Relativity 108a: Schwarzschild Metric - Derivation 

cringe

@@joesmith8288 No u

The joy of learning

​​@@joesmith8288 Don't be jealous of him learning tensor calculus before you. EDIT: Oops. I just read that part about "tears of joy." I've got to retract my statement.

@@dritemolawzbks8574 Ok sissy🤣🙋🙏

Cringe

Amen : )

100% agree for how important the kerr solution is for astrophysics it's poorly described in popular media

I agree it's really needed & important but I dunno if it's possible to do it without e.g. NoahExplainsPhysics derivation of time translation symmetry = energy conservation: it's done in a very formula heavy way that means he only has 2.9k views. So PLEASE do the Kerr metric as no one else does it but please try to present it in a more palatable, mainstream way? I dunno how one would go about doing this.

the thing is that the derivation of the kerr metric is absolutely diabolic, it took literally decades to figure it out, and considering that as you say, it is important, that is pretty scary. So no, of course you won't find "mainstream" derivations, such thing does not exist, I would look the original papers if you really still want a derivation.

A lot of viewers are more interested in retracing Einstein's path to learning Special relativity, equivalence principle, tensor analysis, and using the weak-field approximation to derive Newtonian gravity. The complete GR solutions from Schwarzschild and Friedman are very useful, but it's well above the mathematical abilities of a person not enrolled in a physics graduate program. The more advanced gravitational models with frame dragging, measurable gravitational waves, and black hole rotations are done by computer programs. The first approximation used by Einstein to recover Newton's theory of gravity and confirm the precession of Mercury's orbit, the gravitational deflection of light, and a gravity law that propagates at the speed of light were enough for Einstein and Hilbert to publish their field equations in 1915.

@@alwaysdisputin9930 if you "don't know" then how the hell can you be advising him you dodo??! Keep your mouth shut🙋🙏

That honestly slipped my mind. I'll have to think about that. I think the difference in r-coordinates only becomes apparent close to the Schwarzschild radius. For the earth the Scharzschild radius is only 1cm or so, so there's no real difference. The sun's r_s is make 3km, so again no real difference. I'll talk about that in 108b.

@@eigenchris Thank you! I never really got a satisfactory explanation for that.

This is the first I've heard of this concept. I'm not familiar with it at all. I probably won't be making a video on it. Do you know where I can read more?

@@eigenchris The metric can have curvature or torsion or both of them. If you wish, I can email you a series of resources and articles in this context.

My plan is to do that in 108c (not for earth specifically, but any mass in a Schwarzschild spacetime). I'm pretty sure the solution is analytical. But the solution is very similar to the Newtoniam result unless you're very close to the mass. Basically the gravitational potential gets an extra 1/r^3 term that goes to zero very quickly as r gets big.

I've never heard of that one, so I'm not going to be much help. I'm focusing on my current series on spinors for now.

That will be in the Relativity 110 videos that I hint at near the beginning.

If you look back at the Einstein Field Equations at the top of the slide, and plug in R=0, Lambda=0, Tuv=0, you get Ruv=0.

It's common when solving differential equations to specify "boundary conditions". Sometimes this means sepcifying the field at infinity. I think there would be similar requirements for, say, Maxwell's equations.

For EM field inverse square law was experimentally tested to a high accuracy. For gravity some people propose MOND, and while experiments are still inconclusive, Newtonian gravity better fits the data.

It basically means we can rotate the space part of spacetime around the center of mass and see no change.

@@eigenchris Thank you for your quick answer!

You could take a look at Sean Carroll's notes on General Relativity. www.preposterousuniverse.com/grnotes/

“The Final Theory: Rethinking Our Scientific Legacy “,Mark McCutcheon.

@@eigenchris There is a typo on timestamp 15:50 and corresponding slide #54 from the presentation. In the bottom line g22 is non-zero. Thank you for wonderful lecture!

That's right. I talk about the relationship between the r-coordinate and physical distance in the next video.

Sorry, I don't follow. I would have figured you'd need to already know the christoffel symbols to use the geodesic equation in any meaningful way.

@@eigenchris When I did my exams: I was told of a method of computing the connection coefficients that didn’t involve using the formula: it involved writing out the Lagrangian and then computing the geodesic equation using the Euler Lagrange equations: the connection coefficients would then be deduced via the geodesic equations My teacher always preferred this method then using the general formula, they thought it was much less fidly

@@cameronspalding9792 I'm not so familiar with Lagrangian methods. If you have a link to a page that explains this method, I could take a look.

people.uncw.edu/hermanr/GRcosmo/euler-equation-geodesics.pdf

In theory, it's possible to have a geometry where R=0 but R_μν is not zero. But if we are constrained by the Einstein field equations, and are in a vacuum where T_μν=0 (and also take the cosmological constant to be zero), then R=0 and R_μν = 0 are equivalent conditions.

