I have the follow equations : $$\mathbf{v}_l = \frac{1}{(1+A^2)}[\mathbf{B}-\mathbf{A}\times\mathbf{B} + (\mathbf{A} \cdot \mathbf{B} ) \mathbf{A} ] $$
where :
$$\mathbf{A}=\frac{D_0}{(\rho_n+\rho_s)\kappa}s' $$ $$ \mathbf{B}=\frac{\rho_s\kappa}{(\rho_n+\rho_s)\kappa}(\mathbf{V}_s+\mathbf{v}_s)+ \frac{\rho_n\kappa}{(\rho_n+\rho_s)\kappa}\mathbf{v}_n+\frac{D_0}{(\rho_n+\rho_s)\kappa}s'\times\mathbf{v}_n+\frac{\chi}{(\rho_n+\rho_s)\kappa}s' $$
This comes from a fluid dynamics paper, that says that $$\mathbf{v}_l = f(\mathbf{v}_s, \mathbf{V}_s,s',\mathbf{v}_n,\kappa) $$ and using this relation : $$\mathbf{v}_l \cdot s' =0 $$ is possible (and is what I'm looking for) find the expressions for which $$\chi =0$$
All, the variable written in bold (\mathbf) are 3d vector .
The problem is that I have no license for using mathematica or other software for symbolic calculation in my university, could somebody help me to find the passage for getting this ? Thanks in advance. I have to calculate $\mathbf{v}_l \cdot s'$ from the equation in this equation appear the constant $\chi$ (the variable are also $ \mathbf{v}_s, \mathbf{V}_s,s',\mathbf{v}_n,\kappa$ the others parameter are know) so i have to calculate the expression that give me $\chi =0$ since $\mathbf{v}_l\cdot s' =0 $ it is clear ? if not let me know
EDIT : Thank you very much ! just to be sure can I ask you another little (but very Important for me) help ?
lets : $$\mathbf{v}_l = \mathbf{\dot{X}}_s \quad ; \quad s'=\mathbf{X'}_s $$
How can arrive to this equations ?
$$ \mathbf{\dot{X}}_s = \mathbf{V}_s + h_1 \mathbf{X'}_s \times (\mathbf{V}_n - \mathbf{V}_s) + h_2 \mathbf{X'}_s\times \left[\mathbf{X'}_s \times (\mathbf{V}_n -\mathbf{V}_s) \right] $$
where
$$ h_1 = \frac{\rho_s \kappa D_0}{D^2_0+(\rho_n\kappa + \rho_s\kappa)^2} $$ $$ h_2 = \frac{-\rho_n\kappa(\rho_n\kappa + \rho_s\kappa)-D^2_0 }{D^2_0+(\rho_n\kappa + \rho_s\kappa)^2}$$
In the above equations only $ \mathbf{\dot{X}}_s = \mathbf{v}_l $ and $\mathbf{X'}_s = s'$ change the rest of the terms are the same ! If you can help me in this last step will be wonderful. But I'm sure that for this mathematica is necessary ! Thanks again !
EDIT Your right this equations came from one other paper. In the original paper (where the equation that you solving fo $\chi$) the paper says :
$$\mathbf{v}_l = \frac{1}{(1+A^2)}[\mathbf{B}-\mathbf{A}\times\mathbf{B} + (\mathbf{A} \cdot \mathbf{B} ) \mathbf{A} ] $$ where $A^2 =\mathbf A \cdot \mathbf A $ it is useful to decompose $\mathbf v_n $ and $\mathbf V_s + \mathbf v_s$ into components parallel to $s'$ and components traverse to it ; for example we write $$ \mathbf{v}_n = v_{n\parallel} s' + \mathbf{v}_{n\perp} $$ where $v_{n\parallel} = \mathbf v_n\cdot s'$ and $\mathbf v_{n\perp} = -s'\times s'\times\mathbf{v}_n $. The arbitrary constant $\chi$ is then chosen so that the friction does not create any component odf $\mathbf v_l$ along $s'$ itself. We find :
$$(eq1) \qquad \mathbf{v}_l = h_1(\mathbf V_s +\mathbf v_s) + h_2 s' \times (\mathbf v_n -\mathbf{V}_s - \mathbf{v}_s)+h_3 \mathbf{v}_{n\perp}$$
where :
$$ h_1 = \frac{\rho_s\kappa(\rho_n\kappa+\rho_s\kappa)}{(\rho_n\kappa+\rho_s\kappa)^2+D_0^2} $$ $$h_2 = \frac{\rho_s\kappa D_0}{(\rho_n\kappa+\rho_s\kappa)^2+D_0^2} $$ $$ h_3 = \frac{D^2_0 + (\rho_n\kappa +\rho_s\kappa) \rho_n \kappa}{(\rho_n\kappa+\rho_s\kappa)^2+D_0^2} $$
The equation (eq1) above should be the same as the equation of $\mathbf{\dot{X}_s}$
so could you try to figure out how to arrive to the equation (eq1) and looks if there is some similitude with the equation of $\mathbf{\dot{X}_s}$ Thanks in advance Marco.
