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I became confused while reading this article for the following reason:

For $p=1$ we have strings such that the Nambu-Goto action is proportional to the area of the worldsheet embedded by the maps $X^\mu$ such that $$S_{NG}[X] := -T \int_\Sigma \text{d}^2 \sigma \sqrt{-\det(\gamma_{ab})} \tag1$$

where $T$ is string tension, $\gamma_{ab} := \eta_{\mu\nu}\partial_a X^\mu \partial_b X^\nu$ are the induced metric components and $\Sigma = (\tau_0, \tau_1) \times (0,l) \subset \mathbb{R}^2$ have no additional extra structure besides the usual smooth manifold definition. Since there's no assumption about $\Sigma$ embedding independent metric teor, the area of the worksheet domain $\Sigma$ must be the integral

$$A(\Sigma) = \int_\Sigma \text{d}\tau\text{d}\sigma = (\tau_1 - \tau_0) \cdot l \tag2$$

However if we equip $\Sigma$ with some Lorentzian metric tensor $h$ without referring to any embedding function, it becomes a Lorentzian manifold by itself. The area of this domain, as I understand, must be the integral

$$A(\Sigma) = \int_\Sigma \text{d}\tau\text{d}\sigma \sqrt{-\det{(h_{ab})}} $$ wich is not useful since makes no reference to Minkowski space embedding, and has no dynamical string description.

Imn the referred paper, in order to get an action equivalent to $S_{NG}$ with the $\sqrt{-\det{(h_{ab})}}$ term, the author uses 3 lagrange multipliers $\Lambda^{ab} = \Lambda^{ba}$ to construct in eq.(5)

$$S^{(1)}_{p=1} = -T \int_\Sigma \text{d}\tau\text{d}\sigma \left(\sqrt{-\det{(h_{ab})}} +\Lambda^{ab}(\gamma_{ab} - h_{ab}) \right)$$ which according to my understanding should be actually

$$ -T \int_\Sigma \text{d}\tau\text{d}\sigma \sqrt{-\det{(h_{ab})}} \left( 1 +\Lambda^{ab}(\gamma_{ab} - h_{ab}) \right)$$ since the area element $\text{d}\tau\text{d}\sigma \sqrt{-\det{(h_{ab})}}$ should contain multiply all the integrand.

However, the above action does not lead to the correct Polyakov action wich is derived in the linked paper. Can someone clarify this to me?

REMARK: the author of linked paper makes a development for general $p$-branes, but I made the computations with $p=1$ for simplicity.

EDIT: One "obvious" way to proceed that I didn't noticed is impose the constraint not as $\gamma_{ab} - h_{ab}$ but as $\frac{\gamma_{ab} - h_{ab}}{\sqrt{-\det{(h_{ab})}}}$. I think this is ok because it leads to the same constraint as long as $\sqrt{-\det{(h_{ab})}} \ne 0$, and also it allows to write

$$S^{(1)}_{p=1} = -T \int_{\Sigma} \text{d}\tau\text{d}\sigma \sqrt{-\det{(h_{ab})}} \left[1 + \Lambda^{ab}\left(\frac{\gamma_{ab} - h_{ab}}{\sqrt{-\det{(h_{ab})}}}\right)\right]$$ which is basically the paper's eq.(5) but written in a more explicit way. Is this correct?

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In action (1) you're just integrating an area in flat space-time, that's all you're doing.

It's basically exactly the same as in special relativity where in the analogous situation you're just integrating (a scaled version of) the arc-length of the world-line a point particle traces out in space-time.

If we had to, for some reason, add the 'square root' every time we add some fields, then in the $p = 0$ case of special relativity, when adding an electromagnetic field, the action would have to be (with $q$ the charge of the particle) $$S \stackrel{?}{=}- m \int d \tau \sqrt{\dot{x}^2} - q \int d \tau \sqrt{\dot{x}^2} A_{\mu} \dot{x}^{\mu}$$ but of course the actual action is $$S = - m \int d \tau \sqrt{\dot{x}^2} - q \int d \tau A_{\mu} \dot{x}^{\mu}.$$ Physically it just makes no sense to add the area or arc length to the other terms when you're talking about e.g. the interaction of a point particle moving in space-time with an electromagnetic field existing all though that space-time.

The point of the calculation below equation (3) in section 2 of the paper is to re-interpret the flat space-time surface-area integral as an equivalent curved-space-time 'volume' integral intrinsic the curved space-time of the surface itself.

To appreciate the method by which they [2] derive the Polyakov action it's useful to consider the usual analogous point particle case of this process and then how it compares to their method in this case. I will do it in the $(+,-,-,-)$ metric (which I also used above) to force you to re-do it in the $(-,+,+,+)$ metric as a check.

