General Relativity - Contravariant and Covariant Vectors: aaargghh

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Homework Help Overview

The discussion revolves around the properties of contravariant and covariant vectors within the context of general relativity, specifically focusing on the relationship between these vectors and the concept of scalars. The original poster expresses confusion regarding the proof that the product of a contravariant vector and a covariant vector results in a scalar.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive a relationship by multiplying transformation equations for contravariant and covariant vectors but struggles to understand the implications of their results.
  • Participants question the validity of index usage in the equations presented, particularly regarding the occurrence of indices.
  • Some participants suggest exploring the transformation properties of scalars to clarify the original poster's confusion.
  • There is a discussion about the simplification of expressions involving Kronecker deltas and the implications for scalar transformation.

Discussion Status

The discussion is ongoing, with participants providing insights into the transformation rules for scalars and the nature of tensor indices. Some guidance has been offered regarding the transformation properties, but the original poster still expresses uncertainty about the conclusions drawn.

Contextual Notes

There is a noted difficulty with understanding the implications of index notation and transformations, which may be compounded by the original poster's self-identified struggle with the material.

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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