Can Einstein Index Notation Help Me Solve Equations in Continuum Mechanics?

AI Thread Summary
The discussion revolves around using Einstein index notation to verify the identity matrix in continuum mechanics. The user seeks clarification on whether to change index letters when multiplying matrices A and its inverse. Participants emphasize the importance of maintaining three indices for proper matrix multiplication and confirm that the expression for matrix C can be derived from A and B. The conversation also touches on the relevance of tensors in fluid mechanics. The user expresses gratitude for the assistance received in completing their calculations.
Theta_84
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Hello
I am doing some exercises in continuum mechanics and it is a little bit confusing. I am given the following equations ## A_{ij}= \delta_{ij} +au_{i}v_{j} ## and ## (A_{ij})^{-1} = \delta_{ij} - \frac{au_{i}v_{j}}{1-au_{k}v_{k}}##. If I want to take the product to verify that they give the identity matrix (its components maybe is more accurate), should I change in one of the expressions the index letters and proceed(change the free indices I mean)? Is this the correct approach ## (\delta_{ij} +au_{i}v_{j})(\delta_{mn} - \frac{au_{m}v_{n}}{1-au_{k}v_{k}}) ## and do the calculations? Does this term make sense ## \delta_{ij}\delta_{mn}##?

Thanks lot
 
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Hello ##\Theta##, :welcome:

You don't want four indices, but three: one to sum over and the other two are the indices of the product matrix.

Consider matrices A, B and C: what is the expression for ##C_{ij}## in terms of ##A_{..}## and ##B_{..}## ?

PS do you mean ##A^{-1}_{ij} ## as in ##(A^{-1})_{ij} ## ?
 
BvU said:
Hello ##\Theta##, :welcome:

You don't want four indices, but three: one to sum over and the other two are the indices of the product matrix.

Consider matrices A, B and C: what is the expression for ##C_{ij}## in terms of ##A_{..}## and ##B_{..}## ?

PS do you mean ##A^{-1}_{ij} ## as in ##(A^{-1})_{ij} ## ?
I believe it is ## C_{ij}=A_{ik}B_{kj}## ? Yes you are correct it is ## A_{ij}^{-1}## my mistake. So if I keep three free indices I have something like : ## C_{ij}=(\delta_{im} +au_{i}v_{m})(\delta_{mj} - \frac{au_{m}v_{j}}{1-au_{k}v_{k}})= \delta_{im}\delta_{mj}- \delta_{im}\frac{au_{m}v_{j}}{1-au_{k}v_{k}} +\delta_{mj}au_{i}v_{m} -au_{i}v_{m}\frac{au_{m}v_{j}}{1-au_{k}v_{k}} ## ?
 
Einstein notation... Sorry mate it's all about the lathes, spanners and Sir Issac around here.
 
Theta_84 said:
I believe it is ## C_{ij}=A_{ik}B_{kj}## ? Yes you are correct it is ## A_{ij}^{-1}## my mistake. So if I keep three free indices I have something like : ## C_{ij}=(\delta_{im} +au_{i}v_{m})(\delta_{mj} - \frac{au_{m}v_{j}}{1-au_{k}v_{k}})= \delta_{im}\delta_{mj}- \delta_{im}\frac{au_{m}v_{j}}{1-au_{k}v_{k}} +\delta_{mj}au_{i}v_{m} -au_{i}v_{m}\frac{au_{m}v_{j}}{1-au_{k}v_{k}} ## ?

Yes, that's the idea.
 
PeroK said:
Yes, that's the idea.
I think I completed it. thank you.
 
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xxChrisxx said:
Einstein notation... Sorry mate it's all about the lathes, spanners and Sir Issac around here.
I think fluids is a specialty of Mech,Eng.? and you need tensors.
 

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