## View Poll Results: What do you think about my work and article "Dilation as field"? (post #56 and links)

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• Your work looks like at least interesting, keep developing - I wish you good luck

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# Thread: Kepler's third law rules the Gravity

1. ## Kepler's third law rules the Gravity

Hello all.

I would like to show something that I hope, you may like.
You will see, that Kepler was perfectly right, but in some unexpected way.

I provide references as links in text.
In whole post we assume that c=1 to simplify calculations.

We take 3rd Keppler law for some rotation in R distance where T is period. We may transform it to form of:

Let our constans for considered rotation will be equal to Schwarschild radius R_s, then:

Let us derive time dilation factor gamma for such move:

As you probably see, it is the same factor that is present in Schwarzschild metric, where for geodesics we have:

Let us differentiate our gamma factor by R. Surprise! - we obtain gravitational acceleration...

If you do not believe it is correct, look at the reference, formula (25), keeping in mind, that:

Looks interesting?
Take a look at this. Now, we derive rest of Schwarschild metric...

At first we recall and transform simply Newtonian formula for accelerated move for our case:

as it is easy to calculate if we differentiate above by "R" we obtain:

thus:

Now, we define observer resting in R distance to source of gravity (f.e. we on Earth).
His proper time formula we may denote as:

Thanks to above observer, Schwarzschild metric may be rewritten for "every particular observer" (without any bad looking "dt" in infinity). For geodesics Schwarzschild metric will be in form of:

Let us rewrite above as:

Now, we show, that difference between Schwarzschild and Minkowski is equal to t_g:

If we denote spatial increment as dx

Schwarzschild metric appears to be consequence of some Kepler rotation in Minkowski timespace...

If you are interested why it works this way, I explain it in my article draft and partly in my previous posts.
Last edited by pogono; 2011-Dec-11 at 06:47 PM. Reason: mistake corrected, thanks to grapes

2. Originally Posted by pogono
Let us differentiate our gamma factor by R. Surprise! - we obtain gravitational acceleration...

If you do not believe it is correct,
I admit I have a doubt, but I may be making a mistake, or misunderstanding a step. Shouldn't there be another factor of R in the denominator? I've added the exponent 2:

3. Originally Posted by grapes
I admit I have a doubt, but I may be making a mistake, or misunderstanding a step. Shouldn't there be another factor of R in the denominator? I've added the exponent 2:

Yes, it should be this way, thank you grapes.
I have corrected the post.

I was too hurry rewriting my post from my notebook.
As you may easy see, it does not bring mistake to next formulas.

4. BTW, I hope you may see that I have improved my skills since first posts here ;-)

So, how do you like above ATM idea?

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Originally Posted by pogono
Let us differentiate our gamma factor by R. Surprise! - we obtain gravitational acceleration...
I wonder, why are you surprised with that?

You have two concepts, gravitational time dilation (gamma factor) and gravitational force (acceleration).
What if you consider a "middle concept", gravitational potential, such that you relate time dilation <-> potential <-> force?
If you notice that time dilation is directly proportional to the gravitational potential, they are interchangeable, and gravitational potential is nothing more than the integral of the force over the distance, or in other words the force/acceleration is the gradient of the potential (ie the derivative to R), then it might not seem so surprising.

6. Hello caveman1917!
I am glad to hear you! :-)

Originally Posted by caveman1917
You have two concepts, gravitational time dilation (gamma factor) and gravitational force (acceleration).
What if you consider a "middle concept", gravitational potential, such that you relate time dilation <-> potential <-> force?
If you notice that time dilation is directly proportional to the gravitational potential, they are interchangeable, and gravitational potential is nothing more than the integral of the force over the distance, or in other words the force/acceleration is the gradient of the potential (ie the derivative to R), then it might not seem so surprising.
Exactly.

Time dilation acts here as scalar potential. So, we might even say, that dilation IT IS gravitational potential, the more because ratio between dilation and gravitational potential is just equal to "c"...

