don't really understand that in a way that would satisfy myself but I can take it on faith for now...

The whole thing is just a constant; like that A I used when I gave you some of the differentiation rules. This is another case where the importance of context arises. That came up earlier when we were differentiating “c” (the speed of light) with respect to [imath]\dot{\tau}[/imath]. When you are doing math, you have to be aware of what you are talking about. Sorry about that; but you will get better with practice and, meanwhile, I'll keep you straight.

Hmmm, I can't figure out how to get that last expression... :( [imath]sin (\theta)[/imath] doesn't seem to correspond to [imath]\sqrt{1-cos^2 (\theta)}[/imath] at all... What am I missing?

In a word, trigonometry.

 

Check out the definitions of sine and cosine in that reference. The Pythagorean theorem, requires [imath]sin^2+cos^2=1[/imath] (if they are evaluated at the same angle). Actually that reference is a pretty good source for a number of important trigonometric relationships.

Phew, I'm starting to feel cloudy, I'll have a pause here... I guess I should wait for your response to the above questions anyway...

Actually, you are doing delightfully well considering your lack of experience with mathematics.

Well Qfwfq doesn't strike me as someone deliberately trying to confuse matters... I remember he said once something to the effect of not really wanting to work out through the whole thing...

Yeah, I know that; but considering his training in mathematics together with Occam's razor, “don't make any unnecessary presumptions” it seems to me should find my stuff rather fascinating. It is only his “religious belief” in the correctness of modern physics and the necessity of their presumptions that blocks him from seeing the earth shaking consequences of my discovery. So long as he avoids looking at the details, he will never comprehend just how significant the issue is.

 

Since my original paper is no longer available, it might be nice to quote my preface in order to point out the essence of my work.

Any scientific field may be seen as a body of assumptions (what I am referring to here are those things taken to be true without any examination) together with postulated relationships (and here I mean those things specifically held forth as the basis of the field including any specified assumptions) and the logical deductions which may be obtained from those relationships. Errors may occur in any of those three areas; however, the character and consequences of those errors vary quite considerably.

 

It should be clear, even to the uninitiated, that errors in the logical deductions only occur when an attack is newborn and are quickly eliminated by careful examination of those deductions. Errors in deduction are the easiest to eliminate and, in fact, seldom persist long enough to pervade the field. Certainly, if any idea survives long enough to be part of the body of knowledge passed from one generation to another, one can expect to find few if any errors in the deductions; too many people will have been led through those deductions to allow anything but extremely subtle errors to stand for long.

 

The category I refer to as postulated relationships are usually referred to as theories. Inventing theories and developing their logical deductions is the central work of the most esteemed in the scientific society. Errors in those theories are discovered through comparison to reality: i.e., experimentation. The process of designing and performing the experiments critical to a theory may take time but it is, none the less, a well understood process and sufficient diligence will eventually discover those errors.

 

That brings us to the errors in assumptions. Errors in these assumptions are a completely different issue. The primary problem with finding errors in the kind of assumptions I am referring to here is that the scientist usually has no idea of what they are. Remember, the kind of assumptions I am referring to here are those things which he assumes are true without thinking about them at all. If one reviews the history of science one will find that most of the major breakthroughs can be seen as flowing from the realization that their predecessors had made some subtle unexpressed assumption which was actually without foundation. Errors in these kinds of assumptions usually betray their presence by allowing seemingly contradictory results to be well defended.

My work concerns the issue of representing the problem to be solved (explaining anything) in a form which makes NO assumptions whatsoever. I have managed to do that and the results are absolutely fascinating (at least to Anssi and I).

 

Have fun -- Dick

[Qfwfq]

Well I guess I'm just a lazy bum that doesn't deserve even to point out that something doesn't match up with me, even granting the benefit of doubt in my wording.

 

P. S. I guess I'm just a sinister character, not rectus at all. :P

[Bombadil]

Constructive interference inherent in wave phenomena is the central theme behind analysis of “refraction”.

 

Take a look at the first image on this page. Each wave can be constructed by looking at the constructive interference of waves emanating from the previous wave.

 

So is essentially the path of an object in your four dimensional geometry going to be defined by where the probability wave being refracted encounters constructive and destructive interference with itself? And an object will follow the path of constructive interference? Or is this a different application of refraction and not the same as the one you are using?

