All right, I know there must be an elementary proof of this, but I am not sure why I never came across it before.

Adding a total time derivative to the Lagrangian (or a 4D divergence of some 4 vector in field theory) does not change the dynamics because the variation can be assumed to be zero on the boundary and integrated away.

But I don't see why any arbitrary function (as long as it is well behaved, no discontinuities, etc.) can't be written as a total derivative (or 4D divergence). In fact, I know that any nice scalar function in 3D can be written as a 3D divergence of some vector field, since for any 3D charge distribution, there exists an electric field whose divergence is equal to the charge function because of Gauss' Law.

But if I can write any function as a total derivative (or divergence of some vector) than I can add any function to the lagrangian and get the same dynamics, which means the lagrangian is completely arbitrary, which makes no sense at all.

So my question is, why can't an arbitrary function (as long as it is well behaved, no discontinuities, etc.) be written as a total derivative of some other function (or divergence of a vector)?


Lagrangian is a functional of time, generalized coordinates, and time derivative of generalized coordinates. Obviously many scalars are not total time derivatives; $q^2$ for example.

As for Lagrangian density, keep in mind that it is the functional of field variables $\phi_i(x^\mu)$ and their derivatives $\partial_\mu \phi_i(x^\mu)$. It is not a composite function of coordinates $\boldsymbol{x^\mu}$. So the arbitrary scalar function, in the form of a divergence, indeed does not matter, because the function is of the coordinates, which is independent from the field variables.

Proof that $q^2$ cannot be rewritten as total time derivative:

The total time derivative of any function $$F(q,\dot q,t)=\frac{\mathrm{d}}{\mathrm{d}t}f(q,t)=\frac{\partial f}{\partial q}\dot{q}+\frac{\partial f}{\partial t},$$

automatically satisfies Euler-Lagrangian equation (easy to prove by substitution) $$\frac{\partial F}{\partial q}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial F}{\partial \dot{q}}=0.$$

$q^2$ does not satisfy the above condition, so it can't be written as total time derivative.

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    $\begingroup$ Why is q squared obviously not a total time derivative of some other function? That is my question. And isn't the Lagrangian density implicitly a function of space and time coordinates since it is a function of fields and the fields are functions of space and time? Anyways I am pretty sure whatever argument will hold for the Lagrangian will hold for the Lagrangian density, so we could limit ourselves to just talking about just a 1 dimensional Lagrangian if we like. $\endgroup$ – David Santo Pietro Mar 12 '12 at 16:27
  • $\begingroup$ @DavidSantoPietro: You can try to find something whose total time derivative is $q^2$ and it won't work. " And isn't the Lagrangian density implicitly a function of space and time coordinates since it is a function of fields and the fields are functions of space and time? " No. $\endgroup$ – Siyuan Ren Mar 12 '12 at 17:00
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    $\begingroup$ I tried to find a total derivative of q squared and couldn't think of one, but that is not a proof. When you say "No", are you saying that the fields are not functions of the coordinates? Besides, what does the fact that the Lagrangian is a function of functions that depend on the coordinates add to the argument anyways? $\endgroup$ – David Santo Pietro Mar 12 '12 at 17:26
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    $\begingroup$ OK, point well taken. I was being sloppy with my wording (functional vs. function), and maybe with my reasoning. But still, I feel that we are going off track here. Forget field theory, this question can be completely posed with a 1D Lagrangian which is a functional of functions of time. $\endgroup$ – David Santo Pietro Mar 13 '12 at 5:22
  • $\begingroup$ Comment to the answer(v2): In conventional physics terminology, the action $S=S[q]$ is a functional, while the Lagrangian $L=L(q,v,t)$ is a function. $\endgroup$ – Qmechanic Mar 15 '12 at 21:03

Let us for simplicity consider just classical point mechanics (i.e. a $0+1$ dimensional world volume) with only one variable $q(t)$. (The generalization to classical field theory on an $n+1$ dimensional world volume with several fields is straightforward.)

Let us reformulate the title(v1) as follows:

Why can't the Lagrangian $L$ always be written as a total derivative $\frac{dF}{dt}$?

