how does conversion of unit vectors work

[wheat plover]

i have cartesian co ordinates with unit vectors ax, ay, az but i need to transform them to cylindrical or spherical

• in the question i was solution they did this

[wheat plover]

YES

HELP HAS ARRIVED

HE TYPED

IM SAVED

[cinder nymph]

,tikz[scale=3]
{[thick, -Latex]
\draw (0,0) coordinate(O) -- node[midway, left]{$\vec a_z$} (0,1) coordinate(A);
\draw (O) -- node[midway, below right]{$(\vec a_R)\cos\theta$} ({cos(30)*cos(60)},{cos(30)*sin(60)}) coordinate (B);
\draw (B) -- node[midway, above right]{$(-\vec a_\theta)\sin\theta$} (A);
{[shift={(1,0.5)}]
\draw (0,0) -- node[midway, below right]{$\vec a_R$} ({cos(60)},{sin(60)});
\draw (0,0) -- node[midway, below left]{$\vec a_\theta$} ({-cos(150)},{-sin(150)});
}
}
\draw pic[draw, angle eccentricity = 2, "$\theta$"]{angle=B--O--A} pic[draw]{right angle=O--B--A};

[tawny berry]

cloud

[wheat plover]

so the trigonometry is the unit vectors???

that makes alot alot of sense

man this piture

makes so much sense

thank you so so much 😭

so so so very much

😭

[wheat plover]

@cinder nymph what about the third co oridnate?

(R,angle1,angle2) why is angle 2 zero?

[cinder nymph]

the unit vector $\vec a_\phi$ points in the direction of increasing $\phi$. this makes it entirely horizontal

[tawny berry]

cloud

[wheat plover]

is it not perpendicular to ar and a(angle)?

sorry idk latex

[cinder nymph]

it is, yes

but it also happens to be perpendicular to $\vec a_z$

[tawny berry]

cloud

[wheat plover]

OH

OHHH

its parallel to ax?

[cinder nymph]

no, but it's a combination of ax and ay

[wheat plover]

wait let me try to note take this

1s

[cinder nymph]

the unit vectors point in the direction you would go if you increased that coordinate and held the other coordinates constant

[wheat plover]

so in the ar and a(angle) drawing its poiting towards me?

[cinder nymph]

it depends on exactly how you set up your coordinate system. if it looks something like this then it would point directly into the page

[wheat plover]

is x in terms of all 3?

like is ex = er + e(angle1) + e(angle2)

or is it ex = Ar.er + Aangle1 . e(angle1) + Aangle2 .e(angle2)

[wheat plover]

no this is definetly Ax.ex

[wheat plover]

so i suppose its this

well then

i think i may have got it

TYSM CLOUD

🫶

[cinder nymph]

yes, it is in terms of all 3 angles

wikipedia has a list of all the conversions in matrix form: https://en.wikipedia.org/wiki/Vector_fields_in_cylindrical_and_spherical_coordinates

[wheat plover]

THIS IS SO HELPFUL

whats ur major if u dont mind me asking or is this all self taught?!?!?

cuz ur so smart 😭 like wow 🫶

[pine crescent]

This is all a lot easier to understand if you take linear algebra first. This is just a linear transformation made by composing two rotation matrices that represent phi and theta.

[wheat plover]

they didnt give me linear algebra aaaaa

they just gave me this physics course after matrices and calc 4

wait did u say 2 matrices

[wheat plover]

do you mean this?

i dont get where the cos and sin came from i cannot lie

[cinder nymph]

that is a rotation matrix

[wheat plover]

[wheat plover]

professor said to memorize it

[wheat plover]

we did NOT take a spherical matrix either

professor praying for our downfall

[pine crescent]

I'm not going to pull the matrices apart but you can see the terms like sin(theta)cos(phi) come from a matrix product of two rotations

[wheat plover]

so i dont need to know how the matrix is formed? i can actually just memorize?

its for gauss law

[wheat plover]

imma save this and try applying it on the questions

tysm

both you and cloud

🫶

[cinder nymph]

specifically you rotate $\vec a_x$ and $\vec a_y$ by angle $\phi$ about the $\vec a_z$-axis to obtain $\vec a_r$ and $\vec a_\phi$ which obtains the cartesian-to-cylindrical coordinate transformations.

you also rotate $\vec a_z$ and $\vec a_r$ about the $\vec a_\theta$-axis to obtain $\vec a_R$ and $\vec a_\theta$, this defines the cylindrical-to-spherical transformation.

composing the two transformations (multiplying the matrices) gives the spherical transformation matrix