 

”t” isn't a coordinate, it is no more than a parameter expressing the evolution of path length in the Euclidean geometry used to display the information of interest. On a microscopic scale, the interactions produce “crooked paths” which, when looked at from a macroscopic scale of the “object” in an average perspective (as embedded in an object) appear to be slower: i.e., cover less distance than if the interactions weren't there.

 

It is then, the length of the paths as seen on the macroscopic scale that you are interested in as on the microscopic scale there is no way that we can consider the interactions taking place due to how many there are. However when viewed on a macroscopic scale the object moving slower will result in it being refracted.

 

I do not understand your question.

 

From what I am understanding you are integrating the path length over a geodesic and using n as a scaling factor needed due to the length contraction that is taking place due to not being in an inertial reference frame.

 

This integral will be dependent on the scaling factor and the path being integrated over. So are we looking merely for the scaling factor or are we looking for the shortest path to integrate over?

 

There is one other thing that is kind of puzzling me, I assume that when you say…

 

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.

 

…you are referring to the path length however since you are setting the integral equal to zero and you are calling this the “variation of the path integral” I have to wonder if you have something else in mind. The problem is I just haven’t been able to find any definition for the “variation of the path integral” so I don‘t know if there is something else that you have in mind here.

 

Why not? It's nothing more than a parameter defining the distance an element moves in the Euclidean geometry being used.

 

I just want to make sure that the definition that you are using for time and the definition that is used in applying the Euler-Lagrange equation is the same. That is, you are in fact looking for a point where the path length has a stationary point . That is, the partial derivative of the path length to t is zero.

 

This looks like what you are looking for but I just can’t quite understand how path length can have a stationary point to time when time is in fact the path length. Unless I’m just not properly considering just how you are defining and finding path length.

 

We are simply talking about any interaction representable by particle exchange. The only important factor there is the one over r dependence which is purely a function of the geometry (the surface area of the exchange probability at a given distance). I gave an earlier argument as to the fact that the interaction had to be proportional to M (the mass of the source object) thus, in the final analysis, kappa is nothing more than a parameter setting the strength.

 

The point being that the value of this function must be proportional to the probability of interactions taking place, since the total number of particle at any given radius is a constant the density of partials at any given radius must be proportional to the radius and the probability of an interaction taking place must also be proportional to the radius from the source of the elements that would cause the interactions?

[Doctordick]

Or is this a different application of refraction and not the same as the one you are using?

No, it is no different. Anytime anything is described by wave phenomena, wave propagation is always essentially defined by interference phenomena. This Wikipedia article on the Double-slit experiment gives an excellent representation of the consequences of interference phenomena. You should read it.

It is then, the length of the paths as seen on the macroscopic scale that you are interested in as on the microscopic scale there is no way that we can consider the interactions taking place due to how many there are.

That statement is clearly not true as a decent reading of the Wikipedia article should confirm. It is exactly the microscopic analysis which explains refraction.

…you are referring to the path length however since you are setting the integral equal to zero and you are calling this the “variation of the path integral” I have to wonder if you have something else in mind. The problem is I just haven’t been able to find any definition for the “variation of the path integral” so I don‘t know if there is something else that you have in mind here.

There is a whole field of mathematics called “the calculus of variations”. As I have comment to you before, if you want to really understand mathematics, you need to get yourself into a good college or university and take some mathematics courses. The subject is in no way trivial and you really only have two choices; either just take what people say as valid without proof or take the subject seriously and get professional training.

 

I definitely get the impression that your understanding of calculus is somewhat limited. One of the first things one learns when they study calculus is the fact that “differential equations” essentially define functions in much the same manner as functions define how variables change with respect to arguments.

 

That symbol “[imath]\delta[/imath]” is the symbol used to denote a differential “variation” and is not the symbol of a partial derivative. The integral is a “path integral” meaning that the value of the integral depends on the path; which is not a given in the problem. What is given is how dl is dependent upon your position in the coordinate system: i.e., n is a scaling factor on dl given as a function of position in the coordinate system.

[math]\delta\int^{P_2}_{P_1}ndl=0[/math]

 

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.

 

That is exactly what the Euler-Lagrange equation is all about. As I said, the whole thing is a field of mathematics all to itself.

Unless I’m just not properly considering just how you are defining and finding path length.

I think that is exactly the difficulty. I am using the Euler-Lagrange relationship to define the path length of interest. And solving that Euler-Lagrange equation to find the actual path. The whole thing is essentially quite analogous to the definition of a straight line as the shortest distance between two points.

