# Transformation to locally flat coordinates

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1. Feb 13, 2016

### Whitehole

I'm reading A. Zee's GR book and I'm in the section in which he is showing how to transform coordinates to be locally flat in a neighborhood of a point.
He said that we can always choose our neighborhood to be locally flat for any space of any dimension D.

"Look at how the metric transforms when you go to a new set of coordinates:

$g'_{λσ}(x') = g_{μν}(x) \frac{∂x^μ}{∂x'^λ} \frac{∂x^ν}{∂x'^σ}$

Within reason, you could choose any x you want, and for each choice, you get a new form for the metric. You have a lot of freedom to massage the metric into the form you want. The proof simply amounts to counting how much freedom you have on hand. So, look at our space around a point P. First, for writing convenience, shift our coordinates so that the point P is labeled by x = 0. Expand the given metric around P out to second order:

$g_{μν}(x) = g_{μν}(0) + A_{μν,λ}x^λ + B_{μν,λσ}x^λx^σ + ...$

As always, if you get confused, you should simply refer to the sphere. Thus, let the coordinates of the point P be $(θ_∗, ϕ_∗)$, so that $x^1 = (θ − θ_∗)$, $x^2 = (ϕ − ϕ_∗)$. (Of course, in this simple case, nothing depends on $ϕ_∗$). What we just wrote down is then simply, for example, $g_{ϕϕ} = \sin^2θ = sin^2θ_∗ + 2sinθ_∗cosθ_∗x^1 + . . .$ so that $A_{ϕϕ,1} = 2sinθ_∗cosθ_∗$ and $A_{ϕϕ,2} = 0$. Nothing profound at all. Change coordinates according to $x^μ = K^μ_ν x'^ν + L^μ_{νλ} x'^νx'^λ + M^μ_{νλσ}x'^νx'^λx'^σ + . . .$. Again, nothing profound: K , L, M, . . . are just a bunch of coefficients to be determined. At the point P, the new metric is given by

$g'_{λσ}(0) = g_{μν}(0) K^μ_λK^ν_σ$

Regard this as a matrix equation $g = K^TgK$, where T denotes transpose. Since $g_{μν}(0)$ is symmetric and real, there always exists a matrix K that will diagonalize it. After $g_{μν}(0)$ becomes diagonal (with positive diagonal elements—we will take that as a definition of space), we could scale each coordinate, one by one, by an appropriate factor, so that the diagonal elements become 1. We end up with the Euclidean metric $g_{μν}(0) = δ_{μν}$"

I don't understand how to really transform the coordinates to be flat. One exercise in his book is to find the locally flat coordinates on the Poincaré half plane.

Poincare metric: $ds^2 = \frac{dx^2 + dy^2}{y^2}$

Following the first part of his explanation,

$ds^2 = \frac{dx^2 + dy^2}{y^2}$ → $g_{xx} = g_{yy} = \frac{1}{y^2}$

The metric is already diagonal so K is just the identity matrix, but how do I find L (for the second order term)? Generally, what is the exact way to find the locally flat coordinates. The explanation is kinda fuzzy. I can't point the exact method to find those.

2. Feb 14, 2016

### Staff: Mentor

Since you've already diagonalized, all you have to do is rescale the $x$ and $y$ coordinates so that $g_{xx} = g_{yy} = 1$. How would you do that?

3. Feb 14, 2016

### Whitehole

Just multiply every component by $y^2$, but how does that transform the coordinates?

4. Feb 14, 2016

### Staff: Mentor

You can't just multiply every diagonal metric coefficient by $y^2$ arbitrarily to make them 1; you have to transform the coordinates so that the diagonal metric coefficients become 1.

In other words, you want the metric to look like $ds^2 = dx'^2 + dy'^2$, where I have used $x'$ and $y'$ to denote the new coordinates. That means you want to transform the coordinates such that

$$dx' = \frac{1}{y} dx$$

$$dy' = \frac{1}{y} dy$$

Integrating these equations gives you the coordinate transformations.

