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Electromagnetic tensor

  1. May 9, 2009 #1
    I understand that after writing down this:
    [tex]F^{ \mu v} = \partial^{\mu} A^v- \partial^v A^{\mu} [/tex]
    We can get a nice matrix connecting E and B vectors. But I just wonder what we need this matrix for? I am a little bit confused about all this relativity in electromagnetism...


    And another question I have is:
    How do these equations
    (1)[tex]\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0} [/tex]
    (2)[tex]
    \vec{\nabla} \times \vec{B} - \frac{1}{c^2} \frac{ \partial \vec{E}}{\partial t} = \mu_0 \vec{J}
    [/tex]
    reduce to this one:
    (3)[tex]\partial_{\beta} F^{\alpha\beta} = \mu_0 J^{\alpha} [/tex]


    I know that after Lorenz gauge we get:
    [tex] \square A^ {\mu} = - \mu_0 j^ {\mu} [/tex]
    And I think we can derive easily this Poisson equation for vector potential also from (2). But is such proof sufficient? And is it possible to get from (3) Poisson equation for scalar potential analogical like for vector one?
     
  2. jcsd
  3. May 9, 2009 #2

    jtbell

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    Staff: Mentor

    You can use this matrix to transform the electromagnetic field from one reference frame to another, by multiplying twice by a matrix that represents the Lorentz transformation:

    [tex]F^{\prime \alpha \beta} = F^{\mu \nu} \Lambda^\alpha_\mu \Lambda^\beta_\nu[/tex]
     
  4. May 9, 2009 #3
    Thanks for your answer but I do not fully understand it.
    Lets say we have four-vector w^a and we want to have vector w^a'.
    We can write:
    [tex]w^{\alpha'} = \Lambda^{\alpha'}_{\alpha} w^{\alpha}[/tex]
    So here lambda is a transformation (matrix - tensor), right?

    So why we mulitply that twice?
     
  5. May 9, 2009 #4

    dx

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    Homework Helper
    Gold Member

    You multiply it twice because that's how tensors of the second rank transform.
     
  6. May 9, 2009 #5
    Sorry, I feel like I am too stupid to understand it :(
    I would say I need only one matrix (transform operator) to transform 4x4 matrix(tensor) to another 4x4 one.
     
  7. May 9, 2009 #6

    dx

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    Homework Helper
    Gold Member

    You're not allowed to choose how to transform the tensor. Four dimensional second rank tensors by definition transform in the way given by jtbell.
     
    Last edited: May 9, 2009
  8. May 9, 2009 #7
    Ok, thank you, so I think I need more theoretical knowledge about tensors. I am also curious when we need to change refrence frame of EM field? In standard homework excercises it is rather not used? And I also would be very gratefull if you could help me with my second question presented in my first post in this thread.
     
  9. May 9, 2009 #8

    dx

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    Gold Member

    You just to write it out in components. For example, if you set β = 0 you get

    [tex] \partial_{\alpha} F^{0\alpha} = \partial_{0}F^{00} + \partial_{1} F^{01} + \partial_{2} F^{02} + \partial_{3} F^{03} [/tex]

    [tex] = - \frac{1}{c}(\partial_x E_x + \partial_y E_y + \partial_z E_z) = \mu_{0} J^{0} = \mu_{0} c \rho [/tex] ​

