Transformation of the Line-Element

Anamitra

The situation in the last post has improved considerably.It my request to the audience to consider the inconvenience caused [before editing].

JDoolin

[tex]0=f1\frac{\partial p}{\partial x}\frac{\partial p}{\partial y}+{f2}\frac{\partial{q}}{\partial x}\frac{\partial{q}}{\partial y}[/tex]
----------- (8)

I used notepad search-and-replace to get rid of all the extraneous {} signs. I'm not sure which {} signs are specifically causing the LaTeX compiler to trip up, but getting rid of all the extras seemed to work.

DaleSpam

You have two unknown functions, F1 and F2, and three first order PDE's that they must satisfy. Such a system is called overdetermined and generally doesn't have a solution.

PAllen

To paraphrase, you are asking the equivalent of: under what circumstances can I make 1 + 1 = 0 instead of 2? The fact that valid coordinate transforms (including correct transform of tensors) cannot change anything about the intrinsic geometry of the manifold is the core starting point (really, almost more like a definition; that is definitions were constructed to ensure that this is the case) of differential geometry. You need to at least read some elementary intro to differential geometry before you proceed further.

Anamitra

From equations (6) and (7) of Post 221 we have,
Eqn (1) below
[tex]\frac{\partial q}{\partial x}{=}\sqrt{\frac{1}{f2}{[}{1}{-}{f1}{(}\frac{\partial p}{\partial x}{)}^{2}{]}}[/tex]
Eqn (2) below:

[tex]\frac{\partial q}{\partial y}{=}\sqrt{\frac{1}{f2}{[}{1}{-}{f1}{(}\frac{\partial p}{\partial y}{)}^{2}{]}}[/tex]

Using the above two results in eqn (8) of Post #221 and simplifying we have,
Equation (3):
[tex]{(}\frac{\partial p}{\partial x}{)}^{2}{+}{(}\frac{\partial p}{\partial y}{)}^{2}{=}\frac{1}{f1}[/tex]
Squaring eqn (1) and using eqn(3) on the RHS of squared (1) we have,
Eqn (4) below,
[tex] {(}\frac{\partial q}{\partial x}{)}^{2}{=}\frac{f1}{f2}{(}\frac{\partial p}{\partial y}{)}^{2}[/tex]
Squaring eqn (2) and using eqn(3) on the RHS of squared (2) we have[eqn (5) below],
[tex] {(}\frac{\partial q}{\partial y}{)}^{2}{=}\frac{f1}{f2}{(}\frac{\partial p}{\partial x}{)}^{2}[/tex]
Adding (4) and (5) and using (3),we have equation(6) below,
[tex] {(}\frac{\partial q}{\partial x}{)}^{2}{+}{(}\frac{\partial q}{\partial y}{)}^{2}{=}\frac{1}{f2}[/tex]

Finally we have a pair of PDEs[(3) and (6) :
They have been written below again
[tex]{(}\frac{\partial p}{\partial x}{)}^{2}{+}{(}\frac{\partial p}{\partial y}{)}^{2}{=}\frac{1}{f1}[/tex]
[tex] {(}\frac{\partial q}{\partial x}{)}^{2}{+}{(}\frac{\partial q}{\partial y}{)}^{2}{=}\frac{1}{f2}[/tex]

DaleSpam

Anamitra, I think your best bet is to actually work it out for a couple of concrete examples. I would try polar coordinates on a plane and standard coordinates on a sphere.

