C H A R G E

According to the way Heisenberg did it, in quantum mechanics, both the position and the momentum come out as probabilities whenever you perform a measurement. The math states that these two probability distributions are Fourier Transforms of each other. Oddly enough, there is a theorem of the Fourier Transforms [of each other] that says the product of their standard deviations remain constant. Please note that the Heisenberg Uncertainty Principle states nothing about how precise a measurement can be, however, you can make as precise a measurement as I want in regards to a particle’s position and afterwards, I can make as precise a measurement of its momentum as you’d want.

Now, if you were to measure a particle’s position to absolute precision (for example, with zero uncertainty), it automatically makes momentum completely unknown. Subsequently, it’ll measure the momentum to absolute precision rendering the particle’s initial position invalid as a negative consequence for attempting to make predictions. An example of this would be measuring the position a second time–after the momentum’s measurement–and this will become evident as being different from your earlier prediction (i.e., assumption) or earlier measurement. What is another way of examining what we consider the electron? Well, for one, it is a spin-1/2 quantum object. Measuring the vertical spin of a spin-1/2 quantum object, you’ll come to the observational pause that it is either spin-down or spin-up. Now, that’s the vertical spin; as far as its horizontal spin–it is completely unknown. Measuring the horizontal spin, you’ll come to see that it is spin-left or spin-right but now it’s vice versa with the vertical spin as it’s now entirely unknown. That’s the trade-off between the horizontal spin and vertical spin of an electron (i.e., spin-1/2 quantum object). The two cannot be known at the same time, the same way both the position and momentum of a particle cannot be known as the same time.

In quantum mechanics, this is essential to measurements; if one were to make a precise measurement of a particle’s position, a wave function is incredibly narrow in position space in consistency with that measurement, therefore, the measurement space wave function is very wide and would contain little information. See, the process of measuring a particle’s momentum won’t exactly change the particle’s position. H O W E V E R, the precision of a single measurement of a particle’s position and momentum is seemingly limited by the progression of our very own technology. That’s an assumption. One thing to know in order to encounter such a claim would be to know that you’re not measuring position and momentum simultaneously in experiment, which alludes back to what was stated previously in regards to an electron’s vertical spin/horizontal spin. Rather, you’d be measuring them sequentially via two separate measurements (the “slit” measures the position and your charge-coupled device detector measures another position). So, how would you correctly infer the momentum? Well, it would be from this second measurement if you were to assume the electron’s behavior was likened to a more classical macroscopic object between the two measurements. Understand that measurements are separated in time, thus, they are not occurring simultaneously.

The uncertainty principle talks about our ability to predict the results of measuring a particle’s position and momentum. I mean, if a particle has a specified wave function…

$\Psi (x, t)$

…then, we can predict, for a given time…

$t$, the range $\langle x \rangle \pm \Delta x$

…within which you’ll have an approximate 2/3 probability of measuring the value of…

$x$

…along with the range of…

$\langle p \rangle \pm \Delta p$

…in which you’ll have an approximate 2/3 probability of measuring the value of…

$p$.

With that said, the uncertainty principle tells you that…

$\Delta x$ and $\Delta p$

…are indeed related by…

$\Delta x \Delta p \ge \hbar / 2$

This is all the uncertainty principle will tell you about the particle within the ranges of position and momentum (as well as the position and momentum values) in attempt to interpret the predictions of the uncertainty principle. This should not confuse you with the uncertainty principle itself (hopefully).

Now, onwards to the Heisenberg Hamiltonian, which is

$H=\sum_i \sum_{j \ne i} J_{ij} S_i \cdot S_j$

Understand that this is what’s known as the exchange of energy between the two spins (horizontal and vertical spin of an spin-1/2 quantum object). Also, take note that $latex J$ is not an operator/matrix. Rather, it’s just a number representing the strength of the coupling.

