Meaning of kinetic part in the Lagrangian density?

Question

What is the physical meaning of the kinetic term in the classical scalar field Lagrangian $$\mathcal{L}_{kin}~=~\frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi)~?$$ It gives how does the field change from one space-time point to the other. Right? What is the significance of the relative minus sign between the square of the time gradient and square of the space gradient?

Answer 1

Think of a two dimensional horizontal elastic sheet in three dimensions. Suppose we specify the height everywhere with $\phi(x,y)$.

Then if the sheet moves up and down there is kinetic energy. This kinetic energy is proportional to $\dot{\phi}^2$ because $\dot{\phi}$ is telling you the velocity of that point on the sheet.

Also, suppose we want to pull one point on the sheet up while keeping a neighboring point fixed. This will cost elastic potential energy. The amount of potential energy is proportional to $(\nabla\phi)^2$.

Thus we get a contribution to the lagrangian of the form $(\partial_t \phi)^2 - (\nabla\phi)^2 = \partial_\mu \phi \partial^\mu \phi$. I googled and found a pdf (titled week 7 lecture: concepts of QFT by some one named Andrew Forrester) which does this in more detail. I haven't read it though, so I can't vouch for it too much. You would only need to read the second and third pages.

Answer 2

The equation of motion corresponding with $\mathcal{L}_{kin}$ is $$(∂_{t}^2-{∂_{\mathbf{x}}}^2)φ=0,$$ the Klein-Gordon equation, which has its origin in relativistic field theory. The minus sign is essential for relativistic invariance and leads to propagating solutions (waves). With$$φ(\mathbf{x},t)=\text{exp}[iωt]ψ(\mathbf{x}).$$ we obtain$$ω^2ψ(\mathbf{x})=-∂_{\mathbf{x}}^2ψ(x)$$ Since $-∂_{\mathbf{x}}^2$ is a non-negative operator, $ω^2≥0$. Thus, if the sign was the opposite, there would be no oscillating solutions. More generally, there would be no propagating solutions.

Answer 3

First let's rewrite the kinetic term $\mathcal{L_{kin}}$ such that the metric $\eta^{\mu \nu}$ is explict $$\mathcal{L_{kin}}=\frac 12 \eta^{\mu \nu}\partial_\mu \phi\partial_\nu\phi$$

In this answer we will restrict our attention to 4-dimensional Minkowski space-time $M^4$, the flat "arena" of special relativity, and set the speed of light $c=1$. The metric corresponding to this space is often written $$\eta^{\mu \nu}=(+,-,-,-)$$ which represents the matrix

$$\ \left( \begin{array}{ccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 &-1\end{array} \right)\ $$ Alternatively, it is important to note that some authors choose to define the metric in the form $(-,+,+,+)$. However, the metric $\eta_{\mu\nu}$ is written to be in accord with your question. The importance of the metric is the notion of distance between points in Minkowski space-time being meaningful. Contrast this metric with the standard Euclidean metric and note the positivity of the diagonal elements in the latter.

The answer to the kinetic term $\mathcal{L_{kin}}$ is rather rough. But you can think of the term as representing the energy of motion of the field. Moreover, as you have written an explicit kinetic term you have assumed that the Lagrangian has a natural splitting into kinetic energy and potential terms. The gradient term gives a contribution to the potential energy in addition to the potential $U(\phi)$. It is important that the potential energy is bounded from below, otherwise the dynamics is liable to produce singular fields. Now we are led to the justification of our choice of sign in front of the gradient term.

To delve further into QFT I would recommend reading David Tong's notes. Here is a link corresponding to the classical field theory portion of the class, which touches on your question http://www.damtp.cam.ac.uk/user/dt281/qft/one.pdf