Radius vector in cylindrical coordinates

[Oliver321]

I am starting to learn classical physics for my own. One exercise was, to calculate the vector r (see picture: 1.47 b). The vector r is r=z*z+p*p.
I don’t understand this solution. My problem is: in a vector space with n dimensions there are n basis vectors. In the case of cylindrical coordinates there have to be 3 basis vectors: z,p,phi. (How this basis vectors are defined, is listed in Wikipedia)Every vector in a vector space can be built from a minimum of 3 basis vectors. So we also need all three basis vectors to construct r. But why are we using online two of them? r could possibly point in any direction?

I appreciate every help! Thank you!

 

[kuruman]

The statement $\vec r =z~\hat z+ \rho ~\hat \rho$ in plain English says "Vector $r$ is the same as the sum of vector $z$ and vector $\rho$." This is undeniably true if you look at the drawing.

 

[BvU]

The vector $\ \vec r\$ is $\ \ \vec r =z*\hat z +\rho *\hat\rho$

Correct. But just like with polar coordinates in the 2-D plane you need an extra coordinate to designate the direction of $\hat \rho\$, namely $\phi$

 

[Oliver321]

The statement $\vec r =z~\hat z+ \rho ~\hat \rho$ in plain English says "Vector $r$ is the same as the sum of vector $z$ and vector $\rho$." This is undeniably true if you look at the drawing.

Thank you!

But how is this consistent with the mathematical concept of basis vectors. Of course it is true if I look at the graph, but it’s not obvious for me in mathematical sense.

 

[Oliver321]

Correct. But just like with polar coordinates ijn the 2-D plane you need an extra coordinate to designate the direction of $\hat \rho\$, namely $\phi$

Thanks!
But where is this vector $\phi$ ? In the equation of r it is not present. Or aren’t I am looking close enough?

 

[kuruman]

Unit vector $\phi$ is perpendicular to the plane shown in the figure and points in the direction of increasing angle. You can break down $\hat \rho$ in Cartesian coordinates and write it as $\hat \rho=\cos \phi~\hat x+\sin \phi~\hat y$ Then you will have $\vec r$ in the Cartesian basis.

 

[BvU]

Thanks!
But where is this vector $\phi$ ? In the equation of r it is not present. Or aren’t I am looking close enough?

So there is a unit vector $\hat\phi$, nicely perpendicular to the other two -- as it should be.

But it is not so that there is a term $\phi * \hat\phi$... but nobody promised there would be.

 

[Oliver321]

View attachment 242090
So there is a unit vector $\hat\phi$, nicely perpendicular to the other two -- as it should be.

But it is not so that there is a term $\phi * \hat\phi$... but nobody promised there would be.

I thought there has to be a term like $\phi * \hat\phi$ . Cause otherwise the vector is not fully described in there dimensions.
If I am getting this right: r could only be in a plane with constant phi but not ‚everywhere‘ ?
I am sorry if my questions are obvious! But it confuses me a lot.

 

[kuruman]

You have to learn to live with the fact that while $\hat z$ is a constant unit vector, $\hat \rho$ is not; its direction depends on $\phi$. Note that while $\frac{d}{dt}\hat z=0$ because this unit vector is fixed in space, $\frac{d}{dt}\hat \rho \neq 0$. So if you want to find what it is equal to, you need first to write $\hat \rho$ in the Cartesian representation (see post #6) and then take the time derivative. The answer is quite simple, but you need to find what $\hat \phi$ is in the Cartesian representation before you can simplify.

 

[BvU]

I am sorry if my questions are obvious! But it confuses me a lot.

No need to apologize. We are so well acquainted with Cartesian coordinate systems that a different way feels weird at first (it will get worse with spherical coordinates ).

r could only be in a plane with constant phi but not ‚everywhere‘

Once $\phi$ is established, that designates a unique plane through the z-axis and $\vec r$. In this plane you still need two coordinates $\rho$ and $z$ (as 2-D Cartesian coordinates we usually call $x$ and $y$).

For the vector $\vec r$ to be uniquely determined you clearly need three coordinates $\rho$, $\phi$ and $z$.

