Inverse of a Vector: Find the Correct Form

In summary: The notation is explained on page 293 equations 8.107 and 8.108.$$\mu_0 e : \nabla_S \dfrac{\partial v}{\partial t}=\mu_0 \sum_k e_{ik}\nabla_{kj} \dfrac{\partial v_j}{\partial t}$$and similar for the dot which seems to be ordinary matrix multiplication.
  • #1
chowdhury
36
3
What would the correct form of this?
$$\stackrel{\leftrightarrow}{A} \cdot \vec{b} = \stackrel{\leftrightarrow}{C} : \vec{d}$$

I'd like to know which one is correct form
1.) $$ \vec{b} = (\stackrel{\leftrightarrow}{A})^{-1} ( \stackrel{\leftrightarrow}{C} : \vec{d} ) $$
2.) $$ \vec{b} = (\stackrel{\leftrightarrow}{A})^{-1} \cdot ( \stackrel{\leftrightarrow}{C} : \vec{d} ) $$
3.) None of the above, in this case what would be the correct form?
 
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  • #2
You need to explain the notations.

1) and 2) look the same with 1) having the dot suppressed
 
  • #3
Vectors do not have a (multiplicative) inverse. They usually form an additive group, and your inverse requires a multiplicative group, which they are not.
 
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  • #4
Are A and C supposed to be matrices?
 
  • #5
Office_Shredder said:
Are A and C supposed to be matrices?
Judging by the notation, I would guess so. A right arrow is often used for vectors. A left arrow is rarely used for covectors. And a double arrow, i guess, is for operators, which can be viewed as elements of ##V^*\otimes V##.
 
  • #6
And then I guess the other question is what the heck does the colon mean?
 
  • #7
I haven't seen a two head arrow in such a context at all. The lack of intuition already forbids its use.
 
  • #8
Office_Shredder said:
And then I guess the other question is what the heck does the colon mean?
I don't have a guess for that.
 
  • #9
fresh_42 said:
I haven't seen a two head arrow in such a context at all. The lack of intuition already forbids its use.
The intuition is what I wrote in #5.
 
  • #10
martinbn said:
The intuition is what I wrote in #5.
A bit far-fetched, isn't it? One uses asterisks for covectors, and definitely no double arrows for dyads.
 
  • #11
fresh_42 said:
A bit far-fetched, isn't it?
Yes.
fresh_42 said:
One uses asterisks for covectors, and definitely no double arrows for dyads.
I have seen left arrows for covectors as ##l=\overleftarrow{l}##, so that for the evaluation on a vector ##v=\overrightarrow{v}## the arrows would compensate each other (like forces) and produce a quantity without any arrows i.e a scalar. We'll wait and see what the OP says to clear things up.
 
  • #12
martinbn said:
I have seen left arrows for covectors as ##l=\overleftarrow{l}## ...
I'd rather associate projective limits with it. :wink:
 
  • #13
fresh_42 said:
I'd rather associate projective limits with it. :wink:
Ah, for me that would be ##\underleftarrow{\text{lim}}##.
 
  • #14
martinbn said:
Ah, for me that would be ##\underleftarrow{\text{lim}}##.
nitpicking ...:biggrin:
 
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  • #15
martinbn said:
You need to explain the notations.

1) and 2) look the same with 1) having the dot suppressed

1.) For my special case,
A is a (mxm) matrix, b is a (mx1) vector
C is a (mxn) matrix, d is a (nx1) vector

2.) In fact,
A is a (mxn) tensor, b is a (nx1) vector
C is a (mxnxL) tensor, d is a (nxL) tensor

3.) : means the double summ notation in tensor

I hope this clarifies my query.
 
  • #16
Why don't you use the usual notations? And where did you get these obscure notations from?
chowdhury said:
1.) For my special case,
A is a (mxm) matrix, b is a (mx1) vector
##A## and ##\vec{b}##
chowdhury said:
C is a (mxn) matrix, d is a (nx1) vector
##C## and ##\vec{d}##
chowdhury said:
2.) In fact,
A is a (mxn) tensor, b is a (nx1) vector
##A=\sum_k \vec{x_k}\otimes \vec{y_k}## and ##\vec{b}##
chowdhury said:
C is a (mxnxL) tensor, d is a (nxL) tensor
##A=\sum_k \vec{x_k}\otimes \vec{y_k}\otimes \vec{z_k}## and ##d=\sum_k \vec{x_k}\otimes \vec{y_k}##
chowdhury said:
3.) : means the double summ notation in tensor
?
chowdhury said:
I hope this clarifies my query.
Not really.
 
