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I Maxwell's equations and tensor notation

  1. Dec 7, 2016 #1
    In one of our lectures we wrote Maxwell's equations as (with ##c=1##)
    ##\partial_\mu F^{\mu \nu} = 4\pi J^\nu##
    ##\partial_\mu F_{\nu \rho} + \partial_\nu F_{\rho \mu} + \partial_\rho F_{\mu \nu} = 0##

    where the E.M. tensor is
    ##
    F^{\mu \nu} = \begin{pmatrix}
    0 & -B_3 & B_2 & E_1\\
    B_3 & 0 & -B_1 & E_2\\
    -B_2 & B_1 & 0 & E_3\\
    -E_1 & -E_2 & -E_3 & 0
    \end{pmatrix}##

    and ##J^\nu = (\mathbf J ,\rho)##,
    ##\partial_\mu = (\nabla, \frac{\partial}{\partial t} )##
    (allowing the notation with the equal sign between the component and the 4-vector.)

    Now from this I'm trying to recreate our classic Maxwell's equations.
    If I set ##\nu = 4## in the first equation we get the L.H.S. as ##4\pi \rho##
    The R.H.S. becomes
    ##\partial_\mu F^{\mu 4} = -\frac{\partial}{\partial x}E_1-\frac{\partial}{\partial y}E_2-\frac{\partial}{\partial z}E_3 = -\nabla \cdot E##.
    Now this is a sign error since we should get ##\nabla \cdot E = 4\pi \rho##.

    So I guess maybe I shouldn't put a minus sign before those but that goes against what I learned by using the Minkowski metric, ##g_{\mu \nu} = diag(-1,-1,-1,1)##. Similarly I get the same sign error in the continuity equation. Should I always get + signs?
     
  2. jcsd
  3. Dec 7, 2016 #2

    vanhees71

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    You don't need the metric here, because the index of the four-gradient is already a lower index (covariant components of a vector), i.e.
    $$\partial_{\mu}=\frac{\partial}{\partial x^{\mu}}.$$
    So you get the correct Gauss law (in Gaussian units)
    $$\vec{\nabla} \cdot \vec{E}=4\pi \rho.$$
    The same is true for the continuity equation:
    $$\partial_{\mu}j^{\mu}=\vec{\nabla} \cdot \vec{j}+\partial_t \rho=0.$$
     
  4. Dec 7, 2016 #3
    Then I end up with another question. Consider 4-momentum vectors
    ##P^\mu = (P_x,E_1)##
    ##Q^\mu=(Q_x,E_2)##
    Then ##(P^\mu+Q^\mu)^2= (E_1+E_2)^2-(Q_x+P_x)^2 = E_1^2+E_2^2-Q_x^2-P_x^2+2E_1E_2-2Q_xP_x##.
    However we also have
    ##(P^\mu+Q^\mu)^2 = P^2+Q^2+2P_\mu Q^\mu = E_1^2-P_x^2+E_2^2-Q_x^2+2(E_1E_2+P_xQ_x)##

    So In this case it's obviously a minus sign. Why?
    So it should be something like this i guess
    ##(P+Q)^2 = P^2+Q^2+2g_{\mu \nu}P^\nu Q^\mu##?
    But ##g_{\mu \nu}P^\nu = P_{\mu}## by lowering of index.
     
  5. Dec 7, 2016 #4

    Nugatory

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    Staff: Mentor

    Set ##U=(P+Q)## and remember that ##(P+Q)^2=U^2=g_{\mu\nu}U^{\mu}U^{\nu}##

    Write out the summation explicitly if that doesn't make it clear.
     
  6. Dec 7, 2016 #5
    Yes that's what I did in the "first calculation".

    However a lot of calculations rely on using the second approach looking at ##(P+Q)^2 = P^2+Q^2+\dots##. This approach is something that showed up a lot in our class for solving dynamics problem and when deriving things like Compton scattering or threshold energies.
     
  7. Dec 8, 2016 #6
    Perhaps this can be thought of as it's always the normal scalar product but we have
    ##U^\mu = (\mathbf p , E)## and with lowered index
    ##U_\mu = g_{\mu \nu} U^\nu = (-\mathbf p,E)##.
    In that case the normal scalar product of these two always gives the Lorentzian scalar product.

