General Relativity - Contravariant and Covariant Vectors: aaargghh

[tetris11]

Homework Statement

I know this is an easy question, I just can't seem to grasp what I am actually doing:

Let M be a manifold.
Let V^a be contravariant, and W_a be covariant.

Show that $$\mu$$=V^aW_a

Homework Equations

(couldn't get Latex to work consistently, sorry)
(1) V ^'a = (dx ^'a / dx^b) V^b
(2) W ^'_a = (dx^b / dx ^'a) W_b

(3) A^cD_c = Sum of A^cD_c from c=1...n

The Attempt at a Solution

Well I just multiplied (1) and (2) together to get:
V ^'aW ^'_a = (dx ^'a / dx^b) (dx^b / dx ^'a) V^bW_b

The brackets cancel out and I get:
V ^'aW ^'_a =V^bW_b = Sum of V^bW_b from b=1...n (from (3))

How this proves anything is scalar is beyond me.
Please help!

 

[cristo]

Note that on the right hand side of your equation you have the same index occurring four times-- this is not allowed. An index may appear either exactly once or exactly twice.

In order to answer this question you will need to know what it means to be a scalar: how does a scalar change under a coordinate transformation?

 

[tetris11]

Scalars don't transform under co-ordinate transformations, but I still don't see the solution.

Also, which equation have I used the same index four times?
I can only think of the bottom last equation I wrote, and I think I have only used the b index exactly twice (right?)

 

[cristo]

In this equation "V 'aW 'a = (dx 'a / dxb) (dxb / dx 'a) VbWb", b occurs four times. You are correct that a scalar doesn't transform. Thus, you want to calculate (V_aW^a)' and show that it is equal to V_aW^a. So,

$$(V_aW^a)'=\frac{dx'^a}{dx^b}V^b\frac{dx^c}{dx'^a}W_c$$

What does the right hand side simplify to? [Hint: it will involve a Kronecker delta]

 

[tetris11]

(dx ^'a / dx^b) (dx^c / dx ^'a) V^b W_c = (dx ^c / dx^b) V^b W_c = (with indexes vertically aligned) delta^'c_'b V^b W_c

'c = 'b => delta = 1
'c =/= 'b => delta = 0

So answer is either 0 or V^b W_c.
Unfortunately I am still not seeing it...
Why is this result scalar?

 

[cristo]

$$(V_aW^a)'=\frac{dx'^a}{dx^b}V^b\frac{dx^c}{dx'^a}W_c =\frac{dx^c}{dx^b}V^bW_c=\delta^c{}_b V^b W_c$$

Then, as you say, the delta is only 1 when c=b, so
$$(V_aW^a)'=V^c W_c=V^a W_a$$

Do you see that this is the transformation rule for a scalar?

 

[tetris11]

Kind of.

I take it that as long as the tensor's V and W have the same number of indexes in the same positions on both sides, then no transformation has taken place - only the size of index has changed, which is valid?