Agree

This is a good one, too: kzfaq.info

Yes. It really is. This type of high-quality educational content has been one of the few benefits of the pandemic.

This is the best use of the word literally that I've ever seen.

@@dritemolawzbks8574 it existed before the pandemic in french. This is the translated chanel.

Me

It's just so classic to set the mood 😎

yeah, the spacetime music always does its thing ...

Music sounds like from the Social Network OST by Trent Reznor and Ross Atticus

@@TNTsundar I'd make a wild guess that the music is by Alessandro Roussel himself, too. He has a few electronic tracks on his channel. This track is much more subdued, but this should have been expected if it had been created as a background music for a video.

It hasn’t even premiered yet...

He couldn't even find any solutions to his own equations. The first (non-trivial) solution, the Schwartzschild metric, was found by Karl Schwartzschild in 1916, the year after Einstein initially published his equations

thanks a lot !

so satisfying :P

Yes...

Could not agree more, Rajarshi! To bring in a comparison of didactic approaches, there is also an excellent series of 5 lectures on the introductory diffgeo and GR, 1-1.5 hours each, read by Kip Thorne to the grad class, which covers it all (kzfaq.info/sun/PL1XfECM855xnXjlacd6UkPzKo5IqRgYpP; there are more advanced ones at astro-gr.org). If asked in which order to watch, I would, honestly, stumble. I (used to?) know the subject well enough, but I'm rusty--I dropped out of academia 25 years ago. I'd certainly go to ScienceClic for a quick refresher rather than the classic Wheeler, Thorne et. al. _Gravitation_, at least to begin with. For a noob autodidact, on the other hand, these videos may be a bit too fast paced and too condensed, so it needs good thoughtful pauses and rewinds, but it will help create a good plan for the future deeper self-study. In all, the overall presentation of the material is very well thought through: I can literally _feel_ the hours of work behind every minute of the video. It's exactly as simple as it only could be to remain true, and not simpler.

yeah the DrPhysicsA videos are great, too. but this is def more accessible

That is not a derivation, the only way this can be derived is from the action principle, which not rly is a formal derivation.

haha i just read it after seeing this comment... i feel like a useless human being now hahaha

Imagine Einstein thought about this kind of stuff in his head without no computer or no help from anyone.

Yes

As you said, G is chosen so that we can recover the Newtonian equations when spacetime is static and almost flat. Let me just add that this may not be the last piece of the puzzle: there is a set of theories that aim to modify general relativity in order to get a better description of, say, inflation, dark matter and dark energy. In some of those theories, the gravitational constant G is not actually a constant, but described by an additional dynamical field

@@leastaction_224 i dont get it; is G chosen? what about merging quantum mechanics and general relativity

In Newtonian physics, we describe gravity as an attractive force acting between bodies with mass. Newton said that this force must be proportional to the product of their masses and inversely proportional to the square of their distance. He called G the constant of proportionality. The value of G was later measured many times and we are now pretty sure it is indeed a constant. Now, Albert Einstein enters the chat. He described gravity as the effect of the curvature of spacetime itself, making use of the Einstein field equation(s) described in this video. Now, the video correctly says that this equation cannot be proven: it is a postulate of the theory. Therefore, in a way, it is "chosen", as you said. If you are interested, let me elaborate a little bit more. The Einstein field equation is written keeping in mind that we want it to have a certain set of properties. In particular, we want to relate the curvature of spacetime (described by the metric tensor, the Christoffel symbols and the various curvature tensors) with the energy and momentum of the "stuff" in the universe. We want this relation to be valid in all possible frames of reference (this is a fundamental requirement of every physical law) and we want to make sure that it does not violate physical laws we already know, like conservation of energy. These requirements alone fix the form of the equation as you see them in 1:53, except for the fraction in the left-hand side (the one multiplying the energy-momentum tensor). That coefficient is chosen so that, when gravity is weak and static (i.e. when the metric tensor looks like the one of flat spacetime) we recover the newtonian description. This requires the presence of the factor G in the numerator. Now, as for merging quantum mechanics with general relativity... that's a really complicated topic. The main idea is that we cannot simply apply the rules of quantum mechanics to the gravitational field (as we know we can do with the electromagnetic one). The main problem is that quantities that should be finite number turn out to be infinite, and we have no way to "get rid" of these infinities. Various strategies have been proposed, none of them has ever been confirmed by an experiment.

