2

An exercise in the continuum mechanics book by Gurtin says:

A spatial tensor field $A$ is indifferent if, during any change in observer, A transforms according to $A^{*}(x^{*},t) = Q(t)A(x,t)Q(t)^{T}$.

We have that $x^{*}(p,t) = Q(t)x(p,t)$.

The book does not give a definition for $A^*.$

What does "transforms according to" mean?

Joe
  • 628
  • $A^*$ is the same tensor field as $A$ only expressed in the basis of the new reference frame. – Lythia Oct 15 '15 at 22:06
  • So is there no nice way of writing down something concretely? – Joe Oct 15 '15 at 22:15
  • I'm not sure if I quite get what you mean by concretely in this context. Examples of indifferent tensor fields include $\boldsymbol{D}$ and $\boldsymbol{B}$. – Lythia Oct 15 '15 at 22:33
  • I think that the definition of $A^$ should somehow take into account a functional dependence on the motion of the body. $A^$ should be the tensor field computed when you superpose a rigid motion. In section 25 it becomes clearer. – Gaff Jul 14 '16 at 15:07

2 Answers2

1

Perhaps this will do for your wish for concreteness?

Let $\{\boldsymbol{e}_i, \boldsymbol{e}_j, \boldsymbol{e}_k\}$ be a basis for the space. Under a change of reference frame, this basis becomes $$\{\boldsymbol{e}^*_i, \boldsymbol{e}^*_j, \boldsymbol{e}^*_k\}=\{\boldsymbol{Q}\boldsymbol{e}_i, \boldsymbol{Q}\boldsymbol{e}_j, \boldsymbol{Q}\boldsymbol{e}_k\}.$$ Suppose two different observers are attempting to study the same tensor field. In the reference frame defined by the basis $\{\boldsymbol{e}_i, \boldsymbol{e}_j, \boldsymbol{e}_k\}$ call this tensor field $\boldsymbol{A}$. In the other reference frame call this tensor field, $\boldsymbol{A}^*$. Then $\boldsymbol{A}$ and $\boldsymbol{A}^*$ can be written in components. $$\boldsymbol{A} = A_{ij}\boldsymbol{e}_i\otimes\boldsymbol{e}_j,\\ \boldsymbol{A^*} = A^*_{ij}\boldsymbol{e}^*_i\otimes\boldsymbol{e}^*_j.$$ But using the relation for $\{\boldsymbol{e}^*_i, \boldsymbol{e}^*_j, \boldsymbol{e}^*_k\}$ in terms of the basis $\{\boldsymbol{e}_i, \boldsymbol{e}_j, \boldsymbol{e}_k\}$, then \begin{align} \boldsymbol{A}^* &= A^*_{ij}\boldsymbol{Q}\boldsymbol{e}_i\otimes\boldsymbol{Q}\boldsymbol{e}_j\\ &= \boldsymbol{Q}\left(A^*_{ij}\boldsymbol{e}_i\otimes\boldsymbol{e}_j\right)\boldsymbol{Q}^{T}. \end{align} Now if the tensor field under study is indifferent, then by your definiton \begin{align} \boldsymbol{A}^* &= \boldsymbol{Q}\boldsymbol{A}\boldsymbol{Q}^{T}\\ &= \boldsymbol{Q}\left(A_{ij}\boldsymbol{e}_i\otimes\boldsymbol{e}_j\right)\boldsymbol{Q}^{T}. \end{align} Comparing the two ways of writing $\boldsymbol{A}^*$, it is seen that for indifferent tensor fields $A^*_{ij}=A_{ij}$.

