quiet

On 3/8/2018 at 3:01 AM, Achilles said:

Is there anything left to discover in electromagnetism?

Like someone unexpectedly hitting a column when turning at the end of a corridor, Planck came across quantization when he analyzed the blackbody radiation. The Nobel Prize did not dissipate Planck's concern, because he thought that quantization should not be put as a postulate. Planck's conviction was that quantization must be deduced from Maxwell's equations. He spent a lot of time researching that. And never in his life was he content to introduce that as a postulate.

By 1930 the physics community invited Planck to scientific congresses, more as a historical hero than as a promoter of new research. Hardly anyone thought it possible to deduce quantum properties from Maxwell's equations. Maybe you like to know that. At the same time, nobody has shown until today that this task is impossible. What do you suppose? It could be possible ? If you answer affirmatively, your list of missing discoveries already has something to begin with.

The girl started asking what part of the Newtonian formula fails at great speeds. I answered the following. The equation of kinetic energy is not established by choice. It follows from the structure of the theory. Are you checking that this equation does not work for high speeds? Then the whole theory, with all its structure, does not work for great speeds. He judged that response as evasive and did not accept any further dialogue. That's why I wanted to present the relativistic formula in a way that the girl could associate with the Newtonian formula.

Today I showed her the same thing that I wrote in this note. He loved it and said that now he could understand what is the failed part of the Newtonian formula. Instead of contradicting her I preferred to ask what is the failed part. It's 2, he replied. And she was very happy, although as ignorant as before regarding the characteristics of physical theories. The important thing is that the dialogue was reopened. Now only lack she understands the structural flank.

Let's go slowly.

1. Classical electrodynamics, as far as it has been developed, does not cover quantum behavior.

2. The phenomena included in the part that has been developed have been clearly explained and the theoretical results have agreed perfectly with the practice.

3. Has everything that classic electrodynamics can give has been developed? Or are some developments still missing?

4. If the theme is the propagation of light in a vacuum, we are thinking of developments that involve the wave equation in a vacuum, with all the solutions it admits.

5. How many wave functions can we formulate, referring to the electromagnetic propagation of the vacuum? Let's start with the two most remanded wave functions, corresponding to the electric field and the magnetic field. There is more ? Yes, one for each field and one for each density that the electromagnetic theory formulates. Vector potential wave, energy density wave, linear momentum density wave, are some examples ... And one that corresponds to what Maxwell was forced to take into account and analyze to establish a complete, coherent and consistent theory . That field is the electric displacement. Then we have an electric displacement wave, waiting for it to be formulated and analyzed explicitly. That is to say that we formulate and analyze something fundamental, whose omission dismantles the system and detracts from it.

6. In the simplest case, the electromagnetic wave equation in vacuum admits two solutions in terms of real numbers and a complex solution of exponential type. We have devoted a lot of attention to the two real solutions. Have we done the same with the complex solution?

7. Let's write the complex solution of the wave equation for electric displacement.

[math]D = \hat{D} \ e^{i \left( \omega t - kx  \right)}[/math]

[math]\hat{D} \ \ \ \rightarrow[/math] module of [math] \vec{D} [/math]

Let's write the identity of De Moivre for the case that concerns us.

[math]e^{i \left(\omega t - kx \right)} = cos\left(\omega t - kx \right) + i \ sin\left(\omega t - kx \right)[/math]

Applying that identity we have the following.

[math]D = \hat{D} \ \left[ cos\left(\omega t - kx \right) + i \ sin\left(\omega t - kx \right) \right] [/math]

8. The vector field [math] \vec{D} [/math], which has two components and module [math] \hat{D} [/math], corresponds to a plane electromagnetic wave propagating in the direction of the axis [math] x [/math]. If those components were not mutually perpendicular, they could not correspond to a complex number. They are mutually perpendicular and correspond to different axes of the coordinate system. Which axes?

