4
$\begingroup$

The lagrangian of the Real Scalar Field $\phi$ is given by

\begin{equation} \mathcal{L} = \frac{1}{2}\eta^{\mu \nu} \partial _\mu \phi \partial _{\nu} \phi - \frac{1}{2} m^2 \phi^2 \end{equation} \begin{equation} \eta^{\mu \nu} = diag(+1,-1,-1,-1) \end{equation} But what is the real motivation behind this equation?

I've read in several books this equation. In some books it is "defined", in others is given a completely unsatisfactory derivation for this equation , like "let's make an analogy with a coupled oscillators system". I've understood that the natural units makes the Lagrangian have dimension of $[mass]^4$ and the fact that the Lagrangian must be a Lorentz Invariant so the derivatives of the field must be $\partial _\mu \phi$, but how can i get the "1/2" coefficient in this Lagrangian and where does come from this "$m^2 \phi^2$? The "$m$" coefficient is the mass of the field or its just a constant with dimension of mass? Please help with this issue.

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ What is the motivation behind any lagrangian? $\endgroup$ Commented Jan 10, 2020 at 0:56

2 Answers 2

Reset to default
2
$\begingroup$

As you say, when building a Lagrangian for a theory, we almost always wish it to be a Lorentz scalar. Obviously, we can take any power of the scalar field, $\phi$, that is, we can have terms of the form,

$$\mathcal L = \dots + \sum_{n\geq 0} c_n \phi^n + \dots$$

with coefficients of our choosing. This means we could consider a term like $\sin \phi$ since upon Taylor expansion, at least at the level of the Lagrangian, it amounts to simply adding an infinite number of $\phi^n$ terms (this results in Sine-Gordon theory, incidentally).

We can also consider derivatives of the field $\partial_\mu \phi$ and indeed combining two of them, we can produce the scalar $\partial^\mu \phi \partial_\mu \phi$. A Lagrangian with just this term gives us the wave equation, $\square \phi = 0$, which as you should be aware is fundamental and ubiquitous in physics. This alone is sufficient motivation to consider such a Lagrangian for the scalar.

Adding a mass term gives the Klein-Gordon equation; that it is a "mass" term (distinguished from other powers of $\phi$ one could add) is shown by quantising the theory and considering scattering. Finally the overall factor of $\frac12$ is a convenience. We could put any constant in front, it won't change the physics.

Improve this answer
$\endgroup$
1
$\begingroup$

The Euler-Lagrange equation following from this lagrangian is the Klein-Gordon equation, which is the field theoretical equivalent of the special relativistic energy-mass-momentum equation.

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.