PDA

View Full Version : Playing with the Cristoffel symbol


redstone
Nov9-10, 04:54 PM
I'm learning about the Christoffel symbol and playing around with it, so I'm curious... Does the math work below, or have I done something wrong?

\Gamma^{j}_{cd}=\Gamma^{j}_{cd}
g^{cd}\Gamma^{j}_{cd}=g^{cd}\Gamma^{j}_{cd}
\frac{1}{2}g^{cd}g^{jm}\left(g_{md,c}+g_{mc,d}-g_{cd,m}\right)=\frac{1}{2}g^{cd}g^{jm}\left(g_{md ,c}+g_{mc,d}-g_{cd,m}\right)
\frac{1}{2}g^{cd}g^{jm}\left(g_{md,c}+g_{mc,d}-g_{cd,m}\right)=\frac{1}{2}g^{ab}g^{jm}\left(g_{mb ,a}+g_{ma,b}-g_{ab,m}\right)
\frac{1}{2}g^{jm}\left(g_{md,c}+g_{mc,d}-g_{cd,m}\right)=\frac{1}{2}g_{cd}g^{ab}g^{jm}\left (g_{mb,a}+g_{ma,b}-g_{ab,m}\right)
\Gamma^{j}_{cd}=g_{cd}g^{ab}\Gamma^{j}_{ab}
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

fzero
Nov9-10, 05:54 PM
No, the problem is that you're summing over both c,d in


g^{cd}\Gamma^{j}_{cd}=g^{ab}\Gamma^{j}_{ab}


and there's no operation that we can do to move g^{cd} over to the RHS. There's no free index available to use

g_{ec} g^{cd} = \delta^d_e

or similar identities.

redstone
Nov11-10, 11:38 AM
OK, so I guess there's something wrong with the following too then?

1) g_{ab}A^{ab}=g_{cd}A^{cd}
2) g_{ab}g^a_eA^{eb}=g_{cd}A^{cd}
3) g_{ab}g^a_eA^{eb}=g^e_ag^a_eg_{cd}A^{cd}
4) g_{ab}A^{eb}=g^e_ag_{cd}A^{cd}
5) g^f_bg_{af}A^{eb}=g^e_ag_{cd}A^{cd}
6) g^f_bg_{af}A^{eb}=g^f_bg^b_fg^e_ag_{cd}A^{cd}
7) g_{af}A^{eb}=g^b_fg^e_ag_{cd}A^{cd}
8) g^{af}g_{af}A^{eb}=g^{af}g^b_fg^e_ag_{cd}A^{cd}
9) g^a_aA^{eb}=g^{af}g^b_fg^e_ag_{cd}A^{cd}
10) A^{eb}=g^{af}g^b_fg^e_ag_{cd}A^{cd}
11) A^{eb}=g^{ab}g^e_ag_{cd}A^{cd}
12) A^{eb}=g^{eb}g_{cd}A^{cd}
13) A^{ab}=g^{ab}g_{cd}A^{cd}

Where did I go wrong here? Also, maybe you know of a good reference that gives a good overview of what is and isn't allowed in index gymnastics?

fzero
Nov11-10, 12:55 PM
OK, so I guess there's something wrong with the following too then?

1) g_{ab}A^{ab}=g_{cd}A^{cd}


This is just a relabeling of indices and is correct.


2) g_{ab}g^a_eA^{eb}=g_{cd}A^{cd}


Here g^a_e = g^{af}g_{fe} = \delta^a_e so this equation is correct.


3) g_{ab}g^a_eA^{eb}=g^e_ag^a_eg_{cd}A^{cd}


The LHS is g_{ae} A^{eb}=A^b_a, while the RHS involves the trace of A, so this equation is not generally correct.


4) g_{ab}A^{eb}=g^e_ag_{cd}A^{cd}
5) g^f_bg_{af}A^{eb}=g^e_ag_{cd}A^{cd}
6) g^f_bg_{af}A^{eb}=g^f_bg^b_fg^e_ag_{cd}A^{cd}
7) g_{af}A^{eb}=g^b_fg^e_ag_{cd}A^{cd}


These are all equivalent to 3 and incorrect.


8) g^{af}g_{af}A^{eb}=g^{af}g^b_fg^e_ag_{cd}A^{cd}
9) g^a_aA^{eb}=g^{af}g^b_fg^e_ag_{cd}A^{cd}


For both, the LHS involves the trace of the metric, while the RHS involves the trace of A, this is incorrect in general.


10) A^{eb}=g^{af}g^b_fg^e_ag_{cd}A^{cd}
11) A^{eb}=g^{ab}g^e_ag_{cd}A^{cd}
12) A^{eb}=g^{eb}g_{cd}A^{cd}
13) A^{ab}=g^{ab}g_{cd}A^{cd}


These are all the same equation and are incorrect.



Where did I go wrong here? Also, maybe you know of a good reference that gives a good overview of what is and isn't allowed in index gymnastics?

Any undergrad relativity tex should spend some time explaining index notation. You should know that g^{ab} is the inverse of g_{bc} as a matrix so that g^{ab}g_{bc}=\delta^a_c. You should also know that you generally never repeat an index twice in an expression.

Finally you could have actually thought about what your expressions look like in components as a sanity check. For instance, equation 13 above is telling you that

A^{11} = g^{11} \left( g_{11}A^{11} + g_{12}( A^{12} + A^{21}) + \cdots \right).

The RHS is drastically different from the LHS and would not be satisfied by an arbitrary tensor A.

Do lots of exercises and the formalism will start to sink in.

redstone
Nov11-10, 02:31 PM
g_{ab}g^a_eA^{eb}=g^e_ag^a_eg_{cd}A^{cd}

The LHS is g_{ae} A^{eb}=A^b_a, while the RHS involves the trace of A, so this equation is not generally correct.


This confuses me a bit. The only difference between this step and step 2, is that here the RHS includes g^e_a and g^a_e
which I thought were both just identity matrices? so doesn't that mean it should be the same as step 2?

fzero
Nov11-10, 02:58 PM
This confuses me a bit. The only difference between this step and step 2, is that here the RHS includes g^e_a and g^a_e
which I thought were both just identity matrices? so doesn't that mean it should be the same as step 2?

OK, I did make a mistake there. The LHS is


g_{ab}g^a_eA^{eb}= g_{eb} A^{eb}


but the RHS is


g^e_ag^a_eg_{cd}A^{cd} = \delta^a_a g_{cd}A^{cd}.


\delta^a_a = D, the dimension of the space, so this is still incorrect.