I'm considering a symmetry transformation on a Lagrangian

$$ \delta A = \int L(q +\delta q, \dot{q} + \delta \dot{q} , \ddot{q} + \delta \ddot{q}) dt $$

the general variation takes the form

$$ \delta A = \int \frac{ \partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \delta \dot{q} + \frac{\partial L}{\partial \ddot{q}} \delta \ddot{q} dt $$

Now, the second term inside the integral is normally handled as:

$$ \int \frac{\partial L}{\partial \dot{q}} \delta \dot{q} = \frac{\partial L}{\partial \dot{q}} \delta q - \int \frac{\partial}{\partial t}( \frac{\partial L}{\partial \dot{q}} ) \delta q $$

the third terms requires some more work, I'm having it as:

$$ \int \frac{\partial L}{\partial \ddot{q}} \delta \ddot{q} = \frac{\partial L}{\partial \ddot{q}} \delta \dot{q} - \frac{\partial}{\partial t}( \frac{\partial L}{\partial \ddot{q}} ) \delta q + \int \frac{\partial^2}{\partial^2 t}( \frac{\partial L}{\partial \ddot{q}} ) \delta q $$

So my variation (that in the case of symmetry must be zero up to boundary terms is)

$$ \delta A = \int \Big \{ \frac{\partial L}{\partial q} - \frac{\partial}{\partial t}( \frac{\partial L}{\partial \dot{q}} ) + \frac{\partial^2}{\partial^2 t}( \frac{\partial L}{\partial \ddot{q}} ) \Big \} \delta q dt + \Big \{ \frac{\partial L}{\partial \dot{q}} - \frac{\partial}{\partial t}( \frac{\partial L}{\partial \ddot{q}} ) \Big \} \delta q + \frac{\partial L}{\partial \ddot{q}} \delta \dot{q} $$

Now, I'm taking both boundary terms to be conserved currents:

$$ \frac{\partial L}{\partial \dot{q}} - \frac{\partial}{\partial t}( \frac{\partial L}{\partial \ddot{q}} ) $$


$$ \frac{\partial L}{\partial \ddot{q}} $$

But if the second is a conserved current, then its derivative is zero, and the conserved current becomes trivially identical to the first-order case

What the error in my derivation?


1 Answer 1


Comments to the question (v2):

  1. Let there be given a Lagrangian $$\tag{1} L(q,v,a,t), \qquad v^i~:=~\dot{q}^i,\qquad a^i~:=~\dot{v}^i,\qquad \jmath^i~:=~\dot{a}^i, $$ that depends on up to second time derivative.

  2. Let $$\tag{2} \delta q^i~=~\varepsilon Y^i(q,v,a,t) ,$$ be a (global, vertical) quasi-symmetry of the Lagrangian, i.e. there exists a function $f(q,v,a,\jmath,t)$ such that $$\tag{3} \delta L ~=~ \varepsilon \frac{df}{dt}.$$ Here $\varepsilon$ is an infinitesimal constant parameter.

  3. Noether's (first) theorem states that a single quasi-symmetry (3) corresponds to a single on-shell conservation law$^1$ $$\tag{4} \frac{dQ}{dt}~\approx~0. $$

  4. The corresponding (full) Noether charge is in this case $$\tag{5} Q~:=~\left(\frac{\partial L}{\partial v^i} - \frac{d}{dt}\frac{\partial L}{\partial a^i}\right)Y^i +\frac{\partial L}{\partial a^i}\frac{dY^i}{dt} - f. $$

  5. The error in OP's derivation seems to be that it is basically non-existing.


$^1$ Here the $\approx$ symbol denotes equality modulo Euler-Lagrange (EL) equations. Note that in order to have well-defined EL eqs. it is necessary to impose appropriate boundary conditions.

  • $\begingroup$ "The error in OP's derivation seems to be that it is basically non-existing" I don't understand, what you mean by this? what exactly is non-existing? $\endgroup$ Jun 3, 2015 at 16:34
  • $\begingroup$ It means that there doesn't seem to be any proof that both the two boundary terms [which are suggested in the question (v2)] are conserved currents in the first place, partly because the quasi-symmetry has not been explicitly specified, and hence no actual claims to discuss. $\endgroup$
    – Qmechanic
    Jun 3, 2015 at 16:43
  • $\begingroup$ would $f$ (which I understand to be pure boundary terms) be a different function for different symmetry transformations? Is not clear to me how to compute the $f$ that goes at the end of the $Q$ expression in the general case. Would it be different for translations, rotations or time shifts? $\endgroup$ Jun 4, 2015 at 18:52
  • $\begingroup$ I guess my doubt amounts to this: you have to compute the variation of the action in all cases to obtain $f$? will $f$ be simply equal to all the boundary terms on the final variation? $\endgroup$ Jun 4, 2015 at 18:54
  • $\begingroup$ It is not possible to provide a general formula for $f$. It is only possible to calculate $f$ for concretely given Lagrangians and quasi-symmetry transformations. $\endgroup$
    – Qmechanic
    Jun 4, 2015 at 19:19

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