Lorentz Force and Stress tensor

$F_{ij} =\begin {pmatrix} 0 & D_1 & D_2 & D_ 3\\ -D_1 & 0& H_3 & -H_2 \\ -D_2 & - H_3 & 0 & H_1 \\ -D_3 & H_2 & -H_1 & 0 \end {pmatrix}$

$G_{ij} =\begin {pmatrix} 0 & B_1 & B_2 & B_ 3\\ -B_1 & 0& -E_3 & E_2 \\ -B_2 & E_3 & 0 & -E_1 \\ -B_3 & -E_2 & E_1 & 0 \end {pmatrix}$

The field equation are:

$\partial /\partial x_i F_{i j } = - J_i$

$\partial /\partial x_i G_{i j } =0$

That is the result from last time.

the conservation of charge is:

$\partial/\partial x_i J_i = 0$

thus the 4-Laplacian of the F-field is :

$\partial^2/\partial x_i^2 F_{ij} = 0$

The physical meaning of the simplification of the field equation by the field tensor is, a gradient in the tensor field is equation to the minus of 4-current, or zero. recall that the gradient in 3-D vector space, the conservation of charge density is :

$\nabla \cdot \vec{J} = - \partial \rho/\partial t$

we have the same form in the 4-D tensor space. the creation of field is conservation of the 4-charge displacement, if we integrate the 4-current. i dun know what physical meaning of the G-field. personally, i believe that the F-field and G-field can be related by some transform.

we have another interesting things. we can write the Lorentz force into the field tensor:

$\vec{f}_j = \frac{d\vec{P}_j}{d\tau}=q \frac{d\vec{X}_j}{d\tau}F_{ij}$

the reason why we can write this, i don’t know. any physical meaning? i don’t know. may be we can think in this way, the force depends on the motion of the 4-vector and the field and the charge. thus, it is natural to multiple them together to get the force. But why not the G field? never the less, the field tensor reduce the number of Field qualities into 2.

the Lorents Force can be more simple

$\vec{f}_j = J_i F_{ij}$

that the force is created by the field and the current.

The Electromagnetic stress tensor can also be related with the 4-force by :

$\vec{f}_j = - \partial/\partial x_i T_{ij}$

Thus, combined with the Field tensor:

$\partial/\partial x_iT_{ij}+ J_iF_{ij} = 0$

Magnetic field strength and Proton escape energy

The paper has talked about the strong magnetic field will trap the low energy proton. So, how is the field strength and the proton energy relationship?

the proton moving radius can be formulated by:

$R = \frac {m v}{e B}$

and according the special relativity

$v = c \frac{p c}{E} = c \sqrt { 1- \left( \frac {m c^2}{E} \right) ^2 }$

Thus,

$R = \frac { m c} { e B } \sqrt { 1- \left(\frac {m c^2}{E} \right)^2 }$

Sub all the constant

$R = 3.129738 \frac{1}{B} \sqrt { 1- \left(\frac {938 MeV }{E} \right)^2 }[m][T]$

Thus, for slow proton, say 50MeV, the radius is

$R = 983.02 / B [mm][T]$

even for 1MeV

$R = 144.40 / B [mm][T]$

but for 1keV

$R = 4.57 /B [mm][T]$

a review on Hydrogen’s atomic structure

I found that most of the book only talk part of it or present it separately. Now, I am going to treat it at 1 place. And I will give numerical value as well. the following context is on SI unit.

a very central idea when writing down the state quantum number is, is it a good quantum number? a good quantum number means that its operator commute with the Hamiltonian. and the eigenstate states are stationary or the invariant of motion. the prove on the commutation relation will be on some post later. i don’t want to make this post too long, and with hyperlink, it is more reader-friendly. since somebody may like to go deeper, down to the cornerstone.  but some may like to have a general review.

the Hamiltonian of a isolated hydrogen atom is given by fews terms, deceasing by their strength.

$H = H_{Coul} + H_{K.E.} + H_{Rel} + H_{Darwin} + H_{s-0} + H_{i-j} + H_{lamb} + H_{vol} + O$

the Hamiltonian can be separated into 3 classes.

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Bohr model

$H_{Coul} = - \left(\frac {e^2}{4 \pi \epsilon_0} \right) \frac {1}{r}$

is the Coulomb potential, which dominate the energy. recalled that the ground state energy is -13.6 eV. and it is equal to half of the Coulomb potential energy, thus, the energy is about 27.2 eV, for ground state.

$H_{K.E.} = \frac {P^2}{ 2 m}$

is the non-relativistic kinetic energy, it magnitude is half of the Coulomb potential, so, it is 13.6 eV, for ground state.

comment on this level

this 2 terms are consider in the Bohr model, the quantum number, which describe the state of the quantum state, are

$n$ = principle number. the energy level.