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  • #17
@fresh_42 :
Thanks for your feedack. I obtained it from the book

B. A. Auld, Acoustic fields and waves in solids, vol. 1, 1973.

https://www.amazon.ca/dp/0898747821/

My equation is little bit different

$$0 = -\mu_{0} \nabla \negthinspace \cdot \negthinspace \left( \stackrel{\leftrightarrow}{\epsilon}^{S} \cdot \nabla \phi \right) + \mu_{0} \nabla \negthinspace \cdot \negthinspace \left( \stackrel{\leftrightarrow}{e} : \nabla_{s} \vec{u} \right) $$

Then
$$\nabla \phi = \left( \stackrel{\leftrightarrow}{\epsilon}^{S} \right)^{-1} \color{red}{?} \left( \stackrel{\leftrightarrow}{e} : \nabla_{s} \vec{u} \right)$$
Screenshot (16).png


Please see a screen shot of equation from the acutal book.

My query is what would the proper operator in the question mark.
 
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  • #18
chowdhury said:
@fresh_42 :
Thanks for your feedack. I obtained it from the book

B. A. Auld, Acoustic fields and waves in solids, vol. 1, 1973.

https://www.amazon.ca/dp/0898747821/

My equation is little bit different

$$0 = -\mu_{0} \nabla \negthinspace \cdot \negthinspace \left( \stackrel{\leftrightarrow}{\epsilon}^{S} \cdot \nabla \phi \right) + \mu_{0} \nabla \negthinspace \cdot \negthinspace \left( \stackrel{\leftrightarrow}{e} : \nabla_{s} \vec{u} \right) $$

Then
$$\nabla \phi = \left( \stackrel{\leftrightarrow}{\epsilon}^{S} \right)^{-1} \color{red}{?} \left( \stackrel{\leftrightarrow}{e} : \nabla_{s} \vec{u} \right)$$View attachment 297241

Please see a screen shot of equation from the acutal book.

My query is what would the proper operator in the question mark.
The notation is explained on page 293 equations 8.107 and 8.108.
$$
\mu_0 e : \nabla_S \dfrac{\partial v}{\partial t}=\mu_0 \sum_k e_{ik}\nabla_{kj} \dfrac{\partial v_j}{\partial t}
$$
and similar for the dot which seems to be ordinary matrix multiplication. And where did you get the double arrows from?
 
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  • #19
Office_Shredder said:
Are A and C supposed to be matrices?
A
Office_Shredder said:
Are A and C supposed to be matrices?
@Office_Shredder: In my case, A is a (mxm) tensor, b is a (mx1) vector
C is a (mxnxL) tensor and d is a (nxL) tensor.

We could collapse the third order tensor C to a second order matrix using Voigt notation and d to a vector. To be specific, C is (3x6) matrix and d is (6x1) in contracted notation.

I do not know how to write it properly, using the tensor notation and the matrix notation.
 
  • #20
martinbn said:
Judging by the notation, I would guess so. A right arrow is often used for vectors. A left arrow is rarely used for covectors. And a double arrow, i guess, is for operators, which can be viewed as elements of ##V^*\otimes V##.
@martinbn : I used the upper half arrow for vector, and to denote a tensor use the two-sided arrow. Please let me know, if these are the ways, they are denoted usually. Thanks.
 
  • #21
chowdhury said:
@martinbn : I used the upper half arrow for vector, and to denote a tensor use the two-sided arrow. Please let me know, if these are the ways, they are denoted usually. Thanks.
As already mentioned above, the double arrow is
a) very unusual (anywhere but especially for tensors).
b) even the book you quoted doesn't use it,
c) you probably study physics (concluding from the book) and then you need every tiny space around letters for indices. There is no room for superfluous signs,
d) basically every quantity in physics is a tensor, including scalars, vectors, and matrices.
 
  • #22
fresh_42 said:
The notation is explained on page 293 equations 8.107 and 8.108.
$$
\mu_0 e : \nabla_S \dfrac{\partial v}{\partial t}=\mu_0 \sum_k e_{ik}\nabla_{kj} \dfrac{\partial v_j}{\partial t}
$$
and similar for the dot which seems to be ordinary matrix multiplication. And where did you get the double arrows from?
@fresh_42 :

Your feedback is definitely appreciating. I do not know the proper notation, to be frank, and how to reconcile the two.

