4
$\begingroup$

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:

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

and

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

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ "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? $\endgroup$ – diffeomorphism Jun 3 '15 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 '15 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? $\endgroup$ – diffeomorphism Jun 4 '15 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? $\endgroup$ – diffeomorphism Jun 4 '15 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 '15 at 19:19

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.