@@eigenchris Thanks! I see. There's one more step: after R vanishes from the EFE, all the terms go to zero except for the first term. Thus R_μν must also vanish.

The T=0 used in this video is only for the vacuum *surrounding* the earth, as I point out at 3:15. Since T is the generalization of the mass density, you can think of this as seeing mass density to zero outside the earth. Newtonian gravity is brought in at 24:47, when I force the potential from Newtonian gravity to match up with the results predicted in GR. If you're up for watching more, you can check out my "Relativity 107d" video on Newton-Cartan theory. This is a theory of Newtonian gravity that uses curved spacetime, and it makes the jump to GR much easier. The Einstein Equations are a generalization of Poisson's Equation, where the metric g replaces the gravitational potential, and the energy momentum tensor T replaces the mass density. If you want a solution that takes the interior of a star into account, you can google the "interior schwarzschild solution". I don't know if "weak" has a formal definition. It's basically a limit where relativistic effects are negligeable. So there is no time dilation either from large velocities, and we don't need to take non-Newtonian GR effects into account from large masses. Rotations are considered objective in relativity, because rotating bodies have centripetal acceleration (velocities are relativity but an rotating object fundamentally has non-inertial motion).

Usually metrics output real values so that we can compare two distances as being shorter/longer. You can't make a comparison like this if the the metric outputs complex numbers. But in math you're generally free to define whatever you like as long as it doesn't lead to contradictions. So you could define the concept of a complex metric if you wanted.

@@eigenchris thanks for the reply

Some people call it the "metric perturbation". Sorry for the delay. I guess I missed the notification for your comment.

Thanks. It might be out by the end of October. Most of the slides are done, but there are some concepts I still don't understand, so I'll need to spend more time learning and fact-checking.

dA/dr = (-A+1)/r dA/dr/(-A+1) = 1/r ∫ = dA/dr/(-A+1) dr = ∫ 1/r dr Evalute the integrals we get : -log(-A(r) +1) = log(r) + C_1 => A(r) = -e^(-C_1)/r + 1 A(r) = -k/r +1 = 1 - k/r

I was a bit "loose" with the indices here. To be technically correct, I would need a kronecker delta on one of the terms to raise/lower the index appropriately, so that everything matches. The kronecker delta doesn't actually change any of the values in the formula, it just puts the indices in the right place, so I left it out for simplicity. Sorry if that's confusing.

Thank you for your answer Chris. Clear as usual. I'm getting great marks in GR thanks to you.

The curvature of spacetime only depends on the derivatives of the metric, not the actual metric value. So I don't think small additive numbers like this change the physics.

​@@eigenchris we can eliminate this constant by selecting a different scale. So no new physics, like you said.

@@nenhard I should maybe be a bit more careful... the metric itself does appear in the Einstein Field Equations, but as long as we ignore the cosmological constant, and the Ricci scalar is zero, we don't need to worry about the constant.

Equations are nonlinear and contain product of metric and its derivates.

The "mathematical reason" is that, if we take a constant time, and constant radius, we want the resulting shell surface to have the same geometry as an ordinary 2D spherical surface, by spherical symmetry. I don't think I have a better reason.

@@eigenchris you are the best

I didn't get into the details much... it might become more clear in later videos. A rotating black hole causes additional gravitational effects in the direction of rotation (you can google "ergosphere" to learn more). If we reverse the direction of time, these gravitational effects reverse. So if we force spacetime to be the same in both time directions, it guarantees no rotation.

Also, we multiply time by c to give it units of distance, to put it on the same level as x,y and z.

Is 10:41 the correct timestamp? I'm having a hard time seeing what you mean.

You can look these coefficients up on wikipedia or any GR textbook. en.m.wikipedia.org/wiki/Derivation_of_the_Schwarzschild_solution#Calculating_the_Christoffel_symbols

I was thinking that proving A = 1/B requires the assumption that spacetime becomes flat as r goes to infinity. When you say "lorentz manifolds are the vacuum solutions of the EFE", I was under the impression that a "lorentz manifold" is just a manifold with a metric that can give negative values, and therefore all of spacetime (vacuum or not) would be a lorentz manifold. Empty space is definitely not always Minkowski... since the empty space around the earth is still curved due to its proximity to earth... The EFE only tell us that empty space is "Ricci Flat" (with a zero Ricci Tensor) but the Riemann Tensor can still be non-zero in the vacuum. From my point of view, the idea that spacetime is flat far away from mass is an extra assumption, but I haven't investigated this deeply.