​@@leastaction_224 I see; thanks for clearing up my doubt

This is an excellent question, which I've asked myself a long time ago. The answer is: Einstein's theory doesn't need 8piG/c^4 but just a constant to connect curvature and energy-momentum in the field equation (this constant has a name: the Einstein constant). However, in order for Newtonian gravity to be reproduced in the low-gravity, low-speed approximation, that constant needs to be set this way. Fascinatingly, it turns out that, once you set this constant this way, the equation also works in the high-gravity, high-speed scenarios. Why? Nobody knows. But the equation describes the data. It's part of the theory.

For the man who discovered the first black hole solution, Schwarzschild ("black shield" in german) is quite appropriate :p

The 8th / last episode will be about 3 examples of this type ;) The precession of Mercury is a bit more tricky though so I keep this idea in mind for a later video maybe. Glad you like the series ! Also, the origin of gravity is the bending of spacetime. It's only as a first approximation that it also corresponds to the gradient of time dilation

@@ScienceClicEN perfect! Thanks :)

ScienceClic English Fun Fact: Einstein used the precession of Mercury in order to prove his theory in 1915. However then there was no Schwartzschild metric or any other ‚tool‘ at hand.

Glad you liked it ! The Einstein equation is not really derived, it is a postulate of the theory, the only equation that we take as an "axiom" in the theory. However you might wonder why Einstein supposed this equation and not something else ? There is quite a simple way to find this equation. The idea is that all theories in physics can be expressed as "principles of least action". Basically, any theory can be reduced to the postulate that : "the universe minimizes a certain quantity, S, called the action". This quantity must be a number. In our model, the only number that we have derived was the Ricci scalar R. Therefore we can simply suppose that S=R (or rather the integral of R). This corresponds to postulating that "the universe minimizes the average curvature of spacetime". Adding matter to this, and using the equations that describe the principle of least action (Euler-Lagrange equations), this yields Einstein's equation. So basically the Einstein equation comes from the postulate that the universe minimizes the Ricci scalar.

@@ScienceClicEN Although we mathematically take it as an “axiom”, it surely is not a mere lucky guess, that the action (or I suppose one of the terms in GR Lagrangian) depends on the Ricci scalar. Is there any physical intuition that could lead to this idea?

I think it's because the Ricci scalar is virtually the only scalar quantity that one can construct from curvature tensors. My personal physical intepretation would be that spacetime "wants" to be flat, just like a fabric with tension that would try to minimize its curvature

@@ScienceClicEN and maximize the time between?

OK, great that makes sense, it's like the Fermats principle which states that light always takes the path which requires least time or lagrangian mechanics. In principle it's the same equation just instead of the lagrangian you take the Ricci scalar.

What is a metric cancer?

It's quite similar yes. Energy is the equivalent of the electric charge but for the curvature of spacetime. And much like the electric / magnetic fields, you can say that static energy is the source for gravity, whereas moving energy is the source for "gravitomagnetism", most usually known as frame-dragging : en.wikipedia.org/wiki/Gravitoelectromagnetism

It's a creation for the channel, you can find it on my SoundCloud : soundcloud.com/aroussel

Partial answer / suggestion: Have you checked out episode 6/8 in this series about Energy Fluxes? Rough idea - in the simplest case you model everything as a "fluid", you only need to know a few things about that fluid. For example, the density, pressure and momentum (although ideally that would be at every point in space). The distribution of matter is given with the co-ordinates you are using and it doesn't matter how you measure lengths (including time) or angles in those co-ordinates, so the metric isn't required at that stage. For example, a density could be described as 1Kg of the fluid in every volume element. The volume element can be a wonky-shaped region of space given by 1 x-direction unit by 1 y-direction unit by 1 z-direction unit. It doesn't matter if the x,y, z directions were skewed, sheared, stretched or curvi-linear. Whatever wonky co-ordinates you specify the parameters with (density, pressure etc.) you will get the metric expressed in those same wonky co-ordinates. Hope this helps.

