Covariant Electrodynamics and the Field Tensor · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
En anglais

One tensor for E and B

Science · Electricity & Magnetism · ages 22-24
Name ______________________   Date ____________

  1. At an event in S, B=0 and a nonzero E points along +y. S' moves along +x at a nonzero speed below c, with parallel spatial axes. Which fields does S' measure there?

    • Only E': the magnetic field stays zero
    • Both E' and B': each field is nonzero
    • Only B': the electric field becomes zero

  2. In SI, F is antisymmetric: F^{0i}=-Ei/c, F^{12}=-Bz, F^{23}=-Bx, F^{31}=-By (1=x,2=y,3=z). For E/c=(2,0,0) T and B=(0,0,5) T, which matrix represents F? Rows and columns follow (0,x,y,z); semicolons separate rows.

    • [0, 2, 0, 0; -2, 0, -5, 0; 0, 5, 0, 0; 0, 0, 0, 0] T
    • [0, -2, 0, 0; 2, 0, 5, 0; 0, -5, 0, 0; 0, 0, 0, 0] T
    • [0, -2, 0, 0; 2, 0, -5, 0; 0, 5, 0, 0; 0, 0, 0, 0] T

  3. In SI units, J^mu = (c rho, jx, jy, jz) combines charge density and current density into one four-vector.

    Circle one:   True   False

  4. Use SI, x^mu=(ct,x,y,z), J^nu=(c rho,jx,jy,jz), and ∂_mu=∂/∂x^mu. F is antisymmetric with F^{0i}=-Ei/c, F^{12}=-Bz, F^{23}=-Bx, F^{31}=-By (1=x,2=y,3=z). Which equation gives Gauss's electric and Ampere-Maxwell laws for general nonzero sources?

    • ∂_mu F^{mu nu} = mu0 J^nu
    • ∂_mu F^{mu nu} = -mu0 J^nu
    • ∂_mu F^{mu nu} = 0

  5. Using the lesson's conventions, which tensor equation contains both div B=0 and curl E + ∂t B=0?

    • ∂_lambda F_{mu nu} + ∂_mu F_{nu lambda} + ∂_nu F_{lambda mu} = 0
    • ∂_lambda F_{mu nu} - ∂_mu F_{nu lambda} + ∂_nu F_{lambda mu} = 0
    • ∂_lambda F_{mu nu} + ∂_mu F_{nu lambda} - ∂_nu F_{lambda mu} = 0

  6. F is an antisymmetric 4 by 4 tensor. How many independent components does it have?

    Answer: ______________

  7. F is known in S, and the Lorentz matrix Lambda from S to S' is given. Which is a valid method for finding E' and B'?

    • Transform only one tensor index, as for a four-vector
    • Transform both tensor indices, then read off E' and B'
    • Keep the components of F fixed and rename the frame

  8. At an event, S measures E≠0 and B=0. A student claims every nonzero Lorentz boost must give B'≠0. Which conclusion follows from the boost formula?

    • Yes; every nonzero boost creates a magnetic field
    • Yes; a boost can turn pure E into pure B alone
    • No; v parallel to E still gives B' = 0
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Answer key

For grown-ups. Fold this page away before handing over the rest.

One tensor for E and B W1-mt_RoxM9z6DfK-s1

  1. Both E' and B': each field is nonzero · E is transverse to v, so E'=gamma E is nonzero. Also v cross E points along +z, so B' points along -z and is nonzero.
  2. [0, -2, 0, 0; 2, 0, -5, 0; 0, 5, 0, 0; 0, 0, 0, 0] T · Ex/c=2 T gives F^{01}=-2 T. Bz=5 T gives F^{12}=-5 T. Their reversed entries have opposite signs. All remaining entries are zero.
  3. True · The four-current has c times charge density as its time component and the three current-density components as its spatial components.
  4. ∂_mu F^{mu nu} = mu0 J^nu · The sourced equation is ∂_mu F^{mu nu}=mu0 J^nu. Its time component gives div E=rho/eps0; its spatial components give curl B-(1/c²)∂t E=mu0 j.
  5. ∂_lambda F_{mu nu} + ∂_mu F_{nu lambda} + ∂_nu F_{lambda mu} = 0 · The homogeneous equation adds all three cyclic derivatives with plus signs. Spatial indices give div B=0; combinations containing a time index give Faraday's law.
  6. 6 · An antisymmetric n by n tensor has n(n-1)/2 independent entries. For n=4, this is 4×3/2=6.
  7. Transform both tensor indices, then read off E' and B' · F has two tensor indices, so F'=Lambda F Lambda^T. Its transformed time-space and space-space entries give E' and B'.
  8. No; v parallel to E still gives B' = 0 · With B=0, B'=-gamma(v cross E)/c². A nonzero v parallel to E has v cross E=0, so B'=0.
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