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### #46AnssiH

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Posted 30 June 2010 - 12:11 PM

Doctordick said:

Right!
You are thinking of the gravitational force field (almost all “fields” referred to in physics are “vector force fields”). The idea of a potential field arises from the fact that most force fields (remember, force is a vector) can be expressed via the expression

$\vec{F}=\vec{\nabla}G(x,y,z)\equiv \frac{\partial G}{\partial x}\hat{x}+\frac{\partial G}{\partial y}\hat{y} +\frac{\partial G}{\partial z}\hat{z}$

Or, if the function G has only a radial dependence,

$\vec{F}=\vec{\nabla}G®\equiv \frac{\partial G}{\partial r}\hat{r}.$

In other words, the force is proportional to the derivative of the potential and the derivative of $\frac{1}{r}$ is $-\frac{1}{r^2}$: i.e., if the force is proportional to $\frac{1}{r^2}$, the potential must be proportional to $-\frac{1}{r}$.

Ah, right... Okay, that's clear now.

Quote

This is nothing more than dl expressed in radial coordinates (radial coordinates for the x and y directions). See
Polar Coordinates -- from Wolfram MathWorld

The differentials dr and d theta are always orthogonal to one another so the distance dl can easily be expressed in either set. The only difficulty to be handled is the fact that the actual distance moved when theta is changed is proportional to “r”. I think you can figure the thing out. If you still have problems, let me know and I will clear it up in detail.

Hmmm, ahha, okay, while I don't understand that completely, I can take it on faith, being that that Wolfram page about polar coordinates states that a line element is given by $ds^2 = dr^2 + r^2d\theta^2$.

(If you want to expand on it, feel free to, but I'm pretty convinced it is valid anyway)

Doctordick said:

It follows that the metric to be minimized to yield gravity consistent with a pseudo force is $dl=nds$ which can be written

$dl={\left[1-\frac{2\kappa M}{c^2r}\right]^{-\frac{1}{2}}}\left[(d\tau)^2+(dr)^2+r^2(d\theta)^2\right]^{\frac{1}{2}}$

Hmm, right, looks like $\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$ so yes I understand how you got that expression.

Quote

Using the standard Euler-Lagrange notation, the path integral over which the variation is to vanish (“J” in equation #1) is given by

$J =\int f(t,y,\dot{y})dt=\int^{P_2}_{p_1}\frac{1}{\sqrt{1-\frac{2\kappa M}{c^2r}}}\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}dt$

Since “t” in my reference frame is actually a path length measure (essentially a reference as to where one is on the referenced path) I have changed the “ds” into “dt” thus converting the original expression given for ds, $\left[(d\tau)^2+(dr)^2+r^2(d\theta)^2\right]^{\frac{1}{2}}$, into $\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}dt$.

Hmm, I'm struggling here a bit... So I take it that $d\tau \equiv \dot{\tau}$; that that's just a matter of notation.

Is "t" actually referring to the evolution parameter of your universal notation? What does it mean that it's a path length measure? I don't really understand why it appears in the euler lagrange equation; does it always have something to do with time evolution of something? Needless to say, I'm a bit confused... :I

-Anssi
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### #47Doctordick

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Posted 02 July 2010 - 01:21 PM

I’m somewhat puzzled by how we can measure the distance that the light travels and conclude that the light in the hole is traveling slower then the light outside of the hole because I don’t see how we can establish the horizontal measures and know that the distance at the top will be the same as that in the hole.
The thesis here is that the observed force is due entirely to a distortion of the coordinate system thus the consequences of that acceleration is exactly what would be expected of any other acceleration. Acceleration is a change in velocity. That change in velocity results in Lorentz-Fitzgerald contraction in the direction of that velocity change but no change in measures orthogonal to that acceleration. Thus, so long as the acceleration is uniformly in the same direction (that is, the source of the gravitational field is sufficiently far away as to make the sides of the gravitational well essentially parallel), the distance across the well is the same at the top as it is at the bottom. The clock consists of two mirrors on the opposite sides of the well and thus a slow clock requires light to move slower in the well as the distance moved in a cycle is the same at both heights.