An electron has no intrinsic parity, meaning that it cannot be assigned “left” or “right”-handed helicity since an observer can be moving either slower or faster than that electron making its velocity appear in one direction to one observer and in the complete opposite direction to another observer. The Heisenberg Hamiltonian is in reference to magnetism (not exclusively electromagnetism). Magnetism simply steers a particle around so it’ll appear to be all right for the field to look different for a number of observers. However, if you’re able to alter or change the reference frames and force particles (i.e., neutrinos, in this case) as the “wrong” helicity, you, in theory, can make the weak force associated with it disappear. For particles that have to undergo a weak decay, this is problematic. In other words, for one observer, they would see a decay while for other observers, they would not see a decay. For neutrinos, it would be acceptable [since neutrinos do not decay] but they do get pushed around by the weak force in the same vein that electrons get pushed around by the magnetic field. Momentum of an electron, however, does not travel in trajectory, crossing intermediate points in the field. Atomically speaking, the momentum of an electron is kinetic, exhibiting effects of special relativity as its momentum increases as well as exhibiting dynamical effects of many-body motion. [Core] electrons–in heavy elements–maintain a high momentum in efforts of achieving significant relativistic mass which, in turn, affects the electrons’ momentum as well as their behavior and neighboring electrons absorb the momentum and behavior of the former electrons.

Relating to charges, you’d notice the interaction of a charged particle moving through a magnetic field, that magnetic field will exert a force in the normal-direction (90°). Folks, this is known as magnetic-magnetic interaction, force of magnetic field interaction is at its maximum when they’re parallel. So when the charged particle is moving in the direction, in perpendicular fashion, to the external magnetic field (concentric magnetic field is parallel to the external field). The force on the charge is perpendicular to both the external field and the direction of the electron.

Above, you see an object (the box) entering a homogenous magnetic field. Understand that the magnetic-magnetic interaction approach will require an integration over the magnetic field energy density of the combined vector fields. The vector fields adds on one side and on the other side, they subtract. Using Lorentz [F = qv x B] force method, it comes off a bit simpler. Doing a Lorentz transform into the charged particle rest frame, the static magnetic field transforms into an electric field, the magnetic fields tend to disappear.

Let’s say that you’re trying to determine the speed an electron needs in order to maintain its circular orbit (an electron moves in a circular path with a radius of r = 4 centimeters in a space between two concentric cylinders). The inner cylinder is a positively-charged wire with a radius of a = 1 millimeter and the outer cylinder is a negatively-charged hollow cylinder with a radius of b = 5 centimeters. The potential difference between the inner and outer cylinders is 120 V with the wire at a higher potential. The electric field in the region between the cylinders is [radially] outward with a magnitude of

E = {\frac{V_a_b}}{r\ln \frac{b}{a}}

Speaking of magnetic fields, let me expand a little (nh) on the Earth’s magnetic field. If you want to determine the speed an electron would need in order to maintain its circular orbit, it’s imperative to include the effect of the Earth’s magnetic field to find the axis of symmetry of the aforementioned cylinders–positioned parallel to the magnetic field of Earth. With that said, at what speed must the electron move to maintain the same circular orbit, assuming the magnetic field of the earth has a magnitude of

$1.3 x 10^{-4}$?

For a, you’ll use…

$\sum F = m a$

…which will give you…

$q E = \frac{m v^2}{r}$.

This, in turn, evolves into…

$v = \sqrt{\frac{r q E}{m}}$.

The end result was…

$5.39 * 10^{12 m/s}$.

For b,

$F_b = q v \times B$

…with the inclusion of Newton’s second law…

$\sum F = F_B + F_E = q(E + v \times B = m a = \frac{m v^2} {r}$.

At this point, you’ll have made the decision of the electron’s direction in both the electric and magnetic field and will account for the sign. The equation is now

{m v^2} – {r e B v} – {r q E] = 0

Within that very same equation, you would need to add the magnitude of the charge of the electron. The result will become positive.

The Magnitude of The Charge of The Electron is {(5.39 x 10(-12)}. Let’s just assume that the magnitude of the charge of the electron is no more than 0.00004%.

0.00004 * 1/100 * {(5.39 * 10^(-12)}
= 2.156 x 10^{-18}

Please note that {5.39 * 10^(-12)) is the magnitude of the charge of an electron. The approximate charge of an electron is

$-1.6021773 * 10^{(-19)}$.