 

[Oliver321]

No need to apologize. We are so well acquainted with Cartesian coordinate systems that a different way feels weird at first (it will get worse with spherical coordinates ).Once $\phi$ is established, that designates a unique plane through the z-axis and $\vec r$. In this plane you still need two coordinates $\rho$ and $z$ (as 2-D Cartesian coordinates we usually call $x$ and $y$).

For the vector $\vec r$ to be uniquely determined you clearly need three coordinates $\rho$, $\phi$ and $z$.

Thank you very much!:)
So I am right that r=p*p+z*z is not complete? Why is it legitim to use this equation in this case?

You have to learn to live with the fact that while $\hat z$ is a constant unit vector, $\hat \rho$ is not; its direction depends on $\phi$. Note that while $\frac{d}{dt}\hat z=0$ because this unit vector is fixed in space, $\frac{d}{dt}\hat \rho \neq 0$. So if you want to find what it is equal to, you need first to write $\hat \rho$ in the Cartesian representation (see post #6) and then take the time derivative. The answer is quite simple, but you need to find what $\hat \phi$ is in the Cartesian representation before you can simplify.

It is clear for me that the vector is not constant. I did this exercise also! Thanks! Nevertheless it is not clear to me why a not constant base should mean that I only would need two basis vectors instead of three. :)

 

[kuruman]

In physics it is quite common to have problems that have "cylindrical symmetry". In such problems all quantities, including vectors do not depend on $\phi$. You can see then that the vectors are really two-dimensional in that only $\hat \rho$ and $\hat z$ are sufficient to describe them.

 

[Oliver321]

In physics it is quite common to have problems that have "cylindrical symmetry". In such problems all quantities, including vectors do not depend on $\phi$. You can see then that the vectors are really two-dimensional in that only $\hat \rho$ and $\hat z$ are sufficient to describe them.

So in principle I need a third vector, but in most cases there is no phi dependence so we treat the problem two dimensional?

 

[kuruman]

So in principle I need a third vector, but in most cases there is no phi dependence so we treat the problem two dimensional?

I wouldn't say that "in most cases there is no phi dependence"; I would say "in cases where there is no phi dependence." The choice of coordinate system is yours to make. A good choice of coordinate system is one that reflects the symmetry of the problem. By this is meant that it might be possible to reduce the number of coordinates needed to describe the problem.

 

[Oliver321]

I wouldn't say that "in most cases there is no phi dependence"; I would say "in cases where there is no phi dependence." The choice of coordinate system is yours to make. A good choice of coordinate system is one that reflects the symmetry of the problem. By this is meant that it might be possible to reduce the number of coordinates needed to describe the problem.

Ok. Slowly I am understanding. I have one last question I am not shure about.
In the picture you can see the actual exercise. Is it clear from this context, that it is possible to reduce the number of basis vectors? Would the solution to c be any different if i include the therm with phi?

 

[BvU]

I did this exercise also!

Can you show us how you did this part (c) ?

 

[Oliver321]

Can you show us how you did this part (c) ?

Of course I will. But in my country it is 11pm. So I will show it to you tomorrow!

 

[BvU]

Same time zone. Vacationing till Tuesday (with a cruddy old Ipad )

 

[kuruman]

Ok. Slowly I am understanding. I have one last question I am not shure about.
In the picture you can see the actual exercise. Is it clear from this context, that it is possible to reduce the number of basis vectors? Would the solution to c be any different if i include the therm with phi?

It looks like you don't understand what part (c) is all about. In Cartesian coordinates tha acceleration vector can be written as
$\vec a=\ddot x~\hat x+\ddot y~\hat y+\ddot z~\hat z$
In cylindrical coordinates the same acceleration vector is written as
$\vec a=(\ddot \rho-\rho \dot \phi^2)~\hat \rho+(\rho \ddot \phi+2\dot \rho \dot \phi)~\hat \phi+\ddot z~\hat z$
Which description you use depends on the problem you are treating.
Example 1. A car accelerates in a straight line in along the $x$-axis with constant acceleration $a$.
In the Cartesian representation, you have to solve the differential equation
$\vec a=a~\hat x$
In the cylindrical representation "straight line" means $\phi=const.$
$\vec a=\ddot \rho~\hat \rho$ with the additional provision that $\phi=0$ in which case $\ddot \rho=\ddot x=a$ and $\hat \rho=\hat x$. All this (in a more roundabout way) gives the same equation $\vec a=a~\hat x$.

