I What is the purpose of the vector α in the rotational work integral?

AI Thread Summary
The discussion centers on confusion regarding the vector α in the context of the method of images. Participants clarify that α is not a traditional polar vector but rather an axial pseudovector representing angular position relative to the z-axis. The integral's last line introduces a term dα, which appears to be a vector, leading to further debate about its representation. The consensus suggests that α should be understood as describing the rotation of the dipole around the z-axis. Overall, the conversation emphasizes the distinction between traditional vector representations and the specific use of pseudovectors in this context.
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This brief worked example from a textbook section on the method of images is confusing me.

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Specifically I am confused about the vector α in the integral on the last line.

When α (or θ) is an angle, I've only ever seen the vector quantity α (or θ) as a polar vector in the plane. But here, that can't be right because the dot product of the unit vectors α ⋅ z = 1, so they are clearly parallel.

What is the correct understanding of the vector α?

It seems like maybe it is supposed to be an axial pseudovector whose magnitude gives a sort of angular position in the plane normal to the vector. But I don't think I have ever seen a vector used this way and I am not sure it makes sense.
 
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It's not a vector but the angle of your real dipole with the negative direction of the ##x##-axis, as written in the solution. The virtual "image dipole" is the reflection of the dipole on the ##yz## plane.
 
vanhees71 said:
It's not a vector but the angle of your real dipole
I agree about ##\alpha##, but in the last equation there is a ##d\mathbf{\alpha}## that appears in ##d\mathbf{\alpha}\cdot\mathbf{N}##, which certainly looks like a vector.

I must say I don't recall seeing an angle represented that way, but since angular velocity is a pseudovector presumably ##\mathbf{\omega}\Delta t## is one too.

Note: the alpha in ##d\mathbf{\alpha}## is supposed to be a mathbf alpha, but at least for me it doesn't render noticeably different from a regular alpha. Ditto the omega.
 
Sorry, I've not read the solution carefully enough. It's of course the axial vector ##\alpha \hat{z}##, describing the rotation of the dipole around the ##z##-axis.
 
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