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In the Schwarzschild Metric as the spacetime interval between two points in spacetime approaches $0$ for any ratio between the length of time and space the spacetime interval between the points in spacetime is described by the equation $${\Delta}s^2=\frac{{\Delta}r^2}{1-\frac{2GM}{c^2r}}-\left(1-\frac{2GM}{c^2r}\right)c^2{\Delta}t^2+r^2\left({\Delta}\theta^2+\sin^2\theta{\Delta}\phi^2\right)$$ I understand that sometimes $\sin^2ab$ could mean $\sin^2(ab)$, but it could also mean $b\sin^2a$. So in the Schwarzschild Metric does $\sin^2\theta{\Delta}\phi^2$ mean ${\Delta}\phi^2\sin^2\theta$, or $\sin^2\left(\theta{\Delta}\phi^2\right)$? Also is the $r$ term in the Schwarzschild Metric defined by the equation $A=4{\pi}r^2$?

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    $\begingroup$ You’ve asked a lot of questions about the meanings of basic notation, suggesting that you want to learn physics by just jumping to a couple famous formulas. No offense, but not much of value is going to be gained by that even if we answer your question. It’s like trying to learn a new language from scratch by reading an epic novel written in it. $\endgroup$ – knzhou May 23 at 22:53
  • $\begingroup$ A $\Delta$ indicates a finite interval. This metric is for an infinitesimal interval and should be written in terms of differentials. $\endgroup$ – G. Smith May 24 at 3:59
  • $\begingroup$ You seem to be confused with the notation because you are trying to learn physics just by looking at the equations without understanding them. To understand the physics you need to derive all the equations that you encounter. $\endgroup$ – Richard May 24 at 8:49
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It is $\sin^2(\theta) \mathrm d\phi^2$, as it does not make sense in this context to take $\sin(\mathrm d\phi)$. If you Taylor expand this, you get a power series in $\mathrm d\phi$ but $\mathrm ds^2$ is proportional to $\mathrm d \phi^2$ on dimensional grounds (or e.g., $\mathrm d\phi \mathrm d\theta$).

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  • $\begingroup$ Thank You, for the information! $\endgroup$ – Anders Gustafson May 26 at 22:37
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  1. I basically think you're confused because of a lack of parentheses on the sine function. The final term should read $$r^2 \sin^2{(\theta)} \Delta \phi^2$$ that is to say that $\theta$ is the only argument of the sine function. The $\Delta \phi$ is a separate variable, and since it is scalar you can commute it with $\sin{\theta}$ if you wish.

  2. It depends what area $A$ you are talking about, but generally no. The $r$ in the metric is the radial distance from the centre of the coordinates, which is usually taken to be the centre of the spherically symmetric body of mass you are describing. Then $A$ might be the area of the circle swept out by, maybe, a planet orbitting the star you are describing if you are saying that the orbit is a perfect circle. One thing you should remember it that $r$ is not the radius of the object you are describing, unless you want it to be. Instead $r$ is a general radial distance. It could be on the surface of the mass, or it could be infinitely far away. It cannot, however, be inside the mass, nor can we have values or $r$ where $$r\le\frac{2GM}{c^2}$$ since the metric is undefined in Schwarzschild coordinates for such values.

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  • $\begingroup$ “It depends what area $A$ you are talking about, but generally no.“ In this metric, the area of a sphere of radius $r$ is indeed $4\pi r^2$. Just integrate over the angular variables. $\endgroup$ – G. Smith May 24 at 3:56
  • $\begingroup$ Why are you saying that, "the metric is undefined in Schwarzschild coordinates" inside the horizon? $\endgroup$ – safesphere May 24 at 8:01
  • $\begingroup$ @G.Smith you may be able to link r to an area, but what I was trying to make clear is that that area is not the area of the spherically symmetric mass, since r is variable. So r is not defined from A, that is what I was trying to get across. $\endgroup$ – Ollie113 May 24 at 9:22

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