This is exactly my idea that I prove in my article (that is why article is called "Dilation as field").

Moreover...
If we treat dilation factor as scalar potential we might derive some vector fields based on it.

Maxwell removed charges while keeping Electric field. That is how we obtain Electromagnetic field idea without charge as source of field.

We may do the same with mass. We remove mass from described rotation keeping time dilation as field and we obtain... field equations that I have linked in my previous post (and in my article).

Originally Posted by caveman1917
I wonder, why are you surprised with that?
I am not surprised! :-)
It is exactly what I show by my article. But most of physicists are surprised and I have a lot of troubles with asking them to act as reviewers for my article. The best description of this surprise is here:

Originally Posted by Celestial Mechanic
Dilation Field Theory Dialogues -- Part One: Vocals

The heat wave broke (at least here in Wisconsin) and once again Celestial Mechanic comes out of the woodwork long enough to order up some doughnuts, brew up some coffee, and invite his two friends (who are now grad students), Virginia and Jimmy K. over to ponder over an ATM physics paper.

Jimmy K.: "What have you got for us this time?"

Celestial Mechanic: "Coffee, doughnuts, the usual. Oh, and a paper by Piotr Ogonowski who wants to raise time dilation to the status of a field."

Virginia: "The interesting thing about this paper is that the author actually ventures to write a Langrangian. Most ATM authors have never heard of Lagrangian and Hamiltonian dynamics. Of course his Lagrangian is all wrong."

CM: "True, true. Lately I have been thinking a lot about Langrangians in the context of celestial mechanics and the arguments (not proof!) for the plausibility of the ones that we use. I'll have more to say on that later. Let's start critiquing his paper from the beginning, at the abstract."

V: "Well, the very first sentence is completely wrong on several counts. Ogonowski writes:
In this paper I show that time dilation may be treated as a field ...
V: "... which is wrong because the time dilation measured at a point is a function of velocity (the SR, special relativity part) and the gravitational potential (the GR, general relativity part). It is more sensible to treat velocity as a field, for example a rotating body in orbit. The individual particles of the body have different velocities that can be analyzed into orbital and rotational components. I won't comment on the rest of the sentence."

....
As you may see, it is not easy to pass review process with above idea.
Last edited by pogono; 2011-Dec-06 at 09:36 PM.

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Originally Posted by pogono
At first we recall and transform simply Newtonian formula for accelerated move for our case:

Remember that your is the tangential speed, where is a centripetal acceleration.
So better would be

Where does the come in, what does it mean?

8. Originally Posted by caveman1917
Where does the come in, what does it mean?
Good question.
Here we go with explanation.

Hanging Observer theoretically stands in one place.
But in fact he have to accelerate, to keep his position.

All around him, fall down to source of gravity... So, indeed he moves, relative to the carpet that he stands on - "Minkowski time-space". The same time, by symetry, whole time-space moves around him.

Thus we may consider HO in two frames:

Frame A)

HO velocity relative to the observer in infinity (dt) will be consequence of regular Minkowski, because HO moves relative to time-space:

1)

2)

Frame B)

But the same time, whole Minkowski time-space around him is accelerating, achieving v_rot velocity.

But HOW to note velocity for time-space????

So, we recall back Newtonian v = at.

For "moving time-space" we would write:

3)

now, we make derivative by "dr" as it was shown in previous post thus we obtain:

4)

so we may see that relative to HO, whole time space-time moves as follows:

5) where:

6)

Joining equation 2) and 5) we obtain Schwarzschild Metric for geodesics.

P.S.
What all of above has to do with rotation?

It is not obvious, but we have described both ways rotation between HO and time-space.
Last edited by pogono; 2011-Dec-10 at 09:55 PM.

9. BTW,
it is also easy to see, that:

Interesting coincidences.
Do not you think?

