## View Poll Results: Pogono equations for Field (#1), GR (#2), Schw. transform (#60 and #58) seems to be:

Voters
21. You may not vote on this poll
• Mainly correct, most of math is ok, but nothing seems to confirm the theory

0 0%
• Main idea is interesting, but there are math mistakes and main problem is: contradicting experiment

2 9.52%
• Equations are full of errors and the idea is rather cloudy

4 19.05%
• Total crap, waste of time

15 71.43%

# Thread: Speed of light lower than "c" - field equations

1. Let us call Hypothesis 2 that my Field equation is correct.
Originally Posted by pogono

thus:

I will prove it answering this way MQ23.

Proof:

Step 1.
As we know by definition:
Originally Posted by pogono
Let us define at first three versors n_R, n_x, n_y. For any conductive vector R:

Let us define scalar field and two related vector fields:

As we can show:

Let us define auxiliary scalar field equal (related to angular momentum) and two auxiliary vector fields U and .

Let us also show, that:
Step 2.
From above we obtain:

We also know, that vector direction is: because it is opposite to field A vector by definition.

We also know by definition, that: has direction because it comes from angular momentum for R distance.

Step 3.
Now, we may utilise property:

What ends the proof.

2. Originally Posted by macaw
Nonsense, the RHS is scalar, the RHS is a vector. Besides, you demonstrate (again) that you do not know basic differentiation.
Not exactly.
You have just demonstrated, that you cannot differ incompetence from obvious mistake.
Corrected now. You may read again.

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Originally Posted by pogono

Proof:

Step 1.
Let us consider move on circle with line velocity T equal to:
Every time you try covering a mistake, you make a new one. Light travels in a straight line, so and cannot be collinear. To make matters worse, you keep "forgetting" that your is a function of the Schwarzschild radial coordinate.

4. Originally Posted by macaw
Every time you try covering a mistake, you make a new one. Light travels in a straight line, so and cannot be collinear. To make matters worse, you keep "forgetting" that your is a function of the Schwarzschild radial coordinate.
macaw,
every time I explain something and ask you to read carefully, you post after few seconds calling your lack of understanding as my mistake.

T is the vector for time flow.
How did you get Schwarzschild radius in it - I have no idea....

I describe Photon as disturbance in time flow, traveling through time-space, remember?

If you have no time for carefully reading my post, or this logic is to hard for you - ok, just say it.
But do not call it immediately as my mistake.

I explain things in a lit bit different way that SR/GR - as you call it "backward".
So, please, open mind for backward thinking.

From my proof comes out, that:

- Rotaion of Time Flow vector is related to line velocity.

- Rotaion of Line Velocity vector is related to acceleration.

- Changes of acceleration are related to Time Flow.

It is E-M field. Just physical structure of time-space.

5. Originally Posted by caveman1917
So far that's ok.

Is that really so?

You rely on . But note that if the curve is not a straight line anymore, you have an observer under acceleration. Does the equality hold for an observer under acceleration or only for an inertial observer?

Using a simple case, what is the metric for an observer under constant acceleration? It is not
Hello caveman1917,
thank you for comment.

You are right, for accelerated move this metric should not be valid anymore. But... for accelerated move, we look only from Moving Observer point of view. From his perspective this metric still works, what comes out from integral.

Whole Quantum Mechanic is buid on assumption, that things are considered only for particular observer. So, I did not do nothing wrong here. QM works. My "trick" - also works.

As you may see, from below integral, in Moving Observer reference frame the metric still works. Calculation of integral still results with right value for Resting Observer time flow:

Originally Posted by pogono
Resting Observer Time is at the moment Length of the curve.
As we know using trace calculation formula Observer's time flow is equal to:

Let me show example.
Twin Brother escaping to space in Twin Paradox may use above integral, to calculate predicted Time Flow for his brother Resting on Earth.

- Moving Brother proper time
- Resting "On Earth" Brother time

Resting Twin Brother - can not count times using this metric.
Moving Twin Brother - can, and will obtain right result.
Last edited by pogono; 2011-Sep-29 at 11:45 AM.

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Originally Posted by pogono

Let me show example.
Twin Brother escaping to space in Twin Paradox may use above integral, to calculate predicted Time Flow for his brother Resting on Earth.

- Moving Brother proper time
- Resting "On Earth" Brother time

Resting Twin Brother - can not count times using this metric.
Moving Twin Brother - can, and will obtain right result.
That means that your ATM is incorrect. In mainstream physics , both "brothers" can obtain the correct result. You have just illustrated another failure of your ATM.

7. ## The shortest explanation

Maybe I will try to explain my main idea shorter.
It will be easier to understand me.

Let us start from question:
"what should be photon velocity in its own reference frame?"