As far as the interaction between the Earth’s magnetic field and charged particles goes, here on Earth, the atmosphere protects life by shielding it from damaging radiation, including ultraviolet radiation and ionizing radiation such as gammas and X-rays. There is a process known as Compton Scattering. This is what protects you and I. Harmful particles emitted from the Sun collide with electrons in the atmosphere and disperse, reflect or dissipate most of their harmful energy. Those same electrons are ejected from their nuclear orbits which results in the formation of the ion. However, during this process, a portion of the atmosphere, specifically the thermosphere, becomes heavily ionized and that portion of the atmosphere is colloquially known as the “ionosphere”. Understand that when such ionization is extremely intense, it will result in a “black out”–a “black out” of all electromagnetic waves that are left of infrared on the electromagnetic spectrum. This occurs after a nuclear weapon is detonated or when there are X-Class solar flares. A nuclear weapon in excess of 300 kilotons gets detonated in the upper atmosphere or an X-Class solar flare gets accompanied by a massive proton storm–what happens? What happens is the Compton Scattering process produces a magnetic field in conjunction with all of the ionized energy, it destroys all electrical devices and any other device that utilizes or contains capacitors, transistors, computer chips, etc. This part of the Compton Scattering process is what’s known as an EMP (electromagnetic pulse).

The charge of an electron is exactly equal in magnitude to that of a proton (only two-up quarks plus down quark). It’s a simple exercise to derive based on the Gauss law that physical states must be charge-neutral. This applies to QED (quantum electrodynamics) and to QCD (quantum chromodynamics) as well. The Gauss law must not be interpreted as an operator equation as this would violate the operator algebra/commutation relations. It’s translated into a constraint equation for the physical sector of the theory:

$G (x) = \nabla E (x) + \rho (x)$

$G (x) |\text{phys}\rangle = 0$

Integrating these equations:

$\int_V d^3x G (x) = \oint_{\partial V} dA E (x) + Q$

$[Q - \text{surface charge}] | [\text{phys}\rangle = 0$

To oversimplify a bit, a charge is defined as the divergence of the electric field rather than a distinct physical quantity that just so happens to be exactly equal, everywhere and at all times, to the divergence of…

$E$.

This is one of the reasons why they can’t be identical and that’s why the equation “divergence electric field = fermionic charge density” no longer holds as an operator equation; it remains valid as an equation acting on physical states which are defined as the states by which the Gauss law operator vanishes [these would be the eigenstates of the Gauss law with eigenvalue 0 = the kernel of the Gauss law].

In reference to the electron, electrons are neither small balls orbiting the nucleus nor are they “clouds”. They exist as a wave form with a probability of being detected in certain spots around the nucleus based on the energy level they currently occupy.

If particles are electromagnetically charged, then classical electromagnetism says that they will emit electromagnetic waves, those particles will lose energy and hence approach each other until they are joined together. However, quantum theory states that energy can only be lost in discrete amounts and in particular, there is a minimum energy a system can have so the electrons do not fall into the nucleus. Speaking of nucleus, using classical mechanics and electromagnetism, let’s work out the speed that an electron would have to have in order for it to orbit a hydrogen nucleus at a distance of…

$10^{-12m}$:

$F_c = \frac{m_ev^2}{r} = \frac{kg_e^2}{r^2}$

$v = \sqrt{\frac{kg_e^2}{mr}}$

$\text{where}$

$r = \text{radius of orbit} = 1e-12 m$

$k = 9e9 Nm^2/C^2$

$q_e = 1.602e-19 C$

$m_e = 9.1e-31 kg$

This would obviously be the minimum orbital radius permitted by relativity.

Anyways, the interaction of the electrical/magnetic force with the field of the charge causes radiation. Since the charge has a mass, the charge accelerates as well. Yet, the acceleration itself is not the cause of the radiation.

So when does a charge radiate [and by “radiate”, I mean when does a charge emit electromagnetic radiation, for instance]?

Well, the notion that an accelerated charge should produce electromagnetic radiation is not required by Maxwell’s equations; it is derived from the notion that an accelerated charge must interact with its own field. There have been controversial attempts to explain why gravitational acceleration of a charge should not produce radiation by showing that radiation is a function of the third time derivative of position, such as non-uniform acceleration. The fact is, however, as anyone who tries to make electrons go in a circle using magnets knows that uniformly changing the direction of a moving electron with magnetic force produces radiation.