Example 2. A car goes around a circle of radius $R$ at constant speed $v$.
Finding the components of the acceleration in the Cartesian representation without using the cylindrical (polar) representation is a mess. In the polar representation, $\rho = R$, $\ddot \phi = const.$ which gives
$\vec a=-R \dot \phi^2~\hat \rho$
This is just the centripetal acceleration.

I hope these examples illustrate to you what's going on.

 

[Chestermiller]

In cylindrical coordinates, the component of the position vector in the circumferential direction is zero. So there is the possibility of a component in the circumferential direction, but, in this particular coordinate system, for the position vector, the magnitude of that component is zero.

 

[PeroK]

So in principle I need a third vector, but in most cases there is no phi dependence so we treat the problem two dimensional?

The basis vectors in polar, spherical and cylindrical coordinates are different from the basis vectors you have in linear algebra. This is trickier than many introductory texts would imply.

In particular, at the origin there is a coordinate singularity. The vectors $\hat{r}$ and $\hat{\phi}$ are not defined at the origin. This means you cannot specifiy a position vector in the way you are expecting.

In polar coordinates you have coordinates $(r, \phi)$, which specifies a position. You find this position by taking distance $r$ along the positive x-axis and then rotating that line through an angle $\phi$. You do not find the position by using polar unit vectors, as these are not defined at the origin.

Once you have a point that is not the origin, then you can define unit vectors $\hat{r}, \hat{\phi}$ at that point. This gives a local orthonormal basis at that point. If you take something like the velocity of a particle at that point, then that can be described as $\vec{v} = v_r \hat{r} + v_{\phi} \hat{\phi}$.

You can also see that the position vector from the origin to the point can now be described as $\vec{r} = r \hat{r}$. So, because the vectors $\hat{r}, \hat{\phi}$ are changing with position the position vector of the point (in the local polar coordinate basis) is always described in terms of the radial vector alone.

Note that in Cartesian coordinates $\hat{x}, \hat{y}$ are the same everywhere, so you can move vectors around from point to point. For example, if two particles A and B have the same velocity (same magnitude and direction), then you can see this directly in Cartesian coordinates, as $\vec{v_A} = v_{Ax} \hat{x} + v_{Ay} \hat{y} = \vec{v_B} = v_{Bx} \hat{x} + v_{By} \hat{y}$.

But, that would not be the case in polar coordinates. Two particles at different positions with the same velocity will have different components in polar coordinates. Note also that you cannot specify the velocity of a particle at the origin in polar coordinates, as you have the aforementioned coordinate singularity there.
Finally, note that in physics generally vectors are more complicated and versatile than the vectors you get in a vector space from your linear algebra studies.

 

[Oliver321]

Thanks to all of you for your help!

Can you show us how you did this part (c) ?

Here is my solution. Could be possible that I wrote down something wrong because I erased a lot and may have forgotten to erase something.

It looks like you don't understand what part (c) is all about. In Cartesian coordinates tha acceleration vector can be written as
$\vec a=\ddot x~\hat x+\ddot y~\hat y+\ddot z~\hat z$
In cylindrical coordinates the same acceleration vector is written as
$\vec a=(\ddot \rho-\rho \dot \phi^2)~\hat \rho+(\rho \ddot \phi+2\dot \rho \dot \phi)~\hat \phi+\ddot z~\hat z$
Which description you use depends on the problem you are treating.
Example 1. A car accelerates in a straight line in along the $x$-axis with constant acceleration $a$.
In the Cartesian representation, you have to solve the differential equation
$\vec a=a~\hat x$
In the cylindrical representation "straight line" means $\phi=const.$
$\vec a=\ddot \rho~\hat \rho$ with the additional provision that $\phi=0$ in which case $\ddot \rho=\ddot x=a$ and $\hat \rho=\hat x$. All this (in a more roundabout way) gives the same equation $\vec a=a~\hat x$.