Last edited: Feb 14, 2016
5. Feb 14, 2016

### Whitehole

Let $du = \frac{dx}{y}$ and $dv = \frac{dy}{y}$. Integrate with initial conditions $(0, y*)$

$u = \int \frac{1}{y} \, dx = \frac{x}{y} + c_1$
$v = \int \frac{1}{y} \, dy = ln(y) + c_2$

At $(u,v) = (0, 0)$ → $u = \frac{x}{y}$ and $v = ln(y) - ln(y*) = ln(\frac{y}{y*})$

6. Feb 14, 2016

### Staff: Mentor

Yes, and you can set $y* = 1$ so that it disappears from the equations; this amounts to making $v \rightarrow - \infty$ as $y \rightarrow 0$, which is fine since the $x, y$ chart is only valid for $y > 0$ anyway (that's why the underlying manifold is called the Poincare half-plane).

7. Feb 14, 2016

### Whitehole

I understand what you're saying but I think based on the discussion in the book, Zee wants the reader to find the locally flat coordinates up to the second order correction. That's what's bothering me. I can't see explicitly how Zee did that.

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8. Feb 15, 2016

### Staff: Mentor

I think you may be mixing up different things. The "second order correction" is not the L term in the transformation. "First order", "second order", etc. refer to terms in the expansion of the metric around point P in a power series (the discussion immediately following equation 8 in the text). The point is that, by choosing a suitable coordinate transformation, you can make the metric itself equal to the flat metric at a chosen point P, and you can make the first order correction zero at that chosen point (which is equivalent to making all the first derivatives of the metric zero at point P). But you cannot make all the second order corrections zero at the chosen point, unless the underlying manifold is globally flat (i.e., a Euclidean plane, Euclidean 3-space, etc.). The presence of second order corrections that cannot be eliminated is a manifestation of intrinsic curvature of the manifold; so a "locally flat" metric will not include them, and will only be valid in a region around the chosen point P that is small enough that the second order corrections are negligible.

9. Feb 15, 2016

### Whitehole

I understand that it should be the metric but looking at the answers at the back of the book got me confused. Why is there second order terms in the coordinate transformation? The image is attached below.

10. Feb 15, 2016

### Staff: Mentor

Try writing down the expressions for $dx$ and $dy$ that are derived from the coordinate transformation, and then substituting those expressions into the metric $ds^2$ to get an expression for the metric in terms of $du$ and $dv$.

(Also check your expression for $y$; if $v = 0$ it should give $y = y*$, but it doesn't, so you might have miscopied it.)

Last edited: Feb 15, 2016
11. Feb 16, 2016

### Whitehole

The expression for y is correct since $v = ln(\frac{y}{y*})$ so if $v = 0$, $~0 = ln(\frac{y}{y*}) ~$⇒ $~ e^0 = \frac{y}{y*}~$ ⇒ $~ y = y^*$

Also, $u = \frac{x}{y}$, $v = ln(\frac{y}{y*})~$ ⇒ $~x = yu$, $y = y^*e^v~$ ⇒ $~dy = y^*e^vdv$, $~dx = udy + ydu = y^*ue^vdv + y^*e^vdu = y^*e^v(udv + du)$

$ds^2 = \frac{dx^2 + dy^2}{y^2} = \frac{y^{*2} e^{2v} (u^2 dv^2 + 2ududv + du^2) + y^{*2} e^{2v} du^2}{y^{*2} e^{2v}} = u^2 dv^2 + 2du^2 + 2udu dv$

12. Feb 16, 2016

### Staff: Mentor

Sorry, it somehow hadn't clicked that the expression I was referring to was in an image from Zee's book, not something you wrote. I think the book must have a typo; his expression is:

$$y = y* \left( v + \frac{1}{2} \left( v^2 - u^2 \right) \right)$$

Substituting $u = 0$, $v = 0$ into this gives $y = 0$, not $y = y*$. So either there is a typo, or Zee meant for the point $(x, y) = (0, y*)$, i.e., point P, to have coordinates $(u, v) = (0, v*)$, not $(0, 0)$, where $v*$ is some nonzero number. That would be very unusual; point P is supposed to be at the origin of the locally flat coordinates. So I think Zee made a typo, and the correct formula for $y$ should be

$$y = y* \left( 1 + v + \frac{1}{2} \left( v^2 - u^2 \right) \right)$$

Notice that this has no effect on the formula for $dy$ in terms of $du$ and $dv$, since the $1$ term is constant.

