Simple dot product in polar coordinates

  • #1
joshmccraney
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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##.

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

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  • #2
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?
 
  • #3
joshmccraney
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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##.
 
  • #4
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##.

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.
 
  • #5
joshmccraney
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Thanks!
 
  • #6
joshmccraney
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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?
 
  • #7
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.

You don't need the formula for ##\hat{r}##. Instead, you should start with:

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

Now, can you relate ##a_x## etc. to your polar coordinates?
 
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  • #8
joshmccraney
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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.
 
  • #10
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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.
 
  • #11
joshmccraney
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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!!! :doh:
 
  • #12
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The polar caps have frozen many an unprepared explorer!
 
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  • #13
joshmccraney
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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##?
 
  • #14
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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.
 
  • #15
joshmccraney
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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.
 
  • #16
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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.
 
  • #17
joshmccraney
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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).
 
  • #18
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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?
 
  • #19
joshmccraney
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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##.
 
  • #20
mathwonk
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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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  • #21
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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##.

Does anyone know of a formula for taking the dot product in other non-rectangular coordinate systems, or should I always convert to rectangular?
Regardless of how two vectors are represented, their dot product is defined as the product of their magnitudes times the cosine of the angle between them. The dot product of your example using their polar coordinate form is ## r^2 \cos \frac \pi 2 = 0 ##. The result agrees with the fact that the vectors are orthogonal.
 
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