# Euler-Lagrange equation (EOM) solutions - hairy lagrangian

I'm going through Zwiebach Chapter 6 on relativistic strings to try to solve a
similar problem. I got all the way to my equation of motion
\begin{eqnarray*}
\delta S & = & [ p' \delta \theta]_{z 0}^{z 1} + \int_{z 0}^{z 1} d z \left(
p - \frac{\partial ( p')}{\partial z} \right) \delta \theta\\
& & \\
\Longrightarrow p - \frac{\partial ( p')}{\partial z} & = & 0\\
& & \\
{where} : & & \\
p & = & \frac{\partial L ( z, \theta, \theta')}{\partial \theta}\\
p' & = & \frac{\partial L ( z, \theta, \theta')}{\partial \theta'}
\end{eqnarray*}
The Lagrangian I have is \begin{array}{lll}
L ( z, \theta, \theta') & = & \frac{\cos^3 \theta}{z^5} \sqrt{1 + z^2
\theta'^2}
\end{array}, and I know that this equation is the Euler Lagrange equation.

\begin{eqnarray*}
\frac{\partial L}{\partial \theta} - \frac{\partial}{\partial z} \left(
\frac{\partial L}{\partial \theta'} \right) & = & 0\\
& & \\
{where} : & & \\
\theta' ( z) & = & \frac{\partial \theta ( z)}{\partial z}\\
L ( z, \theta, \theta') & = & \frac{\cos^3 \theta}{z^5} \sqrt{1 + z^2
\theta'^2}
\end{eqnarray*}
Just putting the expression for L into the equation gives me a mess:
\begin{array}{lll}
\frac{- 3 \sin \theta \cos^2 \theta}{z^5} \sqrt{1 + z^2 \theta'^2} -
\frac{\sqrt{2} \theta'' ( 2 \theta'^2 z^2 + 3) \cos^3 \theta}{z^4 (
\theta'^2 z^2 + 2)^{3 / 2}} & = & 0
\end{array}
I don't think I'm supposed to brute force this. I know what I want is to show that
$$\theta = \arcsin ( z)$$ but I can't see how to get it. I'm thinking that there is something I'm missing about the equation of motion.

oh and here is a link to the same question I posted on stackexchange

ChrisVer
Gold Member
What is the meaning of that Lagrangian?
Or in other words, where in Zwiebach did you get it?

