# What do the indices in 4-vector notation indicate?

What do the indices in 4-vector notation indicate? Is it vector components or the vector itself? I have a little confusion after this Wikipedia article. Since the indices are summed over how can the left hand side have any index? Could you please explain?

## Notation

The notations in this article are: lowercase bold for three-dimensional vectors, hats for three-dimensional unit vectors, capital bold for four dimensional vectors (except for the four-gradient), and tensor index notation.

## Four-vector algebra

Four-vectors in a real-valued basis

A four-vector A is a vector with a "timelike" component and three "spacelike" components, and can be written in various equivalent notations:

{\displaystyle {\begin{aligned}\mathbf {A} &=(A^{0},\,A^{1},\,A^{2},\,A^{3})\\&=A^{0}\mathbf {E} _{0}+A^{1}\mathbf {E} _{1}+A^{2}\mathbf {E} _{2}+A^{3}\mathbf {E} _{3}\\&=A^{0}\mathbf {E} _{0}+A^{i}\mathbf {E} _{i}\\&=A^{\alpha }\mathbf {E} _{\alpha }\\&=A^{\mu }\end{aligned}}}

(Source)

• Note that it is very common in physics to make little to no distinction between a tensor and its components. It is common to refer to $A^\mu$ as "a vector" or $g_{\mu\nu}$ (in GR) as "the metric" when what we really mean is "the components of the vector $A$" and "the components of the tensor $g$". Commented Feb 27, 2021 at 11:39
• I have encountered with what you talk about a few times. Sometimes they say vectors, sometimes vector components and it really confuses me. Thank you for your answer. But I have another question. If I multiply a four vector by something with the same index would I always get a scalar? Commented Feb 27, 2021 at 11:47
• $A^\mu X_\mu$ would be a scalar, where we use the summation convention and strictly have one index up one index down (i.e. $A^\mu X^\mu$ is not a scalar, in fact it would be a completely invalid thing to write). These rules are explained in any text that includes an introduction to tensor index notation (usually first encountered in special relativity or electromagnetism). Commented Feb 27, 2021 at 12:01
• Basis vectors are special, a basis is a collection of indexed vectors. $\vec{e}_i$ does not mean "the components of the vector $e$, it means "the $i$th basis vector". Using the summation convention again $X^\mu \vec{e}_\mu$ is a vector expanded in a basis. The fact that we use indices and the summation convention for both components and basis vectors is a bit confusing at first. Commented Feb 27, 2021 at 12:17
• Not necessarily, invariant quantities are scalars. But something like $T^{\mu\nu}X_{\mu}=V^\nu$ is not a scalar, the result is a vector (this is usually called "tensor/index contraction"), note again that I am calling $V^\nu$ a vector when technically I shouldn't. Something like $X^\mu V_\mu$ is a scalar though, and that is a truly invariant quantity. Be careful though because (as usual) in physics we often abuse terminology and call any tensorial quantity "invariant" (as opposed to covariant, the more correct word). Commented Feb 27, 2021 at 13:37

The confusion stems from the fact that one commonly wants to consider what are called passive transformations as opposed to active transformations. The idea can be seen in three dimensions and then generalized from there.

One should not write $${\bf u} = u^a$$, it is an abuse of notation. But one can use a symbol without indices, such $$\bf u$$ or $$U$$, in more than one way, as I shall explain.

Suppose $$\bf u$$ is a vector in three dimensions. Then we can write $${\bf u} = u^1 \hat{\bf e}_1 + u^2 \hat{\bf e}_2 +u^3 \hat{\bf e}_3$$ where $$u^i$$ are the components of the vector and $$\hat{\bf e}_i$$ are some set of basis vectors. Let's suppose those basis vectors are aligned along the coordinate axes of some rectangular system of coordinates.

Now consider the effect of a rotation $$R$$. We can rotate the vector $${\bf u}$$ so as to get some other vector $${\bf v} = R {\bf u}$$. This is called an active transformation. The vector changes to a different vector.