I claim, that in its own reference frame it should be:

thus:

This is something we used to call "time flow".

So now, let us consider photon, as rotating "some field" vectors.
Thus we may utilise Special Relativity Centroid move description, that is:

And that's all folks. Rest is just math.

For such rotation we can easy show, that:

then we can utilize derivative property making velocity:

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Originally Posted by pogono
Maybe I will try to explain my main idea shorter.
It will be easier to understand me.

Let us start from question:
"what should be photon velocity in its own reference frame?"

There is no such thing as . Instead of making up stuff, you should spend your time learning.

I claim, that in its own reference frame it should be:

...which is a nonsensical claim give that it contradicts experment (c is constant)

thus:

This is something we used to call "time flow".
This is something that we actually call "basic mistake"

9. Originally Posted by macaw
There is no such thing as . Instead of making up stuff, you should spend your time learning.
Should be:

I am used to use "s" for "road". exchanged to "x"
corrected. Take a look now.

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Originally Posted by pogono
Should be:

I am used to use "s" for "road". exchanged to "x"
corrected. Take a look now.
Just as wrong. Do you understand that experiment falsifies your claims?

11. Originally Posted by pogono
Maybe I will try to explain my main idea shorter.
It will be easier to understand me.

Let us start from question:
"what should be photon velocity in its own reference frame?"

I claim, that in its own reference frame it should be:

But isn't a function of v, specifically vphoton? And as vphoton = c, . So, by your reasoning, vphoton = 0 ?

I believe this is why there is no valid frame of reference for a photon.

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Originally Posted by Strange
But isn't a function of v, specifically vphoton? And as vphoton = c, . So, by your reasoning, vphoton = 0 ?

I believe this is why there is no valid frame of reference for a photon.
Ahh, you spoiled the fun, I was waiting for pogono to incorporate his "discovery" in his "paper". He never fails to include the new errors in his paper. You can check his daily updates, he's just incorporated this new mistake into the "paper".

I was just about to write essentially the same thing as you did:

Applies for massive particles but does NOT apply to the photon as pogono keeps posting. Why doesn't the above apply to the photon, pogono?

13. Isn't "c" defined as the speed of light in a vacuum and confirmed by observation, how can it be slower than itself?

And aren't current theories of electromagnetic propagation through a vacuum derived from observation and confirmed by further observation and not the other way around?

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Originally Posted by starcanuck64
Isn't "c" defined as the speed of light in a vacuum and confirmed by observation, how can it be slower than itself?

15. Just a second....
photon has no reference frame?

Are you guys, trying to tell me, there may be body carrying energy, without its reference frame?

Maybe we should check, then, where does Minkowski Metric come from.

And now, please tell me, what is time and what is space, when:

P.S. "c" is just the constants. It works as good as now when we say it is the LIMIT of light speed.
Last edited by pogono; 2011-Sep-29 at 09:06 PM.

16. You claim, that centroid move cannot be used for photons.

I claim, that we can do it for photon, and show, that:

Isn't it exactly what we know about photon momentum right now?

17. Originally Posted by pogono
"c" is just the constants. It works as good as now when we say it is the LIMIT of light speed.
"c" is the actual velocity measured for the propagation of electromagnetic waves through a vacuum, it's not a theoretical concept.

18. Originally Posted by starcanuck64
"c" is the actual velocity measured for the propagation of electromagnetic waves through a vacuum, it's not a theoretical concept.
Fantastic!
How can be different?

As we can see from my field equations, photon is disturbance in time-space structure, traveling thorough this structure. It means photons defines local time-space, and always, in every place structure of this time-space is:

*in local reference frame X

But when we consider OBSERVER LOCATED IN INFINITY (what is exactly what I do), he can not confirm, looking at his clock, that photon has speed of light in our, small, local part of time-space, because he is measuring our photon speed relative to his (infinite observer) time increment "dt".

Einstein called above: "time-space curvature".
I call it: new explanation for E-M field, showing, that time-space is the medium for E-M wave, and E-M wave is disturbance in this time-space.
Last edited by pogono; 2011-Sep-29 at 09:24 PM.

19. Originally Posted by pogono
Fantastic!
How can be different?

As we can see from my field equations, photon is disturbance in time-space structure, traveling thorough this structure. It means photons defines local time-space, and always, in every place structure of this time-space is:

But when we consider OBSERVER LOCATED IN INFINITY, he can not confirm, looking at his clock, that photon has speed of light in our, small, local part of time-space, because he is measuring our photon speed relative to his (infinite observer) part of time-space.