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Originally Posted by pogono
HO velocity relative to the observer in infinity (dt) will be consequence of regular Minkowski, because HO moves relative to time-space:

1)

2)
A hanging observer will not have any velocity relative to the observer at infinity, that's why he's a "hanging observer" to use your term.
He certainly will not have velocity since that is the tangential speed of a rotating object, you derived that quantity from (and you don't seem to be treating them as vectors).

11. Originally Posted by caveman1917
A hanging observer will not have any velocity relative to the observer at infinity, that's why he's a "hanging observer" to use your term.
Yup. Hanging observer hangs at one distance "R" to source of gravity.
But, we should rather not call above position "hanging in one place" because all "places" around him are free falling.
So indeed, he moves relative to all around him.

We may also show, that any observer to be not affected by gravity have to move with "Escape velocity" that is accidentally equal to:

what in infinity goes to 0.

So, observer in infinity moves with above velocity or stands?
Because it seems to be the same.

Above v_rot velocity also appears to be free-falling speed.

Originally Posted by caveman1917
He certainly will not have velocity since that is the tangential speed of a rotating object, you derived that quantity from (and you don't seem to be treating them as vectors).
I never said it is tangential.
I wrote, that it acts as some rotation but in quite unexpected way...
By analyzing bodies behavior we should rather say, it is coordinate velocity.
Last edited by pogono; 2011-Dec-11 at 01:57 PM.

12. ## Review!! I have recieved REVIEW!!

I got review!! I have just received the REVIEW for my article!!! :-)))))

Let me cite the beginning:

"Dear Mr. Ogonowski,

First, let me say that I am duly impressed with the level of physics understanding displayed by someone not trained formally in physics. (I checked your background online.) I have reviewed a number of manuscripts by authors without such formal training, and you are easily the most accomplished of the bunch. Excellent learning effort.
(...)
1) The abstract should not be a section, but a paragraph on its own on a title page. Also, technical articles generally do not have a table of contents. I personally disagree with this practice, but it is nearly universal.

2) You use unfamiliar terms and ways to represent otherwise more familiar items. Getting this manuscript into publishable form would need someone with the appropriate background to work with you extensively in order to get it into a form most relativists would feel comfortable with.
(...)
"
Can you see it???
I am jumping up to my chandelier!!!!!!!!!!! :-)))))

Today you all drink at my expense!!! :-D

Incredible. I was waiting for this moment for YEARS!!!!
Wow!

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Congratulations. I hope the advice they give is helpful in getting this into a peer reviewed publication.

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Originally Posted by pogono
Yup. Hanging observer hangs at one distance "R" to source of gravity.
But, we should rather not call above position "hanging in one place" because all "places" around him are free falling.
So indeed, he moves relative to all around him.

We may also show, that any observer to be not affected by gravity have to move with "Escape velocity" that is accidentally equal to:

what in infinity goes to 0.
But escape velocity is the velocity you should have so that you, without any further acceleration, can "barely" reach infinity. Having a constant velocity is not enough to stay stationary at some distance from the source, you need some constant acceleration, if you let him have escape velocity he will...escape

Above v_rot velocity also appears to be free-falling speed.
Of course, the escape speed at some point equals the speed some object that "fell in" from infinity would have at that point, they are the same concept.

I never said it is tangential.
I wrote, that it acts as some rotation but in quite unexpected way...
Do you mean to say that there is a relation between the tangential velocity of a rotating object at some distance and the escape speed at that distance?

If so, yes. From newton: and (taking the orbit to be circular) so they are related by a factor of

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Originally Posted by pogono
I got review!!
Congratulations.

On the note of the use of unfamiliar terms, you might for example want to change "hanging observer" to "stationary observer".

16. Originally Posted by caveman1917
Congratulations.
On the note of the use of unfamiliar terms, you might for example want to change "hanging observer" to "stationary observer".
Originally Posted by Shaula
Congratulations. I hope the advice they give is helpful in getting this into a peer reviewed publication.
Thank you.
It is thanks to all of you here and your questions, forcing me to think and to learn.

