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Electromagnetic Field Tensor in Curvilinear Coordinates

  1. Jul 6, 2012 #1
    How to express electromagnetic field tensor in curvilinear coordinates, that is given a curvilinear coordinates [itex](t,\alpha,\beta,\gamma) [/itex]with metric tensor as follows:
    [tex]
    n_{\mu \nu }=
    \left[ \begin{array}{cccc}h_0^2& 0 & 0 & 0 \\ 0 & -h_1^2 & 0 & 0 \\ 0 & 0 & -h_2^2 & 0 \\ 0 & 0 & 0 & -h_3^2 \end{array} \right]
    [/tex]

    How do we express electromagnetic field tensor [itex]F_{\mu \nu}[/itex] in terms of [itex]E_\alpha , E_\beta , E_\gamma , B_\alpha , B_\beta , B_\gamma[/itex]

    I found in the internet that the [itex]F_{\mu \nu}[/itex] is given by:
    [tex]
    F_{\mu \nu }=
    \left[ \begin{array}{cccc} 0 & -\frac{E_{\alpha}}{h_0 h_1} & -\frac{E_{\beta}}{h_0 h_2} & -\frac{E_{\gamma}}{h_0 h_3} \\ \frac{E_{\alpha}}{h_0 h_1} & 0 & \frac{B_{\gamma}}{h_1 h_2} &-\frac{B_{\beta}}{h_3 h_1} \\\frac{E_{\beta}}{h_0 h_2} & -\frac{B_{\gamma}}{h_1 h_2} & 0 & \frac{B_{\alpha}}{h_2 h_3} \\\frac{E_{\gamma}}{h_0 h_3} & \frac{B_{\beta}}{h_3 h_1} & -\frac{B_{\alpha}}{h_2 h_3}& 0 \end{array} \right]
    [/tex]

    Is it correct and how to derive it?

    Thanks.
     
    Last edited: Jul 6, 2012
  2. jcsd
  3. Jul 6, 2012 #2
    This is easily done using tetrads. There is a tetrad matrix [itex]{h_a}^b[/itex] that converts back and forth between the tetrad basis (which lives in a flat space) and the coordinate basis.

    Let [itex]\tilde x^b = {h_a}^b x^a[/itex], where the tilde denotes a coordinate component, instead of a tetrad component. You can see that [itex]{h_0}^0[/itex] is your [itex]h_0[/itex] and so on.

    The Faraday tensor in the tetrad basis is just the usual smattering of electric and magnetic field components:

    [tex]F_{ab} = \begin{bmatrix}
    0 & -E_1 & -E_2 & -E_3 \\
    E_1 & 0 & B_3 & -B_2 \\
    E_2 & -B_3 & 0 & B_1 \\
    E_3 & B_2 & -B_1 & 0
    \end{bmatrix}[/tex]

    Just act on the tetrad EM tensor [itex]F_{ab}[/itex] with the tetrad field to convert it into the coordinate basis: [itex]\tilde F_{cd} = {h^a}_c {h^b}_d F_{ab}[/itex]. Mind all the usual subtleties of index notation--[itex]{h^a}_b[/itex] is the inverse transpose of [itex]{h_a}^b[/itex]. Or I think it is. I'm not too confident when it comes to index notation myself. The notation is easier (in my mind) to understand in a geometric algebra form, as is the physical nature of the EM tensor.
     
  4. Jul 7, 2012 #3
    It's a little unclear what you are sayin/asking.

    The components of the Fuv tensor still match E and B the same way would in a Minkowski space, so that formula is not correct for what you are actually asking.

    What seems to be going on is they are saying that if you right the Fuv tensor in terms of the E's and B's for a Minkowski situation, if you make a coordinate transformation to a situation where the metric is what you gave, the new Fuv tensor in terms of the *old* E's and B's (not the new E's and B's) will be of the form they gave.

    There actually is still a problem in that the coordinate transformation to go for one metric to another is not unique. There is a degeneracy because the Lorentz transformation (and its generalization for non-Minkowskian situations) leaves the metric unchanged. So for example, suppose we have a situation where we have a Minkowski metric, and just an electric field with no magnetic field. Suppose we made a Lorentz Transformation (boost). We now have a different Fuv because there is now a magnetic field. But the metric is unchanged. So if, for example, you make a transformation from a situation with a non-Minkowski metric to one where the metric is Minkoskian, the Fuv is not really uniquely determined, because different coordinate transformations (resuling in different Fuv tensors) can produce the same Minkoski metric endpoint.
     
  5. Jul 7, 2012 #4
    Thanks. But I am new to this tetrad thing. How to calculate the [itex]{h_a}^b[/itex] actually? From [itex]\tilde x^b = {h_a}^b x^a[/itex] I would think that [itex]{h_a}^b[/itex] is a transformation matrix which seems to be incorrect. Can you give me an example how to calculate the [itex]{h_a}^b[/itex], for example in cylindrical coordinates.