@@eigenchris Thanks for your reply. A Lorentzian Manifold has a signature with one opposite sign. And I suppose the simplest one would be Minkowski. But does "flat" have to mean Minkowski? Or can we get a solution from assuming Euclidian space? I think my question is if we can get Minkowski spacetime as a solution of the EFE without assuming that it must be the result. Can we get Special Relativity from General Relativity? Or did we observe that our universe follows the framework of special relativity and use that as a boundary condition? One of the quotes that draw me to the elegance of GR is that it "assumes no prior geometry", so is assuming Minkowski spacetime as a "basis geometry" a feature of _our_ spacetime or of GR in general?

@@narfwhals7843 I think "flat" and "minkowski" spacetimes are the same thing. But I have to admit I don't know of a way of "deriving" flat spacetime from the EFE... I've only seen it added on top of it as an additional assumption where (far from mass = flat spacetime).

@@eigenchris I'll keep thinking about this. I suppose you always have to make _some_ assumptions. Thanks for your thoughts :) On another note I find it absolutely baffling that we get away with describing spacetime with a set of equations that completely ignores half of the Riemann Tensor. _All_ of the curvature information in the Schwarzschild spacetime is in the part that isn't even in the equation. Shouldn't vacuum solutions use the Weyl Tensor instead? Solving 0=0 and getting a useful result seems so weird...

@@narfwhals7843 While the EFE are more complicated, conceptually I don't see them as being too different from Poisson's Equation or any of the 4 Maxwell equations, where the left side is a "field" and the right side is the "source". We don't expect vacuum spacetime to have zero curvature (since spacetime around the earth or any star is still curved), so it makes sense that a zero source term shouldn't "kill" all the curvature... it just kills that part that cause immediate volume changes (but still allows for the Weyl part that does the squeezing/squashing of tidal forces). That's just my intuition, but I haven't studied the Weyl tensor at all and can't say much else about the mathematics behind it.

I'm not sure what I would say about this topic. Did you have a particular application in mind?

@@eigenchris Mainly the differences between complex manifolds and real manifolds

I plan to make them eventually, but I've covered that content in my previous "tensors for beginners" videos so I've been less motivated to make them. Do you have any particular questions you want answered?

@@eigenchris Maybe not, but I just remember looking forward to the next one after having seen 106a. But I can endure for a bit longer! :-)

The density of the mass/energy of a planet or star is much higher than the space around it. The fact that you can argue that an atom is mostly empty space isn't so relevant. It's the density of mass that matters. Planets don't have a black hole inside them because the schwrazschild radius is smaller than their physical radius. If you look at a point inside the planet, all the mass outside the radius of that point basically "cancels out" with itself in terms of gravity, so you can no longer act like the schwarzschild radius for the entire planet's mass is relevant once you go inside. There's a solution called the "Interior Schwarzschild metric" that you can look at for this case.

First of all I must point out to your credit that you managed to ignite the flame fire within me. You do sacred work in making knowledge accessible in a consistent, clear and brilliant manner. Unlike the way knowledge is obtained from professors in academia. And you know it. For a long time I wanted to deepen and take the study of relativity seriously, beyond my degree in electrical engineering. In addition, I am pleasantly surprised by the speed of response from you. This is completely non-trivial. And that's commendable on my part. And I said just about the tip of the iceberg. Thank you very much. In short, welcome to the Elite Club who do not receive enough appreciation and attention for their work. I sincerely hope that for you it will come soon. Because it's really just a matter of time and perseverance. The PhysicswithElliot channel is also in your state more or less I would say. So far I have come across inspirational channels for laymen + like: 1). ru-vid.comvideos 2). ru-vid.comvideos 3). ru-vid.comvideos 4). ru-vid.comvideos But unfortunately this is also a mixed blessing (like Veritasium) because it makes you think you understand them in depth and in fact - compromise on them without any practical calculation ability. Now regarding what we talked about earlier. I understood the first paragraph (or rather - I took it upon myself) because I do not really know what the deep meaning of "mass" or "charge" is, beyond the fact that they are actually primarily responsible for the bending of space-time. So it's not clear to me why you recorded that mass density or energy is what matters in the context that they cancel each other out of the radius. Maybe I should think about it a little more. Regarding the second paragraph. It is not a good argument in my eyes to disqualify what I have suggested, arguing that the Schwarzschild radius is smaller than their physical radius. This is because you assume a hidden assumption that the density of the stars is more or less the same from the shell to the core. We have good reason to believe this is not the case. And in this case I suggested - a Schwarzschild radius can actually fit for a high density, which is likely to be, in the core region. In addition, you also imply that the order of events is that: first the star and its dimensions are formed and only then does the black hole in the core appear, miraculously. But I said the opposite. I also have other good reasons to believe this is the case (such as: the strength of the magnetic field in the shell of the Earth, which is relatively high in relation to the distance from the core - by the way, a first order calculation also shows that if you aim the radius to the core area then the magnetic field is very huge) Or, rapid plasma bursts from the sun's escape velocity most surprisingly crash back inward in the form of an arc. In a high distribution in areas of sunspots, etc .. One way or another, I currently have no computational knowledge that can justify my argument. I've only been learning the subject for the last two weeks from you. So I have nothing smart to say beyond that. Just right now nice to talk about it with you. How about posting a video that explains "Interior Schwarzschild metric" and no less, also about its applications? PS: If this is acceptable to you, how can you be reached on social networks (Facebook, Instagram, gmail)? I want to show you a calculation I made and ask you about its meaning (search for: GM=tc^3) To be honest, I do not have a professional friend to develop a conversation with on these topics or even interested in it. In my high-tech job, it does not interest anyone and I have never been able to develop an interesting conversation with any purpose on these topics. Sorry if I extended too much in words.

