Timeline for Unit Vectors in Polar Coordinate System [closed]
Current License: CC BY-SA 4.0
12 events
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| Nov 22 at 17:54 | comment | added | Ted Shifrin | We're not trying to add (tangent) vectors at different points. That makes no sense. The point of the moving coordinate basis $\hat r,\hat\theta$ is that we see the physics/geometry in a more intrinsic fashion; $\hat i,\hat j$ is quite artificial, and the physics/geometry is independent of coordinates. This situation is quite similar to the Frenet frame, which is used to study the differential geometry of (space) curves in a powerful way. | |
| Nov 22 at 8:43 | comment | added | Prasad B | Hi @TedShifrin, I guess you are bringing out the difference between the 'position vector' of a location ($r$, $\theta_1$) and a vector that represents a physical quantity (such as electric field) acting at that location. If two forces F1, and F2 are acting at the same location, in the same direction, using the same two unit vectors (of the physical quantity) for all locations will not suffice, as vector addition of F1 and F2 will give a polar angle that will be twice that of F1 or F2. So, unit vectors of a physical quantity must be different at different locations. Is that correct ? | |
| Nov 20 at 19:31 | comment | added | Ted Shifrin | Now you need to be careful to distinguish between points in the plane and (tangent) vectors. When physicists work with $\hat r,\hat\theta$ to write vectors at the given point $(r,\theta)$, they are interested in the vectors (e.g., velocity and acceleration), and are not changing the base point. | |
| Nov 20 at 19:00 | comment | added | CyclotomicField | @PrasadB and the scalar multiplication also doesn't work because of $-1(r,\theta)$ which will give us a negative radius. So it's not just the angle term that's problematic. The main assumption about of a vector space is that they are defined over a field which is a strong algebraic constraint. It will eliminate many examples you'll find in math and physics. For example mass can't be a vector because it's non-negative. Even for things like position you have to choose an origin before you can turn the pairs $(x,y)$ it into a vector space so there is some work to be done. | |
| Nov 20 at 17:11 | comment | added | Prasad B | Hello @CyclotomicField, thank you for that nudge. Two vectors ($r_1$, $\theta_1$), ($r_2$, $\theta_1$), both along the same line when added has an angle that is twice the original -- which is clearly wrong. | |
| Nov 20 at 16:49 | history | closed | Another User José Carlos Santos Dermot Craddock Harish Chandra Rajpoot Ben Steffan | Needs details or clarity | |
| Nov 20 at 16:07 | review | Close votes | |||
| Nov 20 at 16:56 | |||||
| Nov 20 at 16:06 | comment | added | CyclotomicField | Vectors must satisfy scalar multiplication and vector addiction. Can you see why those properties fail for the pair $(r,\theta)$ of radius and angle? | |
| Nov 20 at 15:57 | review | Low quality posts | |||
| Nov 20 at 16:50 | |||||
| Nov 20 at 15:43 | comment | added | CommunityBot | Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. | |
| S Nov 20 at 15:40 | review | First questions | |||
| Nov 20 at 15:43 | |||||
| S Nov 20 at 15:40 | history | asked | Prasad B | CC BY-SA 4.0 |