Zee's formula for $x$ looks fine:

$$x = y* \left( u + uv \right)$$

If you use these formulas (not the ones we wrote earlier) to obtain formulas for $du$ and $dv$, you should be able to show that $ds^2 = du^2 + dv^2$ to first order in $u$ and $v$.

13. Feb 17, 2016

### Whitehole

Oh, I was also wondering about the formula for y. But the bottomline is, how did he arrived with those expressions for x and y?

14. Feb 17, 2016

### Staff: Mentor

I should also say something about those formulas we wrote earlier. Consider what happens if we substitute Zee's expressions (with my correction) into our earlier formulas:

$$u = \frac{x}{y} = \frac{y* \left( u + uv \right)}{y* \left( 1 + v + \frac{1}{2} \left( v^2 - u^2 \right) \right)}$$

$$v = \ln \left( \frac{y}{y*} \right) = \ln \left( 1 + v + \frac{1}{2} \left( v^2 - u^2 \right) \right)$$

If we drop second order and higher terms, and note that, to first order, $1 + v = e^v$, we find that the above expressions reduce to $u = u$ and $v = v$. So to first order, Zee's expressions (corrected) and the ones we wrote earlier are the same.

15. Feb 17, 2016

### Whitehole

I think this should do,

$x = yu = y^*ue^v y^*u(1 + v + . . .) = y^*(u + uv + . . .)$
$y = y^*e^v = y^*(1 + v + \frac{v^2}{2} + . . .)$

The only difference is that in Zee's $\frac{1}{2} (v^2 - u^2)$ Why is that?

16. Feb 17, 2016

### Staff: Mentor

Well, the $- \frac{1}{2} u^2$ term is clearly necessary for the metric to work out right; if you substitute $dx = y* \left( du + v du + u dv \right)$, $dy = y* \left( dv + v dv - u du \right)$ into the equation for $ds^2$ you will see that the $- u du$ term in $dy$ has to be there, otherwise you don't get $ds^2 = du^2 + dv^2$.

As far as how you can get the $- \frac{1}{2} u^2$ term from our formulas, consider that at point P, $u = 0$, so the transformation for $y$ is just as if that term were absent. But what happens as you move away from point P?

17. Feb 17, 2016

### Whitehole

Then it wouldn't be flat anymore, higher order terms cannot be neglected.

18. Feb 17, 2016

### Staff: Mentor

Once again, you need to be careful about what "higher order terms" you are talking about.

The higher order terms that show the presence of curvature--i.e., the ones that make the "locally flat" approximation invalid--are "second order" in the sense that they involve second derivatives of the metric. (Or, if you do an expansion of the metric about point P as Zee does, they are the $B$ terms in the expansion.) Those terms will be negligible as long as we are close enough to point P (where how close "close enough" is depends on how accurate we want our "locally flat" approximation to be). And all of this depends on finding coordinates in which all of the $A$ terms in the expansion--i.e., all of the "first order" terms, the ones involving only first derivatives of the metric--are zero at point P. Those coordinates are the "locally flat" coordinates.

However, in order to make the above true, i.e., in order to have "locally flat" coordinates at all, you will, in general, have to have "higher order" terms in the coordinate transformation from whatever arbitrary coordinates you started with to the desired locally flat coordinates. The $uv$ term in Zee's equation for $x$, and the $v^2 - u^2$ term in Zee's equation for $y$, have to be there in order for the $(u, v)$ coordinates to be locally flat, i.e., for the first derivatives of the metric at point P, i.e., at $u = v = 0$, to be zero. So the "higher order terms" in this sense are present even at point P.

