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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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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

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

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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

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 said:How are you defining the dot product?

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joshmccraney said:

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.

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Thanks!

- #6

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?

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joshmccraney said:

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

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$$\hat{r}(0)=\hat{i}_x$$

$$\hat{r}(\pi/2)=\hat{i}_y$$

$$\hat{r}(\pi/2)=\hat{i}_y$$

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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.

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Yep, it all makes perfect sense! Can't believe I was parameterizing a vector in polar incorrectly!PeroK said:By the way, hopefully, once you get the answer it might look familiar.

- #12

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The polar caps have frozen many an unprepared explorer!

- #13

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##?

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You left off the unit vectors on n.joshmccraney said:

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##?

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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.Chestermiller said:You left off the unit vectors on n.

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

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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).Chestermiller said:I don't know what you are saying.

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

- #19

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

- #20

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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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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.joshmccraney said:

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

Last edited:

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.

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.

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.

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.

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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