Anamitra

We again consider the three equations:
[tex]{1}{=}{[}{f1}{(}\frac{\partial p}{\partial x}{)}^{2}{+}{f2}{(}\frac{\partial q}{\partial x}{)}^{2}{]}[/tex]--------------------------------- (1)
[tex]{1}{=}{[}{f1}{(}\frac{\partial p}{\partial y}{)}^{2}{+}{f2}{(}\frac{\partial q}{\partial y}{)}^{2}{]}[/tex]------------------------------- (2)
[tex]{0}{=}{[}{f1}\frac{\partial p}{\partial x}\frac{\partial p}{\partial y}{+}{f2}\frac{\partial q}{\partial x}\frac{\partial q}{\partial y}{]}[/tex] ------------- (3)
We choose the following two numbers:
f1/(f1+f2) and f2/(f1+f2)
Their sum is 1.
We write:
[tex]{f1}{(}\frac{\partial p}{\partial x}{)}^{2}{=}\frac{f2}{f1+f2}[/tex]
[tex]{f2}{(}\frac{\partial q}{\partial x}{)}^{2}{=}\frac{f1}{f1+f2}[/tex]
[tex]{f1}{(}\frac{\partial p}{\partial y}{)}^{2}{=}\frac{f1}{f1+f2}[/tex]
[tex]{f2}{(}\frac{\partial q}{\partial y}{)}^{2}{=}\frac{f2}{f1+f2}[/tex]
Equations (1) and (2) are clearly satisfied.
Now we write:
[tex]\sqrt{f1}\frac{\partial p}{\partial x}{=}\sqrt{\frac{f2}{f1+f2}}[/tex]
[tex]\sqrt{f2}\frac{\partial q}{\partial x}{=}\sqrt{\frac{f1}{f1+f2}}[/tex]
[tex]\sqrt{f1}\frac{\partial p}{\partial y}{=}\sqrt{\frac{f1}{f1+f2}}[/tex]
[tex]\sqrt{f2}\frac{\partial q}{\partial y}{=}{-}\sqrt{\frac{f2}{f1+f2}}[/tex]
The above four equations satisfy (1),(2) and (3) simultaneously.

DaleSpam

Again, I recommend working out a couple of concrete examples. Polar coordinates on a plane, where we know there is a solution, and standard coordinates on a sphere, where we know there is not a solution.

Anamitra

A Modification over my Previous post[Post #228]

We again consider the three equations:
[tex]{1}{=}{[}{f1}{(}\frac{\partial p}{\partial x}{)}^{2}{+}{f2}{(}\frac{\partial q}{\partial x}{)}^{2}{]}[/tex]--------------------------------- (1)
[tex]{1}{=}{[}{f1}{(}\frac{\partial p}{\partial y}{)}^{2}{+}{f2}{(}\frac{\partial q}{\partial y}{)}^{2}{]}[/tex]------------------------------- (2)
[tex]{0}{=}{[}{f1}\frac{\partial p}{\partial x}\frac{\partial p}{\partial y}{+}{f2}\frac{\partial q}{\partial x}\frac{\partial q}{\partial y}{]}[/tex] ------------- (3)
We choose the following two functions F1 and F2 and consider the expressions:
F1/(F1+F2) and F2/(F1+F2)
Their sum is always 1.
We write:
[tex]{f1}{(}\frac{\partial p}{\partial x}{)}^{2}{=}\frac{F2}{F1+F2}[/tex]
[tex]{f2}{(}\frac{\partial q}{\partial x}{)}^{2}{=}\frac{F1}{F1+F2}[/tex]
[tex]{f1}{(}\frac{\partial p}{\partial y}{)}^{2}{=}\frac{F1}{F1+F2}[/tex]
[tex]{f2}{(}\frac{\partial q}{\partial y}{)}^{2}{=}\frac{F2}{F1+F2}[/tex]
Equations (1) and (2) are clearly satisfied.
Now we write:
[tex]\sqrt{f1}\frac{\partial p}{\partial x}{=}\sqrt{\frac{F2}{F1+F2}}[/tex]
[tex]\sqrt{f2}\frac{\partial q}{\partial x}{=}\sqrt{\frac{F1}{F1+F2}}[/tex]
[tex]\sqrt{f1}\frac{\partial p}{\partial y}{=}\sqrt{\frac{F1}{F1+F2}}[/tex]
[tex]\sqrt{f2}\frac{\partial q}{\partial y}{=}{-}\sqrt{\frac{F2}{F1+F2}}[/tex]
The above four equations satisfy (1),(2) and (3) simultaneously.
It is important that we take care of the following:
[tex]\frac{{\partial}^{2}{p}}{\partial x \partial y}{=}\frac{{\partial}^{2}{p}}{\partial y \partial x}[/tex] ----------- (4)
[tex]\frac{{\partial}^{2}{q}}{\partial x \partial y}{=}\frac{{\partial}^{2}{q}}{\partial y \partial x}[/tex] ----------- (5)

