# Understanding Einstein notation in special relativity

I'm entering the realm of special relativity and amazingly the hardest part about it is the notation! I'm confused on exactly how to intuitively build an understanding, and this may be hindering the way I think. For example, I understand a vector $$x^v$$ can transform to another vector $$\bar{x}^v$$ via matrix multiplication: $$\bar{x}^v = \Lambda^v_\mu x^\mu$$ I tried to reason with covariant vectors as 'row vectors' such that the relation $$\bar{x}_v = \Lambda^{\mu}_v x_{\mu}$$ I think this interpretation breaks down in the covariant transformation: $$v_\mu = g_{\mu v} v^v$$ where $$g_{\mu v}$$ is the Minkowski metric. I'm struggling to see how this relationship makes any sense since I see a 'row vector' on the LHS and a matrix times a vector on the RHS. I think part of the confusion is that the metric is NOT a matrix, but I would love assistance on better understanding this.

It is best to drop always thinking in terms of the row and column vectors because this method breaks down for objects with more than one index. There is no direct way to distinguish between the matrix representations of a covariant and a contravariant object if they have more than one index. For example, a contravariant object $$A^{\mu\nu}$$ will be represented via a $$4\times 4$$ matrix just like its covariant version $$A_{\alpha\beta}$$. And there is no direct trick like associating the "row" version of the matrix with the square matrix of a covariant object and "column" version of the matrix with the square matrix of a contravariant object.

A suggested way of thinking about such objects is in light of their definition--in particular, in light of how they transform and as to with which objects they combine to give scalars. For example, gradients (naturally constructed as $$\frac{\partial \phi}{\partial x^{\mu}}$$) are covariant and the way to think about their covariance is to think about the fact that they combine with the contravariant coordinate displacements ($$dx^\mu$$) to give scalars ($$d\phi$$). In other words, a contravariant vector can be thought of as a mapping which takes in a covariant vector and maps it onto a scalar--and vice versa.

Regarding the specific example that you brought up, there is no fatal issue in considering a contravariant or a covariant object with only one index as a column or a row vector respectively. In particular, $$g_{\mu\nu}$$ can certainly be treated as a matrix so that it can be multiplied with a column vector (a contravariant tensor) to produce another column vector which would be the transpose of the row vector which is supposed to be the desired covariant vector. But, for the reasons expressed earlier, it is recommended to only occasionally think in this mode and not generally.

• This is what bothers me; you said that $g_{\mu v}$ can be treated as a matrix that can be multiplied with a column vector to produce a row vector, but this is impossible? How could such a multiplication work? – Ayumu Kasugano Mar 10 at 20:34
• @AyumuKasugano Thanks for your comment, I see what you are saying. I have deleted the last paragraph of my answer which was not perfectly correct. I will try to expand on the issue of whether you can treat $g_{\mu\nu}$ as a matrix or not later today. – Dvij Mankad Mar 10 at 21:23
• @AyumuKasugano Okay, I have added the last paragraph again with the required editing. – Dvij Mankad Mar 10 at 21:28

You should stop thinking in terms of column vectors, row vectors, and matrices, because tensor components can have any number of indices, not just one or two. For example, $$\epsilon_{\mu\nu\lambda\kappa}$$ has four.

You should start thinking in terms of how physical quantities are tensors of various ranks; how they have contravariant, covariant, or mixed components; how the metric tensor can convert between the various forms of the components; how tensors can be added, multiplied, or contracted to form other tensors; and how tensor components transform under Lorentz transformations.

Tensors of higher rank are not as intuitive as vectors and matrices, but they obey simple rules that you can learn and become comfortable with.

If you need a more intuitive way to think about tensors of rank greater than one, think of them as products of vectors... one vector for each index. The “product” of $$\mathbf{u}$$ and $$\mathbf{v}$$ is not a scalar product $$\mathbf{u}\cdot\mathbf{v}$$ or a cross product $$\mathbf{u}\times\mathbf{v}$$ like you are used to in three dimensions; it is an “outer product” $$\mathbf{u}\otimes\mathbf{v}$$ where in component language the product of $$u_i$$ and $$v_j$$ is just $$u_i v_j$$. You can visualize this as a “multivector” consisting of two vectors, pointing in two different directions. Higher-rank tensors are multivectors of three or more vectors. This kind of product works in all dimensions, unlike the cross product which makes sense only in certain dimensions like 3D.

By the way, higher-rank tensors are not something that you only encounter in relativity, or in four dimensions. For example, in good old three-dimensional Newtonian gravity, the octupole moment of a mass distribution is a tensor with three indices, the hexadecapole moment is a tensor with four indices, etc. Another example is in linear elasticity theory, where the stiffness tensor relating the strain to the stress is a tensor with four indices.

• The product whose result is a 2nd rank object with components $u_i u_j$ is not correctly called a tensor product. It is an outer product or a dyadic product. The outer product of two rank-1 objects is of rank 2. The tensor product of two rank-1 objects is of rank 1 but in a space of more dimensions called the tensor product space (this is much used in quantum mechanics). – Andrew Steane Mar 10 at 21:42
• Thanks for the correction! – G. Smith Mar 10 at 22:27
• @AndrewSteane I edited ny answer and changed “tensor product” to “outer product”. But I’m a bit confused because Wikipedia’s article “Outer product” says “The outer product on tensors is typically referred to as the tensor product.” – G. Smith Mar 10 at 23:38
• I did some research on this too (a while ago). I think it is a case where a distinction that I think worth preserving is being lost as human language evolves. The tensor product commonly used in quantum theory has no other name; the outer product of vectors does have a name, and a good one. Here is inner, outer, tensor product: $\langle a | b \rangle$, $|a\rangle \langle b |$, $| a \rangle \otimes |b \rangle$. – Andrew Steane Mar 11 at 8:37

I too find that the index notation is unnecessarily distracting for learning the main elements of special relativity (SR). It can be avoided for large parts of SR by adopting the notation of matrices and vectors. In matrix notation and vector notation for SR, one proceeds by only writing column vectors and matrices whose components are contravariant in the first instance, with the sole exception the metric tensor. Then when you want covariant things you can obtain them by multiplying by the metric tensor. You can get a long way like this without ever needing to write any indices!

This approach only postpones rather than avoids the use of indices, but it makes it possible to understand the index methods more readily, in my opinion.

The way to think here is to regard all rank-1 objects as column (not row) vectors, be they contravariant or covariant, but you must simply learn which combinations of these vectors require the use of a transpose in order that standard matrix multiplication rules map correctly to the corresponding indexed expression.

One should not self-advertise on this site without good cause, but in this instance I think there is good cause because I will mention a book which was written directly to handle this very question. This is an undergraduate text book by myself with OUP, called Relativity made relatively easy. I there deliberately adopt the policy of avoiding indices at the start, and introducing them only once the reader is already developing some expertise, and then taking some trouble to spell out how the notation corresponds to matrix notation insofar as it does.