I am a bit confused with regards to the concept of solid angle.

Why is the solid angle which is defined as $\sin \theta {\rm d}\phi\, {d\rm }\theta$ equal to $\sin\theta\,{\rm d}\theta {\rm d}\phi = {\rm d}\cos\theta{\rm d}\phi$


The identity

$$ \sin\theta\,{\rm d}\theta {\rm d}\phi = {\rm d}\cos\theta{\rm d}{\phi} $$

comes from the fact that

$$ {\rm d}\cos\theta = \frac{{\rm d}\cos\theta}{{\rm d}\theta}{\rm d}\theta = -\sin \theta {\rm d}\theta $$

  • $\begingroup$ but that doesn't account for the negative sign which is what bothered me about that ! $\endgroup$ – David Abraham Apr 27 '18 at 1:03
  • 1
    $\begingroup$ @DavidAbraham Remember that the integral goes accompanied by the absolute value of the Jacobbian, so the minus sign disappears $\endgroup$ – caverac Apr 27 '18 at 1:14

use the following trick $$\frac {d(cos(θ))}{dt}=-\sin(\theta) \frac {d \theta}{dt}$$ multiply by $dt$ $$ {d(cos(θ))}=-\sin(\theta){d \theta}$$


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