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pogono
2011-Jul-18, 05:59 PM
Hello,
I am looking for support needed to finish the theory attached. The idea is based on time dilation effect - I analyse dilation as the field itself.

As yet theory predicts delay of high-energy photons from gamma flashes vs. the visible light compliant to results obtained with observations through the FERMI telescope (flashes from the GRB090510 source). It also passes OTW tests. Bibliography in the article.

If you are physicist or you know physicist who may give some support - please contact me at piotrogonowski.pl@gmail.com

At www.tp-theory.net/eng you also may see explanation for non physicists.

Thank you in advance
Piotr Ogonowski

tusenfem
2011-Jul-18, 06:26 PM
Dear pogono!

Welcome to BAUT. Please be informed that this part of the forum is to present your ATM, not to ask for help to develop it.

You will be asked questions about your theory (and looking at the first page I can see which are going to come), so please read the rules of the board, specifically those dealing with ATM. You can find a link to the rules in my signature.

Have fun here, and maybe you will get your theory working.

captain swoop
2011-Jul-18, 06:29 PM
beat me to the drop, duplicate Modding removed :)

macaw
2011-Jul-18, 07:28 PM
Hello,
I am looking for support needed to finish the theory attached. The idea is based on time dilation effect - I analyse dilation as the field itself.

As yet theory predicts delay of high-energy photons from gamma flashes vs. the visible light compliant to results obtained with observations through the FERMI telescope (flashes from the GRB090510 source). It also passes OTW tests. Bibliography in the article.

If you are physicist or you know physicist who may give some support - please contact me at piotrogonowski.pl@gmail.com

At www.tp-theory.net/eng you also may see explanation for non physicists.

Thank you in advance
Piotr Ogonowski

Welcome to BAUT

I read your paper and I had to stop after the first line, you claim that the "speed of the photon" (speed of light) is:

v_f=c \sqrt {1- t_p \omega_f}

There is ample experimental proof that falsifies the above. Sorry, your theory fails from the first line.

pogono
2011-Jul-19, 03:49 PM
omega_f}[/tex]

There is ample experimental proof that falsifies the above.

Dear macaw,

Every experiment has its sensitivity and mistake estimation. Please be so kind and calculate suggested velocity for photon energies used in mentioned experiments. You will see then, that the experiments were not sensitive enough to measure the difference between “v” and “c” because even for MeV energies “v-c” is much lower than Planck scales!!

To measure the difference we should look for high energy photons traveling distances counted with “part of universe” scales.

Then please check on FERMI homepage and many per-reviewed publications for lots of experimental data proving, that for high-energy gamma flashes traveling billions of light years distances we can measure unexplained delay.

And now calculate the shift predicted in theory for such circumstances.
Predicted shift comply with experimental shift.

Regards
Piotr

Strange
2011-Jul-19, 04:41 PM
To measure the difference we should look for high energy photons traveling distances counted with “part of universe” scales.

http://www.universetoday.com/43510/einstein-still-rules-says-fermi-telescope-team/

“This measurement eliminates any approach to a new theory of gravity that predicts a strong energy dependent change in the speed of light,” Michelson said. “To one part in 100 million billion, these two photons traveled at the same speed. Einstein still rules.”

So exactly how much delay does your theory predict?

macaw
2011-Jul-19, 04:50 PM
Dear macaw,

Every experiment has its sensitivity and mistake estimation. Please be so kind and calculate suggested velocity for photon energies used in mentioned experiments.

According to the rules of this forum, I do not need to calculate anything. You, on the other hand, have to defend your non-mainstream theory. Now, this experiment:


Brown et al., Phys. Rev. Lett. 30 no. 16 (1973), pg 763.

For visible light and 7 GeV gammas the speed of light differs by at most 6 parts in 106. The speed of 11 GeV electrons is within 3 parts in 106 of the speed of visible light.

clearly falsifies your claim. The higher the energy, the lower the error , in contradiction to your claims. There are more such experiments.

pogono
2011-Jul-19, 05:15 PM
http://www.universetoday.com/43510/einstein-still-rules-says-fermi-telescope-team/
So exactly how much delay does your theory predict?

I will also quote part of mentioned article:


"Of the many gamma ray photons Fermi’s LAT detected from the 2.1-second burst, two possessed energies differing by a million times. Yet after traveling some seven billion years, the pair arrived just nine-tenths of a second apart.

And now, supprise! My theory predicts 0,2-1 sec delay for this source for 20-200GeV energies range (there is no exact energy values in the article).

Please, read also http://arxiv.org/abs/0909.0016
You can find there values of energies and if you compare it with the theory equation: supprise again! Prediction is correct! You can even find exact calculation inside my theory: www.tp-theory.net/tpt_eng.pdf

Try to falsify the equation with real data. I tried many times and could not do it.

Regards
Piotr

pogono
2011-Jul-19, 06:11 PM
Dear macaw,
below you can find list of articles proving, that we can measure small delay dependent of photon energy traveling for very long distances. The shift is too small for any known quantum gravity theory (or candidate).

The same time the shift is the exactly the result of equation derived from my theory. Calculations on www.tp-theory.net/tpt_eng.pdf page 12.

http://arxiv.org/abs/0909.0016
http://arxiv.org/abs/1006.1867
http://www.springerlink.com/content/f344037h6882k4k5/
http://arxiv.org/abs/0911.0046

In experiments you have mentioned my equation derive difference between “c” and “v” with scale smaller than 10^-40
Immeasurable with these experiments.

macaw
2011-Jul-19, 08:48 PM
Dear macaw,
below you can find list of articles proving, that we can measure small delay dependent of photon energy traveling for very long distances. The shift is too small for any known quantum gravity theory (or candidate).

The shift is too small to measure because it falls within the error bars of the experiment. You need to learn the notions associated with error bars. What you take as signal, it is actually noise.


The same time the shift is the exactly the result of equation derived from my theory. Calculations on www.tp-theory.net/tpt_eng.pdf page 12.

Your "theory" is just fitting numbers, you are simply "measuring" noise because you don't understand the notion of error bars. Besides, the calculations in your "paper" are incorrect. Even worse, your equations (5.8)-(5.11) are incorrect. Try posting your calculations here, the data is:

1. Distance traveled by gamma bursts =L
2. Difference in arrival time \Delta t =0.2sec
3. Pulsation of high energy gamma burst \omega >>1

Let's see how you do the calculations.





In experiments you have mentioned my equation derive difference between “c” and “v” with scale smaller than 10^-40
Immeasurable with these experiments.

Precisely my point. Meaning that you are just "measuring" noise.

Tensor
2011-Jul-20, 03:20 AM
As yet theory predicts delay of high-energy photons from gamma flashes vs. the visible light compliant to results obtained with observations through the FERMI telescope (flashes from the GRB090510 source). Bibliography in the article.

No it doesn't. According to the paper you use to claim verification, there are three different possible values for the delay, and you only picked the one that you think matched (they don't). That's called cherry picking the data. Besides that, there are some open questions and some problems with your claim:

The assumptions for the delay time in the paper on GRB090510 are based on the h0 = 0.71, ΩΛ = 0.73, ΩM = 0.27 What parameters are you using for your universe model?

They use a Quantum Gravity mass in their calculations. What specific term does that represent in your equations? If you don't have that term in your equations, how do you convert that term into one you can use in your equations, so you can make a comparison based on the same terms in the calculations?

You don't use error bars. In your paper, you state that the expected delay for a 30 GeV photon is approximately .27 s. What are the error bars on that calculation? The highest value (and the most conservative) in the paper is .22 s. Which means the values do not match (they are off by .05) and your claim is invalid. What's worse, the values only go down from there, and could conceivably be as low as .15s. Well off your claim, still making your claim invalid. Worse still, the higher value is based on a Local Lorentz Invariance (LLI) violation. From your paper, I get the idea that you're trying to keep and simplify GR, which requires LLI. If LLI is invalid, then GR is invalid. Which from what I gather, makes your paper invalid.

Finally, the is also the possibility that the fluxes are associated with different components, with the photons being emitted at different times. Meaning, that the different arrival times have nothing to do with delays in travel time, but more to do with being emitted latter than the other. Of course, there is the possibility that some of the timing difference, in a two component scenario, is time delay, but in that case, it would be a value even lower than the .15 that is the current lower bound.

Besides all that, what values do you get, and how do you calculate them for the decrease in orbital period of the binary pulsar PSR 1913+16?

pogono
2011-Jul-20, 06:16 AM
Try posting your calculations here (...)

Precisely my point. Meaning that you are just "measuring" noise.

Dear macaw,
I suppost to be against the mainstream, not you. Delay in gamma-ray bursts is a fact, measured and documented. It is not noise.

I post calculations below:
1. Frequency of 1 GeV photon is f= 2,418*10^23 Hz
2. Pulsation of 30 GeV photon is then w = f*30*2*pi= 4,558*10^25 Hz
3. Planck time (tp) equals tp = 5,391*10^-44
4. Delta velocity (difference to "1") according to my theory equals du =1-sqrt(1-tp*w)= 1.22*10^-18
5. For photons traveling for L=7*10^9 light years we may expect delay:
t=du * L = 1.22*10^-18 * 7*10^9 = 8,5 * 10^-9 light years
6. As we know light year equals 3,15*10^7 sec
7. Delay equals then: t=8,5*10^-9 * 3,15*10^7 = 0,267 sec

I hope, it will convice you to read more then one line of the theory.

macaw
2011-Jul-20, 02:24 PM
Dear macaw,
I suppost to be against the mainstream, not you. Delay in gamma-ray bursts is a fact, measured and documented. It is not noise.

This is not what I was telling you. What I was telling you is that your theory has no error bars. So you can't tell signal from noise.


5. For photons traveling for L=7*10^9 light years we may expect delay:
t=du * L = 1.22*10^-18 * 7*10^9 = 8,5 * 10^-9 light years

Nope,

\Delta t=\frac{L}{v_f}-\frac{L}{c}



6. As we know light year equals 3,15*10^7 sec
7. Delay equals then: t=8,5*10^-9 * 3,15*10^7 = 0,267 sec

I hope, it will convice you to read more then one line of the theory.

The above calculation is incorrect. The delay is between the high energy burst (that according to your claims travels at v_f=c \sqrt{1-t_p \omega_f}) and normal energy burst, that travels at c. So, you will need to redo your calculations, what you wrote above is incorrect. I even wrote the correct formula for you.

pogono
2011-Jul-20, 02:50 PM
Dear Tensor, you wrote (enumeration is mine):




(1) The assumptions for the delay time in the paper on GRB090510 are based on the h0 = 0.71, ΩΛ = 0.73, ΩM = 0.27 What parameters are you using for your universe model?

(2) They use a Quantum Gravity mass in their calculations. What specific term does that represent in your equations?

(3) You don't use error bars.

(4) Finally, the is also the possibility that the fluxes are associated with different components, with the photons being emitted at different times.

(5) Besides all that, what values do you get, and how do you calculate them for the decrease in orbital period of the binary pulsar PSR 1913+16?

I reply in order:
ad (1). To measure the delay in gamma burst we do make assumptions. We can just read the data an find the same patterns of flases shifted with few parts of second. Let me quote part of abstract of mentioned article:


The peak of the high energy flux is delayed by 0.2 s with respect to the main MeV
pulse detected by the Fermi Gamma Burst Monitor (GBM)

ad (2). My theory explains gravity with time dillation that appears with slowing down photons arround masses (when photon increase energy, its velocity decrease). We do not need to know what shape of univers is to make calculations.

ad (3). That is true. I use exact values and do not need to put error bars. This is not observation. Approximation is mentioned because of rounding values in few equation. I will fix it ang put error bars to it.

ad (4) Yes, there is possibility. But when you see he same flash patters with some delay, you would rader expect, that it was one, the same flash but part of it was slown down somehow. When you hear sounds of hoof, you expect a horse, not a Zebra.

ad (5) Ok, I will calculate prediction based on my theory and post it here. It will take day or two. I have already found required data at http://www.johnstonsarchive.net/relativity/binpulsar.html

P.S. I will use Lagrangian from my theory - euation (3.12). This Lagrangian gives right predictions for all planets in solar systems for orbits and peryhelium shifts per century.

pogono
2011-Jul-20, 03:07 PM
The delay is between the high energy burst and normal energy burst, that travels at c.

Dear macaw,
yes, you are right. But delay for visible light (lets say 0,1 MHz photons) has scale 10^-24 (very easy to calculate, just other source data).

It is 10^-6 smaller than 30GeV photon shift, what makes it negligible for final compartion.

macaw
2011-Jul-20, 04:20 PM
Even then, your calculation is incorrect. I gave you the correct formula, yours is not correct.

macaw
2011-Jul-20, 04:25 PM
P.S. I will use Lagrangian from my theory - euation (3.12). This Lagrangian gives right predictions for all planets in solar systems for orbits and peryhelium shifts per century.


Your lagrangian is also incorrect, I could direct you to mainstream books that give the correct expression, like Rindler , for one. Since you made the above claim, there is no derivation of the advancement of Mercury perihelion anywhere in your paper, would you care to write it down in THIS thread? thanks.

pogono
2011-Jul-20, 04:29 PM
Hi Macaw, you wrote:


Nope,
\Delta t=\frac{L}{v_f}-\frac{L}{c}


Right. You are right. I recalculated. Thank you.
But... it almost did not changed the result. (1/u-1) also equals 1,22*10^-18
I exchange equation in theory PDF. Thank you.
What now... Will you read it?

amazeofdeath
2011-Jul-20, 05:01 PM
But delay for visible light (lets say 0,1 MHz photons) This could be a nit-pick, but visible light is nowhere near 0.1 MHz. Try 500 THz instead.

pogono
2011-Jul-20, 05:32 PM
This could be a nit-pick, but visible light is nowhere near 0.1 MHz. Try 500 THz instead.

Thank you amazeofdeath, now it is even better :-)
Recalculated for 1 eV. You may see at: www.tp-theory.net/tpt_eng.pdf page 12-13
Visible light has completly negliglible shift to "c", about 10^-30

(sorry for external equations, but I somehow still can not put nice equations into my posts)

pogono
2011-Jul-20, 05:34 PM
Should be 0,1 MeV, my mistake

macaw
2011-Jul-20, 06:15 PM
Hi Macaw, you wrote:


Right. You are right. I recalculated. Thank you.
But... it almost did not changed the result. (1/u-1) also equals 1,22*10^-18
I exchange equation in theory PDF. Thank you.
What now... Will you read it?

I read it from end to end, it is just as wrong as before. Care to show the calculations I asked you to?

pogono
2011-Jul-20, 07:02 PM
Hello macaw.

You wrote:

Since you made the above claim, there is no derivation of the advancement of Mercury perihelion anywhere in your paper, would you care to write it down in THIS thread? thanks.

First - short explanation.

From my theory equations we got, that orbiting bodies experience time dilation in space around source of gravity (quite like in GR).

Then orbiting bodies moves just the way to keep constant time flow relative to time flow in surrounding space. That is exactly what my Lagrangian sais.

It (at least) drives to conclusion, that orbit angle shift schould be dependent on time dillation only what means: dependent on Lagrangian only! Does not mater what is the shape of orbit.

Next conclusion: for orbiting body system, entire rotation seems to come with different time than in external, "far away observer" system.

If we take sit of the "far away" observer we would call it angle dillation (entire rotation angle = 2pi seems to be diffrent in both system from external observer point of view).

For simple star systems with big central mass (like Solar System) it is quite easy to calculate orbit anomaly per rotation. It equals (with negliglible error, scales away):

anomaly = (Lagrangian)/(Planet Rest Energy) * (2*pi)^2

If we would like to transform it from radius to "angle seconds" we have to mutiply it by 648000/Pi

If we need value for century, we have to mutiply it by revolutions per century.

Here you can find XLS with callculations for whole Solar System. It works for all planets: www.tp-theory.net/Anomalies.xls

Have fun.

macaw
2011-Jul-20, 09:17 PM
Hello macaw.



anomaly = (Lagrangian)/(Planet Rest Energy) * (2*pi)^2




Not the correct lagrangian, not the correct answer. No real calculations to speak of. Can you do the calculations?

Tensor
2011-Jul-21, 05:40 AM
Dear Tensor, you wrote (enumeration is mine):
I reply in order:
ad (1). To measure the delay in gamma burst we do make assumptions. We can just read the data an find the same patterns of flases shifted with few parts of second. Let me quote part of abstract of mentioned article:

The peak of the high energy flux is delayed by 0.2 s with respect to the main MeV
pulse detected by the Fermi Gamma Burst Monitor (GBM)

I'll have more to say on this in item #3. But for now, let's just say, I note you don't even mention the cherry-picking of your data. Nor the point that your paper predicts .27, and the quote you chose shows your prediction doesn't fall within the data range of the observations.



ad (2). My theory explains gravity with time dillation that appears with slowing down photons arround masses (when photon increase energy, its velocity decrease). We do not need to know what shape of univers is to make calculations.

GR also has a time dilation component. Is your's in addition to, or in place of the GR dilation? I find it interesting that your dilation factor is the inverse of the Gravitational dilation factor in the Schwartzchild metric. Yeah, but if you don't know the amount of matter in the universe or the density, how do you know how much of an effect there should be? And are you saying there is no expansion in your model of the universe?


ad (3). That is true. I use exact values and do not need to put error bars. This is not observation. Approximation is mentioned because of rounding values in few equation. I will fix it ang put error bars to it.

You're the one that put the approximately equal sign in the equation, not me. You said that it was because of rounding. Which means it could be higher, right? Or, it could be lower, right? If that is the case, how can it be an exact value? Just because it's a calculation and not an observation doesn't mean it can't have a range or error bars.

Now, on to some other items. Since you seemed to ignore other problems I brought up, since they weren't questions, here we go.

Your value for the time delay is ~.27 seconds. The highest value in the paper you are using for support is .22 seconds. The lower bound is .15 seconds. Your value is not within the possible time delay range, how can you claim the paper supports your idea, when your value doesn't fall within the observed time delay values?

In addition, the higher value, the .22, requires a Local Lorentz Invariance (LLI) violation. The problem here, is that your paper is based on GR equations. But, wait, GR requires no LLI violations. If there are LLI violations, GR, and hence, your paper are not valid. Based on that, how can you claim the GRB paper for support, if the values in that paper, are based on principles that invalidate the very foundations of your paper?


ad (4) Yes, there is possibility. But when you see he same flash patters with some delay, you would rader expect, that it was one, the same flash but part of it was slown down somehow. When you hear sounds of hoof, you expect a horse, not a Zebra.

On GeV photon from GRB 090510, From Gao et al, 2009:
The internal shocks take place at a rather larger radius (δt /0.1 s), where δt is the detected variability i,3 timescale of the prompt emission. In the comoving frame of
the emitting region, the seed/photosphere photons are moving along the radial direction and are highly anisotropic. In such a case, the strongest IC radiation is from an angle ∼1/Γi relative to the line of sight (Fan & Piran 2006b). The arrival of the GeV photons will be delayed by a time ∼ δt and the GeV radiation duration will be extended, in agreement with the observation.

Note, an explanation that does not involve time delay based on travel time. And with the comments about different possibilities in the your support paper (and you might want to check out some of those other papers listed in your support paper for even more ways the delay can be happening). It's obvious that you didn't bother investigating whether there was any other possibilities for an inferred time delay. Or, if you did, you intentionally disregarded the other possibilities, because it didn't fit with your claim of the paper providing support. Either way, it's sloppy work.

As for the hooves, lt depends on where you are at. If you're on the Serengeti in Africa, those hooves are more likely to be zebras than horses. In this case, I think you saw something that was close to the right answer, claimed it as correct, and didn't bother to see if there were any counter observations or whether the observations you claimed were good or bad.


ad (5) Ok, I will calculate prediction based on my theory and post it here. It will take day or two. I have already found required data at http://www.johnstonsarchive.net/relativity/binpulsar.html

P.S. I will use Lagrangian from my theory - euation (3.12). This Lagrangian gives right predictions for all planets in solar systems for orbits and peryhelium shifts per century.

Yeah, just remember, you can't use any of your equations that have the Schwarzchild solution in it. Remember, it's only valid in a static space, it's not valid where objects have a rapid spin rate. Something those pulsars have a lot of.

And, basing your entire proof on one paper isn't going to get it done. No one is going to take you seriously if all you have for support is one paper, whose value doesn't match your prediction.

Your paper is confusing, you don't explain all your variables when you introduce them, you freely go back and forth between classical and quantum physics (for instance, you talk about Maxwell's Equations, then introduce the photon), why? Why not just talk about the photon? The photon has nothing to do with Maxwell's equations, it's strictly a quantum concept.

pogono
2011-Jul-21, 08:16 AM
Not the correct lagrangian, not the correct answer. No real calculations to speak of. Can you do the calculations?

Macaw, I may agree, that above equation looks too simple - even for me. But I was derived from my theory and drives to pedictions of orbit anomalies even more complient with observations then GR does. For all solar system planets it works. How you explain it - accident?

Look at xls I linked in last post.

Calculations for non-rotating solar system with centrall big mass (example for Mercury):

1. You take Rest Energy of the orbiting planet. For Mercury it is mc^2 = 3.33020*10^23 * c^2

2. You calculate Mechanical Energy (Lagrangian). The easiest way is to calculate it in Aphelium or Peryhelium (does not matter, Mechanical Energy is constant).

