I have recently been wondering, with respect to which quantities (covariant or contravariant) one should vary QFT Lagrangians and whether there is some rule regarding this. Let me give an example which will hopefully clarify what my problem is. Let's take the source-free EM Lagrangian: $$ \mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu} $$ One can now compute: $$ \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}=F^{\nu\mu}\\ \frac{\partial\mathcal{L}}{\partial(\partial^\mu A^\nu)}=F_{\nu\mu}\\ \frac{\partial\mathcal{L}}{\partial(\partial^\mu A_\nu)}=F^{\nu}{}_{\mu}\\ \frac{\partial\mathcal{L}}{\partial(\partial_\mu A^\nu)}=F_{\nu}{}^{\mu} $$ If I want to arrive at the correct equation of motion now, I have to choose $\partial _\mu$ or $\partial ^\mu$ according to how I varied $\mathcal{L}$ (btw: is "varied" the correct word here?). My question is: is there a "correct" way to do this? Can I write the equation of motion in a way, which will work for all Lagrangians without me having to "manually" pick the $\partial$? Most textbooks seem to have: $$ \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}-\frac{\partial\mathcal{L}}{\partial \phi}=0 $$ The example above seems to suggest, that varying w.r.t. a covariant quantity returns a contravariant quantity so to say and vice versa. Is that true in general? I do unfortunately not have any theoretical background on this so far and am being thrown into the cold water.

  • $\begingroup$ Isn't the Lagrangian supposed to be of the form $\mathcal{L}(\phi,\partial_{\mu}\phi)$? In that case you see you should vary wrt. $\partial_{\mu}$ $\endgroup$ – jinawee Dec 27 '13 at 18:57

Let us suppose that we are in a Minkowski (flat) space-time with a diagonal (symmetric) metrics $\eta_{\mu\nu} = Diag(-1, 1, 1, 1)$, and $\eta^{\mu\nu} = Diag(-1, 1, 1, 1)$,

The standard Euler-Lagrange equations for a Lagrangian $\mathcal L(A_\nu, \partial_\mu A_\nu)$, is : $\partial_\mu \dfrac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}-\dfrac{\partial\mathcal{L}}{\partial A_\nu}=0$, because the field variables are $A_\mu$, and its first derivatives $\dfrac{\partial A_\nu}{\partial x^\mu} = \partial_\mu A_\nu$ .

The relation between contravariant ($v^\nu$) and covariant coordinates ($v_\mu$), is simply $v_\mu = \eta_{\mu\nu}v^\nu$, or $v^\nu = \eta^{\nu\mu} v_\mu$. This is the natural relation between "lower" and "upper" indices, and you may generalise it to tensors also, so, for instance your Lagrangian could be written :

$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}= -\frac{1}{4} \eta^{\mu\alpha}\eta^{\nu\beta} F_{\alpha\beta}F_{\mu\nu}$

Now, you may transform the Euler-Lagrange relations in : $\partial_\mu \dfrac{\partial\mathcal{L}}{\partial(\partial_\mu A^\nu)}-\dfrac{\partial\mathcal{L}}{\partial A^\nu}=0$ or $\partial^\mu \dfrac{\partial\mathcal{L}}{\partial(\partial^\mu A_\nu)}-\dfrac{\partial\mathcal{L}}{\partial A_\nu}=0$ or $\partial^\mu \dfrac{\partial\mathcal{L}}{\partial(\partial^\mu A^\nu)}-\dfrac{\partial\mathcal{L}}{\partial A^\nu}=0$, if you want, but this will bring nothing new or useful to you, so it is better to stay with the "standard" expression of the Euler-Lagrange equations.

The "standard" expression of the Euler-Lagrange equations is more natural too, because $A_\mu$ (as a "connnection") has a "natural" covariant nature (like $p_\mu$), and a coordinate $x^\mu$ has a natural contravariant nature, so a derivative relative to $x^\mu$, that is $\partial_\mu$ has a natural covariant nature. This will become important in curved space-times, where the relation between contravariant and covariant quantities is more complex.

  • $\begingroup$ Thanks! Is there an easy way to see, why the field and the momentum are naturally covariant? Or if you can recommend an introductory text on the topic, I'll gladly read it up myself. $\endgroup$ – user35915 Dec 28 '13 at 11:40
  • $\begingroup$ Think to quantum mechanics. In Heisenberg representation, the momentum operator $P_\mu$ is associated to the differential operator $-i \partial_\mu = -i \frac{\partial}{\partial x^\mu}$, so $P_\mu$ is naturally covariant. Think now to the Schrodinger equation in presence of the electromagnetic field, and you know that you have to do the substitution $P_\mu \to P_\mu + e A_\mu$. It is a sign that $A_\mu$ is naturally covariant. The true reason is that $A_\mu$, is mathematically, a connection. $\endgroup$ – Trimok Dec 28 '13 at 12:18
  • $\begingroup$ If you have a matter field coupled to the electromagnetic field (Quantum electrodynamics), the total Lagrangian would be ${L}=\bar \psi(i\gamma^\mu\nabla_\mu -m)\psi \mathcal-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$, where $\nabla_\mu = -i\partial_\mu+ eA_\mu$ corresponds to a covariant derivative. $\endgroup$ – Trimok Dec 28 '13 at 12:18

Varied is the correct word. Finding the equations of motion, as you have, is done by applying calculus of variations to the action and demanding that the variation be an extremum. If you were to do precisely that then you would not encounter choices and the confusion you highlight.

However you have applied the concept at the level of the Lagrangian. That is fine and you will never get anything totally wrong but you may run into little caveats exactly as you did.

Varying the action involves an extra step where you apply integration by parts and that correctly chooses for you the covariant or contravariant derivative based on which type of field you varied with respect to.


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