In a way yes, we say that this metric is "asymptotically flat at infinity"

The fact that Tμν=0 at a given point tells us that there is no Ricci curvature at this same point, i.e. that Rμν=0. But this doesn't tell us that gμν=0, it only tells us that some combinations of second derivatives of gμν are zero, at the given point. This is a restriction on the metric, but it still leaves a lot of degrees of freedom undetermined. At a given point, many different metrics satisfy the condition Rμν=0 (Minkowski, Schwarzschild, Kerr, etc). The thing is, in the Schwarzschild metric we have Tμν≠0 at r=0, at the central singularity. And this is what allows us to have a whole family of solutions. All Schwarzschild metrics (with different values of M) satisfy the condition that Rμν=0 everywhere except at r=0. The parameter M comes from integrating the curvature along the whole spacetime.

@@ScienceClicEN Thank you!

If you can move backwards in time. A closed continuously differentiable world line that is timelike everywhere is not possible. You need "kinks" in the world line so that you do not get any spacelike sections. This means that you have an infinitely large curvature at one of the kinks. Which means that you need an infinitely high energy density at these kinks.

A negative energy density would mean that spacetime would be curved in a different direction. E.g. for warp drive (movement apparently faster than light speed. But this is allowed, since one does not move relative to space with faster-than-light speed. You move space along with you. You surf on a gravitational wave, so to speak), you must be able to bend space-time in both directions for that. For this, you would need a negative energy density. But I don't believe that you could travel back in time with it. But I'm not sure about this point. The consequence would be something like antigravity, i.e. you would be repelled by negative energy concentrations.

@@sevisymphonie5666 This kind repulsive gravity is exactly what happens bellow the Cauchy horizon of a Reissner-Nordström or a Kerr black hole, and I am pretty shure that real time travel exists in the Kerr case, since it has a name, the Carter time machine.

In the Reissner-Nordström solution, the tension in the strong radial electric field appears as a negative pressure/diagonal elements in the Maxwell stress tensor, which causes a negative spacetime curvature. In the Kerr case, it is the centrifugal force.

@@milihun7619 I haven't gone that deep into the subject yet. I come from the field of experimental solid-state physics and nanophysics or it is the direction I want to go. The only solution to Einstein's field equation that I have seen so far was the Schwarzschild solution. But yes, the other cases are not uninteresting either. Through the energy of charges, one can create something like negative energy density and according to the Einstein field equation, closed world lines are possible. But with journeys into the past, as you know, there would be some paradoxes. In other words, inconsistencies in cause and effect. But I understand your question a little better. Let's say you have a world line that runs straight with v in positive x0 or t direction and positive x1 direction. Now you want to reverse the x0 direction. that is dv0/d(tau)=-2*v0 and dv1/d(tau)=0. Ok, at this point it would even be possible without a kink, but we are only interested in if T00 is 0. If I insert this into the geodesic equation, I get an underdetermined system of equations. I will now solve it so that there are as few non-vanishing elements as possible for the Christofel symbol. That would be all zero, except the gamma000=2/v0. This then works with the metric, if g00=-1, g11=0, g10=4*x0/v0, g01=1. But it can also be everything with exactly a different sign. This gives me a Ricci tensor that is 0 everywhere except for R01=+/-inf. And Ricci-Scalar is thus =+/-inf and thus T00=-/+inf. The same with dv0/d(tau)=+2*v0 and dv1/d(tau)=0, dv0/d(tau)=0 and dv1/d(tau)=-2*v0 and dv0/d(tau)=0 and dv1/d(tau)=+2*v0. Then one has infinitely high energy densities at 4 points which can be +/-inf. But maybe I have also calculated a total muck-up.

