# Simple geometrical methods on scalar field

Some prerequisites: some exposures with classical field theory would help

1. Introduction

For definiteness, let us study scalar field on flat ${{\mathbb R}^4.}$ Physicists often call a quantity ${{\phi}(t,\vec{x})}$ as scalar field. because it is a function of space and time. To mathematicians’ eyes, I guess [I am not a mathematician, so I can only assume this] that ${{\phi}(t,\vec{x})}$ is viewed merely as a number. Instead, the scalar field is given by a scalar function ${{\phi}:{\mathbb R}^4\rightarrow{\mathbb R}.}$ The arguments ${(t,\vec{x})}$ are in the domain whereas ${{\phi}(t,\vec{x})}$ takes the value in the codomain.

As physicists, we should already be familiar with viewing ${{\phi}(t,\vec{x})}$ as a scalar field. But in this post, let us take a peek on the alternative point of view. I believe this point of view is preferred not only by mathematicians, but also by some physicists.

2. Klein-Gordon Lagrangian

To describe

$\displaystyle {\phi}(t,\vec{x}), \ \ \ \ \ (1)$

let us define a map

$\displaystyle {\phi}_t:{\mathbb R}^3\rightarrow{\mathbb R} \ \ \ \ \ (2)$

such that ${{\phi}_t(\vec{x}) = {\phi}(t,\vec{x}).}$ There are other maps which are also useful. Define

$\displaystyle {\phi}_t^2:{\mathbb R}^3\rightarrow{\mathbb R} \ \ \ \ \ (3)$

such that

$\displaystyle \begin{array}{rl} {\phi}_t^2(\vec{x}) &={\phi}_t(\vec{x}){\phi}_t(\vec{x})\\ &=({\phi}(t,\vec{x}))^2. \end{array} \ \ \ \ \ (4)$

In general, a multiplication between two functions ${f,g:{\mathbb R}^3\rightarrow{\mathbb R}}$ is given by the function ${fg:{\mathbb R}^3\rightarrow{\mathbb R}}$ such that

$\displaystyle (fg)(\vec{x}) = f(\vec{x})g(\vec{x}). \ \ \ \ \ (5)$

This is a typical definition of multiplication of scalar functions in the context of differential geometry. Let us also define some differentiations. The function

$\displaystyle \dot{\phi}_t:{\mathbb R}^3\rightarrow{\mathbb R} \ \ \ \ \ (6)$

is defined as [I hope I do not need to make use of ${{\epsilon}-{\delta}}$]

$\displaystyle \dot{\phi}_t \equiv \lim_{h\rightarrow 0}\frac{{\phi}_{t+h}-{\phi}_t}{h}. \ \ \ \ \ (7)$

Sometimes, it is more convenient to write ${{\partial}_0{\phi}_t}$ instead of ${\dot{\phi}_t.}$ But in any case, we will avoid (at least in this post) writing “${{\partial}_t{\phi}_t.}$” This is because ${t}$ is a parameter, and can be replaced by any number. So using this choice for example would lead to

$\displaystyle \textrm{}\left({\partial}_4{\phi}_4 = \lim_{h\rightarrow 0}\frac{{\phi}_{4+h}-{\phi}_4}{h}\right)\textrm{''}, \ \ \ \ \ (8)$

whose LHS is mentally irritating. Let us also define

$\displaystyle {\partial}_1{\phi}_t:{\mathbb R}^3\rightarrow{\mathbb R} \ \ \ \ \ (9)$

such that

$\displaystyle {\partial}_1{\phi}_t(x^1,x^2,x^3) =\lim_{h\rightarrow 0}\frac{{\phi}_t(x^1+h,x^2,x^3)-{\phi}_t(x^1,x^2,x^3)}{h}. \ \ \ \ \ (10)$

The functions ${{\partial}_2{\phi}_t, {\partial}_3{\phi}_t}$ can also be defined in a similar way. The “trace” of a function ${f:{\mathbb R}^3\rightarrow{\mathbb R}}$ is given by

