# Direction of an infinitesimal rotation?

I had a question from the magnetic dipole thread that was posted earlier today, but it's a bit more mundane. The torque on a magnetic dipole, using a right handed cross product is ##\vec{\tau} = \vec{\mu} \times \vec{B}##. The work done during a rotation is $$W = \int \vec{F} \cdot d\vec{r} = \int \vec{F} \cdot d\vec{\theta} \times \vec{r} = \int \vec{r} \times \vec{F} \cdot d\vec{\theta} = \int \vec{\tau} \cdot d\vec{\theta}$$Now the result of this depends on which direction we take the vector ##d\vec{\theta}## to be in, and there is only one correct result for work! We know it has to be parallel to the axis of rotation, but get the right answer ##d\vec{\theta}## also has to be in the opposite direction to ##\vec{\tau}##, i.e. if ##\vec{\tau} = \tau \hat{z}## then ##d\vec{\theta} = -d\theta \hat{z}##.

My question is, in the general case, how is the direction of the infinitesimal rotation determined, out of the two possible choices? I apologise if I am missing something obvious.

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Mentor
Dont you still follow the right hand rule? And the fact that the integral goes from one value of theta to and another value of theta.

When no limits are shown we assume the integral is positive and that the thetas range from low to high Ie counterclockwise.

Calling @fresh_42

• etotheipi
Dont you still follow the right hand rule? And the fact that the integral goes from one value of theta to and another value of theta.

It's difficult for me to tell . If we take ##d\vec{\theta}## in the opposite direction to the torque then we get $$W = \int_{\theta_1}^{\theta_2} -\mu B \sin{\theta} d\theta$$and I suppose this choice of direction for ##d\vec{\theta}## would correspond to measuring the angle ##\theta## from ##\vec{B}## to ##\vec{\mu}##, as opposed to from ##\vec{\mu}## to ##\vec{B}##.

• vanhees71