While a clock could be designed that had this built into it, do we need to build it into it? ... If this is not an effect of the acceleration then are you in fact saying that the physics are different for an accelerating object and so the original clock must be redesigned?
Of course this kind of clock has to be redesigned; I am talk about that very fact when I bring up the issue of the size of the clock (the clock would be any clock dependent upon electromagnetic harmonic resonance).

Doctordick said:

The error is clearly a function of the physical size of the clock. This effect can be eliminated by defining our clock's size to be small enough to make the error negligible. I point this out so that we can discuss further ramifications of those instantaneous relativistic effects.
The issue is that the path of a photon is influenced by gravity.

But won’t this scaling result in the faces of the mirror being parallel in this frame and so it would make sense that the clock is the same as one not in the gravity well?
All this says is that, if we make our clock so large that we have to take into account the gravitational refraction of the photon path, we have to include that secondary problem. Think about that! Essentially it says that if the gravitational field is more than the Sun’s, your clock should be somewhat smaller than the sun (that is the circumstance which made this refraction measurable)! If your clock is less than a foot across, the acceleration necessary to make this a problem would be utterly beyond your interest; human beings simply couldn’t survive in such an environment.

AnssiH said:

Hmm, I'm struggling here a bit... So I take it that $d\tau\equiv \dot{\tau}$; that that's just a matter of notation.
Not quite. The "dot" signifies the derivative with respect to "t" so $\dot{\tau}$ is equivalent to $\frac{d \tau}{dt}$; slightly different from what you have stated. In my geometry, $\frac{d \tau}{dt}$ is essentially the velocity of an object in the tau direction just as $\frac{dx}{dt}$ would be the velocity in the x direction.

AnssiH said:

Is "t" actually referring to the evolution parameter of your universal notation? What does it mean that it's a path length measure? I don't really understand why it appears in the euler lagrange equation; does it always have something to do with time evolution of something? Needless to say, I'm a bit confused... :I
In this case, we are looking at how physics looks inside a gravitational field when observed from outside that gravitational field (or perhaps from a different place in that field). The physical process being observed is in freefall: i.e., there are actually no forces on it other than the pseudo force due to the supposed geometric distortion caused by the gravitational potential. A clock at that point in space will run exactly as defined by that potential in its own reference frame. Of course, as seen by the outside observer, it must also be corrected by velocity effects (special relativistic effects); but I have already shown that all clocks actually measure changes in tau along their path in my four dimensional space. Thus dt (as measured by a clock following a geodesic path) and $d\tau$, path length in my geometry, are the same thing (the fact that everything is momentum quantized in the tau direction is immaterial except for projecting out that tau direction when it comes to actual observations). Think of it this way, $d\tau$ in Einstein's picture is what is called the “invariant interval”: i.e., everyone gets the same value when they calculate that term. But dx=dy=dz=0 in the rest frame of a moving clock thus the clock measures exactly the same value everyone else gets for $d\tau$: i.e., the clock measures exactly that invariant path length.

In the Euler-Lagrange equation, all the mathematical terms may be regarded as abstract concepts: i.e., they are talking about abstract solutions to abstract equations. If one can interpret one’s equation in such a way that it conforms to the Euler-Lagrange equation then the solutions to that equation are solutions to his equation. It is as simple as that. That is the great power of mathematics.

Have fun -- Dick
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### #48AnssiH

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Posted 05 July 2010 - 11:42 AM

Doctordick said:

Not quite. The "dot" signifies the derivative with respect to "t" so $\dot{\tau}$ is equivalent to $\frac{d \tau}{dt}$; slightly different from what you have stated. In my geometry, $\frac{d \tau}{dt}$ is essentially the velocity of an object in the tau direction just as $\frac{dx}{dt}$ would be the velocity in the x direction.