$\displaystyle \lbrack f\rbrack =\int d^3\vec{x} f(\vec{x}). \ \ \ \ \ (11)$

With the above ingredients, the Klein-Gordon Lagrangian is given by

$\displaystyle L_t = \left\lbrack{\dfrac{1}{2}}\dot{\phi}_t^2 - {\dfrac{1}{2}}({\partial}_i{\phi}_t)({\partial}_i{\phi}_t) - {\dfrac{1}{2}} m^2{\phi}_t^2\right\rbrack. \ \ \ \ \ (12)$

It is a functional of ${{\phi}_t}$ and ${\dot{\phi}_t.}$ So

$\displaystyle \begin{array}{rl} {\delta} L_t =\left\lbrack \dfrac{{\delta} L_t}{{\delta}\dot{\phi}_t}{\delta}\dot{\phi}_t + \dfrac{{\delta} L_t}{{\delta}{\phi}_t}{\delta}{\phi}_t \right\rbrack. \end{array} \ \ \ \ \ (13)$

On the other hand, direct computation gives

$\displaystyle \begin{array}{rl} {\delta} L_t &=\left\lbrack \dot{\phi}_t{\delta}\dot{\phi}_t - {\partial}_i{\phi}_t{\delta}({\partial}_i{\phi}_t) - m^2{\phi}_t{\delta}{\phi}_t\right\rbrack\\ &\\ &=\left\lbrack \dot{\phi}_t{\delta}\dot{\phi}_t + {\partial}_i{\partial}_i{\phi}_t{\delta}{\phi}_t - m^2{\phi}_t{\delta}{\phi}_t \right\rbrack - \left\lbrack {\partial}_i({\partial}_i{\phi}_t{\delta}{\phi}_t)\right\rbrack, \end{array} \ \ \ \ \ (14)$

where we used the identity ${{\delta}({\partial}_i{\phi}_t) = {\partial}_i({\delta}{\phi}_t),}$ which can be shown directly from the definition. The last term on RHS of eq.(14) gives surface integral at spatial infinity. Let us assume that this vanishes. So we can now read off

$\displaystyle \dfrac{{\delta} L_t}{{\delta}\dot{\phi}_t} =\dot{\phi}_t,\qquad \dfrac{{\delta} L_t}{{\delta}{\phi}_t} ={\partial}_i{\partial}_i{\phi}_t - m^2{\phi}_t. \ \ \ \ \ (15)$

Euler-Lagrange’s equation

$\displaystyle {\partial}_0\dfrac{{\delta} L_t}{{\delta}\dot{\phi}_t} -\dfrac{{\delta} L_t}{{\delta}{\phi}_t} = 0 \ \ \ \ \ (16)$

gives

$\displaystyle \ddot{\phi}_t - {\partial}_i^2{\phi}_t + m^2{\phi}_t = 0, \ \ \ \ \ (17)$

which is Klein-Gordon equation.

Note the version of Euler-Lagrange’s equation (16). This version is often seen in Classical Mechanics but rarely appeared in Classical Field Theory. Physics students are often introduced to Classical Field Theory along the line of “Like in Classical Mechanics, there is also Euler-Lagrange’s equation. But we have to proceed differently to get Euler-Lagrange’s equation

$\displaystyle \textrm{ }{\partial}_{\mu}\dfrac{{\partial}{\cal L}}{{\partial}{\partial}_{\mu}{\phi}} - \dfrac{{\partial}{\cal L}}{{\partial}{\phi}} = 0\textrm{ ''} \ \ \ \ \ (18)$

which is tailored-made for Classical Field Theory”. From my direct experience, this approach makes Classical Field Theory to look far more complicated than Classical Mechanics and that there is an unforeseeable gap between them which cannot easily be filled in.

So hopefully our use of Euler-Lagrange’s equation (16) should make the connection between Classical Mechanics and Classical Field Theory a bit clearer.

