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I A question about Noether theorem

  1. Feb 21, 2017 #1
    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
  2. jcsd
  3. Feb 21, 2017 #2


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    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.
  4. Feb 21, 2017 #3


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    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.)
  5. Feb 21, 2017 #4
    There must be some relation but i can't imagine any. Thank you for your answer
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