# Simple dot product in polar coordinates

Gold Member
Let's take two orthogonal curves in polar coordinates of the form ##\langle r,\theta \rangle##, say ##\langle r,0\rangle## and ##\langle r,\pi/2\rangle##. Cleary both lines are orthogonal, but the dot product is not zero. This must be since I do not have these vectors in the form ##\langle x,y\rangle##.

Does anyone know of a formula for taking the dot product in other non-rectangular coordinate systems, or should I always convert to rectangular?

PeroK
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Let's take two orthogonal curves in polar coordinates of the form ##\langle r,\theta \rangle##, say ##\langle r,0\rangle## and ##\langle r,\pi/2\rangle##. Cleary both lines are orthogonal, but the dot product is not zero. This must be since I do not have these vectors in the form ##\langle x,y\rangle##, but can someone explain conceptually where my mistake is?

How are you defining the dot product?

Gold Member
How are you defining the dot product?
Sorry, I changed the question stem slightly. I'm defining it as ##(a \hat x + b\hat y)\cdot (c \hat x + d\hat y) = ac + bd##.

PeroK
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2021 Award
Let's take two orthogonal curves in polar coordinates of the form ##\langle r,\theta \rangle##, say ##\langle r,0\rangle## and ##\langle r,\pi/2\rangle##. Cleary both lines are orthogonal, but the dot product is not zero. This must be since I do not have these vectors in the form ##\langle x,y\rangle##.

Does anyone know of a formula for taking the dot product in other non-rectangular coordinate systems, or should I always convert to rectangular?

I'd convert to rectangular. It's fairly easy to generate the formula for polar coordinates by doing this in general.

Gold Member
Thanks!

Gold Member
Okay, so maybe I'm making a dumb mistake; could you check my work?

Lets say I have a vector of the form ##r \hat r + \theta \hat \theta##. We know ##\hat r = \cos\theta \hat x + \sin\theta\hat y## and ##\hat\theta = -\sin\theta\hat x + \cos\theta\hat y##. Substituting these expressions into ##r \hat r + \theta \hat \theta## yields ##(r\cos\theta-\theta\sin\theta)\hat x + (r\sin\theta+\theta\cos\theta)\hat y##. But then if I take the two lines ##r \hat r ## and ##r \hat r + \frac{\pi}{2} \hat \theta## and dot them using the above formula, I do not get zero. What am I doing wrong?

PeroK
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Okay, so maybe I'm making a dumb mistake; could you check my work?

Lets say I have a vector of the form ##r \hat r + \theta \hat \theta##. We know ##\hat r = \cos\theta \hat x + \sin\theta\hat y## and ##\hat\theta = -\sin\theta\hat x + \cos\theta\hat y##. Substituting these expressions into ##r \hat r + \theta \hat \theta## yields ##(r\cos\theta-\theta\sin\theta)\hat x + (r\sin\theta+\theta\cos\theta)\hat y##. But then if I take the two lines ##r \hat r ## and ##r \hat r + \frac{\pi}{2} \hat \theta## and dot them using the above formula, I do not get zero. What am I doing wrong?

A vector in polar coordinates is given by ##r \hat{r}##. There's no ##\hat{\theta}## involved.

##\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y##

Now, can you relate ##a_x## etc. to your polar coordinates?

• joshmccraney
Gold Member
A vector in polar coordinates is given by ##r \hat{r}##. There's no ##\hat{\theta}## involved.
Wow, obviously!!!!! Thanks so much, I'm good to go! I can finish what you've written if you'd like but it makes perfect sense now.

Chestermiller
Mentor
$$\hat{r}(0)=\hat{i}_x$$
$$\hat{r}(\pi/2)=\hat{i}_y$$

• joshmccraney
PeroK
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Wow, obviously!!!!! Thanks so much, I'm good to go! I can finish what you've written if you'd like but it makes perfect sense now.

By the way, hopefully, once you get the answer it might look familiar.

Gold Member
By the way, hopefully, once you get the answer it might look familiar.
Yep, it all makes perfect sense! Can't believe I was parameterizing a vector in polar incorrectly!!! jedishrfu
Mentor
The polar caps have frozen many an unprepared explorer!

• joshmccraney
Gold Member
Can you tell me if this now looks correct: given some function ##f(r,\theta) = r^2\theta^3##, I'd like to compute ##\nabla f \cdot \hat n## where ##\hat n## is a normal vector to a line passing the origin with angle, say ##\beta##. We know $$\nabla f = 2r \theta^3 \hat r + 3r\theta^2 \hat \theta = 2r \theta^3 (\cos\theta \hat x + \sin\theta\hat y) + 3r\theta^2 (-\sin\theta\hat x + \cos\theta\hat y).$$ We also know any normal line to ours takes the form ##\langle x,-\cot \beta x + b \rangle##, which implies its unit tangent vector is ##\langle 1,-\cot\beta \rangle \implies \hat n = \langle \sin\beta,-\cos\beta \rangle##.

At this point can't I just directly take $$\langle \sin\beta,-\cos\beta \rangle \cdot \left( 2r \theta^3 (\cos\theta \hat x + \sin\theta\hat y) + 3r\theta^2 (-\sin\theta\hat x + \cos\theta\hat y) \right)$$ and then set ##\theta = \beta##?

Chestermiller
Mentor
Can you tell me if this now looks correct: given some function ##f(r,\theta) = r^2\theta^3##, I'd like to compute ##\nabla f \cdot \hat n## where ##\hat n## is a normal vector to a line passing the origin with angle, say ##\beta##. We know $$\nabla f = 2r \theta^3 \hat r + 3r\theta^2 \hat \theta = 2r \theta^3 (\cos\theta \hat x + \sin\theta\hat y) + 3r\theta^2 (-\sin\theta\hat x + \cos\theta\hat y).$$ We also know any normal line to ours takes the form ##\langle x,-\cot \beta x + b \rangle##, which implies its unit tangent vector is ##\langle 1,-\cot\beta \rangle \implies \hat n = \langle \sin\beta,-\cos\beta \rangle##.

At this point can't I just directly take $$\langle \sin\beta,-\cos\beta \rangle \cdot \left( 2r \theta^3 (\cos\theta \hat x + \sin\theta\hat y) + 3r\theta^2 (-\sin\theta\hat x + \cos\theta\hat y) \right)$$ and then set ##\theta = \beta##?
You left off the unit vectors on n.

Gold Member
You left off the unit vectors on n.
I think the unit normal vector is correct but good call, the "unit" tangent vector I list is actually a tangent vector.

Chestermiller
Mentor
I think the unit normal vector is correct but good call, the "unit" tangent vector I list is actually a tangent vector.
I don't know what you are saying.

Gold Member
I don't know what you are saying.
I mentioned a unit tangent vector in post 13 but it's actually just a tangent vector (not with norm 1).

Chestermiller
Mentor
I mentioned a unit tangent vector in post 13 but it's actually just a tangent vector (not with norm 1).
You’re not trying to het the normal derivative of f, are you?

Gold Member
You’re not trying to het the normal derivative of f, are you?
No, ##\hat n## is a unit normal to a line with slope ##\tan \beta##.

mathwonk
Homework Helper
have you tried just setting x = rcos(t) and y = rsin(t), and then using the usual formula for dot product? i.e. then the dot product of (r1,t1) and (r2,t2) is

just (r1r2)(cos(t1)cos(t2 + sin(t1)sin(t2)) = r1r2.cos(t2 - t1) = the product of the lengths of the vectors by the cosine of the angle t2-t1 between them, which is always the dot product, in any coordinate system.

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