But we can also consider a rotation of the coordinate axes, without rotating our vector $$\bf u$$. This is called a passive transformation because $$\bf u$$ does not change. If we rotate the coordinate axes then we will find the vectors along the new coordinate axis directions are not the same as the ones along the old coordinate axis directions, so let's use a prime to indicate this distinction: $$\hat{\bf e}'_i = R \hat{\bf e}_i.$$ From this it follows that $$\hat{\bf e}_i = R^{-1} \hat{\bf e}'_i.$$ This fact can be useful to note, but for present purposes it is more useful merely to note that the old basis vectors $$\hat{\bf e}_i$$ can themselves be written in terms of the new basis vectors as a linear sum. So each $$\hat{\bf e}_i$$ is equal to some sum of $$\hat{\bf e}'_i$$ and we shall find $${\bf u} = u'^1 \hat{\bf e}'_1 + u'^2 \hat{\bf e}'_2 +u'^3 \hat{\bf e}'_3$$ where $$u'^i$$ are a new set of coefficients, called components. The point being that the vector $$\bf u$$ has not changed but the components $$u'^i$$ are different from the components $$u^i$$ because the basis vectors $$\hat{\bf e}'_i$$ are different from the basis vectors $$\hat{\bf e}_i$$.

An example of this fact, now in 4 dimensions, is the widely used relation $$u'^a = \Lambda^a_{\;\mu} u^\mu$$ where $$\Lambda^a_{\; b}$$ is the Lorentz transformation and I am now adopting the Einstein summation convention. In general relativity this gets generalised to $$u'^a = \frac{\partial x'^a}{\partial x^\mu} u^\mu.$$

So far we have maintained a strict distinction between a vector $$\bf u$$ and its components $$u^a$$ or $$u'^a$$. But often it is useful to find a less cluttered notation, where indices (whether superscript or subscript) are not needed. So to this end let's define $$U = \{ u^a \}.$$ This equation asserts that the symbol $$U$$ refers to the set of components of the 4-vector $$\bf u$$ with respect to the unprimed coordinate basis. Notice I am now using $$U$$ for the set of components and $$\bf u$$ for the 4-vector. Next let's define $$U' = \{ u'^a \}.$$ This asserts that the symbol $$U'$$ refers to the set of components of the 4-vector $$\bf u$$ with respect to the primed coordinate basis. Now if we are doing special relativity where the transformation is a Lorentz transformation, then we can conveniently write the relationship between $$U$$ and $$U'$$ as: $$U' = \Lambda U$$ where it is understood that the components of $$\Lambda$$ are gathered together in a $$4 \times 4$$ matrix, and the lists of components $$U$$ and $$U'$$ are to expressed as the components of column vectors, and $$\Lambda U$$ is the ordinary matrix multiplication operation. This is quite a convenient notation so it is widely used, but it leads to the confusion between the set of components and the 4-vector itself. In a passive transformation, such as a change of inertial frame of reference, the 4-vector itself does not change, but the set of components changes from $$U$$ to $$U'$$. So strictly one should not call $$U$$ here 'the 4-vector' but rather 'the set of components of the 4-vector with respect to the unprimed inertial frame', and $$U'$$ is the set of components of the 4-vector with respect to the primed inertial frame.

If you don't like this index-free notation you don't have to use it. For special relativity I think it is quite a nice notation, but I would not employ it in G.R. where I find it clearer to stick to the use of indices when one is referring to components. This does not prevent one from referring directly to the 4-vectors or other tensors when one wishes.

I agree that the wikipedia article is confusing. $$A^\mu$$ represents the components of the vector $$\mathbf{A}$$ with respect to the basis vectors.

• Thank you for your answer, so if I were to multiply it with basis vectors I would get A which is a vector? I heard somewhere that if you sum over the same index you would get a scalar. Is it correct? Because above we summed over the same index and got a vector. Sorry if my questions are stupid I am a little new to the subject... Commented Feb 27, 2021 at 11:34
• @BruceWayne If you sum the components of the vectors you'd get a scalar. In that sum, you're basically taking a linear combination of some basis vectors with certain coefficients Commented Feb 27, 2021 at 15:31
• I'm oversimplifying but that's the idea Commented Feb 27, 2021 at 15:32

The notation used for a four vector in this Wiki article seems to be putting $$\textbf{A}\equiv\textsf{A}^{\mu}$$.

where, the Greek index runs over the four space-time indices 0,1,2,3 and each represents the components of this 4-vector $$\textrm{A}$$ i.e. {$$A^0,A^1,A^2,A^3$$} with respect to the set of mutually orthogonal basis vectors $${\mathbf{E}_{\mu}}$$ i.e. {$$E_0,E_1,E_2,E_3$$}.

This is in direct analogy with the usual vector representation in 2D or 3D as a column\row vector or a row vector say $$\vec{a}=\{a_1,a_2,a_3\}$$ taken in some orthogonal unit basis $$\hat{e}=\{\hat{e}_1,\hat{e}_2,\hat{e}_3\}$$.

The difference in the 4-vector space-time formulation are the notions of covariant and contravariant indices and their algebra.