3. For Mercury Mechanical Energy equals L= -1.27494909 * 10^(-8) x mc^2

4. Than you get Dilation D = L/mc^2 = -1.27494909 * 10^(-8) ***

5. Anomaly per rotation equals (2pi)^2 * D

6. To obtaint anomaly for century and with angle seconds, you have to multiply it by number of revolutions per century and by 648000/Pi

*** why I call it dilation?
Mechanical Energy means Kinetic Energy in infinity (in infinity there is no gravity, so the only part of the Lagrangian remains Kinetic Energy). If you divide it by Rest Energy, you gey relativistic dillation factor (gamma-1)

macaw
2011-Jul-21, 02:02 PM
For simple star systems with big central mass (like Solar System) it is quite easy to calculate orbit anomaly per rotation. It equals (with negliglible error, scales away):

anomaly = (Lagrangian)/(Planet Rest Energy) * (2*pi)^2

Not the correct derivation for the anomaly, it isn't equal to the ratio between the lagrangian and mc^2. Trying to divide the lagrangian is absurd in first place.
Not the correct expression for the lagrangian, either. For example, the correct lagrangian is:

L=\alpha (\frac{dt}{ds})^2-\frac{1}{\alpha}(\frac{dr}{ds})^2-r^2 (\frac{d \phi}{ds})^2

with

\alpha=1-\frac{2m}{r}

As you can see, the lagrangian is a differential expression, you cannot divide it by mc^2 in order to obtain a number (the advancement of a planet perihelion).

pogono
2011-Jul-21, 04:06 PM
Not the correct derivation for the anomaly, it isn't equal to the ratio between the lagrangian and mc^2


Dear macaw,
writing "bad Lagrangian" or "bad anomaly" do you mean "not equal to lagrangian derived from GR?". That is true. I have my own Lagrangian in the theory leading to difrent formula for orbit and orbit anomaly that leads to CORRECT RESULTS compliant with observations for as orbits as anomalies (I have also checked numericaly, that my Lagrangian almost equals that one from GR up to 2*Shwarchild Radius).

In my Lagrangian, orbiting bodies care only about constant time flow. My Lagrangian (for non rotating big Central Mass system) equals:

L=E_{0}(\frac{1}{\sqrt{1-\frac{v^2}{c^2}})}-1)-E_{0}(\frac{1}{\sqrt{1-\frac{R_{Schw}}{R}})}-1)

If you calculate this Lagrangian for all planets in Solar System and then use my equation for anomaly mentioned in last post, you will get correct anomalies for each planet. That is what I put into XLS at:
http://www.tp-theory.net/Anomalies.xls

By the way: The idea behind above Lagrangian is simple. As Newton says, if there is no force, body stands or moves lineary with constant velocity. But it also means - its time flow is constant!

I say - let's care about time flow. We was thinking for centuries, that kinetic energy depend on move in space. Einstein say: not true. It depends on time flow / time dilation. So I say: analysing field we should also say, that time is more important than something called "gravity force", time flow is the "cause" - moving is just visible effect of that cause.

So if there is a field, described as time flow disturbance in space around mass (dilation as field) body moves such way, to keep constant time flow relative to time flow disturbance in space. In orbiting body system it is the linear move. That is exactly what my Lagrangian sais.

Check the XLS, please, or calculate solar system yourself - you will see it works.

macaw
2011-Jul-21, 04:11 PM
Dear macaw,
writing "bad Lagrangian" or "bad anomaly" do you mean "not equal to lagrangian derived from GR?". That is true. I have my own Lagrangian in the theory leading to difrent formula for orbit and orbit anomaly that leads to CORRECT RESULTS compliant with observations for as orbits as anomalies (I have also checked numericaly, that my Lagrangian almost equals that one from GR up to 2*Shwarchild Radius).

Lagrangians are NOT numbers, so you cannot "check them numerically". What you are doing is numerology, not physics.

pogono
2011-Jul-21, 04:52 PM
Macaw,
what I do is math with physical explanation (given above) based on time flow analyses. Of course, that we can calculate star systems with algebraical or numerical methods! If you only try to calculate my equations you would see, that these equations are very hard to calculate algebraically.

P.S. If you do not belive there may be non-mainstream theory, that helps us to understand the world better, why you discuse with me on Against the Mainstream?

I do not say, my theory is the "only truth".
I found something that amazed me - explanation of the field based on time dilation that passes experimental tests.
At the beginning of this thread I asked for help to develop it. It will be enough if at least there is anything interesting, what may push the world a milimeter forward.

Garrison
2011-Jul-21, 07:03 PM
P.S. If you do not belive there may be non-mainstream theory, that helps us to understand the world better, why you discuse with me on Against the Mainstream?

There are several reasons for others to continue debating you:

Others may learn interesting things about the mainstream theory
You may come to understand why your theory is wrong and cease wasting energy on it
There may actually be something of value somewhere in your ideas that can be exposed and refined, though I have to say I've not seen that happen so far in any ATM thread


I do not say, my theory is the "only truth".
I found something that amazed me - explanation of the field based on time dilation that passes experimental tests.
At the beginning of this thread I asked for help to develop it. It will be enough if at least there is anything interesting, what may push the world a milimeter forward.

But its been pointed out that your inspiration is a flawed theory and your own interpretation does not conform to observed data. However amazing the idea may have been there comes a point where you have to let it go.

slang
2011-Jul-21, 07:28 PM
Finally, the is also the possibility that the fluxes are associated with different components, with the photons being emitted at different times.

Isn't there an implicit assumption of the photons being emitted by a point source as well? Or is it too small to account for (part of) any delay?

pogono
2011-Jul-21, 07:55 PM
Ok, Garrison. Good point. So let me then ask forum members society.

I wrote few words about the idea, that we may do the same mistake with potential energy that we did with kinetic energy. Maybe we should look for time dilation effects, to explain the "field phenomenon", as well for gravity as well for electromagnetism.



I say - let's care about time flow. We was thinking for centuries, that kinetic energy depends on move in space. Einstein says: not true. It depends on time flow / time dilation. So I say: analyzing field we should also say, that time is more important than something called "gravity force", time flow is the "cause" - moving is just visible effect of that cause.

So if there is a field, described as time flow disturbance in space (...) body moves such way, to keep constant time flow relative to time flow disturbance in space.


Is there any one, who would "put a cent" on such idea?
Does it look like something that is worth to be developed somehow?
If no one say even “maybe”, I will quit it.

Tensor
2011-Jul-21, 08:33 PM
Isn't there an implicit assumption of the photons being emitted by a point source as well? Or is it too small to account for (part of) any delay?

Yes, there is. At that distance, a short GRB (thought to be a merger event) would look like a point source. The distances from the merger event, based the start of the emission of the second component ~45,000-66,000 km. Which is quite reasonable as the paper interprets the second component as the forward shock wave.

macaw
2011-Jul-21, 10:13 PM
Macaw,


P.S. If you do not belive there may be non-mainstream theory, that helps us to understand the world better, why you discuse with me on Against the Mainstream?

In order to point out your errors and to show you how the problems are solved for real in the mainstream world.




I do not say, my theory is the "only truth".

Your theory contains no truth.



I found something that amazed me - explanation of the field based on time dilation that passes experimental tests.

It doesn't you are fiddling the numbers in order to fit.



At the beginning of this thread I asked for help to develop it. It will be enough if at least there is anything interesting, what may push the world a milimeter forward.

Errors cannot push the world, mainstream physics does.

macaw
2011-Jul-21, 11:06 PM
Dear macaw,
writing "bad Lagrangian" or "bad anomaly" do you mean "not equal to lagrangian derived from GR?". That is true. I have my own Lagrangian in the theory leading to difrent formula for orbit and orbit anomaly that leads to CORRECT RESULTS compliant with observations for as orbits as anomalies (I have also checked numericaly, that my Lagrangian almost equals that one from GR up to 2*Shwarchild Radius).

In my Lagrangian, orbiting bodies care only about constant time flow. My Lagrangian (for non rotating big Central Mass system) equals:

L=E_{0}(\frac{1}{\sqrt{1-\frac{v^2}{c^2}})}-1)-E_{0}(\frac{1}{\sqrt{1-\frac{R_{Schw}}{R}})}-1)

This can't be. I already gave you the correct expression of the lagrangian in GR (see Rindler, for example). An alternative expression, that clearly contradicts yours can be found here (http://en.wikipedia.org/wiki/Lagrangian#General_relativistic_test_particle).

Shaula
2011-Jul-22, 05:36 AM
Can you provide proof that your new Lagrangian approximates the GR one well? How does the R/Rs term work in, for example, a synchrotron? Or even just free space? You seem to be saying you need one Lagrangian for solar system type problems and another for the other problems. What is the changeover point?

pogono
2011-Jul-23, 09:32 AM
Hello Shaula, you say (enumaration is mine):



1. Can you provide proof that your new Lagrangian approximates the GR one well?
2. How does the R/Rs term work in, for example, a synchrotron?
3. Or even just free space?
4. You seem to be saying you need one Lagrangian for solar system type problems and another for the other problems. What is the changeover point?


1. Algebraicaly I can easy prove that my Lagrangian is approxsimation for Newtonian gravity law. Using Maclaurin's expansion we obtain:

E_{0}(\frac{1}{\sqrt{1-\frac{v^2}{c^2}})}-1)\approx E_{0} \frac{v^{2}}{2c^2}=\frac{m_{0}v^{2}}{2}

E_{0}(\frac{1}{\sqrt{1-\frac{R_{schw}}{R}})}-1) \approx E_{0} \frac{R_{schw}}{2R}= E_{0} \frac{GM}{Rc^2}=G\frac{mM}{R}

To prove that it is approximation of GR Lagrangian I used software (Derive) and numerical metods. It was just formula "GR energy on orbit"-"my proposal of energy on orbit". It was arround 0 up to 2*Rsch distance.

2. How does the R/Rs term work in, for example, a synchrotron?
In LHC we still use to small energies for any measurable effects of my equations. But theoreticaly it is easy to show, that there is something wrong with mainstream theory using synchrotron. Present photon Energy formula (commonly used) lets us to expect photons with pulsation bigger than Planck pulsation!

3. Or even just free space? - we can observe photons traveling very close to the edge of Schwarschild Radius of f.e. big black holes. My theory predicts stronger photon track abberation then GR. And my theory has no “singularity” phenomenon resulting from the GR.

4. You seem to be saying you need one Lagrangian for solar system type problems and another for the other problems. What is the changeover point?

If we have non-spinning central big mass, then center of system mass is in one of them, and second mass is rotating around. We may throu out small-mass dilation effect and spinnings.

But two spinning bodies with similar mass rotating around center of mass (in my theory it is the point, where photons has the same delay**) force to calculate all of the effects coming from both spins and dilations around mass. It makes calculation harder.

Shaula
2011-Jul-23, 11:05 AM
1) But GR's Lagrangian supplants that of Newton. Which was why I asked about the GR version. If you cannot replicate the GR equations then you have a lot more proving to do as you have to show that your versions somehow come out with the same results from different equations.

2) I was unclear - I guess what I really wanted was a better definition of R and Rs for free space solutions. With no gravity around what happens to the term Rs/R - R seems to me to be undefined or at least potentially badly behaved.

macaw
2011-Jul-26, 12:54 AM
Hello Shaula, you say (enumaration is mine):



1. Algebraicaly I can easy prove that my Lagrangian is approxsimation for Newtonian gravity law. Using Maclaurin's expansion we obtain:

E_{0}(\frac{1}{\sqrt{1-\frac{v^2}{c^2}})}-1)\approx E_{0} \frac{v^{2}}{2c^2}=\frac{m_{0}v^{2}}{2}

E_{0}(\frac{1}{\sqrt{1-\frac{R_{schw}}{R}})}-1) \approx E_{0} \frac{R_{schw}}{2R}= E_{0} \frac{GM}{Rc^2}=G\frac{mM}{R}

To prove that it is approximation of GR Lagrangian I used software (Derive) and numerical metods. It was just formula "GR energy on orbit"-"my proposal of energy on orbit". It was arround 0 up to 2*Rsch distance.

This is not the correct way of proving that you constructed a correct lagrangian. You need to prove that your lagrangian satisfies the general Euler-Lagrange equation:

\frac{d}{dt}(\frac{\partial L}{\partial v})-\frac{\partial L}{\partial r}=0

pogono
2011-Jul-26, 11:02 AM
Dear Shaula, I will answer your questions in two posts, each for one question.

Yes, I know, that GR is in place of Newton, so I try to show, that there is big common part of my theory and GR.
F.e if you take dilation as field it makes exactly the same time-space curvature than General Relativity predicts, but coming from field analyses. Short explanation below:

Let R_{X} will be some constant and dilation equals:
\gamma =\frac{dt}{d\tau} = \frac{1}{\sqrt{1-\frac{R_{X}}{R}}}
Let:
U=- \bigtriangledown \gamma = \frac{R_{X}}{2R^2} \gamma^3
therfore:
\triangledown \cdot U=\frac{R_{X}}{V}\gamma^3
We define vector potential A and B and express U by A (just like for Maxwell eq.)
B=\triangledown \times A

We can thus define U as:
U=-\frac{1}{c}\cdot \frac{\partial A}{\partial \tau}= -\frac{\gamma }{c}\cdot\frac{\partial A}{\partial t}
\triangledown \cdot U=-\frac{\gamma }{c}\cdot\frac{\partial (\triangledown \cdot A)}{\partial t}=\frac{R_{X}}{V}\gamma^3

If we take Schwarzschild radius as our constant R_{X} then (what is easy to derive) the relation between zero-element of stress-energy Tensor and A is:
-\frac{4\pi G}{c^4} \cdot T^{00}= \frac{1 }{c^2} \cdot\frac{\partial (\triangledown \cdot A)}{\partial t}=-\frac{R_{Schw}}{V}\gamma^2

And main GR equation simplifys then, because: G_{uv}=2 \cdot -\frac{4\pi G}{c^4} \cdot T_{uv}

If we define tensor Y based on gamma and A it seems that we have gravity defined just like electromagnetism: with “Gamma” as potential and joined “A-vector”.

GR main equation equals now: G_{uv}=2 \cdot Y_{uv}

I know how to calculate Lagrangian for specific conditions from above, but It is an issue to derive general form of Lagrangian coming from above. It only looks nice and simple. That is why I asked for physicist support at the beginning of this thread. I will sit down now, and try to calculate what macaw have just proposed – test for Lagrange function.

BTW, if we take Planck’s length as R_{X} we will get equations for electromagnetism. Velocity of a wave produced by such field is close to C, what you can see in Delambertian and seems to works for delayed photons in gamma flashes:

\frac{\gamma ^2}{c^2}\cdot \frac{\partial^2 U}{\partial t^2}-\triangledown ^2 U =0

The above produces the speed of light that we have started this post with.

tusenfem
2011-Jul-26, 12:02 PM
If U is the gradient of gamma, why does your expression for U seem like a scalar and not a vector?
How did you get the expression for the divergence of (vector?) U and what is V?
The part of the vector potential A, where does that come from, should you not derive A from B(=U?) = curl(A)?
This looks all rather shady.

pogono
2011-Jul-26, 12:02 PM
Dear Shaula, here I reply for question about “free space”.

Things seems to be more complicated, because there is nothing like free space and Shwartshild radius is something a bit different we probably think. Let me show:

(index P means “Planck units”, for approximation we use Maclaurin approximation)

In my theory free space is full of electromagnetic field. We may analyze this filed as set of particles (photons) each with energy:
E=2E_{P}\cdot \left (\frac{1}{\sqrt{1-t_{P}\cdot\omega}}-1 \right ) \approx 2E_{P}\cdot \frac{t_{P}}{2 \omega} = \not{h}\cdot \omega

Photons (under specific conditions) may split in two, what we note as transition:
2E_{P} \left (\frac{1}{\sqrt{1-t_{P}\cdot\omega}}-1 \right ) \mapsto 2 \cdot E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}-1 \right)

Where:
E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}-1 \right) \approx E_{P}\frac{R_{Schw}}{2l_{P}} = \frac{l_{P}\cdot c^4}{G}\cdot \frac{R_{Schw}}{2\cdot l_{P}} = mc^2

And now, let me explain what Schwarzschild Radius may be and why I use it even for elementary particles.

Vectors of field rotation make helix in Minkowski time-space. We may show that if velocity equals “c” on circle then:
Tc = 2\pi R_{\omega } \to \omega = \frac{2\pi}{T} = \frac{c}{R_{\omega}}

For photons, the Curvature of this rotating field vectors helix equals:
\frac{1}{l_{P}}=\frac{R_{\omega}}{R_{\omega}^2-\left (\frac{R_{\omega}}{\gamma} \right )^2}

If we will change the way that vectors rotate, we change helix curvature and obtain particle:
\frac{1}{R_{Schw}}=\frac{l_{P}}{l_{P}^2-\left (\frac{l_{P}}{\gamma} \right )^2}

It means, that Schwarzschild Radius is not what we suspect it was. It is just the measure of some curvature in space-time. We can use it even for a photons or elementary particles.

pogono
2011-Jul-26, 03:01 PM
Hi tusenfem. You wrote (my enumeration):



1. If U is the gradient of gamma, why does your expression for U seem like a scalar and not a vector?

2. How did you get the expression for the divergence of (vector?) U and what is V?

3. The part of the vector potential A, where does that come from, should you not derive A from B(=U?) = curl(A)?


So,
1. It is a vector. We schould rather write it the way representing its real meaning (time flow) according to my above post and Gauss law.

We define:
T_{X}=2\pi\cdot \frac{R_{X}}{c}
then
U=\frac{cT_{X}}{4\pi R^2}\cdot \gamma ^3

2. V is the element of Volume.
The density in my theory is "time flow density" in place of mass density in GR. We can write the time flow density as:
\rho _{t}=\frac{T_{X}}{V}
then:
\triangledown \cdot U=\rho _{t}c\cdot \gamma ^3

3. We use potential and A-vector in fourvectors where A is helpfull for sourceless field. I put B only to show that is exactly the same reasoning that in Tensor Maxwell equations.

macaw
2011-Jul-26, 03:56 PM
Let R_{X} will be some constant and dilation equals:
\gamma =\frac{dt}{d\tau} = \frac{1}{\sqrt{1-\frac{R_{X}}{R}}}

\frac{\gamma ^2}{c^2}\cdot \frac{\partial^2 U}{\partial t^2}-\triangledown ^2 U =0

The above produces the speed of light that we have started this post with.

No, it doesn't, it produces v_f=c/\gamma=c \sqrt{1-R_x/R} which is different from your other incorrect claim that v_f=c \sqrt{1-t_p \omega_f} that I have already proven wrong at the beginning of this thread, in post 4.

tusenfem
2011-Jul-26, 05:00 PM
We define:
T_{X}=2\pi\cdot \frac{R_{X}}{c}
then
U=\frac{cT_{X}}{4\pi R^2}\cdot \gamma ^3

2. V is the element of Volume.
The density in my theory is "time flow density" in place of mass density in GR. We can write the time flow density as:
\rho _{t}=\frac{T_{X}}{V}
then:
\triangledown \cdot U=\rho _{t}c\cdot \gamma ^3


This is wrong, because you do a d/dR as part of the divergence (assuming that the d/dtheta and d/dphi are zero) and you forget that gamma is still a function of R.

pogono
2011-Jul-26, 05:01 PM
No, it doesn't, it produces v_f=c/\gamma=c \sqrt{1-R_x/R} (...)

Yes, it does. Let me show:
I also wrote:

(...) if we take Planck’s length as R_x we will get equations for electromagnetism (...)

Lets then write gamma for:
-gravity:
\gamma =\frac{1}{\sqrt{1-\frac{R_{Schw}}{R}}}

-electromagnetism:
\gamma =\frac{1}{\sqrt{1-\frac{l_{P}}{R}}}

Above equation defines field. But we may also perceive field as set of particles. We localise particle

- for gravity on planck lenght:
\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}

- for photon on distance, where field vectors rotate with velocity "c", that gives:
\frac{1}{\sqrt{1-\frac{l_{P}}{R_{\omega}}}}

But we know already, that we may use for such circumstances:
\omega =\frac{c}{R_{\omega }}

therfore it will be:
c\cdot \sqrt{1-l_{P}\cdot\frac{\omega}{c}}=c\cdot \sqrt{1-t_{P}\cdot \omega }

pogono
2011-Jul-26, 08:15 PM
This is wrong, because you do a d/dR as part of the divergence (assuming that the d/dtheta and d/dphi are zero) and you forget that gamma is still a function of R.

ok tusenfen, but what exactly is wrong. I do not get it.

Where is the mistake in it?:
\triangledown \cdot U = \lim_{V \to 0}\frac{1}{V}\iint_{\partial V}^{ } \frac{cT_{X}}{4\pi R^2} \cdot \gamma ^3\: dS=c \cdot \rho _{t}\cdot \gamma ^3

macaw
2011-Jul-26, 09:02 PM
Yes, it does. Let me show:
I also wrote:


Lets then write gamma for:
-gravity:
\gamma =\frac{1}{\sqrt{1-\frac{R_{Schw}}{R}}}

-electromagnetism:
\gamma =\frac{1}{\sqrt{1-\frac{l_{P}}{R}}}

Nope, in electromagnetism

\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

You can check any book on the subject.


Above equation defines field. But we may also perceive field as set of particles. We localise particle

- for gravity on planck lenght:
\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}

- for photon on distance, where field vectors rotate with velocity "c", that gives:
\frac{1}{\sqrt{1-\frac{l_{P}}{R_{\omega}}}}

But we know already, that we may use for such circumstances:
\omega =\frac{c}{R_{\omega }}

therfore it will be:
c\cdot \sqrt{1-l_{P}\cdot\frac{\omega}{c}}=c\cdot \sqrt{1-t_{P}\cdot \omega }

I knew that you were going to write this. What you wrote is equivalent to t_p \omega_f=R_x/R which is pure numerology, not physics.

pogono
2011-Jul-27, 01:38 AM
Dear macaw, you wrot once again "numerology". I would rather expect some rational arguments here.
But ok. I admire your belief in mainstream physics, so lets talk about mainstream and try to find how it explains things I try to explain.

1. Rest mass. What is it? But do not write, please, that it is the number assigned to the body (numerology?). Explain. Why does it equal mc2?
2. Why in the center of the black hole we have infinity, that Einstein called "singularity problem"? Is there real infinity? Yes or No?
3. What is it photon and why its energy equals exactly E=hv? But please do not write, that we do not have to know (numerology?) becasue it was proposed some day this way and works...
4. Photon energy is kinetic energy. But kinetic energy of what rest energy? Do you believe, there may be kinetic energy without rest energy?...
5. Where is this photon located? Everywhere, because probability function says that and we have discovered wave-corpuscular nature of the nature... How then we may localise photon and why it works this way?
6. why Standard Model needs 13 constans to keep things together? What are these constans coming from? Just numbers?... (pure numerology, but much more complicated...)
Morover,
dark matter, dark energy?
7. Why thousand of people look for non-mainstream theories? They do not know physics? Or maybe just mainstream physics does not aswer to some fundamental questions?...