Well I'm working in $$A d S_5 \times S^5$$
\begin{eqnarray*}
- d s^2 & = & g_{\alpha \beta}^{{induced}} d X^{\mu} d X_{\mu}\\
& & \\
& = & \{ G_{\mu \nu} \partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu} \}
d X^{\mu} d X_{\mu}\\
& & \\
& = & \frac{1}{z^2} ( - d t^2 + d \vec{x}^2 + d z^2) + d \theta + \cos^2
\theta d \Omega_3^2 + \sin^2 \theta d \psi^2
\end{eqnarray*}
\begin{eqnarray*}
{spacetime} {indicies} : & \mu, \nu & \{ d = 0 \ldots 9 \}\\
{world} {brane} {indicies} : & \alpha_1, \alpha_2, \alpha_3 &
\{ p = 0 \ldots 7 \}\\
{spatial} {indicies} {on} {brane} : & i, j & \{ p = 0
\ldots 7 \}\\
{spatial} {indicies} {normal} {to} {brane} : & a, b
& \{ ( p - 1) \ldots d \}\\
{generic} {index} {for} {transverse} {coordinates}
: & & I \longrightarrow ( i, a)
\end{eqnarray*}
and we have:
\begin{eqnarray*}
\mu = 0 & t & \\
1 & x^1 & \\
2 & x^2 & \\
3 & x^3 & \\
4 & z & \\
5 & \alpha 1 & \\
6 & \alpha 2 & \\
7 & \alpha 3 & \\
8 & \theta & \\
9 & \psi &
\end{eqnarray*}
working in the static gauge so instead of $\begin{array}{ll} \mu, \nu & \{ d = 0 \ldots 9 \} \end{array}$we have $$\mu = 0 \ldots 7$$. I do alot of ansatzing
\begin{eqnarray*}
'' {string} {coordinate}'' & : & X^8 ( x_1 \ldots x_7) \Rightarrow
{ansatz} \Rightarrow X^8 \left( \not{t}, \not{\vec{x}}, z,
\not{\alpha_1}, \not{\alpha_2}, \not{\alpha_3} \right) = \theta \left(
\not{t}, \not{\vec{x}}, z, \not{\alpha_1}, \not{\alpha_2}, \not{\alpha_3}
\right) = \theta ( z)\\
& & X^9 ( x_1 \ldots x_7) \Rightarrow {ansatz} \Rightarrow X^9 \left(
\not{t}, \not{\vec{x}}, z, \not{\alpha_1}, \not{\alpha_2}, \not{\alpha_3}
\right) = \psi \left( \not{t}, \not{\vec{x}}, z, \not{\alpha_1},
\not{\alpha_2}, \not{\alpha_3} \right) = \psi ( z)\\
& & {ansatz} : \psi ( z) = 0
\end{eqnarray*}
\begin{eqnarray*}
g_{\alpha \beta}^{{induced}} & = & g_{{Emil}} + G_{8 8}
\partial_{\alpha} X^8 \partial_{\beta} X^8 + G_{99} \partial_{\alpha} X^9
\partial_{\beta} X^9\\
& & \\
& = & \left(\begin{array}{cccccccc}
G_{00} & 0 & & & & & & 0\\
0 & G_{11} & 0 & & & & & \\
& 0 & G_{22} & 0 & & & & \\
& & 0 & G_{33} & 0 & & & \\
& & & 0 & G_{44} + G_{88} \left( \frac{\partial \theta}{\partial z}
\right)^2 & 0 & & \\
& & & & 0 & G_{55} & 0 & \\
& & & & & 0 & G_{66} & 0\\
0 & & & & & & 0 & G_{77}
\end{array}\right)\\
& & \\
& = & \left(\begin{array}{cccccccc}
\frac{1}{x_4^2} & & & & & & \ldots & 0\\
& \frac{1}{x_4^2} & & & & & & \vdots\\
& & \frac{1}{x_4^2} & & & & & \\
& & & \frac{1}{x_4^2} & & & & \\
& & & & \frac{1}{x_4^2} + 1 \left( \frac{\partial \theta}{\partial z}
\right)^2 & & & \\
& & & & & \cos^2 x_8 & & \\
\vdots & & & & & & \cos^2 x_8 \sin^2 x_5 & \\
0 & \ldots & & & & & & \cos^2 x_8 \sin^2 x_5 \sin^2 x_6
\end{array}\right)
\end{eqnarray*}
where I know that
\begin{eqnarray*}
L & = & \det \sqrt{g_{\alpha \beta}^{{induced}}}\\
& & \\
L & = & \frac{\cos^3 \theta}{z^5} \sqrt{1 + z^2 \left( \frac{\partial
\theta}{\partial z} \right)^2} = L ( z, \theta, \theta')
\end{eqnarray*}

The action ends up working as follows:
\begin{eqnarray*}
S & = & \int d x_4 L ( X^8 ( x_1 \ldots x_7), X^9 ( x_1 \ldots x_7))\\
& & \\
S & = & \int d z L ( \theta ( t \ldots \alpha_3), \psi ( t \ldots
\alpha_3))\\
& & \\
& \Rightarrow & {ansatz} !\\
& & \\
S & = & \int d z L ( \theta ( z), 0) = \int d z \frac{\cos^3 \theta}{z^5}
\sqrt{1 + z^2 \left( \frac{\partial \theta}{\partial z} \right)^2}
\end{eqnarray*}
where I want to apply the principle of least action.

ChrisVer
Gold Member
Sorry I don't think I can help..
However to check whether you did it right or wrong, I'd say to try and insert your solution to the Euler Lagrange equation you derived and check whether it's correct or not. In fact I don't see where the $\sin ^2 x_5 \sin x_6$ coming from your $det \sqrt{g_{ab}}$ went...
you could as well try computational method?

This could be done with maple or mathematica.

Take another look at the second term of the "mess" ;-) .