# Subscripts and superscripts in four vectors

1. Feb 3, 2015

### stephen cripps

1. The problem statement, all variables and given/known data
I'm having trouble with understanding four vectors in particle physics.
$v^\mu= \begin{pmatrix} \mu_0 \\ \mu_1 \\ \mu_2 \\ \mu_3 \end{pmatrix}$
and
$v_\mu= \begin{pmatrix} \mu_0 & \mu_1 & \mu_2 & \mu_3 \end{pmatrix}$
But the raising indices equation gives
$v^\nu= \eta^{\mu\nu} v_\mu$
Which doesn't work if I do matrix multiplication with
$\eta^{\mu\nu}= \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$
as it gives $v^\nu$ as a 1x4 not a 4x1 as I would expect from the initial definitions, which leads me to think my initial definitions of the superscripts and subscripts are wrong.

Could anybody help me out?
2. Relevant equations
n/a

3. The attempt at a solution
n/a

2. Feb 3, 2015

Where did you get these two formulas from? I do not see them at the quoted Wikipedia page.

3. Feb 3, 2015

### stephen cripps

I assumed them from the vector representation section

4. Feb 3, 2015

### TSny

Most people write the raising operation as $v^\nu= \eta^{\nu\mu} v_\mu$ (with the indices on $\eta$ switched from the way you wrote it.) It won't make any difference since the $\eta$ matrix is symmetric, but I thought you might want to know how it is usually written.

I think of the equation as simply giving me the recipe for finding the value of $v^\nu$. So, to me, the left hand side is not a 4-vector, but just the $\nu$th component of the 4-vector. When I use the equation, I'm not thinking in terms of matrix multiplication. I just look at the right side and remember the Einstein summation convention for repeated indices. So,

$v^\nu= \eta^{\nu\mu} v_\mu = \eta^{\nu 0} v_0 + \eta^{\nu 1} v_1 + \eta^{\nu 2} v_2 + \eta^{\nu 3} v_3$. The right hand side is just a sum of certain terms and I don't need to visualize it as matrix multiplication.

Of course you can get the same result by matrix multiplication. But then, you would need to write your covariant vector $v_\mu$ as a column vector rather than a row vector. Or, you could keep $v_\mu$ as a row vector that multiplies the $\eta$ matrix from the left.

5. Feb 3, 2015

### stephen cripps

Ok I see how that works now I think. As there's no implicit sum its just the component of a vector. Am I right in thinking though that a subscript usually indicates a row vector, and a superscript a column?

6. Feb 3, 2015

You wrote "I assumed them from the vector representation section". But there you must be careful. There are two different bases in this section. Unfortunately it is not clearly explained there what are these bases.

7. Feb 3, 2015

"Am I right in thinking though that a subscript usually indicates a row vector, and a superscript a column?"

That's correct.

8. Feb 3, 2015

But the basis $e_i$ is a basis of column vectors, while the basis $e^i$ is a basis of row vectors! Can you understand the difference between components of a vector and the index numbering basis vectors? It is important.

9. Feb 3, 2015

### Fredrik

Staff Emeritus
If we use the convention that the number on row $\mu$, column $\nu$ of a matrix $M$ is denoted by $M^\mu{}_\nu$, then the definition of matrix multiplication is $(AB)^\mu{}_\nu=A^\mu{}_\rho B^\rho{}_\nu$. It can be convenient to use the notation $M^\mu$ for the $\mu$th row of $M$ and $M_\mu$ for the $\mu$th column of $M$. But $x^\mu$ isn't the $\mu$th column of a $4\times 4$-matrix $x$. It's just a number, the $\mu$th component of the 4-tuple $x$. $x_\mu$ is the $\mu$th component of the 4-tuple $\eta x$, where $\eta$ denotes the matrix with $\eta_{\mu\nu}$ on row $\mu$, column $\nu$. (The convention described above isn't used for $\eta$ or $\eta^{-1}$).

Last edited: Feb 3, 2015
10. Feb 3, 2015

### stephen cripps

Yeah I think I understand that now, thanks a lot for your help, I see I need to be a lot more careful with what is just a component and what is a matrix. Hopefuly that will come with more practice of this notation.

11. Feb 3, 2015

### Fredrik

Staff Emeritus
These earlier posts may be useful: