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The Final Piece Of The Puzzle!


Doctordick

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You need to take a quick look at the definition of an indefinite integral. Expression #1 is what is commonly called an “indefinite integral”; indefinite in that its value is not known. As is commented there, it is also called an “antiderivative”; which would be the opposite of a derivative.

 

Essentially, what that means is that if [imath]\int f(x)dx = F(x)[/imath] then [imath] \frac{d}{dx}F(x)=f(x)[/imath] by definition: i.e., one operation is the inverse of the other.

 

Okay, after playing around with Wolfram Alpha a bit, I take it then that the indefinite integral [imath]\int f(x)dx[/imath] can be interpeted as the total area between 0 and [imath]x[/imath]...

 

The definite integral is when the integral sign has specific limits: i.e., [imath]\int^b_a f(x)dx [/imath]. In that case the (now “definite”) answer is F(b)-F(a).

 

...in which case that also makes perfect sense.

 

Funny, I actually did see Wikipedia mention that there's a connection between integration and derivatives, and spent some time scratching my head on that exact statement [imath]\int^b_a f(x)dx=F(b)-F(a)[/imath], but just couldn't figure it out on my own, and thought I was completely lost by then :D

 

So I think I get it now, in purely algebraic sense it's easy to just accept it, and just to get some sort of mental idea of it, [imath]\int f(x)dx = F(x)[/imath] means the function [imath]F(x)[/imath] can be seen as a function returning the value of "total area between 0 and x" (in integration sense of "total area"), and the derivative of such function can always be expressed as (or found via) the original function [imath]f(x)[/imath]. (sorry if my terminology is a bit alien to what mathematicians usually use, I'm just stating what I'm thinking so you can be check if I'm picking things up correctly :D)

 

But let's get back to your question. Changing the variable “x” to “t”, if [imath]\int f(t)dt = F(t)[/imath] then it must be true that [imath]\int \frac{d}{dt}\left\{F(t)\right\}dt = F(t)[/imath]; correct?

 

i.e. the indefinite integration of the derivative of [imath]F[/imath] just gets us right back to mere [imath]F[/imath]...

 

Since [imath]\frac{d}{dt}\left\{F(t)\right\}=f(t)[/imath], then by trivial substitution [imath]

\int \frac{d}{dt}\left\{F(t)\right\}dt = \int f(t) dt = F(t)[/imath]

 

yes so it must be. It's easy when it's all spelled out.

 

But what happens to that expression if we look at the differential representation [imath] \frac{d}{dt}F(t)=f(t)[/imath] and, instead of looking at the differential of F(t), we look at G(t)=F(t)+C where “C” is any constant. Since the differential of a constant is zero, we have the fact that [imath] \frac{d}{dt}G(t)=f(t)+0\equiv f(t)[/imath].

 

Yes...

 

This implies an interesting fact. If F(t)=0, then the “indefinite integral” is indefinite to the extent of some arbitrary constant: i.e., [imath]\int \frac{d}{dt}\left\{F(t)\right\}dt = 0 + C[/imath]. That is exactly the reason behind that adjective “indefinite” attached to indefinite integrals.

 

Right, so in other words, the (indefinite) integral of [imath]f(x)[/imath] can always be seen as [imath]F(x) + C[/imath] where F is the anti-derivative and C can be arbitrarily chosen. I.e. "any constant can be added to the anti-derivative and it will still correspond to the same integral (via a derivation)".

 

Okay, I think I pretty much get the idea.

 

The Euler-Lagrange equations of interest in this case (see equation #3) are,

[math]\frac{\partial f}{\partial \tau}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)=0 [/math]

 

[math]\frac{\partial f}{\partial \theta}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)=0 [/math]

 

and

 

[math]\frac{\partial f}{\partial r}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{r}}\right)=0[/math]

 

The first two are trivial; if the differential with respect to t is zero one can integrate the function and obtain a constant...

 

I.e. putting together the facts that;

 

[math]\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)=0[/math]

 

and

 

if [math]F(t)=0[/math], then [math]\int \frac{d}{dt}\left\{F(t)\right\}dt = 0 + Constant[/math]

 

we know that;

[math]

\int \frac{d}{dt}\left\{ \frac{\partial}{\partial \dot{\tau}}f(t) \right\}dt = Constant

[/math]

 

With that in mind...

 

Substituting c for [imath]\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}[/imath], one has the following two first integrals:

[math] \int \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}\right\}\right)dt = \left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\frac{\dot{\tau}}{c}=\;[/math] a constant [math]=\;\frac{1}{cl}[/math]

 

...I can see how;

[math]\int \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}\right\}\right)dt

= Constant[/math]

 

But I don't understand how you got that middle expression;

[math]

\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\frac{\dot{\tau}}{c}

[/math]

 

In particular the [imath]\frac{\dot{\tau}}{c}[/imath] in there... The closest I got to it myself was;

[math]

\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}c\right\}

[/math]

 

which I realize is problematic because there's no [imath]\tau[/imath] in the function at all. I guess that issue has got something to do with that middle step, what you are doing there is preserving the [imath]\tau[/imath] variable in there... So trivial substitution of c for [imath]\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}[/imath] cannot be done...? Hmmm, well I'm not entirely sure what happens there... Help!

 

-Anssi

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Okay, after playing around with Wolfram Alpha a bit, I take it then that the indefinite integral [imath]\int f(x)dx[/imath] can be interpeted as the total area between 0 and [imath]x[/imath]...
Actually that is not a very good interpretation. A better statement is that [imath]\int f(x)dx[/imath] yields a functional relationship which tells one how that total area changes as a function of x, not what the total area actually is.
Funny, I actually did see Wikipedia mention that there's a connection between integration and derivatives, and spent some time scratching my head on that exact statement [imath]\int^b_a f(x)dx=F(b)-F(a)[/imath], but just couldn't figure it out on my own, and thought I was completely lost by then :D
You should note that, if you replace F(x) with G(x)=F(x)+C, G(b)+C - (G(a) +C) = G(b)-G(a) : i.e., the constant C drops out. That is, the answer to the definite integral is exactly the difference between the evaluations of the indefinite integral at the boundaries.
I'm just stating what I'm thinking so you can be check if I'm picking things up correctly :D)

Well, except for that phrase, "total area between 0 and x", which I have already commented on, I think you have things pretty straight.
Okay, I think I pretty much get the idea.
Yeah, you are getting there.
...I can see how;

[math]\int \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}\right\}\right)dt

= Constant[/math]

 

But I don't understand how you got that middle expression;

[math]

\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\frac{\dot{\tau}}{c}

[/math]

 

In particular the [imath]\frac{\dot{\tau}}{c}[/imath] in there... The closest I got to it myself was;

