Lagrangian for scalar field in terms of Klein Gordon equation

Question

I am studying Peskin and Schroeder, at page 287 , Lagrangian for scalar field is

$$L={1\over 2}(\partial _\mu \phi )^2-{1\over 2}m^2 \phi^2.$$

It can be rewritten as

$$L={1\over 2} \phi (-\partial^2-m^2)\phi . $$

I don't understand how it is rewritten, Your little helps may save my lot of time.

Compute the difference and consider what effect that difference has in the action. — G. Smith, May 13 '20 at 20:17
Related: Does a four-divergence extra term in a Lagrangian density matter to the field equations? — G. Smith, May 13 '20 at 20:24
You should take a momentary step back from QFT, review the principle of stationary action for a classical particle, and understand why a total time derivative added to $L$ doesn’t affect the dynamics. In field theory this generalizes to a four-divergence added to the Lagrangian density being irrelevant. — G. Smith, May 14 '20 at 00:46

score 3 · Accepted Answer · answered May 13 '20 at 20:35

3

Let us look at the first term in your Lagrangian:

$$\partial_\mu \phi \partial^\mu \phi$$

Now, for a general derivative $ \partial_x $ we have:

$$\partial_x \left(y \partial_x y\right) = (\partial_x y)^2 + y \partial^2_x y$$ $$\Rightarrow (\partial_x y)^2 = \partial_x \left(y \partial_x y\right) - y \partial^2_x y$$

If you substitute the above for $\partial_\mu \phi \partial^\mu \phi$ in your Lagrangian, you get two terms, one of which is of the form $\partial_\mu (...)$. This term is a total derivative and does not contribute to the equations of motion. This is because in the action,

$$S = \int d\vec{x} dt L$$

the total derivative term will integrate out to give constant. And when we calculate equation of motion, that constant will be differentiated to zero.

answered May 13 '20 at 20:35

Feynman's Cat

375

Why total derivative term will integrate out to give constant @Feynman's Cat – ROBIN RAJ May 14 '20 at 14:42
Let's say that your total derivative term is $\partial_x f(\phi)$. While calculating action, that term gives $\int_{x_1}^{x2} dx \partial_x f = f(\phi(x_2))-f(\phi(x_1))$ which only depends on initial and final coordinates, $x_1$ and $x_2$. So when we calculate the equations of motion, we calculate change in action $\delta S$ due to a change in coordinate (path) $\delta \phi(x)$, and we extremize it while taking $\delta \phi(x) = 0$ at the end points. So, $\delta \phi(x_1) = 0 =\delta \phi(x_2)$. Then $\delta(f(\phi(x_2))-f(\phi(x_1)))=0$ and this term is not present in equations of motion. – Feynman's Cat May 14 '20 at 23:12
Ok this helped ,thanks @Feynman's Cat – ROBIN RAJ May 15 '20 at 13:51

Noone · Answer 2 · 2020-05-13T20:30:49.030

2

The two Lagrangians are the same modulo total spacetime divergence terms. If you vary with respect to the scalar field the action constructed from the lagrangian you will obtain the Klein-Gordon equation. People usually write the kinetic term of the scalar field as $(\partial_{μ} \phi)^2$. You can write it as $\phi \partial^2 \phi$. Both these two expressions after variation will yield $\partial^{2}\phi$.

($\phi \partial^2 \phi$ contains second derivatives of the dynamical variable. Be careful with Euler Lagrange equations.)

edited May 13 '20 at 20:30

answered May 13 '20 at 20:21

Noone

2,538

Answer is not clear – ROBIN RAJ May 13 '20 at 20:22

Lagrangian for scalar field in terms of Klein Gordon equation

2 Answers2