Page 126 - Intro to Tensor Calculus

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EXAMPLE 1.4-10. (Generalized velocity and acceleration) Let t denote time and let x = x (t)
1
N
for i =1,... ,N, denote the position vector of a particle in the generalized coordinates (x ,...,x ). From
the transformation equations (1.2.30), the position vector of the same particle in the barred system of
2
N
1
coordinates, (x , x ,... , x ), is
1
2
N
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i
x = x (x (t),x (t),...,x (t)) = x (t), i =1,... ,N.
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The generalized velocity is v = dx i ,i =1,... ,N. The quantity v transforms as a tensor since by definition
dt
i
dx i ∂x dx j ∂x i j
i
v = = = v . (1.4.51)
dt ∂x dt ∂x j
j
Let us now find an expression for the generalized acceleration. Write equation (1.4.51) in the form
∂x j
i
j
v = v i (1.4.52)
∂x
and differentiate with respect to time to obtain
i
2 j
dv j i ∂ x dx k dv ∂x j
= v + i (1.4.53)
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dt ∂x ∂x k dt dt ∂x
The equation (1.4.53) demonstrates that dv i does not transform like a tensor. From the equation (1.4.7)
dt
previously derived, we change indices and write equation (1.4.53) in the form
" #
a
j
dv j i dx k σ ∂x j j ∂x ∂x c ∂x dv i
= v σ − + .
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dt dt ik ∂x ac ∂x ∂x k ∂x dt
Rearranging terms we find
i
c
j
j
j
∂v dx k j ∂x a i ∂x dx k ∂x ∂v dx k σ i ∂x dx k
+ v = + v σ or
i
∂x k dt ac ∂x i ∂x k dt ∂x ∂x k dt ik ∂x dt
" #
∂v j a dx ∂v σ i dx ∂x
j k σ k j
+ v = + v σ
∂x k ak dt ∂x k ik dt ∂x
σ
δv j δv ∂x j
= σ .
δt δt ∂x
The above equation illustrates that the intrinsic derivative of the velocity is a tensor quantity. This derivative
is called the generalized acceleration and is denoted
2 i
δv i i dx j dv i i m n d x i dx m dx n
i
f = = v ,j = + v v = + , i =1,... ,N (1.4.54)
δt dt dt mn dt 2 mn dt dt
To summarize, we have shown that if
i
i
x = x (t), i =1,... ,N is the generalized position vector, then
dx i
i
v = , i =1,... ,N is the generalized velocity, and
dt
δv i i dx j
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f = = v ,j , i =1,... ,N is the generalized acceleration.
δt dt