Master Theory (edition 2)

That's just not the way it works.

\( \begin{eqnarray} \oint _S \vec{D} \cdot d\vec{s} & = & \varepsilon_0 \iint _{\partial V} \vec{E} \cdot d\vec{s} \\ & = & \varepsilon_0 \iint _{\partial V} \left( - \vec{\nabla} \varphi - \frac{ \partial \vec{A} }{ \partial t } \right) \cdot d\vec{s} \\ & = & \varepsilon_0 \iiint _V \vec{\nabla} \cdot \left( - \vec{\nabla} \varphi - \frac{ \partial \vec{A} }{ \partial t } \right) dv \\ & = & \varepsilon_0 \iiint _V \left( - \nabla^2 \varphi - \frac{ \partial }{ \partial t } ( \vec{\nabla} \cdot \vec{A} ) \right) dv \\ & = & \varepsilon_0 \iiint _V \left( \frac{1}{c^2} \frac{ \partial^2 }{ \partial t^2 } \varphi - \nabla^2 \varphi \right) dv \\ & = & \varepsilon_0 \iiint _V \frac{\rho}{\varepsilon_0 } dv \\ & = & \iiint _V \rho dv = Q \end{eqnarray}\)

The differential laws work in integral form for the same reason that the continuity equation for current is related to the time integral of charge.

\( \frac{\partial \rho}{\partial t} = - \vec{\nabla} \cdot \vec{j} \)

Nothing changes for the volume integration. What changes is when you have two comoving coordinate systems, the limits of integration also change. But we didn't begin to address this and how Maxwell's equations hold up in this form, because you didn't argue for your claim. You merely asserted it.

http://ru.wikipedia.org/wiki/Лоренц-ковариантность

 

Log in or Sign up to hide all adverts.

This is true for quasi-static fields only.

Gauss's and Stokes's theorems are correctly applies to quasi-static fields only.

These theorems do not take into account the time delay from the V-volume to the S-surface.

The result of S-integration will depend on the shape of the surface in the general case.

 

Log in or Sign up to hide all adverts.

Where is the fact-based argument to support this? You have asserted it, but not given a reason to believe it.

Where is the fact-based argument to support this? You have asserted it, but not given a reason to believe it.

Where is the fact-based argument to support this? You have asserted it, but not given a reason to believe it.

How can there be a "time delay" from the V-volume to the S-surface when the S-surface is the limits of the V-volume, \(S = \partial V\), and both are arbitrary and imaginary volumes and surfaces conjured by the mind of the physicist, not the laws of nature?

Absolutely I agree if we were talking about the general case, but we aren't talking about the general case; we are talking about the surface integral of a flux and the volume integral of a divergence of that flux.

Where is the fact-based argument to support your viewpoint? You have asserted it, but not given a reason to believe it.

 

Log in or Sign up to hide all adverts.

It is necessary to recall the proofs of (Gauss's and Stokes's) theorems.

They use a static field or fields, a speed of which is infinite.

Spend a thought experiment:

1. Electric current in a conductor changed.

2. Do L-contour need time to learn about this event, if L-contour located at some distance from the conductor?

 

For sample

If:
Real coordinates:

http: //masterov.qptova.ru/MasterTheory/Formuls/sanmple1.png

Then:
Visual coordinates:

http: //masterov.qptova.ru/MasterTheory/Formuls/sanmple2.png

Next:
Visual speed (with Dopler's effect):

http: //masterov.qptova.ru/MasterTheory/Formuls/sanmple3.png
http: //masterov.qptova.ru/MasterTheory/Formuls/sanmple4.png

 

Common case

Non inertial reference frame

Real coordinates:
http: //masterov.qptova.ru/MasterTheory/Formuls/comn1.png

Real speed:

http: //masterov.qptova.ru/MasterTheory/Formuls/comn2.png
http: //masterov.qptova.ru/MasterTheory/Formuls/comn3.png

Visual speed:

http: //masterov.qptova.ru/MasterTheory/Formuls/comn4.png

Visual coordinates:
http: //masterov.qptova.ru/MasterTheory/Formuls/comn5.png
Visual time:
http: //masterov.qptova.ru/MasterTheory/Formuls/comn6.png

Inertial reference frame

Real coordinates:
http: //masterov.qptova.ru/MasterTheory/Formuls/comn7.png

Visual coordinates:
http: //masterov.qptova.ru/MasterTheory/Formuls/comn8.png
http: //masterov.qptova.ru/MasterTheory/Formuls/comn9.png - Dopler's effect.
http: //masterov.qptova.ru/MasterTheory/Formuls/comna.png
http: //masterov.qptova.ru/MasterTheory/Formuls/comnb.png

 

masterov

Your are not posting the URLs as links and the server continues to come up as not found. The last two post are useless if their content is unavailable.