In short, it is because:

  1. In physics, the action functional $S[q]$ should be local, i.e. of the form $S[q]=\int dt~L$, where the $L$ is a function of the form $$L~=~L(q(t), \dot{q}(t), \ddot{q}(t), \ldots, \frac{d^Nq(t)}{dt^N};t),$$ and where $N\in\mathbb{N}_{0}$ is some finite order. (In most physics applications $N=1$, but this is not important in what follows. Note that the Euler-Lagrange equations get modified with higher-order terms if $N>1$.)

  2. Similarly, we demand that $F$ is of local form $$F~=~F(q(t), \dot{q}(t), \ddot{q}(t), \ldots, \frac{d^{N-1}q(t)}{dt^{N-1}};t),$$ We stress that $L$ and $F$ only refer to the same time instant $t$. In other words, if $t$ is now, then $L$ and $F$ does not depend on the past nor the future.

  3. The special intermediate role played by the $q$ variable in between $L$ and $t$. Note that there can be both implicit and explicit time-dependence of $L$ and $F$.

Counterexample: Consider

$$L~=~-\frac{k}{2}q(t)^2.$$ Then we can write $L=\frac{dF}{dt}$ as a total time derivative by defining

$$F=-\frac{k}{2}\int_0^t dt'~q(t')^2. $$

($F$ is unique up to a functional K[q] that doesn't depend on $t$.) But $F$ is not on local form as it also depends on the past $t'<t$.

Finally, let us mention that one can prove (still under assumption of locality in above sense plus assuming that the configuration space is contractible, due to an algebraic Poincare lemma of the so-called bi-variational complex, see e.g. Ref. 1) that

$$ \text{The Lagrangian density is a total divergence} $$ $$\Updownarrow$$ $$\text{The Euler-Lagrange equations are identically satisfied}. $$


  1. G. Barnich, F. Brandt and M. Henneaux, Local BRST cohomology in gauge theories, Phys. Rep. 338 (2000) 439, arXiv:hep-th/0002245.
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The reason for this is pretty simple. Let me consider the simple case of one dimensional motion (the generalization being trivial). The Lagrangian is a function of the generalized coordinates and (possibly) time $L=L(q,\dot{q},t)$. The equations of motion are obtained by making the action $S$ extremal

$\delta S[q]=0$,

where $S[q]=\int_{t_{1}}^{t_2} L(q,\dot{q},t) dt$.

Now let's add to the Lagrangian any arbitrary function $G=G(q,t)$ such that


If the function $G(q,t)$ is Riemann-integrable [bounded and continuous except in a set of measure zero] then you can always find such $F(q,t)$. This is case of most of the functions which are interesting for physicists. Hence


$S'[q]=S[q] + \int_{t_1}^{t_2}G(q,t) = S[q] +F(q(t_2),t_2)-F(q(t_1),t_1)$,

since when making a variation we impose that $q(t_1)$ and $q(t_2)$ are fixed. So we find that we have added a constant term to the Lagrangian and therefore

$\delta S'[q]=\delta S[q]$,

so the equations of motion are left invariant.

You say:

"...which means the lagrangian is completely arbitrary, which makes no sense at all..."

Indeed, the Lagrangian function is "arbitrary" in the sense that, respecting certain symmetries, it has to give you the correct equations of the motion when the action functional is extremal.

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  • $\begingroup$ So, are you saying that we can indeed add any well mannered function whatsoever to the Lagrangian and get back the same dynamics? $\endgroup$ – David Santo Pietro Mar 12 '12 at 9:14
  • $\begingroup$ That would not make sense since any potential term could be put in the form of a total derivative, and the free system has to have different dynamics than an interacting system. $\endgroup$ – David Santo Pietro Mar 12 '12 at 9:33
  • $\begingroup$ Basically yes, as long as you had the right Lagrangian at the beginning! Appart from symmetries, the only thing that fixes your Lagrangian are equations of motion. $\endgroup$ – DaniH Mar 12 '12 at 9:37
  • $\begingroup$ @David Santo Pietro: I did not say that any potential term can be written in terms of a total time derivative, below there is an example that there are counterexamples. However, if you are able to write an extra term for your Lagrangian which can be written as a total time derivative the new action will have the same extrema than the old one. $\endgroup$ – DaniH Mar 12 '12 at 13:34
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    $\begingroup$ Below it isn't really a counterexapmle. He is just saying that the harmonic oscillator term couldn't be written as a total derivative cause we "know" the dynamics would have to be different. But that is not a proof. Why can't we write the harmonic interaction term as a total derivative? (other than that it would not make sense since we "know" the interacting theory has to be different) $\endgroup$ – David Santo Pietro Mar 12 '12 at 16:18

