Relativistic invaraiance of a simple electrodynamic quantity

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ak99
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Homework Statement


Under what conditions is the quantity [tex]E^2+B^2[/tex] relativistically invariant?


Homework Equations


[tex]E\cdot B[/tex] and [tex]E^2-c^2B^2[/tex] are invariant.

(Also Lorentz transformation laws for E and B which I won't type here.)

The Attempt at a Solution



I think you can just subtract multiples [tex]E^2-c^2B^2[/tex] from the given quantity to show that [tex]E^2[/tex] and [tex]B^2[/tex] must also be separately invariant if the given quantity is invariant, which I think happens only if E and B are both zero. But I have been told this is wrong by someone.
 
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(Technical point: You may set c=1 in the above expressions.)
 
I don't think there are any such conditions, except if each is zero.
 
ak99 said:

Homework Statement


Under what conditions is the quantity [tex]E^2+B^2[/tex] relativistically invariant?


Homework Equations


[tex]E\cdot B[/tex] and [tex]E^2-c^2B^2[/tex] are invariant.

(Also Lorentz transformation laws for E and B which I won't type here.)

The Attempt at a Solution



I think you can just subtract multiples [tex]E^2-c^2B^2[/tex] from the given quantity to show that [tex]E^2[/tex] and [tex]B^2[/tex] must also be separately invariant if the given quantity is invariant, which I think happens only if E and B are both zero. But I have been told this is wrong by someone.

E^2 + B^2 is not invariant in general since it is the energy-density of the field and thus transforms as the 00 component of a rank 2 tensor... but... how about rotations?
 
olgranpappy said:
E^2 + B^2 is not invariant in general since it is the energy-density of the field and thus transforms as the 00 component of a rank 2 tensor... but... how about rotations?
It is invariant under rotations.