 

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

 

Have fun -- Dick

[AnssiH]

You are missing a simple mathematical relationship known by anyone who works with algebra long enough to use it regularly; you have to be careful here. Notice that, in your expression, you can factor out one of those [imath]\frac{1}{r_0}[/imath] factors which, when you take it outside the curly brackets, becomes [imath]\frac{1}{r^2_0}[/imath]. You need to work this out in detail. I think you agree with expression immediately after I cancel out the two epsilon squared terms: i.e.,

[math]\frac{\epsilon^2}{r^2_0}-\frac{1}{r^2_0}\left\{1-\frac{r_0}{r}\right\}^2[/math].

 

In this case, it is convenient to see that product between squared factors in an alternate way:

[math]\frac{1}{r^2_0}\left\{1-\frac{r_0}{r}\right\}^2\equiv \frac{1}{r_0}\frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}\left\{1-\frac{r_0}{r}\right\}\equiv \frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}\frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}[/math].

 

[math]\equiv\left(\frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}\right)\left( \frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}\right)\equiv\left(\frac{1}{r_0}\left\{1-\frac{r_0}{r}\right\}\right)^2\equiv\left\{\frac{1}{r_0}-\frac{1}{r_0}\frac{r_0}{r}\right\}^2[/math]

 

Clearly the two [imath]r_0[/imath] in the second term cancel out and you get exactly what I got.

 

Ahha, of course... My mistake was that I just assumed [imath]a(b+c)^2 = (ab+ac)^2[/imath] without thinking it through... :P That also explains why I got the correct result the first time I tried it; I just happened to get there via different route (I worked out the squaring first in a really obscure way)

 

So let's get back to:

 

Substituting that expression into our differential equation, we now have

[math]\frac{1}{r^4}\left(r'\right)^2=\frac{\epsilon^2}{r^2_0} \left(1-\frac{1}{\epsilon^2}\left\{1-\frac{r_0}{r}\right\}^2\right)= \frac{\epsilon^2}{r^2_0}-\frac{\epsilon^2}{r^2_0} \frac{1}{\epsilon^2}\left\{1-\frac{r_0}{r}\right\}^2= \frac{\epsilon^2}{r^2_0}-\frac{1}{r^2_0}\left\{1-\frac{r_0}{r}\right\}^2[/math]

 

[math]=\frac{\epsilon^2}{r^2_0}-\left\{\frac{1}{r_0} -\frac{1}{r}\right\}^2=\frac{\epsilon^2}{r^2_0}-\frac{1}{r^2_0}+ \frac{2}{r_0r}-\frac{1}{r^2}=-\frac{1}{r^2_0}(1-\epsilon^2)+\frac{2}{r_0r}-\frac{1}{r^2}[/math]

 

Walked through the rest of it reeaaaally carefully and now it looks all right to me...

 

So now I do a little adjusting to make that look more like Schwarzschild's solution. If we multiply this through by [imath]h^2[/imath] and rearrange the terms (then, as a final step, add one to each side) the above equation can be rewritten as:

[math]1-\frac{2h^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].

 

...as does that.

 

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...

 

which implies

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

 

...but yes that is what it would imply....

 

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]

 

Looks valid to me...

 

...setting this equal to [imath]cl[/imath] and solving for epsilon I get, via the following algebra,

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

 

Hmmm.. What does [imath]cl[/imath] correspond to again...? [imath]l[/imath] had something to do with path length...

 

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

 

Hmmm, I guess that latter one is supposed to read [math](cl-1)\frac{c^4h^2}{\kappa^2 M^2}=-(1-\epsilon^2)[/math] (I.e. that l should be 1)?

 

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].

 

Shouldn't that be [math] \epsilon = \left\{ 1 + (cl-1) \frac{c^4h^2}{\kappa^2 M^2} \right \}^\frac{1}{2}[/math]?

 

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].

 

Hmmm, okay, I think I am supposed to work with these:

 

[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]

 

and

 

[math]

\frac{2h^2}{r_0} = \frac{2\kappa M}{c^2}

[/math]

 

and

 

[math]

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

[/math]

 

But I can't make anything out of planting the epsilon in there... I.e. I don't know how to end up with just [imath]c^2l^2[/imath] for that one term... :I My attempt just went haywire from step one, and I started to suspect I am trying to combine wrong steps perhaps... :I

 

Yes it is an advantage when it comes to proof reading. When it comes to doing algebra, when you get confused, the best bet is to do everything one step at a time as I did above.