3. Klein-Gordon Hamiltonian

To obtain Hamiltonian, we first look for conjugate momenta of ${{\phi}_t.}$ It is given by

$\displaystyle \begin{array}{rl} {\pi}_t \equiv\dfrac{{\delta} L_t}{{\delta}\dot{\phi}_t} =\dot{\phi}_t. \end{array} \ \ \ \ \ (19)$

Hamiltonian is then the functional of ${{\phi}_t, {\pi}_t.}$ It is given by

$\displaystyle \begin{array}{rl} H_t &=\lbrack{\pi}_t\dot{\phi}_t\rbrack - L_t\\ &\\ &=\left\lbrack{\dfrac{1}{2}}{\pi}_t^2 + {\dfrac{1}{2}}({\partial}_i{\phi}_t)({\partial}_i{\phi}_t) + {\dfrac{1}{2}} m^2{\phi}_t^2\right\rbrack. \end{array} \ \ \ \ \ (20)$

Let us next compute Poisson’s bracket. In the case of Classical Field Theory, we are often taught to start from the identity

$\displaystyle \{{\phi}_t(\vec{x}),{\phi}_t(\vec{y})\} = 0,\qquad \{{\phi}_t(\vec{x}),{\pi}_t(\vec{y})\} = {\delta}^{(3)}(\vec{x}-\vec{y}),\qquad \{{\pi}_t(\vec{x}),{\pi}_t(\vec{y})\} = 0, \ \ \ \ \ (21)$

where ${{\delta}^{(3)}(\vec{x}-\vec{y})}$ is 3d Dirac’s delta function. However, there is an alternative way to avoid direct manipulation with Dirac’s delta function.

For this, let us view ${{\phi}_t, {\pi}_t}$ as coordinates of some manifold ${{\cal W}}$ [NB: physics students exposed to General Relativity would tend to think of manifold as associated only to spacetime. However, this viewpoint is too limited. It is better to think of Manifold as some mathematical concepts which has wide applications. The description of spacetime as manifold is only one application in physics.]. So ${{\delta}{\phi}_t, {\delta}{\pi}_t}$ are one-forms on ${{\cal W},}$ whereas

$\displaystyle \dfrac{{\delta}}{{\delta}{\phi}_t},\qquad \dfrac{{\delta}}{{\delta}{\pi}_t}, \ \ \ \ \ (22)$

are vectors on ${{\cal W}.}$ The one-forms and vectors are dual in the sense that

$\displaystyle \lbrack f{\delta}{\phi}_t\rbrack\left(\dfrac{{\delta}}{{\delta}{\phi}_t}\right) =f,\qquad \lbrack f{\delta}{\phi}_t\rbrack\left(\dfrac{{\delta}}{{\delta}{\pi}_t}\right) =0, \ \ \ \ \ (23)$

$\displaystyle \lbrack f{\delta}{\pi}_t\rbrack\left(\dfrac{{\delta}}{{\delta}{\phi}_t}\right) =0,\qquad \lbrack f{\delta}{\pi}_t\rbrack\left(\dfrac{{\delta}}{{\delta}{\pi}_t}\right) =f, \ \ \ \ \ (24)$

where ${f:{\mathbb R}^3\rightarrow{\mathbb R}}$ is a well-behaved arbitrary function. Next, let us define a linear map ${\Omega}$ which changes a one-form to a vector as follows

$\displaystyle \Omega\left({\delta}{\phi}_t\right) =\dfrac{{\delta}}{{\delta}{\pi}_t},\qquad \Omega\left({\delta}{\pi}_t\right) =-\dfrac{{\delta}}{{\delta}{\phi}_t}. \ \ \ \ \ (25)$

Let us define a vector field (called Hamiltonian vector field) of a scalar function ${F}$ on ${{\cal W}}$ as

$\displaystyle X_F\equiv\Omega({\delta} F). \ \ \ \ \ (26)$

So it can be shown that Poisson’s bracket of scalar functions ${F,G}$ on ${{\cal W}}$ is given by