[math]

\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}c\right\}

[/math]

You are making a very simple mistake. You are jumping from the fact that the differential of a constant is zero to the assumption that the “partial derivative” of a constant is zero. That is simply not true. Just because a circle can be defined by [imath]r=\sqrt{x^2+y^2}= [/imath] a constant does not imply that the partial with respect to x vanishes. What it really says is that any change in x must be related to a change in y such that the net result for r remains the same.
So trivial substitution of c for [imath]\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}[/imath] cannot be done...? Hmmm, well I'm not entirely sure what happens there... Help!
Sure, it can be done but doing so hides the important dependences here. In this case, we are looking at two important partial differentials: one with respect to [imath]\dot{\tau}=\frac{d \dot{\tau}}{dt}[/imath] and the other with respect to [imath]\dot{\theta}=\frac{d \dot{\theta}}{dt}[/imath]. Since

[math]c=\sqrt{\dot{\tau}^2+\dot{r}^2+r^2\dot{\theta}^2} =\left[\dot{\tau}^2+\dot{r}^2+r^2\dot{\theta}^2\right]^{\frac{1}{2}}[/math],

 

it follows that

[math]\frac{\partial c}{\partial \dot{\tau}}=\frac{1}{2}\left[\dot{\tau}^2+\dot{r}^2 +r^2\dot{\theta}^2\right]^{-\frac{1}{2}}2\dot{\tau}[/math]

 

and

 

[math]\frac{\partial c}{\partial \dot{\theta}}=\frac{1}{2}\left[\dot{\tau}^2+\dot{r}^2 +r^2\dot{\theta}^2\right]^{-\frac{1}{2}}2r^2\dot{\theta}[/math]

 

Essentially, the partial of c with respect to [imath]\dot{\tau}[/imath] is [imath]\dot{\tau}[/imath] divided by c and the partial with respect to [imath]\dot{\theta}[/imath] is [imath]r^2\dot{\theta}[/imath] divided by c. Multiplying through by that other term, we can conclude that

[math]\dot{\tau}=\frac{1}{l}\sqrt{1-\frac{2\kappa M}{c^2r}}\quad\quad and \quad\quad \dot{\theta}=\frac{h}{r^2l}\sqrt{1-\frac{2\kappa M}{c^2r}}[/math]

 

As I said, I chose the two constants (“l” and “h”) in the form I did in order to bring my solution into exactly the same form as the standard Schwarzschild solution to the Einsteinian field equations.

 

I hope this clarifies a few issues and don't worry about your problems; they are pretty easy to clarify.

 

Have fun -- Dick

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Actually that is not a very good interpretation. A better statement is that [imath]\int f(x)dx[/imath] yields a functional relationship which tells one how that total area changes as a function of x, not what the total area actually is.

 

Ahha, of course...

 

You should note that, if you replace F(x) with G(x)=F(x)+C, G(b)+C - (G(a) +C) = G(b)-G(a) : i.e., the constant C drops out. That is, the answer to the definite integral is exactly the difference between the evaluations of the indefinite integral at the boundaries.

 

Yup.

 

You are making a very simple mistake. You are jumping from the fact that the differential of a constant is zero to the assumption that the “partial derivative” of a constant is zero. That is simply not true. Just because a circle can be defined by [imath]r=\sqrt{x^2+y^2}= [/imath] a constant does not imply that the partial with respect to x vanishes. What it really says is that any change in x must be related to a change in y such that the net result for r remains the same.

 

Ahha... Well once again I was trying things out with Wolfram Alpha, and it tells me that [imath]\frac{\partial}{\partial x} ( \sqrt{x^2+y^2} ) = \frac{x}{ \sqrt{x^2+y^2} }[/imath]. I was able to follow the steps of algebra in between, so with that I can understand that;

 

[math]

\frac{\partial}{\partial \dot{\tau}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}

= \frac{\dot{\tau}}{\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}}

= \frac{\dot{\tau}}{c}

[/math]

 

And [imath]\dot{\tau}[/imath] doesn't appear anywhere in [imath]\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}[/imath] so that's why you can write that middle step as [math]\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\frac{\dot{\tau}}{c}[/math]

 

The partial derivative of [imath]\dot{\theta}[/imath] is a bit different case but once again I was able to follow the steps of Wolfram Alpha's derivation and indeed it looks like;

 

[math]

\frac{\partial}{\partial \dot{\theta}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}

= \frac{ r^2 \dot{\theta}}{c}

[/math]

Sure, it can be done but doing so hides the important dependences here. In this case, we are looking at two important partial differentials: one with respect to [imath]\dot{\tau}=\frac{d \dot{\tau}}{dt}[/imath] and the other with respect to [imath]\dot{\theta}=\frac{d \dot{\theta}}{dt}[/imath]. Since

[math]c=\sqrt{\dot{\tau}^2+\dot{r}^2+r^2\dot{\theta}^2} =\left[\dot{\tau}^2+\dot{r}^2+r^2\dot{\theta}^2\right]^{\frac{1}{2}}[/math],

 

it follows that

[math]\frac{\partial c}{\partial \dot{\tau}}=\frac{1}{2}\left[\dot{\tau}^2+\dot{r}^2 +r^2\dot{\theta}^2\right]^{-\frac{1}{2}}2\dot{\tau}[/math]

 

and

 

[math]\frac{\partial c}{\partial \dot{\theta}}=\frac{1}{2}\left[\dot{\tau}^2+\dot{r}^2 +r^2\dot{\theta}^2\right]^{-\frac{1}{2}}2r^2\dot{\theta}[/math]

 

Yup, that is actually very unobvious algebraic step for me but I realize it is just another way of writing the same thing as what I wrote above.

 

Essentially, the partial of c with respect to [imath]\dot{\tau}[/imath] is [imath]\dot{\tau}[/imath] divided by c and the partial with respect to [imath]\dot{\theta}[/imath] is [imath]r^2\dot{\theta}[/imath] divided by c.

 

Check.

 

And then;

[math] \int \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}\right\}\right)dt = \left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\frac{\dot{\tau}}{c}=\;[/math] a constant [math]=\;\frac{1}{cl}[/math]

 

and

 

[math]\int \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\theta}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}\right\}\right)dt = \left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}r^2\dot{\theta}\frac{1}{c}=[/math] a constant [math]=\;\frac{h}{cl}[/math].

 

I have chosen to represent the two constants as I have, in terms of the constants l and h, because this notation will bring my solution into exactly the same form as the standard Schwarzschild solution to the Einsteinian field equations (with one very important difference).

 

I figure the "l" and "h" are just whatever constants would be appropriate to yield valid results. (Are they equal to some parameters that have got a defined meaning in the standard GR?)