Example 2. A car goes around a circle of radius $R$ at constant speed $v$.
Finding the components of the acceleration in the Cartesian representation without using the cylindrical (polar) representation is a mess. In the polar representation, $\rho = R$, $\ddot \phi = const.$ which gives
$\vec a=-R \dot \phi^2~\hat \rho$
This is just the centripetal acceleration.

I hope these examples illustrate to you what's going on.

Thank you. This examples help me a lot!

In cylindrical coordinates, the component of the position vector in the circumferential direction is zero. So there is the possibility of a component in the circumferential direction, but, in this particular coordinate system, for the position vector, the magnitude of that component is zero.

Thanks for the awnser! Ok now it’s clear to me that there must not be a phi component. But nevertheless I don’t know why there should not be one because r could point in any direction so the phi component has not to be zero for every vector?

The basis vectors in polar, spherical and cylindrical coordinates are different from the basis vectors you have in linear algebra. This is trickier than many introductory texts would imply.

In particular, at the origin there is a coordinate singularity. The vectors $\hat{r}$ and $\hat{\phi}$ are not defined at the origin. This means you cannot specifiy a position vector in the way you are expecting.

In polar coordinates you have coordinates $(r, \phi)$, which specifies a position. You find this position by taking distance $r$ along the positive x-axis and then rotating that line through an angle $\phi$. You do not find the position by using polar unit vectors, as these are not defined at the origin.

Once you have a point that is not the origin, then you can define unit vectors $\hat{r}, \hat{\phi}$ at that point. This gives a local orthonormal basis at that point. If you take something like the velocity of a particle at that point, then that can be described as $\vec{v} = v_r \hat{r} + v_{\phi} \hat{\phi}$.

You can also see that the position vector from the origin to the point can now be described as $\vec{r} = r \hat{r}$. So, because the vectors $\hat{r}, \hat{\phi}$ are changing with position the position vector of the point (in the local polar coordinate basis) is always described in terms of the radial vector alone.

Note that in Cartesian coordinates $\hat{x}, \hat{y}$ are the same everywhere, so you can move vectors around from point to point. For example, if two particles A and B have the same velocity (same magnitude and direction), then you can see this directly in Cartesian coordinates, as $\vec{v_A} = v_{Ax} \hat{x} + v_{Ay} \hat{y} = \vec{v_B} = v_{Bx} \hat{x} + v_{By} \hat{y}$.

But, that would not be the case in polar coordinates. Two particles at different positions with the same velocity will have different components in polar coordinates. Note also that you cannot specify the velocity of a particle at the origin in polar coordinates, as you have the aforementioned coordinate singularity there.
Finally, note that in physics generally vectors are more complicated and versatile than the vectors you get in a vector space from your linear algebra studies.

That’s helping me a lot!
So Is it right to argue as follows (But maybe I am completely wrong):
In every 3D vector space there have to be three basis vectors to describe every possible vector. In the case of cylindrical coordinates the phi and roh basis vectors are both phi dependent. If we choose a fixed phi for both basis vectors, I need all three to describe every possible vector r. But in our case we don’t have a static phi. We choose phi always in that way, that the component of the phi-basis vector is zero (so the phi-basis vector does not appear). If this is true it would also be possible to construct every vector only with the roh-basis vector. (except in the origin).

 

[PeroK]

That’s helping me a lot!
So Is it right to argue as follows (But maybe I am completely wrong):
In every 3D vector space there have to be three basis vectors to describe every possible vector. In the case of cylindrical coordinates the phi and roh basis vectors are both phi dependent. If we choose a fixed phi for both basis vectors, I need all three to describe every possible vector r. But in our case we don’t have a static phi. We choose phi always in that way, that the phi-basis vector is zero. If this is true it would also be possible to construct every vector only with the roh-basis vector. (except in the origin).

Polar basis vectors are defined locally; whereas, Cartesian basis vectors are defined globally. In particular, owing to the singularity at the origin, polar basis vectors cannot be used in the way basis vectors are used in linear algebra: where every vector is essentially a position vector from the origin. You must stop trying to do this.