1.) From here and few other places, I noticed that people used double arrow notation for tensor. That's my I used it. I just picked double arrow and used it. I also need to use Fourier Transform late ron, I saw people used underscore or tilde sign below for the Fourier coefficients. These are the considerations, I took to pick and choose a double arrow sign for the tensor.
References:
a.) https://commons.wikimedia.org/wiki/File:Stress_tensor_-_indicial_notation.png
b.) https://www.researchgate.net/figure/1-Direct-tensor-notation_tbl1_235177604

2.) I have this equation to solve for the electric field
$$\epsilon^{S} \cdot \nabla \phi = e : \nabla_{s} u $$

Here,
  • $$\epsilon^{S} $$ is the (3x3) permittivity matrix.
  • $$\nabla \phi $$ is the negative of the electric field.
  • $$e$$ is the third rank tensor.
  • $$u$$ is the (3x3) displacement matrix.
Query: At this point, do I need to use appropriate notation to say that I have used the contracted notation to change the e tensor to a (3x6) matrix, u to a (6x1) vector? What I mean is that do I need to invoke anything at this point to say that I am using full tensor notation, not contracted notation, meaning manipulation of vectors and matrices.

2.) Thanks for pointing the appropriate Equation numbers from the referenced book, ie., Eq.(8.107) and Eq.(8.108); this is using indices.

Query:
  • How would I represent it using the matrix notation? $$ \nabla \phi = \left( \stackrel{\leftrightarrow}{\epsilon}^{S} \right)^{-1} \color{red}{?}\left( \stackrel{\leftrightarrow}{e} : \nabla_{s} \vec{u} \right). $$
  • To the best your study, shall I drop the doube arrow sign on the top for second rank and higher rank tensors?
  • What would I use the question mark place? dot product or matrix multiplication?

3.) There is problem, to me, to the notations used in the book of B. A. Auld, Acoustic fields and waves in solids, vol. 1, 1973.

$$\frac{\partial T}{\partial t} = {c}^{E} : \nabla_{s} v - e \cdot \frac{\partial E}{\partial t},$$
This is Eq.(8.102) this e should be transpose of e, neither the author indicated it. This is NOT obvious from the notation. Anyone trying to get some numbers out will see this problem.

Please provide your suggestion.
 
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  • #23
chowdhury said:
@fresh_42 :

Your feedback is definitely appreciating. I do not know the proper notation, to be frank, and how to reconcile the two.

1.) From here and few other places, I noticed that people used double arrow notation for tensor. That's my I used it. I just picked double arrow and used it. I also need to use Fourier Transform late ron, I saw people used underscore or tilde sign below for the Fourier coefficients. These are the considerations, I took to pick and choose a double arrow sign for the tensor.
References:
a.) https://commons.wikimedia.org/wiki/File:Stress_tensor_-_indicial_notation.png
b.) https://www.researchgate.net/figure/1-Direct-tensor-notation_tbl1_235177604
I haven't seen this before and still wouldn't call it usual. Usual is that people know which letter they used to abbreviate their tensor and work with the letter alone. It's more likely to write ##\sigma_{ij}## than the notation with the arrow is.
chowdhury said:
2.) I have this equation to solve for the electric field
$$\epsilon^{S} \cdot \nabla \phi = e : \nabla_{s} u $$

Here,
  • $$\epsilon^{S} $$ is the (3x3) permittivity matrix.
  • $$\nabla \phi $$ is the negative of the electric field.
  • $$e$$ is the third rank tensor.
  • $$u$$ is the (3x3) displacement matrix.
Query: At this point, do I need to use appropriate notation to say that I have used the contracted notation to change the e tensor to a (3x6) matrix, u to a (6x1) vector? What I mean is that do I need to invoke anything at this point to say that I am using full tensor notation, not contracted notation, meaning manipulation of vectors and matrices.
I'm not sure what you are actually asking. It would probably be better to ask questions about indexing in a physics forum, or at least in a separate thread. Or have a look at the explanations on Wikipedia:
https://en.wikipedia.org/wiki/Einstein_notation
https://en.wikipedia.org/wiki/Raising_and_lowering_indices
https://en.wikipedia.org/wiki/Tensor_contraction
https://en.wikipedia.org/wiki/Covariant_transformation
https://en.wikipedia.org/wiki/Tensor_field

A tensor for me as a mathematician is any expression
$$
\sum_{i_1,i_2,\ldots,i_n} v_{i_1}\otimes v_{i_2}\otimes \ldots \otimes v_{i_n}
$$
where I don't even bother which are vectors and which are covectors, or how the various vectors ##v_{i_k}## are indexed by themselves.

As you are reading a physics textbook that makes heavy use of tensor notations, I recommend taking the time to study the Wikipedia articles. How did you make it on page 300 without?

chowdhury said:
2.) Thanks for pointing the appropriate Equation numbers from the referenced book, ie., Eq.(8.107) and Eq.(8.108); this is using indices.