EDIT basically till this passage the things are the same of the first answer ? $$\begin{align}\mathbf v_l \cdot \mathbf A &= \frac 1{1+A^2}[\mathbf B-\mathbf A\times\mathbf B + (\mathbf{A \cdot B})\mathbf A]\cdot \mathbf A\\&=\frac 1{1+A^2}[\mathbf B \cdot \mathbf A + (\mathbf{A \cdot B})(\mathbf{A \cdot A})]\\&=\frac 1{1+A^2}(\mathbf{A \cdot B})(1 + A^2)\\&= \mathbf{A \cdot B}\end{align}$$ right ?
edit here : $$\mathbf v_n = (\mathbf v_n \cdot s')s' + (-s' \times(s'\times\mathbf v_{n}))$$ did you confuse $\mathbf{v}_n$ with $\mathbf{v}_l$ ?
You don't need a symbolic manipulation package to handle this. The first thing to do is relabel all the complicated constants with simple labels. This will make the problem much easier to follow.
Let $$p = \frac{\rho_s\kappa}{(\rho_n+\rho_s)\kappa}$$ and $$q = \frac{D_0}{(\rho_n+\rho_s)\kappa}$$ and noting that $$\mathbf s' = \frac 1q \mathbf A$$ We have $$\mathbf{B}= p(\mathbf{V}_s+\mathbf{v}_s) + (1-p)\mathbf{v}_n+\mathbf A\times\mathbf{v}_n+\frac{\chi}{D_0}\mathbf A$$ Also $\mathbf{v}_l \cdot s' =0$ is equivalent to $\mathbf{v}_l \cdot \mathbf A = 0$. But $$\begin{align}\mathbf v_l \cdot \mathbf A &= \frac 1{1+A^2}[\mathbf B-\mathbf A\times\mathbf B + (\mathbf{A \cdot B})\mathbf A]\cdot \mathbf A\\&=\frac 1{1+A^2}[\mathbf B \cdot \mathbf A + (\mathbf{A \cdot B})(\mathbf{A \cdot A})]\\&=\frac 1{1+A^2}(\mathbf{A \cdot B})(1 + A^2)\\&= \mathbf{A \cdot B}\end{align}$$ (Making use of the fact that $\mathbf u \cdot (\mathbf u \times \mathbf v) = 0$.) So if $\chi = 0$, then $$0 = \mathbf{A \cdot B} = \mathbf A\cdot [p(\mathbf{V}_s+\mathbf{v}_s) + (1-p)\mathbf{v}_n]$$ The computation is reversible (assuming the scalars are not $0$), so $$ \chi = 0 \iff \mathbf A\cdot [p(\mathbf{V}_s+\mathbf{v}_s) + (1-p)\mathbf{v}_n] = 0$$ or, $$ \chi = 0 \iff \mathbf s'\cdot [p(\mathbf{V}_s+\mathbf{v}_s) + (1-p)\mathbf{v}_n] = 0$$ which has the advantage of being true even when $D_0 = 0$, unlike the version with $\mathbf A$.
Edit addressing the additional questions:
The first thing I note in the additional information you posted: Nowhere do they say that $\chi = 0$. Instead, they are saying that they are setting $\chi$ to whatever value is necessary to make $\mathbf v_l \cdot s' = 0$. So basically, my calculation above is not applicable to their paper - other than saying that when the condition I gave happens to be true, then the value they are setting $\chi$ to will turn out to be $0$.
Next, the decomposition $$\mathbf v_n = (\mathbf v_n \cdot s')s' + (-s' \times(s'\times\mathbf v_{n}))$$ only works when $\|s'\| = 1$ (otherwise, it would give $\|s'\| ^2\mathbf v_n$). So apparently, this has been established elsewhere in the paper.
So what we actually can calculate is that $$0 = \mathbf s'\cdot [p(\mathbf{V}_s+\mathbf{v}_s) + (1-p)\mathbf{v}_n] + \frac{\chi q}{D_0}$$ and so $$\chi = -(\rho_n+\rho_s)\kappa \left( \mathbf s'\cdot [p(\mathbf{V}_s+\mathbf{v}_s) + (1-p)\mathbf{v}_n]\right)$$
I think to get the formulas for $h_1, h_2, h_3$ that they show, I'd have to know more about the definitions of the various variables and how they relate to each other.