Ah, okay.

Quote

...but I have already shown that all clocks actually measure changes in tau along their path in my four dimensional space. Thus dt (as measured by a clock following a geodesic path) and $d\tau$, path length in my geometry, are the same thing (the fact that everything is momentum quantized in the tau direction is immaterial except for projecting out that tau direction when it comes to actual observations).

Right okay, and I suppose then that...

Doctordick said:

Since “t” in my reference frame is actually a path length measure (essentially a reference as to where one is on the referenced path) I have changed the “ds” into “dt” thus converting the original expression given for ds, $\left[(d\tau)^2+(dr)^2+r^2(d\theta)^2\right]^{\frac{1}{2}}$, into $\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}dt$.

...that move is based on the fact that every element is moving at constant velocity, so any change in "t" also entails a similar change in $\sqrt{\tau^2 + r^2 + r^2\theta^2}$

Doctordick said:

Following the standard Euler-Lagrange attack, if the expression

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

satisfies the Euler-Lagrange equation, the variation of the integral over the implied path will vanish. In this case we have three independent variables to deal with (note that “y”, in this case, stands for the variable of interest) which leads to three independent differential equations.

Just to be absolutely sure, by "three independent variables" you refer to $\tau$, $r$ and $\theta$... (We are not concerned of $M$)

Quote

However, we have one additional constraint which simplifies the problem considerably: $(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2=c^2$.

Yup.

Quote

The partials of f with respect to tau and theta vanish

Hmmm, I should probably understand by now why that is, but I'm a bit lost apparently... You are saying that changing only $\tau$ or $\theta$ does not change the output of the function $f$? But I don't quite understand how do we know that... :/

...little help...?

-Anssi
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### #49Doctordick

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Posted 06 July 2010 - 11:25 AM

Hi Anssi, I want you to know that I really appreciate your attempts to understand each and every step in my presentation. If some of these other people would make that kind of examination, this whole thing would have been settled years ago. Again, mathematics is not an easy subject because sometimes one overlooks what would be utterly obvious if one were used to the manipulations being performed. It's a learning thing. I know you will kick yourself when you see what you overlooked.

AnssiH said:

...that move is based on the fact that every element is moving at constant velocity, so any change in "t" also entails a similar change in $\sqrt{\tau^2 + r^2 + r^2\theta^2}$
First, that statement is not exactly correct. The actual fact is that a change in “t” entails a change in tau, r and theta (these are the polar coordinates of the position). The problem is that $\hat{r}$ and $\hat{\theta}$ although orthogonal to one another point in varying directions during a change. If we were in rectilinear coordinates, x,y, tau (z being immaterial by the fact that the motion must be in the x,y plane), $\sqrt{\tau^2 + x^2 + y^2}$ would be the distance from the origin by the Pythagorean theorem. That theorem depends very much on the fact that $\hat{\tau}$, $\hat{x}$ and $\hat{y}$ continue to point in the same directions as the defined position changes. Opposed to this, $\sqrt{\tau^2 + r^2 + r^2\theta^2}$, is not an easily definable measure. On the other hand, $\sqrt{\dot{\tau}^2 + \dot{r}^2 + r^2\dot{\theta}^2}$ and $\sqrt{\dot{\tau}^2 + \dot{x}^2 + \dot{y}^2}$ are essentially the same thing: i.e., they are the magnitude of the instantaneous velocity.

What you should have said is that “a change in t also entails a similar change in position”; however, $\sqrt{\tau^2 + r^2 + r^2\theta^2}$ is not the position.

AnssiH said:

Just to be absolutely sure, by "three independent variables" you refer to $\tau$, $r$ and $\theta$... (We are not concerned of $M$)
That is correct!

But let's get back to your real problem.

