I am self-learning QFT as a hobby, which requires me to first learn special relativity. I have come across the notion of a spacetime vector and I am unclear on a few aspects regarding it.

What I do understand is that if $x = (x_0, x_1 ,x_1 x_3)$ is a spacetime vector, then $x_0 = ct \in \mathbb{R}$ represents time and $(x_1, x_2, x_3) \in \mathbb{R}^3$ represents space. Vectors like these are often represented as $x^\mu$ or $x_\mu$. My first question is why do we use $\mu$ here, does $\mu$ itself mean anything?

Secondly, I am confused on upper vs lower indices. Mathematically I know that a subscript represents a covector and superscript is an ordinary vector, but physically why do we distinguish between the two if we may identify the dual of $\mathbb{R}^4$ with itself?

Lastly, is $x^\mu = (x_0, x_1, x_2, x_3)$ and $x_\mu = (x_0, -x_1, -x_2, -x_3)$ or is it the other way around? In either case, why is it defined this way?


1 Answer 1


My first question is why do we use 𝜇 here, does 𝜇 itself mean anything?

It helps to think about ordinary vectors in three dimensions. You can pick a basis and write the vector in terms of its components like $\vec{r} = (r_1, r_2, r_3)$. Then you could refer to the components collectively as $r_j$, where $j=1$, $2$, or $3$. In the same way, $x^{\mu}$ and $x_{\mu}$ refer to the components of a four-vector or four-covector, respectively, with $\mu =0$, $1$, $2$, or $3$. It is conventional to use greek indices like $\mu$, $\nu$, $\lambda$, etc. to refer to spacetime indices running from 0 to 3 and latin indices like $i$, $j$, $k$, etc. to refer to spatial indices running from 1 to 3. So the $\mu$ in $x^{\mu}$ really means the same thing as the $j$ in $r_j$. By abuse of notation, we often refer to a four-vector $x$ by its components $x^{\mu}$ simply to remind us that we are dealing with a four-vector and not some other kind of object. Similarly we can refer to a four-covector by its components $x_{\mu}$.

Secondly, I am confused on upper vs lower indices ... why do we distinguish between the two if we may identify the dual of $\mathbb{R}^4$ with itself?

Because in special relativity, we're not really dealing with $\mathbf{R}^4$ with the usual Euclidean distance formula. If we were in Euclidean $\mathbf{R}^4$, we'd say that the length of a vector $\vec{v} = (v_1, v_2, v_3, v_4)$ is $|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2 + v_4^2}$, and we'd have a dot product between vectors that looks like $\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3v_3 + u_4v_4$. When we do special relatively, we use a different way of calculating lengths in spacetime and taking "dot products" between vectors that singles out one of the components of the vector as special and gives us the whole notion of causality. Specifically, the "dot product" between two four vectors is \begin{align} x \cdot y = -x^0y^0 + x^1y^1+x^2y^2 + x^3y^3 \end{align} and the "squared length" of a vector is \begin{align} x\cdot x = -(x^0)^2 + (x^1)^2 + (x^2)^2 + (x^3)^2 \end{align} It should hopefully be obvious that this is not really the square of any length, since it can be negative — the language is just an analogy with ordinary $\mathbf{R}^4$.

The way that we identify ordinary vectors with dual vectors in ordinary $\mathbf{R}^4$ is via the dot product. One way to see this is to write the dot product in terms of a metric tensor $\delta_{ij}$ that takes take vectors as input and outputs their dot product: \begin{align} \vec{u}\cdot\vec{v} = \sum\limits_{i,j=1}^{4} \delta_{ij} u^i v^j \end{align} (I'm using upper indices for ordinary Euclidean vectors now, because we want to be careful about what's a vector and what's a dual vector.) Thinking about $\delta$ as a matrix, $\delta$ would just be the identity matrix for ordinary $\mathbf{R}^4$. (All $1$s along the diagonal, and all off-diagonal elements zero.) We can get the components of the dual vector to a vector $\vec{v}$ by applying this matrix: \begin{align} v_i = \sum_{j=1}^4 \delta_{ij}v^j \end{align} Since $\delta$ is the identity, the dual vector will have the same components as the original vector $\vec{v}$, so we can think about them as the same vector.

In spacetime, our metric tensor $\eta$ is no longer the identity. Instead, we have $\eta_{00} = -1$ and $\eta_{11} = \eta_{22} = \eta_{33} = 1$. (The off-diagonal components are zero just like in the Euclidean case.) We can still get the components of a dual vector by contracting with the metric tensor, but now the components of the dual vector are not the same as the components of the original vector! \begin{align} x_{\mu} = \sum_{\nu=0}^{3}\eta_{\mu \nu}x^{\nu} \rightarrow x_0 = -x^0, x_1 = x^1, x_2 = x^2, x_3 = x^3 \end{align} Note the extra minus sign in front of the $0$-component. So in spacetime, vectors and dual vectors are different.

Lastly, is $x^\mu = (x_0, x_1, x_2, x_3)$ and $x_\mu = (x_0, -x_1, -x_2, -x_3)$ or is it the other way around? In either case, why is it defined this way?

Following the previous discussion, and abusing notation to use the components $x^{\mu}$ to refer to a four-vector and components $x_{\mu}$ to refer to its dual four-covector, \begin{align} x^{\mu} = \left(x^0, x^1, x^2, x^3\right) = \left(-x_0, x_1, x_2, x_3\right) \end{align} and \begin{align} x_{\mu} = \left(x_0, x_1, x_2, x_3\right) = \left(-x^0, x^1, x^2, x^3\right) \end{align}

Note: You appear to be using the "mostly minus" convention where the metric tensor has components $\eta_{00} = 1$, $\eta_{11} = \eta_{22} = \eta_{33} = -1$, whereas I've used the "mostly plus" convention. This is purely a matter of convention. It doesn't change any of the physics whatsoever and just shifts some minus signs around. So for example, in your convention, the spacetime "dot product" would be \begin{align} x \cdot y = x^0y^0 - x^1y^1-x^2y^2 - x^3y^3 \end{align} I will leave it as an exercise for you to work out how other formulas change when you switch to this convention.

  • 1
    $\begingroup$ This answer was incredible, I wish I could give it more than one upvote. Thank you very much! $\endgroup$
    – CBBAM
    Nov 4, 2022 at 23:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.