The usual way of doing this procedure is the following. Given the above point particle action, it's canonical momentum is $$p_{\mu} = - \frac{\partial L}{\partial \dot{x}_{\mu}} = \frac{m \dot{x}_{\mu}}{\sqrt{\dot{x}^2}}$$ (very explicitly using the $(+,-,-,-)$ metric here). This Lagrangian results in a zero canonical Hamiltonian, $$H_c = 0,$$ as you should check and hopefully understand the reason why this happens. It also possesses the primary constraint $$p^2 - m^2 = 0,$$ so we can form the constrained Hamiltonian using a Lagrange multiplier $e$, $$H_T = H_c - \frac{e}{2}(p^2 - m^2)$$ resulting in the equivalent first order Lagrangian $$L = - p \cdot \dot{x} + \frac{e}{2}(p^2+m^2).$$ Here we of course treat $e$ and $x$ and $p$ as independent variables. If we eliminate $p$ using it's equation of motion we find $$e p_{\mu} = \dot{x}_{\mu}.$$ Note as a side comment that comparing to the canonical momenta above we can see that we should have $$e = \frac{1}{m} \sqrt{\dot{x}^2}.$$ On inserting the expression for $p_{\mu}$ in terms of $e$ and $\dot{x}_{\mu}$ into $S$ we find the action reduces to $$S = - \frac{1}{2} \int d \tau e (e^{-2} \dot{x}^2 + m^2)$$ which is the (second order) point particle analog of the Polyakov action. This shows that $e^2$ can be interpreted as the metric of the one-dimensional 'surface' generated by the arc length of the point particle, $d \tau \sqrt{e}$ the invariant 'volume element', and $e^{-2}$ the inverse metric, i.e. $(1,0)$-dimensional general relativity with the $(+)$ metric. In this form we can send $m \to 0$ so that the action also accounts for a massless particle, which the previous version doesn't directly do and is often dealt with as a limiting case of the equations of motion without pointing out the issue with the action.

Note also that this is the 'einbein' form of the $(0,1)$ metric (i.e. the vierbein formalism of GR), i.e. if we set $$e_{\tau \tau}^2 = e_{\tau}^a e_{\tau}^b \eta_{ab} = \tilde{h}_{\tau \tau} = \frac{1}{m^2} \dot{x}^{\mu} \dot{x}_{\mu} = \frac{1}{m^2} \partial_{\tau} x^{\mu} \partial_{\tau} x_{\mu}$$ then the following holds trivially $$e = \det{e_{\tau \tau}} = e_{\tau \tau} = \sqrt{\det(\tilde{h}_{\tau \tau})} = \sqrt{\tilde{h}} = \frac{1}{m} \sqrt{\dot{x}^2} = \frac{1}{m} \sqrt{\det(\partial_{\tau} x^{\mu} \partial_{\tau} x_{\mu})}.$$ Note there is no minus sign, i.e. $e \stackrel{?}{=} \sqrt{-\tilde{h}}$ is not the case, only because it's $(1,0)$ relativity in the $(+)$ metric, in the $(-)$ metric you are going to check it will look similar. This whole approach is extremely common. Indeed the string Polyakov action is sometimes directly motivated as the natural $(1,1)$ analog of this action [1].

Note by construction the $e$ equation of motion in the first-order formalism just eliminates results in the constraint $p^2=m^2$, eliminating the $e$ variable from the action, thus the action reduces back to the arc length action (assuming you started from the action, if you think too hard about this point you'll end up doing Dirac's constrained systems). We can similarly expect eliminating $e$ from the second order action via it's equation will again just reproduce the arc length form of the action. Indeed the $e$ equation of motion from this last second order action directly results in the expression $e = \frac{1}{m} \sqrt{\dot{x}^2}$ we expected to find. Inserting this into the Polyakov action of course results in the original arc length action. Thus you can see why, in the string case, if we take the Polyakov action as our starting point as a direct generalization of this last action (as some books do), the energy-momentum tensor equation of motion is expected to reproduce something that of course turns out to be the Nambu-Goto action. Of course all this is classical which is important to note.

