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 Exercise [19.09] 
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Joined: 12 Jul 2010, 07:44
Posts: 154
Post Exercise [19.09]
Physically the only relevant (measurable) quantity is F.

Since F=2dA, adding d\Theta to A gives:

A' = A + d\Theta

F' = 2dA' = 2dA + 2d(d\Theta)

... but d(d(anything)) is zero (by the definition of the 'd' operator - see p. 232) so:

F' = 2dA = F

...i.e. adding d\Theta leaves F unchanged!


19 Mar 2012, 07:20
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