Originally Posted by caveman1917
Do you mean to say that there is a relation between the tangential velocity of a rotating object at some distance and the escape speed at that distance?
It was not my point.
Please, try to see it this way:

Photons (light) are free-falling to source of gravity achieving hypothetically v_rot speed.
In reference frame assigned to "falling photons" it is HO who moves, what drives to:

(1)

But... the same time, by analogy in HO reference frame: photons are falling, what is shown here:

(2)

where from accelerated move v=at we have:

(3)

Joining (1) + (2) you will obtain Schwarzschild metric for geodesics.

Please, make your own calculation of above, to convince yourself that it works.

To obtain make derivative of formula (3) by dR.
Last edited by pogono; 2011-Dec-13 at 09:46 PM.

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You can't take a reference frame for a photon, so i'll assume you mean some massive body free-falling in from infinity.

For reference the (reduced) schwarzschild form

Let's first take the frame of a hanging observer, it is related to the schwarzschild form by

1)

Now for the freefaller, he will relate to the frame of the hanging observer by a normal Lorentz transformation.
As you say his velocity relative to the hanging observer is

doing the lorentz transformation

putting this in terms of the original schwarzschild coordinates

Substituting in the schwarzschild form we then have the freefalling frame as
2)

Having all the correct frames at our disposal, can you say what exactly the point is you're trying to make? I'm still not understanding what you are trying to do.
For example i still don't see how the newtonian comes in, or where comes from.

18. Originally Posted by caveman1917
You can't take a reference frame for a photon (...)
It was Einstein who said: "Everyone new, it is impossible. Someone did not know, and he did it" ;-)

Originally Posted by caveman1917
Having all the correct frames at our disposal, can you say what exactly the point is you're trying to make? I'm still not understanding what you are trying to do.
For example i still don't see how the newtonian comes in, or where comes from.
Let us start from this edge.

We consider accelerating body in some r distance to observer.
Let say, that:

If we know that:

and we know also that:

then we could write:

Now, let us write Minkowski where spatial increment is in polar coordinates:

This way we have obtained Schwarzschild metric for geodesics, where t_obs is "hanging observer" proper time.

Do you agree?

19. the first step is already mathematically wrong.

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Originally Posted by pogono
It was Einstein who said: "Everyone new, it is impossible. Someone did not know, and he did it" ;-)
It is also Einstein who used the Lorentz transformations which tell you that you can't take the reference frame of a photon (not that an argumentum ad verecundiam means much ).
Try transforming from any inertial frame to a frame with relative speed of c, and check what does.

We consider accelerating body in some r distance to observer.
Which observer, the hanging observer, the freefalling one or the one at infinity?

Let's stop there. You should divide by coordinate time to get coordinate acceleration (since v is coordinate velocity). So that should be
And remember that acceleration is defined as

21. Originally Posted by caveman1917
Let's stop there. You should divide by coordinate time to get coordinate acceleration (since v is coordinate velocity). So that should be
And remember that acceleration is defined as
Originally Posted by tusenfem
the first step is already mathematically wrong.
I am afraid this step is correct, because it is part of Rindler's transformation.

You may see it f.e. in this reference: Relativistic acceleration, page 4, formula (2.3) and page 9, formula (3.4)

Tau is proper time of a co-moving body (with velocity v), for infinite small moment being at the same place with the same speed that accelerated body.
Last edited by pogono; 2011-Dec-15 at 09:15 PM.

22. that is not the problem you can easily say that v = a t, no problem at all, we do it all the time.

however your step to t = v/a and then dt = d(v/a) (which is what I call the first step) is too simplistic as a is defined as dv/dt, this is just a mathematical step that you cannot do, point. Nor does Knorr do that on page 4 or page 9.