 

Multiplying through by that other term, we can conclude that

[math]\dot{\tau}=\frac{1}{l}\sqrt{1-\frac{2\kappa M}{c^2r}}\quad\quad and \quad\quad \dot{\theta}=\frac{h}{r^2l}\sqrt{1-\frac{2\kappa m}{c^2r}}[/math]

 

I suppose the small "m" is a typo and doesn't signify any new definition or anything (it appears in the OP too, and it has propagated to another equation two steps down there)

 

Yup, I was able to walk through the algebra between [imath]\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\frac{\dot{\tau}}{c} = \frac{1}{cl}[/imath] and [imath] \dot{\tau} = \frac{1}{l} \sqrt{ 1 - \frac{ 2 \kappa M } {c^2r}} [/imath]

 

And likewise for [imath] \left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}r^2\dot{\theta}\frac{1}{c} = \frac{h}{cl} [/imath] I got the exact same result as you [imath]\dot{\theta} = \frac{h}{r^2 l}\sqrt{1-\frac{2\kappa M}{c^2r}} [/imath]

 

Using [imath](\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2=c^2[/imath], we can write

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

 

Did the substitution and managed to walk my way to that same exact result; looks valid to me.

 

This differential equation (relating r and t) can be transformed to the actual differential equation of interest (that would be r verses [imath]\theta[/imath]) through the following well known replacement

Since [math]\quad r'=\frac{dr}{d\theta}=\frac{\dot{r}}{\dot{\theta}}, \quad \dot{r}=\dot{\theta}r'=\frac{h}{r^2l}r'\sqrt{1-\frac{2\kappa m}{c^2 r}},[/math]

 

and the equation for the geodesic can be directly written as

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

 

(trivial typo there, verses/versus)

 

In terms of pure algebra, that all looks quite valid to me. But what does [imath]r'[/imath] stand for here exactly? I.e. I'm not at all sure what you are saying there with [imath]\quad r'=\frac{dr}{d\theta}=\frac{\dot{r}}{\dot{\theta}}, \quad \dot{r}=\dot{\theta}r'=\frac{h}{r^2l}r'\sqrt{1-\frac{2\kappa m}{c^2 r}},[/imath]

 

I'm probably missing something fairly obvious... :I

 

-Anssi

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I figure the "l" and "h" are just whatever constants would be appropriate to yield valid results. (Are they equal to some parameters that have got a defined meaning in the standard GR?)
Constants are constants. Actually what I obtained could have been written

[math]\left[\frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\dot{\tau}=cA_1 \quad\quad and \quad\quad \left[\frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}r^2\theta^2=cA_2[/math]

 

where A1 and A2 denote the original constants required by the Euler-Lagrange solution. In that case Schwarzschild's l is the inverse of my A1. Since my A2 could certainly be written as some number times my A1 one could say A2=hA1 which would, in fact, yield exactly Schwarzschild's constants.

 

I am sure that there are reasons for Schwarzschild's use of those parameters but I did this derivation many years ago and I have no memory as to what his rational was. I could not find my copy of the book I quoted so I am at something of a loss to explain his parameters. Perhaps some authority on GR here could clarify the issue. Qfwfq, if you are around, you might explain that to us since you are so familiar with the classical representation of things. :banghead:

I suppose the small "m" is a typo and doesn't signify any new definition or anything (it appears in the OP too, and it has propagated to another equation two steps down there)
I have fixed the OP and my previous post. I have also fixed the verses/versus typo. Spell check doesn't catch misuse of a word. :shrug:

 

On the other hand, my work is sometimes shear poetry itself isn't it? :lol: :bow_flowers: :friday:

In terms of pure algebra, that all looks quite valid to me. But what does [imath]r'[/imath] stand for here exactly?
In mathematics, the “prime” stands for the differential with respect to the “other” significant coordinate. As such, it really can only be used when there are only two coordinates: i.e., something like y=f(x) in which case, y' is the derivative of f(x) with respect to x. One can also see the same thing represented as f'(x). What is significant here is the fact that planetary orbits are usually represented in terms of polar coordinates: i.e., r and theta. If you want to plot these orbits you need a function of the form [imath]r=f(\theta)[/imath].
I.e. I'm not at all sure what you are saying there with [imath]\quad r'=\frac{dr}{d\theta}=\frac{\dot{r}}{\dot{\theta}}, \quad \dot{r}=\dot{\theta}r'=\frac{h}{r^2l}r'\sqrt{1-\frac{2\kappa m}{c^2 r}},[/imath]
Well, first of all, the definition of r' is [imath]r'=\frac{dr}{d\theta}[/imath] and, since [imath]\dot{x}[/imath] is defined to be the derivative of x with respect to time, [imath]\dot{r}[/imath] is, by definition, [imath]\frac{dr}{dt}[/imath] and [imath]\dot{\theta}[/imath] is likewise [imath]\frac{d\theta}{dt}[/imath]. Since the symbol “dx” is defined to be the limit of a change in x as that change goes to zero, one can use common algebra to cancel out the associated “dt” entries and it follows that [imath]\frac{\dot{r}}{\dot{\theta}}[/imath] is identical to [imath]\frac{dr}{d\theta}=r'[/imath]. From this (if we multiply r' by [imath]\dot{\theta}[/imath]) we can assert that [imath]\dot{r}=\dot{\theta}r'[/imath]. But we have already discovered that

[math]\dot{\theta}=\frac{h}{r^2l}\sqrt{1-\frac{2\kappa M}{c^2r}}[/math]

 

so substituting into the above representation of [imath]\dot{r}[/imath] we come to the conclusion that [imath]\dot{r}^2[/imath] must be given by

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

 

When you make that substitution into our expression for c2 as a function of [imath]\dot{r}[/imath] you will obtain exactly the result shown.

I'm probably missing something fairly obvious... :I
Obvious is in the eye of the observer. You simply haven't had much experience doing calculus. The more you do it, the more obvious the procedures become. I apologize for being sloppy with my presentation, but I think anyone really familiar with calculus would indeed find the steps pretty obvious.

 

Thanks again Anssi, I really wish some of those other “authorities” on this forum would take the trouble to read this stuff.

 

Have fun -- Dick

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As an aside, it is quite clear that the proper adjustment to our geometry which will yield the standard concept of gravity as a pseudo force is a change in the presumed measure of that geometry: i.e., instead of seeing the speed of light as slower in a gravitational field we could just as well see the speed as unchanged and the distances as increased. After all, once time is defined, distances are reckoned via the speed of light. Though that satisfies the original goal expressed above, the idea of refraction (the speed of light being slowed in a gravitational field) is a much simpler expression of the solution. It is certainly most convenient method of finding the proper geodesics. In fact, there is a very simple view of the situation which will yield exactly that result.

