# Circular motion in one or two dimensions

I and my teacher argued whether a uniform circular motion in polar coordinates is considered to be a motion in one dimension or it's a motion in two dimensions.

Dale
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It generally isn’t a good idea to argue with your teacher. (Especially on topics that make no difference)

Last edited:
russ_watters, fresh_42 and nasu
fresh_42
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I and my teacher argued whether a uniform circular motion in polar coordinates is considered to be a motion in one dimension or it's a motion in two dimensions.
Take two, this way you always have an excuse: You can write ##E+E''=0## as ##E=E(\varphi, d\varphi)## or ##E=E(x,y)##.

Take two, this way you always have an excuse: You can write ##E+E''=0## as ##E=E(\varphi, d\varphi)## or ##E=E(x,y)##.
So it can be considered as a motion in one dimension

It generally isn’t a good idea to argue with your teacher. (Especially on topics that make no difference)
Both of us really love debates

fresh_42
Mentor
So it can be considered as a motion in one dimension
You can choose time or angle ##\varphi##, given a fixed radius and uniform motion, which is one dimension, or you can choose position ##(x,y)## in which case you shouldn't write ##x=\cos \varphi\; , \;y=\sin \varphi##, which introduced a third variable, a parameterization, and made it rather difficult. As a differential equation, here of second degree, you can always argue, that the differentials belong to the equation, in which case you'll have even more variables: ##E=E(\varphi,d\varphi,d^2\varphi)## or ##E=E(x,y,dx,dy,d^2x,dxdy,d^2y)##.

So all in all, there is nothing to add to
It generally isn’t a good idea to argue with your teacher. (Especially on topics that make no difference)

You can choose time or angle ##\varphi##, given a fixed radius and uniform motion, which is one dimension, or you can choose position ##(x,y)## in which case you shouldn't write ##x=\cos \varphi\; , \;y=\sin \varphi##, which introduced a third variable, a parameterization, and made it rather difficult. As a differential equation, here of second degree, you can always argue, that the differentials belong to the equation, in which case you'll have even more variables: ##E=E(\varphi,d\varphi,d^2\varphi)## or ##E=E(x,y,dx,dy,d^2x,dxdy,d^2y)##.

So all in all, there is nothing to add to
So it's all about perspectives, it can be a one dimensional motion with one variable, and it can also be with more than 3 variable in the case of differential equations.

Dale
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2020 Award
Both of us really love debates
But in the end only one of you will be graded by the other. It is a bad idea. Furthermore, by arguing on a pointless topic you are robbing yourself from learning something that matters.

Classification of this type is completely pointless. Whether you call it 1D or 2D doesn’t change the physics. Go back to learning physics, debate is for debate club not physics class.

But in the end only one of you will be graded by the other. It is a bad idea. Furthermore, by arguing on a pointless topic you are robbing yourself from learning something that matters.

Classification of this type is completely pointless. Whether you call it 1D or 2D doesn’t change the physics. Go back to learning physics, debate is for debate club not physics class.
It's knowledge

jbriggs444
Homework Helper
It's knowledge
I agree with @Dale. It is not knowledge. It is pointless classification. Like knowing whether a glass is half empty or half full. Just drink the thing.

jtbell, russ_watters and Dale
Dale
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2020 Award
It's knowledge
It really isn’t.

You can call a handheld light a “torch” or a “flashlight”. Either way it works the same.

nasu
Just a thought!

Cartesian coordinates, (x and y), are said to be 2-dimensional, to describe a 2-D space. Polar coordinates,
(r and theta) are also thought of as two dimensional and define a 2-D space. Of Cartesian and Polar
descriptions, Cartesian is superior. Cartesian can do one thing Polar cannot. The 2-D space spanned by
Polar Coordinates has no meaning for "r = 0." This is to say Cartesian 2-D Space does not map into
Polar 2-D Space.

Khashishi

So it can be consider as a two dimensional motion in polar coordinates

jbriggs444
Homework Helper
So it can be consider as a two dimensional motion in polar coordinates
It can be considered motion in a two dimensional space (the plane in which the circle is embedded) or in a one-dimensional sub-space (the circle).

Coordinates are irrelevant -- they just determine how you parameterize the space. The plane is still a two dimensional space whether you use cartesian coordinates, polar coordinates or something else. The circular sub-space has only one dimension no matter how you parameterize it.

Though with only one dimension, there are not a lot of choices for how to parameterize the points on a circle. About the only choices you have are origin and scaling.

can we say it's a one dimensional motion because there's only a change in theta " angular direction"

Khashishi
Would you prefer if I said both answers are wrong? Why are you still asking this question? The answer makes no difference.

Dale
jbriggs444
Homework Helper
can we say it's a one dimensional motion because there's only a change in theta " angular direction"
What is the space that the motion is taking place in? Is it in a circle? Is it in a plane? Is it both?

Again, whether you use polar coordinates or not is irrelevant. The dimensionality of a [vector] space is invariant with respect to choice of coordinate system. It is the minimum number of elements needed in a basis for the vector space. The dimensionality of a circle, considered as a vector space of angular displacements is one. The dimensionality of a plane considered as a vector space of linear displacements is two.

fresh_42
Mentor
I think the subject is a bit overworked now. I found at least three posts with a clear answer and the rest isn't contradicting either.
It generally isn’t a good idea to argue with your teacher. (Especially on topics that make no difference)