Einstein called above: "time-space curvature".
I call it: new explanation for E-M field, showing, that time-space is the medium for E-M wave, and E-M wave is disturbance in this time-space.
I always thought a photon was a self-propagating interaction between an electrical and magnetic field which travels at a verifiable velocity of approx. 300,000km/s through a vacuum. Why complicate it if it works and is verified by observation?

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Originally Posted by pogono
Just a second....
photon has no reference frame?
Correct, it has no reference frame. It has been known for about 106 years.

Are you guys, trying to tell me, there may be body carrying energy, without its reference frame?

Maybe we should check, then, where does Minkowski Metric come from.

....not from any frame of reference attached to a photon, that's for sure.
Until two days ago, you couldn't even write down the Minkowski metric correctly. You still can't, the correct form is:

You don't understand the difference between and .

And now, please tell me, what is time and what is space, when:
has dimension of length and has dimension of ....time, so, your question shows that you don't understand basic physics.
Last edited by macaw; 2011-Sep-29 at 09:38 PM.

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Originally Posted by pogono

The above equation is FALSE. Photons follow null geodesics, as in:

giving:

contrary to your incorrect equation above.

Einstein called above: "time-space curvature".
MQ24: How could Einstein make such a blunder when the above equations are in FLAT spacetime?
MQ25: Provide a reference that "Einstein called above: time-space curvature" or retract your false claim.

22. Originally Posted by starcanuck64
I always thought a photon was a self-propagating interaction between an electrical and magnetic field which travels at a verifiable velocity of approx. 300,000km/s through a vacuum. Why complicate it if it works and is verified by observation?
I do not complicate, I explain what it means.

Originally Posted by macaw
Correct, it has no reference frame. It has been known for about 106 years.
Even if we know for 1,000y that Earth stands on turtles, it is good to be questioning, that may be different.

Originally Posted by macaw
Until two days ago, you couldn't even write down the Minkowski metric correctly. You still can't, the correct form is:

You don't understand the difference between and .
has dimension of length and has dimension of ....time, so, your question shows that you don't understand basic physics.
Of course I understand. What you do not understand is, there are some local common names.
I am used to use "ds" as "road" as many people here. But you have shown many times, that it is easier for you to call people incompetent than try to understand them.
Moreover, I would say, you do not understand physics, but just calculate numbers.

Let me show:

Originally Posted by macaw
The above equation is FALSE. Photons follow null geodesics, as in:

giving:

contrary to your incorrect equation above.
So tell me, please, why do you call it geodesics, when "dt" is time for observer located in infinity. Why photon should be so kind for this particular observer denying our measurements completely, that it has c speed?

I agree, that photon follow geodesics, but since we say, that time-space is only locally flat, it should be:

*for any place X
what is exactly my equation.

I will try to help you: under that radius, space increment "ds" is exchanged with time increment .
what means, that on that radius for any observer:

Cannot be?
Then check it in your textbook.

23. Otherwords, macaw,
using your own words (assuming c=1):

Originally Posted by macaw

Light follows a null geodesic. In the radial direction so

giving the solution I just showed you, where is the time of the distant observer. Textbook stuff.
What means:

what means, that we should measure light speed as:

But since for us, "Hanging Observer", standing in one, solid place X (like us on Earth):

it should mean:

But as we all know we measure light speed as:

So?
Are you still laugting?

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Originally Posted by pogono

Of course I understand. What you do not understand is, there are some local common names.
I am used to use "ds" as "road" as many people here.
Once again, is not any "road". Contrary to what you keep claiming is not interchangeable with

MQ25: Can you define correctly ?

So tell me, please, why do you call it geodesics, when "dt" is time for observer located in infinity. Why photon should be so kind for this particular observer denying our measurements completely, that it has c speed?
Because, as I have shown you, the NULL geodesic, (not any geodesic) is obtained by setting in the metric. You can consult any textbook, they will show the same thing I have shown you.

I agree, that photon follow geodesics, but since we say, that time-space is only locally flat, it should be:

*for any place X
what is exactly my equation.
Minkowski metric describes GLOBALLY flat spacetime, not LOCALLY flat spacetime as you claim. You can consult any textbook on the subject.

You are confusing flat spacetime (Minkowski) with curved spacetime (Riemann) and you are mixing up the metrics.

I will try to help you: under that radius, space increment "ds" is exchanged with time increment .
what means, that on that radius for any observer:
That would mean . I have shown you the correct form early on, why do you persist in writing incorrect stuff?

Cannot be?
Then check it in your textbook.
MQ26: Please provide a reference that "under that radius, space increment "ds" is exchanged with time increment". If you cannot, admit that you made up the above.

