1. [20 points] Evaluate the following equations; a) ( delta2...

1. [20 points] Evaluate the following equations; a) ( delta_{2 j} b_{j}= ) ? b) ( delta_{i j} varepsilon_{i j k}= ) ? c) ( delta_{2 j} A_{i j}= ) ? d) If ( dot{mathbf{u}}_{1}=omega imes mathbf{u}_{1} ) and ( dot{mathbf{u}}_{2}=omega imes mathbf{u}_{2} ) where ( mathbf{u}_{1} ) and ( mathbf{u}_{2} ) are displacement vector and ( omega ) is angular velocity vector. Show that ( frac{d}{d t}left(mathbf{u}_{1} imes mathbf{u}_{2} ight)=omega imesleft(mathbf{u}_{1} imes mathbf{u}_{2} ight) ) Hint: ( mathbf{a} imes(mathbf{b} imes mathbf{c})=mathbf{b}(mathbf{a . c})-mathbf{c}(mathbf{a} . mathbf{b}) )

Transcript

00:01 In this problem, we have four questions regarding index manipulation.

00:05 The first one, delta 2j bj.

00:11 So the role of the kronecker delta tensor is to replace indices.

00:17 So it will replace the j by 2 everywhere.

00:22 So we will get b2, namely the second component of this b vector.

00:28 Next, we have delta ij epsilon ijk.

00:34 Again, the kronecker delta will replace j by i or i by j and you will obtain epsilon i i k.

00:44 So the levy -chivita tensor is a completely antisymmetric object and we are summing over two indices.

00:52 So it will be like epsilon 1 1 k plus epsilon 2 2 k plus epsilon 3 3 k.

01:03 So these are all 0 because one of the properties of the levy -chieftain tensor is that whenever the two indices are the same, it will return 0.

01:17 Next, we have delta2j aij.

01:24 So we have a tensor here but it doesn't matter.

01:27 The chronicle delta will still do the same work.

01:31 So it will be aij2.

01:35 Okay next and finally we have the following so if u1 vector dot is equal to omega cross u1 and u2 vector dot is equal to omega cross u2 where u1 and u2 are displacement vectors and and omega is the angular velocity vector, we are going to show that d over dt u1 cross u2 is equal to omega cross u1 cross u2.

02:20 Okay, let us do it.

02:21 And we are also given some hint.

02:27 A cross b cross c is equal to b times a dot c minus c times a dot b okay now let us consider these equations component by component so we have u1 i dot equal to so we have a cross product on the right hand side so we are going to use levichivita epsilon i j k omega j u 1 k and similarly for the u2 vector u2 i dot equal to epsilon i j k omega j u2 k now let us consider the ith component of this object so i have du dt epsilon i j k u1 j u2 k.

03:41 Okay, now these are just numbers.

03:44 So we can proceed with the usual rules of derivative.

03:49 So epsilon i j k is constant.

03:52 And we will use the product rule of derivative...

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