Notation difficulties with metric and four vector

1. Jan 3, 2016

tomwilliam2

I'm reading an introduction to relativity which uses different notation to the standard indices used in my college course.
I came across:
L(\nu)gL(\nu)g = 1
Where L is the Lorentz transformations four-vector and g is the metric. Without the indices, I'm a little lost. Is there some convention I'm not aware of?
Also (I'm trying to remember my undergrad relativity) can someone remind me why the Minkowski metric has the sign system it does? I know you can choose (-,+,+,+) or (+,-,-,-) but I forget why the ct dimension has to be a different sign to the space dimensions.

2. Jan 3, 2016

Staff: Mentor

The Lorentz transformation is not described by a 4-vector; it's described by a 4 x 4 matrix. So I'm confused about what this equation is supposed to be saying. Can you give a reference to the actual textbook and page number?

More precisely, the squared length of a timelike interval has to be opposite in sign to the squared length of a spacelike interval. The simplest answer is that if you do it any other way, the theory won't work; it will make incorrect predictions. There isn't a single reason; it's the whole structure of the theory.

3. Jan 3, 2016

andrewkirk

It's because the speed of light has to be the same in all inertial frames, which requires a lightlike vector to be null (have zero magnitude).

A lightlike velocity vector pointing along the x axis will have components (1, 1, 0, 0) because the light travels one unit of distance in one normalized unit of time. For the magnitude of this to be zero we require $g^{00}$ and $g^{11}$ to be equal in magnitude and opposite in sign.

4. Jan 4, 2016

tomwilliam2

Thanks for the replies. The paper is
"An Introduction to Relativistic Quantum Mechanics
I. From Relativity to Dirac Equation"
By M. De Sanctis
Page 8.

I did this rather late last night and didn't notice that in brackets it was a v not a \nu, which solves one of the issues I had. I still don't understand exactly what equation 2.5 (below) implies, nor how it results in equation 2.7:

gμρL(v)ρνLμσ(v)=gνσ

I realise now that L(v) is not the Lorentz matrix (that is denoted by lower case l, I think, but is L just some example four-vector?

5. Jan 4, 2016

Fredrik

Staff Emeritus
I haven't looked at the paper, but $L(v)^\mu{}_\nu$ must be the number on row $\mu$, column $\nu$ of the Lorentz transformation matrix L(v). The equality in post #4 is just the $\mu,\nu$ component of the matrix equation $L(v)^TgL(v)=g$. (Edit: I typed $\nu,\mu$ when I should have typed $\mu,\nu$. I have corrected it now).

A Lorentz transformation can be defined as a 4×4 matrix $\Lambda$ such that $\Lambda^T g\Lambda =g$. So the equality in post #4 is saying that L(v) is a Lorentz transformation, nothing more, nothing less.

Last edited: Jan 4, 2016
6. Jan 4, 2016

tomwilliam2

Thanks, that clears it up. I would prefer there to be an explicit T to denote the transpose, but I guess the author is explaining that it's not needed.

7. Jan 4, 2016

Fredrik

Staff Emeritus
The definition of matrix multiplication is $(AB)_{ij}=A_{ik}B_{kj}$. So if you encounter something like $A_{ki}B_{kj}$, you will have to use the definition of the transpose before you use the definition of matrix multiplication:
$$A_{ki}B_{kj} = (A^T)_{ik}B_{kj} = (A^TB)_{ij}.$$ Alternatively,
$$A_{ki}B_{kj}=B_{kj}A_{ki}=(B^T)_{jk}A_{ki} =(B^TA)_{ji} =((B^TA)^T)_{ij}.$$ It's clear from the order of the indices in the original expression that there's a transpose involved.