## How to prove that isotropic materials have only two independent constants by tensor operation rules

I want to use tensor method to prove that there are only two independent elastic constants for isotropic materials. But I can’ t get two independent results after reading related posts.

The following resources can be referenced:

1.wikipedia

That is to find the invariant linear transformation f under $$SO(3)$$. For any C belonging to $$SO(3)$$, 3 * 3 matrix A. There is $$C.f(A).C^T=f(C^T.A.C)$$, then f can be decomposed into the linear combination of identity, transpose and trace, that is, f = C1 * Identity + C2 * Transpose + C3 * I3 * Trace, I3 is the third-order unit matrix.

2.SE community related posts

Length[indeps = SymmetrizedIndependentComponents[{3, 3, 3, 3}, sym]] sym = {{{2, 1, 3, 4}, -1}, {{3, 4, 1, 2}, 1}}; Length[indeps = SymmetrizedIndependentComponents[{3, 3, 3, 3}, sym]] 

How can I get 2 independent variables?

## How to use Mathematica to prove that isotropic materials have only two independent parameters

Posts on related issues can be found from here or here.

Index symmetries:

A stiffness tensor $$C$$ is a fourth-order tensor with components $$c_{ijkl}$$ which maps symmetric second-order tensors into symmetric second-order tensors, i.e., $$\sigma_{ij} = c_{ijkl} \varepsilon_{kl}$$ (linear elastic law), $$\sigma$$ (stress) and $$\varepsilon$$ (strain) being arbitrary symmetric second-order tensors. Due to the symmetry of the second-order tensors, $$C$$ is allowed to be minor symmetric, i.e., $$c_{ijkl} = c_{jikl} = c_{ijlk}$$. The not minor symmetric part of $$C$$ is irrelevant for the elastic law and is dropped. If the stress $$\sigma$$ is to be related to an elastic energy potential $$W$$ (referred to as hyperelastic behavior), i.e., $$\sigma = \partial W / \partial \varepsilon$$, then, due to Schwarz’s theorem, the stiffness tensor $$c_{ijkl} = \partial^2 W / \partial \varepsilon_{ij} \partial \varepsilon_{kl}$$ has to possess the major symmetry, i.e., $$c_{ijkl} = c_{klij}$$.

Material symmetry:

A material with stiffness $$C$$ is said to possess the material symmetry group $$G$$ (e.g., triclinic, orthotropic, transversally isotropic, …) if

$$$$C = Q \star C \qquad Q \in G$$$$

holds, where $$Q$$ are second-order tensors, referred to as symmetry transformations of $$C$$. The product $$\hat{C} = Q \star C$$ (referred to here as Rayleigh product) is defined in components as

$$$$\hat{c}_{ijkl} = Q_{im}Q_{jn}Q_{ko}Q_{lp}c_{mnop}$$$$

For solids, $$G$$ is a subset of the orthogonal group. In solid mechanics, if suffices to consider rotation matrices $$Q$$ from the rotational group $$SO(3)$$. If $$G = \{I\}$$, $$I$$ being the identity matrix, then $$C$$ is said to triclinic. If $$G$$ possesses more than the identity transformation, then different material classes can be defined (different anisotropy types). If $$G = SO(3)$$, the $$C$$ is said to be isotropic (no direction dependency).

I want to use Mathematica to get the number of independent parameters needed for the fourth-order tensor to make $$C = Q \star C (Q \in SO(3))$$ under the rotation of group $$SO(3)$$.

At present, I can only get 30 independent variables using the following method:

SymmetrizedIndependentComponents[{3, 3, 3, 3},    Symmetric[{1, 2, 3}]] // Length 

However, I still can’t use the rotation of group $$SO(3)$$ to further reduce the number of independent variables. What should I do?

## Can the Revised Ranger’s Beast Companion be an independent mount?

The mounted combat rules say that you can either control a mount, making it move on your turn — but its action can only be Dash, Disengage, or Dodge — or have it be independent, meaning it can do its actions but you have no control over it.

Now, if a ranger uses their Beast Companion as a mount, could they direct it as they usually do when unmounted and still have its full array of actions, since the beast still acts on their own initiative?

