dune gom jabbar scene but it's a hard integral
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- Опубликовано: 29 фев 2024
- The integral in this video is for measuring radial lengths near a Schwarzschild black hole: • Relativity 108b: Schwa...
In general relativity, gravity "bends" space so that it's "bigger on the inside", so volumes near black holes contain more space than you'd expect for a given boundary.
Credit to a combination of Wolfram Alpha and my boyfriend for helping solve it.
POV: you learned about the swarzschild metric and you're trying to calculate your height around a black hole (you wasn't excpecting THAT pain)
Fucking relatable, man…
xD@@RuosongGao
not even spinning... not even charging...
Hey, at least there was a closed-form solution, lucky
I bring up this movie when my family complains about how unkempt and bushy my eyebrows are getting. I Google a picture of Thufir Hawat and say I'm trying to be a mentat. It doesn't work.
But do you also have sapho juice stains on your lips? ;)
@@bjornfeuerbacher5514 Sadly, no. However, my birthday is coming up. I'll ask for some.
yes i study math physics and science
also me during my finals:
0:37 There is a more straightforward way
- Use the (hyperbolic) double-angle formula sinh(2x) = 2sinh(x)cosh(x):
sinh(2*arcosh(y)) = 2*y*sinh(arcosh(y))
- Use the identity cosh^2(x) - sinh^2(x) = 1:
sinh(arcosh(y)) = sqrt(cosh^2(arcosh(y)) - 1) = sqrt(y^2 - 1)
And thus sinh(2*arcosh(y)) = 2*y*sqrt(y^2 - 1)
Yeah, I thought this too, theres usually a smarter way to do things
I’m taking a GR course this term and every time we get to an integral the prof’s instructions are “use substitution to make it a dimensionless integral, now use maple”😂 The integrals are always so gross
As soon as I saw the integral I was laughing my ass off
When the y became the solution to a quadratic equation in e^x, I felt that pain
Just a few days ago I tried looking up how much volume GR "adds" to the inside of the earth. One person said ~100 km^3, but I don't know enough GR math to confirm it myself
Naww it would be a negligible amount surely. I heard the earth in orbiting around the sun emits 400 watts of energy in gravitational waves.
Doing a quick back-of-the-envelope calculation, I could believe that.
The formula you see right at the end around 1:08 shows how much lengths will increate based on a spherical body's Schwarzschild radius r_s. If r_s=0, the formula just reduces to the standard "r" value. For earth, r_s is about 6cm, which is very small, but it does result in increasing the earth's radius by a few cm. It's not much, but given how huge the earth is (about 6000km radius), an extra few cm around the surface does translate into a volume increase of 100 km^3 or so.
This is by far the best math video I have ever seen.
We will use the formula for the integral of an inverse function. Given an inverse function f^-1(y), integral f^-1(y) dy = yf^-1(y) - F(f^-1(y)) + C, where F(y) is the antiderivative of the original function. Now, the inverse function of sqrt(r/(r-r_s)) is r_s r^2/(r^2 - 1). This function has a partial fraction decomposition, since r^2/(r^2 - 1) = 1/2 * r/(r - 1) + 1/2 * r/(r + 1) = 1 + 1/2 * 1/(r - 1) - 1/2 * 1/(r + 1), which means in our case F(r) = r_s * r + r_s/2 * ln(r - 1) - r_s/2 * ln(r + 1) = r_s * r + r_s/2 * ln((r - 1)/(r + 1)). Putting this in the formula and simplifiying the integral becomes: r * sqrt(r/(r-r_s)) - r_s * sqrt(r/(r-r_s)) - r_s / 2 * ln((sqrt(r/(r-r_s)) - )/(sqrt(r/(r-r_s)) + 1)) + C = sqrt(r) * sqrt(r - r_s) + r_s/2 * ln((sqrt(r) + sqrt(r - r_s))/(sqrt(r) - sqrt(r - r_s))) + C
Reintroducing the exponent into the logarithm and multipliying by the conjugate reveals the final answer, which differs from yours by an additive constant.
It just kept going
I'm pretty sure that for integrands that are rational functions of x and sqrt((ax+b)/(cx+d)) you can just substitute the latter with t. Also in this specific case you can divide numerator and denominator by r and recognize that the integrand is the inverse function of a rational function with quadratic denominator, and use the formula of integral of an inverse function.
I asked wolfram alpha to integrate it and it expressed it in terms of inverse sinh instead of log.
that's equivalent, indeed
Why must you give me GR ptsd memories.
I do math for "fun"
One of my favorite movies! 😂
You should make the same thing with the remake.
Not believable. Just think how much better Wolfram Alpha will be in the year 10,191.
This is heinous. You look at it and it doesn't seem that hard. Then you try to do it and it impales you on a burning stake. Where in the seven hells did all this difficulty come from?
Watch it through and I felt nothing.
Have I gone too far? Am I a lost cause
not that dissimilar from my doctoral thesis
yesssssssss
do one of Feynman's :D
Where does that integral come from?
I left an explanation in the description.
@@eigenchriswait you have a boyfriend? I thought you were straight. Why didn’t you go for a girl? Wouldn’t you be more happier?
And now the easy part; let's verify: You start with 1:09 and then take the derivative. Indeed you find sqrt( r / (r-rs) ). But the question remains: why would you want to calculate this. Certainly numerical integration does this very well at least on the real axis. Is it just the pleasure of getting to apply a goniometric transformation whenever you see a possibility of substitutions reaching sqrt(x^2 +- 1) in the integrand?
I left an explanation in the description. Closed-form solutions are useful because they can be reused easily for different cases.
@@eigenchris Sorry I did not read the description: my fault. I agree that closed-form expressions do have their uses and are the most general and give the highest performance if you need to calculate many of them. Many people have worked on closed-form expressions, written massive books with integral tables (Abramovitch, Gradshteyn etc.) and nowadays we have software that is pretty smart as well.
It is just that, years ago, back in the 1990's, some people in university thought integrals were the only important things, while nowadays it seems that almost nobody cares anymore. To me the balanced approach is to attack on three fronts: 1) look at a graph to see if there is something nasty to be expected, 2) check if some calculus or Wolfram Alpha can do it, 3) try numerically whether there are any issues.
Thanks for your reply.
isn't it... an incomplete beta function?
I mean, it's much less fun this way, but I feel that would be kinda more practical
I have no idea what an incomplete beta function is. I tend to stay as far away from integrals as I can.
@@eigenchris It is possible to hide from them, you just have to be very discrete
Your boyfriend solved it.
Me: It can't be that bad.
Five substitutions later: F*ck it. Numerical approximation will do. 🗿
I'm an engineer btw.