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How do I calculate a sum based on the variable $v$ in this expression $\sqrt{\frac{\hbar(v) }{4G}} r$ which changes in increments of 1 with $v_i= 1$ to $v_f = 6.25E34$?

Straight forward it would look something like this: $$\sqrt{\frac {\hbar{((1)/(6.25E34))} }{4G}} r + \sqrt{\frac {\hbar{((2)/(6.25E34))} }{4G}} r + \sqrt{\frac {\hbar{((3)/(6.25E34))} }{4G}} r +...+ \sqrt{\frac {\hbar{((6.25E34)/(6.25E34))} }{4G}} r = \text{total mass}$$

Tivity
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  • Both the square root and the integral can pass constants through, as $\sqrt{Cx}=\sqrt C\sqrt x$ and $\int Kf(y)\text dy=K\int f(y)\text dy$ (for constants $C\ge 0,K$). But the h-bar reference looks a bit like a function, so you may want to give that piece a closer inspection. – abiessu Dec 03 '21 at 14:32
  • I'm sorry but what you contributed is difficult for me to decipher. I should mention that math is not my strong suit. Also, the h-bar is a constant as well. It's actually Planck's constant. – Tivity Dec 03 '21 at 22:07
  • You may find it useful to consult the Wikipedia entry for Planck units, as it appears your initial definition of a $1$kg mass was not accurate... – abiessu Dec 15 '21 at 17:27

1 Answers1

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Both the square root function and the integration operation handle multiplicative constants nicely. In particular, if you have non-negative real numbers $a,b$ then the identity $\sqrt{ab}=\sqrt a\sqrt b$ holds. So for your title quantity we can say $$\sqrt{\frac{\hbar (v)}{4G}}r=r\sqrt{\frac{\hbar}{4G}}\sqrt v=C_0\sqrt v$$ where we will use $C_0$ to refer to that constant portion moving forward. Then we can write the listed integral as $$\int_{v_i}^{v_f}C_0\sqrt v\text dv=\frac 23C_0\left(v^{\frac 32}\right|_{v_i}^{v_f}$$ since a multiplicative constant may be moved outside an integral. Evaluation of the integral result can be addressed separately.

Edit: from the new listed sum of terms, the individual sum term would be expressed more like

$$\sqrt{\frac{\hbar v}{4Gv_f}}r$$

and the proposed sum can be approached with

$$\sqrt{\frac{\hbar}{4Gv_f}}r\sum_{v=1}^{v_f}\sqrt v$$

but this has a similar problem to the original in that if $v_f\to\infty$ then the result of the sum will approach infinity as well. See this question Conversion of Riemann Sum to Integral with Square Root for a very similar problem and how much of a $dx$ is required to make the sum converge...

Edit 2, corrected:

The area under a square root curve $f(x)=c\sqrt x$ in an interval $[a,b]$ is always $\frac 23c(b^{\frac 32}-a^{\frac 32})$. Summing over a large set of consecutive integers plugged in for $x$ in this function will closely approximate this result, especially as the number of values increases. So for the listed value of $v_f$ we should expect that the result of the listed sum will be within the best possible error tolerances of $\frac 23C_0v_f=\frac 23v_fr\sqrt{\frac{\hbar}{4G}}$.