We are all teached historical way of physics.
Coulomb law was discovered before Maxwel's concept of the field, so Coulomb uses "charge" idea. What is it charge and why it is discreet? Is it just a number assigned to non localised wave function?
So does e-m field need a charge or not? What was first on the world: photon or electrons?
In laboratory: electrons, that is why we made phisics with elementary charges in equations...

I would realy like to see your answers based on physics - not on the "numerology".

macaw
2011-Jul-27, 02:07 AM
Dear macaw, you wrot once again "numerology". I would rather expect some rational arguments here.
But ok. I admire your belief in mainstream physics, so lets talk about mainstream and try to find how it explains things I try to explain.

1. Rest mass. What is it? But do not write, please, that it is the number assigned to the body (numerology?). Explain. Why does it equal mc2?

Rest mass does not equal mc^2. Where is it going to? I pointed out some basic mistakes in your posts, why don't you address them?

Tensor
2011-Jul-27, 05:47 AM
But ok. I admire your belief in mainstream physics, so lets talk about mainstream and try to find how it explains things I try to explain.

1. Rest mass. What is it?

The amount of mass an object has in it's rest frame.


But do not write, please, that it is the number assigned to the body (numerology?). Explain. Why does it equal mc2?

It doesn't. Rest mass is the m in that formula. As such, you're saying m = mc2. Which, would be good if c was 0, however, c is a constant which makes that equation numerology.


2. Why in the center of the black hole we have infinity,

Using the General Relativity equations, either show that there is an infinity at the center of a black hole, or retract this.


that Einstein called "singularity problem"? Is there real infinity? Yes or No?

No, there isn't. Again, if you think so, you need to study General Relativity a bit more.


3. What is it photon and why its energy equals exactly E=hv? But please do not write, that we do not have to know (numerology?) becasue it was proposed some day this way and works...

A photon is the quanta of EM radiation. And E = hv is what is MEASURED. If it was just because it was proposed, and didn't work, that would be numerology.


4. Photon energy is kinetic energy.

But kinetic energy of what rest energy?

Hmmmmm, you know, this makes no sense. But, I'll assume you mean rest mass, so the statement does make sense. Well, let's see. The equation for relativistic energy is E2 = ((mc2 ))2 + (pc)2. Since a photon's rest energy is 0, the Equation becomes E2 =(pc)2. There you go.


Do you believe, there may be kinetic energy without rest energy?...

Well, yes, see above. It would be helpful if you would use common physics terms. As in there can be kinetic energy without rest mass. For that question, Newtonian models require mass (which is in motion) to determine the kinetic energy. However, quantum models do not require this.


5. Where is this photon located? Everywhere, because probability function says that and we have discovered wave-corpuscular nature of the nature... How then we may localise photon and why it works this way?

Actually, you're wrong here also. We can localize a photon quite precisely, we just can't then determine it's momentum. Mainly due to the uncertainty principle, and the fact that the


6. why Standard Model needs 13 constans to keep things together?

Actually, it's 18.


What are these constans coming from? Just numbers?... (pure numerology, but much more complicated...)

Actually measured. You can keep calling it numerology, but all the things you've called numerology have actually been measured.


Morover, dark matter, dark energy?

What about it? There are observations we can't explain. The cause of those observations are just called Dark Matter or Dark Energy. Once we actually figure it out, they have actual names. This has happened before you know. In they 1840s, there were observations that couldn't be explained. They gave the cause a temporary name. When they actually found the cause, they gave it a name, they called it Neptune. In the 1930s, they just went ahead and gave the cause a name, even though they hadn't observed the object yet, the neutrino.


7. Why thousand of people look for non-mainstream theories? They do not know physics?

Well, in my experience, the majority of those that have posted here. Especially those that make elementary physics mistakes, mistakes in the theories they think they are replacing, or misuse common physics terms, nope, they don't know physics.


Or maybe just mainstream physics does not aswer to some fundamental questions?...

The fundamental questions are of a philosophical nature, not a physics nature. Philosophy is down the hall.


We are all teached historical way of physics.
Coulomb law was discovered before Maxwel's concept of the field, so Coulomb uses "charge" idea. What is it charge and why it is discreet? Is it just a number assigned to non localised wave function?

Actually, charge isn't a wave function. You might want to study the standard model also.


I would realy like to see your answers based on physics - not on the "numerology".

Yeah, numerology, well as I said, everything you asked for has been actually measured. Unlike your equations. So, how about you explain how measured quantities are numerology?

How about your claim of support on those GRB bursts? You know, the one that doesn't fall within the measured limits. You haven't retracted that one yet.

tusenfem
2011-Jul-27, 07:35 AM
ok tusenfen, but what exactly is wrong. I do not get it.

Where is the mistake in it?:
\triangledown \cdot U = \lim_{V \to 0}\frac{1}{V}\iint_{\partial V}^{ } \frac{cT_{X}}{4\pi R^2} \cdot \gamma ^3\: dS=c \cdot \rho _{t}\cdot \gamma ^3

The problem is that you forget that \gamma = \gamma(R)
and the divergence is most easily calculated as (\partial/\partial x, \partial/\partial y, \partial/\partial z) insted of the integral form and as you have \gamma = \gamma(R) = (1 - R/Rs)^{(-1/2)} and as the divergence in polar coordinates is can be found in e.g. Jackson (classical electrodynamics) you can see what you do wrong. Don't use the unusable integral form but use the differential form. Also, as you apparently only do the \partial/\partial R part of the calculation and thus forget about the \partial/\partial \theta and \partial/\partial \phi you are assuming that there is spherical symmetry of the problem or you are looking at a 1 dimensional space.


ETA: Also the definition of your divergence in integral form is incorrect, as the volume V should be inside the integrals and ofcourse you are only calculating the divergence at "0" here.

pogono
2011-Jul-27, 09:52 AM
Tensor, thank you for taking a glove. But in science we try to not to say: "zebras are painted grid because it was measured" (BTW, poor zebras). We try to find "why?".
I am trying to say, there are no answers for this questions in mainstream and what I propose is to echange "well establisched belief and numerology" with some explanation...

Answering your explanations:
1. You define rest mass by rest frame...
2. My question is: WHY rest energy equals mc^2?
It was measured that it is, but how do you know, that you do not measure:

E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}-1 \right) \approx E_{P}\frac{R_{Schw}}{2l_{P}} = \frac{l_{P}\cdot c^4}{G}\cdot \frac{R_{Schw}}{2\cdot l_{P}} = mc^2

It is undiscernible.

3. I said "why" photon energy is E=hv? Show me any relation to EM field. I do...

4. Excellent, we may say, that "pc" exists when m=0. But the problem is, that: p = mv* gamma
Of course, for photons pc = hv, I forgot. But why E=hv? - because it was measured. We take this explanation and forget about it.

5. 18 constans in Standard Model - but it is even worse...
We could take all known fermions and quarks and treat them as constans. Would be less than 18 constans!

I propose to build physics on most fundamental constans called Planck's units. We may treat them as "unity". It is coherent, and makes sense.
I am not able to create this whole theory myself, that is why I am looking for support at the begining of this thread.

And answering for GBR experiment question:
Show me any mainstream concept that will be so close to meashured GBR burst delay that I propose.
Mainstream says, there should be no delay. It means, that I am 0,26 sec closer to thruth than mainstream...

Geo Kaplan
2011-Jul-27, 11:37 AM
I am trying to say, there are no answers for this questions in mainstream and what I propose is to echange "well establisched belief and numerology" with some explanation...

I propose to build physics on most fundamental constans called Planck's units. We may treat them as "unity". It is coherent, and makes sense.
I am not able to create this whole theory myself, that is why I am looking for support at the begining of this thread.

And answering for GBR experiment question:
Show me any mainstream concept that will be so close to meashured GBR burst delay that I propose.
Mainstream says, there should be no delay. It means, that I am 0,26 sec closer to thruth than mainstream...

But you arrived at your number through erroneous mathematics. That's why your result is numerology, not physics. I note that you still have refused to answer many of Tensor's questions, including an important one concerning error bars.

I could arrive at an even closer value by taking the square root of the delay and multiplying by NG, the nanognome factor. Less work, and I can make the result even closer to what you are calling the measured value. Is that physics? Your approach is more mathematically elaborate (baroque, to be honest), but it's philosophically the same as the nanognome approach, viz, it yields a value I want, and it makes sense to me. Therefore I declare it to be correct ("amo ergo est").

macaw
2011-Jul-27, 04:11 PM
U=\frac{cT_{X}}{4\pi R^2}\cdot \gamma ^3

So, by the way you defined it, U is a scalar potential (http://en.wikipedia.org/wiki/Scalar_potential), right?




then:
\triangledown \cdot U=\rho _{t}c\cdot \gamma ^3

You are attempting to calculate the divergence of a scalar. Therefore you are getting nonsense.
Besides, your elementary math is also wrong, you have "forgotten" that \gamma is also a function of R as tusenfem already pointed out to you.

macaw
2011-Jul-27, 04:14 PM
The problem is that you forget that \gamma = \gamma(R)
and the divergence is most easily calculated as (\partial/\partial x, \partial/\partial y, \partial/\partial z) insted of the integral form and as you have \gamma = \gamma(R) = (1 - R/Rs)^{(-1/2)} and as the divergence in polar coordinates is can be found in e.g. Jackson (classical electrodynamics) you can see what you do wrong. Don't use the unusable integral form but use the differential form. Also, as you apparently only do the \partial/\partial R part of the calculation and thus forget about the \partial/\partial \theta and \partial/\partial \phi you are assuming that there is spherical symmetry of the problem or you are looking at a 1 dimensional space.


ETA: Also the definition of your divergence in integral form is incorrect, as the volume V should be inside the integrals and ofcourse you are only calculating the divergence at "0" here.


Actually, it is much worse than that, he's attempting to calculate the divergence of a scalar.

pogono
2011-Jul-27, 05:27 PM
So, by the way you defined it, U is a scalar potential, right?
No.
\gamma is scalar potential.
U= -\triangledown \gamma is its gradient.



(...) it is much worse than that, he's attempting to calculate the divergence of a scalar.

No. I am affraid it only points, that you probably do not waste your time to the theory you want to revile.

But tusenfem may be right. I recalculate divergence now to check it.
I appreciate your review, that is exactly the support I was looking for.

macaw,
If there are mistakes in the reasoning - I want to find them. If the theory is wrong, I would realy like to know that.
But assumption at the beginning that non-mainstream theory has to be wrong, has nothing to do with discussion.

pogono
2011-Jul-27, 05:40 PM
(...)
1) I note that you still have refused to answer many of Tensor's questions, including an important one concerning error bars.
(...)
2) I could arrive at an even closer value by taking the square root of the delay and multiplying by NG, the nanognome factor.

Hello Geo Kaplan,
1. I will answer soon, I work it over. Esspecialy pulsar is great idea to check how my theory works in more complicated system than Mercury-Sun. Thank you Tensor
2. There is nothing like NG in my theory - only Planck Units. I do not use such way to derive solutions. Please, read pdf with theory. If you find a mistake somewhere or anything like NG - point it. I will answet to that.

macaw
2011-Jul-28, 12:16 AM
No.
\gamma is scalar potential.
U= -\triangledown \gamma is its gradient.

The way you wrote it, the RHS is a scalar, so you are contradicting yourself.





No. I am affraid it only points, that you probably do not waste your time to the theory you want to revile.

Since you have so many errors, "debunk" is the correct term, I do not "revile" your "theory".


But tusenfem may be right. I recalculate divergence now to check it.

You are still not convinced that you missed the fact that \gamma is a function of R?





macaw,
If there are mistakes in the reasoning - I want to find them. If the theory is wrong, I would realy like to know that.


You already know about the mistakes:

1. Your formula for v_f is contradicted by experiment
2. Your calculations for GRB delay are wrong
3. Your lagrangian L is incorrect since it does not satisfy the Euler-Lagrange equation. (Did you try checking this?)
4. Your potential U is incorrect and your attempt to calculate its divergence is fraught with errors.

pogono
2011-Jul-29, 11:06 AM
(...)
Don't use the unusable integral form but use the differential form.
(...)
Also, as you apparently only do the \partial/\partial R part of the calculation and thus forget about the \partial/\partial \theta and \partial/\partial \phi you are assuming that there is spherical symmetry of the problem or you are looking at a 1 dimensional space.


Hi tusenfem.
Defined dilation field has spherical symetry. For me it seems, that it will be easier to use Ostrogradski-Gauss theorem and the Flux. Reasoning below.
\vec{U}=-\triangledown \gamma

then U we can express through T_x:
U=\frac{R_{X}}{2R^2}\gamma ^3=\frac{2\pi R_{X}}{4\pi R^2}\gamma ^3=\frac{cT_{X}}{4\pi R^2}\gamma ^3
\Phi =\oint_{S}^{ } \vec{U}d\vec{S}=4\pi R^2\cdot U(R)=cT_{X}\cdot \gamma ^3

From Gauss-Ostrogradski we got:
\oint_{V}^{ } \triangledown \vec{U}dV =\oint_{S}^{ } \vec{U}d\vec{S}=\Phi

Let's define "time flow density" as:
T_{X}=\oint_{V}^{ } \rho _{t}\: dV

Then we obtain:
\oint_{V}^{ } \triangledown \vec{U}dV =c\gamma ^3\cdot \oint_{V}^{ } \rho _{t}\: dV

I will recalculate and post new divergence.

pogono
2011-Jul-29, 11:18 AM
You already know about the mistakes:



Hello macaw.
Accually my tasks are:

1. Add error bars and check, if the photon formula comply with GRB experiments (Tensor's remark)
2. Check if my Lagrangian satisfy the Euler-Lagrange equation (your remark)
4. Recalculate the divergence (tusenfem's remark)
5. Calculate pulsar PSR 1913+16 system and check if orbit anomally match the experiment (Tensor's remark)

Thank you.
I will be back here with some results.

tusenfem
2011-Jul-29, 12:38 PM
sheeeeeeesh that takes time for something so simple

assuming your definition for gamma and spherical symmetry you get for the divergence:

\nabla \cdot \vec{U} = \frac{R_x}{2R^3} \gamma^3 + \frac{R_x^2}{4 R^4} \gamma^5

macaw
2011-Jul-29, 01:48 PM
Hello macaw.
Accually my tasks are:

1. Add error bars and check, if the photon formula comply with GRB experiments (Tensor's remark)

Your formulas don't comply, they are falsified by experiment as already explained.





2. Check if my Lagrangian satisfy the Euler-Lagrange equation (your remark)

It doesn't.



4. Recalculate the divergence (tusenfem's remark)

You are trying to calculate the divergence of a scalar. Even in this case, your calculation is clearly incorrect since you failed to notice that \gamma is a function of R as per your own definition.

pogono
2011-Jul-30, 08:53 PM
Hello macaw, Hello tusenfem,
It seems that the divergence remains as it was. It is not so easy with integral when we have space contraction, explanation below.

We start from equation:
\oint_{V}^{ } \left (\triangledown \cdot \vec{U} \right )dV =c\gamma ^3\cdot T_{X}

As we know V=dxdydz

Now we go to observer that fell space contraction (also time dilation). But from his point of view all space is contracted with one, constant gamma factor, that depends on where observer is located. Whole his space is cotracted with one constant gamma factor! We can write his "prim" space dimensions as:
V{}'=\frac{dx}{\gamma}\frac{dy}{\gamma}\frac{dz}{\ gamma}=V\frac{1}{\gamma ^3}

Whole volume with one gamma, so it makes, that gamma bahave as constant under this integral.

Thus we may write:
\frac{1}{c}\cdot \frac{1}{\gamma^3}\oint_{V}^{ } \left (\triangledown \cdot \vec{U} \right )dV=\frac{1}{c}\cdot \oint_{V'}^{ } \left (\triangledown \cdot \vec{U} \right )dV'=T_{X}

BTW, as you may see, interpreting above divergence, in the theory Time seems to flow through the space like it was a liquid...

We can define:
T_{X}=\oint_{{V}'}^{ } {\rho}' _{t}\: d{V}'=\oint_{V}^{ } \rho _{t}\: dV

Thus:
\triangledown \cdot \vec{U}={\rho}' _{t}c=\rho _{t}c\cdot \gamma ^3

macaw
2011-Jul-30, 10:08 PM
Hello macaw, Hello tusenfem,
It seems that the divergence remains as it was. It is not so easy with integral when we have space contraction,

No , it doesn't. The divergence calculation has nothing to do with any "space contraction". You are just compounding your errors.

pogono
2011-Jul-30, 10:32 PM
macaw,
please, read what is written farther.


The divergence calculation has nothing to do with any "space contraction".

Maybe in math - I agree.
In physics we count divergence, rotation, etc. in real space and real time. They have their real interpretation.
Up to Lorentz and Einstein we all used to think, that time and space must be one for every observer.
We do not think this way anymore, remember?

I define field as dilation factor.
It is space contraction factor as well.

macaw
2011-Jul-31, 12:12 AM
macaw,
please, read what is written farther.

I read, it is all wrong. See here:


But from his point of view all space is contracted with one, constant gamma factor, that depends on where observer is located. Whole his space is cotracted with one constant gamma factor! We can write his "prim" space dimensions as:
V{}'=\frac{dx}{\gamma}\frac{dy}{\gamma}\frac{dz}{\ gamma}=V\frac{1}{\gamma ^3}

Error 1: There is no contraction in the direction perpendicular to the motion
Error 2: You are still making the error of forgetting that \gamma is a function of R in your definition, so you cannot take it out from under the integral.

pogono
2011-Jul-31, 08:57 AM
macaw,
in SR - agree.

But in GR bodies near big mass feel time dilation just from beeing near source of gravity. So these bodies also fell space contraction. Which particular direction then?

All directions - whole space is shrinking.

Gamma factor is the attribute of the space dimentions that we multiply by in the integral.
It acts like constant.

tusenfem
2011-Jul-31, 10:17 AM
\triangledown \cdot \vec{U}={\rho}' _{t}c=\rho _{t}c\cdot \gamma ^3

sorry,wrong, see post #63, where I did the calculation for you in spherical coordinates

Why on Earth are you suddenly using carthesian coordinates x y z wheras up to now you have used spherical coordinates. Should you not care about the Jacobian in that case? Furthermore, just following the math that U is the gradient of gamma, then it takes about 2 minutes to write down the divergence of U, as I did in post #63.

The reason why you come up with these weird answers is because you put some mathemagics in your assuptions, creating new variables through integrals, wanting to use integrals and Gauss' theorem to calculate the divergence of a very simple(!!!!!!) variable. If you want to do complicated math, you may also want to use the simple way of doing it to actually check whether your answer is correct.

The triple gamma that you use in your "volume integral" is naturally total nonsense, as the observer (don't know why there is suddenly a moving observer in this problem) will only measure a gradient in gamma as a function of R, so there is only 1 gamma in the whole problem, the other two directions which you yourself have chosen to be invariant (that would be the theta and phi direction) they will NOT give you extra terms of gamma.

The of course as a "deus ex machina" you define the "time flow" through a volume, and then whammy you get your wrong answer for the divergence of U, which actually would be the Laplacian of gamma, but that as an aside.

Nonsense in gets nonsense out.

pogono
2011-Jul-31, 12:57 PM
ok, tusenfem,
show me which step of reasoning is wrong.

1. We define dilation factor as \gamma=\frac{1}{\sqrt{1-\frac{R_{X}}{R}}}=\frac{dt}{d\tau}.
It acts as space contraction factor as well.

2. We have one special observer in infinity, because there we do not feel space contraction and dilation.
\lim_{R \to \infty }\gamma =1

3. It means, that gamma is indeed:
\gamma=\frac{d\tau (\infty)}{d\tau}

4. Whole space around body near source of such field seems to shrink. Tree dimentions (not one as in Special Relativity).

5. We take sit of an observer in infinity and calculate gradient.
So gradient is calculated for one, chosen observer in infinity.

6. Calculating divergence, rotation, etc we have to treat gamma as constant atribute of space/time that we use for integrals. It may be seen f.e. in equations:

U=-\frac{1}{c}\cdot \frac{\partial A}{\partial \tau}= -\frac{\gamma }{c}\cdot\frac{\partial A}{\partial t}

\frac{1}{c}\cdot \frac{1}{\gamma^3}\oint_{V}^{ } \left (\triangledown \cdot \vec{U} \right )dV=\frac{1}{c}\cdot \oint_{V'}^{ } \left (\triangledown \cdot \vec{U} \right )dV'=T_{X}

macaw
2011-Jul-31, 01:31 PM
macaw,
in SR - agree.

But in GR bodies near big mass feel time dilation just from beeing near source of gravity. So these bodies also fell space contraction. Which particular direction then?

In the direction perpendicular to the gravitational field lines. Have you heard about tidal forces? Nothing to do with any "space contraction".




All directions - whole space is shrinking.

Nope, see above.


Gamma factor is the attribute of the space dimentions that we multiply by in the integral.
It acts like constant.

No, it isn't since you , yourself defined it as a function of R.

tusenfem
2011-Jul-31, 01:40 PM
ok, tusenfem,
show me which step of reasoning is wrong.


Oh Golly!

The fact that you take ages to calculate something simple should be indication enough that there is someting rotten in the state of Denver.



1. We define dilation factor as \gamma=\frac{1}{\sqrt{1-\frac{R_{X}}{R}}}=\frac{dt}{d\tau}.
It acts as space contraction factor as well.


fine, but don't forget that you are specifically defining gamma here as a function of R, which means NOT a function of phi or theta, the other two directions in spherical coordinates.



2. We have one special observer in infinity, because there we do not feel space contraction and dilation.
\lim_{R \to \infty }\gamma =1

3. It means, that gamma is indeed:
\gamma=\frac{d\tau (\infty)}{d\tau}


Who cares??????????????? This interjection is useless, does not add a thing, as the ONLY thing you were looking for (and I actually forget even why you want the divergence of U, but I don't care about that eiter).



4. Whole space around body near source of such field seems to shrink. Tree dimentions (not one as in Special Relativity).