Lythia
  • 1,303
  • Lythia, thank you for your reply. But I am still a bit confused. If $A$ is the same as $A^{}$ which you say, then isn't the property of indifference of $A$ equivalent to $A(x^{},t) = Q(t)A(x,t)Q(t)^{T}$ where this is an equality of tensors? – Joe Oct 16 '15 at 01:49
  • To be clear, $\boldsymbol{A}$ and $\boldsymbol{A}^$ are not the same. They measure the same tensor field, but from two different perspectives. If the tensor field is indifferent, then $\boldsymbol{A}$ and $\boldsymbol{A}^$ are related in a nice way (the property you state). – Lythia Oct 16 '15 at 03:01
  • Well then we still have not given a definition for $A^{}$. Consider the case when $A$ is not indifferent so we do not have the last equation below. Now if we agree that $A$ and $A^{}$ are the same tensor then $A = A_{ij} e_i \otimes e_j = A^{}_{ij} e_i^{} \otimes e_j^{}$ and hence we have defined the $A^{}{ij}$ implicitly so we are ok. However, if we do not have this then it seems that you have just written $A^{*}{ij}$ and not defined what it actually is. What am I missing here? – Joe Oct 16 '15 at 03:14
  • What you have to link them is the description of the tensor field in the reference configuration. $\boldsymbol{A}$ and $\boldsymbol{A}^$ can be thought of as descriptions of the same tensor field given two different descriptions of the current configuration. To get from one to the other requires going back to the reference and then forward to the other current configuration. If the tensor field is indifferent, you can skip going back to the reference and convert from one configuration to the other through the rule $\boldsymbol{A}^=\boldsymbol{Q}\boldsymbol{A}\boldsymbol{Q}^T$. – Lythia Oct 16 '15 at 03:26
  • I still don't believe that $A^{}{ij}$ is defined. Surely we must have some equation that involves both $A{ij}$ and $A^{}{ij}$ and I don't see that here unless we are assuming indifference. $A^{*}{ij}$ are still just stray symbols to me unless we assume indifference to connect them to $A_{ij}$. Are you sure they are independently defined here? If so please show an equation relating them to $A_{ij}$. – Joe Oct 16 '15 at 03:43
  • What makes you believe that $A_{ij}$ are defined, but $A^_{ij}$ are not? The key point that I alluded to in the previous note is that you have the equation $$\chi^(\boldsymbol{X},t) = \boldsymbol{y}(t) + \boldsymbol{Q}(t)\left(\chi(\boldsymbol{X},t)-\boldsymbol{o}\right)$$ for any change of reference frame. From this equation, which we get by looking back into the reference configuration first and then going forward using the motions $\chi \text{ and } \chi^*$, you get everything you want, – Lythia Oct 16 '15 at 04:03
  • Because I start with a fixed tensor $A$. If I choose a basis $(e_1,e_2,e_3)$ then the $A_{ij}$ are defined as the matrix entries corresponding to the tensor $A$ under that basis. Then $A$ "transforms" magically to $A^{}$. What is $A^{}$? Why can't we just write $A^* := $(something having to do with $A$). – Joe Oct 16 '15 at 04:11
  • Because the something to with $A$ part will change depending on the type of tensor you have. You have to derive the relationship from the equation I have above which relates the two motions. One of the nicest such relationships that occurs is if the tensor is indifferent. An even nicer relationship occurs if the tensor is invariant. – Lythia Oct 16 '15 at 04:17
  • When he writes $A$ is indifferent if $A$ transforms to $A^{}$ "according to $A^{}(x^{},t) = Q(t)A(x,t)Q(t)^{T}$" he is implying that in general (ie even if that eqn does not hold) we have some association between $A$ and $A^{}$. If I try to relate $A^{}$ and $A$ by the motion as you say, as soon as I write down "$A^{}$" I am lost because I don't know what it is. – Joe Oct 16 '15 at 04:27
  • Suppose we want to describe a tensor field. It is important to note that I have NOT given this tensor field any name yet. We must first define an origin $\boldsymbol{o}$ and a basis $\boldsymbol{e}_i, \boldsymbol{e}_j, \boldsymbol{e}_k$. Only once we have done these two things should we start naming our tensor field. If we choose the name $A$, then we are implicitly saying that $A$ is the name of the tensor field only for the chosen basis and origin. If we choose a different origin and basis, then we should change the name. – Lythia Oct 16 '15 at 04:37
  • If this choice of new basis and origin is related to the old basis and origin by a change of reference, then we name the tensor field $A^$. Both $A$ and $A^$ measure the same thing, just from a different perspective. For a particular tensor field the relationship between $A$ and $A^*$ can be derived through the equation relating the two motions. – Lythia Oct 16 '15 at 04:38
  • So for a fixed $x,t$ are $A^{*}(x,t)$ and $A(x,t)$ the same tensor (perhaps under a different basis) or no? – Joe Oct 16 '15 at 05:02
  • I guess I don't mean tensor but I mean the same linear map. Maybe my confusion is that when I have been saying tensor I mean linear map? – Joe Oct 16 '15 at 05:06
0

It means the following:

Let $x^*$ be a new coordinate system related to $x$ by $x^*(p,t) = Q(t)x(p,t)$. We denote by $A^*$ the spacial vector field in the coordinate $x^*$. Then $A$ is called indifferent if for any change of coordinate system as above we have $$A^*(x^*,t) = Q(t)A(x,t)Q(t)^T.$$

Daniel Robert-Nicoud
  • 29,795
  • I don't understand this can you please write $A^{}(x^{},t)$ := ... and $A^{*}(x,t) := ...$ I mean what are the explicit definitions of the previous terms. Thanks! – Joe Oct 15 '15 at 21:31
  • $A^*$ needs to be a tensor field or this is wrong. – Lythia Oct 15 '15 at 21:50
  • It is a tensor field. It says it in the question. He just wrote vector field by accident I think. Do you know how to give an explicit definition for $A^{*}$? Thanks. – Joe Oct 15 '15 at 21:51
  • The best I can come up with is to define $A^{}(x^{}(x,t),t) = A(x,t)$ but this is not even well defined. – Joe Oct 15 '15 at 22:04