Simple dot product in polar coordinates

It should be the gradient dotted with ##\hat{n}##, not just ##\sin\beta##.In summary, we discussed taking the dot product in non-rectangular coordinate systems and concluded that it is best to convert to rectangular coordinates for simplicity. We also corrected a mistake in parameterizing a vector in polar coordinates and used it to compute the dot product in a specific scenario. We found that the dot product should be taken between the gradient of a function and the unit normal vector to a line, not just a scalar value.
  • #1
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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
joshmccraney said:
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
PeroK said:
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
joshmccraney said:
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
Thanks!
 
  • #6
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
joshmccraney said:
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
PeroK said:
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.
 
  • #9
$$\hat{r}(0)=\hat{i}_x$$
$$\hat{r}(\pi/2)=\hat{i}_y$$
 
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  • #10
joshmccraney said:
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
PeroK said:
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
The polar caps have frozen many an unprepared explorer!
 
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  • #13
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
joshmccraney said:
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
Chestermiller said:
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
joshmccraney said:
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
Chestermiller said:
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
joshmccraney said:
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
Chestermiller said:
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
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
joshmccraney said:
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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Related to Simple dot product in polar coordinates

1. What is a simple dot product in polar coordinates?

The simple dot product in polar coordinates is a mathematical operation used to find the scalar product of two vectors in a polar coordinate system. It is also known as the inner product or the scalar product.

2. How is the simple dot product calculated in polar coordinates?

The simple dot product in polar coordinates is calculated by multiplying the magnitudes of the two vectors and the cosine of the angle between them.

3. What are the benefits of using the simple dot product in polar coordinates?

Using the simple dot product in polar coordinates can simplify complex vector calculations and make them easier to visualize. It also allows for the use of polar coordinates in applications such as physics and engineering.

4. How is the simple dot product related to the cross product?

The simple dot product and the cross product are two different types of vector products. The dot product is a scalar quantity, while the cross product is a vector quantity. They are related through the vector triple product formula.

5. What are some real-life applications of the simple dot product in polar coordinates?

The simple dot product in polar coordinates is commonly used in physics, engineering, and navigation. It can be used to calculate torque, work, and the angle between two forces. It is also used in radar systems and satellite tracking.

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