No, that is not true, because gamma is only a function of R, by your own definition, there is no tree dimensions (there may be tHree, but even then there is nothing to your claim) the gradient in gamma is ONLY in R, nothing more nothing less.



5. We take sit of an observer in infinity and calculate gradient.
So gradient is calculated for one, chosen observer in infinity.


Another useless interjection, has nothing to do with what you want to calculate. Keep christmas decorations in the closet, even though Harrod's may have opened its winterwonderland already.



6. Calculating divergence, rotation, etc we have to treat gamma as constant atribute of space/time that we use for integrals. It may be seen f.e. in equations:

U=-\frac{1}{c}\cdot \frac{\partial A}{\partial \tau}= -\frac{\gamma }{c}\cdot\frac{\partial A}{\partial t}


What nonsense is that? treat gamma as a constant REALLY??? then the gradient of gamma would be zero and the divergence of the gradient would be zero. Could you please THINK before you write something. This may sound real profound, but it is rubbish. And what the frak is A suddenly in your equation??????????????????????????????????????? And now suddenly you define U as the time derivative of an unknown variable/function A, whereas in the beginning of this thread it was that U was the gradient of gamma? Guess you don't care much about being consistent?



\frac{1}{c}\cdot \frac{1}{\gamma^3}\oint_{V}^{ } \left (\triangledown \cdot \vec{U} \right )dV=\frac{1}{c}\cdot \oint_{V'}^{ } \left (\triangledown \cdot \vec{U} \right )dV'=T_{X}

It has already been said by both macaw and me that your third power gamma is incorrect, as length contraction only works in 1 (one, UNO) direction, so you can only get ONE gamma, and for the rest you have not show anything in this message, except for the fact that you are trying to come up with some new physical model, that apparently does not adhere to normal physics, and next to that has its own math, in which variables may sudddenly appear without explanation and nonsense equations are assumed to be a result.

Why not start at the beginning and write in scrupulous detail what exactly you think you are doing? Explaining ALL the variables that you are using, and checking whether what you want to calculate in a difficult integral method, actually agrees what normal mortals calculate with a simple (double) derivative?

tusenfem
2011-Jul-31, 01:54 PM
F.e if you take dilation as field it makes exactly the same time-space curvature than General Relativity predicts, but coming from field analyses. Short explanation below:

Let R_{X} will be some constant and dilation equals:
\gamma =\frac{dt}{d\tau} = \frac{1}{\sqrt{1-\frac{R_{X}}{R}}}
Let:
U=- \bigtriangledown \gamma = \frac{R_{X}}{2R^2} \gamma^3


Just to help you, up to this point there is no error in your math.



therfore:
\triangledown \cdot U=\frac{R_{X}}{V}\gamma^3


Then the second line, the divergence of U is WRONG because you forgot that gamma is a function of R and the correct expression can be found in post #65 (which you strangely seem to be ignoring ........ ) The correct expression is:

\nabla \cdot \vec{U} = \frac{R_x}{2R^3} \gamma^3 + \frac{R_x^2}{4 R^4} \gamma^5



We define vector potential A and B and express U by A (just like for Maxwell eq.)
B=\triangledown \times A


Sure, but what is the vector potential A and what is the vector B?



We can thus define U as:
U=-\frac{1}{c}\cdot \frac{\partial A}{\partial \tau}= -\frac{\gamma }{c}\cdot\frac{\partial A}{\partial t}
\triangledown \cdot U=-\frac{\gamma }{c}\cdot\frac{\partial (\triangledown \cdot A)}{\partial t}=\frac{R_{X}}{V}\gamma^3


I guess for some weird reason you can define U also as the time derivative of a "vector potential" A, but is is useful?
Naturally, the divergence is incorrect, and there should not be a V in the numerator either, somewhere you have put in a factor 4/3 pi.




If we take Schwarzschild radius as our constant R_{X} then (what is easy to derive) the relation between zero-element of stress-energy Tensor and A is:
-\frac{4\pi G}{c^4} \cdot T^{00}= \frac{1 }{c^2} \cdot\frac{\partial (\triangledown \cdot A)}{\partial t}=-\frac{R_{Schw}}{V}\gamma^2


Really? And where does this "easy to derive" relation come from? Probably a book or wiki, but are you sure that your "A" can be used?

I think I will just stop here, as your divergence was already incorrect, I don't see any use in trying to figure out what you want to do further on.

Geo Kaplan
2011-Jul-31, 02:48 PM
ok, tusenfem,
show me which step of reasoning is wrong.

1. We define dilation factor as \gamma=\frac{1}{\sqrt{1-\frac{R_{X}}{R}}}=\frac{dt}{d\tau}.
It acts as space contraction factor as well. {snip}

During the cold fusion fiasco of the late 1980s, Peter Hagelstein presented his theory about how it works. Unfortunately for prof. Hagelstein, the audience consisted of fusion experts. One declared that his equations merely possessed the gestalt of physics, but weren't actually physics.

Something similar is the case here. Your math, as has been pointed out by tusenfem and macaw (and others), is fundamentally flawed. Not a little wrong, but off-the-charts wrong. The fact that you got some numbers to agree roughly with observations (but again, "roughly" needs to be refined, and you have continued to ignore the very basic question about error bars that has been put to you repeatedly) is therefore a purely numerological coincidence, as I said before.

My nanognome theory produces numerical predictions that agree perfectly (after the fact) with any observational data, has no wrong math (because it offers none). It also just "feels" right to me, and therefore I declare it to be superior to your theory.

Note that the NG theory has every bit as much evidentiary support as yours, and does not suffer from mathematical self-inconsistency. This is not a good thing for your theory.

pogono
2011-Jul-31, 06:12 PM
Ok tusenfem, macaw and Tensor. It is clear now.



Really? And where does this "easy to derive" relation come from? Probably a book or wiki, but are you sure that your "A" can be used?

Relation below.
From Einstein's GR we have:

G_{uv}=-\frac{8\pi G}{c^4} \cdot T_{uv}

Zero-element of Stress-Energy tensor equals:

T^{00}=\rho c\gamma ^2=\frac{mc}{V}\gamma ^2

but for symetric solution with Schwartschild Radius we may write down:

\frac{2 G}{c^4}=\frac{R_{Schw}}{mc^2}

We may then write for zero-element righ side of Einstein equation as:

-\frac{8\pi G}{c^4} \cdot T^{00}=-4\pi \frac{R_{Schw}}{mc^2}\cdot \frac{mc}{V}\gamma ^2=-2\cdot 2\pi \frac {R_{Schw}}{c} \frac{1}{V}\gamma^2

If we remain my definition of time-matery relation and "time flow density" definition we may write:

T_{Schw}=2\pi \frac {R_{Schw}}{c}

Thus we may write:

-\frac{8\pi G}{c^4} \cdot T^{00}=2\frac {T_{Schw}}{V}\gamma^2=2\rho _{t}\gamma^2

That is why I get interested in time flow and dilation fields.

pogono
2011-Jul-31, 06:23 PM
you get for the divergence:

\nabla \cdot \vec{U} = \frac{R_x}{2R^3} \gamma^3 + \frac{R_x^2}{4 R^4} \gamma^5

It may be writen:

\nabla \cdot \vec{U} = \frac{R_x}{2R^3} \gamma^3 + \frac{R_x^2}{4 R^4} \gamma^5= \frac{R_x}{4R^3} \gamma^3 \left (2+ \frac{R_x}{R} \gamma^2 \right ) = \frac{R_x}{4R^3} \gamma^3 \left (1+ \gamma^2 \right )

I was using Gauss-Ostrogradski theorem to obtain density in my equations.
Would you write down above with density form? I do not know how to transform it.
Thank you in advance.

And it will probably will end my thread.

tusenfem
2011-Aug-01, 06:30 AM
It may be writen:

\nabla \cdot \vec{U} = \frac{R_x}{2R^3} \gamma^3 + \frac{R_x^2}{4 R^4} \gamma^5= \frac{R_x}{4R^3} \gamma^3 \left (2+ \frac{R_x}{R} \gamma^2 \right ) = \frac{R_x}{4R^3} \gamma^3 \left (1+ \gamma^2 \right )


No, it may not, unless your last = is supposed to be a \neq

I have no idea what density you are talking about, and no I cannot do that. It is YOUR ATM so YOU have to do the work.

Yes, this defnintely ends your thread.

I was using Gauss-Ostrogradski theorem to obtain density in my equations.
Would you write down above with density form? I do not know how to transform it.
Thank you in advance.

And it will probably will end my thread.[/QUOTE]

pogono
2011-Aug-01, 10:04 AM
No, it may not, unless your last = is supposed to be a \neq


I apreciate amount of work you have done to prove, that I am wrong.
But sometimes I am right:

2+ \frac{R_x}{R} \cdot \frac{1}{1-\frac{R_x}{R}} = 1+\frac{1}{1-\frac{R_x}{R}}

thus:

\frac{R_x}{4R^3} \gamma^3 \left (2+ \frac{R_x}{R} \gamma^2 \right ) = \frac{R_x}{4R^3} \gamma^3 \left (1+ \gamma^2 \right )


Wish you best.

tusenfem
2011-Aug-01, 12:58 PM
Here you are correct, my bad.

pogono
2011-Aug-05, 11:21 AM
Hello all,
it seems that I have solved Lagrangian and Divergence probleblems.
I also responce to tusenfen question about "what are the fields I define"

Lets define field as:
\varphi =-\frac{c}{\gamma_{G} }=-c\sqrt{1-\frac{R_{Schw}}{R}}
As we already know, that is negative of what I define as speed of light around mass.

Let:
\vec{A}=-\nabla \varphi
A=c\frac{R_{Schw}}{2R^2}\gamma_{G}=\frac{c}{c^2}\c dot \frac{2GM}{2R^2}\gamma_{G}=\frac{1}{c}\cdot \frac{GM}{R^2}\gamma_{G}

As you see it is correct relativistic gravitational field intensity c\vec{A}=\vec{g} and divergence looks much better now.

Let define vector fields (i explain what are these fields few lines below)
\nabla\times \vec{A}=-\frac{\gamma _{G}}{c}\frac{\partial \vec{P}}{\partial t}
\nabla\times \vec{P}=\frac{\gamma _{G}}{c}\frac{\partial \vec{A}}{\partial t}
\nabla\times \vec{U}= \vec{P}

We may define now sourceless A with U as:
\vec{A}=-\frac{\gamma _{G}}{c}\frac{\partial \vec{U}}{\partial t}
As we see, we may treat U as field representing Newtonian definition of velocity (where A is centripetal acceleration). P (as we will see futher) descibes Space Expansion

And we obtain Delambetian:
\frac{\gamma ^2_{G}}{c^2}\frac{\partial^2 \vec{A}}{\partial t^2}-\nabla^2 \vec{A}=0

In next post I will put test for Lagrangian.

pogono
2011-Aug-05, 12:20 PM
My Lagrangian defined in system of an resting observer located in infinity is:

L=E_0\left ( \gamma _{kin} -1 \right )-E_0\left ( \gamma _{G} -1 \right )=E_0\left ( \gamma _{kin} - \gamma _{G} \right )=E_0\left ( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} - \frac{1}{\sqrt{1-\frac{R_{Schw}}{R}}} \right )

Let's check now if:
\frac{\partial \frac{\partial L}{\partial v}}{\partial t}-\frac{\partial L}{\partial R}=0

\frac{\partial L}{\partial v}=\frac{E_0}{c^2}\cdot v\cdot \gamma ^3_{kin}=mv\gamma ^3_{kin}

\frac{d\left (mv\gamma ^3_{kin} \right )}{dt}=m\vec{a}\gamma ^3_{kin}

and
\frac{dL}{dR}=mc^2\cdot \frac{R_{Schw}}{2R^2}\gamma ^3_G=G\frac{Mm}{R^2}\gamma ^3_G

\frac{dL}{dR}=m\vec{g}\gamma ^3_G


It means we have to check, if: m\vec{a}\gamma ^3_{kin} = m\vec{g}\gamma ^3_G

Lets use ' ("prim") for relativistic force and make transformation:
c\int\int \vec{F'}\frac{1}{\gamma ^2_{G}} dt^2= c\int\int\vec{F'_G}\frac{1}{\gamma ^2_{kin}}dt^2

c^2\int\int \frac{\vec{F'}}{c}\;\; d\tau^2_G= c^2 \int\int\frac{\vec{F'_G}}{c}\;\; d\tau^2_{kin}=\vec S

Above integral is ACTION, and let's name it "vector field S", then

\frac{\vec{F'_G}}{c}=\frac{1}{c^2}\frac{\partial^2 \vec{S}}{\partial \tau^2_{kin}}=\frac{\gamma^2_{kin}}{c^2}\frac{\par tial^2 \vec{S}}{\partial t^2}

following our definition of gamma_G as gravity contraction factor we obtain:

\frac{\vec{F'}}{c}=\frac{1}{c^2}\frac{\partial^2 \vec{S}}{\partial \tau^2_{G}}=\nabla^2\vec S

So we obtain condition for common ACTION in form of Laplacjan for ACTION:

\frac{\gamma^2_{kin}}{c^2}\frac{\partial^2 \vec{S}}{\partial t^2}-\nabla^2\vec S=0 \;\; \to \;\; \vec{F'}=\vec{F'_G}

Condition is true, if gravity defines space metric (just like in GR) and if we have common ACTION in both observer's systems.

macaw
2011-Aug-05, 01:18 PM
It means we have to check, if: m\vec{a}\gamma ^3_{kin} = m\vec{g}\gamma ^3_G


Meaning that you claim that \gamma_{kin}=\gamma_G, something that is clearly meaningless and impossible, since you are claiming that \frac{v}{c}=\sqrt{\frac{R_X}{R}} as I pointed out in earlier posts (36,49).

pogono
2011-Aug-05, 05:49 PM
macaw,
what is exactly wrong with the velocity in the Lagrangian? For me it works fine.
And please remember, that we are saing about relative force, that for shure schould be equal to gravitational force.

To be more exact let me write mentioned velocity in two limits known as cosmic velocities.

A. For second cosmic velocity we have:
E_{kin}=E_G

thus, from my Lagrangian (two posts up):

\frac{v_{II}^2}{c^2}=\frac{R_{Schw}}{R}

v_{II}=\sqrt{c^2\cdot\frac{2GM}{c^2R}}=\sqrt{\frac {2GM}{R}}

CORRECT!

B. For first cosmic velocity we have from Virial Theorem:
2E_{kin}=E_G

2 \left ( \gamma _{kin}-1 \right )=\gamma _{G}-1

We may aproximate it with Maclaurin expansion for small velocities as:

2\cdot \frac{v_{I}^2}{2c^2}=\frac{R_{Schw}}{2R}

v_{I}=\sqrt{c^2\cdot\frac{2GM}{c^2 2R}}=\sqrt{\frac{GM}{R}}

CORRECT!

And one additional explanation for above.

Light velocity depends on photon energy and equals: c\sqrt{1-t_P\omega }
For grawitation we consider delay to light velocity that is: -c\sqrt{1-\frac{R_{Schw}}{R}}

That let us write Lagrangian for one specific case: for ray of light near mass. Lagrangian will be:
L=E_P\left ( \frac{1}{\sqrt{t_P\cdot \omega}} -1\right )-E_P\left ( \frac{1}{\sqrt{1-\frac{R_{Schw}}{R}}} -1\right )

P.S.
Has anyone data for galaxies with rotation curve anomalies? (dark mater suspected). I will check then, maybe my Lagrangian will work better then GR.

macaw
2011-Aug-05, 08:55 PM
My Lagrangian defined in system of an resting observer located in infinity is:

L=E_0\left ( \gamma _{kin} -1 \right )-E_0\left ( \gamma _{G} -1 \right )=E_0\left ( \gamma _{kin} - \gamma _{G} \right )=E_0\left ( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} - \frac{1}{\sqrt{1-\frac{R_{Schw}}{R}}} \right )

Let's check now if:
\frac{\partial \frac{\partial L}{\partial v}}{\partial t}-\frac{\partial L}{\partial R}=0

\frac{\partial L}{\partial v}=\frac{E_0}{c^2}\cdot v\cdot \gamma ^3_{kin}=mv\gamma ^3_{kin}

\frac{d\left (mv\gamma ^3_{kin} \right )}{dt}=m\vec{a}\gamma ^3_{kin}

and
\frac{dL}{dR}=mc^2\cdot \frac{R_{Schw}}{2R^2}\gamma ^3_G=G\frac{Mm}{R^2}\gamma ^3_G

\frac{dL}{dR}=m\vec{g}\gamma ^3_G


It means we have to check, if: m\vec{a}\gamma ^3_{kin} = m\vec{g}\gamma ^3_G



Lagrangian mechanics requires that \frac{\partial \frac{\partial L}{\partial v}}{\partial t}-\frac{\partial L}{\partial R} is IDENTICALLY equal to zero, what you obtained is a condition tying v and R. This is obviously wrong, as already pointed out since v and R are INDEPENDENT variables in the lagrangian.

macaw
2011-Aug-05, 09:11 PM
macaw,
what is exactly wrong with the velocity in the Lagrangian? For me it works fine.
And please remember, that we are saing about relative force, that for shure schould be equal to gravitational force.

To be more exact let me write mentioned velocity in two limits known as cosmic velocities.

A. For second cosmic velocity we have:
E_{kin}=E_G

thus, from my Lagrangian (two posts up):

\frac{v_{II}^2}{c^2}=\frac{R_{Schw}}{R}

v_{II}=\sqrt{c^2\cdot\frac{2GM}{c^2R}}=\sqrt{\frac {2GM}{R}}

CORRECT!



Nope, INCORRECT! v and R are INDEPENDENT variables, what you obtained is a relation between them. The Euler-Lagrange condition I gave you must be satisfied for ALL v and R.

pogono
2011-Aug-05, 09:52 PM
Nope, INCORRECT! v and R are INDEPENDENT variables, what you obtained is a relation between them.

macaw, please. I've just started to be in doubt with you.
Cosmic velocities, means:
-> v_1 is orbital speed
-> v_2 is escape speed
Cosmic velocities do define relation between this independent values R and v.
That is why I put it here - to show, that my Lagrangian works fine.

Here you may find few words about it:
http://en.wikipedia.org/wiki/Orbital_speed
http://en.wikipedia.org/wiki/Escape_velocity

macaw
2011-Aug-05, 11:51 PM
Cosmic velocities do define relation between this independent values R and v.

Not in the definition of the lagrangian. In the definition (http://en.wikipedia.org/wiki/Lagrangian#.22Cyclic_coordinates.22_and_conservati on_laws) of the lagrangian, v and R are coordinates, independent of each other.


That is why I put it here - to show, that my Lagrangian works fine.

Here you may find few words about it:
http://en.wikipedia.org/wiki/Orbital_speed
http://en.wikipedia.org/wiki/Escape_velocity

You are not addressing your error, the expression needs to be identically zero, what you got is a relationship between v and R. As I explained to you, in the Lagrangian, v and R are independent variables, also called "generalized coordinates" (http://en.wikipedia.org/wiki/Lagrangian#.22Cyclic_coordinates.22_and_conservati on_laws) , so you are compounding the errors.

pzkpfw
2011-Aug-06, 12:18 AM
macaw, please. I've just started to be in doubt with you.

Knock it off. Stick to the facts.

Celestial Mechanic
2011-Aug-06, 04:14 AM
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."

CM: "He proposes two new formulas which he claims he will prove. He proposes a photon velocity formula:

v = c * sqrt(1 - tPlanck * omega)
CM: "and a photon energy formula:

E = 2 * EPlanck(1/sqrt(1 - tPlanck * omega) - 1)
CM: "Both of these formulas have Planck units within them. I want to say a few cautionary words about Planck units.

CM: "I agree that wherever we have three physical quantities whose dimensions are linearly independent we may define an associated length, time and mass, and hence associated velocity, acceleration, energy, momentum, and so on. But there is nothing particularly physical about these quantities. There is no object that is exactly one Planck length long, nor is there a time interval of one Planck time, nor an elementary particle that is one Planck mass in size."

JK: "And what about the constants themselves? Should we use h or should we use h-bar? What about G? Should we use the Newtonian G or the Einsteinian one, which is 8*pi*G?"

CM: "Good question! We aren't privy to the exact definitions of the constants that Nature uses, so all our so-called Planck units are nothing but order of magnitude estimates of two very small (length and time) and one very large (mass, at least for an elementary partile) quantities."

V: "You're not suggesting that Planck units are meaningless, are you?"

CM: "No, I'm just trying to remove some of the meaningless mystique about the Planck units and to urge caution in their use. In the case of this paper, I would say that it is more proper for the Planck time and Planck energy to be replaced with a scale time and a scale energy that are within a few orders of magnitude of the so-called Planck units, quantities that will ultimately be determined by observation.

CM: "The second section, Introduction, is essentially a restatement of the abstract. They should be merged together. The only thing I want to point out is that the first sentence here is also very wrong:

One of the key issues of contemporary theoretical physics is Quantum Gravity and its related problems, e.g.:

* the nature of rest mass,
* the nature of the photon's kinetic energy, and
* the explanation of the observed slowing of high-energy photons in gamma-ray bursts.
JK: "Hey, that's not exactly what's written in the manuscript!"

CM: "I know. I'm taking the liberty of correcting the language to make it more idiomatic and proper. English is a very difficult language to master and I don't want language issues to cloud our discussion.

CM: "Anyway, the 'nature' of rest mass is not an issue of quantum gravity, it is an effect of the appropriate quantum field theories. Rest mass of elementary charged leptons is due to the interaction with the Higgs field. The rest masses of hadrons are a combination of the rest masses of the constituent quarks and/or antiquarks with the energy of the chromodynamic fields. In the case of nucleons, chromodynamic fields are the biggest contributor rather than the quark rest masses. The masses of the neutrinos is due to something called the 'see-saw effect' that I really don't understand yet."

V: "And the photon does not have kinetic energy, it just has energy. Period."

JK: "Well, I'd say that all the energy of a photon is all kinetic because it has no rest mass and hence no rest mass energy."

CM: "Hmm, there's something to both viewpoints. Let's leave it at the photon has energy and momentum appropriate to a particle with no rest mass and leave it at that. The 'nature' really isn't a mystery at all.

