# Einstein Summation Convention / Lorentz Boost

1. Dec 22, 2007

### raintrek

Einstein Summation Convention / Lorentz "Boost"

1. The problem statement, all variables and given/known data

I'm struggling to understand the Einstein Summation Convention - it's the first time I've used it. Would someone be able to explain it in the following context?

Lorentz transformations and rotations can be expressed in matrix notation as

$$x^{\mu'} = \Lambda^{\mu'}\!_{\mu}\:x^{\mu}$$

Coordinates are defined by $$x^{\mu}$$ with $$\mu = 0,1,2,3$$, such that $$(x^{0}, x^{1}, x^{2}, x^{3}) = (ct, x, y, z)$$

I'm seeking clarification on the meanings of the various $$\mu, \mu'$$ indices in the matrix notation equation. Any help would be massively appreciated!

2. Dec 22, 2007

### ophase

Summation rule

$$\mu, \acute{\mu}$$ means that these are all different indices which can have 0,1,2,3 and cannot contract with each other. I mean for example, while $$\mu = 0$$, the other one can be $$\acute{\mu} = 3$$.
Einstein summation rule is that you have to sum the terms with same indices. So you should scan all the values of the indices.
So
$$x^{\acute{\mu}} = \Lambda^{\acute{\mu}}\!_{\mu}\:x^{\mu}$$ means that

$$x^0 = \Lambda^{0}\!_{0}\:x^{0}+\Lambda^{0}\!_{1}\:x^{1}+\Lambda^{0}\!_{2}\:x^{2}+\Lambda^{0}\!_{3}\:x^{3}$$
$$x^1 = \Lambda^{1}\!_{0}\:x^{0}+\Lambda^{1}\!_{1}\:x^{1}+\Lambda^{1}\!_{2}\:x^{2}+\Lambda^{1}\!_{3}\:x^{3}$$
$$x^2 = \Lambda^{2}\!_{0}\:x^{0}+\Lambda^{2}\!_{1}\:x^{1}+\Lambda^{2}\!_{2}\:x^{2}+\Lambda^{2}\!_{3}\:x^{3}$$
$$x^3 = \Lambda^{3}\!_{0}\:x^{0}+\Lambda^{3}\!_{1}\:x^{1}+\Lambda^{3}\!_{2}\:x^{2}+\Lambda^{3}\!_{3}\:x^{3}$$

Don't forget that each term is actually a matrix element. So write the values of the matrix elements above and obtain the Lorentz transformation equations.

3. Dec 22, 2007

### raintrek

Ah, so am I correct in thinking that whenever an expression contains one index as a superscript and the same one as a subscript, a summation over those values is implied?

Also, does the $$\Lambda^{\mu'}\!_{\mu}$$ just mean a matrix with $$\mu'$$ representing the row number and $$\mu$$ representing the column number?

so, in essence,

$$\Lambda^{0'}\!_{0} \Lambda^{0'}\!_{1} \Lambda^{0'}\!_{2} \Lambda^{0'}\!_{3}$$
$$\Lambda^{1'}\!_{0} \Lambda^{1'}\!_{1} \Lambda^{1'}\!_{2} \Lambda^{1'}\!_{3}$$
$$\Lambda^{2'}\!_{0} \Lambda^{2'}\!_{1} \Lambda^{2'}\!_{2} \Lambda^{2'}\!_{3}$$
$$\Lambda^{3'}\!_{0} \Lambda^{3'}\!_{1} \Lambda^{3'}\!_{2} \Lambda^{3'}\!_{3}$$

is the $$\Lambda^{\mu'}\!_{\mu}$$ matrix produced from your equations?

Last edited: Dec 22, 2007
4. Dec 22, 2007

### ophase

Yes you'r right.
But
$$\Lambda^{\musingle-quote}\!_{\mu}$$ matrix cannot produce from above equation. We'r talking about a new representation of Lorentz transforms.

You can find the $$\Lambda^{\musingle-quote}\!_{\mu}$$ matrix here --> http://en.wikipedia.org/wiki/Lorentz_transformation

5. Dec 22, 2007

### siddharth

Yes.

6. Dec 22, 2007

### HallsofIvy

Staff Emeritus
More correctly $\Lambda^{\mu'}_\mu$ means a tensor that can, in a given coordinates system, be represented by such a matrix.