String Theory, Part II: Classical Strings

When discussing strings, it is useful to talk about tension. Consider a small piece of string from (x, y) to (x+dx, y+dy) , with

\displaystyle\left|\frac{\partial y}{\partial x}\right| \ll 1

In the case of transverse oscillations, the longitudinal force is negligible and so it is only neccessary to compute the net vertical force. If we let T_0 be the tension, we find that for a small amount of vertical force

dF_{v} = \left.T_0\frac{\partial y}{\partial x}\right|_{x+dx}-\left.T_0\frac{\partial y}{\partial x}\right|_x \approx T_0\frac{\partial^2 y}{\partial x^2}dx

Since the length of the string is dx and if we denote the mass per unit length by \mu_0, the total mass is \mu_0 dx. According to Newton’s second law,

F = \mathit{ma}\implies T_0\frac{\partial^2 y}{\partial x^2} dx = (\mu_0 dx)\frac{\partial^2 y}{\partial t^2}\implies\frac{\partial^2 y}{\partial x^2}-\frac{\mu_0}{T_0}\frac{\partial^2 y}{\partial t^2} = 0

This is a just wave equation describing a wave with transverse velocity v_0 = \sqrt{T_0\mu_0} .

To compute the action for a nonrelativistic string, we must find the Lagrangian, \mathit{L} = T - V. In this case, the kinetic energy, T is equal to the sum of the kinetic energies of all the infinitesimal pieces of the string:

\displaystyle\mathit{T} =\int_0^a\frac{1}{2} (\mu_0 dx)\left(\frac{\partial y}{\partial t}\right)^2

The potential energy, V can be thought of as the spring potential when a piece of string with ends at x and x+dx is stretched by dx in the x-direction and dy in the y-direction. The change in length \Delta l can be computed as

\displaystyle\Delta l = \sqrt{(dx)^2 + (dy)^2} - dx = dx\left(\sqrt{1+\left(\frac{\partial y}{\partial t}\right)^2}-1\right) \approx dx\frac{1}{2}\left(\frac{\partial y}{\partial t}\right)^2

so our potential becomes

\displaystyle\mathit{V} = \int_0^a\frac{1}{2}(T_0)\left(\frac{\partial y}{\partial t}\right)^2 dx

Our Lagrangian is therefore

\displaystyle\mathit{L} = T - V=\int_0^a\frac{1}{2}\left[\left(\frac{\partial y}{\partial t}\right)^2\\-\frac{1}{2}(T_0)\left(\frac{\partial y}{\partial t}\right)^2 dx\right] dx

and so our action becomes

\displaystyle\mathit{S} = \int_{t_i}^{t_f}L\; dt=\int_{t_i}^{t_f}\int_0^a\frac{1}{2}\left[\left(\frac{\partial y}{\partial t}\right)^2-\frac{1}{2}(T_0)\left(\frac{\partial y}{\partial t}\right)^2 dx\right]dx dt

By varying S by \delta y and using the Euler-Lagrange equation,

\displaystyle\left(\frac{\partial L}{\partial\dot y}\right) - \frac{d}{dt}\left(\frac{\partial L}{\partial y}\right) = 0

we are able to get the equation of motion:

\displaystyle\left(\frac{\partial\mathcal{P}^t}{\partial t}\right) + \left(\frac{\partial\mathcal{P}^x}{\partial y}\right) = 0


\displaystyle\mathcal{P}^t = \frac{\partial L}{\partial\dot y} = \mu\frac{\partial y}{\partial t}\;\text{and}\; \mathcal{P}^x = \frac{\partial L}{\partial y^\prime} = -T_0\frac{\partial y}{\partial x}

Close examination reveals this to be the wave equation.

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4 Responses to String Theory, Part II: Classical Strings

  1. Sean, do you want to comment on the identity of \mathcal{P}^t and \mathcal{P}^x. If memory serves these are momentum densities, right?

  2. I’d like to know more about those in general. In the case of a system with finite degrees of freedom we can use the Lagrangian to create a map (Legendre transform) between the tangent bundle and cotangent bundle. I’ve been interested in what happens if there are an infinite number of freedoms. I’d really like to know the connection between classical differential geometry and the Hilbert spaces where such functions (with infinite number of freedoms) live.

    • seanparrottwolfe says:

      and this is where i get lost having no differential geometry or functional analysis background. There’s obviously more behind the scenes.

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