Home / Math and Physics Files / Math / Curvilinear Systems / Tensor Doc and Support / tensor doc support 2_22_12 thru 3_8_12
Weight changing theorem
DOCX · 18.3 KB
Open DOCX file
A brief working note by Phil (PhL, 2.25.12) from his tensor document support files. It considers a tensor density T of weight W and the object U = k|g|^(w/2) T, and tries to show U is a tensor density of weight W-w. The proof uses g'/g = J^2 and the sign choices for the square root of J. He notes the result is just the outer product rule, and that it was dropped from the main document.
AI-written summary; may contain errors.
Extracted text (machine-read; may contain errors)
Weight changing theorem PhL 2.25.12
This theorem is no longer needed, but I just save it here anyway. It is just an example of adding tensor weights in a particular case, and the two equations are just the covariant results. I then have to renumber the previous item 8 on J=1 to be item 7 and edit all references. Done.
7. Weight Changing Theorem: Suppose T is a tensor density of weight W, and suppose some tensor-like object U is related to T in this manner, where k and w are any constants,
Uabcde = k (sg)w/2 Tabcde = k |g|w/2 Tabcde
U'abcde = k (sg')w/2 T'abcde = k |g'|w/2 T'abcde
The theorem states that U transforms as a tensor density of weight W-w.
One would say according to Section 7 (u) that this tensor density equation is "covariant".
Proof: Start as in item 5 with a generic sample tensor density
T' abcde = J-W Raa' Rbb' Rcc' Rdd' Ree' Ta'b'c'd'e'
Then starting with the second line of the pair of lines above, [ (g'/g)w/2 = (σJ)w = σwJw ]
_____________________________________________________________________
g'/g = J2 > 0 from 5 (k). then this is the key:
(g'/g)1/2 = ± J, two choices. The RHS could be either + or - ! We have to make choices.
Suppose we choose (g'/g)1/2 = +J. The RHS can still be either + or -.
Then we get (g'/g)w/2 = Jw and again J could be + or -. Then theorem is valid!
But I don't need a theorem here! Since |g| is a TD with weight 2, |g|w/2 has weight w and we are just doing our outer product rule, nothing is new here! Very good. s
____________________________________________________________________
U'abcde = k(sg')w/2 T'abcde = k(sg'/sg)w/2 (sg)w/2 T'abcde = k(g'/g)w/2(sg)w/2 T'abcde
= σw Jw k(sg)w/2 {J-W Raa' Rbb' Rcc' Rdd' Ree' Ta'b'c'd'e')
= σw Jw { J-W Raa' Rbb' Rcc' Rdd' Ree' k(sg)w/2Ta'b'c'd'e')
= σw Jw { J-W Raa' Rbb' Rcc' Rdd' Ree' Ua'b'c'd'e')
= σw J-(W-w) Raa' Rbb' Rcc' Rdd' Ree' Ua'b'c'd'e' QED
OUCH! This theorem now fails and I need to do something! [ I rescued it! ]