AnssiH said:

Hmmm, I should probably understand by now why that is, but I'm a bit lost apparently... You are saying that changing only $\tau$ or $\theta$ does not change the output of the function $f$? But I don't quite understand how do we know that... :/

...little help...?
First, look at the line immediately prior to the one you quote: i.e., just before

Doctordick said:

However, we have one additional constraint which simplifies the problem considerably: $\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2=c^2.$
That line is

Doctordick said:

Following the standard Euler-Lagrange attack, if the expression

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

satisfies the Euler-Lagrange equation, the variation of the integral over the implied path will vanish. In this case we have three independent variables to deal with (note that “y”, in this case, stands for the variable of interest) which leads to three independent differential equations.
If $\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2=c^2$, then f must be

$f(t,y,\dot{y})=\frac{c}{\sqrt{1-\frac{2\kappa M}{c^2r}}}:$

i.e., the only argument in “f” is “r” ; there is no dependence upon either tau or theta. (Actually, it is not necessary to make that substitution as all it really does is make f look simpler; however, it does hide an "r" dependence but that has no consequence as we don't solve the third equation.) The Euler-Lagrange equations of interest in this case (see equation #3) are,

$\frac{\partial f}{\partial \tau}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)=0$

$\frac{\partial f}{\partial \theta}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)=0$

and

$\frac{\partial f}{\partial r}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{r}}\right)=0$

The first two are trivial; if the differential with respect to t is zero one can integrate the function and obtain a constant which is exactly what I do (that constant is exactly identical to what occurs in the parenthesis between $\frac{d}{dt}$ and the dt being integrated over).

Quote

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

The third equation is quite complex but I don't need to solve it because I have the fact that $\sqrt{\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2}=c$. I can use that relationship to obtain $\dot{r}$ from the solutions to the first two. Let me know if you have any trouble following the conversion from $\dot{r}$ and $\dot{\theta}$ to $\frac{dr}{d \theta}$.

I know you are enjoying this! You are aren't you? ;) :lol::lol:

Have fun -- Dick
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### #50AnssiH

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Posted 07 July 2010 - 08:31 AM

Doctordick said:

Hi Anssi, I want you to know that I really appreciate your attempts to understand each and every step in my presentation.

Thank you, that is always nice to hear :)

Quote

First, that statement is not exactly correct...
.
.
.
What you should have said is that “a change in t also entails a similar change in position”; however, $\sqrt{\tau^2 + r^2 + r^2\theta^2}$ is not the position.

Ah right! Of course... But yes, what I tried to say was indeed "change in position"

Quote

But let's get back to your real problem.
First, look at the line immediately prior to the one you quote: i.e., just before
That line is
If $\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2=c^2$, then f must be

$f(t,y,\dot{y})=\frac{c}{\sqrt{1-\frac{2\kappa M}{c^2r}}}:$

i.e., the only argument in “f” is “r” ; there is no dependence upon either tau or theta.

Ah right that's what you meant, okay.

By the way, all those parentheses in $\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2=c^2$, I guess they are there because we are taking the square of the derivative i.e. $\left (\frac{dr}{dt} \right )^2$, while writing $\dot{r}^2$ would be taken as $\frac{dr^2}{dt}$? (the derivate of the squared r?) I'm not sure if that even makes any sense, perhaps it would be just ambiguous notation without the parentheses or something?

Quote

(Actually, it is not necessary to make that substitution as all it really does is make f look simpler; however, it does hide an "r" dependence but that has no consequence as we don't solve the third equation.) The Euler-Lagrange equations of interest in this case (see equation #3) are,

$\frac{\partial f}{\partial \tau}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)=0$

$\frac{\partial f}{\partial \theta}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)=0$

and

$\frac{\partial f}{\partial r}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{r}}\right)=0$

Ahha, right I see.

Quote

The first two are trivial; if the differential with respect to t is zero one can integrate the function and obtain a constant which is exactly what I do (that constant is exactly identical to what occurs in the parenthesis between $\frac{d}{dt}$ and the dt being integrated over).

