Answer to Question #90877 in Field Theory for Vasiliy

Question #90877
How does F^{mu nu} tensor is connected with the E and B vectors?
1
Expert's answer
2019-06-18T08:09:00-0400

By definition 


"F^{\\mu\\nu}=\\partial^\\mu A^\\nu-\\partial^\\nu A^\\mu"

where (c=1)


"\\partial^\\mu=(\\frac{\\partial}{\\partial t},-\\nabla)\\,,\\quad\n\\nabla=(\\frac{\\partial}{\\partial x},\\frac{\\partial}{\\partial y},\\frac{\\partial}{\\partial z})"

Connection between the vectors "\\vec{E}" , "\\vec{B}" and 4-vector "A^\\mu=(A^0,\\vec{A})" is


"\\vec{E}=-\\nabla A^0-\\partial^0\\vec{A}\\,,\\quad\n\\vec{B}=\\operatorname{curl}\\vec{A}=\\nabla\\times\\vec{A}"

So

"F^{0i}=\\partial^0 A^i-\\partial^i A^0=\\partial^0 A^i+\\nabla_i A^0=-\\vec{E}_i\\,,\\quad i=1,2,3"

"F^{12}=\\partial^1 A^2-\\partial^2A^1=-\\partial_x A_y+\\partial_y A_x=-[\\nabla\\times\\vec{A}]_z=-{B}_z"

"F^{13}=\\partial^1 A^3-\\partial^3A^1=-\\partial_x A_z+\\partial_z A_x=[\\nabla\\times\\vec{A}]_y={B}_y""F^{23}=\\partial^2 A^3-\\partial^3A^2=-\\partial_y A_z+\\partial_z A_y=-[\\nabla\\times\\vec{A}]_x=-{B}_x"

Taking into account that "F^{\\mu\\nu}=-F^{\\nu\\mu}" and "F^{\\mu\\mu}=0" we have

"F^{\\mu\\nu}=\n\\left(\n \\begin{array}{cccc}\n 0&-E_x&-E_y&-E_y\\\\\n E_x&0&-B_z&B_y\\\\\n E_y&B_z&0&-B_x\\\\\nE_z&-B_y&B_x&0\\\\\n \\end{array}\n\\right)"


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