 

Won’t seeing light as slowed in a gravitational field be equivalent to seeing the distance measured along the [imath]\tau[/imath] axis as being scaled to make up for all objects still having a constant velocity in your x,y,z,[imath]\tau[/imath] geometry, and since light still cant travel along the [imath]\tau[/imath] axis, since it is still massless, in effect we are scaling the [imath]\tau[/imath] axis and saying that objects are moving a shorter distance in the x,y,z,[imath]\tau[/imath] space instead of scaling the x,y,z axis’s and saying that objects are traveling a longer distance as viewed from outside of the gravitational field. Of course since the [imath]\tau[/imath] axis can’t be directly measured and actual location on it makes no difference such an effect could not be noticed.

 

Any physical object (any structure stable enough to be thought of as an object) must have internal forces maintaining that structure. Any interaction with another distant object must be via the virtual particle exchange I just commented about. Thus it is that one would expect the fundamental element of that physical object interacting via that delta function would have its momentum altered, not the whole object; however, that alteration would create a discrepancy in the structure of the object under discussion. Since that object must have internal forces maintaining its structure, it is to be expected that those internal interactions (which are also mediated by delta function exchange forces) will bring the trajectory that interacting fundamental element essentially back to its original path (at least on average).

 

This is the effect that you are referring to as refraction, that is this sort of moving back and forth of a object is causing it to move slower in the [imath]\tau[/imath] direction even though the object really is not going any where as a result the object will travel slower in the [imath]\tau[/imath] direction even though the object is not traveling a long distance in the x,y,z space and can effectively remain in the same spot?

 

If the distant object and the object under observation are not moving with respect to one another (they are moving parallel to one another in the tau direction), the net effect of that refraction is to curve the paths of the two objects towards one another: i.e., there will be an apparent attraction between them. It is also evident that, since the mass of the source object (the source of these bosons external to the object of interest) is proportional to the total momentum of that object, one should expect the apparent density (as seen from the object of interest) should be proportional to its mass: i.e., one should expect the exchange forces to be proportional to mass.

 

I don’t understand why this is unless it’s that since one of the objects is moving slower in the [imath]\tau[/imath] direction the object must move in the x,y,z direction to make up for it. But, didn’t you just say that the object would counter this movement with internal forces, at least on average, and so would not move in the x,y,z directions at least on average. Or is the idea that since this is only on average an object is still capable of moving along the shortest path in the x,y,z,[imath]\tau[/imath] space which would bring the objects closer together due to the maximum velocity as seen from outside of the gravity well being slower closer to the massive object. In which case is this what you are suggesting results in gravity? If this is the case I’m not entirely sure as to why an object would try to follow the shortest path except perhaps that it would move the furthest along it on average.

 

Also, how do we know that the source of these bosons is proportional to the total momentum of the object?

 

Clearly the interaction just discussed arises from differential effects in the basic interactions thus it will amount to a force considerably less than the underlying force standing behind that differential effect. Thus it is that the two forces I have already discussed (the forces due to massless boson exchange: shown to yield electromagnetic effects and the forces due to massive boson exchange: shown to yield fundamental nuclear forces) will end up being split into four forces. Differential effects will yield a correction to both basic forces which correspond quite well with the forces observed in nature. The differential effect on massless boson exchange yields what appears to be a very weak gravitational force (weak when compared to the underlying electromagnetic effects) and the differential effect on massive boson exchange yields what appears to be a very weak nuclear force (weak compared to the underlying nuclear force). What is interesting is that the “weak nuclear force” can be shown to violate parity symmetry whereas the “weak electrical force” (gravity) does not. This is a direct consequence of the fact that the nuclear exchange bosons are massive.

 

Just why is it that this is resulting from the differential effects? Does it have something to do with this effect resulting from how far an object moves rather then an interaction due to the delta function, and in effect, this is a result of conserving the total distance that the object moves rather then an interaction due to the delta function?

 

Also, are you saying that gravity was left out in the derivation of the Dirac and Maxwell equations and that gravity is a correction factor and so in effect there is no so called graviton rather there is a correction to the magnetic field which results in gravity? If so I don’t understand how we know that these forces where left out of the Dirac and Maxwell equations other then that people have looked for such a thing and not found it, or is this some kind of average effect that can only be noticed if a large number of particles that form an object is considered.

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Won’t seeing light as slowed in a gravitational field...
In my picture, all entities propagate through my four dimensional universe at the same arbitrary velocity. Time is not a measurable variable. Clocks actually measure changed in tau, an axis just like any other axis except for the fact that all entities which go to make up any physical object are in a momentum quantized state in the tau direction. The rest of your first paragraph seems to be no more than continued confusion.
This is the effect that you are referring to as refraction...
No it isn't! The point is that, since all entities propagate through the four dimensional universe at the same arbitrary velocity, if they are (as a collection) at rest in the x,y,z space, any disturbance in their individual paths will, in effect, yield an apparent slowing of the component in the tau direction. It is the very small differential variation in this effect which yields refraction: i.e., the apparent velocity in the tau direction varies across the physical object. If you look up refraction you will discover that this is the exact nature of refraction of any wave phenomena.
I don’t understand why this is unless it’s that since one of the objects is moving slower in the [imath]\tau[/imath] direction the object must move in the x,y,z direction to make up for it. But, didn’t you just say that the object would counter this movement with internal forces, at least on average...
You are apparently confusing the meaning of “object” as I am using it. I have defined “a physical object” to be a collection of fundamental entities which can be seen as a single thing propagating through space without significant distortion in its configuration. It is the individual fundamental entities who's velocity in the tau direction varies, “not one of the objects. There is only one object in this discussion and its velocity in the tau direction is the average of all the fundamental entities which go to make it up.
Also, how do we know that the source of these bosons is proportional to the total momentum of the object?
The momentum of each of the fundamental entities making up the object can be seen as essentially the same. Thus the number of fundamental entities making up the object must be proportional to the total momentum in the tau direction. Since the interactions are between the fundamental entities themselves, the net effect must be proportional to the total momentum of the object in the tau direction: i.e., its net mass.
Just why is it that this is resulting from the differential effects? ... are you saying that gravity was left out in the derivation of the Dirac and Maxwell equations and that gravity is a correction factor and so in effect there is no so called graviton rather there is a correction to the magnetic field which results in gravity? If so I don’t understand how we know that these forces where left out of the Dirac and Maxwell equations other then that people have looked for such a thing and not found it, or is this some kind of average effect that can only be noticed if a large number of particles that form an object is considered.
To my knowledge no one has ever made any effort to calculate the net differential effect in electromagnetic effects over a large number of atoms made of charged particles. Look at the relative strengths of the four forces seen in nature. Gravity is down by a factor of 10-36 and the weak nuclear interaction is down by a factor of almost 10-20 from the strong interaction if one takes into account the fact that its range is close to a thousand times as far. The calculation errors far outweigh those differences.