Quote

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

Um, hmmm... I don't understand integration enough to know anything about how or why you can "integrate the function and obtain a constant if the differential with respect to t is zero"... :I

I was looking at those equations in the OP and couldn't really understand the moves you do there, I suspect simply because I don't understand what you mean by that comment exactly...

Quote

The third equation is quite complex but I don't need to solve it because I have the fact that $\sqrt{\left(\dot{\tau}\right)^2+\left(\dot{r}\right)^2+r^2\left(\dot{\theta}\right)^2}=c$. I can use that relationship to obtain $\dot{r}$ from the solutions to the first two. Let me know if you have any trouble following the conversion from $\dot{r}$ and $\dot{\theta}$ to $\frac{dr}{d \theta}$.

Well, one step at a time...

Quote

I know you are enjoying this! You are aren't you? ;) :lol::lol:

Yes! :D

-Anssi
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### #51Doctordick

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Posted 07 July 2010 - 10:39 AM

AnssiH said:

I'm not sure if that even makes any sense, perhaps it would be just ambiguous notation without the parentheses or something?
Actually no; there is no good reason for those parentheses other than to emphasize the existence of the “dot” on top. I am reminded of a math professor I once had who got his Ph.D. in math long before the existence of air conditioners. The “dot” notation was invented by Newton and the $\frac{d}{dt}$ notation by Leibniz. My professor referred to them as “dot-ism” and “deism”. His position was that “deism” was preferable, as with “dot-ism” one often had to contend with flies performing unwanted differentiations. :lol: :lol:

Your idea that $\dot{r}^2$ could be interpreted as the differential of $r^2$ is quite new to me. I suppose it could. Anyway, it does appear that “dot-ism” is fraught with possible erroneousness interpretations. I only used it because the web page on the Euler-Lagrange equation did.

AnssiH said:

Um, hmmm... I don't understand integration enough to know anything about how or why you can "integrate the function and obtain a constant if the differential with respect to t is zero"... :I
Sorry about that. You do so well at interpreting what I say that I forget you don't have a formal education in integral calculus. You need to take a quick look at the definition of an indefinite integral. Expression #1 is what is commonly called an “indefinite integral”; indefinite in that its value is not known. As is commented there, it is also called an “antiderivative”; which would be the opposite of a derivative.

Essentially, what that means is that if $\int f(x)dx = F(x)$ then $\frac{d}{dx}F(x)=f(x)$ by definition: i.e., one operation is the inverse of the other. The definite integral is when the integral sign has specific limits: i.e., $\int^b_a f(x)dx$. In that case the (now “definite”) answer is F(b)-F(a).

You might enjoy reading the Wikipedia entry for calculus terminology, both for its “history of calculus” entry and its excellent “Introduction” to calculating integrals. I know I did.

But let's get back to your question. Changing the variable “x” to “t”, if $\int f(t)dt = F(t)$ then it must be true that $\int \frac{d}{dt}\left\{F(t)\right\}dt = F(t)$; correct? But what happens to that expression if we look at the differential representation $\frac{d}{dt}F(t)=f(t)$ 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 $\frac{d}{dt}G(t)=f(t)+0\equiv f(t)$.

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

Think about it a little and I think you will understand exactly what is going on there.

Have fun -- Dick
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### #52AnssiH

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Posted 08 July 2010 - 04:53 AM

Doctordick said:

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 $\int f(x)dx = F(x)$ then $\frac{d}{dx}F(x)=f(x)$ 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 $\int f(x)dx$ can be interpeted as the total area between 0 and $x$...

Quote

The definite integral is when the integral sign has specific limits: i.e., $\int^b_a f(x)dx$. 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 $\int^b_a f(x)dx=F(b)-F(a)$, 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, $\int f(x)dx = F(x)$ means the function $F(x)$ 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 $f(x)$. (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)

Quote

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

i.e. the indefinite integration of the derivative of $F$ just gets us right back to mere $F$...

