The results I am trying to derive can be found in this paper, appendix B. In class I have only ever dealt with actions that involve a single scalar field so dealing with actions of this form is quite unfamiliar to me. For some context we are in flat $(2+1)$ dimensional Minkovski space and have the following set of totally symmetric bosonic fields $\varphi^i=\{h_{\alpha(2s)},h_{\alpha(2s-4)}\}$ where $s>1\in\mathbb{Z}$ is the spin of the massless field. The notation $h_{\alpha(2s)}\equiv h_{\alpha_1\cdots\alpha_{2s}}=h_{({\alpha_1\cdots\alpha_{2s}})}$ is shorthand for a totally symmetric field with $2s$ spinor indices (i.e. $\alpha=1,2$).

I am trying to derive the equations of motion (EoM's) for both the integer and half integer spin cases. Surprisingly, I am able to derive all 5 equations, however, in order to get them I fudge the method. The fudge works consistently, but I don't know why. I am almost certainly doing this in a very incorrect way, so I would like someone to point out the right method. I will only show how I get one of the EoM's (equation B.4a of the integer spin case in the paper), the rest go in a very similar way. The relevant action is $$\begin{align}\mathcal{S}_s[\varphi^i]=\frac{1}{2}(-\frac{1}{2})^s\int\text{d}^3x\bigg\{h^{\alpha(2s)}\square h_{\alpha(2s)}-\frac{1}{2}s(\partial^{\beta(2)}h_{\beta(2)\alpha(2s-2)})^2-(s-1)(2s-3)\bigg[\\ h^{\alpha(2s-4)}\partial^{\beta(2)}\partial^{\gamma(2)}h_{\beta(2)\gamma(2)\alpha(2s-4)}+4\frac{s-1}{s}h^{\alpha(2s-4)}\square h_{\alpha(2s-4)}+\frac{1}{2}(s-2)(2s-5)(\partial^{\beta(2)}h_{\alpha(2s-6)\beta(2)})^2\bigg]\bigg\}.\end{align} \tag{B.3}$$

Here $\square=\partial^a\partial_a=-\frac{1}{2}\partial^{\beta(2)}\partial_{\beta(2)}$ and notation of the following form means $(\partial^{\beta(2)}h_{\beta(2)\alpha(2s-2)})^2= (\partial^{\beta(2)}h_{\beta(2)\alpha(2s-2)})(\partial_{\gamma(2)}h^{\gamma(2)\alpha(2s-2)})$. It seems to me that one obtains the two given EoM's by varying each field $\varphi^i$ independently. So I will obtain the EoM which corresponds to a variation in $\varphi^1$, for which case, thankfully, the 4th and 5th terms of the integrand do not contribute. So I only need to consider the first three terms, I will do each term seperately and label them $I_i=I_1,I_2,I_3$ for ease of reference. The first fudge comes at the very beginning; $$\delta\big[I_1\big]=\delta\big[h^{\alpha(2s)}\square h_{\alpha(2s)}\big]=\delta h^{\alpha(2s)}\square h_{\alpha(2s)}+h^{\alpha(2s)}\square \delta h_{\alpha(2s)}=2\delta h^{\alpha(2s)}\square h_{\alpha(2s)}.$$ I don't know why(/if) the last equality should hold, but doing the same fudge works for all 5 EoM's. Now for the next term; $$\delta\big[I_2\big]=\delta\big[-\frac{1}{2}s(\partial^{\beta(2)}h_{\beta(2)\alpha(2s-2)})^2\big]=-s(\partial^{\beta(2)}h_{\beta(2)\alpha(2s-2)})(\partial_{\gamma(2)}\delta h^{\gamma(2)\alpha(2s-2)})$$ Here comes the next (and last) fudge, I think the idea is right, but I am not executing it in a rigorous way, at all. The idea is to integrate by parts and kill off the variation at the boundary; $$\begin{align}\int\text{d}^3x\delta[I_2]&=-s\underbrace{\big[\partial^{\beta(2)}h_{\beta(2)\alpha(2s-2)}\delta h^{\gamma(2)\alpha(2s-2)}\big]^{x_f}_{x_i}}_{=~0, \text{variation vanishes at boundaries}}+s\int\text{d}^3x(\partial_{\gamma(2)}\partial^{\beta(2)}h_{\beta(2)\alpha(2s-2)})(\delta h^{\gamma(2)\alpha(2s-2)})\\ &=s\int\text{d}^3x\delta h^{\alpha(2s)}\partial^{\beta(2)}\partial_{\alpha(2)}h_{\beta(2)\alpha(2s-2)}. \end{align}$$ In the second line I have just relabeled and rearranged indices appropriately. To me, the term being evaluated at the boundaries doesn't really make sense since this is the only place so far where there are free indices, but I couldn't get around it. I have read that in such situations it is possible to use stokes/gauss' theorem and assume the fields vanish at the boundary, or something like this. The third term involves both fields $\varphi^1$ and $\varphi^2$, but I think I am justified in only varying the $\varphi^1$ field; $$\begin{align}\delta\big[I_3\big]&=\delta\big[-(s-1)(2s-3)h^{\alpha(2s-4)}\partial^{\beta(2)}\partial^{\gamma(2)}h_{\beta(2)\gamma(2)\alpha(2s-4)}\big]\\ &=-(s-1)(2s-3) h^{\alpha(2s-4)}\partial^{\beta(2)}\partial^{\gamma(2)}\delta h_{\beta(2)\gamma(2)\alpha(2s-4)}. \end{align}$$
Now I integrate by parts twice, and assume that not only the variations at the boundary vanish, but also the derivative of the variations vanish at the boundary (another fudge?); \begin{align} \int\text{d}^3x\delta I_3 &=-(s-1)(2s-3)\int\text{d}^3x h^{\alpha(2s-4)}\partial^{\beta(2)}\partial^{\gamma(2)}\delta h_{\beta(2)\gamma(2)\alpha(2s-4)}\\ &=-(s-1)(2s-3)\bigg\{\underbrace{\big[h^{\alpha(2s-4)}\partial^{\gamma(2)}\delta h_{\beta(2)\gamma(2)\alpha(2s-4)}\big]^{x_f}_{x_i}}_{=~0}-\int\text{d}^3x\partial^{\beta(2)}h^{\alpha(2s-4)}\partial^{\gamma(2)}\delta h_{\beta(2)\gamma(2)\alpha(2s-4)}\bigg\}\\ &=(s-1)(2s-3)\bigg\{\underbrace{\big[\partial^{\beta(2)}h^{\alpha(2s-4)}\delta h_{\beta(2)\gamma(2)\alpha(2s-4)}\big]^{x_f}_{x_i}}_{=~0}-\int\text{d}^3x\delta h_{\beta(2)\gamma(2)\alpha(2s-4)}\partial^{\beta(2)}\partial^{\gamma(2)}h^{\alpha(2s-4)}\bigg\}\\ &=-(s-1)(2s-3)\int\text{d}^3x\delta h^{\alpha(2s)}\partial_{\alpha(2)}\partial_{\alpha(2)}h_{\alpha(2s-4)}. \end{align} Putting all three terms together gives $$\delta\mathcal{S}_s[\varphi^1]=\int\text{d}^3x\delta h^{\alpha(2s)}\bigg\{2\square h_{\alpha(2s)}+s\partial^{\beta(2)}\partial_{\alpha(2)}h_{\beta(2)\alpha(2s-2)}-(s-1)(2s-3)\partial_{\alpha(2)}\partial_{\alpha(2)}h_{\alpha(2s-4)}\bigg\}.$$ Setting the variation in the action equal to zero gives the required EoM: $$\square h_{\alpha(2s)}+\frac{1}{2}s\partial^{\beta(2)}\partial_{\alpha(2)}h_{\beta(2)\alpha(2s-2)}-\frac{1}{2}(s-1)(2s-3)\partial_{\alpha(2)}\partial_{\alpha(2)}h_{\alpha(2s-4)}=0.$$ I feel like the approach I am taking is in the right direction, since it gets me the right EoM in not just this case but the other 4 also, however I think it is very poorly executed and sloppy. Apologies for the length of the post but I wanted to show you how I was approaching the problem so that it might be easier to see where I am going wrong.