 

I think you are not trying to see things from the perspective I have laid out but rather, trying to map those thoughts into the compartmentalized thinking of common science. Essentially I feel you are making the same mistake being made by Erasmus.

 

Have fun -- Dick

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While writing a reply, I spotted something that looks like an error in the OP;

 

Following the standard Euler-Lagrange attack, if the expression

[math]f(t,y,\dot{y})=\frac{1}{\sqrt{1-\frac{2\kappa M}{c^2r}}}\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}[/math]

 

satisfies the Euler-Lagrange equation, the variation of the integral over the implied path will vanish. In this case we have three independent variables to deal with (note that “y”, in this case, stands for the variable of interest) which leads to three independent differential equations. However, we have one additional constraint which simplifies the problem considerably: [imath](\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2=c^2[/imath]. The partials of f with respect to tau and theta vanish leading to the fact that the first integrals of the Euler-Lagrange equation for those variables are trivial. Substituting c for [imath]\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}[/imath], one has the following two first integrals:

[math] \int \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{-\frac{1}{2}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}\right\}\right)dt = \left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{-\frac{1}{2}}\frac{\dot{\tau}}{c}=\;[/math] a constant [math]=\;\frac{1}{cl}[/math]

 

and

 

[math]\int \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\theta}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{-\frac{1}{2}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}\right\}\right)dt = \left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{-\frac{1}{2}}r^2\dot{\theta}\frac{1}{c}=[/math] a constant [math]=\;\frac{h}{cl}[/math].

 

Shouldn't it be [math]\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}[/math] or alternatively [math]\left[ 1-\frac{2\kappa M}{c^2r}\right]^{-\frac{1}{2}}[/math] in those last two equations?

 

 

Constants are constants. Actually what I obtained could have been written

[math]\left[\frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\dot{\tau}=cA_1 \quad\quad and \quad\quad \left[\frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}r^2\theta^2=cA_2[/math]

 

where A1 and A2 denote the original constants required by the Euler-Lagrange solution. In that case Schwarzschild's l is the inverse of my A1. Since my A2 could certainly be written as some number times my A1 one could say A2=hA1 which would, in fact, yield exactly Schwarzschild's constants.

 

Ahha, yup...

 

I am sure that there are reasons for Schwarzschild's use of those parameters but I did this derivation many years ago and I have no memory as to what his rational was. I could not find my copy of the book I quoted so I am at something of a loss to explain his parameters. Perhaps some authority on GR here could clarify the issue. Qfwfq, if you are around, you might explain that to us since you are so familiar with the classical representation of things. :banghead:

 

Yeah it would be interesting to know...

 

I have fixed the OP and my previous post. I have also fixed the verses/versus typo. Spell check doesn't catch misuse of a word. :shrug:

 

On the other hand, my work is sometimes shear poetry itself isn't it? :lol: :bow_flowers: :friday:

 

Yeah it's crate ;)

 

In mathematics, the “prime” stands for the differential with respect to the “other” significant coordinate. As such, it really can only be used when there are only two coordinates: i.e., something like y=f(x) in which case, y' is the derivative of f(x) with respect to x. One can also see the same thing represented as f'(x). What is significant here is the fact that planetary orbits are usually represented in terms of polar coordinates: i.e., r and theta. If you want to plot these orbits you need a function of the form [imath]r=f(\theta)[/imath].

 

Ahha, okay.

 

Well, first of all, the definition of r' is [imath]r'=\frac{dr}{d\theta}[/imath] and, since [imath]\dot{x}[/imath] is defined to be the derivative of x with respect to time, [imath]\dot{r}[/imath] is, by definition, [imath]\frac{dr}{dt}[/imath] and [imath]\dot{\theta}[/imath] is likewise [imath]\frac{d\theta}{dt}[/imath]. Since the symbol “dx” is defined to be the limit of a change in x as that change goes to zero, one can use common algebra to cancel out the associated “dt” entries and it follows that [imath]\frac{\dot{r}}{\dot{\theta}}[/imath] is identical to [imath]\frac{dr}{d\theta}=r'[/imath]. From this (if we multiply r' by [imath]\dot{\theta}[/imath]) we can assert that [imath]\dot{r}=\dot{\theta}r'[/imath]. But we have already discovered that

[math]\dot{\theta}=\frac{h}{r^2l}\sqrt{1-\frac{2\kappa M}{c^2r}}[/math]

 

so substituting into the above representation of [imath]\dot{r}[/imath] we come to the conclusion that [imath]\dot{r}^2[/imath] must be given by

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

 

When you make that substitution into our expression for c2 as a function of [imath]\dot{r}[/imath] you will obtain exactly the result shown.

 

Yup, excellent, that's how I figured, apart from not being sure what meaning the prime held.

 

I'll continue from here later...

 

-Anssi

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In my picture, all entities propagate through my four dimensional universe at the same arbitrary velocity. Time is not a measurable variable. Clocks actually measure changed in tau, an axis just like any other axis except for the fact that all entities which go to make up any physical object are in a momentum quantized state in the tau direction. The rest of your first paragraph seems to be no more than continued confusion.

 

And that the location on the [imath]\tau[/imath] axis has no influence due to the uncertainty of location on the [imath]\tau[/imath] axis. As a result the location on the [imath]\tau[/imath] axis also can’t be measured.

 

This results in the Lorenz transformation being needed to move from a frame that can be considered to be at rest to another frame that can also be considered to be at rest. This is a consequence of the requirements for defining all measurements in a self consistent way however what I don’t understand is how to still take this position when you now point out that.

 

If an experimenter in a gravitational well sets up a clock via a light pulse traveling back and forth between two horizontally displaced mirrors, since we can establish horizontal measure (simple vertical lines carry those measures to different heights in the hole) and his clock must run slow, we must see the apparent velocity of light to be

[math]c'=c\sqrt{1+\frac{2\Phi}{c^2}}[/math]

 

(at least for small gravitational effects). The gravitational red shift (which we are using here) is an instantaneous differential effect whereas the expression above concerns the horizontal path of the light in the clock.

 

Because if we still consider light to be massless then wont it still have the same arbitrary velocity, and it was originally used to define a measure of change in [imath]\tau[/imath] so how can we not conclude that it is traveling the same distance in the gravity well as outside of it? The only other option that I can easily come to is that as seem from outside of the gravity well the light in it has gained mass, but isn’t this by definition impossible?