    So,

    [tex] \nabla \cdot E = - \mu_{0} c^2 \rho = - \frac{\rho}{\epsilon_{0}} [/tex] ​
     
  10. May 9, 2009 #9
    Oh ok, was not that difficult as it looked like :) Thanks.
     
  11. May 9, 2009 #10
    The fact that you need two Lorentz transformations rather than one is a less complicated piece of tensor algebra than you might think. It's for exactly the reason as an ordinary matrix A transforms under a change of basis P as
    [tex]A'=P^{-1}AP[/tex]
    which is in fact the way you would probably do the calculation if you wanted to work things out explicitly. The best book I've found for explaining tensors properly is An Introduction to Differential Geometry by T.J. Willmore.
    As for why you would want to change reference frame:
    Suppose you stand next to a long current-carrying wire such that you measure a magnetic field. Then accelerate instantaneously into the rest frame of the electrons. Suddenly, there's no net motion of charge in your frame any more, so classically you wouldn't predict the same observable consequences of experiments as someone in the rest frame of the wire. SR rescues physics by "mixing up" the E and B fields under a lorentz transformation of the field strength tensor, so that everyone experiences the same physics. You can actually use the field strength tensor to construct a lagrangian for the system entirely using tensorial quantities to show that everyone will percieve the same fundamental dynamics.
     
  12. May 9, 2009 #11
    I see, however I have met with such thing (A'=P^{-1} AP) before only once(in finding min or max of some function while having some other functions as conditions) and that was only connected with examine a quadratic form as positive define or not. Could you then explain me why is such matrix change diffrent than normal A'=PA?

    What about that book, I ll try to look for it, but I predict that it will be a bit difficult to get it in my country. If you know any worth reading free ebook it would be great!

    And your explanation of changing reference frame is just brilliant. Always new things seems to be strange on the begining but after getting know them better they become so obvious :)

    However, according to your example if I would like to egzamine current density I would just have to find [tex]F' ^{\alpha \beta} [/tex] (but how? what lambda operators I am supposed to use?) and use this equation:
    [tex]
    \partial_{\beta} F'^{\alpha\beta} = \mu_0 J^{\alpha}
    [/tex] right?
     
  13. May 10, 2009 #12
    The way a normal matrix transforms under a change of basis is as I wrote above:
    [tex] B = P^{-1}AP [/tex]
    I think you're getting different concepts confused, so let me try and explain what's going on. In elementary linear algebra, a matrix is usually thought of as representing a linear transformation- apply a matrix to a vector, with the usual rules of matrix multiplication, and you get a different vector. For example, rotation matrices take vectors and rotate them around some particular axis. The way in which a matix specifies this transformation is that the columns of the matrix are the results of applying the matrix to each of the basis vectors e.g. the matrix
    [tex] \left ( \begin{array}{ccc}
    cos(\theta) & -sin(\theta) & 0 \\
    sin(\theta) & cos(\theta) & 0 \\
    0 & 0 & 1 \\
    \end{array} \right)
    [/tex]
    which describes a rotation in the xy plane by theta radians anticlockwise would send the vector (1,0,0) to [tex] (cos(\theta),sin(\theta),0)[/tex],etc. Matrix multiplication A=BC is a composition of these linear maps. So, the matrix A represents a linear map which has exactly the same effect as operating on a vector with a matrix C, then operating on the result with the matrix B. Now, suppose the three vectors to which our matrix maps the 3 basis vectors are also a valid basis of our vector space. We want to describe the same transformation, but the components of the matrix will have to change. This is because the vectors on which they act will have different components in the directions of the new basis vectors, e.g. the vector we wrote as [tex] (cos(\theta),sin(\theta),0)[/tex] with respect to the old basis will be written as (1,0,0) with respect to the basis defined by the matrix above! The operation [tex] B = P^{-1}AP [/tex] changes the array of numbers so that the transformation represented by the matrix A with respect to the old basis is represented by B with respect to the new one. For a full explanation of how this works, see page 17 of these notes.
    From a mathematical point of view, most of special relativity is changing the basis of a four-dimensional space with a very strange way of measuring distance!
     
  14. May 10, 2009 #13
    To examine the example I gave above in the "moving" frame, you'd need the lorentz transform describing a change of basis into the reference frame of the electrons, so a boost along the axis of the wire. Bearing in mind the discussion above, you'd also need to remember that the resulting 4-current density vector J would be expressed in components with respect to this new basis. But physically, you've got the right idea. In this particular example it's the principles that are important; the sums are much easier done in the rest frame of the wire! :smile:
     
  15. May 14, 2009 #14
    Thank you! :)
     
  16. May 16, 2009 #15
    You're welcome :smile:
     
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