I A question about Noether theorem

AI Thread Summary
The discussion centers on the relationship between Noether's theorem and the work done by a force perpendicular to velocity. It is established that the work theorem suffices to show that when the force is perpendicular to the velocity, the work done is zero, indicating that the kinetic energy remains constant. Participants express confusion about how this concept relates to Noether's theorem, which addresses the invariance of work under orthogonal transformations. Despite the acknowledgment of a potential connection, the exact relationship remains unclear to the contributors. The conversation highlights the need for further exploration of how these principles interconnect in classical mechanics.
larsa
Messages
47
Reaction score
2
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
 
Physics news on Phys.org
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.
 
  • Like
Likes larsa
vanhees71 said:
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.)
 
  • Like
Likes larsa
fresh_42 said:
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
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'A scenario of non-uniform circular motion'
(All the needed diagrams are posted below) My friend came up with the following scenario. Imagine a fixed point and a perfectly rigid rod of a certain length extending radially outwards from this fixed point(it is attached to the fixed point). To the free end of the fixed rod, an object is present and it is capable of changing it's speed(by thruster say or any convenient method. And ignore any resistance). It starts with a certain speed but say it's speed continuously increases as it goes...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top