Research on accelerated charges speaks on the electromagnetic field (“Radiation from a Uniformly Accelerated Charge”):

Abstract: The electromagnetic field associated with a uniformly accelerated charge is studied in some detail. The equivalence principle paradox that the co-accelerating observer measures no radiation while a freely falling observer measures the standard radiation of an accelerated charge is resolved by noting that all the radiation goes into the region of space time inaccessible to the co-accelerating observer.

Real radiation, such as light, ought to be an observer independent. Its angular momentum is a Lorentz invariant. If [Maxwell] radiation is independent of the observer then there will be an invariant definition that does not involve the observer. Source-free Maxwell equations,

$F = dA$ and $d*F = 0$,

…produce two types of fields. Then again, aren’t Maxwell’s equations about the interaction of charges and fields? One can remove the charge entirely by cutting the point where the charge lives out of the space and be left with a topological charge only. This topological charge will only have a coulomb field and it’s expected to radiate just like any real spherical charge on acceleration.

There are three ways that a charge can be accelerated:

1. Being situated in or moving through an electromagnetic field
2. Being absorbed or emitted from a photon
3. Being situated in or moving through a gravitational field

A charge distribution represents energy and therefore, enter inertia. That inertia would be contained in the field within the volume of space that contains the charge distribution. The electron, representing a point charge, cannot represent a charge distribution. There ain’t no volume to distribute the charge, playa playa! Besides, if you were to think of the electron as a sphere containing its charge as being made up of point charges and if you let the mass of the electron in addition to letting the mass of the electron equal the energy of this notional charge distribution, you’ll end up with the classical electron radius of

$r_{e} = \frac{1}{4\pi\varepsilon_0\frac{e^2}{m_{{e}} c^2}} = 2.8179403267(27) \times 10^{-15} {m}$

…or, 2.8 E-15 m, for short. Maaaaaaan, this would make the electron way too large. Working out the self-energy of the electron as a point charge with the use of classical or quantum principles, the end result would be infinite self-energy of the electron. While a charge distribution has energy and emits photons by itself, that is not the case with an electron. The requirement is that the electron go through some sort of interaction with another object, i.e., magnetic field, in order to emit a photon. However, some [physicists….and I’ll add engineers as well] view the classical electron radius approach in a nonsensical light. In the name of integration, by integrating the energy density of the field down to a radius such that the energy density equals the rest energy of the electron.  When it’s impacted by force, an electron will respond with inertia (m) and the field will take such a long time to reorganize itself. This cause the field’s “spreading out” at c, i.e. the “speed of light”. This “spreading out” of the field is not the acceleration of the field but you would be accelerating the electron itself. Finding a radius so the outside field energy is equal to…

$mc^2$

…is unnecessarily doable but not physical.

-Desmond (DTO™)

References

Radiation from a Uniformly Accelerated Charge

Author: David G. Boulware

Source: Annals of Physics: 124, 169-188 (1980)

Classical Radiation from a Uniformly Accelerated Charge

Authors: Thomas Fulton, Fritz Rohrlich,

Source: Annals of Physics: 9, 499-517 (1960)

Radiation Damping in a Gravitational Field

Author: Bryce S. Dewitt, Robert W. Brehme

Source: Annals of Physics: 9, 220-259 (1960)

Principle of Equivalence

Author: Fritz Rohrlich

Source: Annals of Physics: 22, 169-191 (1963)

Radiation from an Accelerated Charge and the Principle of Equivalence

Authors: A. Kovetz, G.E. Tauber

Source: American Journal of Physics, Vol. 37 (4), April 1969

Radiation from a particle uniformly accelerated for all time

Author: S. Parrott

Source: General Relativity and Gravitation: 27 1463-1472 (1995)

Onanel

Desmond J. Watson is a Co-Founder and the Chief Executive Officer of Hexagon Lavish®, a scientific R&D start-up that specializes in informational interpretation software development and unique IR techniques. He also manages the blog, Arheliean. You can follow him on Twitter, @TheRealOnanel. If you're rich and want to invest in Hexagon Lavish®, then do so.

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