[tawny berry]

cloud

[wheat plover]

its 3 am and its too many words im sorry but i didnt really get this 😭

but well ill keep note of it

and try to read it another time

[cinder nymph]

basically the conversion from cartesian to cylindrical is a rotation that looks like this rotation

\begin{tikzpicture}[scale=3]
{[thick, -Latex]
\draw (0,0) coordinate (O) -- (1,0) node[right]{$\vec a_x$} coordinate (A);
\draw (0,0) -- (0,1) node[right]{$\vec a_y$} coordinate(C);
\draw (0,0) -- ({cos(30)},{sin(30)}) node[above right]{$\vec a_r$} coordinate(B);
\draw (0,0) -- ({cos(90+30)},{sin(90+30)}) node[above right]{$\vec a_r$} coordinate(D);
}
\filldraw[fill=white] (0,0) circle[radius=0.1] node[below left]{$\vec a_z$};
\fill (0,0) circle[radius=0.05];
\draw pic[draw, -Latex, "$\phi$", angle eccentricity = 2, angle radius=20]{angle=A--O--B};
\draw pic[draw, -Latex, "$\phi$", angle eccentricity = 2, angle radius=20]{angle=C--O--D};
\end{tikzpicture}

and has this rotation matrix
[
\mat{\vec a_r \ \vec a_\phi \ \vec a_z} = \mat{\cos\phi & \sin\phi & 0 \ -\sin\phi & \cos\phi & 0 \ 0 & 0 & 1} \mat{\vec a_x \ \vec a_y \ \vec a_z}
]

[tawny berry]

cloud

[cinder nymph]

and similarly the conversion from cylindrical to spherical looks like this
\begin{tikzpicture}[scale=3]
{[thick, -Latex]
\draw (0,0) coordinate (O) -- (1,0) node[right]{$\vec a_r$} coordinate (A);
\draw (0,0) -- (0,1) node[right]{$\vec a_z$} coordinate(C);
\draw (0,0) -- ({cos(-30)},{sin(-30)}) node[above right]{$\vec a_\theta$} coordinate(B);
\draw (0,0) -- ({cos(90-30)},{sin(90-30)}) node[above right]{$\vec a_R$} coordinate(D);
}
\filldraw[fill=white] (0,0) circle[radius=0.1] node[below left]{$\vec a_\phi$};
\fill (0,0) circle[radius=0.05];
\draw pic[draw, Latex-, "$\theta$", angle eccentricity = 2, angle radius=20]{angle=B--O--A};
\draw pic[draw, Latex-, "$\theta$", angle eccentricity = 2, angle radius=20]{angle=D--O--C};
\end{tikzpicture}

and has this rotation matrix

[ \mat{\vec a_R \ \vec a_\theta \ \vec a_\phi} = \mat{\sin\theta & 0 & \cos\theta \ \cos\theta & 0 & -\sin\theta \ 0 & 1 & 0} \mat{\vec a_r \ \vec a_\phi \ \vec a_z} ]

[tawny berry]

cloud

[pine crescent]

Honestly, you are crazy to write an answer to this involving TikZ

You deserve some sort of medal

[wheat plover]

i fully agree

im helping someone out rq ill read these in a bit

tysm again cloud 🫡

[cinder nymph]

from that you can see that
\begin{align*}
\mat{\vec a_R \ \vec a_\theta \ \vec a_\phi} &= \mat{\sin\theta & 0 & \cos\theta \ \cos\theta & 0 & -\sin\theta \ 0 & 1 & 0} \mat{\vec a_r \ \vec a_\phi \ \vec a_z} \
&=\mat{\sin\theta & 0 & \cos\theta \ \cos\theta & 0 & -\sin\theta \ 0 & 1 & 0} \mat{\cos\phi & \sin\phi & 0 \ -\sin\phi & \cos\phi & 0 \ 0 & 0 & 1} \mat{\vec a_x \ \vec a_y \ \vec a_z} \
&= \mat{\sin\theta\cos\phi& \sin\theta\sin\phi & \cos\theta \ \sin\theta\cos\phi& \cos\theta\sin\phi& -\sin\theta \ -\sin\phi & \cos\phi & 0} \mat{\vec a_x \ \vec a_y \ \vec a_z}
\end{align*}

[tawny berry]

cloud

[cinder nymph]

and the inverse matrices work similarly

[wheat plover]

Oh yes we just transpose the matrix

On the left side

[cinder nymph]

rotations have the nice property of being orthogonal, which does mean the transpose is equal to the inverse

[wheat plover]

Yessssss

Im gonna be honest i actually thought inverse and transpose are the same thing symbol wise

U just pointed it out

Wait

Matrix power -1 is inverse or transpose?

Oh ok ok so transpose has a t symbol

And inverse has a -1

[cinder nymph]

if $A$ is a square matrix then $A^{-1}$ is the inverse, defined such that $A A^{-1} = A^{-1} A = I$, and $A^T$ is the transpose, obtained by swapping the rows and columns

[tawny berry]

cloud

[wheat plover]

THANK YOU

[wheat plover]

How do you speak in the language of definitions

This is something id expect to see from wikipedia

Ur just too much smarter than what im used to seeing