Now the functions F1 and F2 are of our choice and we do have a certain amount of freedom over them.If they cater to the requirements in (4) and (5) the process of integration would be favored.
Though we can exert our freedom of choice over F1 and F2 ,two distinct conditions (4) and (5) have to be satisfied.It might be difficult if F1 and F2 are considered as analytical functions.
But we may think in the direction of numerical integration

[What DaleSpam has advised in the previous posting[#229]---- to work out concrete examples is extremely important-- and I do have that in my mind. I have simply placed here what I have done uptill now.]

Anamitra

It is important to remember that there are several issues going on in the thread--and "flattening of the sphere" in the literal sense is one of them.I have been trying it out in #228,#230.

We should always be aware of the fact that we may have curved surfaces in flat space-time and flat surfaces in curved spacetime.[The planetary orbits lie on such planes in curved space.]
[We can always have a straight line[spatial] in curved spacetime: The only thing is that such a line,generally speaking, will not be a spatial geodesic.But in some cases straight lines may correspond to spatial geodesics: for ex: radial lines in Schwarzschild's Geometry]

One of our basic interests is to find a one to one correspondence between curved space time and flat space time,preserving the line element[ds^2 invariant]. I have some examples at following posts[#155 and #164]:

http://physicsforums.com/showpost.ph...&postcount=155

http://physicsforums.com/showpost.ph...&postcount=164
In the second one we have "flattened" r= constant slices in Schwarzschild's Geomtery. These are 3D surfaces which we have flattened

An arbitrary curve[world line or spacetime curve] will lie on some 3D surface embedded in 4D space. The same curve may be contained by several surfaces[the curve lying at the intersection of several surfaces] The curve itself seems to be of paramount importance]

WannabeNewton

Your examples don't show that you have a one to one correspondence between curved and flat space. You can find an embedding map that takes a curved manifold into a submanifold of a flat one but an embedding map of such a kind isn't a diffeomorphism so if you have an embedding [itex]\phi[/itex] that doesn't mean [itex]\phi ^{-1}[/itex] exists. If the mapping isn't a diffeomorphism then [itex]ds^{2}[/itex] doesn't have to remain invariant. Also, the metric for a 2 - sphere doesn't describe flat space; there is a non - vanishing component [itex]R_{\theta \phi \theta \phi }[/itex].

Anamitra

Further Calculations[in relation to Post #230]:

Metric[Line element]

[tex]{ds}^{2}{=}{d}{\theta}^{2}{+}{Sin}^{2}{(}{\theta}{)}{d}{\phi}^{2}[/tex]
-------------- (1)
Equations:

[tex]\frac{\partial \theta}{\partial x}{=}{\psi}[/tex] -------- (2)
[tex]\frac{\partial \theta}{\partial y}{=}{\chi}[/tex] ------------- (3)
[tex]\frac{\partial \phi}{\partial x}{=}\frac{1}{{Sin}{(}{\theta}{)}}{\chi}[/tex] ----- (4)
[tex]\frac{\partial \phi}{\partial y}{=}{-}\frac{1}{{Sin}{(}{\theta}{)}}{\psi}[/tex] ----- (5)


[tex]{\psi}^{2}{+}{\chi}^{2}{=}{1}[/tex] -------- (6)

Exact Differential Conditions:
1. [tex]\frac{\partial \psi}{\partial y}{=}\frac{\partial \chi}{\partial x}[/tex] ------- (7)
2. [tex]\frac{\partial }{\partial y}{(}\frac{1}{Sin{(}{\theta}{)}}{\chi}{)}{=}{-}\frac{\partial }{\partial x}{(}\frac{1}{Sin{(}{\theta}{)}}{\psi}{)}[/tex] -------- (8)

The first condition[in relation to exact differentials] implies:

[tex]{\psi}{=}\int\frac{\partial \chi}{\partial x}{dy}{+}{f}{(}{x}{)}[/tex]
Or,
[tex]{\psi}{=}\int\frac{\partial }{\partial x}\sqrt{{1}{-}{\psi}^{2}}{dy}{+}{f}{(}{x}{)}[/tex]
----------- (9)
If solutions exist for the above integral equation,we write,
[tex]\frac{\partial \theta}{\partial x}{=}{\psi}[/tex] --------- (10)
[tex]\frac{\partial \theta}{\partial y}{=}{\chi}[/tex] --------- (11)
The exact differential condition is satisfied for theta.

