Timeline for Galactic rotation speed
Current License: CC BY-SA 4.0
10 events
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| Oct 27, 2023 at 5:40 | comment | added | xzczd♦ |
@Alex Oops, yeah it's a typo and should be m==s R/e. I mean, in the implementation of trapezoid rule, the Rho[0] term always disappear when m == 0, so the singularity at R==0 doesn't influence my method. (As shown above, The Rho[0] term is not solved in my answer. ) And my method doesn't differentiate the equation, so I think the Integrate[G[m] m^2 Sqrt[1 - e^2]/Sqrt[R^2 - e^2 m^2]^3 , {m, 0, R}] is not related here. My explanation for the numeric error is, the grid is too coarse for small $R$, this also explains why a graded grid improves the accuracy.
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| Oct 27, 2023 at 4:43 | comment | added | Alex Trounev |
Do you mean m=s R/e? Also please pay attention that Rho term is not zero at R->0. You suppose that Integrate[G[m] m^2 Sqrt[1 - e^2]/Sqrt[R^2 - e^2 m^2]^3 , {m, 0, R}]->0 at R->0. But it is not right. :)
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| Oct 27, 2023 at 4:04 | comment | added | xzczd♦ |
@Alex I just took a closer look at the change of variable m == R/e. If I understand it correctly, this won't play a role in numeric solving (at least for my method, I haven't looked into other answers yet), because the Rho term always disappear for R == 0, so I'll only have equations for Rho/@Rest@grid in the end.
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| Oct 27, 2023 at 3:19 | comment | added | xzczd♦ | @AlexTrounev Yeah, I've seen your clever treatment, but still decide to solve this problem in a manner as naive as possible :) . | |
| Oct 27, 2023 at 3:14 | comment | added | Alex Trounev |
I see that you used asymptotic solution as well. We can exclude singularity at R->0 by substitution m=s R/e.
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| Oct 27, 2023 at 3:04 | history | edited | xzczd♦ | CC BY-SA 4.0 |
added 265 characters in body
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| Oct 27, 2023 at 2:56 | comment | added | xzczd♦ |
@AlexTrounev Yeah but the jump area is quite narrow, we can simply treat it as numeric error (actually, I believe it's just numeric error, after checking various settings for points), and use the solasymp to supplement the function value for very small $R$ if necessary.
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| Oct 27, 2023 at 2:51 | comment | added | Alex Trounev |
Jump at R->0 is not clear. We discussed with Ulrich that it should be rho[0]=.025/(1-int[e])=0.179802, where int[e]=Sqrt[1 - e^2]/e^3 Integrate[s^2/Sqrt[1-s^2]^3,{s,0, e}], and therefore int[0.995]=0.860958. But in your solution we see jump from 0.025 to 0.179802, same as in the wrong iterative solution in the first answer @Ulrich :)
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| Oct 27, 2023 at 1:35 | history | edited | xzczd♦ | CC BY-SA 4.0 |
add the missing definition.
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| Oct 26, 2023 at 9:56 | history | answered | xzczd♦ | CC BY-SA 4.0 |