Simple indice question -- metric, traces

  • Thread starter binbagsss
  • Start date
  • #1
1,232
10
Mod note: OP warned about not using the homework template.
Let ##g_{ac}## be a 3-d metric.
So the trace of a metric is equal to its dimension so I get ##g_{ac}g^{ac}=3##

But I'm a tad confused with the expression : ##g^{ac}g_{ad}##=##delta^{c}_{d}##
I thought it would be ##3delta^{c}_{d}###

Because ##delta^{c}_{d}## implies that ##g^{ac}g_{ad}## is only non-zero when c=d, in which case don't you just get: ##g_{ac}g^{ac}=3##

Cheers in advance.
 
Last edited by a moderator:

Answers and Replies

  • #2
140
16
The summation rule applies, in the last expression c is being summed over, but in delta-expression it is not. If you contract the indices in it, you get trace of the identity matrix, which is 3 in three dimensions:

##3=\delta^c_c=g^{ac}g_{ac}\neq\delta^c_d=g^{ac}g_{ad}##

The first is a scalar, the second is a (1,1) tensor. Btw, you need a blackslash \ in front of delta you want to write ##\delta##
 
  • #3
20,867
4,546
Another way of thinking about this is that you are raising an index:
[tex]g^{ac}g_{ad}=g^{ca}g_{ad}=g^c_d=\delta^c_d[/tex]
Chet
 

Related Threads on Simple indice question -- metric, traces

  • Last Post
Replies
2
Views
783
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
5
Views
1K
Replies
2
Views
2K
Replies
28
Views
9K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
744
  • Last Post
Replies
1
Views
5K
Top