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Tony Hau

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- Thread starter Tony Hau
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In summary, the velocity of a particle below is expressed in polar coordinates, with bases e r and e theta. I know that the length of a vector expressed in i,j,k is the square of its components. But here er and e theta are not i,j,k. Plus they are changing as well. Can someone help convince me that the length of v is still the square of its components?

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Tony Hau

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Gaussian97

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$$\hat{e}_r\cdot\hat{e}_r, \qquad \hat{e}_r\cdot\hat{e}_\theta, \qquad \hat{e}_\theta\cdot\hat{e}_\theta$$

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Tony Hau

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e r dot e r is (e r)^2Gaussian97 said:

$$\hat{e}_r\cdot\hat{e}_r, \qquad \hat{e}_r\cdot\hat{e}_\theta, \qquad \hat{e}_\theta\cdot\hat{e}_\theta$$

e r dot e theta is 0

e theta dot e theta is (e theta)^2

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Gaussian97

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And what are the values of ##\hat{e}_r^2## and ##\hat{e}_\theta^2##?

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Tony Hau

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Both are 1?Gaussian97 said:And what are the values of ##\hat{e}_r^2## and ##\hat{e}_\theta^2##?

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Gaussian97

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Exact! Now you can compute directly what is the product

$$v^2 = \vec{v}\cdot \vec{v}$$

$$v^2 = \vec{v}\cdot \vec{v}$$

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Tony Hau

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Gaussian97

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Yes, essentially is this, although it's good to make the computation once in your life :)

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Tony Hau

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May I confirm that if a vector is expressed in orthonormal bases, they will have the same operation as if they are expressed in the standard bases? It is because I got another HW question asking the cross product of two vectors expressed in spherical coordinates.PeroK said:

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For the cross product you also have to think about the right-handedness of your basis vectors. Not just the orthonormality.Tony Hau said:May I confirm that if a vector is expressed in orthonormal bases, they will have the same operation as if they are expressed in the standard bases? It is because I got another HW question asking the cross product of two vectors expressed in spherical coordinates.

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Tony Hau

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Sorry I only learned how to determine the right-hanedness of standard bases. How can you determine the right-handedness of other basis vectors?PeroK said:For the cross product you also have to think about the right-handedness of your basis vectors. Not just the orthonormality.

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Tony Hau said:Sorry I only learned how to determine the right-hanedness of standard bases. How can you determine the right-handedness of other basis vectors?

With your right hand! It's the same idea.

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Gaussian97

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It's the same idea as the scalar product. To compute the scalar product in one base you need to know the scalar product of the basis vectors ##\hat{e}_i\cdot\hat{e}_j##. Then to compute the vector product of a vector you need to know first the vector product ##\hat{e}_i\times\hat{e}_j##. To be right-handed simply means ##\hat{e}_i\times\hat{e}_j = \varepsilon_{ijk}\hat{e}_k##Tony Hau said:Sorry I only learned how to determine the right-hanedness of standard bases. How can you determine the right-handedness of other basis vectors?

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Tony Hau

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Thanks for your replyGaussian97 said:It's the same idea as the scalar product. To compute the scalar product in one base you need to know the scalar product of the basis vectors ##\hat{e}_i\cdot\hat{e}_j##. Then to compute the vector product of a vector you need to know first the vector product ##\hat{e}_i\times\hat{e}_j##. To be right-handed simply means ##\hat{e}_i\times\hat{e}_j = \varepsilon_{ijk}\hat{e}_k##

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It's a simple example for curvilinear orthogonal coordinates. You start from a cartesian basis ##\vec{e}_1## and ##\vec{e}_2##. Then you define polar coordinates ##(r,\theta)## with ##r \in \mathbb{R}_{>0}## and ##\theta \in ]-\pi,\pi]## (of course you can define the angle to be in any semi-open interval of length ##2 \pi##) as

$$\vec{r}=x \vec{e}_1+ y \vec{e}_2=r \cos \theta \vec{e}_1 + r \sin \theta \vec{e}_2.$$

Then it's also convenient to define, at each point, a basis system given by the tangent vectors of the coordinate lines, i.e., the lines given by holding one of the two polar coordinates constant:

$$\vec{b}_r=\partial_r \vec{r}=\cos \theta \vec{e}_1 + \sin \theta \vec{e}_2, \quad \vec{b}_{\theta}=\partial_{\theta} \vec{r}=-r \sin \theta \vec{e}_1 + r \cos \theta \vec{e}_2.$$

It's easy to see that

$$\vec{b}_r \cdot \vec{b}_{\theta}=0,$$

i.e., that the two tangent vectors on the coordinate lines at any point are orthogonal to each other. In this case, usually one chooses the basis for the curvilinear coordinates as normalized. Now ##\vec{b}_r^2=1## and ##\vec{b}_{\theta}^2=r^2##. Thus we define

$$\vec{e}_r=\vec{b}_r = \cos \theta \vec{e}_1 + \sin \theta \vec{e}_2, \quad \vec{e}_{\theta}=\frac{1}{r} \vec{b}_{\theta} = -\sin \theta \vec{e}_1 + \cos \theta \vec{e}_2.$$

Now you have at any point a orthonormal basis system since

$$\vec{e}_{r}^2=\vec{e}_{\theta}^2=1, \quad \vec{e}_r \cdot \vec{e}_{\theta}=0.$$

Now as with any basis you can define vector components with respect to this "curvilinear bases". In your case this was done for the velocity vector. To that end just use the original Cartesian coordinates first and then express the resulting vectors in terms of the new basis vectors:

$$\vec{v}=\dot{\vec{r}}=\mathrm{d}_t (r \cos \theta \vec{e}_1 + r \sin \theta \vec{e}_2) = \dot{r} \vec{e}_r + r \dot{\theta} \vec{e}_{\theta}.$$

Thus you have the components

$$v_r=\dot{r}, \quad v_{\theta}=r \dot{\theta}.$$

Now, since these are components with respect to an orthronormal basis, you get

$$\vec{v}^2=(v_r \vec{e}_r + v_{\theta} \vec{e}_{\theta})^2=v_r^2 \vec{e}_r^2 + v_{\theta}^2 \vec{e}_{\theta}^2 + 2 v_r v_{\theta} \vec{e}_r \cdot \vec{e}_{\theta}=v_r^2 + v_{\theta}^2 = \dot{r}^2 + r^2 \dot{\theta}^2.$$

Polar coordinates are a way of representing points in a two-dimensional plane using a distance (r) from the origin and an angle (θ) from a reference axis.

To convert a vector expressed in polar coordinates (r, θ) to Cartesian coordinates (x, y), you can use the following formulas: x = r * cos(θ) and y = r * sin(θ). These formulas use trigonometric functions to calculate the x and y coordinates based on the given distance and angle.

The length of a vector expressed in polar coordinates is simply the value of the distance (r). This represents the magnitude or size of the vector, regardless of its direction.

No, the Pythagorean theorem only applies to right triangles in Cartesian coordinates. In polar coordinates, the distance (r) represents the hypotenuse of a right triangle, but the angle (θ) is not necessarily 90 degrees, so the Pythagorean theorem cannot be used.

The angle (θ) in polar coordinates represents the direction of the vector. It is measured from a reference axis, usually the positive x-axis, in a counterclockwise direction. To find the angle, you can use inverse trigonometric functions (such as arctan or arcsin) to solve for θ.

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