 

This is not helped by the statement found further down that page where you now say that.

 

Since, in my four dimensional geometry, it has already been shown that clocks actually measure tau, the above also implies any lower object will appear to proceed in the tau direction at a slower rate proportional to exactly that same factor: i.e., the fundamental velocity of any object in my geometry will appear to be slower in a lower gravitational potential. This immediately suggests that we should be using Fermat's principle to establish the metric which will yield the proper geodesics: i.e., we should consider the phenomena of refraction.

 

I’m really at some what of a loss as of how to combine these statements.

 

You are apparently confusing the meaning of “object” as I am using it. I have defined “a physical object” to be a collection of fundamental entities which can be seen as a single thing propagating through space without significant distortion in its configuration. It is the individual fundamental entities who's velocity in the tau direction varies, “not one of the objects. There is only one object in this discussion and its velocity in the tau direction is the average of all the fundamental entities which go to make it up.

 

So in effect the individual elements that make up the object are interacting with the bosons, since this effect is effected by distance the elements that are closest to the source of the bosons will have the greatest interaction with the bosons. However the object that the elements make up will still be held together by other internal forces and can still be considered a single object. This results in the waves that make up the object not having a constant velocity in the [imath]\tau[/imath] direction but instead the object will be refracted just like any other wave is refracted.

 

Now this refraction takes place along the [imath]\tau[/imath] axis in that it is the speed along this axis that is being effected however the object must still have a constant velocity which will result in a change in the movement along the x,y,z axis?

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While writing a reply, I spotted something that looks like an error in the OP
Anssi, when you are right, you are right. :doh: :thanks:

 

I have quoted that exact expression myself a couple of times in this thread without noticing the error. I have fixed the OP and the places where I quote myself (or you) with regard to those expressions. Maybe doing that was an error because it makes some of your complaints not make sense. All I can say is I apologize sincerely.

 

I owe you -- Dick

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And that the location on the [imath]\tau[/imath] axis has no influence due to the uncertainty of location on the [imath]\tau[/imath] axis. As a result the location on the [imath]\tau[/imath] axis also can’t be measured.
That is correct, the location in the tau direction cannot be identified; however, the velocity is another matter as the net velocity through the four dimensional space is fixed.
I don’t understand is how to still take this position when you now point out that [imath]c'=c\sqrt{1+\frac{2\Phi}{c^2}}[/imath]
The original issue was, what kind of non-Euclidean distortion in the coordinate system could yield gravitational effects. What I showed was that allowing the length of the units of measure in the reference frame in the vicinity of the source of the gravitational field would yield exactly the desired effect. I then pointed out that it was much easier to allow that “fixed velocity” mentioned above to decrease in the gravitational field. It simply makes the math easier and yields exactly the same results.
Because if we still consider light to be massless then wont it still have the same arbitrary velocity ...
Yes, its velocity through the four dimensional space is still fixed; but now it is fixed to the new velocity, [imath]c'=c\sqrt{1+\frac{2\Phi}{c^2}}[/imath], and clocks will still define a measure of change in tau, they simply yield a smaller number. It doesn't gain mass because its momentum in the tau direction is still zero: i.e., the photon's velocity in the tau direction is still zero.
So in effect the individual elements that make up the object are interacting with the bosons, since this effect is effected by distance the elements that are closest to the source of the bosons will have the greatest interaction with the bosons. However the object that the elements make up will still be held together by other internal forces and can still be considered a single object. This results in the waves that make up the object not having a constant velocity in the [imath]\tau[/imath] direction but instead the object will be refracted just like any other wave is refracted.
Because the individual elements are not actually traveling directly in the tau direction: their individual paths can be seen as crooked as they are influenced by the boson exchange.
... however the object must still have a constant velocity which will result in a change in the movement along the x,y,z axis?
The objects constant velocity will be the average of all the elements which make it up which will be slower than the standard free velocity applied to individual elements. Refraction is a wave effect due to the fact that the wave velocities in the tau direction are less in the lower gravitational potential.

 

There is another subtle secondary effect here. If an object is heated up, the deviations of the individual elements will increase. This will lead to an apparent slowing of the object in the tau direction. Exactly the same effect one would expect if the mass of the object itself had increased. In fact, I will lay odds that the apparent mass increase of the object due to the heat energy will exactly match E=mc2.

 

Have fun -- Dick

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I have fixed the OP and the places where I quote myself (or you) with regard to those expressions. Maybe doing that was an error because it makes some of your complaints not make sense.

 

Well I decided it would cause the least amount of confusion to fix the error in my posts as well.. :P

 

and the equation for the geodesic can be directly written as

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

 

which can be rearranged to yield:

[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]

 

which, except for the final term in square brackets, is precisely the Schwarzschild solution to Einstein's field equations for a spherically symmetric case (see Adler, Bazin and Schiffer, Introduction to General Relativity, McGraw-Hill Co., New York, 1965, p. 180.).

 

Got the same result, so it looks valid to me. I don't have the Introduction to General Relativity, and I could only find the few first pages from the internet. I tried to look for the Schwarzschild solution, but I could only find it being expressed in somewhat different form so it's hard for me to compare (for instance, the equation 11.40 at http://www.phys.uu.nl/~thooft/lectures/genrel.pdf )

 

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.

 

In order to comprehend exactly what the impact of that extra term is, I would like to compare the Schwarzschild solution to to the classical Newtonian result. Note that, if one differentiates the classical elliptical solution to the Newtonian problem, [imath]\frac{1}{r}=\frac{1}{r_0}(1-\epsilon \;cos (\theta))[/imath], and chooses the appropriate constants; the Newtonian solution can be written:

[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}.[/math]

 

Hmmm, I don't really understand what happens there. I'm not sure what [imath]\frac{1}{r}=\frac{1}{r_0}(1-\epsilon \;cos (\theta))[/imath] says and how [imath]\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}[/imath] relates to it... :I

 

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]

 

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]

 

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.

 

Have you worked out how small the effect is for Mercury?

 

I'll continue from here soon...

 

-Anssi

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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).

 

varified -> verified :)

 

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}=A_1 \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}=A_2[/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]

 

I'm having troubles getting to that result, i.e. I end up with just "1" instead of [imath]c^2[/imath] as the first term on the right side...

