An Inner Product in Non-Euclidean Space

Classical Mechanics Level 1

Two vectors a i = ( 1 , 2 , 3 ) a^i = (1,2,3) and b i = ( 2 , 3 , 4 ) b^i = (2,3,4) live in a space with spatial metric given by the invariant interval

d s 2 = 2 d x 2 + 1 2 d y 2 + d z 2 . ds^2 = 2dx^2 + \frac12 dy^2 + dz^2.

Compute the inner product a i b i = g i j a i b j a_i b^i = g_{ij} a^i b^j .

3 18 19 20

Matt DeCross by Matt DeCross , 27, USA

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3 solutions

Matt DeCross
May 10, 2016

The metric is diagonal. The inner product is therefore, using the Einstein summation notation:

g i j a i b j = g 11 a 1 b 1 + g 22 a 2 b 2 + g 33 a 3 b 3 = 2 ( 1 ) ( 2 ) + 1 2 ( 2 ) ( 3 ) + 1 ( 3 ) ( 4 ) = 4 + 3 + 12 = 19. g_{ij} a^i b^j = g_{11} a^1 b^1 + g_{22} a^2 b^2 + g_{33} a^3 b^3 = 2(1)(2)+\frac12(2)(3) + 1(3)(4) = 4 + 3 + 12 = 19.

11 Helpful 1 Interesting 3 Brilliant 2 Confused

Note that we may write d s 2 = ( d x , d y , d z ) [ 2 0 0 0 1 / 2 0 0 0 1 ] ( d x d y d z ) . ds^2 = (dx, dy, dz)\begin{bmatrix}2 & 0 & 0 \\0 & 1/2 & 0 \\ 0 & 0 & 1 \end{bmatrix} \left(\begin{array}{c}dx\\ dy \\ dz\end{array}\right). From this we get our metric g μ ν = [ 2 0 0 0 1 / 2 0 0 0 1 ] . g_{\mu \nu} = \begin{bmatrix}2 & 0 & 0 \\0 & 1/2 & 0 \\ 0 & 0 & 1 \end{bmatrix}. The rest of the problem is solved as @Matt DeCross did.

8 Helpful 1 Interesting 1 Brilliant 0 Confused
Sean Spence
Mar 18, 2021

Diagonal matrix hence 3 terms with i=j; i,j=1: 2x1x2=4; 2: 0.5x2x3=3; 3: 1x3x4=12; Total 2+3+12=19.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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