I now see why the first 'fudge' holds... when integrating by parts I will now not write the boundary terms since we can assume the fields vanish at infinity. Then the first fudge holds true by the following; \begin{align}\int\text{d}^3x \delta [h^{\alpha(2s)}\square h_{\alpha(2s)}] &=\int\text{d}^3x \bigg\{\delta h^{\alpha(2s)}\square h_{\alpha(2s)}+ h^{\alpha(2s)}\square \delta h_{\alpha(2s)}\bigg\}\\ &= \int\text{d}^3x \bigg\{\delta h^{\alpha(2s)}\square h_{\alpha(2s)}- (\partial^ah^{\alpha(2s)})(\partial_a \delta h_{\alpha(2s)})\bigg\} \\ &=\int\text{d}^3x \bigg\{ \delta h^{\alpha(2s)}\square h_{\alpha(2s)}+(\partial^a\partial_ah^{\alpha(2s)})\delta h_{\alpha(2s)}\bigg\}\\ &=\int\text{d}^3x \bigg\{ \delta h^{\alpha(2s)}\square h_{\alpha(2s)}+(\square h^{\alpha(2s)})\delta h_{\alpha(2s)}\bigg\}\\ &=\int\text{d}^3x \bigg\{2 \delta h^{\alpha(2s)}\square h_{\alpha(2s)}\bigg\}\end{align} where I have integrated by parts twice and Latin indices are Lorentz indices.

Remaining questions for bounty:

I am still unsure of how to write the boundary terms when integrating by parts without having free indices where they shouldn't be. Additionally, I am not sure if the reason for our ability to kill off the boundary terms is that the variation in the fields vanish at infinity or the field itself vanishes at infinity... i think it is the latter since this is now a Lagrangian density and the integral is over all space-time indices from $-\infty$ to $\infty$.


1 Answer 1


OP is essentially asking the following.

Why does the boundary terms (BT) vanish in the variation of the action (B.3)?

First thing first: As literally written, the BTs don't vanish. There are additional BTs that Ref. 1 doesn't write, as explained below.

However, it is not the scope of Ref. 1 to consider the explicit form of boundary terms (BTs) in the action principle (B.3) (as long as they are known to exists).

In fact, the action (B.3) depends on second space-time derivatives. However, it is not hard to figure out the correct action (B.3'), by formal integration by parts, ignoring BTs. The corrected action (B.3') contains only space-time derivatives up to first order.

This corrected action (B.3') serves as the actual action principle for Euler-Lagrange (EL) eqs.

It is not hard to see that all BTs vanish in the variation of the corrected action (B.3') if we impose Dirichlet boundary conditions (BCs).


  1. S.M. Kuzenko & D.X. Ogburn, Off-shell higher spin ${\cal N}=2$ supermultiplets in three dimensions, arXiv:1603.04668.
  • $\begingroup$ Do you know of a resource (Lectures, textbook, papers) which I could learn the proper formalism for this kind of field theory? I guess I am quite naive of the subtleties and techniques in this area. For example, I don't even know the significance of the action containing second derivatives and why you point it out. Nor do I know the proper way to execute integration by parts for such functionals. $\endgroup$ Aug 28, 2017 at 14:07

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