The stess-energy tensor is a tensorial field, meaning that it has different values dependant on where in space-time you evaluate it. It is non-zero where there is mass (like in the center of the earth) but zero where there is none (like outside it). As you have pointed out in the timestamp you provided, this does mean that at those points the Ricci tensor is indeed 0. However, you must remember that the Ricci tensor is a contraction of the Riemannian curvature tensor and while the Ricci tensor is 0 it does not mean that every components of the Riemann tensor are (only that the sum of its appropriate components cancels out).

@@CreeperMaster78 I get that so far, but I don't see in your equation and in the solution where the Earth comes in. I would have expected that to see the stress-energy tensor at the point of the Earth, and how this tensor influences the point the video shows where the stress-energy tensor is 0. Of course there must be a connection, but I fail to see that connection in the formulas of the video.

@@CreeperMaster78 Maybe to clarify more where my problem is. If I understand correctly, then the stress-energy tensor defines the metric (i.e. curves space-time). The metric contains, among other things, the mass M of Earth. However, the stress-energy tensor shown in the video does not contain the mass M. For me it feels that there is an equation missing that connects something containing M to the metric.

That is because the Einstein equation is a differential equation : the Ricci tensor tells you the derivatives of the metric, but not directly the metric itself. To get the metric you need additionnal information, "boundary conditions" in a way (like when you integrate a function) : that's when the Earth comes in. Basically, the Einstein equation at this point tells you that the metric must be a Schwarzschild metric, but it does not tell you what the value of M is (it is an integration constant, it could be any number). To determine M, that's when you need to integrate over space. It's a bit tricky but you can define the notion of mass / energy using the fact that spacetime is flat at infinity, and this way you can show that M must be the mass of the Earth (almost by definition, in a way it's almost as if we define "the mass of the Earth" to be this constant M)

@@ScienceClicEN I see. Still I don't see the influence of Earth in the equations you have given in the example. What makes the difference between a point in space with no surrounding mass, where the stress-energy tensor would be zero at that point, and the situation where there is mass nearby. Still, the stress-energy tensor at that point is 0, as you explained in the video. Where in the equation do I find the earth which requires the solution to be a Schwarzschild metric and not just the Minkowski metric? Wouldn't that mean that at some point in space (e.g. the centre of the Earth or somewhere surrounding it), the stress-energy tensor would be different from 0? If the Schwarzschild metric is determined by the stress-energy tensor field only, then I would expect that somewhere the stress-energy tensor must be non-zero, right? If the stress-energy tensor is everywhere 0 then we would get the Minkowski metric?

That is essentially how the real universe behaves, since the distribution of matter, energy, and pressure throughout the universe is (on large scales) the same everywhere. So if we assume a uniform density and pressure everywhere, then Einstein's equation leads to a simple and very important metric that we use in cosmology, called the Friedmann-Lemaître-Robertson-Walker metric, or FLRW for short, after its discoverers. This metric describes a geometry which can either be flat (no curvature), or have constant positive or negative curvature, and a universe with each of these geometries will evolve in different ways.

Good question, actually kind of. It is derivable from the postulate that the action (in the context of the least action principle) is the most simple one possible : the integral over spacetime of the Ricci scalar. By just postulating this, we get the Einstein equations. I made this video to show the full derivation (it's in French however) : kzfaq.info/get/bejne/adRierKY1q2tk5s.htmlsi=eMTY79WYz5ekWs8r

I do all my animations with After Effects, for the equations it was a bit painful but I found a few tricks to make the process faster. Basically I slice the equations into different layers for each term that I want to move, and then animate everything by hand.