Let $X^\mu (t,\sigma ^1,\ldots ,\sigma ^p)$ be a $p$-brane in space-time and let $g$ be the metric on $X^\mu$ induced from the ambient space-time metric. Then, the Nambu-Goto action on $X^\mu$ is defined to be $$ S:=-T\int dt\, d\sigma \sqrt{-\det (g)}. $$ (We use the convention of space-time signature $(-,+,+,+)$.)

Let us try to compute the Hamiltonian for this theory. The first step is to calculate the conjugate momenta: $$ P_\mu :=\frac{\partial L}{\partial (\partial _tX^\mu)}, $$ where of course $L:=-T\sqrt{-\det (g)}$. It turns out that you are unable to invert this map to write $\partial _tX^\mu$ as a function of $X^\mu$, $P^\mu$, and $\partial _{\sigma ^i}X^\mu$, which means that the image of the Legendre transformation $(X^\mu ,\partial _{\sigma ^i}X^\mu ,\partial _tX^\mu )\mapsto \left( X^\mu ,\partial _{\sigma ^i}X^\mu ,\frac{\partial L}{\partial (\partial _tX^\mu )}\right)$ is not surjective, but instead the image of this map is a proper sub-manifold (of course, in general the image of a manifold under a smooth map will not in general be a manifold, but in our particular case, that should not be a problem) of $T^pT^*M$, and so (locally anyways) will be specified by $N$ constraints $\phi _j(X,\partial _{\sigma ^i}X,P)=0$, where $N$ is the dimension of the kernel of the derivative of the Ledgendre transformation.

In the case of the Nambu-Goto Lagrangian, I have found that $N=1+p$. I found this by calculating the multiplicity of $0$ as an eigenvalue of $\frac{\partial ^2L}{\partial (\partial _tX^\mu )\partial (\partial _tX^\nu )}$. The question is: now that I know how many constraints there should be, how do I systematically find what those constraints actually are?

For what it's worth, I know that one constraint is $$ P^2+T^2\det (k)=0, $$ where $k$ (I have suppressed the time-dependence of $k(t)$ in the notation) is the metric on the space-like sub-manifold $X_t^\mu (\sigma ^1,\ldots ,\sigma ^p):=X^\mu (t,\sigma ^1,\ldots ,\sigma ^p)$ induced from the metric $g$ on $X^\mu$, and that the other $p$ constraints are $$ \partial _{\sigma ^i}X\cdot P=0 $$ for $1\leq i\leq p$. I even know how to verify that these are in fact constraints. What I don't know, however, is how to come up with these constraints without simply pulling them out of my ass.

Is there a systematic, yet computationally feasible way of determining what these $1+p$ constraints should be?

Furthermore, I have a hunch that each of these constraints arises from a corresponding re-parameterization invariance, so if that is indeed the case, it would be wonderful if someone could elucidate this connection for me. (Is there a re-parameterization invaraince-$\Rightarrow$-constraint theorem analogous to Noether's Theorem for symmetries and conserved quantities?)

  • $\begingroup$ Could you explain how did you calculate the multiplicity of 0 as an eigenvalue of that matrix? $\endgroup$
    – Anthonny
    May 15, 2015 at 19:15
  • $\begingroup$ Actually, since it's been a little over two years now since I actually did the computation, I admittedly don't remember exactly what I did, and I can't say for sure unless I just re-do everything. I can, however, tell you the first thing I would try . . . $\endgroup$ May 15, 2015 at 19:28
  • $\begingroup$ Write out the formula for $\det (g)$ using the co-factor expansion along the first column (or row) to obtain an explicit formula for $L$. For concrete-ness, take the space-time dimension to be, say, $D=4$. This way, you will be able to write down an explicit formula for the matrix $\frac{\partial ^2L}{\partial (\partial _tX^\mu )\partial (\partial _tX^\nu )}$ (of course if you want a general proof you will need to do it for arbitrary $D$). The point of doing this is that now we will have a very explicit $4\times 4$ matrix that you can just plug-in to Mathematica (or some other CAS) . . . $\endgroup$ May 15, 2015 at 19:29
  • $\begingroup$ Then you can just have Mathematica row-reduce this matrix so that you can just read-off the multiplicity of $0$ as an eigen-value. If you do this for several different values of $p$, you should see a pattern. This of course gives you the answer. To prove it, you would have to turn this calculation into a proof; I am quite confident that I never actually bothered to do that. $\endgroup$ May 15, 2015 at 19:32
  • $\begingroup$ Honestly, there is probably a much more clever way to do this, but I personally don't think it is a wise use of time to try to find a better solution to something like this. $\endgroup$ May 15, 2015 at 19:33

1 Answer 1


I managed to find a quasi-systematic way to do this. The idea that allowed me to do this was inspired by Noether's Theorem.

