I Is there an algebraic derivation of the area element in polar coordinates?

SamRoss

Gold Member
148
11
Summary
There is a simple geometric derivation of the area element in polar coordinates. Is there an algebraic derivation as well?
There is a simple geometric derivation of the area element ## r dr d\theta## in polar coordinates such as in the following link: http://citadel.sjfc.edu/faculty/kgreen/vector/Block3/jacob/node4.html

Is there an algebraic derivation as well beginning with Cartesian coordinates and using ## x=rcos\theta## and ## y=rsin\theta ## to transform the Cartesian area element ##dx dy##?
 

WWGD

Science Advisor
Gold Member
4,170
1,739
There is a simple geometric derivation of the area element ## r dr d\theta## in polar coordinates such as in the following link: http://citadel.sjfc.edu/faculty/kgreen/vector/Block3/jacob/node4.html

Is there an algebraic derivation as well beginning with Cartesian coordinates and using ## x=rcos\theta## and ## y=rsin\theta ## to transform the Cartesian area element ##dx dy##?
I am not sure what you mean by algebraic derivation, this is usually called a change of coordinates. In this case, between polars and Cartesian.
 

SamRoss

Gold Member
148
11
I am not sure what you mean by algebraic derivation, this is usually called a change of coordinates. In this case, between polars and Cartesian.
Yes, that is what I'm looking for.
 

WWGD

Science Advisor
Gold Member
4,170
1,739
The setup ( generalized to n variables) is usually of the sort:
##x=f(a,b),\text{ so }
dx=f_a(a,b)da##
##y=g(a,b)\text , \text{ so } dy=f_b(a,b)db##
Where ##f_a, f_b ## are the partials with respect to ##a,b ##. In this case we use ##x=rcos\theta##, etc.

Is that helpful?
 
Last edited by a moderator:

pasmith

Homework Helper
1,677
375
The setup ( generalized to n variables) is usually of the sort:
##x=f(a,b),\text{ so }
dx=f_a(a,b)da##
##y=g(a,b)\text , \text{ so } dy=f_b(a,b)db##
Where ##f_a, f_b ## are the partials with respect to ##a,b ##. In this case we use ##x=rcos\theta##, etc.

Is that helpful?
[tex]
dx = f_a(a,b)\,da + f_b(a,b)\,db[/tex] etc...

EDIT: Just expanding dx and dy in the new coordinates and multiplying those expressions will not yield the correct answer; instead you must calculate the exterior product [itex]dx \wedge dy[/itex].
 
Last edited:

WWGD

Science Advisor
Gold Member
4,170
1,739
[tex]
dx = f_a(a,b)\,da + f_b(a,b)\,db[/tex] etc...

EDIT: Just expanding dx and dy in the new coordinates and multiplying those expressions will not yield the correct answer; instead you must calculate the exterior product [itex]dx \wedge dy[/itex].
OP is about change of coordinates, not about finding the total derivative.
 

pasmith

Homework Helper
1,677
375
OP is about change of coordinates, not about finding the total derivative.
The OP wants an algebraic derivation of the expression for the area element, and the relvant algebra is that of 2-forms. The derivation is then to compute [itex]dx \wedge dy[/itex] in terms of [itex]da \wedge db[/itex], for which purpose [itex]dx[/itex] must be expressed as [itex]dx = f_a\,da + f_b\,db[/itex] and similarly for [itex]dy[/itex].

Alternatively the OP could start from [tex]dx\,dy = \left \| \frac{\partial \mathbf{r}}{\partial a} \times
\frac{\partial \mathbf{r}}{\partial b}\right\| da\,db.[/tex]

In any event, I'm not sure how the OP is supposed to get from [itex]dx = f_a\,da[/itex] and [itex]dy = f_b\,db[/itex] to [itex]dx\,dy = |f_a g_b - f_b g_a|da\,db[/itex].
 

WWGD

Science Advisor
Gold Member
4,170
1,739
I referred specifically to the change of coordinates, not to the area element. You expect an answer for a question I did not address. I take things step by step, and start with a change of coordinates. As stated, this is correct. Yes, you did complete the needed steps; I was doing it more gradually.Edit. But, yes, this is correct, the new area is scaled by the norm of the determinant of the change of variables.
 
Last edited:

WWGD

Science Advisor
Gold Member
4,170
1,739
So, yes, to clarify, what pasmith says is correct. The change of area element is given by dadb=dxdy|J(x,y)| , where J(x,y) is the Jacobian of the change of variables matrix.
 
402
65
##x=r\cos\theta, y=r\sin\theta \rightarrow dx=\cos \theta dr + -r\sin\theta d\theta, dy=\sin\theta dr + r\cos\theta d\theta##

thus, ##dx \wedge dy = \cos \theta dr + -r\sin\theta d\theta \wedge sin\theta dr + r\cos\theta d\theta \rightarrow \cos \theta dr \wedge sin\theta dr + \cos \theta dr \wedge r\cos\theta d\theta + -r\sin\theta d\theta \wedge \sin\theta dr + -r\sin\theta d\theta \wedge r\cos\theta d\theta \rightarrow r \cos^2 \theta dr \wedge d\theta + -r \sin^2 \theta d\theta \wedge dr \rightarrow r \cos^2 \theta dr \wedge d\theta + r \sin^2 \theta dr \wedge d\theta ##

Which finally gives us: ## (\cos^2 \theta + \sin^2 \theta) r dr\wedge d\theta \rightarrow r dr\wedge d\theta \rightarrow r drd\theta ##

You may wonder why some terms went to zero and how i can switch the signs, and that would be due to some properties of exterior derivatives (and wedge products), which i suggest you look into. Hopefully this is what you were looking for.
 

WWGD

Science Advisor
Gold Member
4,170
1,739
I believed s/he wanted to go through it gradually , so I first went over the change of variables, but I guess we have to wait and see if they come back and tell us what they wanted.
 

SamRoss

Gold Member
148
11
Hi everybody. Thanks for all the help and sorry it took me so long to reply back. I had attempted the problem using a change of coordinates and was unable to get it to work. Thank you WWGD for bringing up the Jacobian. I had not thought to use that. Thank you Pasmith for bringing up the exterior product. That is a concept I was unfamiliar with and I will be sure to read up on it. And thank you Romsofia for the full solution. I appreciate everyone's assistance with this.
 

Want to reply to this thread?

"Is there an algebraic derivation of the area element in polar coordinates?" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top