1

In Exploring Black Holes, when dealing with the metric for polar coordinates for flat spacetime, he says "This derivation is valid only when dΘ is small - vanishingly small in the calculus sense - so that the differential segment of arc rdΘ is indistinguishable from a straight line." (page 3-2)

So just how small is vanishingly small? Lets say we are working at r = 1,000 metres from the event horizon. Can we reasonable work with a line 1 degree wide? or .1 degree? or .01 degree?

foolishmuse
  • 4,551
  • 4
    The question as written has no answer because the author means it's an infinitesimal, not just a small number, cf. e.g. https://physics.stackexchange.com/q/70376/50583, https://physics.stackexchange.com/q/92925/50583 and their linked questions. (Also: Why would you need to "work" with any particular value for $\mathrm{d}\Theta$?) – ACuriousMind Jan 10 '23 at 18:11
  • 1
    @ACuriousMind because in order to understand how this all works, I want to use real numbers and real physical sizes that I can actually envision; not just work with abstract formulas as it seems most of physics is done. – foolishmuse Jan 10 '23 at 18:19
  • 3
    If you want to do that -- you should think of it as a progression, not a transition past a threshold. In other words, given an approximation, understand what happens to its error as the "step size" gets smaller and smaller. You're not looking for small enough -- you're looking for "you can get as close as you want by making the step as small as you need to" – Alex K Jan 10 '23 at 18:50
  • 1
    @AlexK The trouble is if I do the calculations, all I get is a number. But I have no idea of how wrong the number is. I'm hoping that someone with more experience can provide guidance on the situation. – foolishmuse Jan 10 '23 at 19:10
  • 1
    I actually think you know enough that you can work this out on your own. You are looking for the difference between the arc length of a segment of a circle of radius $r$ that spans an angle $d\theta$, and you are comparing that to the approximation $rd\theta$. The former result is something you can obtain exactly using high school geometry. However I agree with the other comments that the goal here shouldn't be to solve a specific numerical example, but to understand continuity and the concept that the approximation can be made arbitraily good by taking $d\theta$ arbitrarily small. – Andrew Jan 10 '23 at 19:13
  • @Andrew I don't think that method would work. There is no spacetime curvature along a constant radius shell surrounding a black hole. I need to jump from one height to another height at different angles in order to experience spacetime curvature. This would be beyond high school geometry (at least in the dark ages when I was in high school). Unless you can suggest otherwise. – foolishmuse Jan 10 '23 at 21:10
  • @foolishmuse I'm not sure what you mean by "experience spacetime curvature" - curvature isn't really relevant to your question as far as I can tell. What matters is that you are using curvilinear coordinates in which the metric components are not constant. This doesn't mean that the space in question is curved, and indeed the derivation you're asking about refers specifically to flat 2D Euclidean space, which possesses no curvature. – J. Murray Jan 10 '23 at 21:48
  • @J.Murray Yes, I can see my mistake now. But my original erroneous question stands: are we thinking of space the size of a football field or a postage stamp? I'm examining your answer below very carefully, with "Exploring Black Holes" on my lap in front of me. – foolishmuse Jan 10 '23 at 22:12
  • @foolishmuse In order for the approximation to be good, we should have that $\Delta \phi$ and $\Delta r/r$ are much less than $1$. I've added a few lines to my answer to make this more explicit. A postage stamp, a football field, and an entire galaxy could all be considered "small" in the appropriate context, as long as they are sufficiently far away from the origin. – J. Murray Jan 10 '23 at 22:45
  • @foolishmuse The example you've linked to on page 3-2 is about a flat 2 dimensional space; there is no curvature to worry about. – Andrew Jan 11 '23 at 00:15

1 Answers1

4

Your question pertains to the 2D Euclidean metric in polar coordinates. Given that $$x = r \cos \phi \qquad y = r \sin\phi$$ It's easy to show that the distance between any two points $(r_1,\phi_1)$ and $(r_2,\phi_2)$ is $$\Delta s^2 = \Delta x^2 + \Delta y^2 =\big(r_2\cos(\phi_2) - r_1\cos(\phi_1)\big)^2 + \big(r_2\sin(\phi_2) - r_1\sin(\phi_1)\big)^2$$ $$= r_1^2 + r_2^2 - 2 r_1 r_2\cos\big(\phi_2 - \phi_1\big)$$

If we define $\Delta r \equiv r_2-r_1$ and $\Delta \phi\equiv\phi_2-\phi_1$ and drop the subscript on $r_1$, then

$$\Delta s^2 = \Delta r^2 + 2r^2\big(1-\cos(\Delta \phi)\big) + 2 r\Delta r \big(1-\cos(\Delta \phi)\big)$$

This expression is exact for any value of $\Delta r$ and $\Delta \phi$. However, expanding about $\Delta \phi,\Delta r= 0$, this becomes $$\Delta s^2 = \Delta r^2 + r^2 \Delta \phi^2 + \left[ r\Delta r (\Delta \phi)^2 -r^2 \frac{(\Delta \phi)^4}{12} +\ldots\right]$$

The terms outside of the brackets are quadratic in $\Delta r$ and $\Delta \phi$, whereas the terms inside the brackets are cubic and quartic, respectively. If you want to know how good a particular approximation is, then you can evaluate the terms in the bracket and compare them to the terms outside.

Note that we can also rewrite the above expression as

$$\Delta s^2 = \Delta r^2 + r^2 \Delta \phi^2 \left[ 1 + \frac{\Delta r}{r} + \frac{(\Delta \phi)^2}{12} + \ldots \right]$$

The larger point is that if we let $\Delta r,\Delta \phi$ be infinitesimal quantities $\mathrm dr,\mathrm d\phi$, then lowest order terms the quadratic ones and we find that $$\mathrm ds^2 = \mathrm dr^2 + r^2 \mathrm d\phi^2$$ This can be used to compute the distances along any curve $C :[0,1]\ni \lambda \mapsto \big(r(\lambda),\phi(\lambda)\big)$ by integration:

$$\Delta s = \int_{0}^{1} \sqrt{r'(\lambda)^2 + r(\lambda)^2\phi'(\lambda)^2 } \mathrm d\lambda$$

J. Murray
  • 69,036