Since $\frac{d}{dt}\left\{F(t)\right\}=f(t)$, then by trivial substitution $\int \frac{d}{dt}\left\{F(t)\right\}dt = \int f(t) dt = F(t)$

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

Quote

But what happens to that expression if we look at the differential representation $\frac{d}{dt}F(t)=f(t)$ 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 $\frac{d}{dt}G(t)=f(t)+0\equiv f(t)$.

Yes...

Quote

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

Right, so in other words, the (indefinite) integral of $f(x)$ can always be seen as $F(x) + C$ 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.

Doctordick said:

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

$\frac{\partial f}{\partial \tau}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)=0$

$\frac{\partial f}{\partial \theta}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)\equiv -\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\theta}}\right)=0$

and

$\frac{\partial f}{\partial r}-\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{r}}\right)=0$

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;

$\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{\tau}}\right)=0$

and

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

we know that;
$\int \frac{d}{dt}\left\{ \frac{\partial}{\partial \dot{\tau}}f(t) \right\}dt = Constant$

With that in mind...

Doctordick said:

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

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

...I can see how;

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

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

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

In particular the $\frac{\dot{\tau}}{c}$ in there... The closest I got to it myself was;
$\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}c\right\}$

which I realize is problematic because there's no $\tau$ 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 $\tau$ variable in there... So trivial substitution of c for $\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}$ cannot be done...? Hmmm, well I'm not entirely sure what happens there... Help!

-Anssi
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### #53Doctordick

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Posted 08 July 2010 - 12:02 PM

AnssiH said:

Okay, after playing around with Wolfram Alpha a bit, I take it then that the indefinite integral $\int f(x)dx$ can be interpeted as the total area between 0 and $x$...
Actually that is not a very good interpretation. A better statement is that $\int f(x)dx$ yields a functional relationship which tells one how that total area changes as a function of x, not what the total area actually is.

AnssiH said:

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 $\int^b_a f(x)dx=F(b)-F(a)$, 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.

AnssiH said:

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.

AnssiH said:

Okay, I think I pretty much get the idea.
Yeah, you are getting there.

AnssiH said:

...I can see how;

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

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

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

In particular the $\frac{\dot{\tau}}{c}$ in there... The closest I got to it myself was;
$\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}c\right\}$
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 $r=\sqrt{x^2+y^2}=$ 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.

AnssiH said:

So trivial substitution of c for $\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}$ 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 $\dot{\tau}=\frac{d \dot{\tau}}{dt}$ and the other with respect to $\dot{\theta}=\frac{d \dot{\theta}}{dt}$. Since

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

it follows that

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

and

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

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

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

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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### #54AnssiH

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Posted 09 July 2010 - 03:21 AM

Doctordick said:

Actually that is not a very good interpretation. A better statement is that $\int f(x)dx$ 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...

Quote

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.

Quote

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 $r=\sqrt{x^2+y^2}=$ 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 $\frac{\partial}{\partial x} ( \sqrt{x^2+y^2} ) = \frac{x}{ \sqrt{x^2+y^2} }$. I was able to follow the steps of algebra in between, so with that I can understand that;

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

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

The partial derivative of $\dot{\theta}$ 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;

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

Quote

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 $\dot{\tau}=\frac{d \dot{\tau}}{dt}$ and the other with respect to $\dot{\theta}=\frac{d \dot{\theta}}{dt}$. Since

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

it follows that

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

and

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

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.

Quote

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

Check.

And then;

Doctordick said:

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

and

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

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?)

Quote

Multiplying through by that other term, we can conclude that

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

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 $\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\frac{\dot{\tau}}{c} = \frac{1}{cl}$ and $\dot{\tau} = \frac{1}{l} \sqrt{ 1 - \frac{ 2 \kappa M } {c^2r}}$

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

Doctordick said:

Using $(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2=c^2$, we can write

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

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

Quote

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

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

and the equation for the geodesic can be directly written as

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

(trivial typo there, verses/versus)

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

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

-Anssi
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### #55Doctordick

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Posted 09 July 2010 - 10:06 AM

AnssiH said:

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

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

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:

AnssiH said:

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:

AnssiH said:

In terms of pure algebra, that all looks quite valid to me. But what does $r'$ 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 $r=f(\theta)$.

