General Relativity - Contravariant and Covariant Vectors: aaargghh

tetris11
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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 Va be contravariant, and Wa be covariant.

Show that \mu=VaWa



Homework Equations


(couldn't get Latex to work consistently, sorry)
(1) V 'a = (dx 'a / dxb) Vb
(2) W 'a = (dxb / dx 'a) Wb

(3) AcDc = Sum of AcDc from c=1...n


The Attempt at a Solution


Well I just multiplied (1) and (2) together to get:
V 'aW 'a = (dx 'a / dxb) (dxb / dx 'a) VbWb

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

How this proves anything is scalar is beyond me.
Please help!
 
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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?
 
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?)
 
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]
 
(dx 'a / dxb) (dxc / dx 'a) Vb Wc = (dx c / dxb) Vb Wc = (with indexes vertically aligned) delta'c'b Vb Wc

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

So answer is either 0 or Vb Wc.
Unfortunately I am still not seeing it...:frown:
Why is this result scalar?
 
Last edited:
<br /> (V_aW^a)&#039;=\frac{dx&#039;^a}{dx^b}V^b\frac{dx^c}{dx&#039;^a}W_c<br /> =\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
<br /> (V_aW^a)&#039;=V^c W_c=V^a W_a<br />

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