@@eigenchris I'm interested to hear your opinion on the matter :)

Sorry for the late reply. Some times I see a comment but then get distracted and forget to reply. For my 2nd paragraph... the concept of "Schwarzschild radius" is not a very "smart" or "complicated" concept. All it is, is a way to translate mass into a length. It doesn't take into account the distribution of mass over space or density variations. It's just mass-to-length, and that's it. If we have a bunch of mass concentrated in a point/singularity, the Schwarzschild radius is physically meaningful: it is a mathematical surface beyond which light can never escape. This radius is less meaningful if you look at mass distributed over space, where the mass is distributed outside the radius. If we look at the sun, the schwarzschild radius for the Sun's total mass is about 3km. But if we were to look at the mass inside the sun's core with a radius of 3km, that spherical chunk of sun is significantly smaller than the sun's total mass, and so the schwarzschild radius for that spherical chunk of sun will be extremely short. We can repeat that process of looking at smaller "chunks" over and over again, but there's not much physical meaning to it. It doesn't mean there is a black hole inside the sun. Also, are you implying that the black hole appeared first, and then the Sun appeared around it? I'm confused there, so please clarify. I don't have facebook or instagram. Maybe you can link me to a google doc with your calculations?

@@eigenchris The full calculation is at your disposal: (You may view and comment the paper) (Sorry for the inconvenience, though RU-vid is keeps deleting my message because of this PDF address) Just Search for - cryptii nato phonetic alphabet - and copy-paste on the right window text. Thank you for your cooperation. Hotel Tango Tango Papa Sierra (Colon)(Slash)(Slash) Alfa Papa Papa Stop Lima Uniform Mike India November Papa Delta Foxtrot Stop Charlie Oscar Mike (Slash) Victor India Echo Whiskey Echo Romeo (Slash) Six Two Three Foxtrot Five Two Charlie Eight Eight Zero Bravo Nine Delta Echo Eight Four Delta Alfa Delta Seven Foxtrot Five Three One _______________________________________________________________________________________________________________________________________________ GM=tc^3 predicts that the number and size of primordial Black Holes is much larger than previously thought. This leads to the question: where are they? One could be in the last place humans would look-beneath their feet! The Cassini spacecraft found that Saturn’s moon Enceladus gives off far more heat-about 6TW-than can be explained, and it is concentrated at the South Pole. The little moon’s interior is an excellent place to seek a Black Hole. Earth’s internal heat and magnetic field make the planet’s core another likely place for a small Black Hole - barely the size of a grain of sand' but weighing almost as much as the moon. (The earth is surrounded by a black hole - yes. millions of years ago earth would have formed around this black hole, just as a pearl forms around a grain of sand). That black hole would be responsible for generating Earth's internal heat which just causes volcanoes and earthquakes but has also caused our Islands to form and the black hole would also generate the magnetic field of Earth. All of this comes from the enormous amount of radiation (Hawking radiation). When scientists describe a star collapsing and suddenly turning into a black hole, they miss the fact that the black hole has always existed there. The "collapse" results from an imbalance between the internal pressures in the core (radiation) and the curvature in space-time, that is, an attraction towards the event horizon that actually fed from those objects that deviate from their stable trajectory around. it And about the varying G hypothesis that has been promoted by Dirac and others - a change in c is more plausible and fits the data. In reality, scientists attempting to determine the rate of expansion of the cosmos are inadvertently and unintentionally determining the rate of light's slowing down. P.S : When t was tiny c was enormous and the universe expanded in a Bang. Have a nice weekend !

I may need your help again to see the exact problem. After hours of looking at these slides, my brain becomes basically incapable of seeing errors if they exist. At 21:42 I see I'm doing (1/2) - (1/4), so a result of +(1/4) seems right. Is there a sign error earlier on?

Yeah, it's not the most fun. But once I write it out once, I can refer to it any time in the future easily from this video.