CM: "Lastly, we have the effect of slowing photons from GRBs. I will leave it to others to argue the validity of the cited experiments. The moment that the error bars stop enclosing c will be the time to quantify and explain the effect.

CM: "Let's now take up section 3, Time dilation as a field. I want to note that the equation numbers are not in sync with the section number. All the equations in this section are numbered (2.x)."

JK: "The thing that strikes me is the failure to properly distinguish scalar and vector quantities which makes the paper hard to read. His usage of U is also nonstandard, the standard being U, the gravitational potential, a scalar with the sign convention that it is positive. His equation (2.2) really needs some boldface and a couple of r-hats."

V: "Then, after a few notational definitions, the wheels come off in (2.10)."

CM: "How so?"

V: "He declares (slightly corrected):

Let us define the U field using the vector potential A. As according to our reasoning the field is the time dilation, therefore we will use the proper time for diffentiating the vector potential.
U = -delgamma - (1/c)*@A/@tau = -delgamma - (gamma/c)*@A/@t. (2.10).
V: "This is wrong. In general @A/@tau does not equal gamma*@A/@t. They are only equal if A is independent of the space coordinates, that is, A is a function of the time only. I'm sure that isn't what Mr. Pogonowski wants."

JK: "And what about the units? Since gamma is dimensionless, U must be an inverse length and A must also be dimensionless. And what ever happened to B defined in (2.9)? It is never referred to again."

CM: "The problem I have, and I'm sure you're having it too, is that there is no real motivation for adding this extra term using this dimensionless vector potential."

JK: "Equation (2.11) isn't very pretty either. He forgets that his gamma is a field so that his equation (2.11) is missing a term:

-(delgamma/c)*@A/@t.
JK: "The rest of the section is completely wrong from that point on. In particular, his equation (2.13) and his conclusion that the wave moves at speed c/gamma."

V: "And that's sort of the whole point of his paper."

CM: "'Fraid so. Let's take a little break here to fill our cups."

JK: "Say, what was the music you had on while we were talking?"

CM: "The Third Symphony of Henryk Gorecki, called the 'Symphony of Sorrowful Songs'. It is now the best-known piece of vocal music sung in Polish in the English-speaking world. A recording of it actually was in the Top Ten of the British charts!"

JK: "That first movement and its canon is awesome! But don't the string players get tired of sawing on the same two notes for bar after bar in the last movement?"

V: "Have patience and it will come to you."

To be continued ...

pogono
2011-Aug-06, 10:46 PM
Hello all, I am back.
Celestial Mechanic - it is really nice to meet you.

I have just fixed few mentioned problems (f.e. eq. enumeration problems), wrote few new things (test for lagrangian, new dilation field definition, etc) and updated newest theory draft at my www.

I am sure now, that scalar field definition schould be:
\varphi =-\frac{c}{\gamma}

If we make derivative on it we will obtain:

A) relative veloicty for kinetic gamma:
\frac{d\left (- \frac{c}{\gamma_{kin}} \right )}{dv}=\frac{d \left (c \cdot \sqrt{1-\frac{v^2}{c^2}} \right )}{dv}= \frac{1}{c} \cdot v\cdot \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{c} \cdot v \cdot \gamma_{kin}


B) relative acceleration for gravity:
\frac{d\left (- \frac{c}{\gamma_{G}} \right )}{dR}=\frac{d \left (c \cdot \sqrt{1-\frac{R_{Schw}}{R}} \right )}{dR}= \frac{cR_{Schw}}{2R^2} \cdot \frac{1}{\sqrt{1-\frac{R_{Schw}}{R}}}=\frac{1}{2R} \cdot \frac{1}{c} \cdot v_G^2 \cdot \gamma_{G}

where: v_G = c\sqrt{\frac{R_{Schw}}{R}}

BUT!!... it seems to be indeed pulsation, just like we should expect for vector field i have defined. That value comes from:
\vec{\omega} = \frac{1}{2} \cdot \nabla \times \vec{V}

so:
\frac{1}{2R} \cdot \vec{v_G^2} \cdot \gamma_{G} = \nabla \times \vec{V}

because:
R \cdot \omega = V


I ask you all for support necessary to finish the theory.
P.S. It seems to me, that it schould be rewriten into tensor algebra now.

macaw
2011-Aug-06, 10:53 PM
I ask you all for support necessaryto finish the theory.
.
We have already given you all the support pointing out the mistakes in your paper. For example:



so:
\frac{1}{2R} \cdot \vec{v_G^2} \cdot \gamma_{G} = \nabla \times \vec{V}

The LHS is a scalar, the right RHS is a vector!

pogono
2011-Aug-06, 11:31 PM
This is yet another basic mistake since it is very easy to show that your definition results into:
\vec{\omega} =0
which is absurd.

Hello macaw,
2\vec{\omega} = \nabla \times \vec{V}
It is vector identity.

Here you may find an example:
http://muweb.millersville.edu/~jdooley/macro/derive/eulerapd/curl1/curl1.htm

macaw
2011-Aug-06, 11:52 PM
Hello macaw,
2\vec{\omega} = \nabla \times \vec{V}
It is vector identity.

Here you may find an example:
http://muweb.millersville.edu/~jdooley/macro/derive/eulerapd/curl1/curl1.htm

The point was that

\frac{1}{2R} \cdot \vec{v_G^2} \cdot \gamma_{G} = \nabla \times \vec{V}

is just nonsense since you are equating a vector (RHS) with a scalar (LHS).

macaw
2011-Aug-07, 12:01 AM
B) relative acceleration for gravity:
\frac{d\left (- \frac{c}{\gamma_{G}} \right )}{dt}=\frac{d \left (c \cdot \sqrt{1-\frac{R_{Schw}}{R}} \right )}{dt}= \frac{cR_{Schw}}{2R^2} \cdot \frac{1}{\sqrt{1-\frac{R_{Schw}}{R}}}=\frac{1}{2R} \cdot \frac{1}{c} \cdot v_G^2 \cdot \gamma_{G}

Your computation of derivatives is incorrect. The units on the RHS do not match the units for acceleration. The whole calculation is just nonsense.

pogono
2011-Aug-07, 12:16 AM
ok, macaw. Such scalar we used to call pseudo-vector. It is case of notation.

Let me show one more thing.
Folowing my definition of the field vectors we should expect, that mentioned above velocity assigned to the particle equals on Planck Length:

v=v_G^2(l_P)=\frac{R_{Schw}}{l_P}

So its angle velocity (pulsation) for R=l_P will be:

\omega=\frac{R_{Schw}}{l_P \cdot t_P}

But it is exactly what I mentioned in post #43 about photon transition into two particles.
It is exactly the omega we need to photon split in two as we see below.



In my theory free space is full of electromagnetic field. We may analyze this filed as set of particles (photons) each with energy:
E=2E_{P}\cdot \left (\frac{1}{\sqrt{1-t_{P}\cdot\omega}}-1 \right ) \approx 2E_{P}\cdot \frac{t_{P}}{2 \omega} = \not{h}\cdot \omega

Photons (under specific conditions) may split in two, what we note as transition:
2E_{P} \left (\frac{1}{\sqrt{1-t_{P}\cdot\omega}}-1 \right ) \mapsto 2 \cdot E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}-1 \right)

Where:
E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}-1 \right) \approx E_{P}\frac{R_{Schw}}{2l_{P}} = \frac{l_{P}\cdot c^4}{G}\cdot \frac{R_{Schw}}{2\cdot l_{P}} = mc^2

And now, let me explain what Schwarzschild Radius may be and why I use it even for elementary particles.

Vectors of field rotation make helix in Minkowski time-space. We may show that if velocity equals “c” on circle then:
Tc = 2\pi R_{\omega } \to \omega = \frac{2\pi}{T} = \frac{c}{R_{\omega}}

For photons, the Curvature of this rotating field vectors helix equals:
\frac{1}{l_{P}}=\frac{R_{\omega}}{R_{\omega}^2-\left (\frac{R_{\omega}}{\gamma} \right )^2}

If we will change the way that vectors rotate, we change helix curvature and obtain particle:
\frac{1}{R_{Schw}}=\frac{l_{P}}{l_{P}^2-\left (\frac{l_{P}}{\gamma} \right )^2}

It means, that Schwarzschild Radius is not what we suspect it was. It is just the measure of some curvature in space-time. We can use it even for a photons or elementary particles.

As you may see it works!

EigenState
2011-Aug-07, 12:27 AM
Greetings,



Photons (under specific conditions) may split in two, what we note as transition ...


My apologies if this has already been addressed, but exactly what are the conditions under which a photon "may split in two"?

Best regards,
EigenState

pogono
2011-Aug-07, 12:32 AM
Your computation of derivatives is incorrect. The units on the RHS do not match the units for acceleration.

You right, it was mind shortcut. To obtain acceleration, we need to multiply it by "c" what comes out from Lagrangian. It was shown in post #82.

To be more exact we obtain angle velocity, just like I pointed.
\frac{1}{2R}\cdot \frac{1}{c}\cdot v_G^2\cdot \gamma

with units:
\frac{1}{[meter]}\cdot \frac{[sec]}{[meter]}\cdot \left (\frac{[meter]}{[sec]} \right )^2=\frac{1}{[sec]}

If we mutiply it by "c" we would obtain gravitational acceleration.

pogono
2011-Aug-07, 12:46 AM
Greetings,
My apologies if this has already been addressed, but exactly what are the conditions under which a photon "may split in two"?


Probably we all would like to know. You may read about pair production at:
http://en.wikipedia.org/wiki/Pair_production

My concept is that it seems to be a result of strong enough disturbance in e-m vectors rotation.
But to explain how strong is enough, we must explain why it have to have exactly this rotation to become a particle.
So it sounds like "ATM Standard Model" based on my theory. It is to fast for that. Now, I am looking for friendly physicists for the first piece of such thing.

BTW, macaw,
I have already echanged dt to dR for gravity in post #91. It was obvios mistake.

macaw
2011-Aug-07, 01:36 AM
ok, macaw. Such scalar we used to call pseudo-vector. It is case of notation.

This is pure numerology , you are equating a scalar to a vector and you are insisting that this is correct.
The dimensions don't match and they do not match acceleration and you are insisting that what you are doing is correct.
You calculate incorrectly a simple derivative wrt t and you insist that what you are doing is correct.




Let me show one more thing.

Don't show me any more "things", please address the exact objections to your incorrect derivations.

macaw
2011-Aug-07, 01:38 AM
B) relative acceleration for gravity:
\frac{d\left (- \frac{c}{\gamma_{G}} \right )}{dR}=\frac{d \left (c \cdot \sqrt{1-\frac{R_{Schw}}{R}} \right )}{dR}= \frac{cR_{Schw}}{2R^2} \cdot \frac{1}{\sqrt{1-\frac{R_{Schw}}{R}}}=\frac{1}{2R} \cdot \frac{1}{c} \cdot v_G^2 \cdot \gamma_{G}

This is even worse, the LHS does not have dimensions of acceleration (neither does the RHS). So, your "correction" only made the errors worse. Besides, \frac{d\left (- \frac{c}{\gamma_{G}} \right )}{dR} is not the formula for acceleration. Not even close.

pogono
2011-Aug-07, 09:57 AM
This is even worse, the LHS does not have dimensions of acceleration (neither does the RHS). So, your "correction" only made the errors worse. Besides, \frac{d\left (- \frac{c}{\gamma_{G}} \right )}{dR} is not the formula for acceleration. Not even close.

Hello macaw,
I am affraid you are wrong, again.

I understand that ATM Forum rules are, that you may say anything - I have to do the calcucations to prove. But I have already done the calculations.

As you may see below, we are close to gravitational acceleration, to be more exact, we obtain exactly the value of it divided by "c".



(...)
\varphi = - \frac{c}{\gamma_G}
(...)
Let:
\vec{A}=-\nabla \varphi
A=c\frac{R_{Schw}}{2R^2}\gamma_{G}=\frac{c}{c^2}\c dot \frac{2GM}{2R^2}\gamma_{G}=\frac{1}{c}\cdot \frac{GM}{R^2}\gamma_{G}

As you see it is correct relativistic gravitational field intensity c\vec{A}=\vec{g} and divergence looks much better now.
(...)


I have defined: v_G=c \cdot \sqrt{\frac{R_{Schw}}{R}}

thus: v_G^2=c^2 \cdot \frac{R_{Schw}}{R}

That is why:



(...)
B) relative acceleration for gravity:
\frac{d\left (- \frac{c}{\gamma_{G}} \right )}{dR}=\frac{d \left (c \cdot \sqrt{1-\frac{R_{Schw}}{R}} \right )}{dR}= \frac{cR_{Schw}}{2R^2} \cdot \frac{1}{\sqrt{1-\frac{R_{Schw}}{R}}}=\frac{1}{2R} \cdot \frac{1}{c} \cdot v_G^2 \cdot \gamma_{G}
(...)


I hope it is clear, now.

macaw
2011-Aug-07, 12:52 PM
I hope it is clear, now.

Yes, it is clear:

1. There is no such thing as "relative acceleration of gravity" in physics.
2. The expression that you posted does not have dimensions of acceleration, do the dimensional analysis and you will find out.

pogono
2011-Aug-07, 01:08 PM
Yes, it is clear:

1. There is no such thing as "relative acceleration of gravity" in physics.
2. The expression that you posted does not have dimensions of acceleration, do the dimensional analysis and you will find out.

1. There is relativistic force in physics. Really.
http://en.wikipedia.org/wiki/Special_relativity#Force

We may say, that it is relativistic mass multipyed by acceleration, or just "rest mass" multiplyed by relativistic acceleration. As you wish.

And if I try to explain gravity without time-spase curvature, I have to use relativistic force. Can you see another way? But, please, do not say: "there is no such need because GR is correct so we have to leave relativistic force concept and count time-space curvature"


2. Let us check dimentions once again:

\frac{d\left ( \frac{[meter]}{[sec]} \right )}{d[meter]} will produce dimention of ANGULAR VELOCITY \frac{1}{[sec]}
(as I already pointed in few post above)

When you will multiply it by c, you will obtain acceleration dimention:

\frac{[meter]}{[sec]} \cdot \frac{1}{[sec]}=\frac{[meter]}{[sec]^2}
(as I already pointed in few post above)

Geo Kaplan
2011-Aug-07, 01:19 PM
Let us check dimentions once again:


Perhaps you need to do so yet one more time. The RHS of the eqn in post 101 does not readily appear to have the dimensions of acceleration. Would you check/clarify?

macaw
2011-Aug-07, 01:55 PM
1. There is relativistic force in physics. Really.
http://en.wikipedia.org/wiki/Special_relativity#Force

The formula that you linked to disagrees with your formula of acceleration.


We may say, that it is relativistic mass multipyed by acceleration, or just "rest mass" multiplyed by relativistic acceleration. As you wish.

Nope, check the wiki page again. Once again:

1. There is no such thing as "relative acceleration of gravity" in physics.
There is coordinate acceleration and there is proper acceleration.





2. Let us check dimentions once again:

I did, they are still wrong.

EigenState
2011-Aug-07, 02:46 PM
Greetings,




... but exactly what are the conditions under which a photon "may split in two"?

Probably we all would like to know. You may read about pair production at:
http://en.wikipedia.org/wiki/Pair_production

My concept is that it seems to be a result of strong enough disturbance in e-m vectors rotation.
But to explain how strong is enough, we must explain why it have to have exactly this rotation to become a particle.


Thank you for the clarification. I had interpreted your statement to be that a photon could decay spontaneously into two photons.

I will not pursue the hypothesis regarding pair production being the result of a "strong enough disturbance in e-m vectors rotation". I suspect such questions would be deemed tangential to your original ATM hypothesis and thus deemed off-topic.

Best regards,
EigenState

pogono
2011-Aug-07, 02:51 PM
Perhaps you need to do so yet one more time. The RHS of the eqn in post 101 does not readily appear to have the dimensions of acceleration. Would you check/clarify?

Dear macaw and Geo Kaplan.

I know my idea is against the mainstream. I appreciate your attention.
But please, take a deep breath and read what is below very slowly and carefully.

Step 1.

We do not expect, that RHS of the eqn in post 101 should be acceleration. It should be ANGULAR VELOCITY (It is vector rotation of some velocity assigned to particle, that I described in post #91 ).

So it should be angular velocity dimension: \frac{[radian]}{[sec]}=\frac{1}{[sec]}

Let us check RHS of #101: \frac{cR_{Schw}}{2R^2}\gamma_G

C – sped of light, [meter] / [sec]
R_Schw, [meter]
R^2 – [meter]^2
\gamma – no dimention

Thus:
\frac{cR_{Schw}}{R^2}\gamma_G=\frac{[meter]}{[sec]}\cdot \frac{[meter]}{[meter]^2}=\frac{1}{[sec]}

CORRECT!

Step 2.
I also point, that this ANGULAR VELOCITY is the same time the gravitational acceleration divided by “c”.

So now, we will multiply it by “c”, to obtain gravitational acceleration, just like I described in few post above.
\frac{[1]}{[sec]}\cdot \frac{[meter]}{[sec]} = \frac{[meter]}{[sec]^2}

It will be:
c \cdot \frac{cR_{Schw}}{2R^2}\gamma_G= c^2 \cdot \frac{R_{Schw}}{2R^2}\gamma_G=G\frac{M}{R^2}\gamma _G

If you still see, that there is some mistake in dimensions – please, point it.

pogono
2011-Aug-07, 03:07 PM
(...)
Once again:
1. There is no such thing as "relative acceleration of gravity" in physics.
There is coordinate acceleration and there is proper acceleration.
(...)


Dear macaw,
do not we use in physics four vector of velocity? Why then?
There is nothing like relative velocity if we multiply it by mass in momentum four-vector.
So mass is relativistic or velocity is relativistic?

We define such four-vectors, because they are useful in calculations.
I do the same. There is nothing wrong with using relativistic acceleration multiplied by rest mass.
If there is no term "relative acceleration" in your physical dictionary - just define it, as I did.

macaw
2011-Aug-07, 03:25 PM
I also point, that this ANGULAR VELOCITY is the same time the gravitational acceleration divided by “c”.

Why would you divide gravitational acceleration by c? What is the connection with angular velocity? What angular velocity are you talking about? This isn't physics, this is just numerology.


So now, we will multiply it by “c”, to obtain gravitational acceleration, just like I described in few post above.

Why are you multiplying with c? Where is this coming from?




If you still see, that there is some mistake in dimensions – please, point it.

You simply finagled things after the error was pointed to you.

macaw
2011-Aug-07, 03:39 PM
Dear macaw,
do not we use in physics four vector of velocity? Why then?

Because the four-vector formalism, as opposed to yours, works




We define such four-vectors, because they are useful in calculations.
I do the same.

What you do is not useful because it is riddled with errors.


There is nothing wrong with using relativistic acceleration multiplied by rest mass.
If there is no term "relative acceleration" in your physical dictionary - just define it, as I did.

It doesn't work this way since the formula for force contradicts your definition of "relative gravitational acceleration", whatever that may mean in your mind. See here:

\vec{F}=\vec{v}m(\vec{v} \vec{a})\gamma^3/c^2+\gamma m \vec{a}

pogono
2011-Aug-07, 04:05 PM
(...)
See here:
\vec{F}=\vec{v}m(\vec{v} \vec{a})\gamma^3/c^2+\gamma m \vec{a}

Yes, I know it. First part of force is paralel to the move, second is perpendicular. My force is parpendicular. That is all.

macaw
2011-Aug-07, 04:15 PM
Yes, I know it. First part of force is paralel to the move, second is perpendicular. My force is parpendicular. That is all.

Why is your "force perpendicular"? I thought that you were trying to model the gravitational force. Last I checked, contrary to your claims, gravitational force is a radial force. Did you invent a new gravitational force while I was pointing out the errors in your posts?

pogono
2011-Aug-07, 04:34 PM
Enumeration is mine.



1) Why would you divide gravitational acceleration by c?
2) What is the connection with angular velocity?
3) What angular velocity are you talking about?
4) This isn't physics, this is just numerology.
5) Why are you multiplying with c?
6) Where is this coming from?
7) You simply finagled things after the error was pointed to you.

macaw, did you read my posts or theory draft? It is all explained there...

7) - You may check dates/time of my posts on www. I do not edit posts after somebody point an error with quote.
4) - Call it as you wish.

1,2,3,5,6) - You will find explanation in posts where I define a field, mainly: #81 , #82 , #91
In post #96 I also explain relation betwen ANGULAR VELOCITY and particle energy and phenomenon of pair production.

macaw
2011-Aug-07, 04:56 PM
Enumeration is mine.



macaw, did you read my posts or theory draft? It is all explained there...

I checked your draft, you changed the lagrangian after I pointed out that it is incorrect. The new lagrangian is just as incorrect.
Please answer the questions here, in this thread, stop referring us to the "theory draft".

macaw
2011-Aug-07, 05:00 PM
You right, it was mind shortcut. To obtain acceleration, we need to multiply it by "c" what comes out from Lagrangian. It was shown in post #82.

Your lagrangian (both the old one and the new one) isn't a correct lagrangian, so whatever "comes out of the lagrangian" isn't correct either. Even if you gratuitously multiply it with c.

macaw
2011-Aug-07, 05:05 PM
ok, macaw. Such scalar we used to call pseudo-vector. It is case of notation.

Let me show one more thing.
Folowing my definition of the field vectors we should expect, that mentioned above velocity assigned to the particle equals on Planck Length:

v=v_G^2(l_P)=\frac{R_{Schw}}{l_P}


Huh?

v=v_G^2(l_P) ??!!!

Did you forget to check dimensions again?

pogono
2011-Aug-07, 05:11 PM
Huh?

v=v_G^2(l_P) ??!!!

Did you forget to check dimensions again?

Ups. Here you are right. To fast writen.
Should be:
v=\frac{1}{c} v_G^2(l_P)

pogono
2011-Aug-07, 05:46 PM
1) Why would you divide gravitational acceleration by c?
2) What is the connection with angular velocity?
3) What angular velocity are you talking about?
4) This isn't physics, this is just numerology.
5) Why are you multiplying with c?
(...)

Please answer the questions here, in this thread, stop referring us to the "theory draft".



