# Non-uniform circular motion

So my suspicion that you were confusing complex numbers with real numbers was not entirely unfounded. A complex number is not "just a number". It is a pair of real numbers: $(a, b) = a + ib$. A 2D vector is also a pair of real numbers. This is why we can represent 2D vectors as complex numbers and vice versa.

I know that it is similar to a vector. A complex number is "just a number" except that isn't necessarily a real number. I see the similarity with a vector, but my question still holds. I argue that it just can't be the same as a vector no matter how similar. I think the "proof" I posted in the image above is something similar to what I'm looking for. Was my point clear in the image I posted? (I'm sorry if I'm being annoying).

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olivermsun
davidbenari,

Your diagram is perfectly clear. The radial and tangential parts really do come straight out of the chain rule in polar coordinates.

olivermsun: But I think this way of doing things is gonna be very tedious when I consider the second derivative. How can I generalise this or make it simpler?

olivermsun
Well, it isn't so bad considering the usefulness of the polar form. Also, once you've derived it, you have the formula in-hand and don't need to do it again.

If your interest is in elliptical planetary orbits, I also recommend looking through the wikipedia entry on Kepler's[/PLAIN] [Broken] laws.

If it's a more general motion you're looking at, then I'd probably start with vector paths in cartesian coordinates, [x(t), y(t), z(t)], which have very easy derivatives. Or perhaps the TNB frame I linked to earlier.

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I know that it is similar to a vector. A complex number is "just a number" except that isn't necessarily a real number. I see the similarity with a vector, but my question still holds. I argue that it just can't be the same as a vector no matter how similar..
Your argument is invalid. A complex number is a vector, not just similar to one.That is why we need a plane to visualize them. That is why we can use complex numbers to derive statements on 2D vectors, like it was done in your first message. If you really want to argue further, look up the definitions of a complex number and a vector, and show that a complex number does not satisfy the definition of a vector.

sophiecentaur
Gold Member
Your argument is invalid. A complex number is a vector, not just similar to one.That is why we need a plane to visualize them. That is why we can use complex numbers to derive statements on 2D vectors, like it was done in your first message. If you really want to argue further, look up the definitions of a complex number and a vector, and show that a complex number does not satisfy the definition of a vector.
Those two statements contradict each other.

I would rather put it that a complex number is two dimensional - which can be represented as a vector and this is common usage. But nothing 'spacial' is implied by complex numbers.

Those two statements contradict each other.
The second one is not a statement - not my statement at least. So no contradiction as far as I can tell.

I would rather put it that a complex number is two dimensional - which can be represented as a vector and this is common usage. But nothing 'spacial' is implied by complex numbers.
A complex number is a vector because all of the axioms of a vector space are satisfied by the set of complex numbers. It is also a vector in the sense that one can be used to represent little arrows on a flat sheet of paper and vice versa, which may be what you meant by "spatial".

sophiecentaur
Gold Member
I get it now. The two statements a actully disagree but you are saying that (and I missed the subtle construction and the earlier few words) that you (one) won't be able to justify the second.
I agree.

Well to begin with, arithmetical operations are defined differently for them (except to my knowledge, addition). I agree with sophiecentaur in that they are only representations and there's nothing spatial about them. https://www.physicsforums.com/showthread.php?t=458201 post#4 here says this:

"" Of course you can represent the real and imaginary parts of a complex number as a point on a plane (the Argand diagram) and you can do the same for the components of a 2-dimensional vector. Therefore complex numbers and 2-dimensional vectors will have some "geometrical" properties that are similar. But as you go further into using complex numbers in calculus (for example "analytic functions"), and study things like infinite-dimensional vector spaces where the elements of the vectors are not even numbers at all, you will find there are many more differences than similarities. ""

Apart from that I thought of another argument for my REAL question : If complex numbers add like vectors, then it makes sense that those two terms are their actual components in that direction.

olivermsun
Sure, the real and imaginary parts add separately, like orthogonal components of vectors in R2:
Y + Z = Re(Y) + Re(Z) + i(Im(Y) + Im(Z)),
and so on, which allows you to use complex numbers interchangeably with 2-vectors for adding, rotating, etc. As you've already pointed out, however, some of the operations are not the same.

sophiecentaur
Gold Member
We soon get philosophical here. The reason the Maths, in all its forms, 'works' in our physical world is really quite hard to take in. Even just at the level of two beans plus two beans gives four beans. . . .
We look at vectors (2 and 3D) in spatial terms but, once you get more than 3D, you are back to abstractions. No 'direction' at all, even if there is a magnitude.

Well yeah, but my question was not philosophical. I think what I wrote on my diagram with the square shows why the complex plane analysis actually does represent the tangential and radial components. I thought (and still do) there was a more general way of proceeding than my diagram analysis.

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sophiecentaur