Four-vectors YouTube

Position vectors

Up until now, we have written the coordinates of an event in space-time in the form $$(t, x, y, z)$$

Nb. If we set $c \neq 1$ then the correct form is $(ct, x, y, z)$.

From now on, we shall often use component notation, $$\begin{align*}t &\rightarrow X^0\\x &\rightarrow X^1\\y &\rightarrow X^2\\z &\rightarrow X^3\end{align*}$$ and then we can just write $$X^\mu, \text{ where } \mu = 0,1,2,3$$

The $X^\mu$, using superscripts, are called contravariant vector components. We can define a second set, called the covariant components which use subscripts, $$X_\mu, \text{ where } \mu = 0,1,2,3$$

$X^\mu$ and $X_\mu$ are related to each other in the following way: if $$X^\mu \rightarrow \left(t,x,y,z\right)$$ then $$X_\mu \rightarrow \left(t,-x,-y,-z\right)$$

That is, $$\begin{align*}X_0 &= X^0\\X_1 &= -X^1\\X_2 &= -X^2\\X_3 &= -X^3\end{align*}$$

Einstein summation convention

Consider $$X^\mu \rightarrow \left(t,x,y,z\right)$$

Then, $$\begin{align*}\sum_{\mu=0}^{3} X_\mu X^\mu &= X_0 X^0 + X_1 X^1 + X_2 X^2 + X_3 X^3\\&= X^0 X^0 + (-X^1) X^1 + (-X^2) X^2 + (-X^3) X^3\\&= (X^0)^2 - (X^1)^2 - (X^2)^2 - (X^3)^2\\&= t^2 - x^2 - y^2 - z^2\end{align*}$$

Einstein decided that, since summation were going to crop up many times in the equations, he would drop the summation sign altogether, so that the summation is implicit in an expression like $$X_\mu X^\mu$$

We note the following conventions:

  • Greek indices are used for four-vectors, $\mu, \nu, \dots \in \left\{0,1,2,3\right\}$
  • Latin indices are used for three-vectors, $m, n, \dots \in \left\{1,2,3\right\}$
  • Superscripts represent contravariant four-vectors
  • Subscripts represent covariant four-vectors
  • Summation is implied when you have the same symbol for exactly one subscript and one superscript in an expression.

Recall, the full set of symmetric transformations include boosts, rotations and any compositions. We can write these transformations as matrices, $$L = \left[L_\nu^\mu\right]_{\mu, \nu = 0}^3$$

For example, $$L_\nu^\mu \rightarrow \begin{bmatrix}\gamma & -\gamma v & 0 & 0\\-\gamma v & \gamma & 0 & 0\\0&0&1&0\\0&0&0&1\end{bmatrix}$$ where superscripts refer to rows, and subscripts to columns.

Nb. The placement of the indices (superscript or subscript) was the only one consistent with expressions involving four-vectors. We shall see later that $L_\nu^\mu$ forms the components of a rank-2 tensor, which is a generalisation of four-vectors to objects that have two indices.

Then, we can write the Lorentz transformations from the unprimed frame to the primed frame as $$X^{\mu'} = L_\nu^{\mu'} X^\nu$$ where the summation over $\nu$ is implied.

Definition of four-vectors

We can consider position four-vectors $X^\mu$ (coordinate events in space-time) to be the archetypal contravariant four-vectors, since we already know what happens to them under Lorentz transformations.

We say that the quantities, $A^\mu$, are the components of a contravariant four-vector if they transform from one reference frame to another in the same way as the components of a position vector.

So if $A^\mu$ and $A'^\mu$ are the components of a four-vector in frames where the primed frame is moving at speed $v$ in the common $x$-direction relative to the unprimed frame, then $$A^{\mu'} = L_\nu^{\mu'} A^\nu$$

That is, $$\begin{align*}A^{0'} &= \gamma \left(A^0 - vA^1\right)\\A^{1'} &= \gamma \left(- vA^0 + A^1\right)\\A^{2'} &= A^2\\A^{3'} &= A^3\end{align*}$$

We can show that four-vectors do in fact form a vector space. We need to show that if $A^\mu,B^\mu$ are (the components of) four-vectors, then $$C^\mu = \alpha A^\mu + \beta B^\mu$$ is also a four vector, for any $\alpha,\beta \in \mathbb{R}$.

But, $$\begin{align*}C^{\mu'} &= \left(\alpha A^\mu + \beta B^\mu\right)'\\&= \alpha A^{\mu'} + \beta B^{\mu'}\\&= \alpha \left(L_\nu^{\mu'} A^{\nu}\right) + \beta \left(L_\nu^{\mu'} B^{\nu}\right)\\&= L_\nu^{\mu'} \left(\alpha A^{\nu} + \beta B^{\nu}\right)\\&= L_\nu^{\mu'} C^{\nu}\end{align*}$$

Proper length

The (square of the) proper length of a four-vector, $A^\mu$, is $$\begin{align*}A_\mu A^\mu &= A_0 A^0 + A_1 A^1 + A_2 A^2 + A_3 A^3\\&= (A^0) A^0 + (-A^1) A^1 + (-A^2) A^2 + (-A^3) A^3\\&= \left(A^0\right)^2 - \left(A^1\right)^2 - \left(A^2\right)^2 - \left(A^3\right)^2\\&= A^0 (A^0) + A^1 (-A^1) + A^2 (-A^2) + A^3 (-A^3)\\&= A^0 A_0 + A^1 A_1 + A^2 A_2 + A^3 A_3\\&= A^\mu A_\mu\end{align*}$$

We have already seen that the proper length is an invariant quantity - the same in all reference frames, $$A_{\mu'} A^{\mu'} = A_\mu A^\mu$$

Dot product

The dot product between two four vectors, $A^\mu, B^\mu$, is $$A_\mu B^\mu = A^0 B^0 - A^1 B^1 - A^2 B^2 - A^3 B^3 = A^\mu B_\mu$$

We can show that this is invariant too. Since $A^\mu, B^\mu$ are four vectors, then so is $C^\mu = A^\mu + B^\mu$ and the proper length of this is invariant, by definition. Now, $$\begin{align*}C_{\mu'} C^{\mu'} &= \left(A_\mu + B_\mu\right)' \left(A^\mu + B^\mu\right)'\\&= A_{\mu'} A^{\mu'} + 2 A_{\mu'} B^{\mu'} + B_{\mu'} B^{\mu'}\end{align*}$$ and we can re-arrange this so all of the invariant quantities are on the right-hand side, $$A_{\mu'} B^{\mu'} = \frac{1}{2}\left(C_{\mu'} C^{\mu'} - A_{\mu'} A^{\mu'} - B_{\mu'} B^{\mu'}\right)$$ and so the expression on the left-hand side - the dot product - is also invariant.