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The own energy of electromagnatic field

  1. May 18, 2012 #1
    In electrodynamics,the density of energy and momentum of electromagnetic field is expressed by the intensity of electronic field and magnetic field.

    In electrodynamics, if we have know the distribution of electronic charge and electronic current,we can derive the electronic scalar potential and magnatic vector potential from solving the Laplace Equation(but we generally can't derive the intensity of electronic field and magnetic field directly,and must through potential,because Maxwell's Equations are vector equation,can't be solved directly ), and derive the intensity of electronic field and magnetic field, by calculating the gradient of the electronic scalar potential and the rotation of magnatic vector potential .

    So we can see that the potential is seemly most essential,so we should firstly express the energ of electromagnetic field with the two potentials. So the situation will be similar to the Lagrangian dynamics,which is only the equation of energy,and has a clear and high opinion of dynamics of material system. Perhaps someone can derive the Maxwell's Equations from the equation of energy.

    And it gives us a similar picture with electromagnetic field equation , Schrodinger Equation and the field equation in General Relativity. So it is a very deep way of calculating and thinking.It can enlighten us to think the identity of the nature.
     
    Last edited: May 18, 2012
  2. jcsd
  3. May 18, 2012 #2

    Jano L.

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    This is an interesting issue. Already Maxwell introduced and used potentials, but their status has been subject of controversy ever since. Heaviside did not like them and had chosen to avoid them. It is entirely possible. It is possible to use and solve Maxwell's equations without mentioning the potentials at all. Also mechanics works well without them - Newton equations of motion contain only fields E,B.

    However, potentials are interesting mathematically, because they make possible some mathematical developments, such as formulation through Lagrangian or Hamiltonian equations, and also simplify the solution of Maxwell's equations.

    I think regarding potentials as physical quantities is troublesome because there is no unique way to fix them - there are many choices that lead to the same fields and hence the same behavior. I think potentials, although possibly useful, are merely convenient mathematical functions whose values are devoid of any physical significance.
     
  4. May 19, 2012 #3
    But a vector equation have three variables and only one equation,although we know the three variables have some relation because they are parts of a same vector,we can't express the relation ,so we can't solve the equation directly.

    But thank you for leading me to the job of Heaviside,it is very helpful for me.
     
    Last edited: May 19, 2012
  5. May 19, 2012 #4

    Jano L.

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    Equations are not enough - we need also boundary conditions. The six unknowns [itex]\mathbf E,\mathbf B[/itex] are determined by the six equations


    [tex]
    \nabla \times \mathbf E = -\partial_t \mathbf B
    [/tex]

    [tex]
    \nabla \times \mathbf B = \mu_0 \mathbf j + \mu_0 \epsilon_0 \partial_t \mathbf E
    [/tex]

    and the known initial values of fields at the time [itex]t_0[/itex] at all positions [itex]\mathbf x[/itex]:

    [tex]
    \mathbf E (\mathbf x, t_0) , ~~\mathbf B(\mathbf x, t_0)
    [/tex]

    These initial values have to obey the conditions

    [tex]
    \nabla \cdot \mathbf E (\mathbf x, t_0) = \rho(\mathbf x, t_0)/\epsilon_0,~~\nabla \cdot \mathbf B (\mathbf x, t_0) = 0.
    [/tex]
     
  6. May 20, 2012 #5
    Yes,so I must confess that through Maxwell's Equations we can derive the intensity of electronic field and magnetic field directly,although more difficult. And we can't judge whether potential or intensity of field is essential through the procedure of solving the Maxwell's equation.

    But in Aharonov-Bohm effect,we can see that the intensity of magnetic field is not enough,and must consider magnetic vector potential.The electron has an additional momentum which is expressed as (-eA), it makes contribution to the interference of electron wave :

    (2e[itex]\pi[/itex][itex]/[/itex]h)[itex]\oint[/itex]A[itex]\cdot[/itex]dl



    This is the additional subtraction of phase.So in this situation,if we only have known potential A,we can also get intensity B,and calculate the subtraction of phase.But if we only have known intensity B,we can't get A,beacuse is this problem, B=0.We must get A from the distribution of electronic current.

    This experiment is meaningful to the status of potential.
     
    Last edited: May 20, 2012
  7. May 20, 2012 #6

    Jano L.

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    The explanation of the shift of interference fringes by Bohm and Aharonov is in my opinion controversial.

    First, in order to claim that the phase shift is given uniquely by the integral of vector potential A and that this indicates physical status of A, this integral would have to have unambiguous value. However, the magnetic field in the space can be described by vector potential A' that is equal zero everywhere outside the solenoid. Then the line integral above gives zero.

    It is true that the function A' is singular in the centre of the solenoid, but the particles are supposed to never enter the solenoid anyway.

    Thus, since other functions giving correct B are already allowed by gauge invariance, I do not see a compelling reason to reject A' as unphysical. It give B correctly. That is its purpose.


    Second, there are reasons to believe that the shift is caused by the actio of electromagnetic force of the solenoid on the particles. For example, see the papers by Timothy Boyer, such as


    2000 Boyer Does the Aharonov Bohm Effect Exist?
    Foundations of Physics, Vol. 30, No. 6, 2000

    Let me know if you can't get it.
     
  8. May 21, 2012 #7
    Yes, I can't get that paper,can you show me one if you have time ? Thank you!

    Another problem, In Aharonov-Bohm effect ,the magnetic vector potential A is not zero everywhere outside the solenoid. But you say :

    "However, the magnetic field in the space can be described by vector potential A' that is equal zero everywhere outside the solenoid. Then the line integral above gives zero. "

    So I can't understand your opinion.The intensity B is the curl of the vector potential A, at the space outside the solenoid ,B=0,but A≠ 0,this is the fact about the experiment (from my text book).Now I am not able to point out if the description about the experiment by my textbook is questionable,because I have not seen more introduction about it.
     
  9. May 21, 2012 #8

    Jano L.

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    The line integral is nonzero, if you use
    [tex]
    \mathbf A = \frac{B\pi r_0^2}{2\pi r} \mathbf e_{\varphi}
    [/tex]

    outside the soleonoid. But this is not the only possible function. Equally good potential is A' which is, outside the solenoid,

    [tex]
    \mathbf A' = \mathbf 0
    [/tex]

    and non-zero inside (the exact value of A' inside the solenoid can be found from Maxwell's equations. It turns out it has a singularity in the center of the solenoid, but I do not think this invalidates it, since the particles are not supposed to enter the solenoid anyway).

    Jano
     
  10. May 25, 2012 #9

    In the paper <Doer the Aharonov-Bohm effect exist? > by Boyer , he use a formula to calculate the phase subtraction.(The Equation (8) ) :


    equation.GIF


    He first calculate the inluence from electrostatic force,then use the way of analogy to derive the situation of magnetic field.The Equation (8) is the former one. But if deduct it into the latter one directly , I have a question that what is the momentum of electron in the magnetic field ? Why in the Hamilton function we have the additional momentum (-eA) ,but general momentum is not relative to the potential of electronic field ? So the analogy is flawed . In Aharonov-Bohm effect ,the status of electronic one and of magnetic one are different.
     
    Last edited: May 25, 2012
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