 

Here's my attempt:

 

[imath]\dot{\tau}[/imath] and [imath]\dot{\theta}[/imath] are:

 

[math]

\dot{\tau}

=

A_1 c \left(1-\frac{2\kappa M}{c^2r}\right)^{\frac{1}{2}}

\quad\quad and \quad\quad

\dot{\theta}

=

A_2 \frac{c}{r^2} \left(1-\frac{2\kappa M}{c^2r}\right)^{\frac{1}{2}}

[/math]

 

Substituting into [imath](\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2=c^2[/imath]:

 

[math]

A_1^2 c^2 \left(1-\frac{2\kappa M}{c^2r}\right)

+

(\dot{r})^2+

A_2^2 \frac{c^2}{r^2} \left(1-\frac{2\kappa M}{c^2r}\right)

=

c^2

[/math]

 

Since [imath]\dot{r}=\dot{\theta}r'[/imath]:

 

[math]

A_1^2 c^2 \left(1-\frac{2\kappa M}{c^2r}\right)

+

A_2^2 \frac{c^2}{r^4} \left(1-\frac{2\kappa M}{c^2r}\right) (r')^2

+

A_2^2 \frac{c^2}{r^2} \left(1-\frac{2\kappa M}{c^2r}\right)

=

c^2

[/math]

 

And rearrangement:

 

[math]

A_1^2 \left(1-\frac{2\kappa M}{c^2r}\right)

=

\frac{c^2}{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)

+

\frac{A_2^2}{r^4} (r')^2 \left (\frac{2\kappa M}{c^2r} \right )

[/math]

 

Not sure where the error is, but it seems to me you have gotten rid of the [imath]c^2[/imath]'s on all the terms but the one that was only [imath]c^2[/imath] to begin with.

 

-Anssi

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Well I decided it would cause the least amount of confusion to fix the error in my posts as well.. :P
Seems reasonable to me.
Got the same result, so it looks valid to me. I don't have the Introduction to General Relativity, and I could only find the few first pages from the internet. I tried to look for the Schwarzschild solution, but I could only find it being expressed in somewhat different form so it's hard for me to compare (for instance, the equation 11.40 at http://www.phys.uu.nl/~thooft/lectures/genrel.pdf )
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^2\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.

Hmmm, I don't really understand what happens there. I'm not sure what [imath]\frac{1}{r}=\frac{1}{r_0}(1-\epsilon \;cos (\theta))[/imath] says and how [imath]\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}[/imath] relates to it... :I
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].

 

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. 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]

 

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].

 

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]

 

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].

 

That expression has two independent embedded constants,

[math]\frac{2h^2}{r_0} \quad \quad and \quad \quad 1-\frac{h^2}{r^2_0}(1-\epsilon^2)[/math]

 

If we set

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

 

the second constant will become

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

 

If we then set this equal to [imath]cl[/imath] and solve for epsilon, we have

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

 

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] and that there is no [imath]\frac{1}{r^3}[/imath] in the differential equation for the Newtonian elliptical orbit. Since the Schwarzschild solution must reduce to the Newtonian orbit as a approximate solution (approximating that [imath]\frac{1}{r^3}[/imath] term with zero), one can conclude that this particular collection of embedded constants correspond to the Schwarzschild solution of interest.

Have you worked out how small the effect is for Mercury?
Not exactly; however, as I commented in my opening post,
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.
That perhelic shift is observable over a substantial time, certainly not in a single orbit. Now the momentum of Mercury tangent to change in theta is extremely large compared to its momentum in the radial direction. Since both corrections are down by essentially the same factor from the Newtonian result, one should not expect the radial effect to be detectable.

 

In addition, you will note that near the end of the post I point out that the change is essentially equivalent to an erroneous measurement of “r” and I suspect the actual radius of the orbit of Mercury is not known to sufficient accuracy to detect such an effect.

I'm having troubles getting to that result, i.e. I end up with just "1" instead of [imath]c^2[/imath] as the first term on the right side...
You are absolutely correct! My error is in the definition of A1 and A2. As given, they are both unit-less constants but in my algebra I substitute them as if they have units of velocity. When I make the substitution into [imath]c^2=\dot{\tau}^2+\dot{r}^2+r^2\dot{\theta}^2[/imath] the act is is equivalent to presuming c is one.
Not sure where the error is, but it seems to me you have gotten rid of the [imath]c^2[/imath]'s on all the terms but the one that was only [imath]c^2[/imath] to begin with.
Yeah, all those constants A1 and A2 should not have been divided by c. Or another way to look at it is that the "A" terms (under the definitions I gave them) should still have the c attached to them (as you concluded) but that would make the result look different from Schwarzschild's solution (in the source I used): i.e., if that one term, [imath]c^2[/imath] is to be there, all the other terms have to have units of velocity squared also.

 

I have corrected the original post to show them as divided by c in their definitions. In that case, the need for a division by c later is not required and the bogus term [imath]\frac{c^2}{c^2}=1[/imath] remains as it was, [imath]c^2[/imath].

 

Thanks again for your trouble. I would have stopped posting here years ago had you not showed up. Take a peak at the page #4 of the “Life as an accident?” thread. I had expected a little interest from Qfwfq, but he appears to be pretty miffed at me. He and Erasmus both seem to want to show what I am saying can't be true without making the first effort to examine what I am saying. :shrug: And, on top of that; they have the gall to accuse me of assuming I know how the universe works. If that isn't a misrepresentation of the communication problem I don't know what is. I suppose “ignorance is bliss” is the only thing to be learned here.

 

Have fun -- Dick

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He and Erasmus both seem to want to show what I am saying can't be true without making the first effort to examine what I am saying. :ohdear:

 

I have spent lots of my time attempting to discuss with you, and to this day, do not feel as if you ever gave a straight answer to a single question I raised. I've given up this conversation, but I am surprised you doubt, given the amount of time I clearly spent on this, that I gave a good faith effort.

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And, on top of that; they have the gall to accuse me of assuming I know how the universe works. If that isn't a misrepresentation of the communication problem I don't know what is.
Isn't this a misrepresentation of what I said?

 

Dick, you replied to a query of mine saying that no, you don't think the universe is a finite collection, but your knowledge of it is. Isn't that like being inside Plato's cave? And yet you seemed to be implying on grounds of Ramsey considerations that this allows you to say life was an inevitable occurrance...:ohdear:

 

There is a quite good and detailed treatment of the Schwartzschild solution in Weinberg's classic Gravitation and Cosmology or several other good texts on the topic.

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The original issue was, what kind of non-Euclidean distortion in the coordinate system could yield gravitational effects. What I showed was that allowing the length of the units of measure in the reference frame in the vicinity of the source of the gravitational field would yield exactly the desired effect. I then pointed out that it was much easier to allow that “fixed velocity” mentioned above to decrease in the gravitational field. It simply makes the math easier and yields exactly the same results.

 

So rather then use a fixed speed of light and see distance as scaled and try to find the shortest path by distance, 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?

