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You’ve probably heard the word ‘tensor’ before, but if you’re like me, you’ve gone through your entire physics degree without knowing what they are. So today, I’m going to express the relativistic wave equation in the language of tensors and Lagrangians, and show just how powerful these things can be.

Let’s first begin with an example we all know, the dot product. Now, given any two vectors, we can calculate their dot product, $\vec{A}\cdot\vec{B}$. Let’s do some fancy renaming and let

$$\mathbf{g}({\vec{A},\vec{B}})=\vec{A}\cdot\vec{B}.$$

Let’s make some notes about $\mathbf{g}$.

1. It is a function of 2 vectors
2. It returns a real number
3. As we know, since $\mathbf{g}$ is just the dot product, it is bi-linear, meaning it is linear in both arguments

This is in fact an example of what we call a (0, 2) tensor, as it takes as its arguments 2 vectors and returns a real number. This has a matrix representation, namely the 2 by 2 identity matrix. However, in polar co-ordinates the metric would have matrix representation with components given $g_{00}=1$ and $g_{11}=r^2$ with the rest of the entries zero. We can denote any element of of the matrix, and hence elements of the tensor, as $g_{\alpha\beta}$.

Now, you may be thinking, why is it a (0,2) tensor, what does that zero stand for? Well, for that we have to introduce the notion of a covector.

A covector is function which maps vectors into the real numbers, for those of you who are snazzy with their linear algebra, these are elements of the dual space of a vector space. As an example, consider the some useless function $z$ which maps every element in our vector space to $0 \in \mathbb{R}$, this $z$ is a covector, it maps our vectors into the real numbers. Now, we will make an important distinction in notation, and this will come back in a huge way later, for now, memorise it. We represent the components of a vector $\vec{V}$ with raised indices, $V^{\alpha}$ and components of a covector $\tilde{V}$ with lowered indices, $V_{\beta}$. It is important that you know this, say goodbye to arrows and tildes indicating vectors, we know use the position of the index, up is a vector, down is a covector.

Now, an important relationship between vectors and covectors is that a vector can be mapped into a covector, using the Minkowski metric of minkowski space, the metric that has components $\eta_{00}=-1$, $\eta_{11}=\eta_{22}=\eta_{33}=1$, and all other entries zero. Using our notation this is

$$V_{\alpha}=\eta_{\alpha\beta}V^{\beta}.$$

What on earth does this mean? This is a summation over the $\beta$ index, known as the Einstein summation convention, we say that the $\beta$ indices contract. This is true whenever you have the same index appearing lowered and raised. What is important to note is that the index was lowered, it has become a covector. In much of what will follow, we will use the metric to raise and lower indices as we please. We note that an index can be raised by using the inverse of the Minkowski metric, $\eta^{\alpha\beta}$.

Now, back to the zero in (0,2) tensor; it states that our tensor takes 0 covectors as arguments. A (1,1) tensor, lets call it $\mathbf{R}$, would take 1 covector and 1 vector as arguments and return a real number. We can represent the components of this tensor by $\mathbf{R}^{\alpha}_{\beta}$, where $\alpha$ represents the index of the covector and and $\beta$ represents the index of the vector. If $\mathbf{R}$ was instead a (2,1) tensor, it would be indexed by $\mathbf{R}^{\alpha \beta}_{\gamma}$. In general, a (r,s) tensor will have $r$ upper indices representing the number of covectors the tensor takes as input and $s$ lower indices representing the number of vector inputs. It is important to note that any real valued function is a (0,0) tensor. So really, tensors have been in front of your eyes this entire time, you just haven’t realised it.