 

Hehe, yeah, sometimes I just don't realize the simple way of doing some things, and end up performing really obscure moves... :D

 

When you get competent, you can jump across steps easily; and easily make lots of errors. :wave:

 

I can hardly wait... :I

 

The whole thing is just a constant; like that A I used when I gave you some of the differentiation rules. This is another case where the importance of context arises. That came up earlier when we were differentiating “c” (the speed of light) with respect to [imath]\dot{\tau}[/imath]. When you are doing math, you have to be aware of what you are talking about. Sorry about that; but you will get better with practice and, meanwhile, I'll keep you straight.

 

Ahha yeah good. It is hard to remember the context as long as I have to put all my focus into understanding the details... I find myself backtracking a lot to remember what were we talking about again, one example of that is the issue of me not understanding what [imath]cl[/imath] corresponds to, I can't really remember what happened and some backtracking didn't make it any clearer to me yet... (I'm beginning to think it's just another "universal definition" which ends up having some supposed meaning in the context of modern physics?)

 

In a word, trigonometry.

 

Check out the definitions of sine and cosine in that reference. The Pythagorean theorem, requires [imath]sin^2+cos^2=1[/imath] (if they are evaluated at the same angle).

 

ah... and "doh".

 

I'll continue from here soon...

 

-Anssi

[Doctordick]

Ahha, of course... My mistake was that I just assumed [imath]a(b+c)^2 = (ab+ac)^2[/imath] without thinking it through... :P That also explains why I got the correct result the first time I tried it; I just happened to get there via different route (I worked out the squaring first in a really obscure way)

Yeah, that's the problem. Whenever it comes to complex algebra, mistakes are easy to make.

It should be clear, even to the uninitiated, that errors in the logical deductions only occur when an attack is newborn and are quickly eliminated by careful examination of those deductions. Errors in deduction are the easiest to eliminate and, in fact,seldom persist long enough to pervade the field. Certainly, if any idea survives long enough to be part of the body of knowledge passed from one generation to another, one can expect to find few if any errors in the deductions; too many people will have been led through those deductions to allow anything but extremely subtle errors to stand for long.

The problem is that my work has not gone through that process and there are minor insignificant errors all over the place. Actually, sometimes trivial algebraic errors can propagate through scientific papers for years because people just presume the stuff they are reading has been proofed. When I was a graduate student, I became good friends with a new chemistry professor (we were both interested in water pollution issues) who told me a story about a chemical experiment he had intended to demonstrate in a lecture. Prior to the lecture, he did the experiment and got an answer different from the one in the "Handbook of Chemistry and Physics" so he didn't actually do the experiment during the lecture and just talked about it as a demonstration of something (I have forgotten since that was over fifty years ago). But he did the experiment again later and got exactly the same “wrong” answer.

 

It took him about a year of research to finally discover that the problem was a typo in a paper published some twenty years before. The guy had gotten his “incorrect” result quoted so many places that the result had even made the “Handbook of Chemistry and Physics”, the very authority as to what the correct answer to these questions are supposed to be. So errors are hard to get rid of; particularly if a result isn't commonly measured or the algebra isn't commonly done.

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.

Hmmm.. What does [imath]cl[/imath] correspond to again...? [imath]l[/imath] had something to do with path length...

Comparing the expression I obtained from “Introduction to General Relativity” by Adler, Bazin and Schiffer with the expression your found in Hooft, [imath]cl[/imath] corresponds to E; however, if you look at Hooft's algebra (see page 51 of http://www.phys.uu.nl/~thooft/lectures/genrel.pdf ), you will see that he is doing nothing except pointing out that the differential equations require two constants (which he calls E and J). This is exactly the same thing I was doing when I came up with the constants cl and h.

 

Now these constants certainly reflect the fact that every orbit has specific values for these constants but it isn't really trivial to determine how these constants depend upon the commonly used physical constants actually defining the orbits. I'll just leave the issue for authorities to explain. Qfwfq or Erasmus, I am sure either of you can explain these things more clearly than I might. (Anssi, why don't you ask their assistance, if they fail to clear the issues up I will do so in a later post; I just thought that since they complain so much about the clarity of my explanations they might be able to do a better job.)