 

Remove(delite) a spacer after "http:"

http

Please Register or Log in to view the hidden image!

//masterov.qptova.ru/MasterTheory/Formuls/comna.png

 

Ah! An obvious over site on my part.

 

There's a reason that you're not allowed to post links and images before you've accrued a certain number of posts on the site. I'll leave you to speculate on the particulars...

 

Yea, I knew that I had just missed the dead space in his URL until he pointed out the ? In my cut and paste. It would be very difficult to follow the math from a series of images like that. Too much work for me. I have to work to keep up with the stuff these days when it is properly formatted .... and fully explained.

 

That's an issue if you use only part of Maxwell's equations. With all of Maxwell's equations, a time-varying charge requires the existance of currents which are the source of magnetic fields. And the net effect of electromagnetism on physical measurements is the sum of the effects of electric and magnetic fields.

Another way to see this is that electromagnetism has never predicted that signals travel with inifinite velocity, so your claim that E and B respond instantaneously to the charge and current distribution ignores that the charge and current distribution are physically constrained to vary in limited manner.

A third way to see this is to use the 4-dimensional version of the divergence theorem, which gives us:
\(\iiint _{\partial \Omega} \partial_k A^m d\omega^k = \iiint \int_{\Omega} \partial^k \partial_k A^m dV dt = \iiint \int_{\Omega} \partial^k \partial_k A^m dV dt = \iiint \int_{\Omega} \partial^2 A^m dV dt = \mu_0 \iiint \int _{\Omega} J^m dV dt \)

But a college level education in electromagnetism is usually only gained by people going on to actually get a degree in physics or an advanced degree in electrical engineering. So your ignorance is understandable. Your motivations to claim expertise in this field is not understood.

Electric current flows in circuits. So lets make this concrete.

Imagine an eternal loop of charge of radius R, which is large, and linear charge density of \(\ell = \frac{Q}{2 \pi R}\).
Before time t=0, the loop is motionless. After time t=0, the loop rigidly rotates with linear velocity v, which gives a current of \(I = v \ell = \frac{Q v}{2 \pi R}\). Or using the step function, \(I(t) = \frac{Q v}{2 \pi R} \theta(t)\)

Thus Ampère's circuital law suggests that for a loop of constant radius r just outside the charges wire (r << R), that the magnetic field is \(B = \frac{\mu_0}{2 \pi r} I(t) = \frac{\mu_0 Q v}{4 \pi^2 R r} \theta(t)\) But this is not Maxwell's equation, but an equation of magnetostatics misapplied to a case where the current changes.

Maxwell realized that the Ampère's circuital law was wrong, and added a term he called the displacement current to it. The displacement current is not really a current, but it's the standard name for this term.

So since we have a lot of heavy lifting to do, lets review:
\(\vec{E}(\vec{r},t) = -\vec{\nabla}\varphi(\vec{r},t) - \frac{\partial}{\partial t}\vec{A}(\vec{r},t) \, ; \; \vec{B}(\vec{r},t) = \vec{\nabla} \times \vec{A}(\vec{r},t) \, ; \; \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \varphi(\vec{r},t) - \nabla^2 \varphi(\vec{r},t) \varphi(\vec{r},t) = \frac{\rho(\vec{r},t)}{\varepsilon_0} \, ; \; \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \vec{A}(\vec{r},t) - \nabla^2 \varphi(\vec{r},t) \vec{A}(\vec{r},t) = \mu_0 \vec{j}(\vec{r},t)\)

Our plan to solve \(\left( \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2 \varphi(\vec{r},t) \right) Y(\vec{r},t) = X(\vec{r},t)\) is to express X in terms of sums of point functions and then have Y in terms of Green's functions. So we need to solve: \(\left( \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2 \varphi(\vec{r},t) \right) G(\vec{r},t,\vec{r}',t') = \delta(\vec{r}-\vec{r}') \delta(t - t')\). Please verify that \(G(\vec{r},t,\vec{r}',t') = \frac{\delta \left( t - t' - \frac{\left| \vec{r}-\vec{r}' \right| }{c} \right) }{4 \pi \left| \vec{r}-\vec{r}' \right|}\) is a solution.