    Thanks.
     
    Last edited: Jul 7, 2012
  6. Jul 7, 2012 #5
    I am refering to this paper at http://arxiv.org/pdf/gr-qc/0409025.pdf . In the paper, the author write out the field tensor in spherical coordinates in terms of [itex]E_r , E_\theta E_\phi , B_r , B_\theta B_\phi[/itex]. I wonder how the author derived that and how to write it in more general curvilinear coordinates.
     
  7. Jul 7, 2012 #6
    The tetrad field obeys [itex]g_{ab}={h_a}^c h_{cb}[/itex]. This makes it like a square root of the metric. Ideally, you'd be given the tetrad field instead of the metric, but in this case, your metric is diagonal, so the easiest tetrad field to pick is just the square root of the metric's corresponding elements.

    To compute the tetrad field would require some information about the geometry. Do you mean compute the flat space tetrad for some curvilinear coordinates?
     
  8. Jul 7, 2012 #7
    Muprid, there is not a unique tetrad to go from one metric to a new metric in a new coordinate system. There are an infinite number of tetrasds that will do it, each differing from eachother by a (generalized) Lorentz Transformation.
     
  9. Jul 7, 2012 #8
    Indeed, the metric in invariant under generalized rotations while the tetrad isn't.
     
  10. Jul 7, 2012 #9
    ngkamsengpeter, the force law is

    m (d2xu/ds2) [u index is up] = q Fuv [u index up and v index down] Jv [ index is up]

    It appears that he is writing it out in a way that makes it glaring that curvilinear coordinates are used by putting it in the form

    m (d2xu/ds2) [u index is up] = q Fus [u index up and s index up] gsv [both indices down] Jv [ index is up]

    and then is putting the two terms q Fus [u index up and s index up] gsv [both indices down] together.

    I don't think it is a good way to express it.
     
  11. Jul 7, 2012 #10
    How to derive this equation [itex]g_{ab}={h_a}^c h_{cb}[/itex]? Yes. In flat space, how to write the [itex]F_{\mu \nu}[/itex] in curvilinear coordinates. Given the equation [itex]\tilde F_{cd} = {h^a}_c {h^b}_d F_{ab}[/itex], how to compute the flat space tetrad for some curvilinear coordinates? For example, in flat space cylindrical coordinates, the metric tensor is diag(1,-1,-r^2,-1), so the tetrad is just square root of this metric tensor?

    And I am not quite understand what ApplePion is saying, can anyone of you explain in more details?

    Thanks.
     
  12. Jul 7, 2012 #11
    Okay, so there are a couple things going on here.

    I won't prove the relation between metric and tetrad; it's a bit involved, really.

    Calculating the flat space tetrad is easy. Take the equations that convert from your coordinates to Cartesian and find the Jacobian matrix. This is the tetrad field in flat space.
     
  13. Jul 8, 2012 #12
    Let me try an example in cylindrical coordinates, the jacobian matrix I found for transform [itex] (t,x,y,z)[/itex] to [itex] (t, r, \theta,z)[/itex] is
    [tex]
    J = \begin{bmatrix}
    0 & 0 & 0 & 0 \\
    0 & cos \theta& -r sin \theta & 0 \\
    0 &sin \theta & r cos \theta & 0 \\
    0 & 0 & 0 & 0
    \end{bmatrix}
    [/tex]

    So J is the tetrad field? But how to apply to [itex]\tilde F_{cd} = {h^a}_c {h^b}_d F_{ab}[/itex]? I cannot get rid of the cos and sin term.

    Thanks.
     
  14. Jul 8, 2012 #13
    Hm, I think I see now that this is really just a problem about going between the coordinate basis and the orthonormal basis. The only reason the factors of [itex]h_0, h_1, \ldots{}[/itex] need to appear in your EM tensor is if the electric and magnetic fields are components corresponding to unit vectors where the EM tensor's components do not. Is that what you're trying to do?
     
  15. Jul 8, 2012 #14
    What I am trying to do is reproduce the field tensor in spherical coordinates shown in the paper http://arxiv.org/pdf/gr-qc/0409025.pdf at equation 15 which is:
    [tex]
    F_{ab} = \begin{bmatrix}
    0 & E_r & r E_\theta & r sin \theta E_\phi \\
    -E_r & 0 & -r B_\phi & r sin \theta B_\theta \\
    -r E_\theta &r B_\phi& 0 & -r^2 sin \theta B_r \\
    -r sin \theta E_\phi & -r sin \theta B_\theta & r^2 sin \theta B_r & 0
    \end{bmatrix}
    [/tex]

    Yes. I think it is just the problem of changing coordinate basis but I am new to this so dont exactly know how to do this. What I am trying to do is reproduce the field tensor in spherical coordinates as shown above and also in cylindrical coordinates.

    Thanks.
     
