According to Wikipedia, Noether's theorem (for the mechanics of a point particle) says that if the following transformation is a symmetry of the Lagrangian

$$t \to t + \epsilon T$$

$$q \to q + \epsilon Q$$

Then the following quantity is conserved

$$\left( \frac{\partial L}{\partial \dot{q}}\dot{q} - L \right) T - \frac{\partial L}{\partial \dot{q}} Q$$

Now at this point we almost always consider either $T=1$ or $T=0$ --- we might consider some interesting transformation of the spatial co-ordinate, such as $\vec{Q} = \vec{n} \times \vec{q}$ for spatial rotations, but we rarely consider some interesting transformation of time.

Suppose our Lagrangian is given by

$$L = \frac{1}{2}m\dot{q}^2$$

i.e. a simple kinetic Lagrangian. Then can we not make the transformation

$$t \to t' = t + \epsilon t = (1+\epsilon)t$$

$$q \to q' = q + \epsilon q = (1+\epsilon)q$$

i.e. $T=t$ and $Q=q$. This is the simplest example of a time transformation I could think of that wasn't the trivial $T=1$ or $T=0$. Then I would argue that our Lagrangian is invariant under this transformation, since

$$\frac{d q'}{d t'} = \frac{d q'}{d q}\frac{d q}{d t}\frac{d t}{d t'} = (1+\epsilon) \frac{dq}{dt}(1+\epsilon)^{-1} = \frac{dq}{dt}$$

and so in the new co-ordinates, we have the same Lagrangian. Then from the expression at the top of this post, the quantity

$$\left(\frac{1}{2}m\dot{q}^2\right)t - (m\dot{q})q$$

should be conserved. We can trivially show it isn't, however.

Where is my error?

  • 2
    $\begingroup$ Related by OP: physics.stackexchange.com/q/91808 $\endgroup$ – Kyle Kanos Dec 30 '13 at 21:40
  • $\begingroup$ The derivation in Wikipedia does not hold in this case. You should compute the derivative with respect to the scaling $u=1+\epsilon$ of the action $$I(u)=\int_{t_1}^{t_2}L(uq(ut),\dot q(ut),ut)u\,\text dt$$ and try to reduce the integrand to a total derivative. (One useful such is $\frac{d}{dt}(tL)$.) It's also important to note that even for a free particle $I'(u)\neq 0$ when $u=1$. $\endgroup$ – Emilio Pisanty Dec 31 '13 at 0:30
  • $\begingroup$ I get the equation $$\frac{dI}{du}(1)=\left[tL+q\frac{\partial L}{\partial \dot q}\right]_{t_1}^{t_2}-\int_{t_1}^{t_2} \dot q\frac{\partial L}{\partial \dot q}\text dt,$$ but that's about as far as it goes as far as I can see. $\endgroup$ – Emilio Pisanty Dec 31 '13 at 0:33
  • $\begingroup$ Various transformations to prove energy conservation via Noether's theorem is discussed in my Phys.SE answer here. $\endgroup$ – Qmechanic Dec 26 '18 at 10:48

Noether's theorem requires that the action is to be invariant under the transformations and not the Lagrangian. For the transformations that change the measure of integration $dt$ this is different from invariant Lagrangian.

If we want to demand the invariance of the action $$ I = \int \frac{m \dot{q}}{2}\, dt $$ under the transformation which includes rescaling of time, the correct transformation for this case would be $$ t\to t' = t+\epsilon t,\qquad q\to q' = q+\epsilon \frac q2,$$ (notice the factor $1/2$ for the $q$, since we have to compensate only one $(1+\epsilon)$ multiplier in the action by the rescaling of $q$).

Using the wikipedia's definition for the Noether's conserved quantity we get: $$ A = \frac{m \dot{q}^2}2 t - \frac{m \dot{q} q}{2}.$$ It is clearly conserved on equations of motion $\ddot{q}=0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.