Explaining the Context of $\nabla\times A=0$ with a Uniform Y-Component of A

[saravanan13]

I came across in a journal that for a \nabla\times A=0 where A is vector, if y-component of that A is uniform then author claims that y-component of curl of a A is zero?
Can anyone explain the above context?

 

[henry_m]

Are you sure that's what's being claimed? Because it isn't true; the y-component of the curl is independent of the y-component of the vector. Maybe there is something extra assumed about A? Can you link to the article or attach the relevant section?

 

[saravanan13]

how to attach the pdf?

 

[saravanan13]

see Eq. (7) and previous statement connecting to that...

 

[henry_m]

The vanishing of the curl is what is assumed here (it's Ampere's law without currents). Then \partial_z H_x =\partial_x H_z is just taking the y-component of the curl. The uniformity of the y-component of H is a separate conclusion, and comes from the assumption that nothing depends on y ("the wave is not modulated transversely"). Then the x- and z-components of the curl give you z- and x-independence of H_y respectively.

Hope that helps.

 

[saravanan13]

Dear henry_m

Please clearly elucidate this statement "Then the x- and z-components of the curl give you z- and x-independence of Hy respectively."

 

[henry_m]

Please clearly elucidate this statement "Then the x- and z-components of the curl give you z- and x-independence of Hy respectively."

Yeah sure, sorry that wasn't very clear.

We're assuming the curl of H vanishes from the physics. The x-component of this says that \partial_y H_z=\partial_z H_y. But H is y-independent by assumption, so \partial_y H_z=0, and hence \partial_z H_y is zero. This means that H_y is independent of z.

If you do the same for the z-component of the curl, you find that H_y is independent of x.

Finally, we're assuming H is independent of y. So H_y is independent of x,y, and z, and hence uniform.