## How to Tokenize a String and Save Each Substring Separately (Independent Names for Each Substring)

I am given two files one with the name of person and the location that they are from (Evan Lloyd|Brownsville) and one with the name and salary (Evan Lloyd|58697) (the line number that you find the employee on in the first file is not necessarily the line number that find the employee on in the second). The user inputs a location (whole or part). For example if they input “ville” or “Ville” it should include all of the employees in Brownsville, Clarksville, Greenville, etc. I am supposed to join the the name and salary and return them if they are in the city searched for i.e. “ville” or “Ville.” I am attempting to use a vector for both of the files vector(name, address) and vector(name, salary) and later return a vector of a string and tuple (address, (name, salary)) as my output. I don’t know how to tokenize the string for example “Evan Lloyd|Brownsville” into the substrings “Evan Lloyd” and “Brownsville” separately (without the |) and save them separately so that I can push both strings into my vector(name, address). How would I tokenize the strings that way or should I try something else entirely?

## Finding maximal cardinal independent set given oracle

The problem is given an oracle $$O(G, k)$$ that would say if graph G contains IS of size k devise an algorithm for finding independent set of max cardinality that makes poly number of calls to the oracle. My attempt has been that first finding the maximal possible size and then try to find the set of that size by removing vertices one at a time. I understand for a given node it either has to be in or not in the set, then I noticed that that there are multiple overlaps between chain of removals and I could devise a DP algorithm of sorts. But I’m just really stuck after that and was wondering if any hint could be given.

## Finding vertex coverage that is also independent set

Given a graph G and integer k, find a vertex coverage set of size k that is also an independent set. I need to either prove this problem is np-complete or find a polynomial solution.

Any idea ?

Thanks!

## Choosing an independent hash function, given hash function value

Supposed we have a function $$h:U\to [m_1]$$. Given this hash function, can we generate without using randomization or a universal hash collection another hash $$h’:U \to [m_2]$$, which depends on $$h$$ and the values of $$h'(x)$$ are uniform? i.e., for every value $$y\in [m_2]$$ we have $$\Pr_{x} (h'(x)=y|h)=1/m_2$$ ($$x$$ is chosen at random).

I know that if $$U=[2^t]$$, $$m_1=2^{n_1}$$, $$m_2=2^{n_2}$$, $$t>n_1+n_2$$ and $$h(x)$$ is the first $$n_1$$ bits of $$x$$ this problem is easy: we can select $$h'(x)$$ to be the bits $$n_1+1$$ till $$n_2$$ of $$x$$. My question is how to generalized this approche.

## Can a sorcerer make independent spell-effect choices for a Twinned spell?

Can a sorcerer make independent spell-effect choices for a Twinned spell?

For instance, the Polymorph spell can be used on willing and unwilling targets; the unwilling can make a Wisdom save to avoid the effects.

The sorcerer’s Twinned Spell metamagic option creates a second instance of the same spell on 2 viable targets within range.

Does Twinned Spell force me to polymorph both targets into the same thing, or do I get to choose what each instance of polymorph does independently?

In this case I am trying to turn my friend into a T-rex, and an enemy into a goldfish.

## Reduction from Independent set to Restricted independent set and vice versa

Independent Set

Input: undirected graph G = (V, E) and an integer k ≥ 0.

Output: does G have an independent set S ⊆ V of size k (Yes or No).

Restricted Independent Set

Input: undirected graph G = (V, E), a vertex u ∈ V , and aninteger k ≥ 0,

Output: does G have an independent set S ⊆ V of size k that includes u? (Yes or No).

Is it possible to reduce independent set to restricted independent set and is it possible to reduce restricted independent set to independent set?

## KL divergence of two independent variables

I’m missing something obviously this is the way to start this:

$$\sum_{x\in X}^{}p(x)log(p(x)) – \sum_{x\in X}^{}p(x)log(q(x))$$

how to continue from here?

It looks intuitive that if the variables are independent then the sum of all the partial variables ($$p_{1},…,p_{i}$$) expression will be equal to the KL divergence. Is my intuition ok? How can it be shown in a mathematical way?