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EQUATIONS OF ELASTIC HYPERSURFACES

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2. OUTLINE <strong>OF</strong> DIFFERENTIAL GEOMETRY 27<br />

this will replace all vectors u 2 in the determinant by the vectors v 2 without altering the determinant.<br />

We continue the procedure, based on the formulae (2.84) and replace successively<br />

all u 1 , . . . , u k by v 1 , . . . , v k without altering the determinant. Since v 1 , . . . , v k are orthogonal<br />

(see (2.85)), the matrix is diagonalized and maintains the value.<br />

Remark 2.28 If u 1 , . . . , u k are linearly independent, the system<br />

is even orthonormal<br />

w j := 1<br />

|v j | v0 j j = 1, . . . , k , (2.90)<br />

〈w j , w m 〉 = δ jk , j, m = 1, . . . , k , (2.91)<br />

where δ jk = 0 for k ≠ j and δ kk = 1 is the Kroneker symbol.<br />

Let<br />

P(u 1 , . . . , u k ) =<br />

{<br />

}<br />

α 1 u 1 + . . . + α k u k : 0 ≤ α j ≤ 1 , j = 1, . . . , k<br />

denote the hyper-parallelepiped in R n spanned by the vectors u 1 , . . . , u k ∈ R n .<br />

Lemma 2.29 Let u 1 , . . . , u n−1 ∈ R n . The Volume Vol P(u 1 , . . . , u n−1 ) of the hyper-parallelepiped<br />

coincides with the length of the vector product<br />

(cf. Lemma 2.25.h).<br />

Proof: Clearly,<br />

Vol P(u 1 , . . . , u n−1 ) = |u 1 ∧ . . . ∧ u n−1 | = √ G (u 1 , . . . , u n−1 ) (2.92)<br />

P(u 1 ) ⊂ P(u 1 , u 2 ) ⊂ · · · ⊂ P(u 1 , . . . , u n−1 )<br />

and Vol P(u 1 ) = |u 1 |, Vol P(u 1 , u 2 ) = |u 1 |h 2 , where h 2 = |u 2 | cos θ 2 is the projection of<br />

the vector u 2 on the axes orthogonal to u 1 and θ 2 is the angle between the vector u 2 and this<br />

orthogonal axes. By proceeding further we find that<br />

Vol P(u 1 , . . . , u n−1 ) = Vol P(u 1 , . . . , u n−2 )h n−1 = |u 1 |h 2 . . . h n−1 , (2.93)<br />

where h k = |u k | cos θ k = |v k | is the altitude of the hyper-parallelepiped P(u 1 , . . . , u k )<br />

with the base P(u 1 , . . . , u k−1 ) and θ k is the angle between u k and v k ((see Fig. 1; due to<br />

(2.87) the vector v k is orthogonal to the frame P(u 1 , . . . , u k−1 ) ⊂ L (u 1 , . . . , u k−1 ) for all<br />

k = 1, . . . , n − 1). Then<br />

|u 1 | = |v 1 |, h 2 2 = 〈u 2 , v 2 ) = |v 2 | 2 , . . . , h 2 n−1 = 〈u n−1 , v n−1 〉 = |v n−1 | 2<br />

(cf. (2.84), (2.87)) and we get the final result<br />

(cf. (2.86), (2.93)).<br />

Vol P(u 1 , . . . , u n−1 ) = |v 1 | |v 2 | . . . |v n−1 | = √ G (u 1 , . . . , u n−1 )

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