$\displaystyle \{F,G\} = {\delta} G(X_F). \ \ \ \ \ (27)$

We want to compute ${\{{\phi}_t,H_t\}, \{{\pi}_t,H_t\}.}$ For this, we first compute

$\displaystyle {\delta} H_t =\left\lbrack {\pi}_t{\delta}{\pi}_t + (m^2{\phi}_t - {\partial}_i{\partial}_i{\phi}_t){\delta} {\phi}_t\right\rbrack. \ \ \ \ \ (28)$

So

$\displaystyle \{{\phi}_t,H_t\} ={\delta} H_t\left(\dfrac{{\delta}}{{\delta}{\pi}_t}\right) = {\pi}_t, \ \ \ \ \ (29)$

$\displaystyle \{{\pi}_t,H_t\} ={\delta} H_t\left(-\dfrac{{\delta}}{{\delta}{\phi}_t}\right) = {\partial}_i{\partial}_i{\phi}_t-m^2{\phi}_t, \ \ \ \ \ (30)$

4. Solution to equation of motion

This section is not so related to the title of the post. This means that we do not make use of any geometrical methods in the analysis [It might be possible to use one, but I still do not have enough knowledge to do so or to see if it is really possible]. However, I include this section just to make this post a bit more complete.

From either Lagrangian or Hamiltonian analyses, we obtain equation of motion for ${{\phi}_t:}$

$\displaystyle \ddot{\phi}_t - {\partial}_i{\partial}_i{\phi}_t + m^2{\phi}_t = 0. \ \ \ \ \ (31)$

Recall that LHS of this equation is in fact a function ${{\mathbb R}^3\rightarrow{\mathbb R}.}$ So let us apply it on ${\vec{x}\in{\mathbb R}^3.}$ For this, let us consider the Fourier transform

$\displaystyle {\phi}_t(\vec{x}) =\int\dfrac{d^3\vec{k}}{(2{\pi})^3}e^{i\vec{k}\cdot\vec{x}}\tilde{\phi}_t(\vec{k}). \ \ \ \ \ (32)$

Substituting this into the equation of motion (31) (whose LHS is already applied to ${\vec{x}}$) gives

$\displaystyle \int\dfrac{d^3\vec{k}}{(2{\pi})^3} e^{i\vec{k}\cdot\vec{x}}(\ddot{\tilde{\phi}}_t(\vec{k}) + (\vec{k}^2+m^2)\tilde{\phi}_t(\vec{k})) =0. \ \ \ \ \ (33)$

By inverse Fourier transforming, we obtain

$\displaystyle \ddot{\tilde{\phi}}_t(\vec{k}) + (\vec{k}^2+m^2)\tilde{\phi}_t(\vec{k}) =0. \ \ \ \ \ (34)$

For each ${\vec{k},}$ this equation is just SHO equation. So it is easy to see (by high school or beginning undergraduate physics students in Thailand) that the solution is given by

$\displaystyle \tilde{{\phi}}_t(\vec{k}) =A_{\vec{k}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + B_{\vec{k}}e^{i\sqrt{\vec{k}^2+m^2}\;t}. \ \ \ \ \ (35)$

But reality condition ${{\phi}_t(\vec{x})^* = {\phi}_t(\vec{x})}$ implies that ${\tilde{\phi}_t(\vec{k}) = \tilde{\phi}_t(-\vec{k})^*,}$ which in turn implies ${B_{\vec{k}} = A_{-\vec{k}}^*,}$ and hence

$\displaystyle \tilde{{\phi}}_t(\vec{k}) =A_{\vec{k}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + A_{-\vec{k}}^*e^{i\sqrt{\vec{k}^2+m^2}\;t}. \ \ \ \ \ (36)$

This gives

$\displaystyle \begin{array}{rl} {\phi}_t(\vec{x}) &=\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}e^{i\vec{k}\cdot\vec{x}}\left(A_{\vec{k}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + A_{-\vec{k}}^*e^{i\sqrt{\vec{k}^2+m^2}\;t}\right)\\ &\\ &=\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\left(A_{\vec{k}}e^{i\vec{k}\cdot\vec{x}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + A_{\vec{k}}^*e^{-i\vec{k}\cdot\vec{x}}e^{i\sqrt{\vec{k}^2+m^2}\;t}\right)\\ &\\ &=\displaystyle 2\textrm{Re}\int\dfrac{d^3\vec{k}}{(2{\pi})^3}A_{\vec{k}}e^{i\vec{k}\cdot\vec{x}}e^{-i\sqrt{\vec{k}^2+m^2}\;t}. \end{array} \ \ \ \ \ (37)$

Sometimes, it is preferable to write RHS in terms of relativistic quantities. Based on the insight from Special Relativity, one could expect to put ${t}$ and ${\vec{x}}$ on equal footing. Similarly, one would expect to put ${{\omega}}$ and ${\vec{k}}$ on equal footing. For this, let us make use of the trick of Dirac’s delta function to get

$\displaystyle \begin{array}{rl} {\phi}_t(\vec{x}) &=2\textrm{Re}\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\int_0^\infty d{\omega} A_{\vec{k}}e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}{\delta}({\omega}-\sqrt{\vec{k}^2+m^2}). \end{array} \ \ \ \ \ (38)$

The exponent is already in the relativistic form: ${-i{\omega} t+i\vec{k}\cdot\vec{x} = -ik^{\mu} x_{\mu}.}$ However, the argument of the Dirac’s delta function is still not. So let us make use of the identity

$\displaystyle {\delta}({\omega}^2-\vec{k}^2 - m^2) =\dfrac{{\delta}({\omega} -\sqrt{\vec{k}^2+m^2}) + {\delta}({\omega} +\sqrt{\vec{k}^2+m^2})}{2\sqrt{\vec{k}^2+m^2}}. \ \ \ \ \ (39)$

So by using the fact that ${{\delta}({\omega} +\sqrt{\vec{k}^2+m^2})}$ vanishes for ${{\omega}\in[0,\infty),}$ we obtain

$\displaystyle \begin{array}{rl} {\phi}_t(\vec{x}) &=2\textrm{Re}\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\int_0^\infty d{\omega} A_{\vec{k}}e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}2\sqrt{\vec{k}^2+m^2}{\delta}({\omega}^2-\vec{k}^2-m^2). \end{array} \ \ \ \ \ (40)$

The coefficient would look nicer if we set

$\displaystyle c_{\vec{k}} \equiv 2\sqrt{\vec{k}^2 + m^2}A_{\vec{k}}, \ \ \ \ \ (41)$

[NB: This choice has no physical motivation. It just only makes the presentation a bit nicer. The choice which often appears in Classical Field Theory is ${a_{\vec{k}} \sim (\vec{k}^2 + m^2)^{1/4}A_{\vec{k}}.}$ In order to discuss this choice, I will need to include other discussions. But I do not intend to do it in this post.] giving

$\displaystyle \begin{array}{rl} {\phi}_t(\vec{x}) &=2\textrm{Re}\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\int_0^\infty d{\omega} c_{\vec{k}}e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}{\delta}({\omega}^2-\vec{k}^2-m^2)\\ &\\ &=4{\pi}\textrm{Re}\displaystyle\int\dfrac{d^4k}{(2{\pi})^4} \Theta({\omega})c_{\vec{k}}e^{-i k^{\mu} x_{\mu}}{\delta}(k^{\mu} k_{\mu}-m^2). \end{array} \ \ \ \ \ (42)$

Perhaps this form is the most relativistic that one could get. As a cross-check, one could integrate out ${{\omega}}$ to obtain the original form

$\displaystyle {\phi}_t(\vec{x}) =\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\frac{1}{2\sqrt{\vec{k}^2+m^2}}\left(c_{\vec{k}}e^{i\vec{k}\cdot\vec{x}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + c_{\vec{k}}^*e^{-i\vec{k}\cdot\vec{x}}e^{i\sqrt{\vec{k}^2+m^2}\;t}\right). \ \ \ \ \ (43)$

Let us substitute the solution into the Hamiltonian (20). For this, it is convenient if we first rewrite

$\displaystyle {\phi}_t(\vec{x}) =\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\frac{1}{2\sqrt{\vec{k}^2+m^2}}\left(c_{\vec{k}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + c_{-\vec{k}}^*e^{i\sqrt{\vec{k}^2+m^2}\;t}\right)e^{i\vec{k}\cdot\vec{x}}. \ \ \ \ \ (44)$

Then we compute

$\displaystyle \begin{array}{rl} {\pi}_t(\vec{x}) &=\dot{\phi}_t(\vec{x})\\ &\\ &=\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\frac{i}{2}\left(-c_{\vec{k}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + c_{-\vec{k}}^*e^{i\sqrt{\vec{k}^2+m^2}\;t}\right)e^{i\vec{k}\cdot\vec{x}}. \end{array} \ \ \ \ \ (45)$

Any well-behaved functions ${f,g:{\mathbb R}^3\rightarrow{\mathbb R}}$ can be written as a Fourier transform

$\displaystyle f(\vec{x}) =\displaystyle\int \dfrac{d^3\vec{k}}{(2{\pi})^3}e^{i\vec{k}\cdot\vec{x}}\tilde f(\vec{k}), \ \ \ \ \ (46)$

$\displaystyle g(\vec{x}) =\displaystyle\int \dfrac{d^3\vec{k}}{(2{\pi})^3}e^{i\vec{k}\cdot\vec{x}}\tilde g(\vec{k}). \ \ \ \ \ (47)$

Direct computation gives

$\displaystyle \lbrack fg\rbrack =\displaystyle\int \dfrac{d^3\vec{k}}{(2{\pi})^3}\tilde f(\vec{k})\tilde g(-\vec{k}). \ \ \ \ \ (48)$

Using this identity, it is easy to show that

$\displaystyle \lbrack{\pi}_t^2\rbrack =-\dfrac{1}{4}\int\dfrac{d^3\vec{k}}{(2{\pi})^3} (c_{\vec{k}}c_{-\vec{k}}e^{-2i\sqrt{\vec{k}^2+m^2}\;t} -2|c_{\vec{k}}|^2 + c_{\vec{k}}^*c_{-\vec{k}}^*e^{2i\sqrt{\vec{k}^2+m^2}\;t}), \ \ \ \ \ (49)$

$\displaystyle \lbrack{\partial}_i{\phi}_t{\partial}_i{\phi}_t\rbrack =\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\dfrac{\vec{k}^2}{4{\omega}_{\vec{k}}^2} (c_{\vec{k}}c_{-\vec{k}}e^{-2i\sqrt{\vec{k}^2+m^2}\;t} +2|c_{\vec{k}}|^2 + c_{\vec{k}}^*c_{-\vec{k}}^*e^{2i\sqrt{\vec{k}^2+m^2}\;t}), \ \ \ \ \ (50)$

$\displaystyle \lbrack{\phi}_t^2\rbrack =\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\dfrac{1}{4{\omega}_{\vec{k}}^2} (c_{\vec{k}}c_{-\vec{k}}e^{-2i\sqrt{\vec{k}^2+m^2}\;t} +2|c_{\vec{k}}|^2 + c_{\vec{k}}^*c_{-\vec{k}}^*e^{2i\sqrt{\vec{k}^2+m^2}\;t}). \ \ \ \ \ (51)$

Combining these results give

$\displaystyle H_t ={\dfrac{1}{2}}\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3} |c_{\vec{k}}|^2. \ \ \ \ \ (52)$

So Hamiltonian of the system is time-independent.

Let us simply end this post here.