Query:
  • How would I represent it using the matrix notation? $$ \nabla \phi = \left( \stackrel{\leftrightarrow}{\epsilon}^{S} \right)^{-1} \color{red}{?}\left( \stackrel{\leftrightarrow}{e} : \nabla_{s} \vec{u} \right). $$

Firstly, you have to choose the matrix entries for ##(e^{S})^{-1}.## As far as I can tell, the dots and colons are simple matrix multiplications.

chowdhury said:
  • To the best your study, shall I drop the doube arrow sign on the top for second rank and higher rank tensors?
  • What would I use the question mark place? dot product or matrix multiplication?

3.) There is problem, to me, to the notations used in the book of B. A. Auld, Acoustic fields and waves in solids, vol. 1, 1973.

$$\frac{\partial T}{\partial t} = {c}^{E} : \nabla_{s} v - e \cdot \frac{\partial E}{\partial t},$$
This is Eq.(8.102) this e should be transpose of e, neither the author indicated it. This is NOT obvious from the notation. Anyone trying to get some numbers out will see this problem.

Please provide your suggestion.

I haven't found any other definition of ##a:b## than ##a:b=\left(\sum_{k=1}^n a_{ik}b_{kj}\right)_{ij}\,.##
 
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  • #24
fresh_42 said:
I haven't seen this before and still wouldn't call it usual. Usual is that people know which letter they used to abbreviate their tensor and work with the letter alone. It's more likely to write ##\sigma_{ij}## than the notation with the arrow is.

I'm not sure what you are actually asking. It would probably be better to ask questions about indexing in a physics forum, or at least in a separate thread. Or have a look at the explanations on Wikipedia:
https://en.wikipedia.org/wiki/Einstein_notation
https://en.wikipedia.org/wiki/Raising_and_lowering_indices
https://en.wikipedia.org/wiki/Tensor_contraction
https://en.wikipedia.org/wiki/Covariant_transformation
https://en.wikipedia.org/wiki/Tensor_field

A tensor for me as a mathematician is any expression
$$
\sum_{i_1,i_2,\ldots,i_n} v_{i_1}\otimes v_{i_2}\otimes \ldots \otimes v_{i_n}
$$
where I don't even bother which are vectors and which are covectors, or how the various vectors ##v_{i_k}## are indexed by themselves.

As you are reading a physics textbook that makes heavy use of tensor notations, I recommend taking the time to study the Wikipedia articles. How did you make it on page 300 without?Firstly, you have to choose the matrix entries for ##(e^{S})^{-1}.## As far as I can tell, the dots and colons are simple matrix multiplications.
I haven't found any other definition of ##a:b## than ##a:b=\left(\sum_{k=1}^n a_{ik}b_{kj}\right)_{ij}\,.##
 
  • #25
@fresh_42 I have read many chapters from the book, however, I found problem, meaning inconsistency in notation. That is why I am enquiring mathematically how can we write the equation

$$\epsilon^{S} \cdot \nabla \phi = e : \nabla_{s} u $$

Then what would be the gradient of voltage?
$$\nabla \phi = (\epsilon^{S})^{-1} \color{red}{?} (e : \nabla_{s} u) $$
Queries:
1.) Clearly you can see it should be transpose of e in the above equation. Isn't it? Putting aside physics, mathematically, don't you think I am correct? But the book does not show transpose.
2.) If you are familiar with Matlab, there is a difference between array multiplication operator (.*) and matrix multiplication (*). Where I am getting at is in the question mark place, what would be proper, per mathematics say, symbol?

$$\nabla_{s} u = S$$

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FAQ: Inverse of a Vector: Find the Correct Form

What is the inverse of a vector?

The inverse of a vector is a vector that has the same magnitude but opposite direction as the original vector.

How do you find the inverse of a vector?

To find the inverse of a vector, you can multiply the vector by -1 or use the formula (-1) * v, where v is the original vector.

Why is finding the inverse of a vector important?

Finding the inverse of a vector is important in many applications of vector algebra, such as solving systems of linear equations, finding the direction of a force, or calculating the reflection of a vector.

Can all vectors have an inverse?

No, not all vectors have an inverse. Only non-zero vectors can have an inverse. A zero vector, which has a magnitude of 0 and no direction, does not have an inverse.

Is the inverse of a vector unique?

Yes, the inverse of a vector is unique. This means that for any given vector, there is only one vector that is its inverse. This is because the inverse is determined by the magnitude and direction of the original vector.

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