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Question on how Einstein derived the EM transformations

  1. Oct 1, 2015 #1
    I was reading Einstein's seminal work: http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf


    Go to this part:

    II. ELECTRODYNAMICAL PART § 6. Transformation of the Maxwell-Hertz Equations for Empty Space. On the Nature of the Electromotive Forces Occurring in a Magnetic Field During Motion

    Einstein says "If we apply to these equations the transformation developed in § 3, by referring the electromagnetic processes to the system of co-ordinates there introduced, moving with the velocity v, we obtain the equations"

    But I'm not sure how applying the transformations obtains those equations.

    Can someone show me the intermediate steps?
     
  2. jcsd
  3. Oct 1, 2015 #2

    vanhees71

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    Thanks to Minkowski, it's much simpler nowadays. Einstein (and Lorentz, Poincare et al before him) had to do this by hand.

    The trick is to use Minkowski's ingenious covariant formalism. The electromagnetic field is described by an antisymmetric four-tensor (the Faraday tensor), which derives as a four-curl from a vector potential:
    $$F_{\mu \nu}(x) = \partial_{\mu} A_{\nu}(x)-\partial_{\nu} A_{\mu} (x).$$
    The four-potential vector field transforms as a vector field, i.e., for an arbitrary Lorentz transformation ##{\Lambda^{\mu}}_{\nu}##, which transforms the coordinates from one inertial frame to another via
    $$\tilde{x}^{\mu} = {\Lambda^{\mu}}_{\nu} x^{\nu}, \quad \tilde{x} = \hat{\Lambda} x$$
    you have
    $$\tilde{A}^{\mu}(\tilde{x}) = {\Lambda^{\mu}}_{\nu} A^{\nu}(x)={\Lambda^{\mu}}_{\nu} A^{\nu}(\hat{\Lambda}^{-1} \tilde{x}).$$
    Then taking the derivatives of the new tensor components with respect to the new spacetime coordinates, using this transformation formula, you get the transformation of the field components written down in Einstein's paper.
     
  4. Oct 1, 2015 #3
    haha I don't understand the Minkowski method.

    what did Einstein use? no need to write the full working, just give me some pointers, thanks.
     
  5. Oct 1, 2015 #4

    jtbell

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    Basically, you have to convert the partial derivatives of X, Y, Z, L, M, N with respect to the original coordinates x, y, z, t (e.g. ##\partial X / \partial t##) to partial derivatives with respect to the new coordinates ##\xi, \eta, \zeta, \tau## (e.g. ##\partial X / \partial \tau##), by using the chain rule for partial derivatives. It's a tedious slog.

    I seem to remember participating in a previous thread about this... ah, here it is!

    https://www.physicsforums.com/threads/transforming-maxwells-equations-in-special-relativity.667526/
     
    Last edited: Oct 1, 2015
  6. Oct 1, 2015 #5

    vanhees71

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    Well, this Einstein describes in detail in his famous paper of 1905. First he uses the field components in the initial frame but writes the Maxwell equations in terms of the spacetime coordinates of the boosted frame. Then he argues, that there should be transformations of the field components to new field components such that for these new components the Maxwell equations hold in the same form as for the original components in the original frame, and he derives these equations. It's not so difficult in principle, but a bit tedious to do by hand. Of course, modern vector notation should simplify the task significantly. Because that tedious exercise is a good motivation for you to learn the manifestly covariant four-vector formalism, because then you don't need this tedious calculation, Einstein had to do 110 years ago :-).
     
  7. Oct 2, 2015 #6
    Thanks, that thread is very useful. But how do we go from this:

    [tex]
    \begin{align}
    t' = & \gamma \left( t-\frac{v}{c^2}x \right) \\
    x' = & \gamma \left( x-vt \right) \\
    y' = & y \\
    z' = & z
    \end{align}
    [/tex]

    to this:
    [tex]
    \begin{align}
    \frac{\partial }{\partial t}\equiv & \gamma \left( \frac{\partial }{\partial t'} - v \frac{\partial }{\partial x'}\right) \\
    \frac{\partial }{\partial x}\equiv & \gamma \left( \frac{\partial }{\partial x'} - \frac{v}{c^2} \frac{\partial }{\partial t'}\right) \\
    \frac{\partial }{\partial y}\equiv & \frac{\partial }{\partial y'} \\
    \frac{\partial }{\partial z}\equiv & \frac{\partial }{\partial z'}
    \end{align}
    [/tex]
     
  8. Oct 2, 2015 #7
    I also find it interesting that the y and z components don't transform in the same manner. one is +v and the other is -v.
     
  9. Oct 2, 2015 #8

    vanhees71

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    You just use the chain rule
    $$\partial_t=\frac{\partial t}{\partial t'} \partial_{t'} + \frac{\partial t}{\partial x'} \partial_{x'}.$$
    The inverse Lorentz boost needed reads
    $$t=\gamma(t'+v/c^2 x'), \quad x=\gamma(x+v t).$$
    Thus you have
    $$\frac{\partial t}{\partial t'}=\gamma, \quad \frac{\partial t}{\partial x'}=\frac{\gamma v}{c^2}.$$
    The same you have to do for ##\partial_x##.
     
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