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

In summary, the vector α in the integral on the last line is not a polar vector as it is commonly seen, but rather an axial pseudovector whose magnitude gives an angular position in the plane normal to the vector. This is confirmed by the fact that the dot product of the unit vectors for α and z is equal to 1, indicating that they are parallel. Additionally, the virtual "image dipole" is the reflection of the dipole on the yz-plane. However, in the last equation, there is a dα vector that appears in dα⋅N, which may cause confusion. This is actually the axial vector αz, describing the rotation of the dipole around the z-axis. Similarly, angular
  • #1
pherytic
7
0
This brief worked example from a textbook section on the method of images is confusing me.

Untitled.png


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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • Like
Likes Ibix

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

What is a vector?

A vector is a mathematical quantity that has both magnitude (size) and direction. It is typically represented by an arrow pointing in the direction of the vector, with the length of the arrow representing the magnitude.

What is rotational work?

Rotational work is the work done by a force that causes an object to rotate about an axis. It is calculated by multiplying the magnitude of the force by the distance the object moves in a circular path.

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

The vector α, also known as the angle of rotation, is used in the rotational work integral to calculate the work done by a force that is not acting in the same direction as the displacement of the object. It allows us to take into account the angle between the force and the displacement, which affects the amount of work done.

How is the vector α related to the rotational work integral?

The vector α is used in the rotational work integral as part of the equation for calculating the work done by a force. It is typically multiplied by the force and the displacement to account for the angle between them.

Why is it important to consider the vector α in the rotational work integral?

Considering the vector α allows us to accurately calculate the work done by a force that is not acting in the same direction as the displacement of the object. Without taking into account the angle between the force and displacement, the calculated work would not be accurate.

Back
Top