What about phi?
By the differentiation of Equation (8) we have,
[tex]{-}\frac{Cos\theta}{{sin}^{2}{\theta}}\frac{\partial \theta}{\partial y}{\chi}{+}\frac{1}{Sin \theta}\frac{\partial \chi}{\partial y}{=}\frac{Cos \theta}{{sin}^{2}\theta}\frac{\partial \theta}{\partial y}{\psi}{-}\frac{1}{{Sin}\theta}\frac{\partial \psi}{\partial y}[/tex]
Or
[tex]\frac{{Cos}{(}{\theta}{)}}{{Sin}^{2}{(}{\theta}{)}}{(}{\chi}^{2}{+}{\ps i}^{2}{)}{=}\frac{1}{{Sin}{(}{\theta}{)}}{(}\frac{\partial \psi}{\partial x} {+}\frac{\partial \chi}{\partial y}{)}[/tex]
[(10) and (11) have been used in obtaining the above relation]
Finally:
[tex]{Cot}{(}{\theta}{)}{=}{(}\frac{\partial \psi}{\partial x} {+}\frac{\partial \chi}{\partial y}{)}[/tex] ----- (12)
Or,
[tex]{Cot}{(}{\theta}{)}{=}{[}\frac{{\partial}^{2}{\theta}}{{\partial x}^{2}}{+}\frac{{\partial}^{2}{\theta}}{{\partial y}^{2}}{]}[/tex] ----- (13)

On relation (13) we may apply the relations (10) ,(11) and (6) to go in the backward direction and arrive at the second exact differential condition that is, relation (8)

[The values of del(psi)/del(x) and del(chi)/del(y} as obtained from the results of the integral equation[ relation (9) should satisfy relation (12). This implies that at least certain solutions of (13) will cater to our requirement]]

DaleSpam

Such as which solutions? Any one will do.

Anamitra

An Analytical Endeavor:
Equation:
[tex]{Cot}{(}{\theta}{)}{=}{[}\frac{{\partial}^{2}{\theta}}{{\partial x}^{2}}{+}\frac{{\partial}^{2}{\theta}}{{\partial y}^{2}}{]}[/tex] ----- (1)

We look for a solution of the form:
[tex]\theta{=}{A}{(}{-}{ln}{Sin}{x}{-}\frac{1}{2}{x}^{2}{)}{+}{B}{(}{-}{ln}{Sin}{y}{-}\frac{1}{2}{y}^{2}{)}{+}{\lambda}{x}^{2}[/tex]------------- (2)
A and B are constants Lambda is a parameter independent of x and y.

Using the above trial in relation (1) we have

[tex]{A}{Cot}^{2}{x}{+}{B}{cot}^{2}{y}{+}{2}{\lambda} {=}{Cot}\theta[/tex]--- (3)

We may eliminate lambda between (2) and (3) to find Particular Integrals[surfaces theta=f(x,y)]
[Even if one has to use numerical methods, the equation will be an ordinary one instead of a“differential Equation”]
Instead of (2) one may use a trial of the form below, to obtain a greater variety of particular integrals:
[tex]\theta{=}{A}{f1}{(}{x}{)}{+}{B}{f2}{(}{y}{)}{+}{\lambda}{x}^{2}[/tex]------------- (4)
f1 and f1 are known ,arbitrary functions--well behaved ones of course,in terms of continuity ,differentiability etc.

The homogeneous part of equation (1) is simply Laplace’s equation in two dimensions. We have the familiar solutions.

[This posting will be revised]

DaleSpam

So plug your transformation equation back into the metric and calculate the curvature.