# I How do we decide dimension of motion?

1. May 28, 2017

### Pushoam

How do we decide dimension of motion?
Consider a particle moving along $\hat x$ direction.
This motion is known as one dimensional motion as only one coordinate i.e. x is changing with respect to time.
Consider a particle having circular motion in X-Y plane.
In Cartesian Coordinate system, both x and y are changing with respect to time and hence the motion is two dimensional.
But in Polar coordinate system(r,θ) ,only one coordinate i.e. θ is changing with respect to time and hence the motion is one dimensional.
So, does the dimension of motion depend on the coordinate system?
Or,
Do we always use Cartesian Coordinate system to determine the dimension of motion ?

2. May 28, 2017

### Staff: Mentor

If the motion is in a straight line, it is one dimensional. If it is not in a straight line, but is occurring in a plane, it is two dimensional. If it is not in a straight line, and is not occurring in a plane, it is three dimensional.

3. May 28, 2017

### Pushoam

According to this ,
no.of dimension of motion = no.of Cartesian coordinates (needed to describe the motion) changing with respect to time

Can you please give me some reference for it?

4. May 28, 2017

### Staff: Mentor

In some contexts, 1D means that the problem can be described using 1 spatial independent variable, 2D means that the problem can be described using 2 spatial independent variables, and 3D means that the problem must be described using 3 spatial independent variables.

5. May 28, 2017

### Pushoam

I want to know these contexts.
Because for circular motion,2 spatial independent variables are needed in Cartesialn Cordinate system, while only one spatial independent variable is needed in Polar Coordinates.

6. May 28, 2017

### DennisN

Hmm, you may want to rethink that...
Who says that r is fixed?
r is of course not necessarily fixed to a specific value, and it is independent of θ, thus the motion can also occur in two dimensions (assuming polar coordinates).

EDIT: Also, when r is fixed and θ is changing (using polar coordinates), both x and y are changing in the rectangular (Cartesian) coordinate system.

See e.g. Coordinate Systems (HyperPhysics)

7. May 28, 2017

### Pushoam

In circular motion , r remains constant.

8. May 28, 2017

### DennisN

Ok... then, who says the motion has to be circular? Note: I promise I am not trying to annoy you .

9. May 28, 2017

### Pushoam

I have taken circular motion as an example.
This reference doesn't relate coordinate system and dimension.

10. May 28, 2017

### phinds

@Pushoam perhaps it would be helpful to you to think of things from the point of view of a traveler. A traveler on a circle can go forward and backward but although he is not able of his own volition to vary the DEGREE to which he moves left/right, he does none-the-less have to also move left/right AS he goes forward/backward, thus 2D motion. In a straight line, this is not the case, so 1D motion. In a helix, he would have to move left/right, forward/backward, and up/down, thus 3D motion.

11. May 28, 2017

### Pushoam

From this ,too, the following can be concluded:
Isn't the above statement right?
Can you please give me an example where the above statement won't work?

12. May 28, 2017

### hilbert2

A vector $\mathbf{x}$ or a set of vectors $\mathbf{x}_i$ in 2D can be obtained with a projection operation from higher-dimensional vectors of arbitrary dimensionality, so it isn't really useful to debate whether circular motion happens in 1D or 2D.

13. May 28, 2017

### phinds

I don't know about others here but I have no idea what you just said. Is it possible to express it in English?

14. May 28, 2017

### hilbert2

Well, a vector $(x,y)$ can be obtained from $(x,y,0)$, $(x,y,0,0)$, $(x,y,0,0,0)$ and so on by just ignoring the dimensions after the second.

15. May 28, 2017

### phinds

I still have no idea how this relates to the subject at hand.

16. May 28, 2017

### hilbert2

I mean that it doesn't really have any practical significance whether we call motion of a particle, constrained on a 2-dimensional plane of a 3-dimensional space, two- or three-dimensional.

17. May 28, 2017

### phinds

OK, thanks. That at least I do understand but find it hard to agree with. By that logic, motion along a straight line (in 3D space) could be considered 1D, 2D, or 3D (or perhaps you mean it should always be considered 3D if it's in a 3D space. I've always understood that motion on a straight line is considered 1D motion.

18. May 28, 2017

### pixel

I think in the case of a particle constrained to move in a circle, we would say there is one degree of freedom, but the motion is 2-dimensional.

19. May 28, 2017

### Pushoam

By this ,you meant :
There is no need to worry about the dimension of the motion.
When working in polar coordinates system, there will be only one equation of motion as only one coordinate is changing w.r.t. time.
When working in Cartesian coordinates system, there will be two equations of motion as two coordinates are changing w.r.t. time.
Solve the equation of motion, get the answer and forget all about the dimension of motion.

This approach simply removes the problem
, and at the same time it makes the concept of dimension useless, doesn't it?

20. May 28, 2017

### Staff: Mentor

It might be better to say that we don't have to consider more than one dimension if we can describe the motion with one number, don't have to consider more than two dimensions if we can describe the motion with two numbers and so forth.

However, that's "don't have to", not "must not". Depending on the problem, I may find it easier to think of uniform circular motion as one number on a (curved) one-dimensional path or two numbers in the Euclidean plane with the constraint that #x^2+y^2## is constant.

21. May 28, 2017

### Pushoam

Then, can it be concluded that dimension of motion will depend on the no. of variables changing w.r.t.time in a given co-ordinate system?

22. May 28, 2017

### Staff: Mentor

I am in the "it doesn't matter" camp.

If you want you can use Newtonian mechanics and analyze it using 2 variables. If you want you can use Lagrangian mechanics and analyze it using 1 variable.

In the end you get the same answer.

23. May 28, 2017

### phinds

Fair enough

24. May 28, 2017

### Pushoam

O.K.
Thanks to all.