  16. Jul 8, 2012 #15
    Yeah, I think this has to do with the difference between how we usually do things in vector calculus--using unit vectors--with how things are done in GR, using non-unit coordinate basis vectors because things are generalized. Really, all you're seeing here is the difference between the two. For example, look at the [itex]rE_\theta[/itex] component. This arises just because the coordinate basis vector [itex]g^\theta[/itex] has magnitude [itex]1/r[/itex]. [itex]E_\theta[/itex] describes the [itex]\hat g^\theta = r g^\theta[/itex] component of the electric field, but the EM tensor expects the [itex]g^\theta[/itex] component.

    This is my distinct impression, at least. I too would appreciate any clarification others can provide.
     
  17. Jul 9, 2012 #16

    Mentz114

    User Avatar
    Gold Member

    The problem is that changing from rectilinear coordinates to spherical polar coords is not a linear transformation. To express Fmn in new coordinates might require going back to
    [tex]
    F_{mn}=\partial_{m}A_n-\partial_{n}A_m
    [/tex]
    transforming the potential, and using the appropriate differential operators.

    [Edit]
    I had occassion to find F in cylindrical coords (t,z,r,∅) some time ago, and I see that for A0 being a function of z,r,∅ I used the operator (1/r2) ∂. This F gave the correct EMT.
     
    Last edited: Jul 9, 2012
  18. Jul 9, 2012 #17
    ######################
    The problem is that changing from rectilinear coordinates to spherical polar coords is not a linear transformation. To express Fmn in new coordinates might require going back to

    F mn =∂ m A n −∂ n A m

    transforming the potential, and using the appropriate differential operators
    ##########################

    That seems a bit more complicated than necessary. Rather than doing it that way, one should just use the rule for transforming a second rank tensor.
     
  19. Jul 9, 2012 #18
    Yeah, this is a moment where I find index notation really cumbersome. Using traditional vector notation is eschewed because it lacks an obvious way to talk about tensors, but there's a way around that, and I think it makes the physical content of GR a lot clearer to see.

    Let there be an algebra of unit bivectors in Minkowski space, denoted [itex]e_a \wedge e_b[/itex]--meaning the plane spanned by the vectors [itex]e_a[/itex] and [itex]e_b[/itex]. This wedge product generates higher ranking tensors from vectors in this general fashion.

    In a local frame, we can write the EM tensor as [itex]F = F_{01} e^0 \wedge e^1 + F_{02} e^0 \wedge e^2 + F_{03} e^0 \wedge e^3 + F_{12} e^1 \wedge e^2 + F_{23} e^2 \wedge e^3 + F_{31} e^3 \wedge e^1[/itex]. The tetrad field is called [itex]\overline h[/itex] and converts the local tetrad basis vectors to the coordinate basis vectors, which are called [itex]g_a[/itex]. So [itex]\tilde F = F_{01} g^0 \wedge g^1 + \ldots{}[/itex]. The only issue here is that the vectors [itex]g^a[/itex] are not necessarily unit. In spherical coordinates, we know some of them aren't, but their magnitudes can be easily found.

    This question is one entirely of the correspondence between the measured electric and magnetic fields and the convention used to describe the components of the Faraday bivector. It seems to me that what's expected are the components corresponding to the (not necessarily unit) coordinate basis.

    In short, one traditionally writes the electric and magnetic fields in terms of unit vectors even in curvilinear coordinates. That is, [itex]E = E_{\hat r} g^{\hat r} + E_{\hat \theta} g^{\hat \theta} + E_{\hat \phi} g^{\hat \phi}[/itex]. I think this is the issue: the Faraday bivector expects components in terms of [itex]g^a[/itex], not [itex]g^{\hat a}[/itex].

    To me, this became obvious to see when I could shuffle the normalization factor around between the components and the basis vectors. In index notation, it would've required what, an additional change of basis matrix and a lot of extra indices to introduce? Very yucky, to me, but I know I do things differently from most.
     
  20. Jul 9, 2012 #19
    I am not quite understand what are you saying. Can you suggest some good references or some books that I can refer to on the tetrads thing?

    Besides that, another question just pop into my mind, how about the EM field tensor in some medium? Is it I just replace the B and E by H and D?

    Thanks.
     
  21. Jul 9, 2012 #20

    Mentz114

    User Avatar
    Gold Member

    Just busking, but this looks promising

    from

    x = r sin(θ) cos(∅)
    y = r sin(θ) sin(∅)
    z = r cos(θ)

    we get

    dx = sin(θ) cos(∅) dr + r cos(θ) cos(∅) dθ - r cos(θ) sin(∅) d∅
    dy = sin(θ) cos(∅) dr + r cos(θ) cos(∅) dθ + r cos(θ) cos(∅) d∅
    dz = cos() dr - r sin(θ) dθ

    which is a linear transformation of the differential basis. If these terms are arranged into a tetrad L, it looks as if the transformation L.F.transpose(L) will get the F of equation (15)
     
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