The polar coordinates specify a point. But, to do that, they require the x-y-z axes, as I described above.

Instead, you can think of a different vector space at every point. At the point $(1, 1)$, for example, the local polar basis defines a vector space in which all local vectors are defined (velocity, momentum, acceleration, the electic field at that point etc.)

In any case, you must move away from the idea that polar basis vectors are there to define a position vector from the origin to a set of points in a plane or in 3D. They are not designed for this.

This is also a good time to move from the "abstract linear algebra" view of vectors to the "physical" view of vectors that you require for physics.

In fact, to jump ahead slightly but hopefully give you some motivation: in GR (General Relativity) the position vector disappears altogether and vectors become exclusively a local quantity!

 

[PeroK]

@Oliver321 Let me give you an example of what I'm talking about. Let's say we have a particle of charge $q$ at some point $(x, y)$. The particle is moving with velocity $\vec{v} = (v_x, v_y)$. And, let's assume that there is a magnetic field at this point $\vec{B} = (B_x, B_y)$.

The Lorentz force law that says that the force on the particle is $\vec{F} = q(\vec{v} \times \vec{B})$.

Now, we have three vectors: force, velocity and magnetic field.

As these are specified in Cartesian coordinates, we could move all the vectors to the origin and imagine they are defined as vectors in a vector space, centred at the origin. But, that's not a very physical approach. The particle is at some point $(x, y)$, the velocity represents a physical quantity at that point and the force certainly represents something physical acting at that point.

So, it's more physical to imagine a local Cartesian system centred on the point. And, as the particle moves, the local Cartesian system moves with it. The vectors we are interested in are defined at the point where the particle happens to be and are not vectors emanating from some globally defined origin.

Now, if we are using polar coordinates, in fact we cannot move all our vectors to the origin. First, the definition of $\hat{r}$ and $\hat{\phi}$ changes if we move position; and, they are not even defined at the origin. We must, therefore, stay in the local system, defined at the point where the particle is.

At that point, the local polar basis vectors form a nice, orthonormal basis and we can do all our vector algebra in the usual way.

One point about the position vector (from the origin to our point). We can think about moving this vector from the origin to our point. Then, we can treat that vector as a local vector, defined at our point as $r \hat{r}$. So, instead of thinking about the position vector as a vector starting at the origin, we can look at it as a vector that has been moved to our local system.

In summary, if you use Cartesian coordinates, you can think about moving all vectors to the origin; although, it may be better to start thinking in terms of having a local Cartesian basis at every point. But, if you use polar, cylindrical or spherical coordinates, you must start thinking in terms of a local set of basis vectors at every point; and, if the position vector is needed, then you must think of it being moved to your local system, where it will necessarily be $r \hat{r}$. Or, $\rho \hat{\rho} + z \hat{z}$ in cylindrical coordinates.

 

[Oliver321]

I appreciate your help very much! Thank you for spending so much time.
I have read your posts 3 times to be shure I am understanding everything. Now I have one (probably dumb) question.
You said:

You can also see that the position vector from the origin to the point can now be described as →r=r^rr→=rr^\vec{r} = r \hat{r}. So, because the vectors ^r,^ϕr^,ϕ^\hat{r}, \hat{\phi} are changing with position the position vector of the point (in the local polar coordinate basis) is always described in terms of the radial vector alone.

I don’t fully get why in a local system I need only r \hat{r}. My guess: the local system is rotated about an angle ϕ, so we do not need this component? So in particular it is not quite right to think of the position vector r to point from the origin to a particular point ( like it’s printed in the sketch of my first post). And it is not the case in polar coordinates, that the acceleration vector is parallel to the position vector (like in cartesian)?

 

[PeroK]

I don’t fully get why in a local system I need only r \hat{r}. My guess: the local system is rotated about an angle ϕ, so we do not need this component?

Even more directly, the local system is defined so that one of the basis vectors is in the direction of the position vector to/from the origin.

