Lagrange's Equation Generalized Coordinates

  • Thread starter Bashyboy
  • Start date
  • #1
1,421
5

Main Question or Discussion Point

Hello,

I am currently reading about the topic alluded to in the topic of this thread. In Taylor's Classical Mechanics, the author appears to be making a requirement about any arbitrary coordinate system you employ in solving some particular problem. He says,

"Instead of the Cartesian coordinates r = (x,y,z), suppose that we wish to use some other coordinates. These could be spherical polar coordinates [itex](r, \theta, \phi)[/itex], r cylindrical coordinates [itex]\rho , \phi , z)[/itex], or any set of "generalized coordinates" q1, q2, q3, i the property that each position specifies a unique value of (q1, q2, q3); that is,

[itex]q_i = q_i(\vec{r})[/itex] for [itex]i = 1,2~and~3[/itex],

[itex]\vec{r}=\vec{r}(q_1,q_2,q_3)[/itex]

I understand that [itex]\vec{r}[/itex] is our position vector function, and that requiring coordinates (q1, q2, 3) in our arbitrary coordinate system to be able to be represented by a position vector makes sense, although I wouldn't mind someone telling me the exact reason we require this. Again, it also makes sense that we would require (q1, q2, q3) and the position vector function representing that point to specify one unique point--if this weren't true, then a particle could occupy to positions at the same time, although I wouldn't mind someone telling me the exact reason we require this. But I don't understand the requirement [itex]q_i = q_i(\vec{r})[/itex], what does this mathematical statement mean?
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,759
6,271
The reason for introducing non-cartesian "generalized coordinates" is that sometimes a problem is way easier to solve in such coordinates than in cartesian ones. E.g., if there is rotation symmetry around a given point, it's advantageous to introduce spherical coordinates.

Usually one has only the one-to-one mapping of a part of the Eucildean space. E.g., standard spherical coordinates, which are related to the cartesian coordinates via
[tex]x=r \cos \vartheta \cos \varphi, y=\quad r \cos \vartheta \sin \varphi, \quad z=r \sin \vartheta,[/tex]
have a singularity along the [itex]z[/itex] axis (i.e., the polar axis of the spherical coordinates). The ranges for the generalized coordinates are in this case [itex]r>0, \quad \vartheta \in (0,\pi), \quad \varphi \in [0,2 \pi)[/itex].
 
  • #3
hilbert2
Science Advisor
Insights Author
Gold Member
1,366
431
One other reason for using generalized coordinates is that if there are some constraints in the system (e.g. a particle is constrained to move on a certain surface), these constraints can be "built-in" implicitly in the coordinate system you choose to use, instead of introducing them with the method of undetermined multipliers.

Strictly speaking, a given point in space does not have to correspond to a unique set of coordinates, for example in cylindrical coordinates the points ##(r,\theta,z)## and ##(r,\theta + 2\pi,z)## describe the same point in space.
 
  • #4
1,421
5
I am not sure either of you have answered my question. I understand why we employ coordinate systems other than the cartesian (rectangular). What I would like to understand better is the reason for the two conditions we must place upon any arbitrary coordinate system.
 
Last edited:
  • #5
489
188
Well it basically says that you have a one-to-one map between the systems. Those 2 statements together that is.
As you can see clearly by noting that the composition of the function [itex]q_i[/itex] and [itex]\vec{r}[/itex] on the right hand side of your equations results in the identity map.

I'm going to use a little different notation now. You have the function [itex]q_i(\vec{r})[/itex]. But the coordinate has the same name. Let's name this function that maps the vector r to the coordinates [itex]q_i[/itex], [itex]f_i:\vec{r}\rightarrow q_i[/itex].
Now the same reasoning holds for the vector r as a function of the [itex]q_i[/itex]. The vector function [itex]\vec{g}\equiv\vec{r}:(q_1,q_2,q_3)\rightarrow \vec{r}[/itex].

Can you see that a generalised vector function [itex]F(\vec{r}) :\vec{r}\rightarrow(q_1,q_2,q_3)[/itex] bundles all these expressions together. What happens if you take a look at the composition of F and g? Do you see what the constraints on the transformations mean now?

Also look at the comment by vanhees. Although those singularities, indefinite values are mostly ignored as far as I know.


* I hope this is a little bit clear, sometimes I'm having trouble writing this stuff down in a comprehensible way.
 

Related Threads on Lagrange's Equation Generalized Coordinates

Replies
3
Views
1K
  • Last Post
Replies
4
Views
830
Replies
6
Views
500
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
1
Views
345
Replies
10
Views
7K
Top