So for the neighborhood near the wire at r'=0, we have a source term which is \(\mu_0 \vec{j}(\vec{r},t) = \frac{\mu_0 Q v}{2 \pi R} \theta(t) \delta(r_x)\delta(r_y)\) (in the z-direction) which gives rise to vector potential \(\vec{A}(\vec{r},t) = \frac{\mu_0 Q v}{4 \pi^2 R} \frac{\theta \left(t - \left| \vec{r} \right| /c \right)}{\left| \vec{r} \right|}\) (in the direction of the current) which gives us \(\vec{B}(\vec{r},t) = \vec{\nabla} \times \vec{A}(\vec{r},t) = \frac{\mu_0 Q v}{4 \pi^2 R \left| r \right|} \theta \left(ct - \left| \vec{r} \right| \right)\) (in the direction expected ).

So the expectation is that the B field "turns on" at a later time depending on how far away you are from the current.

For further reading, I direct you to any college textbook for a year-long course in electromagnetism or http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html (especially the section on the Time-Dependent Maxwell's equations).

 

Comment

Assume that the relativistic mass-point gives a flash of light at regular intervals (in its reference frame). Then, while visual and visual coordinates can be measured by measuring (by optical instruments) coordinates the flash and time intervals between outbreaks.

Real coordinates are calculated by double integration of acceleration, which may be measuring (in mass-point reference frame).

 

Rpenner, imagine a conductor with alternating current, which passes through the two contours of different radii. And (if speed of propagation of EMF is finite) whether left integral of expression:

Please Register or Log in to view the hidden image!

is the same for both circuits in a t the same time?

 

What I (and everyone who studies electromagnetism since Maxwell) am saying is that if the current, I, which is at the center of the circular surface changes, there is a pulse of electric field flux that originates with the change in time and space and propagates out with the speed of light. So after of time T from the time of the change, the magnetic field B reflects the old value of I if cT < r and the new value of B if cT > r.

This comes from the part of E which comes from the \(- \frac{\partial \vec{A}}{\partial t}\) term.

This is not obvious when a person with limited math education just looks at Maxwell's four equations. But physics education is about actually solving equations, and a changing current results in fields that vary over space and over time. So physics education also strengthens one's math intuition for some types of problems. If you study the link I provided, it gives a guide on how Maxwell's four equations can be turned into solutions for the electric and magnetic fields now in terms of the charges and currents of the past and their time derivatives.

equations 541 and 542 of

http://farside.ph.utexas.edu/teaching/em/lectures/node52.html

\(\vec{E}(\vec{r},t) = \frac{1}{4\pi \varepsilon_0} \iiint \left( \frac{\vec{r} - \vec{r}'}{\left| \vec{r} - \vec{r}' \right| ^3} \rho \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) + \frac{\vec{r} - \vec{r}'}{c \left| \vec{r} - \vec{r}' \right| ^2} \dot{\rho} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) - \frac{1}{c^2 \left| \vec{r} - \vec{r}' \right|} \dot{\vec{j}} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) \right) d V'\)
\(\vec{B}(\vec{r},t) = - \frac{\mu_0}{4\pi} \iiint \left( \frac{\vec{r} - \vec{r}'}{\left| \vec{r} - \vec{r}' \right| ^3} \times \vec{j} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) + \frac{\vec{r} - \vec{r}'}{c \left| \vec{r} - \vec{r}' \right| ^2} \times \dot{\vec{j}} \left( \vec{r}',t - \frac{1}{c} \left| \vec{r} - \vec{r}' \right| \right) \right) d V'\)

I'm recovering from surgery to remove a tumor, so please excuse that I probably made some math mistakes in the prior post. But you seem particularly ill-equipped to criticize either special relativity or Maxwell's electromagnetism.

 

I did not understand your text, but your formula look like to true. And it (your formula) is not look like to the integral Maxwell equations. (This is what I mentioned above.)

I repeat: my specialty - nonlinear dynamics (attractors, bifurcations, auto-wave, etc.)

