I understand that $k$ describes positive, negative, or no curvature. However, why can't there be, for example, +0.5 (semi-positive) curvature, etc?
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$\begingroup$ Which values $k$ can or cannot take depends on its exact definition, as hinted at e.g. in the Wikipedia article. Please state your definition of $k$ and what exactly you don't understand about its values range. $\endgroup$– ACuriousMind ♦Commented Apr 2, 2018 at 20:59
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$\begingroup$ I don't think this is seriously unclear. The definition of $k$ is clear from the fact that OP knows it can only be $0$ or $\pm 1$. $\endgroup$– JavierCommented Apr 2, 2018 at 22:34
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$\begingroup$ @Javier Okay, well, it is unclear. I asked my cosmology professor and he said, "Interesting idea, you should look into that." But, I clearly don't understand enough advanced maths to read the equation ACuriousMind linked. As stated, I understand that +1 is positive curvature. Why isn't +0.1, +0.95, etc? Not everyone on SE has masters degrees. I don't understand why I'm constantly talked down to for not understanding and braving to ask. $\endgroup$– Rubellite FaeCommented Apr 3, 2018 at 19:28
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$\begingroup$ Sorry, I was responding to the above comment. I was saying that the question shouldn't be closed for being unclear, because it is not. It is a good question IMO. $\endgroup$– JavierCommented Apr 3, 2018 at 19:30
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1$\begingroup$ @Javier I see. Perhaps I have gotten touchy from past experiences. I thought you meant the answer is obvious, not the question. Apologies $\endgroup$– Rubellite FaeCommented Apr 3, 2018 at 19:48
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1 Answer
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The continuum of curvatures does exist, but we find it more convenient to put it elsewhere. The crucial part of the metric that encodes the curvature is a factor $1-K r^2$, where $r$ is the radial coordinate (using any point as the origin), and $K$ is any real number, positive or negative. By dimensional analysis, there is some length $L$ such that either $K=1/L^2$ or $K=-1/L^2$, so we can rewrite our formula as $1 - k (r/L)^2$, where $k$ is just the sign of $K$, or $0$ if $K=0$. The case $k=0$ corresponds to $L$ equal to infinity.
This new variable $k$ can only have the values $\pm 1$ or $0$, but that's okay because $L$ still can be any length, so we have the whole range of curvatures. $L$ is known as the radius of curvature of the universe, and a larger $L$ implies a smaller curvature. $k$ determines whether this curvature is positive or negative.
Now, and this is a bit of a technical point, we can make that $L$ go away if we measure our coordinate $r$ in units of $L$. In our formula, we can set $x = r/L$ to get just $1-kx^2$. The continuum of curvatures is still there but it is hidden inside of $x$, because the physical interpretation of $x$ depends on $L$: $x$ is how many times $L$ fits in your distance. So if for example $L = 1\ \text{light-year}$, $x=2$ is a distance of $2$ light-years, but if $L = 3$ light-years then $x=2$ is actually a distance of $6$ light-years. Mathematically, the price we pay is that $L$ now shows up elsewhere in the formulas, in the part we use to calculate lengths.
To sum up, the curvature can indeed take any value: the closer it is to zero, the closer space is to being flat. The fact that $k$ can only be $\pm 1$ or zero is just a matter of convenience: we use $k$ to label the three qualitatively different scenarios of positive/negative/zero curvature. $L$ just sets the size scale for the universe.
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$\begingroup$ Great! That helped so much. So in $(ds)^2 = \frac{[R(t)]^2}{1 + \frac{kr^2}{4}} [(dx)^2 + (dy)^2 + (dz)^2] - c^2 (dt)^2$ where is our L or "x?" The r next to k ? $\endgroup$ Commented Apr 5, 2018 at 15:28
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$\begingroup$ @RubelliteFae $L$ is inside $R(t)$, which is $L$ times an adimensional function $a(t)$ which tells you how the universe expands over time. Also, to make LaTeX display, enclose it between dollar signs. $\endgroup$– JavierCommented Apr 5, 2018 at 16:10
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$\begingroup$ Aha! We learned that $R(t)$ is the scale factor, but that $R(t)=\frac{l(t)}{r}$ not $R(t)=L*a(t)$. But, you are helping me make more sense of it. I'll reread my previous chapter when I get some time. Thanks so much $\endgroup$ Commented Apr 5, 2018 at 17:09