Hmmm, I guess that latter one is supposed to read [math](cl-1)\frac{c^4h^2}{\kappa^2 M^2}=-(1-\epsilon^2)[/math] (I.e. that l should be 1)?

Absolutely correct. I have fixed it and I have also fixed the sign error you found in the next expression.

Hmmm, okay, I think I am supposed to work with these:

 

[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]

 

and

 

[math]

\frac{2h^2}{r_0} = \frac{2\kappa M}{c^2}

[/math]

 

and

 

[math]

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

[/math]

Yes and no! You need to start with

[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]

 

solve the second, [imath]\frac{2h^2}{r_0} = \frac{2\kappa M}{c^2}[/imath], for r₀, obtaining

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

 

Then substitute that expression for r₀ into the start equation. (Since you obtained the expression by setting

[imath]\frac{2h^2}{r_0}=\frac{2\kappa M}{c^2}[/imath], you had should obtain the result

[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]

 

Now, if the factor [imath]1-\frac{\kappa^2 M^2}{c^4h^2}(1-\epsilon^2)[/imath] were to be [imath]c^2l^2[/imath] that equation would become exactly what we were looking for: the [imath]c^2l^2[/imath] term in Schwarzschild's solution. That is why we solved that equation for the value of epsilon necessary to make that term equal to the term in Schwarzschild's solution. If we made no errors in that solution for epsilon, the substitution has to yield back [imath]c^2l^2[/imath].

 

The real point here is that the specific elliptical orbit was specified by two constant parameters, r₀ and epsilon. The Schwarzchild' (or Hooft's) solution is also entirely specified by two constant parameters. All we actually showed was that there existed a transformation which would relate those parameters across the transformation from one representation (the Newtonian elliptical orbit) to an approximation to the other (the Einsteinian relativistic orbit). It just gives an easy way to compare the two solutions.

(I'm beginning to think it's just another "universal definition" which ends up having some supposed meaning in the context of modern physics?)

Not really. We are just looking at the specific form of constant terms in both solutions from a perspective which gives us an idea as to how relativity has altered the mathematical representation of the orbit. For large enough r and small enough M, the two orbits become identical and that is exactly what GR tells us is supposed to happen. Those are exactly the factors which require things to be close to the sun before we can detect the GR consequences.

 

I appreciate your efforts immensely and Happy birthday! I know the last four years seemed to go by awful fast. Watch out, the next ten will seem about that long and the twenty after that will go by just as fast. As my wife says, “time flies when your having fun”.

 

Have fun – Dick

[C. michael Turner]

Au contraire. I regard tenets of theology and theosophy to be fiction, not able (even in principle) to provide support. There is no way to falsify something that can't be tested. Let's hope that your "tau axis" doesn't fall into the same category.

 

 

 

 

I guess the key word there is "almost". There is where you might be able to test a prediction.

 

Obviously I disagree with your statement: "modern physics is itself a mere tautology: i.e., it tells us nothing about reality and is thus a religion."

 

For example, Einstein's general relativity tells us many things about the physical universe, 'reality' (and is thus not a religion).

 

 

 

 

Yet two hours ago in another thread ("the nature of time") you wrote "I am in the process of writing out my proof of my fundamental equation."

 

 

 

 

Your hypothesis is not solely that "an explanation must be internally self consistent!" There is more involved than just that.

 

 

 

 

I thought this thread was supposed to be The final piece of the puzzle!

 

 

CC

Okay guys, remember me, C. Michael Turner. I am helping out with this final piece

So that the author is then correct.

There are Two missing pieces to understand, Deductively Deduct, how the universe runs. I will dumb it down because it will help to see.

1). Spacetime is relative because the principles of all relativity are the same. Sound relativity exists as a Point of origin energy transfer into a wave with a fixed Speed, frequency and wavelength from a source. Key point - a source!!!

 

So space-time is generated by all matter and energy as a natural decay process. Each piece of energy is generating it's OWN space-time.

 

2). You are missing a fundamental wave interaction. The backaction of wavefront formation or

Wave tension or ready- "Gravity", yes gravity is the tension reaction dipole and monopole wave collision. Gravity is a Monopole wave collision back action tension that is relieved by the sources generating the waves spacially moving closer.

Ta -Da.

 

 

Time, space, and Gravity are actions of wave generation and wavefront formation. (including backaction)