I refer you to my posts, nr. #81 , #82 , #91, #96 where it is explained.
But ok, if you wish I explain it here, in this post.

4) Let's start from multipling by "c" called numerology.
If multiplying by "c" is numerology, what would you say to Einstein, multiplying G with 8\pi in his GR equation - pure numerology ;-)


1), 5) - Below I explain where this "c" comes from.
Relations between fields I define are:


Lets define field as:
\varphi =-\frac{c}{\gamma_{G} }=-c\sqrt{1-\frac{R_{Schw}}{R}}
As we already know, that is negative of what I define as speed of light around mass.

Let:
\vec{A}=-\nabla \varphi
A=c\frac{R_{Schw}}{2R^2}\gamma_{G}=\frac{c}{c^2}\c dot \frac{2GM}{2R^2}\gamma_{G}=\frac{1}{c}\cdot \frac{GM}{R^2}\gamma_{G}

As you see it is correct relativistic gravitational field intensity c\vec{A}=\vec{g} and divergence looks much better now.

Let define vector fields (i explain what are these fields few lines below)
\nabla\times \vec{A}=-\frac{\gamma _{G}}{c}\frac{\partial \vec{P}}{\partial t}
\nabla\times \vec{P}=\frac{\gamma _{G}}{c}\frac{\partial \vec{A}}{\partial t}
\nabla\times \vec{U}= \vec{P}

We may define now sourceless A with U as:
\vec{A}=-\frac{\gamma _{G}}{c}\frac{\partial \vec{U}}{\partial t}

Rest is just math, I propose to use some software. I use Derive 6.1

2),3) - One answer for both.
Rotation of a velocity vector field means ANGULAR VELOCITY - it is one of vector identity. Relation is (but do not talk again about "pulsation may not be called pseudo-field, please):




B) relative acceleration for gravity:
\frac{d\left (- \frac{c}{\gamma_{G}} \right )}{dR}=\frac{d \left (c \cdot \sqrt{1-\frac{R_{Schw}}{R}} \right )}{dR}= \frac{cR_{Schw}}{2R^2} \cdot \frac{1}{\sqrt{1-\frac{R_{Schw}}{R}}}=\frac{1}{2R} \cdot \frac{1}{c} \cdot v_G^2 \cdot \gamma_{G}

where: v_G = c\sqrt{\frac{R_{Schw}}{R}}

BUT!!... it seems to be indeed pulsation, just like we should expect for vector field i have defined. That value comes from:
\vec{\omega} = \frac{1}{2} \cdot \nabla \times \vec{V}

so:
\frac{1}{2R} \cdot \vec{v_G^2} \cdot \gamma_{G} = \nabla \times \vec{V}

because:
R \cdot \omega = V
(...)


But, if we count what V is (from field relations) we obtain:



Folowing my definition of the field vectors we should expect, that mentioned above velocity assigned to the particle equals on Planck Length:

v=v_G^2(l_P)=\frac{R_{Schw}}{l_P} ***

So its angle velocity (pulsation) for R=l_P will be:

\omega=\frac{R_{Schw}}{l_P \cdot t_P}

But it is exactly what I mentioned in post #43 about photon transition into two particles.
It is exactly the omega we need to photon split in two**** as we see below.

*** with small correction 1/c of your last post
**** should be written: "into two particles".

Then let's explain what is the relation with particle energy:


(...)
In my theory free space is full of electromagnetic field. We may analyze this filed as set of particles (photons) each with energy:
E=2E_{P}\cdot \left (\frac{1}{\sqrt{1-t_{P}\cdot\omega}}-1 \right ) \approx 2E_{P}\cdot \frac{t_{P}}{2 \omega} = \not{h}\cdot \omega

Photons (under specific conditions) may split in two, what we note as transition:
2E_{P} \left (\frac{1}{\sqrt{1-t_{P}\cdot\omega}}-1 \right ) \mapsto 2 \cdot E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}-1 \right)

Where:
E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}-1 \right) \approx E_{P}\frac{R_{Schw}}{2l_{P}} = \frac{l_{P}\cdot c^4}{G}\cdot \frac{R_{Schw}}{2\cdot l_{P}} = mc^2

And now, let me explain what Schwarzschild Radius may be and why I use it even for elementary particles.

Vectors of field rotation make helix in Minkowski time-space. We may show that if velocity equals “c” on circle then:
Tc = 2\pi R_{\omega } \to \omega = \frac{2\pi}{T} = \frac{c}{R_{\omega}}

For photons, the Curvature of this rotating field vectors helix equals:
\frac{1}{l_{P}}=\frac{R_{\omega}}{R_{\omega}^2-\left (\frac{R_{\omega}}{\gamma} \right )^2}

If we will change the way that vectors rotate, we change helix curvature and obtain particle:
\frac{1}{R_{Schw}}=\frac{l_{P}}{l_{P}^2-\left (\frac{l_{P}}{\gamma} \right )^2}

It means, that Schwarzschild Radius is not what we suspect it was. It is just the measure of some curvature in space-time. We can use it even for a photons or elementary particles.

pogono
2011-Aug-07, 05:58 PM
And one word more. Why the curvature looks this way?

If we check Helix curvature, you will see, that for parametrical function based on "a, b" parameters, Helix curvature equals:
\kappa = \frac{a}{a^2+b^2}

In Minkowski time-space thanks to its metric tensor using "minus" we obtain, that curvature is:
\kappa = \frac{a}{a^2-b^2}

Quite simple, indeed.

pogono
2011-Aug-07, 06:20 PM
It is a year now, since I try to publish above Idea in some per-reviewed publications.

Most of replies are sort of macaw ones.
And I do some mistakes - it is immature idea.

I have decided to find some friendly physicist that may help with puting pieces together, rewriting to tensor algebra and wish to publish it as common work.
Again, I ask you for support.

macaw
2011-Aug-07, 06:27 PM
Ups. Here you are right. To fast writen.
Should be:
v=\frac{1}{c} v_G^2(l_P)

Just as bad since you have :

v=v_G^2(l_P)=\frac{R_{Schw}}{l_P}

How can anybody take your "paper" seriously when it is riddled with silly mistakes?

pogono
2011-Aug-07, 07:16 PM
Just as bad since you have :

v=v_G^2(l_P)=\frac{R_{Schw}}{l_P}

How can anybody take your "paper" seriously when it is riddled with silly mistakes?

Yes, macaw. You are right. Silly mistakes.
I work alone and post quicly. Really sorry for such silly mistakes.

Should be:

v=\frac{1}{c} \cdot v_G^2(l_P)=c \cdot \frac{R_{Schw}}{l_P}

\omega = \frac{c}{R_{\omega}} = \frac{R_{Schw}}{l_P \cdot t_P}

Geo Kaplan
2011-Aug-08, 12:09 AM
Step 1.

We do not expect, that RHS of the eqn in post 101 should be acceleration. It should be ANGULAR VELOCITY (It is vector rotation of some velocity assigned to particle, that I described in post #91 ).

I agree that the equation has dimensions consistent with an angular velocity. What was confusing was that you introduced the equation with the following words:


B) relative acceleration for gravity:

It is therefore natural for a reader to expect that the equation following that introductory text would, in fact, have the dimensions of acceleration, not velocity, yes?

macaw
2011-Aug-08, 01:18 AM
Yes, macaw. You are right. Silly mistakes.
I work alone and post quicly. Really sorry for such silly mistakes.

Should be:

v=\frac{1}{c} \cdot v_G^2(l_P)=c \cdot \frac{R_{Schw}}{l_P}

\omega = \frac{c}{R_{\omega}} = \frac{R_{Schw}}{l_P \cdot t_P}

So you got the math finally right but the physics is as meaningless as at the beginning of the thread.

pogono
2011-Aug-08, 07:16 AM
I agree that the equation has dimensions consistent with an angular velocity. What was confusing was that you introduced the equation with the following words:
(...)
It is therefore natural for a reader to expect that the equation following that introductory text would, in fact, have the dimensions of acceleration, not velocity, yes?

Yes, I should write it more clearly. I take care next time.
I treat "c" quite like "1" and write to fast. That is where the problem comes from.

I am happy that you confirm that calculations are corect.

pogono
2011-Aug-08, 08:43 AM
So you got the math finally right but the physics is as meaningless as at the beginning of the thread.

Not exactly. Beside movement in gravitational field description, above theory does few things that GR do not.

1. It explains wave-particle duality, showing that:
a) photon is a pair of E-M field energy quants:


E=2E_{P}\cdot \left (\frac{1}{\sqrt{1-t_{P}\cdot\omega}}-1 \right ) \approx 2E_{P}\cdot \frac{t_{P}}{2 \omega} = \not{h}\cdot \omega

b) particle rest energy is a quant of gravitational field energy:


E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}-1 \right) \approx E_{P}\frac{R_{Schw}}{2l_{P}} = \frac{l_{P}\cdot c^4}{G}\cdot \frac{R_{Schw}}{2\cdot l_{P}} = mc^2


c) Lagrangian condition is the same time Delambertian condition for common action denoted as “S”:


\frac{\partial \frac{\partial L}{\partial v}}{\partial t}-\frac{\partial L}{\partial R}=0 \;\; \to \;\; \frac{\gamma^2_{kin}}{c^2}\frac{\partial^2 \vec{S}}{\partial t^2}-\nabla^2\vec S=0 \;\; \to \;\; \vec{F'}=\vec{F'_G}


d) It uses field equation showing, that gravity (A vector), velocity of a bodies (U vector) and expansion (rotation of P vector) seems to work together like fields related to each other.

2. It explains what field really IS (time dilation factor) in a lit bit different way than GR does and may be used for other fields.

All of it digs a tunnel between Big Scales Theories (like GR) and Quantum Mechanics.

3. It assigns particle with some rotation on helix. This way it opens possibilities to find out what kind of rotation makes what kind of particles. It may let us leave Standard Model with its 18 parameters, and just find truly geometrical explanation for particle families.



For photons, the Curvature of this rotating field vectors helix equals:
\frac{1}{l_{P}}=\frac{R_{\omega}}{R_{\omega}^2-\left (\frac{R_{\omega}}{\gamma} \right )^2}

If we will change the way that vectors rotate, we change helix curvature and obtain particle:
\frac{1}{R_{Schw}}=\frac{l_{P}}{l_{P}^2-\left (\frac{l_{P}}{\gamma} \right )^2}


4. It changes our imagination of a black holes and beginning of universe. In my theory if you increase photon/particle energy higher and higher it SLOWS DOWN!
If we reverse that film – sounds promising…

and so on…

All of it works fine IF we will observe photon’s delays related to energy and will have to through away GR (that we all are afraid of).

IF… then we should look closer to above idea.

macaw
2011-Aug-08, 01:12 PM
Not exactly. Beside movement in gravitational field description, above theory does few things that GR do not.

1. It explains wave-particle duality, showing that:
a) photon is a pair of E-M field energy quants:

MQ1: The above equation, (put together by hand), results into a variable speed of light according to your theory, v_f=c \sqrt{1- t_p \omega}. This has been falsified by experiment, so how do you explain this away?


b) particle rest energy is a quant of gravitational field energy:

MQ2: The above equation, (put together by hand), results into a variable speed of light according to your theory, v_f=c \sqrt{1- R_s/R}. This has been falsified by experiment, so how do you explain this away?




c) Lagrangian condition is the same time Delambertian condition for common action denoted as “S”:

MQ3: Nope, your Lagrangian forces a relationship between what should be independent variables v and R. How do you explain this recurring error away?



d) It uses field equation showing, that gravity (A vector), velocity of a bodies (U vector) and expansion (rotation of P vector) seems to work together like fields related to each other.

MQ4: If it did (it doesn't really), you would have no problem calculating the speed of an object v as a function of the radial coordinate R for an object dropped from radial distance R_0. Let's see your derivation from your "field equations".




2. It explains what field really IS (time dilation factor) in a lit bit different way than GR does and may be used for other fields.

You got things backwards, experiment show that (gravitational) time dilation is a CONSEQUENCE of the presence of spacetime curvature, not the other way around, as you claim.





All of it digs a tunnel between Big Scales Theories (like GR) and Quantum Mechanics.

Only because you put in by hand t_p \omega = R_s/R?








4. It changes our imagination of a black holes and beginning of universe. In my theory if you increase photon/particle energy higher and higher it SLOWS DOWN!
If we reverse that film – sounds promising…

MQ5: What "slows down"?




All of it works fine IF we will observe photon’s delays related to energy

MQ6: What "photon delays related to energy"? Whose energy?



and will have to through away GR (that we all are afraid of).

Not in favor of your "theory". Not a chance, the way it is riddled with mistakes.




IF… then we should look closer to above idea.

I did, all I saw is a lot of elementary mistakes.

pogono
2011-Aug-08, 03:46 PM
MQ1: The above equation, (put together by hand), results into a variable speed of light according to your theory, v_f=c \sqrt{1- t_p \omega}. This has been falsified by experiment, so how do you explain this away?
Experiments you mentioned are not sensitive enough to check if it is truth. Difference between sped of photon and "c" may be measured only for VERY long distances and high photon energies (Gamma Ray Bursts). F.e. if you take an experiment done for visible light (about eV energy) we have:
\omega_{eV}=1,51955 \cdot 10^{15}
t_{P}=5,391 \cdot 10^{-44}
c-v_f=c\cdot \left ( 1-\sqrt{1-8,1918\cdot 10^{-29}} \right )=c\cdot 4,0959\cdot 10^{-29}
Show me your experiment sensitive enough…



MQ2: The above equation, (put together by hand), results into a variable speed of light according to your theory, v_f=c \sqrt{1- R_s/R}. This has been falsified by experiment, so how do you explain this away?
v_f=c \sqrt{1- R_s/R} is a mesure of time delay around mass coming out from GR that already has been proven. I only say - photons are our way to count time flow.



MQ3: Nope, your Lagrangian forces a relationship between what should be independent variables v and R. How do you explain this recurring error away?
Nope, it is not relationship you think of. Derivative of my Lagrangian says:
Force acting on body equals Gravitational Force.
If we want to use relativistic force, we have to use kinetic gamma factor in it. That's all.



MQ4: If it did (it doesn't really), you would have no problem calculating the speed of an object v as a function of the radial coordinate R for an object dropped from radial distance R_0. Let's see your derivation from your "field equations".

Where you see such problem? My Lagrangian works.
You get proper planet orbits, it results with corect cosmic velocities, corect orbit anomalies, and so on. If you test it for photons, you obtain correct value of photon's track cuvature on Sun's dick.
What more would you want?




MQ5: What "slows down"?
MQ6: What "photon delays related to energy"? Whose energy?

Photons slows down when we increase photons energy.



I did, all I saw is a lot of elementary mistakes.
Elementary mistakes with counting or not by "c" does not negate the idea behind.

macaw
2011-Aug-08, 03:56 PM
Experiments you mentioned are not sensitive enough to check if it is truth. Difference between sped of photon and "c" may be measured only for VERY long distances and high photon energies (Gamma Ray Bursts). F.e. if you take an experiment done for visible light (about eV energy) we have:
\omega_{eV}=1,51955 \cdot 10^{15}
t_{P}=5,391 \cdot 10^{-44}
c-v_f=c\cdot \left ( 1-\sqrt{1-3,5477\cdot 10^{-59}} \right )=c\cdot 1,7738\cdot 10^{-59}
Show me your experiment sensitive enough…

I have already corrected your misconception in post 10. Why are you going around in circles?





v_f=c \sqrt{1- R_s/R} is a mesure of time delay around mass coming out from GR that already has been proven.

MQ7: please provide proof that it is "proven". Any citation from peer-reviewed paper?




Nope, it is not relationship you think of. Derivative of my Lagrangian says:

We've been over this several times, you don't understand the basics of lagrangian mechanics.



Force acting on body equals Gravitational Force.
If we want to use relativistic force, we have to use kinetic gamma factor in it. That's all.

We've been over this mistake (http://www.bautforum.com/showthread.php/118332-Dilation-as-field?p=1922551#post1922551) as well. Are you still claiming that gravitational force is transverse?




Where you see such problem? My Lagrangian works.
You get proper planet orbits, it results with corect cosmic velocities, corect orbit anomalies, and so on. If you test it for photons, you obtain correct value of photon's track cuvature on Sun's dick.
What more would you want?

MQ8: prove it. Derive

-light bending by the Sun

-the advancement of Mercury perihelion

Both of the above require a knowledge of lagrangian mechanics and a correct lagrangian. You have neither.







Photons slows down when we increase photons energy.

MQ9: Prove it. With math. Experiment falsifies your claim. We have been over this before as well.





Elementary mistakes with counting or not by "c" does not negate the idea behind.

There are many more (http://www.bautforum.com/showthread.php/118332-Dilation-as-field?p=1922551#post1922551) that you refuse to admit to.

pogono
2011-Aug-08, 04:04 PM
Hi macaw,
sorry for that, by the same time I edited my post and you have quoted my post. In first part of photon light speed is magnitude of 10^(-29)

macaw
2011-Aug-08, 04:22 PM
Hi macaw,
sorry for that, by the same time I edited my post and you have quoted my post. In first part of photon light speed is magnitude of 10^(-29)

Just as wrong (for reasons I have already explained to you in post 10). Tensor explained the same exact thing to you in post 11. Why are you going around in circles?

pogono
2011-Aug-08, 06:47 PM
MQ7: please provide proof that it is "proven" (***). Any citation from peer-reviewed paper?
(*** - time dilation around mass as higher, as I gues)

Time dilation around mass is proven and it is exactly the equation I use. It is a base for GPS time calculations, f.e.
http://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere
If you do not belive, I will post some per-reviewed citation.
The idea of slowing down (curve track of) photons in GR takes place of Potential Energy.



We've been over this several times, you don't understand the basics of lagrangian mechanics.
We've been over this mistake (http://www.bautforum.com/showthread.php/118332-Dilation-as-field?p=1922551#post1922551) as well. Are you still claiming that gravitational force is transverse?

Gravitational force is radial. But it means transverse to rotation velocity axis. I use polar metric system.



MQ8: prove it. Derive
-light bending by the Sun
-the advancement of Mercury perihelion

Correct advancement of peryhelions of all planets in Solar System you may find at:
http://tp-theory.net/Anomalies.xls
I have also explained it in post #26 (citing with editorial corection)



(...)
Calculations for non-rotating solar system with centrall big mass (example for Mercury):
1. You take Rest Energy of the orbiting planet. For Mercury it is mc^2 = 3.33020 \cdot 10^{23} \cdot c^2

2. You calculate Mechanical Energy (value of Lagrangian). The easiest way is to calculate it in Aphelium or Peryhelium (does not matter, Mechanical Energy is constant).

3. For Mercury Mechanical Energy equals L= -1.27494909 \cdot 10^{-8} \cdot mc^2

4. Than you get Dilation D = \frac{L}{mc^2} = -1.27494909 \cdot 10^{-8} ***

5. Anomaly per rotation equals \bigtriangleup\alpha = (2\pi)^2 \cdot D

6. To obtaint anomaly for century and with angle seconds, you have to multiply it by number of revolutions per century and by \frac{648000}{\pi}

*** why I call it dilation?
Mechanical Energy means Kinetic Energy in infinity (in infinity there is no gravity, so the only part of the Lagrangian remains Kinetic Energy). If you divide it by Rest Energy, you gey relativistic dillation factor (\gamma-1)

I will put here calculations for light bending by the Sun and for Pulsar PSR B1913+16 system (Tensor's remark that I am working over).
Light bending I have made as numerical simulation in Derive. This is realy hard to derive in simple form - Lagrangian for photons is here:


L=E_P\left ( \frac{1}{\sqrt{t_P\cdot \omega}} -1\right )-E_P\left ( \frac{1}{\sqrt{1-\frac{R_{Schw}}{R}}} -1\right )




MQ9: Prove it(***). With math. Experiment falsifies your claim. We have been over this before as well.

(*** - that photons slows down when we increase its energy)

I am working over error bars for measuring GRB delay (Tensor's and your remark).
I will post it here, soon.

macaw
2011-Aug-08, 08:49 PM
(*** - time dilation around mass as higher, as I gues)

Time dilation around mass is proven and it is exactly the equation I use. It is a base for GPS time calculations, f.e.
http://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere
If you do not belive, I will post some per-reviewed citation.
The idea of slowing down (curve track of) photons in GR takes place of Potential Energy.

This is NOT what you claimed. Let me remind you one more time what you claimed:


v_f=c \sqrt{1- R_s/R} is a mesure of time delay around mass coming out from GR that already has been proven. I only say - photons are our way to count time flow.

MQ7 is asking you to show proof of the above, NOT of gravitational time dilation.



Gravitational force is radial. But it means transverse to rotation velocity axis. I use polar metric system.

Actually you are trying to cover up your misunderstanding (http://www.bautforum.com/showthread.php/118332-Dilation-as-field?p=1922548#post1922548) of force in special relativity.






Correct advancement of peryhelions of all planets in Solar System you may find at:
http://tp-theory.net/Anomalies.xls
I have also explained it in post #26 (citing with editorial corection)

This is not a calculation, you are putting in the results by hand. Please do a derivation starting from your (incorrect) lagrangian . Let's see the calculations






I will put here calculations for light bending by the Sun and for Pulsar PSR B1913+16 system (Tensor's remark that I am working over).
Light bending I have made as numerical simulation in Derive. This is realy hard to derive in simple form

No, it isn't, it is a simple exercise solved by grad students. So, I'd like to see you using your theory to answer both challenges.






- Lagrangian for photons is here:

Nope, this is not a lagrangian, it does not satisfy the Euler-Lagrange condition, please stop copying the same error. Besides, it proves nothing about :




(*** - that photons slows down when we increase its energy)

Try to treat my questions seriously, rather than posting invalid answers.




I am working over error bars for measuring GRB delay (Tensor's and your remark).
I will post it here, soon.

You are a little late, it is obvious that the values coming from the GRB "experiments" are (much) smaller than the error bars, i.e. they are NOISE.

pogono
2011-Aug-08, 10:30 PM
You are a little late, it is obvious that the values coming from the GRB "experiments" are (much) smaller than the error bars, i.e. they are NOISE.

If you are so sure that GRB measurements are noise, there is no way to convince you that may be anything interesting in my theory. I start from a fact, that GRB are delayed what was measured.