The paper is basically just doing the same thing done above but directly from the arc length form of the action. By defining $$h_{\tau \tau} = \partial_{\tau} x^{\mu} \partial_{\tau} x_{\mu}$$ in the point particle action $$S = - m \int d \tau \sqrt{\dot{x}^2} = - m \int d \tau \sqrt{\det(\partial_{\tau} x^{\mu} \partial_{\tau} x_{\mu})} = - m \int d \tau \sqrt{\det(h_{\tau \tau})} = - m \int d \tau \sqrt{h} $$ we would just like to try and re-interpret the arc length integral through in flat space-time as a curved space-time 'volume integral' in the 'space-time' of the curved world-line arc with metric $h_{\tau \tau}$. We do this by considering the following action $$S = - m \int d \tau [\sqrt{\det(g_{\tau \tau})} + \lambda (h_{\tau \tau} - g_{\tau \tau})] $$ The $\lambda$ equation of motion just enforces $g_{\tau \tau} = h_{\tau \tau}$ obviously reproducing the original action. Thus we just reproduce the arc length integral we started with, so just on a basic level the extra term we added to the action is no different to adding the electromagneic potential term above, as long as it's dimensionally consistent etc... there is no issue. Re-writing it as $$S = - m \int d \tau \sqrt{\det(g_{\tau \tau})}[1 + \lambda \frac{(h_{\tau \tau} - g_{\tau \tau})}{\sqrt{\det(g_{\tau \tau})}}] $$ makes it look more like a 'general relativity' action, but the term as added above already stands on it's own, and the Lagrange multiplier $\lambda$ is also an independent variable accounting for the dimensions. The re-write is consistent but it's at least unnecessary and distracts from the point trying to be made. In fact one could determine how $\lambda$ should transform under a 'general coordinate transformation' so that things are consistent, this is in fact important to check for $e$ in the previous example as can be seen e.g. in ([1], eq. (2.1.8)).

However instead 'eliminating' the $g_{\tau \tau}$ variable just as we eliminated $\lambda$ or e.g. $p_{\mu}$ previously by taking it's equation of motion (in all cases, since there are no time derivatives of this variable, it's equation of motion is just a constraint), we can obviously use this to instead eliminate the $\lambda$ (this can be done directly in the first-order point particle action above also and it's important to understand this can be done while still not completely fixing the 'gauge' in the full 'constrained system' interpretation of the above discussion) $$\lambda = \frac{1}{2\sqrt{g_{\tau \tau}}} = \frac{1}{2} \sqrt{g_{\tau \tau}} g^{\tau \tau}.$$ Eliminating $\lambda$ in $S$ thus gives $$S = - m \int d \tau [ \sqrt{g_{\tau \tau}} + \frac{1}{2} \sqrt{g_{\tau \tau}} g^{\tau \tau}(h_{\tau \tau} - g_{\tau \tau})] = - \frac{1}{2} m \int d \tau \sqrt{g_{\tau \tau}} (g^{\tau \tau} h_{\tau \tau} + 1) $$ and so using $h_{\tau \tau} = \partial_{\tau} x^{\mu} \partial_{\tau} x_{\mu}$ this reads as $$S = - \frac{1}{2} m \int d \tau \sqrt{g_{\tau \tau}} (g^{\tau \tau} \partial_{\tau} x^{\mu} \partial_{\tau} x_{\mu} + 1) $$ As written it looks like it does not work with the massless limit, however if we set $g_{\tau \tau} = m^2 e^2$ (i.e. although it's just a definition we're of course tracing back from the constraint $g_{\tau \tau} = h_{\tau \tau}$ and $h_{\tau \tau} = m^2 \tilde{h}_{\tau \tau} = m^2 e^2$ to get $g_{\tau \tau} = m^2 e^2$) this becomes $$S = - \frac{1}{2} m \int d \tau m e (\frac{1}{m^2} e^{-2} \partial_{\tau} x^{\mu} \partial_{\tau} x_{\mu} + 1) $$ so it still describes the same massless limit and agrees with the previous discussion despite the tiny change in notation but drastic change in approach.

The huge virtue of the approach in [2] is how relatively easy it is to directly get the Polyakov brane action from the Nambu-Goto action directly. If one tries the higher analog of the Hamiltonian approach it really is a mess.

References:

  1. Green, Schwarz, Witten, "Superstring Theory", Vol. 1.
  2. Nieto, "Remarks on Weyl invariant p-branes and Dp-branes".
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    $\begingroup$ Thanks for your answer, I really appreciated it. So, basically, the last edit in my post is wrong, or at least unnecessary? $\endgroup$ Sep 18, 2021 at 23:01
  • $\begingroup$ It's unnecessary because the added term as stated in the paper has to be consistent simply due to the way it was written, it's just assumed that the Lagrange multiplier ensures everything is consistent as is commonly the case when adding such terms. $\endgroup$
    – bolbteppa
    Sep 18, 2021 at 23:14
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    $\begingroup$ Thanks again. At the end, my problem was all about the formal definition of integrals on metric manifolds via volume forms. In your example, the initial point particle action considers $\mathbb{R}$ only as a set. However, after your derivation paralell to ref.[2], your intrinsic, metric $g_{\tau \tau}$ would create a metric manifold $(\mathbb{R}, g_{\tau\tau})$, of course, implicitly assuming all the assumptions to the construction of a manifold with the real line. My edit was about this technical problem in my reasoning. $\endgroup$ Sep 18, 2021 at 23:29

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