Free boson Equation motion from action

So in David tongs notes we have

$$S=\frac{m}{8\pi}\int d^2x\partial_i\varphi\partial^i\varphi$$

and he finds that the equation of motion is

$$[\partial_{t}^2-v^2\partial_{x}^2]\varphi=0$$

now my problem is, how he derived this eom, because the metric etc. was not expliticly given, so whats the metric of this action and how do you get this eom, from there.

• The metric can be guessed from the context. If it is not in the context of gravity/GR, then it is certainly not a (Pseudo)-Riemannian metric. Furthermore, if it is not in the context of Newton's classical physics, the 3-dim metric neither. As it is an action of a field, it is rather clear that it must be the Minkowski metric. To treat a field with an infinite propagation speed in modern textbooks would seem awkward. But $\varphi$ could be a field-like displacement in a cristal. Then the propagation speed is not necessarily speed of light. Check over which values the index $i$ runs. – Frederic Thomas Apr 10 at 10:08
• i runs over x and t varphi is free boson. it is flat minkovski metric but, we probably have v instead of c. however, still i don't know how that v^2 comes. if this is given in open form it is easy to solve el equations. – physshyp Apr 10 at 10:25
• $\partial_i \varphi= \frac{\partial \varphi}{\partial x^i}$. So if $x^i$ runs over "t" and "x", then all components $x^i$ have to have the same dimensions, so that's the reason why the velocity comes in. The velocity is not part of the metric. $\eta_{Minkowski}=diag(1,-1,-1,-1)$ or shortened to $diag(1,-1)$ (other conventions are possible which change the total sign, but not relative signs). The velocity is in the coordinates $x^i$. – Frederic Thomas Apr 10 at 10:34
• would it be possible for you to derive that eom starting from that action with the assumptions in your comment, i would be really glad. – physshyp Apr 10 at 10:38

The EOMs are obtained from the action by requiring the action to be stationary with respect to the variation of the field we are interested in ($$S = \int dx^2 L$$ with $$L$$ called Lagrangian, summation is implicitly applied over double appearing indices, here $$i$$):

$$0=\frac{\delta S [\varphi]}{\delta \varphi } =\frac{\partial L}{\partial \varphi } - \frac{\partial}{\partial x^i}\left( \frac{\partial L}{\partial(\partial_i \varphi)}\right) = - \frac{\partial}{\partial x^i}\left( \frac{\partial L}{\partial(\partial_i \varphi)}\right)$$.

We benefit here from the fact that the Lagrangian only depends on the derivatives of the field and not the field itself.

So we only have to compute the derivative of $$\frac{\partial L}{\partial(\partial_i \varphi)}$$:

$$\frac{\partial L}{\partial(\partial_i \varphi)} = \frac{m}{4\pi} \partial^i \varphi$$

so

$$\frac{\partial}{\partial x^i}\left( \frac{\partial L}{\partial(\partial_i \varphi)}\right)= \frac{m}{4\pi} \partial_i\partial^i \varphi$$

According to the requirement of the stationarity of the action this expression is zero, so :

$$0=\partial_i\partial^i \varphi\equiv \frac{\partial^2 \varphi}{v^2\partial t^2} - \frac{\partial^2 \varphi}{\partial x^2}$$

Here the following definitions of the coordinates were used: $$x^i=(vt, x)$$ and $$x_i=(vt, -x)$$.

• thanks a lot. final question in action whats the form of integrant do we have $d^2x=dx^1dx^2=vdtdx$ – physshyp Apr 10 at 11:25
• Yes. One could of course wonder about the Lorentz invariance of such a volume element for $v\neq c$, but, I think, it cannot be required for a field propagating through a medium making $v<c$. – Frederic Thomas Apr 10 at 12:31