Question

dr - 0 dz 1 dx P x² + y² + 하공 8. OP 11 위공 d츠 1 ㅇ d² 0= 이

Answer 1

(a) The scalar product of the given two tensors can be obtained as

.

2 2 هوا + 2 مهر ماه اوه يه = نوه خود ہے (- ) - ( + 4(- 4) + 3.1 _ 6 - = 3+ و- = 3 + 8 - 1 - =

(b) The Jacobian for transformation of Cartesian coordinates into cylindrical polar coordinates is given by

.

т Отә O2 Әт Эк — Әу Ә. ох де Әу 22 Ә2 Ә2 Ә 22 ә2 Әу ( ) т ут х» үр = Jху O — у/ О О o x=2 Cose Cos o — Sino Cosa 0 9 = msine - Sino 6 О

Hence the transformed tensors can be obtained as

( رو رع) ذیه ک = (2 ر2 ,2) اما Cos o Sino - 4 :) -Sino Cose O - Coso - 4 Sind Sing - 4 Cos

and

.

2(3, 0, 2) = 20; (x, y, 2) Ć Coso o Sino 1 - - -Sino los o O 2. 3 — 0 0 -Cos o t 2 Sino ll G + Sino +2 Cos o 3

(c) The dot product in the cylindrical polar coordinates is

re; o = (- Cose +2 Sino) (Cos 0 - 9Sino) +(+ Sino +2 Cos 0) (-sino-4 los e) +3 Cos² + 2 sine los at 4 sino lose = -8 Sin²o - sinza 2 Sino los e . - 4 Sino cose - 8 los 26+3 =-9 (Sin²o + cos2o) + 3 = -9 + 3 = -6

which is same as the previous one.

(d) In cylindrical polar coordinates,

,

По а пор

where $g^{\alpha \beta }$ is the metric tensor in cylindrical polar coordinates.