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Originally Posted by pogono
Otherwords, macaw,
using your own words (assuming c=1):

What means:

Wrong, it means

what means, that we should measure light speed as:

You are piling up the errors again, it means:

The above is the COORDINATE speed of light.
The proper speed of light, measured locally, is , of course "1". By definition. This is how the metric is constructed. Any textbook will teach you the same exact thing.

But since for us, "Hanging Observer", standing in one, solid place X (like us on Earth):

it should mean:

Incorrect, see above.

But as we all know we measure light speed as:

So?
Are you still laugting?
Yes. BTW, you copied the same mistakes in your "paper", equations (3.5)-(3.12) faithfully reproduce your errors in this thread. You might want to compare them with an introductory textbook in GR and fix them.
(3.13)-(3.25) reproduce faithfully the errors recently pointed out to you relative to circular motion, again, an introductory textbook (or my corrections in this thread) could fix it.
But , if you fix everything to be correct, your "theories" will amount to nothing, having been replaced with elementary mainstream science. The tradeoff is that you haven't discovered anything but, at least, you have your facts correct and you learned a little relativity. Your choice.
Last edited by macaw; 2011-Sep-30 at 02:32 PM.

26. Originally Posted by pogono
Moreover, I would say, you do not understand physics, but just calculate numbers.

That is enough, pogono, just explain yourself CLEARLY and DETAILED.

27. Originally Posted by macaw
You are piling up the errors again, it means:

The above is the COORDINATE speed of light.
The proper speed of light, measured locally, is , of course "1". By definition. This is how the metric is constructed. Any textbook will teach you the same exact thing.
Ok, I agree, above was silly mistake. I have begun post #203 with citing your formula that had silly mistake inside. I should take a look to textbook, before I posted.

So, we should say: light has "c" velocity.
Einstein was right. macaw is right. (I am sure, you will be glad).

But... we have still something to consider.
And moreover, it seems not to be ATM idea, now ;-) Just regular, mainstream idea.

2. Let us consider Hanging Observer (HO), standing in one place at "r" distance to source of gravity. HO is useful, because any body traveling through space, may be considered as body passing spots where such observers stands. At least we, on Earth, are such HOs.

We denote HO proper time following Schwarzschild metric as:

So, we will denote above ratio as:

3. We may now express Schwarzschild metric relative to such HO observers' proper-times as:

4. So now, let us check geodesic for photon relative to proper time of HO it passes. For every HO we have:

Because HO means just "reference frames assigned to places in space" it means, that we may consider my equations for whole space using above metric.

It does not mean, that light is lover than "c", so my thread name is not valid anymore.
But it also means, that for whole space my field equations should work.
Last edited by pogono; 2011-Sep-30 at 09:22 PM.

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Originally Posted by pogono
Ok, I agree, above was silly mistake. I have begun post #203 with citing your formula that had silly mistake inside. I should take a look to textbook, before I posted.

So, we should say: light has "c" velocity.
Einstein was right. macaw is right. (I am sure, you will be glad).
Yes, meaning that you were WRONG all along.

And moreover, it seems not to be ATM idea, now ;-) Just regular, mainstream idea.
Meaning that your "theory" is FALSE. Has been false from the very beginning.

2. Let us consider Hanging Observer (HO), standing in one place in "r" distance to source of gravity. HO is useful, because any body traveling through space, may be considered as body passing spots where such observers stands. At least we, on Earth, are such HOs.

We denote HO proper time following Schwarzschild metric as:

So, we will denote above ratio as:

3. We may now express Schwarzschild metric relative to such HO observers' proper-times as:

4. So now, let us check geodesic for photon relative to proper time of HO it passes. For every HO we have:

The above is true for RADIAL motion, since you made the assumption that
But your theory is about ROTATING frames, so you cannot make , you need to make .
So, your theory is still as wrong as before.

MQ27: What is the correct expression for the proper light speed when you make ?

It does not mean, that light is lover than "c", so my thread name is not valid anymore.
It is not only the name of the thread that is invalid, it is the whole theory that is invalid.

But it also means, that for whole space my field equations should work.
No, it doesn't. To get the correct Maxwell's equations in GR, you need to replace the standard derivatives with covariant ones. This is something that you missed (though it is found in any textbook). Besides, your whole "derivation" involving potentials has already been shown to be false since you are putting in results by hand, rather than deriving them .

29. Originally Posted by pogono
I do not complicate, I explain what it means.
I don't have the mathematical background to judge what you're trying to describe, I see why it's so important in this section that thread starters be required to provide mathematical proofs of their claims.

30. Originally Posted by pogono
Because HO means just "reference frames assigned to places in space" it means, that we may consider my equations for whole space using above metric.
I may have misunderstood your point here, but isn't this equivalent to saying that because a small area on the surface of a sphere can be approximated as a plane, we can use the plane equation to describe the entire sphere.