I was asked to compute the energy momentum tensor of a string whose Lagrangian is:

$$\mathcal{L} = \frac{1}{2}\rho \dot{y}^2-\frac{1}{2}\tau y'^2 \tag{1}$$

I've used:

$$T_{\mu\nu} = \frac{\partial\mathcal{L}}{\partial(\partial^\mu\phi)}\partial_\nu\phi-g_{\mu\nu}\mathcal{L} \tag{2}$$

To get:

$$T_{00}=\frac{\partial\mathcal{L}}{\partial \dot{y}}\dot{y}-1\cdot\mathcal{L}=\rho \dot{y}^2-(\frac{1}{2}\rho \dot{y}^2-\frac{1}{2}\tau y'^2)=\frac{1}{2}\rho \dot{y}^2+\frac{1}{2}\tau y'^2=\mathcal{H} \tag{3}$$

$$T_{11}=\frac{\partial\mathcal{L}}{\partial y'}y'-1\cdot\mathcal{L}=\tau y'^2-(\frac{1}{2}\rho \dot{y}^2-\frac{1}{2}\tau y'^2)=\frac{3}{2}\tau y'^2-\frac{1}{2}\rho \dot{y}^2 \tag{4}$$

I'm guessing the off-diagonal terms vanish as the metric tensor is just the identity.

My question is - is this true and if so what does $T_{11}$ mean?

  • $\begingroup$ Instead of guessing, why not just calculate the other term? $\endgroup$
    – Javier
    Nov 3, 2020 at 1:52

1 Answer 1


The off-diagonal parts are not zero. The conservation equations are the local energy conservation equation $$ \partial_t {T^0}_0+ \partial_x {T^1}_0= \partial_t\left\{\frac \rho 2\dot y^2 +\frac T 2 y'^2\right\}+\partial_x \{-T\dot y y'\}=0, $$ in which the energy flux ${T^1}_0= -T\dot yy'$ evaluated at a point $x$ is the rate of working by the string to the left of $x$ on the string to the right of $x$, and the local pseudomomentum conservation equation $$ \partial_t {T^0}_1+ \partial_x {T^1}_1= \partial_t\{-\rho \dot yy'\}+\partial_x\left\{\frac \rho 2\dot y^2 +\frac T 2 y'^2\right\}=0. $$ Pseudomomentum is the conserved quantity associated with the translation symmetry of the position of the wave on the string, the string itself not being translated. Pseudomeomentum is conserved if the density $\rho$ is the same everywhere on the string. It is not to be confused with the longitudinal Newtonian momentum. The latter is associated with the translation symmetry of the wave and string together, and is indeterminate in the usual transverse motion approximation because the logitudinal motion of the string depends on the Young's modulus of the string in addition to its tension. The longitudinal waves do not move at the same speed as the transverse waves.

I'm not sure that you have computed ${T^1}_1$ correctly. I will check that my second equation follows from the wave equation.

Checked: my second equation follows from the wave equation, so $$ {T^1}_1= \left\{\frac \rho 2\dot y^2 +\frac T 2 y'^2\right\} $$ There is no minus sign. The same is true for $T_{11}$. You probably slipped up because $\partial^\mu y= g^{\mu\nu}\partial_\nu y\ne \partial_\mu y$.


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