23. Originally Posted by tusenfem
however your step to t = v/a and then dt = d(v/a) (which is what I call the first step) is too simplistic as a is defined as dv/dt, this is just a mathematical step that you cannot do, point. Nor does Knorr do that on page 4 or page 9.
Hi tusenfem,
thank you for your post. I appreciate your remarks and many mistakes that you have pointed to me.

But this time, you are wrong.

There is absolutely no mathematical reason to avoid such transformation:

Moreover,
since know that:

and we know that acceleration is (gravitational acceleration) equal to:

thus we know, that:

thus:

Originally Posted by pogono
then we could write:

Now, let us write Minkowski where spatial increment is in polar coordinates:

This way we have obtained Schwarzschild metric for geodesics, where t_obs is "hanging observer" proper time.

Do you agree?
I can agree with you in one thing, that comes out from between lines of your post text.

It SHOULD NOT BE true, because if it is...
... then I have explained GR using SR, and just shown, that impossible is possible - I have shown how to assign reference frame to photons.
Last edited by pogono; 2011-Dec-15 at 09:19 PM. Reason: quotation added

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You seem to be mixing frames, can you rewrite clearly stating which quantities belong to which frames (such as i did a couple of posts ago, use subscripts fully for each different frame)?

25. Originally Posted by pogono
Hi tusenfem,
thank you for your post. I appreciate your remarks and many mistakes that you have pointed to me.

But this time, you are wrong.

There is absolutely no mathematical reason to avoid such transformation:

Moreover,
since know that:

and we know that acceleration is (gravitational acceleration) equal to:

Well yeah if you want to mix apples and oranges, you will still have fruit.
There is still the little thingy that dv/dt = a.
However, as it is noted that you never actually use the "d" in the steps above, you get out of having a problem.
You seem to think that "d" is just a "multiplicative factor" instead of an operator.

ETA:
The step that is taken by you and by Knorr, to say that t = v/a has several implications in it. First it assumes that the acceleration is a constant, second it assumes that the value you take for v is the difference in velocity Δv, from its value before and after acceleration over time t, just some of the things that you forget about in your ideas.
Also, if I take your v and differentiate it by time I cannot get your expression for a.

26. Originally Posted by tusenfem
Well yeah if you want to mix apples and oranges, you will still have fruit.
There is still the little thingy that dv/dt = a.
It is easy to check, that my fruit salad works fine.

Let us check what it is:

Errata: should be

thus:

Errata: should be

Last edited by pogono; 2011-Dec-16 at 04:22 PM. Reason: Errata added

27. Originally Posted by caveman1917
You seem to be mixing frames, can you rewrite clearly stating which quantities belong to which frames (such as i did a couple of posts ago, use subscripts fully for each different frame)?
ok, fine.

Let us define accelerating body P (we know already, it will be "Photon") in some observer "Obs" reference frame.

According to Rindler's transformation we should be able to analyse accelerating "body P" by co-moving body velotity, in every particular place in space. This co-moving body proper time is Tau.

If we note co-moving body proper time in Minkowski metric, then in "Obs" reference frame it will be:

I hoper rest of transformation is obvious.

We only chave to check what is happening if we know that:

- "v" and "a" are considered as gravitational,

- we use polar coordinates:

Have fun ;-)

28. Originally Posted by pogono
It is easy to check, that my fruit salad works fine.

Let us check what it is:

thus:

To actually state that t = dt ??????? dt is an infinitessimally small time unit, t is some macroscopic time.
Ah well ... have fun

29. Originally Posted by tusenfem
To actually state that t = dt ??????? dt is an infinitessimally small time unit, t is some macroscopic time.
Maybe it will help....

http://en.wikipedia.org/wiki/Motion_...nd_derivatives

If you do not agree with my derivation - show me mistake in the transformation.

30. for one, I think you put in two different functions for "a" and "v" for which "a = dv/dt" does not hold (unless I made a mistake in differentiation of v)

secondly you cannot say that t = dt, if anything you can only say that t is the integral from t1 to t2 of dt, unless ofcourse you want to go to a difference equation, but then you should put in your deltas.