2
$\begingroup$

Why does the inner product between the four force (caused by the electromagnetic field tensor) and the four velocity equaling zero imply that the electromagnetic field tensor is antisymmetric?enter image description here This image is from the textbook General Relativity: An Introduction for Physicists by Hobson, Efstathiou and Lasenby

$\endgroup$
  • $\begingroup$ Please describe where you found this claim and include the formulas that you are referring to. You can use latex formatting. $\endgroup$ – my2cts Feb 20 '19 at 9:25
  • $\begingroup$ @my2cts Oh sorry, I don’t really know how to use latex. I found this claim in General Relativity: An Introduction to Physicists by M. P. Hobson, G. Efstathiou and A. N. Lasenby on page 136. $\endgroup$ – Steve Feb 20 '19 at 9:56
  • 2
    $\begingroup$ $F_{mn}u^m u^n = 0, \; F_{nm}u^n u^m = 0$. Add the two expressions and use the fact that $u^m u^n$ is symmetric $\implies (F_{mn} + F_{nm})u^m u^n = 0 \implies F_{mn} = -F_{nm}$. In general, the product of a symmetric and antisymmetric tensor is zero. $\endgroup$ – GodotMisogi Feb 20 '19 at 10:27
  • $\begingroup$ Perhaps you can update the question now, using all this information. $\endgroup$ – my2cts Feb 20 '19 at 12:23
  • $\begingroup$ And the question remains if the reasoning is circular. $\endgroup$ – my2cts Feb 20 '19 at 12:25
3
$\begingroup$

The premise is that $F_{ab}u^a u^b=0$ for all timelike $u$. I'll abbreviate this as $uFu=0$.

Given any two timelike vectors $x$ and $y$, we can multiply one of them by a non-zero scale factor so that $x+y$ is also timelike. Then the premise implies $$ xFx=0 \hskip2cm yFy=0 \hskip2cm (x+y)F(x+y)=0. \tag{1} $$ This combination of equations implies $$ xFy+yFx=0. \tag{2} $$ Since $x$ and $y$ were arbitrary timelike vectors except for the relative scale factor, and since the relative scale vector doesn't affect equation (2), equation (2) holds for any two timelike vectors.

Any spacelike or lightlike vector $x'$ can be written as the difference of two non-zero timelike vectors, and likewise for any spacelike or lightlike vector $y'$, so equation (2) implies $x'Fy'+y'Fx'=0$ for all vectors $x'$ and $y'$. Since this is true in particular for all of the vectors in some orthogonal basis, this immediately implies that $F$ is antisymmetric.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.