# Schutz - A First Course in GR - Simple Summation Question

1. Feb 18, 2010

### CFDFEAGURU

Hello all,

In the book "A First Course in General Relativity" by Schutz (1985 Edition) in chapter 2 there is a problem concerning summation that has me confused.

Note: This is not homework, just an interest of mine.

The given quantities are:

A = (5,0,-1,-6)
B = (0,-2,4,0)

C = [ 1 0 2 3
5 -2 -2 0
4 5 2 -2
-1 -2 -2 0 ]

Find:

A (super alpha) * C (sub alpha, beta); for all beta.

As usual, I apologize for not using LaTex but I can never get it to work right.

My attempt.

The only sum is on alpha because it is the only repeated upper and lower index. I should end up with a set of 4 numbers.

I followed the example given on page 41 of the book and applied it to this problem, but I got all 4 numbers wrong.

For the first value I calculated:

(1)*(5) + (0)*(0) + (2)*(-1) + (3)*(-6) = -15

The answer is given as (7, 1, 26, 17).

Any help on what I am doing wrong would be greatly appreciated.

Thanks
Matt

2. Feb 18, 2010

### Fredrik

Staff Emeritus
Matrix multiplication is defined by $(AB)_{ij}=A_{ik}B_{kj}$ (when we write all indices downstairs), so what you're supposed to calculate is $AC$, not $CA^T$.

3. Feb 18, 2010

### CFDFEAGURU

Thanks Fredrik,

I know that I am supposed to calculate AC, but I keep getting the wrong values. In the example given in the book, Schutz simply multiplies each row of the matrix (C) with the column vector (A) and then sums the values. Very easy and straightforward. What am I doing different that is causing me to calculate the wrong values.

What would make you think that I am trying to calculate A transposed times C?

Thanks
Matt

4. Feb 18, 2010

### Altabeh

You are approaching as follows for $$\beta=1:$$

$$T_{\beta}=C_{\alpha\beta}A^{\alpha},$$

Or in the matrix notation,

T=CA=\left[ \begin {array}{cccc} 1&0&2&3\\ \noalign{\medskip}5&-2&-2&0 \\ \noalign{\medskip}4&5&2&-2\\ \noalign{\medskip}-1&-2&-2&0 \end {array} \right]\left[ \begin {array}{c} 5\\ \noalign{\medskip}0\\ \noalign{\medskip} -1\\ \noalign{\medskip}-6\end {array} \right] .

Now decompose the matrix C into four row vectors and select the first one from the left side and lablel it $$\beta=1.$$ This will give $$T_1$$ as

$$T_1=(1)(5) + (0)(0) + (2)(-1) + (3)(-6) = -15.$$

This is wrong, because in $$C_{\alpha\beta}$$, $$\alpha$$ and $$\beta$$ represent, respectively, the row and column number so that $$C_{\alpha1}$$, would be a column vector, while you choose the row vector to have the matrix relation hold. While this sounds correct that the matrix representation gives the result as yours, but in tensor notation we are encountering number-by-number, i.e. component-by-component, multiplication not vector-by-vector multiplication!

AB

5. Feb 18, 2010

### CFDFEAGURU

Thanks Altabeh, I understand that I am doing something wrong. Could you please show how the first value of 7 is obtained?

Thanks
Matt

6. Feb 18, 2010

### Altabeh

This means that

C_{\alpha1}=\left[ \begin {array}{c} 1\\ \noalign{\medskip}5\\ \noalign{\medskip} 4\\ \noalign{\medskip}-1\end {array} \right].

Multiplying each component by the corresponding component of the vector A gives

$$T_1=(1)(5)+(5)(0)+(4)(-1)+(-1)(-6)=7.$$

Do a similar calculation to get other components of the vector T.

AB

7. Feb 18, 2010

### Rasalhague

The version on Google Books has $C_{31}=-1$ and $C_{32}=-3$ (excercise 2.9, 1), rather than $-2$ and $-2$.

8. Feb 18, 2010

### CFDFEAGURU

Yes, you are correct. I do have the wrong values

Thanks alot for the help. I didn't realize in the example problem you would get the same answer using the column as you would using the row.

Now it all makes sense.

Thanks alot everyone.

Now I know why Fredrik thought I was calculating the tranpose of CA

Matt

9. Feb 18, 2010

### Fredrik

Staff Emeritus
I guess this is a little late, but if A is a 4×1 matrix, what we get from the definition of matrix multiplication is

$$(CA^T)_i=C_{ij}A^T_j=C_{ij}A_j$$

and

$$(AC)_i=A_jC_{ji}$$