AnssiH said:

I.e. I'm not at all sure what you are saying there with $\quad r'=\frac{dr}{d\theta}=\frac{\dot{r}}{\dot{\theta}}, \quad \dot{r}=\dot{\theta}r'=\frac{h}{r^2l}r'\sqrt{1-\frac{2\kappa m}{c^2 r}},$
Well, first of all, the definition of r' is $r'=\frac{dr}{d\theta}$ and, since $\dot{x}$ is defined to be the derivative of x with respect to time, $\dot{r}$ is, by definition, $\frac{dr}{dt}$ and $\dot{\theta}$ is likewise $\frac{d\theta}{dt}$. 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 $\frac{\dot{r}}{\dot{\theta}}$ is identical to $\frac{dr}{d\theta}=r'$. From this (if we multiply r' by $\dot{\theta}$) we can assert that $\dot{r}=\dot{\theta}r'$. But we have already discovered that

$\dot{\theta}=\frac{h}{r^2l}\sqrt{1-\frac{2\kappa M}{c^2r}}$

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

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

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

AnssiH said:

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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Posted 10 July 2010 - 09:16 AM

Doctordick said:

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 $\tau$ axis as being scaled to make up for all objects still having a constant velocity in your x,y,z,$\tau$ geometry, and since light still cant travel along the $\tau$ axis, since it is still massless, in effect we are scaling the $\tau$ axis and saying that objects are moving a shorter distance in the x,y,z,$\tau$ 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 $\tau$ axis can’t be directly measured and actual location on it makes no difference such an effect could not be noticed.

Doctordick said:

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 $\tau$ direction even though the object really is not going any where as a result the object will travel slower in the $\tau$ 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?

Doctordick said:

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 $\tau$ 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,$\tau$ 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?

Doctordick said:

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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### #57Doctordick

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Posted 11 July 2010 - 10:27 PM

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 $\tau$ 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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### #58AnssiH

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Posted 15 July 2010 - 06:42 AM

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

Doctordick said:

Following the standard Euler-Lagrange attack, if the expression

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

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

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

and

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

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

Doctordick said:

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

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

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...

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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...

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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 ;)

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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 $r=f(\theta)$.

Ahha, okay.

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Well, first of all, the definition of r' is $r'=\frac{dr}{d\theta}$ and, since $\dot{x}$ is defined to be the derivative of x with respect to time, $\dot{r}$ is, by definition, $\frac{dr}{dt}$ and $\dot{\theta}$ is likewise $\frac{d\theta}{dt}$. 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 $\frac{\dot{r}}{\dot{\theta}}$ is identical to $\frac{dr}{d\theta}=r'$. From this (if we multiply r' by $\dot{\theta}$) we can assert that $\dot{r}=\dot{\theta}r'$. But we have already discovered that

$\dot{\theta}=\frac{h}{r^2l}\sqrt{1-\frac{2\kappa M}{c^2r}}$

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

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

When you make that substitution into our expression for c2 as a function of $\dot{r}$ 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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Posted 15 July 2010 - 06:26 PM

Doctordick said:

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 $\tau$ axis has no influence due to the uncertainty of location on the $\tau$ axis. As a result the location on the $\tau$ 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.

Doctordick said:

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

$c'=c\sqrt{1+\frac{2\Phi}{c^2}}$

(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 $\tau$ 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.

Doctordick said:

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.

Doctordick said:

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 $\tau$ direction but instead the object will be refracted just like any other wave is refracted.

Now this refraction takes place along the $\tau$ 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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### #60Doctordick

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Posted 15 July 2010 - 06:32 PM

AnssiH said:

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

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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