9c6c8fc5bd-p-78.pdf - 7.7 APENDICE 1 BEMOL Y SOSTENIDO Pn...

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7.7. AP ´ ENDICE 1: BEMOL Y SOSTENIDO 143 Entonces podemos suponer que v = n i =1 a i v i V y v = n j =1 a j φ j . Ahora bien, a j = v ( v j ) = v, v j y a j = v ( v j ) = v, v j = n i =1 g ij a i , (7.3) donde g ij = v i , v j . Por tanto, v = n i,j =1 a i g ij φ j . Si denotamos por ( g ij ) i,j la matriz inversa de ( g ij ) i,j entonces multiplicando ambos miem- bros de la igualdad (7.3) por g jk y sumando en j obtenemos: a k = n j =1 a j g jk En conclusi´on, si φ = n i =1 b i φ i V * y suponemos φ = n j =1 b j v j entonces φ = n i,j =1 b i g ij v j . (7.4) Observaci´on: Si · , · es eucl´ ıdea y B es una base ortonormal entonces a j = a j para todo j ∈ { 1 , . . . , n } . (En el caso lorentziano la ´unica diferencia es a 1 = - a 1 ). Este isomorfismo entre V ( R ) y V * ( R ) induce isomorfismos entre ten- sores tipo (2 , 0), (0 , 2) y (1 , 1) (y, en general, entre tensores tipo ( r, s ) y ( r , s ) con r + s = r + s ). Por ejemplo, supongamos que T es un tensor (2 , 0) y queremos construir a partir de ´ el un tensor ˆ T que sea (0 , 2). Entonces basta con definir ˆ T ( φ, ψ ) := T ( φ , ψ ). Obs´ ervese que
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  • Winter '15
  • pn, Espacio vectorial, Cálculo tensorial, Adición, Producto escalar, Lagrangiano

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