But please remember, that at the beginning we all was sure, that Earth stands on crocodiles. Ancient "science" gave a lot of proves for that fact.
Einstain said about QM: "Good is not playing dices" - now we believe he plays.

We think we are sure until some strange theory will not show, there may be another explanation.
I am not sure about my theory, it is immature idea. It may be wrong, but for me sounds interesting enough to check it.



No, it isn't, it is a simple exercise solved by grad students

Give it to the students as an funny exercise, then.
For me solving my Lagrangian is really hard and time-consuming.

macaw
2011-Aug-08, 11:48 PM
If you are so sure that GRB measurements are noise, there is no way to convince you that may be anything interesting in my theory. I start from a fact, that GRB are delayed what was measured.

But please remember, that at the beginning we all was sure, that Earth stands on crocodiles. Ancient "science" gave a lot of proves for that fact.
Einstain said about QM: "Good is not playing dices" - now we believe he plays.

We think we are sure until some strange theory will not show, there may be another explanation.
I am not sure about my theory, it is immature idea. It may be wrong, but for me sounds interesting enough to check it.


Give it to the students as an funny exercise, then.
For me solving my Lagrangian is really hard and time-consuming.


Would you please answer the questions? Thank you.

Tensor
2011-Aug-09, 12:04 AM
If you are so sure that GRB measurements are noise, there is no way to convince you that may be anything interesting in my theory. I start from a fact, that GRB are delayed what was measured.


But you ignore the fact that the calculations of your idea do not fall within the observed "delay". Your calculation show the delay should be ~.27 seconds. If there is a travel delay, It would be, by the methods of that paper, no less than .15 seconds and no more than .22 seconds. YOUR CALCULATION FALLS OUTSIDE OF THOSE CONSTRAINTS. Which means, your idea fails to match observation. You also ignore the two or three other explanations that can explain the observed difference the time from the start of the GRB and the arrival of the 30 GeV photon. And it's not as if the other explanations are hidden in obscure journals, they are in the paper you provided. You just chose to ignore those explanation.

On top of all that, you still have completely missed the fact that any actual delay would be a violation of Local Lorentz Invariance(LLI). If you have a violation of LLI, then the General Relativity equations need to be corrected . Which means the Schwartzchild solution needs correction. Which means you can't use the Schwartzchild solution, as it is, in your paper(or if you do, it's just something that you put in to make it work, an ad hoc solution). Which means, your entire paper, as it is, does not have the correct equations. So, either way, that paper you linked to, kills your idea.

pogono
2011-Aug-09, 09:29 AM
On top of all that, you still have completely missed the fact that any actual delay would be a violation of Local Lorentz Invariance(LLI). If you have a violation of LLI, then the General Relativity equations need to be corrected . Which means the Schwartzchild solution needs correction. Which means you can't use the Schwartzchild solution, as it is, in your paper(or if you do, it's just something that you put in to make it work, an ad hoc solution). Which means, your entire paper, as it is, does not have the correct equations. So, either way, that paper you linked to, kills your idea.


Not exactly.
I also say, that Schwarzschild Radius is not what we think it is. It is just an curvature of some rotation in Minkowski time-space and may be derived without General Relativity. Below you have explanation:

We take a body rotating on circle. From Angular Momentum Conservation Law we may write down its Angular Momentum as:

\vec{L}_{spin}=\vec{R}\times \vec{p_{rot}}=\vec{R}\times \vec{v_{rot}}\cdot m = const

Then through away the mass from the equation by defining auxiliary constant R_spin with [meter] dimension. For spinning body its value is:

R_{spin}=\frac{L_{spin}}{mc}=R\cdot \frac{v_{rot}}{c}

\frac{v_{rot}}{c}=\frac{R_{spin}}{R}=\frac{1}{R} \cdot \frac{L_{spin}}{mc}

Now, from special relativity we may write:

\left (\frac{v_{rot}}{c} \right )^2+ \left (\frac{1}{\gamma_{rot}} \right ) ^2 =1

thus, we may trasform to form of:

\gamma_{rot}^2=R \cdot \frac{R}{R^2-R_{spin}^2}

But let us see that part of RHS may be understood as Helix curvature in Minkowski time-space:

\kappa= \frac{1}{R_{cur}}=\frac{R}{R^2-R_{spin}^2}

Then:

\gamma_{rot}^2=\frac{R}{R_{cur}}

We may now rewrite special relativity condition as:

\left (\frac{v_{rot}}{c} \right )^2 = 1 - \frac{R_{cur}}{R}

\left v_{rot}=c \sqrt{1 - \frac{R_{cur}}{R}}

It is also explanation for macaw, why I use such kind of velocity related to time dilation for photons.
All we have to see is, that we may use rotating field vectors in place of body.

And from linking above with equations in Post #127 the \; R_{spin}=\frac{R_{\omega}}{\gamma} \; is intermediate measure of internal angular momentum of rotating field vectors of photon/particle, that we used to call SPIN.



For photons, the Curvature of this rotating field vectors helix equals:
\frac{1}{l_{P}}=\frac{R_{\omega}}{R_{\omega}^2-\left (\frac{R_{\omega}}{\gamma} \right )^2}

If we will change the way that vectors rotate, we change helix curvature and obtain particle:
\frac{1}{R_{Schw}}=\frac{l_{P}}{l_{P}^2-\left (\frac{l_{P}}{\gamma} \right )^2}

pogono
2011-Aug-09, 11:13 AM
We should be aware of one more thing. Above “spin concept” as explanation for Schwarzschild Radius works for elementary particles, only.

Why?
We get used to sum masses with simple summing.
Since we believe it is quantum of gravitational field, we have to remember, that we use simple summing only thanks to Maclaurin’s approximation.

\sum_{i}^{ } E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw_i}}{l_{P}}}}-1 \right) \approx E_{P} \sum_{i}^{ } \frac{R_{Schw_i}}{2l_{P}} = \frac{l_{P}\cdot c^4}{G}\cdot \sum_{i}^{ }\frac{R_{Schw_i}}{2\cdot l_{P}} =
= \sum_{i}^{ }m_i \cdot c^2

It means, that Schwarzschild Radius derived for Big Mass does not say anything about its spinning. It is just an approximation of its components' (summarized) Schwarzschild Radiuses

pogono
2011-Aug-09, 12:54 PM
We may now rewrite special relativity condition as:

\left (\frac{v_{rot}}{c} \right )^2 = 1 - \frac{R_{cur}}{R}

\left v_{rot}=c \sqrt{1 - \frac{R_{cur}}{R}}

It is also explanation for macaw, why I use such kind of velocity related to time dilation for photons.
All we have to see is, that we may use rotating field vectors in place of body.
(...)

I hope now you see that my field equations are not taken by hand.
Rotating line velocities of twisted field vectors has exactly mentioned relation.

\left v_{rot}=c \sqrt{1 - \frac{R_{cur}}{R}}=\frac{c}{\gamma }

I ask for support, I ask for some experienced professional to review, finish and help with publishing my strange theory.

macaw
2011-Aug-09, 01:57 PM
We should be aware of one more thing. Above “spin concept” as explanation for Schwarzschild Radius works for elementary particles, only.


\sum_{i}^{ } E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw_i}}{l_{P}}}}-1 \right) \approx E_{P} \sum_{i}^{ } \frac{R_{Schw_i}}{2l_{P}} = \frac{l_{P}\cdot c^4}{G}\cdot \sum_{i}^{ }\frac{R_{Schw_i}}{2\cdot l_{P}} =
= \sum_{i}^{ }m_i \cdot c^2


Actually it does not work at all, it is riddled with mistakes.

MQ9: Why is your summation using multiple Schwarzschild radiuses R_{Schw}_i?
MQ10: Why is the rest energy E_p the same for all the particles when the masses m_i are all [/b]different[/b]?
MQ11: adding up all the Scwarzschild radiuses is physically meaningless, what justifies this?





Why?
We get used to sum masses with simple summing.

MQ12: Experiment says that mass isn't additive, didn't you know that?



Since we believe it is quantum of gravitational field, we have to remember, that we use simple summing only thanks to Maclaurin’s approximation.

More like numerology refuted by experiment again.




It means, that Schwarzschild Radius derived for Big Mass does not say anything about its spinning. It is just an approximation of its components' (summarized) Schwarzschild Radiuses

MQ13: "Components" of what?

macaw
2011-Aug-09, 01:58 PM
I hope now you see that my field equations are not taken by hand.
Rotating line velocities of twisted field vectors has exactly mentioned relation.

\left v_{rot}=c \sqrt{1 - \frac{R_{cur}}{R}}=\frac{c}{\gamma }

MQ14: Where are the field equations in the expression above?



I ask for support, I ask for some experienced professional to review, finish and help with publishing my strange theory.

You got the professional review, several of us spent a lot of time pointing out the many mistakes in your theory. It now behooves on you to go back and try to correct them.

pogono
2011-Aug-09, 03:42 PM
Ok, I am sorry, macaw. I thought it was easy.



MQ10: Why is the rest energy E_p the same for all the particles when the masses m_i are all [/b]different[/b]?


I show, that energy as for leptons as for bozons are quants of field energy (for bosons doubled), what is explained in post #43

I use Planck Units treating them as fundamental units with meaning "one".
E_P is Planck Energy - I show it is the base for all quants.
Here you may read about Planck Units: http://en.wikipedia.org/wiki/Planck_units

In my theory (as I just described in post #138) difference between particles' energies comes from different values of angular momentum of rotating field vectors.

All of it drives to rest mass formula for leptons as quant:


E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}-1 \right) \approx E_{P}\frac{R_{Schw}}{2l_{P}} = \frac{l_{P}\cdot c^4}{G}\cdot \frac{R_{Schw}}{2\cdot l_{P}} = mc^2

As you may see, we obtain mc^2 thanks to Maclaurin's approximation.



MQ13: "Components" of what?

All condensed matter we know is made of rest mass of its components and/or field energy between them (f.e. hadrons are complex). It means, that it is build of fundamental quants of field energy.



MQ9: Why is your summation using multiple Schwarzschild radiuses R_{Schw}_i?

If matter is made of many hadrons, leptons (etc.) then we have to enumerate them summing its masses. I use index "i" to enumerate components of complex matter.
Schwartshild radius is used as a curvature - the result of calculating angular momentum for rotation (explained in post #138). As for bosons as for leptons Plank Energy is a base for these quants. Only angular momentums are different, that is way R_{Schw} is enumerated.



MQ11: adding up all the Schwarzschild radiuses is physically meaningless, what justifies this?

It is only to show, that for big mas we may use Schwarzschild Radius as the effect of Maclaurin's approximation.



MQ12: Experiment says that mass isn't additive, didn't you know that?

What experiments? You mean for hadrons masses?

macaw
2011-Aug-09, 03:46 PM
Ok, I am sorry, macaw. I thought it was easy.



I show, that energy as for leptons as for bozons are quants of field energy (for bosons doubled), what is explained in post #43.

If matter is made of many hadrons, leptons (etc.) then we have to enumerate them summing its masses. I use index "i" to enumerate components of complex matter.

Elementary particles don't have a Schawrazschild radius.

MQ15: When do you plan to stop making up nonsense formulas?


Only angular momentums are different, that is way R_{Schw} is enumerated.

Elementary particles do not have a classical angular momentum, as you seem to believe. The Scharzschild radius doesn't exist for elementary particles and it is not associated with angular momentum anyways.

MQ16: How much longer do you plan to add up mistakes to your "theory"?

pogono
2011-Aug-09, 04:00 PM
MQ14: Where are the field equations in the expression above?

Field equations you may find in post #81


(...)
Let define vector fields (i explain what are these fields few lines below)
\nabla\times \vec{A}=-\frac{\gamma _{G}}{c}\frac{\partial \vec{P}}{\partial t}
\nabla\times \vec{P}=\frac{\gamma _{G}}{c}\frac{\partial \vec{A}}{\partial t}
\nabla\times \vec{U}= \vec{P}
(...)
And we obtain Delambetian:
\frac{\gamma ^2_{G}}{c^2}\frac{\partial^2 \vec{A}}{\partial t^2}-\nabla^2 \vec{A}=0

As you may see they are made of rotating velocities folowing c/\gamma factor derived for rotation.



You got the professional review, several of us spent a lot of time pointing out the many mistakes in your theory. It now behooves on you to go back and try to correct them.
I suppose. I appreciate your time and mistakes you have pointed. I work over mistakes as you may see and I post results here.

But, excuse me, words: "supportive" and “friendly” does not math to your style. You are trying to shoot me, I try to survive. That are ATM forum rules.

I ask for supportive and friendly help of somebody who believe, that the idea may be true. I really need such help.

macaw
2011-Aug-09, 04:03 PM
I ask for supportive and friendly help of somebody who believe, that the idea may be true. I really need such help.

The most friendly advice that I have for you is to start studying, you lack the basics. Because of that , you make a lot of fundamental mistakes.

macaw
2011-Aug-09, 04:06 PM
What experiments? You mean for hadrons masses?

I mean basic stuff that contradicts your claims and that you should know before starting on writing a nonsense theory. See here, why, contrary to your recent claims, mass isn't additive (http://en.wikipedia.org/wiki/Mass_in_special_relativity#The_mass_of_composite_s ystems), for example.

pogono
2011-Aug-09, 04:06 PM
Elementary particles don't have a Schawrazschild radius.


macaw, please once again and very slowly read my posts: #138 , then: #143
I explain there what is it R_Schw for particles.
And this way I anwer to your accusation, that v_f=c \sqrt{1- R_s/R} was taken by hand.

macaw
2011-Aug-09, 04:08 PM
macaw, pleas once again and very slowly read my posts: #138 , then: #143
I explain there what is it R_Schw for particles.
And this way I anwer to your accusation, that it was taken by hand.

MQ17: Do you need a pointer to learning what Schwarzschild radius is and why there is no such thing as a Schwarzschild radius for elementary particles?

pogono
2011-Aug-09, 04:18 PM
MQ17: Do you need a pointer to learning what Schwarzschild radius is and why there is no such thing as a Schwarzschild radius for elementary particles?

macaw, do you read my posts, or you just slightly take a look at it?
I know what Sch. Radius is.
But you please, read my post #43
part of it here:



(...)
In my theory free space is full of electromagnetic field. We may analyze this filed as set of particles (photons) each with energy:
E=2E_{P}\cdot \left (\frac{1}{\sqrt{1-t_{P}\cdot\omega}}-1 \right ) \approx 2E_{P}\cdot \frac{t_{P}}{2 \omega} = \not{h}\cdot \omega

Photons (under specific conditions) may split in two, what we note as transition:
2E_{P} \left (\frac{1}{\sqrt{1-t_{P}\cdot\omega}}-1 \right ) \mapsto 2 \cdot E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}-1 \right)

Where:
E_{P}\left (\frac{1}{\sqrt{1-\frac{R_{Schw}}{l_{P}}}}-1 \right) \approx E_{P}\frac{R_{Schw}}{2l_{P}} = \frac{l_{P}\cdot c^4}{G}\cdot \frac{R_{Schw}}{2\cdot l_{P}} = mc^2

And now, let me explain what Schwarzschild Radius may be and why I use it even for elementary particles.

Vectors of field rotation make helix in Minkowski time-space. We may show that if velocity equals “c” on circle then:
Tc = 2\pi R_{\omega } \to \omega = \frac{2\pi}{T} = \frac{c}{R_{\omega}}

For photons, the Curvature of this rotating field vectors helix equals:
\frac{1}{l_{P}}=\frac{R_{\omega}}{R_{\omega}^2-\left (\frac{R_{\omega}}{\gamma} \right )^2}

If we will change the way that vectors rotate, we change helix curvature and obtain particle:
\frac{1}{R_{Schw}}=\frac{l_{P}}{l_{P}^2-\left (\frac{l_{P}}{\gamma} \right )^2}

It means, that Schwarzschild Radius is not what we suspect it was. It is just the measure of some curvature in space-time. We can use it even for a photons or elementary particles.

macaw
2011-Aug-09, 04:20 PM
Not exactly.
I also say, that Schwarzschild Radius is not what we think it is. It is just an curvature of some rotation in Minkowski time-space and may be derived without General Relativity. Below you have explanation:

We take a body spinning in circle. From Angular Momentum Preservation Principle we may write down its Angular Momentum as:

\vec{L}_{spin}=\vec{R}\times \vec{p_{rot}}=\vec{R}\times \vec{v_{rot}}\cdot m = const

Then through away the mass from the equation by defining auxiliary constant R_spin with [meter] dimension. For spinning body its value is:

R_{spin}=\frac{L_{spin}}{mc}=R\cdot \frac{v_{rot}}{c}


1. Particles don't have classical spin, so the above derivation doesn't apply.
2. MQ19: Why are you dividing by mc? The particle of mass m doesn't have the momentum mc. This looks like numerology again.
3. MQ20: R_{spin}, as defined by you (incorrectly, see point 2) has nothing to do with the Schwarzschild radius.



\frac{v_{rot}}{c}=\frac{R_{spin}}{R}

Only in numerology, not in physics.

pogono
2011-Aug-09, 04:26 PM
Elementary particles don't have a Schawrazschild radius.
MQ15: When do you plan to stop making up nonsense formulas?


When Schrodringer formulated his nonsense formulas for "nobody knows" wave function, some people read it. It was working.

My formulas works. Read it with brain concentrated, please. I have already explained many of your doubts.

And, please, open your mind for non-regular interpretations that you are not used to. It is ATM forum - do not expect from me regular mainstream explanations.

macaw
2011-Aug-09, 04:32 PM
When Schrodringer formulated his nonsense formulas for "nobody knows" wave function, some people read it. It was working.

My formulas works.

MQ21: how can they "work" when thy are falsified by existing experiments, as already shown repeatedly in this thread?

MQ22: How can they "work" when the derivations are riddled with mistakes?

pogono
2011-Aug-09, 04:33 PM
1. Particles don't have classical spin, so the above derivation doesn't apply.
2. MQ19: Why are you dividing by mc? The particle of mass m doesn't have the momentum mc. This looks like numerology again.
3. MQ20: R_{spin}, as defined by you (incorrectly, see point 2) has nothing to do with the Schwarzschild radius.
Only in numerology, not in physics


Follow these equations. If you see mistake - show the mistake. But do not call math numerology.
I may divide by mc or mc^2 to obtain needed values coming from preservation principles.
We have more strange auxiliary variables in mainstream physics.

a.d. 1 - It is example for rotating body, that is used for rotating field vectors at the end

macaw
2011-Aug-09, 04:39 PM
Follow these equations. If you see mistake - show the mistake.

I have already showed you the mistakes.



I may divide by mc or mc^2 to obtain needed values coming from preservation principles.

MQ23: What is "preservation principles"? Did you invent this term?
MQ24: Why divide by mc and not by 2 mc? Or by mv? Please explain

pogono
2011-Aug-09, 04:40 PM
2. MQ19: (...) The particle of mass m doesn't have the momentum mc. This looks like numerology again.


So what is in your opinion momentum of a body rotating on circle with mass "m" and velocity "v_rot"?

macaw
2011-Aug-09, 04:44 PM
So what is in your opinion momentum of a body rotating on circle with mass "m" and velocity "v_rot"?

....that it does not apply to elementary particles, as you insist in your mistakes. Google "spin" and/or "intrinsic angular momentum".

pogono
2011-Aug-09, 04:45 PM
MQ23: What is "preservation principles"? Did you invent this term?

It is "conservation law" :-), sory for other name
Here: http://en.wikipedia.org/wiki/Angular_momentum
My term comes from polish way of calling this law.

pogono
2011-Aug-09, 04:52 PM
....that it does not apply to elementary particles, as you insist in your mistakes. Google "spin" and/or "intrinsic angular momentum".

Do not mix quantum rule. I say about a body, not quantum functions.
For a body rotating on a circle we may consider Angular Momentum Conservation Law for its move. Yes or not?

pogono
2011-Aug-09, 04:59 PM
MQ24: Why divide by mc and not by 2 mc? Or by mv? Please explain

Angular Momentum Conservation Law it is just symetry, it is an attribute of space-time we live that comes from Noether Theorem.
http://en.wikipedia.org/wiki/Noether%27s_theorem#Example_2:_Conservation_of_cen ter_of_momentum

I may divide Momentum by mas if I want.
And I show that it is useful.

pogono
2011-Aug-09, 05:33 PM
Dear all readers.

Has anyone tried to check out equations from post #138

(...)It is also explanation for macaw, why I use such kind of velocity related to time dilation for photons.(...)
and may confirm if they are derived correctly or not?

macaw
2011-Aug-09, 10:31 PM
Do not mix quantum rule. I say about a body, not quantum functions.
For a body rotating on a circle we may consider Angular Momentum Conservation Law for its move. Yes or not?

NO. You are still trying to apply the classical concept of angular momentum to elementary particles. To make the errors even worse, you are mixing in the Schwarzschild radius and the radial coordinate R.

macaw
2011-Aug-09, 10:34 PM
Not exactly.
I also say, that Schwarzschild Radius is not what we think it is. It is just an curvature of some rotation in Minkowski time-space and may be derived without General Relativity. Below you have explanation:

We take a body rotating on circle. From Angular Momentum Conservation Law we may write down its Angular Momentum as:

\vec{L}_{spin}=\vec{R}\times \vec{p_{rot}}=\vec{R}\times \vec{v_{rot}}\cdot m = const

Then through away the mass from the equation by defining auxiliary constant R_spin with [meter] dimension. For spinning body its value is:

R_{spin}=\frac{L_{spin}}{mc}=R\cdot \frac{v_{rot}}{c}

\frac{v_{rot}}{c}=\frac{R_{spin}}{R}=\frac{1}{R} \cdot \frac{L_{spin}}{mc}

Total nonsense, as already (http://www.bautforum.com/showthread.php/118332-Dilation-as-field?p=1923153#post1923153) pointed out.

pogono
2011-Aug-09, 11:22 PM
macaw,
let us forget about particles and Sch. Radius. Let us do it other way.