Master Theory was born out of a different theory. This (other) theory describe a model of a matter-generator. This generator is a Hamiltonian generator of solitons.

My attempts to reconcile this theory with Einstein's theory were futile, but nothing has given me. I had to delve into Einstein's theory.

Einstein gave the absolute cross-scale, but not for time. I'm looking for a specialist in Einstein's theory, which justified this choice of Einstein.

Such a person absent on this forum, unfortunately.

I wish you a speedy recovery and hope to see Maxwell's integral equations (of Master Theory context) in your performance.

 

Your formula:

is an approximation.
It can not be used for derivation of Maxwell's integral equations.

 

I came to the forum not in order to my a knowledges demonstrate, but in order to get answers to questions.
I'm looking for a professional who understands the relativism and can answer the question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis?

Einstein had entitled (bestowed the absoluteness to time), but no did it.

Why?

 

Incorrect -- it is not an approximation to Maxwell's equations; rather it is a general solution to Maxwell's equations, as it tells you how to calculate B and E.
This has been in textbooks since the 1960's.
http://en.wikipedia.org/wiki/Jefimenko's_equations

Incorrect, with the continuity equation for electric charge, one may prove Maxwell's time-invariant equations follow as consequences.
http://rediscoveries.blogspot.com/2011/07/in-2007-statement-and-proof-of.html
http://arxiv.org/abs/0812.4785v1
As for the assertion that the differential and integral forms of Maxwell's equations are inequivalent, the world is still waiting for your physically valid demonstration of this claim of pure mathematics.

Translation: A Lorentz transformation corresponding to relative motion does not affect the projection of distances measured perpendicularly to the direction of relative motion. Or, shorter: a boost in the X direction leaves ΔY and ΔZ unchanged. Why?

Answer 1: Physics tells us c(Δt)²−(Δx)²−(Δy)²−(Δz)² is physically relevant to the extent that observers agree on this value in a boosted environment. c(Δt)²−(Δx)²−(Δy)²−(Δz)² = c(Δt')²−(Δx')²−(Δy')²−(Δz')². The corresponding Euclidean relationship for rotation of axes: (Δx)²+(Δy)²+(Δz)²=(Δx')²+(Δy')²+(Δz')² factors into (Δx)²+(Δy)²=(Δx')²+(Δy')² and (Δz)²=(Δz')² when the rotation is confined the the x-y plane. So it is natural to assume that motion in the t-x plane factors the relativistic invariant as c(Δt)²−(Δx)² = c(Δt')²−(Δx')² and (Δy)²=(Δy')² and (Δz)²=(Δz')² and the only way to continuously maintain (Δz)²=(Δz')² across all possible boost speeds and still identify v=0 with the identity transform is Δz=Δz'. Similarly for Δy=Δy'.

Translation 1 (low-confidence guess): Why is Δt = Δt' not the physics adopted by Einstein.

Answer: Because Einstein understood Maxwell's equations better than you understand Maxwell's equations, and Einstein trusted that they accurately reflected the physics of the universe. Galileo and Newton thought Δt = Δt' because they had no experience with objects moving faster than 0.01% the speed of light, and with their clocks of 300+ years ago didn't have the precision to measure the time effects on the order of 0.00001% or smaller.
Since 1859, as experiments and observations accumulated, more experimental evidence points at Δt ≠ Δt' for high-speed objects.

http://sciforums.com/showpost.php?p=2039656&postcount=28

 

Retarded potentials and Jefimenko's equations studied in other specialization.
I agree to: retarded potentials or Jefimenko's equations can be used for create Maxwell's equations for Master Theory. But it sum no for me. (I not want to deprive somebody of a livelihood and my scientific career had completed eight years ago.)

Gaus and Stokes use infinite velocity of light (or static fields). Used retarded potentials or Jefimenko's equations.

A Lorentz transformation is absonant (is irrational).
Two observers can not see the slowdown of each other simultaneously. If one sees a slowdown, then the second sees the acceleration. (Otherwise causality principle come to incorrect.) But such asymmetry violates the equality of inertial reference frames.
Time can not slow down.

It is not a scientific argument.


_______________________________________________________________

Do I understand your questions: you (just as me) do not see any reason for the absoluteness of the cross-scale (transverse dimensions)?