I A question about Noether theorem

[larsa]

How can I derive that the work of a force perpendicular to velocity is always zero from the theorem of Noether?
I have heard that there is a relation between these two but in Google I found nothing.

Thank you very much

 

[vanhees71]

For that you don't need Noether's theorem. The usual work theorem will do. The Newtonian EoM reads
$$m \ddot{\vec{x}}=\vec{F}.$$
Now multiply with $\dot{\vec{x}}$, and you get
$$m \dot{\vec{x}} \cdot \ddot{\vec{x}}=\frac{\mathrm{d}}{\mathrm{d} t} \frac{m}{2} \dot{\vec{x}}^2 = \vec{F} \cdot \dot{\vec{x}}.$$
Now you if $\dot{\vec{x}} \perp \vec{F}$ the right-hand side is 0, and thus the kinetic energy is constant, i.e., the force doesn't do work.

 

[fresh_42]

For that you don't need Noether's theorem.

Yes, but the question has been, how they are related? I came so far to see that Noether says: work as a function of force and position ($\;dW = F(\vec{x}) \cdot \nabla \vec{x}\;$) is invariant under orthogonal coordinate transformations. But how is this related to the fact, that an orthogonal force doesn't add work? (I just don't see the argument.)

 

[larsa]

Yes, but the question has been, how they are related? I came so far to see that Noether says: work as a function of force and position ($\;dW = F(\vec{x}) \cdot \nabla \vec{x}\;$) is invariant under orthogonal coordinate transformations. But how is this related to the fact, that an orthogonal force doesn't add work? (I just don't see the argument.)

There must be some relation but i can't imagine any. Thank you for your answer