If you do go back to your linear algebra and take any vector $w$. Then, define a basis that consists of $w$ and some other linearly independent vectors. How do you describe $w$ in this basis? Well, $w = 1.w$ in this basis. You don't need the other basis vectors to describe $w$.

So in particular it is not quite right to think of the position vector r to point from the origin to a particular point ( like it’s printed in the sketch of my first post). And it is not the case in polar coordinates, that the acceleration vector is parallel to the position vector (like in cartesian)?

A vector is a vector, independent of any basis or coordinate system. It's how you describe a vector that we are talking about. If you have the origin and another point then the position vector is the vector from one to another. Imagine all the axes removed. It's just a physical vector in space, joining two points. One point marked O and the other P, say.

Now, if you construct a Cartesian system at the origin, with the x-y-z axes wherever you like, you can describe that vector in that coordinate system using the basis vectors at the origin.

But, if you construct a polar coordinate system, there are no basis vectors at the origin. You cannot describe that vector in terms of polar basis vectors defined at the origin. BUT, you can describe that vector in terms of the polar basis vectors defined at the point P. And that's what you are doing when you say that $\vec{OP} = r \hat{r}$.

$\hat{r}$ is a local basis vector defined at P. It's not defined at O.

All I'm saying is that to eliminate any ambiguity you could think of moving the vector to start at the point P, because that is the local system of basis vectors in which you have chosen to specify $\vec{OP}$.

 

[Oliver321]

Even more directly, the local system is defined so that one of the basis vectors is in the direction of the position vector to/from the origin.

If you do go back to your linear algebra and take any vector $w$. Then, define a basis that consists of $w$ and some other linearly independent vectors. How do you describe $w$ in this basis? Well, $w = 1.w$ in this basis. You don't need the other basis vectors to describe $w$.
A vector is a vector, independent of any basis or coordinate system. It's how you describe a vector that we are talking about. If you have the origin and another point then the position vector is the vector from one to another. Imagine all the axes removed. It's just a physical vector in space, joining two points. One point marked O and the other P, say.

Now, if you construct a Cartesian system at the origin, with the x-y-z axes wherever you like, you can describe that vector in that coordinate system using the basis vectors at the origin.

But, if you construct a polar coordinate system, there are no basis vectors at the origin. You cannot describe that vector in terms of polar basis vectors defined at the origin. BUT, you can describe that vector in terms of the polar basis vectors defined at the point P. And that's what you are doing when you say that $\vec{OP} = r \hat{r}$.

$\hat{r}$ is a local basis vector defined at P. It's not defined at O.

All I'm saying is that to eliminate any ambiguity you could think of moving the vector to start at the point P, because that is the local system of basis vectors in which you have chosen to specify $\vec{OP}$.

Tank you! I think I finally got it!

 

[Chestermiller]

Thanks for the awnser! Ok now it’s clear to me that there must not be a phi component. But nevertheless I don’t know why there should not be one because r could point in any direction so the phi component has not to be zero for every vector?

The unit vector in the radial $\rho$ direction is a function of $\phi$. So that is where the phi dependence of r comes i.

 

[vanhees71]

Of course to uniquely define the vector you need three coordinates. In zylinder coordinates these are $\rho$, $\phi$, and $z$. Now by definition
$$\vec{r}=\rho \hat{\rho} + z \hat{z}.$$
Of course you need alos $\phi$, because $\hat{\rho}$ depends on $\phi$. The position vector is always the superposition of only two of the basis vectors, but the basis vectors in curvilinear bases usually depend on the generalized coordinates defining them.

For spherical coordinates $(r,\vartheta,\varphi)$ it looks even more simple:
$$\vec{r}=r \hat{r},$$
but again, of course you need all three generalized coordinates $(r,\vartheta,\varphi)$ to specify this position vector. Indeed $\hat{r}$ dependds on $\vartheta$ and $\varphi$, i.e., you must know these two angles to know $\hat{r}$ and thus (together with the coordinate $r$, being explicitly in the above formula) you need all three spherical coordinates to specify the position vector.

 

[IPTV013]

Oliver321, could you share the title of the book you are using ? Thank you.

 

[IPTV013]

The title of the book is : Classical Mechanics by John R. Taylor.