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Line element 
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#1
Aug3006, 12:35 PM

P: 460

A general orthogonal coordinate system (u,v,w) will have a line elemet of the form:
[tex] ds^2 = f^2 du^2 + g^2dv^2 + h^2dw^2[/tex] I have done a lot of vector calculus, but for some reason I can't figure out what this means! What is a line element? I know about the differential length element and its a vector, but this is a scalar! Also, what's up with all the squares? 


#2
Aug3006, 02:15 PM

P: 212

Think of it as the definition of length (the straight line) of the vector (or "line element") which is defined using components du, dv, dw. Note I'll do a lot of handwaving and putting things in quotes, since I'm being a little imprecise/crude here.
Consider normal Euclidean space with the usual coordinates/basis: [tex](\Delta s)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2[/tex] Which basically encapsulates the pythagorian theorem and distance function, by stating the differential length between two "points" is the sum of the squares of the differential of the individual components. When dealing with such "metrics" it is useful to consider it for an "infinitesimal" element: [tex]ds^2 = dx^2 + dy^2 + dz^2[/tex] for euclidean space with the standard basis. Why is this significant? We can consider alternative spaces: [tex]ds^2 =  dt^2 + dx^2 + dy^2 + dz^2[/tex] is the metric of flat spacetime in special relativity. In other words, the "distance" between two "locationevents" [tex](t, x, y, z)[/tex] and [tex](t + dt, x + dx, y + dy, z + dz)[/tex] is [tex]ds[/tex] Now back to your original question. What is the meaning of [tex] ds^2 = f^2 du^2 + g^2dv^2 + h^2dw^2[/tex]? Well, consider: [tex]ds^2=dr^2+r^2 d \theta\ ^2+ r^2 \sin ^2 d \phi\ ^2 [/tex] Which fits the form of the above, with suitable substitutions for symbols and functions for f, g, and h. This is merely euclidean space, defined exactly as I did above! The "infinitesimal" elements are in polar (or technically, spherical) coordinates and so in order to find the euclidean distance you must change the "metric" to match the change of variables, which then preserves the "nature" of the space in question. 


#3
Aug3006, 02:21 PM

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Here's my take on it...
Take for exemple the spherical case*, where [itex]\vec{r}=r\hat{r}[/itex]. You know that given a parametrized curve [itex]\vec{r}(t)[/itex], its lenght is given by [tex]L=\int_{t_1}^{t_2}\left\Vert\frac{d\vec{r}}{dt} \right\Vert dt[/tex] Well, in spherical coordinates, [tex]\frac{d\vec{r}}{dt} = \frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}+r\sin\theta\frac{d \phi}{dt}\hat{\phi}[/tex] [tex]\Rightarrow \left\Vert\frac{d\vec{r}}{dt} \right\Vert = \sqrt{\left(\frac{dr}{dt}\right)^2+\left(r\frac{d\theta}{dt}\right)^2+\ left(r\sin\theta\frac{d\phi}{dt}\right)^2}[/tex] [tex]\Rightarrow L=\int_{t_1}^{t_2}\sqrt{\left(\frac{dr}{dt}\right)^2+\left(r\frac{d\the ta}{dt}\right)^2+\left(r\sin\theta\frac{d\phi}{dt}\right)^2}dt[/tex] And now we say, let us define the line element dl by (Could we say "Let us implicitely define the function l by its total differential"?) [tex]dl=\sqrt{\left(\frac{dr}{dt}\right)^2+\left(r\frac{d\theta}{dt}\right)^ 2+\left(r\sin\theta\frac{d\phi}{dt}\right)^2}dt[/tex] [tex]\Rightarrow (dl)^2=\left(\frac{dr}{dt}dt\right)^2+\left(r\frac{d\theta}{dt}dt\right )^2+\left(r\sin\theta\frac{d\phi}{dt}dt\right)^2[/tex] But [itex]\frac{du(t)}{dt}dt[/itex] is the definition of the differential du. Hence, we can write, [tex]\Rightarrow (dl)^2=(dr)^2+(rd\theta)^2+(r\sin\theta\phi)^2[/tex] which is of the above form, with f=1,g=r,h=rsinO IMO, the concept of a line element is superfluous and I plain don't like it because it assumes the dt in the integral to be a "living entity" instead of an inert part of the integral notation. *the cartesian case is dull; f,g,h=1. 


