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Basis/Unit vectors in other coordinate systems 
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#1
Dec812, 05:48 PM

P: 784

We all know the ##\vec{i}##,##\vec{j}##,##\vec{k}## unit vectors for Cartesian space. But I've never been shown basis unit vectors in other coordinate systems.
Do basis vectors exist in other coordinate systems? And if so what are they? 


#2
Dec912, 03:32 PM

Sci Advisor
P: 6,031

Any sort of rotation which keeps things orthogonal. Example i' = (i+j)/√2, j' = (ij)/√2, k' = k.



#3
Dec912, 05:35 PM

P: 784

But those rotations make use of the ##\vec{i}##,##\vec{j}##,##\vec{k}## vectors. And I could define ##\vec{i}## as ##(1,0,0)## and the other two similarly.
Can I do something similar for other coordinates? For example writing a unit/basis vector in cylindrical coordinates like ##\vec{a}= (r,\theta, z)## where the three inputs don't depend on the Cartesian basis vectors? 


#4
Dec1012, 03:15 PM

Sci Advisor
P: 6,031

Basis/Unit vectors in other coordinate systems
The main requirement for a set of basis vectors is that they each have a dimension of length. The other requirement is that they be mutually orthogonal.
You can't do it in cylindrical coordinates  angle is nondimensional. It looks like you are confusing basis with coordinate system. 


#5
Dec1012, 05:53 PM

P: 784

Is there an issue with defining three vectors: ##\vec{r}=(1,0,0)##,##\vec{\theta}=(0,1,0)##,##\vec{z} =(0,0,1)## in vector form ##(r,\theta,z)##? Then it seems to me that the magnitude would be ##sqrt(r^2+z^2)## and you could do all the normal things you do with I J and K. I'll admit I believe my vocabulary was off, perhaps it is not basis vectors I'm looking for, just something akin to the I J K vectors in other coordinate systems that do not explicitly depend on I J and K. Edit: I'm having weird latex problems, sorry about that. 


#6
Dec1112, 09:09 AM

P: 350

Hi Vorde,
The construction you describe doesn't work because those don't obey the usual rules for vector addition/scalar multiplication. However, you can use basis vectors that are compatible with cylindrical coordinates. The usual way is to define [itex] \hat{r}, \hat{\theta}, \hat{z}[/itex] to be unit vectors pointing in the radial, angular, and axial directions respectively. You don't express a point as a combination of these vectors. You express vectors as combinations of these vectors. Note that the direction of these vectors depends on their basepoint. In rectangular coordinates, it makes sense and is more or less conventional to consider two vectors the same as long as they point in the same direction and have the same length. But in other coordinate systems, the basepoint of the vector cannot be ignored. The radial vector, for example, changes direction when you change its basepoint. So if you have a vector V and you want to resolve it into its radial, angular and axial components, then those respective components depend on the basepoint of V. For example, suppose [itex] V=\vec{i}[/itex]. Then if the basepoint of V is at (x=1,y=0,z=0), then its radial component is 1 and its angular component is 0, i.e. [itex] V=\hat{r}[/itex]. But if V is based at (x=0,y=1,z=0), then its radial component is now 0 and its angular component is 1, [itex] V= \hat{\theta}[/itex]. The important thing to notice is that these are not free vectors in the sense that you can translate them around the plane. They are bound vectors which are "bound" to their basepoint. The reason is that the direction of [itex]\hat{r}[/itex], for example, depends on the basepoint. If you move from one point to another then the radial direction changes, so [itex] \hat{r}[/itex] changes along with it. 


#7
Dec1112, 12:41 PM

P: 784

That makes total sense to me, and thank you for your response. It does not surprise me that the vectors I listed do not work, but it is true that any attempt to 'do things like I was thing' will fail? And if so, why?
Requiring that vectors all have units of length, which I did not consider, complicates things with a vector of angles, but because (nonrigorously) I can imaging having 'circular' vectors like that, I wonder if it is not possible to define an anglevector. Though I guess then I'd be changing the definition of vector, so my result doesn't really matter. 


#8
Dec1112, 01:11 PM

P: 350

Strictly speaking (from the perspective of algebra), vectors don't have to have "units". But in applications they do. And if the space you are working in is basically geometric space, then they should have units of length (because fundamentally a vector is like a displacement), so mathman's point is like a dimensional analysis check of whether you are doing things wrong.
It might help you if you think of vectors geometrically as linear displacements (like a directed line segment). Then the parallelogram law for vector addition makes sense. And a "circular vector" does not make sense. However, at a given point, a vector that points in the angular direction does make sense. In Euclidean space with Cartesian coordinates, we can get away with identifying a point (x,y,z) with the radial vector xi+yj+zk. In other words, there is no necessity to distinguish between points and vectors. The problem is that this trick does not really make sense in nonlinear coordinate systems. If you want to think of a point (r,theta,z) as a radial vector, that's fine, but the radial vector cannot be expressed in the form r A + theta B+ z C, where A,B,C are some fixed vectors. 


#9
Dec1112, 01:16 PM

P: 784

I guess I'm not satisfied because I can imagine defining a system where what I'm thinking could happen, but I completely accept that it's not possible with vectors.
Thank you very much. 


#10
Dec1112, 01:42 PM

P: 350

I know what you mean. Sometimes (in various contexts) I feel like there is something there under the surface that I can't see, and I can't really ask anyone about it because I can't define exactly what I am looking for with enough precision to get a satisfactory answer. :)
You might find your answer is a book on Lie groups/algebras, or maybe in differential geometry. Good luck. 


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