The Maths of General Relativity (8/8) - Summary and Applications
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- čas přidán 18. 05. 2024
- In this series, we build together the theory of general relativity. This eight and last video gives a summary of how one can solve a physics problem in general relativity. The video is then split into 3 examples, to illustrate the method.
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Alessandro Roussel,
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For the brave who try to solve the 2 exercises I proposed, here are the answers [SPOILERS]:
Exercise 1 :
(1) For a person on Earth, we can use the same formula as for the space station, but replacing the altitude r by the radius of the Earth (6370 km), because the person is on Earth, and the speed v by 0 because the person remains almost motionless (contrary to the station which is in orbit moving pretty fast). When we perform the calculation we get roughly 1.000 000 000 7 (that is 0,7 nanoseconds per second)
(2) To obtain the ratio of time dilation between the astronaut in the station and the person on Earth, we divide one by the other (we want to consider how the former's time evolves compared to the latter's). We hence calculate 1.000 000 001 / 1.000 000 000 7 ~ 1.000 000 000 3. The astronaut in the space station undergoes a time dilation of roughly 0,3 nanoseconds per second with respect to the person on Earth. Even if the astronaut is further from the Earth, their time runs slower because of the speed of the station (special relativistic effect).
Exercise 2 :
[EDIT, the conservation of energy is actually not needed for this]
We use the 2nd geodesic equation : (dvʳ/dτ) = -(GM/r²)(1-2GM/rc²)(vᵗ)² + (GM/r²c²)(vʳ)²/(1-2GM/rc²)
And the norm of the velocity vector : ||v||² = (c²-2GM/r)(vᵗ)² - (vʳ)²/(1-2GM/rc²)
We identify that (dvʳ/dτ) = -(GM/r²)||v||²/c². Since ||v||² = c², we obtain the desired result : (dvʳ/dτ) = -GM/r²
Knowing the answer wont tell you much, but understanding the question would tell you more than the answer will.
Congratulations. The Energy of the relation E = mc²vᵗ(1-2GM/rc²) is measured from a relatively motionless inertial observer A at infinite (as you said). Generally in a free fall from infinite distance will be valid the temporal relation [P(0) at infinite] = mc^2.(1)= const = [P(0) at every point of the orbit] , [ wiht "(0)" in P(0) we symbolize the covariant down index 0 of time t], because from known theorem we get that in your example the metric tensor is temporal static: d[g(μν)]/dt = 0. Therefore, at some K(t, r) point (or four-event) this gives E = U^(0).P^(0).g(00) = P(0)[g(00)]^(-1/2)= mc²(1-2GM/rc²)^(-1/2), because g(00) [U^(0)]^2 =1 and P^(0).g(00) = P(0) and also where we symbolize the contravariant component U^(0) where "^(0)" in U^(0) symbolize the contravariant up index 0 of time t and obviously the P^(0). And where g(μν) is the metric tensor, and t0. We point out that here we use the General System of coordinates where at point K(t,r) its observer is at rest relative to 3-space and so he is defined by a four-velocite that has only the time-component different from 0 and the fact that on the orbit the product of the observer's world velocity U with the body's world momentum P at some point K(t,r) will be invariant (as product of two world vectors) in all the systems of references (with curvilinear coordinates or not). A useful tool in energy calculations that may be correlated with the problem if the sum of positive and negative energy of the universe is zero. For example the quontity - c(PU) + mc^2 is the negative potential energy at the point K(t,r) of the orbit.
Exercise 2!?! Yeah, I'm not a calc freak, so I'm not among the brave to have solved the exercises.
Thanks so much for providing these exercises, they were really nice to do! Interestingly, for exercise 2 I didn't need to use conservation of energy; I just used the expression for the norm of the velocity vector to solve for vᵗ in terms of vʳ and then substituted that back into the 2nd geodesic equation. From there, most of the terms canceled out including all the vʳ terms, so I didn't need to relate it to energy or anything else) leaving behind the equation for Newtonian gravity. Not sure if there's something I'm missing here, but this worked perfectly fine for me and I was wondering if there's any particular reason that conservation of energy has to be used
You don't need the expression for the energy, just multiply the equation for the norm of v with GM/r²c², and add that equation then to the 2nd geodesic equation (lhs1 + lhs2 = rhs1 + rhs2), and you get immediately (dvʳ/dt) + GM/r² = 0 + 0 : )
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The mathematics of quantum field theory now....
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For QFT series I recommend viascience
I also want
@@infinummjb Use "spooky" instead
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@@jx5967 yeah ik but look at 3blue1brown or pbs spacetime for example, they have huge followings
It was not.
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@@hoodio It has few subs because hardly anyone is interested in science and whenever I say something about black holes to start a conversation I'm labelled as " nerd" and what not
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