9. Let's write the vector expression of the electric displacement.

[math]\vec{D}= \vec{P} + \varepsilon \vec{E} [/math]

In the vacuum is [math] \varepsilon = \varepsilon_o [/math]. We apply it.

[math]\vec{D}= \vec{P} + \varepsilon_o \vec{E} [/math]

The components [math] \vec{P} [/math] and [math] \varepsilon_o \vec{E} [/math] are mutually perpendicular. Can they be both cross-sectional? Let's reason. In terms of local results, polarization is a field with colinear symmetry that does not alter the electrical neutrality. That means that, within a finite length segment, there is a pair of equal and opposite vectors, resulting from all local contributions. In the case we are dealing with, could polarization be transversal? Impossible, because two transverse vectors that correspond to different values of [math] x [/math] are not collinear. Two longitudinal vectors corresponding to different [math] x [/math] values are collinear, because both vectors have the [math] x [/math] axis  direction.

10. What does the vacuum do when a wave propagates? Is it inert or participate in any way? The velocity of propagation is determined only by two properties of the vacuum, which are the permeability [math] \mu_o [/math] and the permitivity [math] \varepsilon_o [/math]. That leaves no doubt. The vacuum participates. How do it participate? The expression of displacement leaves no doubt. Participate polarizing. In that way it set the speed [math] C [/math] of propagation. That means that the displacement has a transversal component [math] \varepsilon_o \vec{E} [/math] and a longitudinal component [math] \vec{P} [/math].

[math]\vec{D}= \vec{x} P + \vec{y} \varepsilon_o E[/math]

In terms of the wave function we have the following.

[math] \vec{D} = \vec{x} \hat{D} \ cos\left(\omega t - kx \right) + \vec{y} \hat{D} \ sin\left(\omega t - kx \right) [/math]

[math] \vec{D} [/math] has finite divergence, corresponding to the charge density of the polarization. That divergence has the form of a wave function.

[math]\nabla \cdot \vec{D} = \hat{D} \ k \ sin\left( \omega t - kx \right)[/math] 

Does that mean that some charge travels in a vacuum when the wave propagates? No charge needs to travel to produce that divergence. In the cities there are giant illuminated signs, which are panels populated by thousands of luminous cells, controlled by a programmable device. A program can achieve that the brightness of each cell varies sinusoidally, in the form corresponding to a wave function. You can program two colors, say blue and red. The first cell is initially dark. Then the blue light grows sinusoidally, reaches the maximum and decreases sinusoidally, until the cell becomes dark. Follow the sinusoidal stage of the red light, which does the same. All the cells are immobile on the board, but the program manages to see alternate blue and red areas traveling along the board. The effect is equivalent to colors in movement. At each point of the vacuum, the charge density varies sinusoidally. The signs of the charge do the same as the colors. The effect is equivalent to alternating zones with opposite charges traveling in the direction of propagation, although no infinitesimal or finite charge is actually moving.

Hi. One girl in high school said the following: It bothers me a lot that the relativistic kinetic energy formula has [math] v ^ 2 [/math] tucked into a root that is in the denominator. That prompted me to look for a way to rewrite that relativistic equation in a way that has more Newtonian flavor. Welcome your didactic opinions, or the type that they are. Is the next.

[math] T = m_o \ C^2 \left( \dfrac{1}{ \sqrt{ 1 - \dfrac{v^2}{C^2}} } \ -1\right) [/math]

[math] T = m_o \ \ C^2 \ \dfrac{1- \sqrt{ 1 - \dfrac{v^2}{C^2}} }{ \sqrt{ 1 - \dfrac{v^2}{C^2}} } [/math]

[math] T = m_o \ \ C^2 \ \dfrac{1- \sqrt{ 1 - \dfrac{v^2}{C^2}} }{ \sqrt{ 1 - \dfrac{v^2}{C^2}} } \ \ \dfrac{1+ \sqrt{ 1 - \dfrac{v^2}{C^2}} }{ 1+\sqrt{ 1 - \dfrac{v^2}{C^2}} } [/math]

We solve the difference of squares in the numerator.