I started trying to answer this then found a web site that did a better job than I could, so here it is: http://en.wikibooks.org/wiki/Classical_Mechanics/Lagrange_Theory#Is_the_Lagrangian_unique.3F

This is really a comment, but it got a bit involved to go in a comment box:

Start with the Langrangian for a free particle $L_{Free}$ and add to it a function $G$ defined by:

$$G = L_{SHO} - L_{Free}$$

where $L_{SHO}$ is the Lagrangian for a simple harmonic oscillator. Can $G$ be written as a total time differential. If it can, the action for a free particle isn't changed by adding $G$ to the Lagrangian, and we have to conclude that the action for a free particle is the same as the action for a simple harmonic oscillator. Since this isn't the case that suggests $G$ cannot be written as a total time differential.

The function $G$ is obviously just $-kx^2$. I did try finding a function $F(x, t)$ such that the total time derivative was $-kx^2$ but without any luck, which doesn't necessarily prove anything.

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  • $\begingroup$ I don't normally respond to downvotes, but it seems to me that the link nicely answers David's question. If you think it doesn't please tell me so rather than just do a drive-by downvote. All of us hereabouts are just trying to help. $\endgroup$ – John Rennie Mar 12 '12 at 9:10
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    $\begingroup$ Thanks for the link. It doesn't go as far as I am looking for though. It shows that the Lagrangian is not unique, and gives examples. But I want to know if the Lagrangian is completely arbitrary given that any well mannered function seems like it can be put in the form of a total derivative. (I didn't downvote BTW) $\endgroup$ – David Santo Pietro Mar 12 '12 at 9:17
  • $\begingroup$ Giving an answer that includes only a link without explaining what is being referenced and how it helps answer the question is not really that useful. Links change, they take time to follow, sometimes it's hard to extract the useful information from the target page, and so on. Here's more detail on MSO. In general you can expect link-only answers to be downvoted or deleted (there is some debate on which course of action is appropriate), but you can always edit the answer to include the essentials (even a couple sentences) of the linked content. $\endgroup$ – David Z Mar 12 '12 at 10:13
  • $\begingroup$ Ah, OK. Is it obvious that any function can be written as the total derivative of another function? For example, can the Langrangian for a simple harmonic oscillator be written as the total time derivative of some other function? $\endgroup$ – John Rennie Mar 12 '12 at 11:07
  • $\begingroup$ @John Rennie You have proved by your calculation that not all functions can be written as a total time derivative (one counterexample). Therefore, adding a quadratic potential in your Lagrangian changes the dynamics because it will change the equations of motion. This is not surprising because the condition (total time derivative) to leave the action invariant has not been taken into account. $\endgroup$ – DaniH Mar 12 '12 at 13:30

A function that has only $x$ terms cannot be a total time derivative because, $${dF(x,\dot x,t) \over dt} = \dot x {∂F \over ∂x} +{∂F \over ∂\dot x} \ddot x + {∂F \over ∂t} $$

Here, you can't eliminate the $\dot x$ as long as you choose a $F$ that has only $x$ and no $\dot x$. Because if $F$ the has $x$ then, ${dF\over dt}$ must have $\dot x$.

If you try to eliminate the term by including a $\dot x$ into $F$ you will now have to deal with the $\ddot x$ term now. Similarly as you keep going you'll always have the highest order time derivative sitting at the end of the equation, that you just can't eliminate without including its higher order time derivative.

Since we generally don't see any physical systems that are dependent on time derivatives of order greater than $2$, we can stop here itself. This is just one of the many functions that I believe cannot be written as total derivative of time.

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