#4
Aug3006, 03:48 PM

P: 460

Line element
Both of your guys' explanations make sense.
I have one more question though. In all the texts that I have seen, they all write the length of the infinitesimal length vector like the following: [tex] ds^2 = f^2 du^2 + g^2dv^2 + h^2dw^2[/tex] Why not just write: [tex] ds = \sqrt{f^2 du^2 + g^2dv^2 + h^2dw^2}[/tex] it makes much more sense this way. Also, how does it make sense to have a square of a differential? I mean, aren't you going to get zero if you square such infinitesimal quantity? (sorry if that sounds dumb, because I have never been able to get comfortable with differentials/infinitesimals). 


#5
Aug3006, 04:21 PM

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[tex]df=f'(x)dx[/tex] Notice that df is a function of the two independant variables x and dx. For a fixed x, it is the equation of a line tangeant to f at x. In several variables, df is defined analogously by [tex]df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy[/tex] df is a function of the 4 independant variables (x,y,z,dx,dy). For x,y fixed, it is the equations of a plane tangeant to f(x,y). When dx,dy approach zero, df approaches f(x+dx,y+dy). Anyway, the point is, a differential is not an infinitesimal*; it is a function, so its square is well defined. I think this topic borders on differential geometry/differential forms, a subject I am not familar with, so there may be discrepancies in what I said. I got the info from a very basic calculus book (Stewart). *There is a thing called nonstandard analysis that introduces the field of the hyperreal as an extension to the real and in which infinitesimal numbers exists and are well defined. Nonstarndard analysis is another way of doing calculus in which, I believe, dx do denote a socalled infinitesimal. (Hurkyl will be able to tell you more) 


#6
Aug3006, 04:45 PM

P: 212

This is where the "hand waving" gets too vague to be really called mathematical. Notice I put "infinitesimal" in quotes; in reality, when dealing with modern analysis (excepting the "surreal numbers", etc) there are no "infinitesimals" in the traditional sense of an infinitely small number.
Why [tex]ds^2[/tex]?. While this isn't really an answer, let me give you an idea. The norm of a vector is (usually) defined with the dot product on itself: [tex] \bf{s} \cdot \bf{s} = ds^2 = dx^2 + dy^2 + dz^2[/tex] When dealing with traditional euclideanstyle metrics (i.e., Riemannian), which forces that inner product to be positive definite(excepting zero vectors), we could take the square root of that positive quantity...but then we would be writing this: [tex] \sqrt{ \bf{s} \cdot \bf{s} } = ds = \sqrt{ dx^2 + dy^2 + dz^2 }[/tex] With square roots on both sides. Now, in cases such as relativity, the metric is not euclidean or euclideanstyle at all. Rather, it is considered lorenzian, or more generally, pseudoRiemannian, which does not enforce a condition that the inner product be positive definite; rather, it need only be nondegenerate. What does this mean? It means that vectors in the pseudoRiemannian space can result in either a positive or negative inner product (in the language of relativity, timelike or spacelike) If we were to take the square root of this, we would get an imaginary quantity. While this may be considered just an acceptable interpretation of the mathematics, when dealing with real cases (particularly physics of general relativity) it is less than convienient. Thus, the squared quantity is somewhat preferred. You can also see this in the fact that many people choose to use the metric: [tex]ds^2 =  dt^2 + dx^2 +dy^2 + dz^2[/tex] over the perhaps more "consistant": [tex]ds^2 = + (i dt)^2 + dx^2 +dy^2 + dz^2[/tex] 


#7
Sep506, 10:33 AM

P: 460




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