$l$ = orbital angular momentum. this give the degeneracy of each energy level.

$m_l$ = magnetic angular momentum.

it is reasonable to have 3 parameters to describe a state of electron. each parameter gives 1 degree of freedom. and a electron in space have 3. thus, change of basis will not change the degree of freedom. The mathematic for these are good quantum number and the eigenstate $\left| n, l, m_l \right>$ is invariant of motion, will be explain in later post. But it is very easy to understand why the angular momentum is invariant, since the electron is under a central force, no torque on it. and the magnetic angular momentum is an invariant can also been understood by there is no magnetic field.

the principle quantum number $n$ is an invariance. because it is the eigenstate state of the principle Hamiltonian( the total Hamiltonian )!

the center of mass also introduced to make more correct result prediction on energy level. but it is just minor and not much new physics in it.

Fine structure

$H_{Rel} = - \frac{1}{8} \frac{P^4}{m^3 c^2}$

is the 1st order correction of the relativistic kinetic energy. from $K.E. = E - mc^2 = \sqrt { p^2 c^2 + m^2c^4} - mc^2$, the zero-order term is the non-relativistic kinetic energy. the 1st order therm is the in here. the magnitude is about $1.8 \times 10^{-4} eV$. ( the order has to be recalculate, i think i am wrong. )

$H_{Darwin} = \frac{\hbar^{2}}{8m_{e}^{2}c^{2}}4\pi\left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)\delta^{3}\left(\vec r\right)$

is the Darwin-term. this term is result from the zitterbewegung, or rapid quantum oscillations of the electron. it is interesting that this term only affect the S-orbit. To understand it require Quantization of electromagnetic field, which i don’t know. the magnitude of this term is about $10^{-3} eV$

$H_{s-o} = \left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)\left(\frac{1}{2m_{e}^{2}c^{2}}\right)\frac{1}{r^3} L \cdot S$

is the Spin-Orbital coupling term. this express the magnetic field generated by the proton while it orbiting around the electron when taking electron’s moving frame. the magnitude of this term is about $10^{-4} eV$

comment on this level

this fine structure was explained by P.M.Dirac on the Dirac equation. The Dirac equation found that the spin was automatically come out due to special relativistic effect. the quantum number in this stage are

$n$ = principle quantum number does not affected.

$l$ = orbital angular momentum.

$m_l$ = magnetic total angular momentum.

$s$ = spin angular momentum. since s is always half for electron, we usually omit it. since it does not give any degree of freedom.

$m_s$ = magnetic total angular momentum.

at this stage, the state can be stated by $\left| n, l, m_l, m_s \right>$, which shown all the degree of freedom an electron can possible have.

However, $L_z$ is no longer a good quantum number. it does not commute with the Hamiltonian. so, $m_l$ does not be the eigenstate anymore. the total angular momentum was introduced $J = L + S$ . and $J^2$ and $J_z$ commute with the Hamiltonian.  therefore,

$j$ = total angular momentum.

$m_j$ = magnetic total angular momentum.

an eigenstate can be stated as $\left| n, l, s, j, m_j \right>$. in spectroscopy, we denote it as $^{2 s+1} L _j$, where $L$ is the spectroscopy notation for $l$.

there are 5 degrees of freedom, but in fact, s always half, so, there are only 4 real degree of freedom, which is imposed by the spin ( can up and down).  the reason for stating the s in the eigenstate is for general discussion. when there are 2 electrons, s can be different and this is 1 degree of freedom.

Hyperfine Structure

$H_{i-j} = \alpha I \cdot J$

is the nuclear spin- electron total angular momentum coupling. the coefficient of this term, i don’t know. Sorry. the nuclear has spin, and this spin react with the magnetic field generate by the electron. the magnitude is $10^{-5}$

$H_{lamb}$

is the lamb shift, which also only affect the S-orbit.the magnitude is $10^{-6}$

comment on this level

the hyperfine structure always makes alot questions in my mind. the immediate question is why not separate the orbital angular momentum and the electron spin angular momentum? why they first combined together, then interact with the nuclear spin?

may be i open another post to talk about.

The quantum number are:

$n$ = principle quantum number

$l$ = orbital angular momentum

$s$ = electron spin angular momentum.

$j$ = spin-orbital angular momentum of electron.

$i$ = nuclear spin. for hydrogen, it is half.

$f$ = total angular momentum

$m_f$ = total magnetic angular momentum

a quantum state is $\left| n, l, s, j,i, f , m_f \right>$. but since the s and i are always a half. so, the total degree of freedom will be 5. the nuclear spin added 1 on it.

Smaller Structure

$H_{vol}$

this term is for the volume shift. the magnitude is $10^{-10}$.

in diagram: