9.3 The variational principle with three-dimensional fields

We now generalize to the case where the Lagrangian density depends on a field that is defined in three dimensional space, \(\eta(x,y,z,t)\). The Lagrangian density, in general, will be expressed as:

\begin{align} \mathcal{L}=\mathcal{L}(\eta,\frac{d\eta}{dt},\frac{d\eta}{dx},\frac{d\eta}{dy},\frac{d\eta}{dz},t,x,y,z) \tag{9.27}\end{align}

It should be immediately apparent that the notation is going to be cumbersome if keep all three space coordinates and time as we apply the variational principle. One particularly interesting observation in the previous derivation for the one-dimensional field, is that the variables \(t\) and \(x\) were treated in the same way. There was nothing special about \(t\); it was just another variable that the field depended on. With that in mind, we introduce a new notation where \(t\) is treated exactly as another one of the space coordinates. Let the 4-dimensional vector, \(x^\mu\), have the following components:

\begin{align} x^\mu\equiv(ct,x,y,z) \tag{9.28}\end{align}

where the index \(\mu\) goes from 0 to 3. We have multiplied \(t\) by a constant, \(c\), that has dimension of speed so that \(ct\) has dimensions of length along with the other three components. Additionally, we introduce the convention that when \(x\) is indexed with a roman letter, the indices run from 1 to 3 (the three space coordinates), and when \(x\) is indexed with a greek letter, the indices run from 0 to 3 (all four “coordinates”). The Lagrangian is thus given by:

\begin{align} L=\int \lagd \,dx^i \tag{9.29}\end{align}

where \(dx^i\), with the roman index, stands for \(dx\,dy\,dz\). The action is then given by:

\begin{align} S=\int L dt = \int \int \lagd \,dx^i \,dt =\frac{1}{c} \int \lagd \,dx^\mu \tag{9.30}\end{align}

where \(\mu\) now runs over all four coordinates and we had to divide by \(c\) since \(dt = c dx^0\). Again, note that the Lagrangian does not treat time and the space coordinates in the same way (\(dx^i\) instead of \(dx^\mu\)). However, the equations of motion are obtained from the action, where all four coordinates can be combined.

We thus write the Lagrangian density as:

\begin{align} \lagd(\eta,\frac{d\eta}{dt},\frac{d\eta}{dx},\frac{d\eta}{dy},\frac{d\eta}{dz},t,x,y,z)\to\lagd(\eta,\frac{d\eta}{dx^\mu},x^\mu) \tag{9.31}\end{align}

We also introduce the “Einstein summation convention”, where repeated indices occurring as a product must be summed over. For example, \(x_\mu x^\mu\equiv x_0x^0+x_1x^1+x_2x^2+x_3x^3\), where, at this point, there is no distinction between an index being a subscript or a superscript.

Example 9-2

Use the Einstein summation convention to simplify \(x_\nu y^\mu \delta^\nu_\mu\),where \(\delta^\nu_\mu\) is the “Kronecker delta” (equal to 1 if \(\nu\) = \(\mu\) and 0 otherwise).

Since the Einstein summation convention implies that we sum over repeated indices, we have:

\begin{align*} x_\nu y^\mu \delta^\nu_\mu&=\sum_{\nu=0}^3\sum_{\mu=0}^3x_\nu y^\mu \delta^\nu_\mu\\ &=\sum_{\nu=0}^3(x_\nu y^0 \delta^\nu_0+ x_\nu y^1 \delta^\nu_1+x_\nu y^2 \delta^\nu_2+x_\nu y^3 \delta^\nu_3 )\\ &=(x_0 y^0 \delta^0_0+ x_0 y^1 \delta^0_1+x_0 y^2 \delta^0_2+x_0 y^3 \delta^0_3\\ &+(x_1 y^0 \delta^1_0+ x_1 y^1 \delta^1_1+x_1 y^2 \delta^1_2+x_1 y^3 \delta^1_3\\ &+(x_2 y^0 \delta^2_0+ x_2 y^1 \delta^2_1+x_2 y^2 \delta^2_2+x_2 y^3 \delta^2_3\\ &+(x_3 y^0 \delta^3_0+ x_3 y^1 \delta^3_1+x_3 y^2 \delta^3_2+x_3 y^3 \delta^3_3\\ \end{align*}

Dropping the terms with the Kronecker \(\delta\) equal to zero, we have:

\begin{align*} \therefore x_\nu y^\mu \delta^\nu_\mu&=x_0y^0+x_1y^1+x_2y^2+x_3y^3\\ &=x_\nu y^\mu \end{align*}

