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Coordinate axes are chosen to be perpendicular to each other because it is convenient, calculations are easier. This is what I have read.

But in case that coordinate axes are per example 60 degrees apart could again a physical system exhibit linearity?

The superposition theorem could work?

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    $\begingroup$ Have a look at : en.wikipedia.org/wiki/Skew_coordinates . What do you mean "a physical system exhibit linearity"? Can you be more precise? Physical laws do not depend upon the choice of coordinate systems. $\endgroup$ – SRS Mar 19 at 16:41
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    $\begingroup$ As in many crystals. The base vectors are very often not perpendicular to each other. $\endgroup$ – Pieter Mar 19 at 17:10
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    $\begingroup$ The physical behaviour of a system is independent of the coordinate system you use to model it. That is just about the "zeroth law" of the whole of science! The choice of a coordinate system has no effect on the linear or nonlinear behaviour of what is being modelled, unless the model is wrong. $\endgroup$ – alephzero Mar 19 at 18:05
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    $\begingroup$ yes, as far as they are linearly independent, for instance, you dont want the third axis to be in the same plane than the other two $\endgroup$ – Wolphram jonny Mar 19 at 23:54
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So here's the basic reason why: we like to express vectors in terms of their components, for which we invent the unit vectors $\hat e^{1,2,3}$ and components $v_{1,2,3}$ for a vector $\vec v$ such that $$ \vec v= v_1\hat e^1 + v_2\hat e^2 + v_3 \hat e^3. $$ There is a bilinear combination between two vectors to form a scalar called the dot product, $$\vec u\cdot\vec v =\lvert \vec u\rvert\lvert \vec v \rvert\cos\theta.$$ For orthogonal unit vectors we have the nice property that $$\hat e^m\cdot \hat e^n=\{1\text{ if } m=n\text{ else } 0\},$$ which simplifies this expression to just $$\vec u\cdot\vec v=u_1v_1+u_2v_2+u_3v_3.$$ This simple formula is why we use orthogonal unit vectors in practice.

There is a slightly more complicated formula in skewed coördinate systems, or ones without unit vectors for their basis. To do this you simply have to find a new set of vectors that makes this relationship true, called the dual basis. So if your basis vectors are $\vec b^{1,2,3}$ then your dual basis is $\vec b_{1,2,3}$ such that $$\vec b_{m}\cdot \vec b^n=\{1\text{ if } m=n\text{ else } 0\},$$ in other words for $\vec b^1$, say, you look at the plane spanned by $\vec b^{2,3}$, identify a vector normal to it $\vec n$, and then rescale it until its dot product with $\vec b^1$ is $1$, and that then is $\vec b_1$. So the dual basis for some periodic lattice of points tells you about the planes of that lattice, via their normal vectors.

Then every vector gets two sets of components, the regular and the dual components, $$\begin{align} \vec v&= v_1\vec b^1 + v_2\vec b^2 + v_3 \vec b^3 &=v^1\vec b_1 + v^2\vec b_2 + v^3 \vec b_3, \end{align}$$ at which point the earlier easy property is restored as long as we always pair a lower index with an upper index,$$ \vec u=u_1v^1+u_2v^2+u_3v^3=u^1v_1+u^2v_2+u^3v_3.$$

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enter image description here

The vector $\vec{R}$ can described with the vectors basis $\vec{q}_1\,,\vec{q}_2$ : $$\vec{R}_s= a_1\begin{bmatrix} 1\\ 0\\ \end{bmatrix}+a_2\begin{bmatrix} \cos(\theta)\\ \sin(\theta)\\ \end{bmatrix}=a_1\,\vec{q}_1+a_2\,\vec{q}_2 \tag 1$$

or with the vectors basis $\vec{e}_1\,,\vec{e}_2$

$$\vec{R}_c= c_1\begin{bmatrix} 1\\ 0\\ \end{bmatrix}+c_2\begin{bmatrix} 0\\ 1\\ \end{bmatrix}=c_1\,\vec{e}_1+c_2\,\vec{e}_2 \tag 2$$

both case the superposition theorem is valid

Edit

what's about the EOMs?

I) generalized coordinates are $x$ and $y$ with $c_1=x$ and $c_2=y$ we obtain for the position Vector $\vec{R}_c$ equation (2)

$$\vec{R}_c= x\begin{bmatrix} 1\\ 0\\ \end{bmatrix}+y\begin{bmatrix} 0\\ 1\\ \end{bmatrix} $$

Kinetic energy

$T=m\,\frac{1}{2}\dot{R}_c^T\,\dot{R}_c$ and

Potential energy

$V=-m\,g\,y$

$\Rightarrow\quad$ EOMs

$$ \begin{bmatrix} \ddot{x}\\ \ddot{y}\\ \end{bmatrix}=\left[ \begin {array}{c} 0\\ -g\end {array} \right] \tag 3$$

II) generalized coordinates are $q_1$ and $q_2$

With $a_1=q_1$ and $a_2=q_2$ we obtain for the Position Vector $\vec{R}_s$ equation (1)

$$\vec{R}_s= q_1\begin{bmatrix} 1\\ 0\\ \end{bmatrix}+q_2\begin{bmatrix} \cos(\theta)\\ \sin(\theta)\\ \end{bmatrix}$$

Kinetic energy

$T=m\,\frac{1}{2}\dot{R}_s^T\,\dot{R}_s$ and

Potential energy

$V=-m\,g\,q_2\,\sin(\theta)$

$\Rightarrow\quad$ EOMs

$$ \begin{bmatrix} \ddot{q}_1\\ \ddot{q}_2\\ \end{bmatrix}=\left[ \begin {array}{c} {\frac {g\cos \left( \theta \right) }{\sin \left( \theta \right) }}\\ -{\frac {g}{\sin \left( \theta \right) }}\end {array} \right] \tag 4 $$

The generalized accelerations equation (3) and (4) are not equal. we can transfer equation (4) to get the accelerations in the orthogonal coordinates $\vec{e}_1\,,\vec{e}_2$

$$\begin{bmatrix} \ddot{x}\\ \ddot{y}\\ \end{bmatrix}=\left[ \begin {array}{cc} 1&\cos \left( \theta \right) \\ 0&\sin \left( \theta \right) \end {array} \right]\,\begin{bmatrix} \ddot{q}_1\\ \ddot{q}_2\\ \end{bmatrix}=\left[ \begin {array}{c} 0\\ -g\end {array} \right] \tag 5$$

so the accelerations equation (3) and (5) are now the same.

Conclusion

The equations of motion described with skew vectors basis compare to the equations of motion described with orthogonal vectors basis are not the same!.

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