Now, I told you the notation was going to come back in a huge way, and here it is. Say we have a (1,1) tensor $\mathbf{R}^{\alpha}_{\beta}$. And say we wish to input it a vector, $V^{\beta}$, by our notation and the summation convention, we can simply write

$$\mathbf{R}^{\alpha}_{\beta}V^{\beta}$$

and note that our indices contract naturally. The lowered $\beta$ on $\mathbf{R}$ which represents the vector input is contracted with the raised $\beta$ on $V$ a vector. The same is true of covectors. With this notation now, if you make a mistake, you will know it, and almost always you can just guess how to put things together just by looking at how the indices should contract! It’s powerful stuff.

Now let’s see this stuff in action by finding the equation of motion for the following Lagrangian density

$$L=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^2\phi^2.$$

Note that the summation convention is applied to the indices on the partial derivatives. Now the Euler Lagrange equation for fields is

$$\frac{\partial L}{\partial \phi}-\partial_{\mathbf{\alpha}}(\frac{\partial L}{\partial(\partial_{\mathbf{\alpha}}\phi)})=0,$$

where the summation convention occurs again over the bold $\alpha$ indices, one is raised and one is lowered by being in the denominator of the fraction.

Clearly, $\frac{\partial L}{\partial \phi}=-m^2\phi$. Now for the fun part.

\begin{align*}

\frac{\partial L}{\partial(\partial_{\mathbf{\alpha}}\phi)}

&=\frac{\partial }{\partial(\partial_{\mathbf{\alpha}}\phi)}(\frac{1}{2}\partial_{\mu}\phi\partial^{\bf{\mu}}\phi)\\

\intertext{The first thing we note is that we are differentiating with respect to lowered partial derivatives, while we contain a term with raised partial derivatives, so we use our metric to lower the index that is bolded}\\

&=\frac{\partial }{\partial(\partial_{\mathbf{\alpha}}\phi)}(\frac{1}{2}\partial_{\mu}\phi\eta^{\beta\alpha}\partial_{\beta}\phi)\\

&=\frac{\partial }{\partial(\partial_{\mathbf{\alpha}}\phi)}(\frac{1}{2}\eta^{\beta\alpha}\partial_{\mu}\phi\partial_{\beta}\phi)\\

\intertext{Now using the product rule}

&=\frac{1}{2}\eta^{\beta\alpha} [\frac{\partial \partial_{\mu} }{\partial(\partial_{\mathbf{\alpha}}\phi)}\partial_{\beta}\phi)+\partial_{\mu}\frac{\partial \partial_{\beta} }{\partial(\partial_{\mathbf{\alpha}}\phi)}]\\

\intertext{We now see that these derivatives are either 1 when $\alpha=\beta~~(\mu)$ or zero otherwise. This is just the Kroneker Delta function.}

&=\frac{1}{2}\eta^{\beta\alpha}[\delta^{\alpha}_{\mu}\partial_{\beta}\phi+\delta^{\alpha}_{\beta}\partial_{\mu}\phi]\\

\intertext{ Note the position of the indices, these are the only positions in which the summation makes sense! But, as handy rule, if you have 1 over a lower index, and it becomes a raised index when you bring it to the numerator. So bringing in the metric which will raise the indices on the delta functions}

&=\frac{1}{2}[\delta^{\alpha\beta}\partial_{\beta}\phi+\delta^{\alpha\mu}\partial_{\mu}\phi\\

&=\partial^{\alpha}(\phi).

\end{align*}

Now, we can see that in order for our lagrange equation of motion to make since, this is the only possible position of index  we could have gotten, because now, the summation convection holds for when we put in the final partial derivative from the Euler Lagrange equation of motion, and we get

$$\partial_{\alpha}\partial^{\alpha}(\phi)=-m^2\phi$$

which can be re-written as

$$(\partial_{\alpha}\partial^{\alpha}+m^2)\phi=0.$$

Now, for the physics buffs, since $\phi$ is a function of (t,x,y,z) we have that $\partial_{\alpha}\partial^{\alpha}$ is just the d’Alembert operator, which is the Laplace operator in Minkowski spacetime. So now, the Euler Lagrange equation which we just derived should be recognised as the Klein-Gordon equation.