 

There is another subtle secondary effect here. If an object is heated up, the deviations of the individual elements will increase. This will lead to an apparent slowing of the object in the tau direction. Exactly the same effect one would expect if the mass of the object itself had increased. In fact, I will lay odds that the apparent mass increase of the object due to the heat energy will exactly match E=mc2.

 

What about the case in which all of the elements that form an object becoming very consistent in their movement in the [imath]\tau[/imath] direction compared to each other. Will constructive interference between the elements that make up the object result in any related effects?

 

But let us get back to gravitational effects. It is my position that gravity is a direct consequence of the fact that the presence of an object with momentum in the tau direction yields a secondary differential effect which causes the speed of elements through my four dimensional Euclidean space to slow: i.e., refraction of the wave solution to my fundamental equation occurs. In order to check my proposition, I need to be able to work out the geodesic paths implied by such a notion. To begin with, my assertion is that the four dimensional velocity of all elements can be seen as slowed in the presence of a gravitational potential. Thus it is that the “index of refraction” of a gravitational field is given by the instantaneous value of [imath]n=\frac{c}{c'}[/imath] evaluated a specific point in that field (where c is the velocity of light far removed from any massive influence). Geodesic paths can be obtained by minimizing nds where “ds” is the distance differential along that geodesic path in my four dimensional Euclidean geometry.

 

Looking into Format’s principal the idea looks like, that the path taken by the wave in question has stationary time, that is, it’s derivative to t is zero. In particular we want the shortest such path of all possible paths though the coordinate system. This path is the path that we are looking for. But isn’t there the problem though of when that principle was derived, wasn’t it derived for a three dimensional coordinate system where a clock would then supply the needed time variable. The point being, in your four dimensional coordinate system in which a clock no longer measures time how do we know what to use to calculate the path length. I am assuming that it is the t coordinate you are using for this purpose.

 

Looking at your first equation particularly the equation

[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]

It seems that you are integrating just over the path and using the scaling factor to define the length at any point. If this is the case is it this path that we are looking for and not the actual scaling factor? Either way how do we know that this is going to be zero in the final result, aren’t we really just looking for a minimum of this equation?

 

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.

 

I’m not sure that I follow. From what I am reading any solution to the Euler-Lagrange equation will define a function for which the derivative to time is zero. In our case perhaps saying path length is at a stationary point would be a better thing to say, as I don’t like using the word ‘time’ here until I am sure that it is a zero of the derivative to t that you are looking for and that you are calling this ‘time‘. However, a stationary point will happen at both a minimum and a maximum and any saddle points. I assume there are no saddle points, so after this equation is solved won’t the solution still have to be substituted back into the above integral and the minimum function found?

 

In order to simplify the problem, I would like to turn to a spherically symmetric case. Essentially a point source with a very large momentum in the tau direction and negligible momentum in the other three dimensions. This is in order to look at geodesic paths in the vicinity of a single very massive object where the other properties of the object can be ignored. In such a case, I can write [imath]\Phi(\vec{x})[/imath] in a very simple form consistent with standard notation:

[math]\Phi(\vec{x})=-\frac{\kappa M}{r}[/math]

 

where [imath]\kappa[/imath] is the proper proportionality constant to yield the correct potential generated by the mass M. This expression can be further simplified by changing to cylindrical coordinates (omit z as by symmetry we need only consider motion in the x-y plane).

 

But how do we have any idea of what the dependence of [imath]\Phi(\vec{x})[/imath] on r is. shouldn’t we use something like the term f(x) here until we have some idea of what the dependence will be?

 

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.

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I originally posted this to the wrong thread. The only excuse I have is shear mental incompetentcy :( I am easily confused and have to be very careful or I screw up regularly.

 

Of issue here are the supposed challenges from Erasmus and Qfwfq. These two seriously put themselves forward as rational thinking individuals and seem to really believe they are living up to their own standards. And I am pretty sure they both have the tools to follow my thoughts; however, there are other problems there which seriously confound me. The following is composed for them in an attempt to reach their interests.

 

Gentlemen, I am fully cognizant of exactly what your positions are. And I know you personally think that your approach to this conversation is open and unbiased; however there are two sides to every coin and both of you have concentrated your entire efforts on a single side of my presentation and totally ignored the “other side of the coin”.

 

There are two very different aspects of my presentation. One is the proof which I discovered over fifty years ago and the other consists of interpreting the consequences of that proof; which took me another ten years to uncover. Both of you have concentrated your entire interest on the consequences issue and no time whatsoever, that I am aware of, on the proof itself. (Qfwfq, perhaps you put a little time into it when I first arrived on this forum but you certainly haven't concerned yourself with that proof for quite a while.)

 

Essentially you have both approached the problem from one, very single minded, perspective: "OK how do you show this specific example is a solution to your equation?" The best answer to that question is: “gee guys, I don't have an answer to all your questions; I have only found a few very specific approximate solutions (which are essentially confined to modern physics and economics)”.

 

However, I did mistakenly make many attempts to clarify the real problems with your examples. Absolutely all of the examples the two of you presented contained so little data that no worthwhile predictions could be made anyway. Of course whenever I mentioned that, neither of you could comprehend what I was saying because you were invariably presuming massive amounts of data not explicitly represented in your presentation. (Essentially making it quite obvious that you had no comprehension whatsoever of the proof behind the equation.)

 

That is the nugget of the difficulty I just couldn't get you to look at. I suspect it is a direct consequence of the compartmentalized thinking always used in the exact sciences. Nevertheless, getting you to see around that block was apparently impossible.

I've given up this conversation, but I am surprised you doubt, given the amount of time I clearly spent on this, that I gave a good faith effort.
I don't think you gave the first thought towards trying to understand the proof. Any time spent outside that endeavor was pretty much a waste of time.
And yet you seemed to be implying on grounds of Ramsey considerations that this allows you to say life was an inevitable occurrance...:confused:
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. 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.

 

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. But it was my mistake to presume you understood anything of the proof and I apologize.

The great strength of logical thought is that the conclusions reached through logical thought are guaranteed to be as valid as the premises upon which they are based. The weakness of logical thought is that it is limited to a very small number of premises: i.e., the specific number of factors which can be included in the analytical statement of the problem. This is a seriously small number when compared to the volumes of information available to us ...

 

... Now mathematics and formal logic provide us with a certain respite from that last constraint ...

 

... when a problem can be approached with math and logic, one can be quite sure of the absolute validity of their conclusions. ...

 

if you want to do science, you should remember that even your most cherished squirrel decisions could be wrong. Even you guys who are not "crackpots" should remember that. A lot of science is done in the total absence of logical thought and that has to be so [as done, it is really a “guess and by golly” endeavor]; but scientists should not forget that fact. If they do, science folds over to religion. It may work great, but that does not mean it is valid.

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

 

Have fun -- Dick

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