(Of course, here we have restricted "higher order" to just second order; we are ignoring terms of third and higher order even in the coordinate transformations, at least in Zee's equations. Why that works is probably a separate discussion.)

So we have locally flat coordinates $(u, v)$ in which the following is true at point P, i.e., at $u = v = 0$: $ds^2 = du^2 + dv^2$, and $\partial g_{\mu \nu} / \partial u = \partial g_{\mu \nu} / \partial v = 0$. But those things are exactly true only at point P. As soon as we move away from point P, they are no longer exactly true; they are only approximately true. Let's see how that works by comparing our transformation equations to Zee's.

Our equations, as you point out, can be written as $x = y u$, $y = y^* e^v$; this gives, after some substitution, $dx = y^* e^v ( du + u dv )$, $dy = y^* e^v dv$. Substituting these into the metric gives:

$$ds^2 = \left( du + u dv \right) ^2 + dv^2 = du^2 + 2 u du dv + \left( 1 + u^2 \right) dv^2$$

At $u = v = 0$, you can see that this reduces to $ds^2 = du^2 + dv^2$, as desired. But we also have to check the first derivatives; fortunately $v$ does not appear explicitly anywhere, so we only need to take derivatives with respect to $u$:

$$\frac{\partial g_{uv}}{\partial u} = 2$$

$$\frac{\partial g_{vv}}{\partial u} = 2 u$$

Oops! We have a problem; one of the first derivatives is not zero at point P. But suppose we rewrite the equation for $y$ so that its expansion to second order includes Zee's extra term: $y = y^* e^v ( 1 - \frac{1}{2} u^2 )$. Then $dy = y^* e^v ( dv - u du )$, and the metric becomes

$$ds^2 = \frac{1}{1 - \frac{1}{2} u^2} \left[ \left( du + u dv \right) ^2 + \left( dv - u du \right)^2 \right] = \frac{1 + u^2}{1 - \frac{1}{2} u^2} \left( du^2 + dv^2 \right)$$

The first derivatives now have $u$ in the numerator, so they are zero at $u = 0$, as desired. (Also notice that the $du dv$ cross term has gone away.) So that is where the extra term comes from.

As far as how we missed it when doing the integrations earlier, remember that the equation we integrated was $dv = dy / y$. If $y$ was a function only of $v$, the integral would be $v = ln(y)$ as we had before. But if $y$ can also be a function of $u$, then that is no longer true; we can insert an arbitrary function of $u$ into the formula for $y$ without affecting anything in terms of $v$. To put it another way, if we rewrite the equation as $dy / dv = y$, we can see that $y = e^v$ is a solution, but $y = e^v f(u)$ is also a solution, with $f(u)$ an arbitrary function of $u$. We ignored that possibility before, because we used sloppy notation; we should have written something to the effect of $\partial y / \partial v = y$, which would have made it clear that the equation did not constrain the $u$ dependence of $y$, but unfortunately there is no way to easily express that when writing integrals.

Last edited: Feb 18, 2016
19. Feb 17, 2016

### Whitehole

Ah yes, stupid of me to not notice that. It is indeed because of the sloppiness. So basically, trial and error. ?

20. Feb 18, 2016

### Staff: Mentor

Quite possibly, yes. Sometimes that's all that works. There might be a slicker way to see that the arbitrary function of $u$ needs to be $1 - \frac{1}{2} u^2$, but I haven't found one--I only picked it because I already had Zee's formula in front of me.

21. Feb 18, 2016

### Whitehole

I also think that is the most "practical" way in this situation given that the metric is not too complex, so it works. But I think the coordinate transformation with the Chrostoffel symbol establishes the recipe. Anyways, thanks for all your effort in helping me realize all this!