1. Do you agree, that in clasical mechanic for body moving on circle in potential field, we can define radial force responsible for centryfugal barrier related to angular momentum (denoted as L_spin) equal to:

F_{rad}=\frac{L_{spin}^2}{m\cdot R^3}

2. Thus we may define acceleration equal to:

a=\frac{L_{spin}^2}{m^2\cdot R^3}

As you may see even in clasical mechanic we divide angular momentum with m if we need it. All I do is to name the result of such product.

3. Using my auxiliary constant we can write it as:

a=R_{spin}^2 \cdot \frac{c^2}{R^3}
where:
R_{spin}=\frac{L_{spin}}{mc}

Do you agree for above form now?

macaw
2011-Aug-10, 12:25 AM
macaw,
let us forget about particles and Sch. Radius. Let us do it other way.

1. Do you agree, that in clasical mechanic for body moving on circle in potential field, we can define radial force responsible for centryfugal barrier related to angular momentum (denoted as L_spin) equal to:

F_{rad}=\frac{L_{spin}^2}{m\cdot R^3}

2. Thus we may define acceleration equal to:

a=\frac{L_{spin}^2}{m^2\cdot R^3}

Nope, you are using F=ma. this is not valid at high (relativistic) speeds. Didn't you know that?

EigenState
2011-Aug-10, 01:01 AM
Greetings,


1. Do you agree, that in clasical mechanic for body moving on circle in potential field, we can define radial force responsible for centryfugal barrier related to angular momentum (denoted as L_spin) equal to ...

pogono, you need to be more careful with nomenclature. You have opted to discuss an example from classical mechanics and yet chosen to refer to it as spin which is a purely quantum mechanical phenomenon. The motion might be orbital or rotational, but it most certainly is not spin.

Best regards,
EigenState

Tensor
2011-Aug-10, 03:21 AM
Not exactly.
I also say, that Schwarzschild Radius is not what we think it is. It is just an curvature of some rotation in Minkowski time-space and may be derived without General Relativity.

Curvature of what? Minkowski spacetime? If you have curvature of spacetime, you no longer have Minkowski spacetime. If it is curvature of something else, what exactly has curvature? How exactly do you calculate the curvature if you are not using General Relativity (GR)? Also, if you not using GR, how do you define Minkowski spacetime, much less curvature in it, if Minkowski is a specific exact solution in GR? Since you say it can be derived, I look forward to you providing the derivations of Minkowski spacetime, without using GR.


Below you have explanation:

snip...

Now, from special relativity we may write:

\left (\frac{v_{rot}}{c} \right )^2+ \left (\frac{1}{\gamma_{rot}} \right ) ^2 =1

NO, YOU MAY NOT. Did you do any study of symmetry before diving into this idea? Remember, you are claiming photons moving faster than c are a result of your idea. This means Local Lorentz Invariance (LLI) is not a part of your idea. Hence Relativity equations (either Special or General), as they are now, cannot be part of your idea, because those equations show LLI.

You will have to derive the new equations, that produce the changed relativity equations, then use those in your idea. Basically, you have to come up with an entirely new theory of Relativity, that allows a LLI violation. Until you provide those derivations, anything after, and including this step, in this post, is wrong. And I'm sure there are many other places in your original paper that have to be rewritten, due to the inclusion of Special and General relativity equations or solutions.


thus, we may trasform to form of:

But let us see that part of RHS may be understood as Helix curvature in Minkowski time-space:

I look forward to you providing us with your derivation of Minkowski spacetime, without using General Relativity. Along with your derivations of the Special Relativity equations, in the case of a LLI violation. By the way, what, exactly, are your postulates for the starting point?

Or, are you using one of the other possibilities being considered for the case of a LLI violation? If so, which possibility? And if you are, why are you not using the equations from that possibility?

pogono
2011-Aug-10, 07:25 AM
Curvature of what?

Curvature of Helix, as I wrote farther.
If you twist two vectors rotating around each other, you get double Helix.
BOSONS in my theory are such double Helixes. That is why they split into 2 parts (not 3, 7 or 11) - always into 2 separated helixes with oposite atributes: LEPTONS.
Curvature of that Helix is counted using Minkowski time-space metric.



NO, YOU MAY NOT.
(...)
You will have to derive the new equations, that produce the changed relativity equations, then use those in your idea.
(...)

Lorentz transformation does not need that light has speed of c, indeed.
It only need to EXISTS some constant (we call it "c") as a part of transformation for space-time. We have called it "c" and assigned to ligth as something obvious.

As you may see from my formula for photon speed, if we decrease photon's energy - he speeds up, and the limit of its velocity is "c". So in my theory "c" is the limit of speed of ligth.
Relativity works in my theory. But photon is nothing special - just a boson, but fast.

BTW - as you may see from my equations, every field distributed with bosons may be distributet at least with "c" speed.
And it is not invented - it comes as the result of equations.

pogono
2011-Aug-10, 08:13 AM
Greetings,
pogono, you need to be more careful with nomenclature. You have opted to discuss an example from classical mechanics and yet chosen to refer to it as spin which is a purely quantum mechanical phenomenon. The motion might be orbital or rotational, but it most certainly is not spin.
Best regards,
EigenState
Ok, you are right. I'll rename the constant.


Nope, you are using F=ma. this is not valid at high (relativistic) speeds. Didn't you know that?
Eeeechchch...
macaw, you might also write "Yes, but in non-relativistic physics", I was mentioning classical, non-relativistic case.

Two physicists was in journey to CERN, Geneva for the first time.
Suddenly experimental physicist sais: “look, they are all black cows in Switzerland”
Theoretical physicist sais: “no, we know for sure, there is one black cow in Switzerland”
I am sure you would say: “no, we know for sure, there is one black cow in Switzerland, black on one side”
;-)

But ok, you are right.
I will write it other way, renaming constants following EigenState advice.

1. Do you agree, that in clasical, non-relativistic case for orbiting bodies we define centryfugal barrier related to angular momentum (denoted now as L_ang)?

It is known as Perycentrum R_{perycentrum}

2. Could you agree that in clasical, non-relativistic case we may define another, ultimate barrier denoted now as R_min:

R_{min}=\frac{L_{ang}}{mc}

Above auxiliary constant R_min as we may see from below has a very physical meaning:
\vec{R}_{min}=\frac{\vec{L}_{ang}}{mc}=\vec{R}\tim es \vec{\frac{v}{c}} = const

Then R_min is the distance, where our orbiting body's rotation velocity v_rot would achieve "c" velocity. We know that v_rot is ortogonal to R, then:

v_{rot}=c \;\; \to \;\; R_{min}=R

Do you agree for such auxiliary constant?
P.S. F.E. we may use such auxiliary constant insted of Angular Momentum to shoortern calculations.

pogono
2011-Aug-10, 10:01 AM
Let us go farther,

do you agree, that Relativistic Angular Momentum Conservation Rule force center of mass to move exactly the same velocity that orbiting body does?

If \vec{R} is a distance to center of mass and orbiting body has relativistic momentum p, then:
\frac{d\vec{R}}{dt}=\frac{\vec{p}}{m}

Do you agree?

tusenfem
2011-Aug-10, 11:31 AM
Eeeechchch...
macaw, you might also write "Yes, but in non-relativistic physics", I was mentioning classical, non-relativistic case.

Two physicists was in journey to CERN, Geneva for the first time.
Suddenly experimental physicist sais: “look, they are all black cows in Switzerland”
Theoretical physicist sais: “no, we know for sure, there is one black cow in Switzerland”
I am sure you would say: “no, we know for sure, there is one black cow in Switzerland, black on one side”
;-)



And that is enough with the jokes. Better spend time actually being careful and specific in the things that you write down.

tusenfem
2011-Aug-10, 11:57 AM
1. Do you agree, that in clasical, non-relativistic case for orbiting bodies we define centryfugal barrier related to angular momentum (denoted now as L_ang)?

It is known as Perycentrum R_{perycentrum}


No, that is not he pericentre, the pericentre is defined as: the point in the orbit of a celestial body closest to the central body around which the first body revolves, which is comparable at Earth with the term perigee. Apparently you want to define a radius R for which the centrifugal force is exceeding the gravitational/attractional force for a body of mass m and velocity v. Which would just be the critical radius (a function of both m and v) Rcrit(m,v).



2. Could you agree that in clasical, non-relativistic case we may define another, ultimate barrier denoted now as R_min:

R_{min}=\frac{L_{ang}}{mc}


No. one cannot agree there, because you put in c to define this critical radius, and THUS you will be dealing with relativistic effects, and you will have to take that into account in your equation, whether you like it or not, because the increase of m with high velocities is a well know experimental fact.



Above auxiliary constant R_min as we may see from below has a very physical meaning:
\vec{R}_{min}=\frac{\vec{L}_{ang}}{mc}=\vec{R}\tim es \vec{\frac{v}{c}} = const


Yes, but the relativistic effects in your definitions ......




Then R_min is the distance, where our orbiting body's rotation velocity v_rot would achieve "c" velocity. We know that v_rot is ortogonal to R, then:

v_{rot}=c \;\; \to \;\; R_{min}=R


NO! how can the body reach 'c' when you are dealing with classical mechanics??????????????????????????????????

This is rubbish, nonsense and just mathemagics, saying you want to do classical non-relativistic and then thinking you can get out some critical radius where the velocity would be "c". Naturally, with your loose definitions you may claim now that "c" is not lightspeed, and further blah blah to try to get out of the problems you are creating yourself.




Do you agree for such auxiliary constant?
P.S. F.E. we may use such auxiliary constant insted of Angular Momentum to shoortern calculations.

I guess the overal answer is NO.

tusenfem
2011-Aug-10, 11:58 AM
Let us go farther,

do you agree, that Relativistic Angular Momentum Conservation Rule force center of mass to move exactly the same velocity that orbiting body does?

If \vec{R} is a distance to center of mass and orbiting body has relativistic momentum p, then:
\frac{d\vec{R}}{dt}=\frac{\vec{p}}{m}

Do you agree?

How can we agree when the questions in not understandable?
And why are we suddenly doing things relativistically again?
And are you going to tie this to the non-relativistic result that you posted in the previous message?

pogono
2011-Aug-10, 01:40 PM
tusenfem,
I do not say that we will reach this Radius R_min. It is a definition of auxiliary constant.

But ok,
let us then take relativistic angular momentum and claim that remains constant.

Let us rotate some body on a string and make string schorten. Now our body has only v_rot. Let us see, that R_min is important radius.

\vec{L}_{ang}=\vec{R}\times m\gamma\vec{v} = const

we define:
R_{min}=\frac{L_{ang}}{mc}=R \cdot \frac{v_{rot}}{c}\gamma = const

\frac{v_{rot}}{c}\cdot \frac{1}{\sqrt{1-\frac{v_{rot}^2}{c^2}}}=\frac{R_{min}}{R}

v_{rot}=c\cdot \frac{\frac{R_{min}}{R}}{\sqrt{1+\frac{R_{min}}{R} }}

As you may see below, our defined Radius R_min is important constant for calculations.
R=R_{min}\;\; \to \;\;\frac{v_{rot}}{c}=\frac{1}{\gamma}


Can you agree for definition of such Radius as some important radius for rotating system?

macaw
2011-Aug-10, 02:13 PM
As you may see from my formula for photon speed, if we decrease photon's energy - he speeds up,

MQ25: Experiment falsifies the above statement, I have shown you, so why do you insist in this falsity?



.
Relativity works in my theory. But photon is nothing special - just a boson, but fast.

MQ26: No, it doesn't since you would obtain different values for :

-light bending by the Sun
-Shapiro delay

I have asked you to calculate the above repeatedly, when are you going to do it?

macaw
2011-Aug-10, 02:15 PM
Ok, you are right. I'll rename the constant.


Eeeechchch...
macaw, you might also write "Yes, but in non-relativistic physics", I was mentioning classical, non-relativistic case.

MQ27: Then all your calculations for \gamma go out the window because they only hold for v<<c, right?



v_{rot}=c \;\; \to \;\; R_{min}=R

Do you agree for such auxiliary constant?
P.S. F.E. we may use such auxiliary constant insted of Angular Momentum to shoortern calculations.

Err, no.

MQ28: You are just piling up the errors since you started from F=ma so you cannot turn around and say v_{rot}=c . You know that F=ma holds only for v<<c, right?

macaw
2011-Aug-10, 02:31 PM
tusenfem,
I do not say that we will reach this Radius R_min. It is a definition of auxiliary constant.

But ok,
let us then take relativistic angular momentum and claim that remains constant.

Let us rotate some body on a string and make string schorten. Now our body has only v_rot. Let us see, that R_min is important radius.

\vec{L}_{ang}=\vec{R}\times m\gamma\vec{v} = const

we define:
R_{min}=\frac{L_{ang}}{mc}=R \cdot \frac{v_{rot}}{c}\gamma = const

MQ29: How can be R \cdot \frac{v_{rot}}{c}\gamma "const(ant)" when it depends on v? When will you stop with the metamagics?

pogono
2011-Aug-10, 02:43 PM
MQ29: How can be R \cdot \frac{v_{rot}}{c}\gamma "const(ant)" when it depends on v? When will you stop with the metamagics?

Wait a sec..,
this value also depends on R. And it seems to me, that this is exactly the idea behind Angular Momentum Conservation Law. When we make R shorten, velocity increase.
Isn't it?

macaw
2011-Aug-10, 02:48 PM
Wait a sec..,
this value also depends on R. And it seems to me, that this is exactly the idea behind Angular Momentum Conservation Law. When we make R shorten, velocity increase.
Isn't it?

MQ30: Even worse, how can you claim it is constant? Your mistake starts at the very beginning when you claim that L=const. The correct claim is that L is conserved in the absence of external forces, i.e. \frac{dL}{dt}=0. L is a FUNCTION of BOTH R and v BY DEFINITION. You are piling up the errors.

macaw
2011-Aug-10, 02:56 PM
v_{rot}=c\cdot \frac{\frac{R_{min}}{R}}{\sqrt{1+\frac{R_{min}}{R} }}
As you may see below, our defined Radius R_min is important constant for calculations.
R=R_{min}\;\; \to \;\;\frac{v_{rot}}{c}=\frac{1}{\gamma}



Simple math says that the above is FALSE, care to redo your "math"?

pogono
2011-Aug-10, 03:08 PM
Simple math says that the above is FALSE, care to redo your "math"?

I did math again and deleted part about golden ratio.
But rest of it is ok.
L_ang = constant and has its value named Angular Momentum.
We use its value to calculate orbits.

Strange
2011-Aug-10, 03:13 PM
L_ang = constant and has its value named Angular Momentum.
We use its value to calculate orbits.

So the angular momentum of all the planets, an ice skater, my bicycle wheel, are all the same?

pogono
2011-Aug-10, 03:14 PM
MQ30: Even worse, how can you claim it is constant? Your mistake starts at the very beginning when you claim that L=const. The correct claim is that L is conserved in the absence of external forces, i.e. \frac{dL}{dt}=0. L is a FUNCTION of BOTH R and v BY DEFINITION. You are piling up the errors.

Angular Momentum is function of both V and R and I wrote down its value here.
At first I wrote it in Newtonian physics - you did not agree.
Then I wrote it for relativistic physics - you do not agree.

Do not you agree with angular momentum definition?
Do not you agree, that it has its own value and remain constant?
Or anything I say you do not agree by definition?

pogono
2011-Aug-10, 03:20 PM
So the angular momentum of all the planets, an ice skater, my bicycle wheel, are all the same?
Please... You know that, why do you ask for such things.
L_ang is constant in particular frame.

Please, do not treat me as a freak.
I will have to catch your words, too.

macaw
2011-Aug-10, 03:21 PM
Angular Momentum is function of both V and R and I wrote down its value here.
At first I wrote it in Newtonian physics - you did not agree.

Because you are mixing Newtonian formulas with relativistic formulas.



Then I wrote it for relativistic physics - you do not agree.

Because you are mixing relativistic formulas with Newtonian formulas.



Or anything I say you do not agree by definition?

I disagree with your numerology. I disagree with you never stopping writing nonsense and never admitting to errors. I disagree with you not answering the questions that clearly point out your mistakes.

macaw
2011-Aug-10, 03:23 PM
v_{rot}=c\cdot \frac{\frac{R_{min}}{R}}{\sqrt{1+\frac{R_{min}}{R} }}

As you may see below, our defined Radius R_min is important constant for calculations.
R=R_{min}\;\; \to \;\;\frac{v_{rot}}{c}=\frac{1}{\gamma}


Can you agree for definition of such Radius as some important radius for rotating system?

Still just as FALSE as before you edited your post.

pogono
2011-Aug-10, 03:35 PM
I disagree with your numerology. I disagree with you never stopping writing nonsense and never admitting to errors. I disagree with you not answering the questions that clearly point out your mistakes.

You do not give me time for calculating important answers shooting to everything I post by definition, without carefuly reading my posts.

I have to put a lot of work to show, that not everything I write is wrong.
You are wrong in post #186, like in many others.
Examples: posts about cosmic velocities and field derivation that drives to angular velocity, posts about gravitational acceleration coming out from my equations, and so on.
Now, you claim, that angular momentum has no value...

It takes time. I you want to shoot me anyhow, let's leave it.

I treated you seriously.
I did here a lot of math, that is more ambitious then high school.
If you only want to be a leader in number of posts, do not write here. It has no sense.

I care about my idea.
And if anyone here show me mistakes - I make list of task to do and point after point try to resolve it. Be sure.

pogono
2011-Aug-10, 03:38 PM
Still just as FALSE as before you edited your post.

R_{min}=R\cdot \frac{v}{c}\gamma

R_{min}=R \; \; \rightarrow \; \; \frac{v}{c}=\frac{1}{\gamma}

Right?

macaw
2011-Aug-10, 03:46 PM
v_{rot}=c\cdot \frac{\frac{R_{min}}{R}}{\sqrt{1+\frac{R_{min}}{R} }}

As you may see below, our defined Radius R_min is important constant for calculations.
R=R_{min}\;\; \to \;\;\frac{v_{rot}}{c}=\frac{1}{\gamma}


Can you agree for definition of such Radius as some important radius for rotating system?

You really want your basic errors pointed out to you one by one, right?

When you make R=R_{min} you get \frac{v_{rot}}{c}=\frac{1}{\sqrt{2}}

As an aside, your whole "derivation" is nonsense from beginning to end, there is no valid physical content to your formulas.

Strange
2011-Aug-10, 03:54 PM
Please... You know that, why do you ask for such things.

But one of the problems is that you use terms (in descriptions as well as the math) in a very non-standard way. Constant doesn't mean what you seem to think it does.


L_ang is constant in particular frame.

When you say "L_ang is constant", do you really mean that angular momentum is a constant? Like pi or G or ε0?

And, from this response, it is not clear what you mean by the word "frame".

pogono
2011-Aug-10, 04:07 PM
You really want your basic errors pointed out to you one by one, right?

When you make R=R_{min} you get \frac{v_{rot}}{c}=\frac{1}{\sqrt{2}}

As an aside, your whole "derivation" is nonsense from beginning to end, there is no valid physical content to your formulas.

Hmmm,
great example of what I wrote post ago.

If you do your simple math, you will se, that the only way to:
\frac{v}{c}=\frac{1}{\gamma }

is to be:
\frac{v}{c }=\frac{1}{\gamma }=\frac{1}{\sqrt{2}}

because:
\left (\frac{v}{c} \right )^2+\left (\frac{1}{\gamma } \right )^2=1

I have my mind concentrated.
How about you?

During 4 posts you was making "aura", that I do elementary mistakes.
What now?

macaw
2011-Aug-10, 04:13 PM
\frac{v}{c }=\frac{1}{\gamma }=\frac{1}{\sqrt{2}}



This can't be since it implies:

\gamma=\sqrt{2}

i.e.

v=\frac{c}{\sqrt{2}}

In other words it is NOT valid in general, it is ONLY true for a specific value of v. Physics is not the same thing as numerology.


\left (\frac{v}{c} \right )^2+\left (\frac{1}{\gamma } \right )^2=1

The above is valid for ANY v. Do you now understand the difference between this formula and the error you posted above?


During 4 posts you was making "aura", that I do elementary mistakes.
What now?

You continue to do elementary mistakes. We continue to point them out to you.

pogono
2011-Aug-10, 04:35 PM
This can't be since it implies:
\gamma=\sqrt{2}
i.e.
v=\frac{c}{\sqrt{2}}
In other words it is NOT valid in general, it is ONLY true for a specific value of v. Physics is not the same thing as numerology.
The above is valid for ANY v. Do you now understand the difference between this formula and the error you posted above?
You continue to do elementary mistakes. We continue to point them out to you.

macaw,
in post #174 I wrote, that we may use value R_min instead of Angular Momentum value. This one special radius is the place, where above equation is true. Not in general. It means, that I am right.

But you probably do not even remember what we were talking about it, or you do not want to remember. It does not matter.

In both cases defending my theory from you has no sense.
You will never confess that you was wrong.

If I put here more complicated calculations, will be even worse. You will probably start from shooting me without testing anything and then forget what we were talking about.

Let's leave it.
Thank you for all the real mistakes you have pointed.
I have long list of tasks, so now I go out to work on it.
Howgh.

macaw
2011-Aug-10, 04:41 PM
macaw,
in post #174 I wrote, that we may use value R_min instead of Angular Momentum value. This one special radius is the place, where above equation is true.

You keep repeating the same mistake throughout your paper. R and v are INDEPENDENT variables, you keep tying them together in your metamagics.





Not in general.

Correction: Not EVER.





It means, that I am right.

Of course.


Thank you for all the real mistakes you have pointed.
I have long list of tasks, so now I go out to work on it.

Not so fast, you still have a lot of questions to answer. Please do so.

tusenfem
2011-Aug-10, 05:24 PM
tusenfem,
I do not say that we will reach this Radius R_min. It is a definition of auxiliary constant.

But ok,
let us then take relativistic angular momentum and claim that remains constant.

Let us rotate some body on a string and make string schorten. Now our body has only v_rot. Let us see, that R_min is important radius.

\vec{L}_{ang}=\vec{R}\times m\gamma\vec{v} = const

we define:
R_{min}=\frac{L_{ang}}{mc}=R \cdot \frac{v_{rot}}{c}\gamma = const


But why divide by mc, why not by γmc?
And as already said, you don't seem to know what "conservation of angular momentum" means.