Electromagnetic field tensor: contractions with metric tensor

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 4; Problems 4.8, 4.9.

The electromagnetic field tensor {F^{ij}} is

\displaystyle  F^{ij}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (1)

If we contract {F^{ij}} with the metric tensor in flat space, we get

\displaystyle   \eta_{ij}F^{ij} \displaystyle  = \displaystyle  -\eta_{ij}F^{ji}\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  -\eta_{ji}F^{ji}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  -\eta_{ij}F^{ij} \ \ \ \ \ (4)

In the first line, we used {F^{ij}=-F^{ji}}; in the second, {\eta_{ij}=\eta_{ji}} and in the last line, we swapped the dummy indexes {i} and {j}. Thus the original quantity is equal to its negative, so

\displaystyle  \eta_{ij}F^{ij}=0 \ \ \ \ \ (5)

Now consider {\eta_{ia}\eta_{jb}F^{ij}F^{ab}}. First, since {\eta_{00}=-1} and {\eta_{ii}=+1} for {i=1,2,3}:

\displaystyle  \eta_{jb}F^{ij}=F_{\;\; b}^{i}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ E_{x} & 0 & B_{z} & -B_{y}\\ E_{y} & -B_{z} & 0 & B_{x}\\ E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (6)

This is because the only sign change occurs when {\eta_{jb}=-1}, which is when {j=b=0}, so all elements {F^{i0}} change sign.

Then

\displaystyle  \eta_{ia}F_{\;\; b}^{i}=F_{ab}=\left[\begin{array}{cccc} 0 & -E_{x} & -E_{y} & -E_{z}\\ E_{x} & 0 & B_{z} & -B_{y}\\ E_{y} & -B_{z} & 0 & B_{x}\\ E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (7)

This time, the sign change occurs when {i=a=0}, so elements {F_{\;\; b}^{0}} change sign. Thus lowering both indices in flat spacetime in rectangular coordinates changes the sign of all the electric field entries, but leaves the magnetic field unchanged.

Combining them, we get

\displaystyle   \eta_{ia}\eta_{jb}F^{ij}F^{ab} \displaystyle  = \displaystyle  F_{ab}F^{ab}=-F_{ab}F^{ba} \ \ \ \ \ (8)

To work this out, note that if we first formed the matrix product {F_{cb}F^{bd}} , then {F_{ab}F^{ba}} is the sum of the diagonal elements of this matrix product. That is

\displaystyle   -F_{ab}F^{ba} \displaystyle  = \displaystyle  -\left(F_{0b}F^{b0}+F_{1b}F^{b1}+F_{2b}F^{b2}+F_{3b}F^{b3}\right)\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  -\left(E_{x}^{2}+E_{y}^{2}+E_{z}^{2}\right)-\left(E_{x}^{2}-B_{z}^{2}-B_{y}^{2}\right)-\left(E_{y}^{2}-B_{z}^{2}-B_{x}^{2}\right)-\left(E_{z}^{2}-B_{y}^{2}-B_{x}^{2}\right)\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  2B^{2}-2E^{2} \ \ \ \ \ (11)

7 thoughts on “Electromagnetic field tensor: contractions with metric tensor

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  5. Chris Kranenberg

    Glenn,

    Your solution to Problem 4.9 using the lowering of indices, an economical method, is not addressed in Chapter 4. A direct calculation without index manipulation may be more instructive for those who are seeing GR index notation for the first time. Also, Problem 4.11 asks to use index notation; the posted solution uses the vector notation.

    Thank you.

    Reply
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