We introduce one last notation, equivalent to the dots and apostrophes that we had for the \(t\) and \(x\) derivatives:

\begin{align} \eta_{,\mu}\equiv\frac{d\eta}{dx^\mu} \tag{9.32}\end{align}

The comma is not a mistake! We have the comma to allow for the case when the field, \(\eta\), is a vector field (or more generally a “tensor field”). For example, if \(\eta\) is the electric field, you can think of \(\eta\) as having “components” in each space direction, \(\eta_i\). For example, if \(\eta\) were the electric field, \(\vec E(\vec r,t)\), we would have:

\begin{align} \eta_{1,2}\equiv\frac{dE_x}{dy} \tag{9.33}\end{align}

In fact, in this formalism, electro-magnetism is indeed handled by a four-dimension field, with the first component related to the electric potential, \(\phi\), and the other three components related to the vector potential, \(\vec A\).

Thus, the variational principle requires us to consider:

\begin{align} \delta S=\delta \frac{1}{c} \int \lagd(\eta,\eta_{,\mu},x^\mu) \,dx^\nu \tag{9.34}\end{align}

Since \(c\) is constant it can be ignored as it will not influence the variation. We again consider the case where the varied field, \(\bar\eta(x^\mu)\) is given by:

\begin{align} \bar\eta(x^\mu)&\equiv\eta(x^\mu)+\epsilon\phi(x^\mu)\\ \delta \eta&\equiv\epsilon\phi(x^\mu)\\ \delta \eta_{,\mu}&=\epsilon\phi_{,\mu}(x^\mu) \tag{9.35}\end{align}

The variation of the Lagrangian density is:

\begin{align} \delta \lagd &\equiv \lagd(\eta+\delta\eta,\eta_{,\mu}+\delta\eta_{,\mu},x^\mu)-\lagd(\eta,\eta_{,\mu},x^\mu)\\ &=\die{\lagd}{\eta}\delta\eta+\die{\lagd}{\eta_{,\mu}}\delta\eta_{,\mu} \tag{9.36}\end{align}

where we have used a Taylor series to expand the varied Lagrangian density around the unvaried value and dropped terms that are higher than first order in the variations. Note that the second term contains the product of two terms with the same index, which according to the Einstein summation convention must be summed over. To be explicit:

\begin{align} \die{\lagd}{\eta}\delta\eta+\die{\lagd}{\eta_{,\mu}}\delta\eta_{,\mu}&=\die{\lagd}{\eta}\delta\eta\\ &+\die{\lagd}{\left(\frac{d\eta}{cdt}\right)}\delta\left(\frac{d\eta}{cdt}\right)+\die{\lagd}{\left(\frac{d\eta}{dx}\right)}\delta\left(\frac{d\eta}{dx}\right)+\die{\lagd}{\left(\frac{d\eta}{dy}\right)}\delta\left(\frac{d\eta}{dy}\right)+\die{\lagd}{\left(\frac{d\eta}{dz}\right)}\delta\left(\frac{d\eta}{dz}\right) \tag{9.37}\end{align}

which correctly reduces to the form we had earlier for the one-dimensional field. Also note how the \(c\)’s cancel! Of course, the reason we introduced the notation was to avoid having to expand it out, so we will let the reader convince themselves that the math is correct by expanding out future formulas.

The variation of the action is thus:

\begin{align} \delta S&=\int\left(\die{\lagd}{\eta}\delta\eta+\die{\lagd}{\eta_{,\mu}}\delta\eta_{,\mu}\right)\,dx^\nu\\ &=\epsilon\int\left(\die{\lagd}{\eta}\phi+\die{\lagd}{\eta_{,\mu}}\phi_{,\mu}\right)\,dx^\nu \tag{9.38}\end{align}

Setting that the rate of change of the action with respect to \(\epsilon\) be zero, and then integrating by parts, we have:

\begin{align} \frac{\delta S}{\epsilon}&=\int\left(\die{\lagd}{\eta}\phi+\die{\lagd}{\eta_{,\mu}}\phi_{,\mu}\right)\,dx^\nu\\ &=\int\phi\left(\die{\lagd}{\eta}-\frac{d}{dx^\mu}\die{\lagd}{\eta_{,\mu}}\right)\,dx^\nu \end{align}

Again, this must be true for any choice of \(\phi\), so we obtain the Euler-Lagrange equation for a scalar field in three dimensional space:

\begin{align} \frac{d}{dx^\mu}\die{\lagd}{\eta_{,\mu}}-\die{\lagd}{\eta}=0 \tag{9.39}\end{align}

This result easily generates to the case of multiple fields or multiple components of a single field (i.e. a tensor or vector field).