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, 2013 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 $u=1$. $\endgroup$ Dec 31, 2013 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. $\endgroup$ Dec 31, 2013 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, 2018 at 10:48

1 Answer 1


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 $\ddot{q}=0$.


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