abiessu
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  • Thank you, so it would be $r\sqrt{\frac{\hbar (v)}{4G}}$ * ($({v_f}^{\frac 32})-({v_i}^{\frac 32})$) ? – Tivity Dec 07 '21 at 05:02
  • That's how I would evaluate what you have posted, yes. – abiessu Dec 07 '21 at 06:13
  • Shouldn't it actually be this? $r\sqrt{\frac{\hbar}{4G}}$ * $\frac 23(({v_f}^{\frac 32})-({v_i}^{\frac 32}))$ ? – Tivity Dec 09 '21 at 19:54
  • However in either case, the integral doesn't seem to be functioning properly, If I enter 299792458 into $\sqrt{\frac {\hbar(299792458) }{4G}} × 9.25 × 10^7 = 1,\text{kg}$. However, when I enter 299792458 and the final velocity and 299792457 as the initial velocity in the integral. In either case, it doesn't add up to the correct sum. – Tivity Dec 09 '21 at 20:34
  • You're right about the $\frac 23$, good catch. I don't know enough about the data you're working with to troubleshoot what you're seeing. – abiessu Dec 09 '21 at 22:56
  • What is the $9.25$ quantity referring to, or else what is the $10^7$? Is seems as though you have more data pieces there than what are listed in your question... – abiessu Dec 10 '21 at 01:01
  • $9.25×10^7$ this is the $r$ radius in Planck lengths. Just FYI this is the Schwarzschild radius for a 1 kg mass. – Tivity Dec 10 '21 at 04:35
  • That makes a bit more sense then. Where did the need to integrate over the expression arise from? – abiessu Dec 10 '21 at 05:27
  • If you look that the diagram at the top of this post (the red sphere's). Each sphere has a different density based on $v$ in $\sqrt{\frac{\hbar (v)}{4G}}r$ where $v$ ranges from 0 - 1. Each infinitesimal change in $v$ represents a sphere with a certain density and thus a certain mass. I need to find the total mass of all the spheres. – Tivity Dec 10 '21 at 06:00
  • Did you take a small sum and compare it to the result given by this integral? – abiessu Dec 10 '21 at 06:46
  • Yes, When I enter a range of 1-0 in $\frac 23(({1}^{\frac 32})-({0}^{\frac 32}))$ I get .66. However, if I enter a range 0.1-0 in $\frac 23(({0.1}^{\frac 32})-({0}^{\frac 32}))$ I get 21. I get a bigger sum when I input a lower range. That right there tells me there's an issue with the integral. – Tivity Dec 10 '21 at 07:19
  • Actually that should tell you that your calculator has an issue, because $.1^{\frac 32}\lt .1$. – abiessu Dec 10 '21 at 19:23
  • Good catch, I had to physically add a multiplication sign after the first term $\frac 23(({1}^{\frac 32})-({0}^{\frac 32})) = 0.66$. However, the integral is still incorrect. When I enter $9.25×10^7\sqrt{\frac{\hbar}{4G}} 0.66$ I get $3.83E-5$, which includes the final velocity 1 in the integral. But if I manually add the 1 into the equation $9.25×10^7\sqrt{\frac{\hbar*1}{4G}}$, I get $5.81E-5$. I'm getting a bigger value inputting 1 by itself than using the integral which includes 1 in the initial to the final velocity range, thus the integral should have a greater sum - yet it does not. – Tivity Dec 11 '21 at 08:47
  • Okay, I think I know what's happening. We have integrated over a function to obtain "the area under the curve", but the results you want (or are comparing to) require that each segment has length $1$ instead of the infinitesimal used by the integral. Either your comparison method must change or else the function we integrate over must change to make these match in the way you are expecting. – abiessu Dec 12 '21 at 03:17
  • I've done some research and think I need something like this - https://www.youtube.com/watch?v=m-_uvxWPZUk&list=PLMHtqvGHsradUqUWo5S8KEo-5Hs3qKVwE – Tivity Dec 12 '21 at 03:31
  • The real solution for this particular problem is to set up the sum and take a limit as the difference between $v_i$ values approaches $0$. Of course, such a sum would be expected to be infinite since you expect each value to be positive and not approaching $0$, so you may want to reevaluate your approach. What is the source of the part of the problem which asks for the infinitesimal differences? – abiessu Dec 12 '21 at 14:34
  • You're right about the sum being infinite thus I've reformulated the question. Can you take a look? – Tivity Dec 15 '21 at 06:29
  • I've added some to the end of my answer but at this point it would be a good idea to express in your question where the problem conditions arise from and how you are formulating your sum. More specifically, reference to specific published material is recommended in this case. Also, the referenced speed $v_f$ is rather large... Surely this is theoretical at best? – abiessu Dec 15 '21 at 16:58
  • It wouldn't be $v_f\to\infty$ it would be $v_f=6.25E34$, yes that is certainly a large sum but it's still finite. Also, this is the original problem but instead of an integral from 0-1 which would be infinite. I'm summing in incremental steps of 1 Planck units. But for the moment let's say it's $v_i=1$ to $v_f=10$. How would you solve that? – Tivity Dec 16 '21 at 04:54
  • Took a moment for heuristics, I only counted up to about $700$ terms but it was pretty obvious what the result would be. – abiessu Dec 16 '21 at 15:50
  • Thanks, but again math is not my strong suit, can you elaborate just a bit more? – Tivity Dec 17 '21 at 04:40
  • Not sure what you're looking for, but I put what the final formula would be. – abiessu Dec 17 '21 at 05:09
  • thanks, I'll do my best to figure it out. Can you tell me what the obvious result is that you derived - just so I know I'm on the right track? – Tivity Dec 17 '21 at 05:17
  • Just that the heuristics reminded me that summing square roots always has exactly one result, and it is increasingly accurate to that result by increasing the number of terms summed. – abiessu Dec 17 '21 at 06:24