[math] T = m_o \ \ C^2 \ \dfrac{1-1+\dfrac{v^2}{C^2}}{ \sqrt{ 1 - \dfrac{v^2}{C^2}} \left(1+\sqrt{ 1 - \dfrac{v^2}{C^2}} \right) } [/math]

We simplify and order.

[math] T = \dfrac{m_o \ v^2}{ \sqrt{ 1 - \dfrac{v^2}{C^2}} \left(1+\sqrt{ 1 - \dfrac{v^2}{C^2}} \right) } [/math]

Let's symbolize R to the denominator.

[math] R =  \sqrt{ 1 - \dfrac{v^2}{C^2}} \left(1+\sqrt{ 1 - \dfrac{v^2}{C^2}} \right) [/math]

When the speed tends to zero we have the following.

[math] \lim\limits_{\large{v \to 0}}(R)=2 [/math]

I suppose that in that way the girl will feel less rejection for relativistic kinetic energy. Is that likely or have I complicated everything in a worse way?

[math] e^{i \phi} [/math]

Write something

[math]C^2[/math]

Write something

[math]e^{i\phi}[/math]

Hi. Maybe you can help me. When I joined the forum I put the nickname quite and actually I want it to be quiet. Is there a possibility to change it? (apologize for dyslexia)

13 hours ago, Strange said:

For example https://en.m.wikipedia.org/wiki/Zero-energy_universe

Yes, it's the kind of information I need. My mind does not conceive another kind of universe, only the class whose total net energy is equal to zero. And my mind asks more questions, about the roles that negative energy can play.

Regarding the roles of positive energy, I have fewer questions, because it appears as constituting waves and matter, appears ascribed to movement and, in general, ascribed to phenomena closer to vulgar knowledge.

Net energy equal to zero implies equal amounts of energy of both signs in the whole universe. In other words, it implies an extremely important symmetry, which in one aspect imposes restrictive conditions. In another aspect may involve the freedom to obtain unlimited amounts of useful energy, designing sequences according to symmetry.
 

Thank you Strange and Phi for All for your answers. The focus of my interest is not on religions. It is in evaluating the possibility of emptiness being the source of everything that exists. Does that possibility violate any physical law, or more than one?

Hi. I would like to read opinions regarding the following.

There are old books and there are religions. We know with certainty that religions invoke parts of ancient books to justify the structure of this or that religion. Regardless of that, the old books are full of affirmations of all kinds, including some that we can discuss scientifically today.

Take the Bible as an example, although the same thing appears in other ancient books. Regarding the genesis of what exists, it explains that the formation started from nothing. Do I commit abuse assuming that, in that context, the word nothing equals emptiness? In case of not committing abuse, let's continue. The second affirmation is that light was born from the vacuum. And the third, that light then formed everything that makes up the universe.

Personally, I interpret that sequence as the pure vacuum is unstable and polarizing resolves the  instability, that is producing pairs of linked charges. Each pair of linked charges that appears causes an event that generates radiation. The formation of an electron / positron pair from two photons that collide with each other in a vacuum has been verified experimentally. I have no news of larger mass particles obtained from pure radiation, or from an environment full of photons, electrons and positrons.

That's where I arrived. Opinions?

Thank you very much ! Now I will try it.

[math]e{i\phi}=\cos{\phi}+i\sin{\phi}[/math]

Have this been successful?

Thank you very much ! It will surely help me many times in the future. The first thing I need to know is how to activate LaTex within a post. What should I put? Do I put [tex] [/ tex], or what else?

Hello. This is my first post. I am glad to be a member of this good forum.

I ignore the way to use LaTex in a post. Can someone teach how to do it ? 

Thanks in advance.