Consider a Lagrangian $\mathcal{L}$ which is function of, for example, some vector fields $A^\mu$ and tensor fields $B^{\mu\nu}$. That is, \begin{align} \mathcal{L}=\mathcal{L}(A^\mu, B^{\mu\nu}) \end{align} Then I would like to ask that is it possible to derive from such $\mathcal{L}$ different equations of motion which are not consistent by using Euler-Lagrange equation but with regard to different fields?

By "consistent" I mean for example one equation of motion gives \begin{align} A^{\mu}A^{\nu}+B^{\mu\nu}=0 \end{align} while the other equation of motion writes \begin{align} A^{\mu}A^{\nu}+2B^{\mu\nu}=0 \end{align}

If it is possible, then what does it mean? Does it mean that such Lagrangian is a taboo in constructing?

  • $\begingroup$ your eoms are not in consistent the answer is just zero. $\endgroup$
    – physshyp
    Oct 31, 2020 at 23:32
  • $\begingroup$ btw it feels like you may doing something related to do with gauge theories. is your system have anything to do with electromagnetic fields. $\endgroup$
    – physshyp
    Nov 1, 2020 at 3:57
  • $\begingroup$ Yes, what I am doing contains gauge field. And no, the system is not about electromagnetic fields $\endgroup$
    – Chunhui
    Nov 2, 2020 at 3:59
  • $\begingroup$ if its gauge field you should fix gauge i think your issue is something else than the answer suggests. you probably messing up gauge fixing. $\endgroup$
    – physshyp
    Nov 2, 2020 at 11:42

2 Answers 2


Classical mechanics answer

A simple example of this would be to impose constraints that can't be solved simultaneously, using Lagrange multipliers. For example, let's take a particle in 2 spatial dimensions and require that it is simultaneously on a circle of radius $R$ and a circle of radius $2R$

\begin{equation} L = \frac{m}{2} \dot{\vec r}^2 + \lambda_1(|\vec{r}|-R) + \lambda_2(|\vec{r}|-2R) \end{equation} where $\lambda_1,\lambda_2$ are Lagrange multipliers.

I think the interpretation is that your most fruitful course of action is to give up on this Lagrangian and try again.

I think this answers the question that was asked, but I have added a new section below to clarify what happens quantum mechanically based on some interesting discussion in the comments.

What happens quantum mechanically?

We can construct a path integral from this lagrangian as

\begin{equation} Z[J] = \int \mathcal{D} \vec{r} \mathcal{D} \lambda_1 \mathcal{D} \lambda_2 e^{i \left(\frac{m}{2} \dot{\vec r}^2 + \lambda_1(|\vec{r}|-R) + \lambda_2(|\vec{r}|-2R)\right) + \vec{r} \cdot \vec{J}} \end{equation}

Now, we do the integrals over $\lambda_1$ and $\lambda_2$ explicitly by using the following representation of a delta function \begin{equation} \int {\mathcal D} \xi e^{i C \xi} = \delta[C] \end{equation} This is just a functional version of the 1d integral \begin{equation} \int_{-\infty}^{\infty} {\rm d} x e^{i k x} = \delta(k) \end{equation} Anyway, using this identify, the path integral becomes \begin{equation} Z[J] = \int \mathcal{D} \vec{r} \delta\left[|\vec{r}|-R\right] \delta\left[|\vec{r}|-2R\right] e^{i \left(\frac{m}{2} \dot{\vec r}^2 + \vec{r} \cdot \vec{J} \right)} \end{equation}

Since there is no value of $\vec{r}$ that will make the arguments of both delta functionals zero simultanously, the partition function is identically zero, $Z[J]=0$. Therefore, all transition amplitudes are zero. The theory cannot be unitary, since probabilities have to sum to one but all transition amplitudes are zero, so this is not a sensible quantum theory.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – tpg2114
    Nov 1, 2020 at 5:25

It is probably easy to find a Lagrangian producing inconsistent equations of motion. To this end, one can, e.g., start with a complex Lagrangian. It is not obvious that such Lagrangians are necessarily unsatisfactory, because, while they do not seem to make sense in classical mechanics, the relevant path integral in quantum mechanics may be meaningful. Of course, I cannot be sure that such Lagrangians can be useful.

  • $\begingroup$ This kind of thing actually can be useful. For example, in his 2007 Les Houches lectures arxiv.org/abs/hep-ph/0701129, Goldberger derives a complex effective action describing the motion of test particles after integrating out gravitational degrees of freedom. The complex action leads to non-unitary dynamics which capture dissipation due to the emission of gravitational radiation. (see eg Eq 2.18) $\endgroup$
    – Andrew
    Nov 1, 2020 at 0:14
  • $\begingroup$ @Andrew : the article is long, so could you tell me where to find that? $\endgroup$
    – akhmeteli
    Nov 1, 2020 at 1:19
  • $\begingroup$ Eq 2.18 involves the imaginary part of the effective action $\endgroup$
    – Andrew
    Nov 1, 2020 at 1:25
  • $\begingroup$ @Andrew: Thank you. Interesting. $\endgroup$
    – akhmeteli
    Nov 1, 2020 at 1:44
  • $\begingroup$ I did some reading after our discussion on the other comment thread and this imaginary effective action example requires some care. The EOMs associated with the imaginary part of the effective action are not satisfied. However it's also true that if you start with the complex effective action and directly trying to quantize it, you run into problems (the quantum theory won't be unitary). But, it is possible to use the complex action to describe dissipative, quantum systems using the Schwinger-Keyldish formalism. References include hep-th/9501040 and M. Morikawa, Phys. Rev. D33, 3607 (1986). $\endgroup$
    – Andrew
    Nov 1, 2020 at 4:55

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