    So my confusion is only that every other 4-vector I've seen I'm used to seeing ##U^\mu = (\mathbf p , E)## and ##U_\mu = (-\mathbf p,E)## while for the differentiation operator we defined it the 3-vector part positive for the lowered index.

    I guess the problem I had is that in class we always spoke of
    ##U_\mu U^\mu## or ##A_\mu B^\mu## as the Lorentz scalar product.
     
  8. Dec 8, 2016 #7

    vanhees71

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    The Minkowski product is a bilinear form, and it is always good to distinguish vectors and their components although physists always talk about ##V^{\mu}## is a vector, but really right is to say that these are tensor components with respect to a pseudo-orthogonal basis with four basis vectors ##\boldsymbol{e}_{\mu}##. The vector is
    $$\boldsymbol{V}=V^{\mu} \boldsymbol{e}_{\mu}$$
    with
    $$\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\nu}=\eta_{\mu \nu}.$$
    You have the somewhat unusual convention to have the spatial components as the first three and the temporal component as the fourth component. So in your case, written in a matrix notation you have ##(\eta_{\mu \nu})=\mathrm{diag}(-1,-1,-1,1)## (west-coast convention).

    A Lorentz transformation describes the change of one pseudo-orthogonal basis to another, i.e.,
    $$\boldsymbol{e}_{\mu} = {\Lambda^{\nu}}_{\mu} \boldsymbol{e}_{\nu}',$$
    which implies
    $$\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\rho}=\eta_{\mu \rho} ={\Lambda^{\nu}}_{\mu} {\Lambda^{\sigma}}_{\rho} \boldsymbol{e}_{\nu}' \cdot \boldsymbol{e}_{\sigma}'={\Lambda^{\nu}}_{\mu} {\Lambda^{\sigma}}_{\rho} \eta_{\nu \sigma}.$$
    Now it is conventient to introduce also the inverse metric components via
    $$(\eta^{\mu \nu})=(\eta_{\mu \nu})^{-1}=\mathrm{diag}(-1,-1,-1,1)$$
    and covariant vector components and contravariant basis vectors via
    $$V_{\mu} = \eta_{\mu \rho} V^{\rho}, \quad \boldsymbol{e}^{\mu}=\eta^{\mu \rho} \boldsymbol{e}_{\rho},$$
    and then you have, e.g.,
    $$\boldsymbol{V}=\boldsymbol{e}^{\mu} V_{\mu}=\eta^{\mu \nu} \boldsymbol{e}_{\nu} V_{\mu}$$
    and
    $$\boldsymbol{V} \cdot \boldsymbol{W}=\eta_{\mu \nu} V^{\mu} W^{\nu}=V^{\mu} W_{\mu}=V_{\mu} W^{\mu},$$
    etc.

    For the Lorentz transformation properties of the components you now get in a straight forward way
    $$\boldsymbol{V}=V^{\mu} \boldsymbol{e}_{\mu} = V^{\mu} {\Lambda^{\nu}}_{\mu} \boldsymbol{e}_{\nu}',$$
    and thus since the decomposition of a vector in terms of a basis is unique (i.e., the basis vectors are a linearly independent set of vectors)
    $$V^{\prime \nu}={\Lambda^{\nu}}_{\mu} V^{\mu}.$$
    For the covariant components you find
    $$V_{\rho} ' =\eta_{\rho \nu} V^{\prime \nu}=\eta_{\rho \nu} {\Lambda^{\nu}}_{\mu} V^{\mu} = \eta_{\rho \nu} {\Lambda^{\nu}}_{\mu} \eta^{\mu \sigma} V_{\sigma} = {\Lambda_{\rho}}^{\sigma} V_{\sigma}.$$
    I hope this clarifies your confusion about how to handle all the signs from the pseudo-scalar-product of Minkowski four-vector space!
     
  9. Dec 8, 2016 #8
    Cheers! That was very well explained and helped me a lot!
     
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