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On page 23 of I.E. Irodov's Fundamental laws of mechanics,

The author defines two vectors

$\vec{r}_1 = [ \phi_1 , r]$ and $\vec{r}_2 = [\phi_2,r]$

Then he states that these two vectors can be added so that the angles sum up.

My question how is this vector addition when the $r$-components don't add up?

In other words, why is $\; \vec{r}_1 + \vec{r}_2 \not= [\phi_1 + \phi_2,\, 2r\, ]$?

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  • $\begingroup$ This section of the book is analyzing the specific motion of rotation about a fixed axis that passes through the origin. The radial distance from the origin to all points in such a rigid body is fixed. $\endgroup$ Apr 21, 2020 at 12:33
  • $\begingroup$ Draw and understand the diagram. $\endgroup$ Apr 21, 2020 at 14:01
  • $\begingroup$ But if I have vectors like $\endgroup$
    – Babu
    Apr 21, 2020 at 14:14
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    $\begingroup$ a=3i+4j b=2i+ a+b = 5i +4j See both components add here it doesnt. It is more of mathematial question than physics $\endgroup$
    – Babu
    Apr 21, 2020 at 14:15

2 Answers 2

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When dealing with vectors, we typically choose a coordinate system so that we can represent vectors as a unique tuple of the form $(x_1,x_2,...)$ and work with the numbers. The space of all tuples of this form is a called real coordinate space or $\Bbb R^n$.

Choosing a global basis generates a bijective linear map from a vector space to real coordinate space $(V \rightarrow \Bbb R^n)$. We can represent any vector as a unique linear combination of this basis and simply read off the components.

But we don't have to choose a global basis to define a coordinate system! Curvillinear corrdinates allow us to choose a general smooth map from $(V \rightarrow \Bbb R^n)$ and we can identify the vector as a tuple $(x_1,x_2,...)$.

The only issue is that if the map is not linear (which is the case for polar coords!), then addition of vectors in the different spaces does not commute with your map. In other words if F is the map from your space of position vectors to space of corrdinates, $F: (V \rightarrow \Bbb R^n)$, and F is not a linear map, then: $F(a + b) \ne F(a) + F(b)$.

This means that you can't add your polar coordinates together and hope to get the right answer by converting back!

One way around this is to convert back into a global basis, (i.e. extract the x and y component) and work with these instead.

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  • $\begingroup$ 1. What is a global basis? 2. What is a smooth map? 3. I have read some vector theory, shouldn't the position vectors be linear to even be called a vector? $\endgroup$
    – Babu
    Apr 21, 2020 at 17:47
  • $\begingroup$ 1. A global basis is one where the basis vectors are the same everywhere. In spherical coordaintes, the e_r and e_φ vectors depend on position, so the spherical basis vectors are not global. 2. A smooth map is basically a differentiable and invertible map of multiple variables. They are essentially "nice" functions to work with. (A mathemetician could give you a way better definition) $\endgroup$
    – KierD
    Apr 21, 2020 at 18:00
  • $\begingroup$ 3. A position vector is still a vector, even if you choose to desribe it in a way/coordinate systems which doesnt make it look very vector like. Fundamentally, you can sum two position vectors and get another position vector, you can scale them etc. So they are still vectors. $\endgroup$
    – KierD
    Apr 21, 2020 at 18:00
  • $\begingroup$ Very good answer, thank you $\endgroup$
    – Babu
    Apr 21, 2020 at 18:34
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Any vector can be defined by specifying its magnitude and direction. The sum of two vectors can be represented by a triangle. The given two form two sides of the triangle and the third side represents the sum. The direction for the sum can be equal to the sum of the other two directions only if all three are zero.

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  • $\begingroup$ Ok sure the radius being invariant makes sense if you have the triangle picture in mind but is it not possible to speak of vectors without bringing in the triangle idea? $\endgroup$
    – Babu
    Apr 21, 2020 at 17:12
  • $\begingroup$ It's possible, but any statement about vector sums should be consistent with the triangle representation. $\endgroup$
    – R.W. Bird
    Apr 21, 2020 at 18:18

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