Re-parameterization invariance is a symmetry of the system, a symmetry much stronger than an ordinary global symmetry. Similarly, however, a constraint is also a conserved quantity, but it is something much stronger than that. Knowing that there was some relation between the two, I suspected there might be a way to derive constraints given re-parameterization invariance in a similar way that Noether's Theorem allows you to derive conserved quantities from a known symmetry. I thus managed to hack together a modification of the'proof' of Noether's Theorem that allowed me to calculate the constraints. Unfortunately, however, putting the constraints entirely in terms of $X$, $\partial _{\sigma ^i}X$, and $P$ was not completely systematic, but it was still much more straightforward than simply coming up with the constraints out of thin air. Anyways, here's what I did. For simplicity, I only addressed the case of the string ($p=1$).

For a mapping of the string $X\mapsto X'$ which depends on a parameter $\varepsilon$, I abbreviate $\frac{d}{d\varepsilon}\big| _{\varepsilon =0}$ by $\delta$. This notation is common amongst physicists, but they often do mention exaclty what they mean by it. Under such a transformation of the string alone, I have \begin{align*} \delta L & =\frac{\partial L}{\partial X}\cdot \delta X+\frac{\partial L}{\partial (\partial _tX)}\cdot \delta (\partial _tX)+\frac{\partial L}{\partial (\partial _\sigma X)}\cdot \delta (\partial _\sigma X) \\ & =\frac{\partial}{\partial t}\left[ \frac{\partial L}{\partial (\partial _tX)}\right] \cdot \delta X+\frac{\partial L}{\partial (\partial _tX)}\cdot \delta (\partial _tX)+\frac{\partial}{\partial \sigma}\left[ \frac{\partial L}{\partial (\partial _\sigma X)}\right] \cdot \delta X \\ & +\frac{\partial L}{\partial (\partial _\sigma X)}\cdot \delta (\partial _\sigma X) \\ & =\frac{\partial}{\partial t}\left[ \frac{\partial L}{\partial (\partial _tX)}\cdot \delta X\right] +\frac{\partial}{\partial \sigma}\left[ \frac{\partial L}{\partial (\partial _\sigma X)}\cdot \delta X\right] . \end{align*} Note that I have assumed that the derivatives commute with the transformation (i.e. $\delta (\partial _tX)=\partial _t(\delta X)$ and $\delta (\partial _\sigma X)=\partial _\sigma (\delta X)$). In the case of our re-parametrization invariance, this turns out to be the case, though I did have to check this and I don't see any reason why this should be true in general (though do point it out if you are aware of a reason)).

Assuming that $\delta L$ is of the form $\delta L=\partial _tf+\partial _\sigma g$, we can re-arrange this equation to obtain $$ \frac{\partial}{\partial t}\left[ \frac{\partial L}{\partial (\partial _tX)}\cdot \delta X-f\right]=\frac{\partial}{\partial \sigma}\left[ g-\frac{\partial L}{\partial (\partial _\sigma X)}\cdot \delta X\right] . $$ Thus, under the assumption that the appropriate functions vanish at infinity, the integral over $\sigma$ of the quantity in the time derivative will be conserved. We suspect this might be a constraint. This is enough motivation to write down that quantity to see if it is in fact a constraint (note that the above derivation is only supposed to be motivation to check, not an actual proof).

In the case of time re-parametrization $X(t)\mapsto X(t+\varepsilon \xi )$, $\delta X=\xi \dot{X}$ and $\delta L=\partial _t(\xi L)$ (the first you can see right away, the second I had to actually sit down and calculate). Thus, we suspect that $$ \xi \dot{X}\cdot P-\xi L=\text{const} $$ might be a constraint. In fact, if you calculate $P$ and plug it in, we indeed see that this expression vanishes identically. So indeed it is true that $$ \partial _tX\cdot P=-L. $$ If you do the same thing with $\sigma$ re-parameterization, you find $$ \partial _\sigma X\cdot P=0. $$ Note that in this case $\delta L$ is a $\sigma$ derivative, as opposed to a time derivative as before, and so doesn't show up. Fantastic! The only thing that remains to be down is to eliminate the pesky $\dot{X}$. To do this, we have to actually compute $P$.

It turns out that $$ P_\mu =\frac{T^2}{L}\left( (\partial _tX\cdot \partial _\sigma X)\partial _\sigma X_\mu-(\partial _\sigma X)^2\partial _tX_\mu \right) . $$ The idea is that we can use the time re-parameterization constraint to eliminate $\partial _tX$ from the expression for $P$ by contracting $P$ with itself: $$ P^2=\frac{T^2}{L}\left( (\partial _tX\cdot \partial _\sigma X)\partial _\sigma X\cdot P-(\partial _\sigma X)^2\partial _tX\cdot P\right) =-T^2(\partial _\sigma X)^2. $$

Et voila! There be the sought after constraints!

And now I move on with my life . . .


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