I just binged the ENTIRE Tensor Beginners playlist AND the Relativity one! Too many other videos on relativity made me curious on HOW it works. And these videos laid out quite clearly. Maybe I can’t repeat the math, but I have better appreciation for it. Amazing that Einstein figured it out. But even MORE amazing that Schwarrzchild solved it in just a couple months.
Thanks, I'm glad you liked my videos. I think Einstein had a lot of help when it came to the math of relativity. Minwkoski helped come up with the more "geometrical" aspects of relativity. And Einstein needed help from his friend Grossman to understand the mathematics of curvature that Riemann and Cartan came up with. And yes, I'm sometimes amazed at how quickly progress is made when I read about the early history of relativity and quantum mechanics.
Doing the same here. I did 14 lectures in the MIT Open Course (8.962) Ware on GR then switched to eigenchris to get a visual picture of GR. Dr. Hughes at MIT is a very good professor and I leaned enough to prep me for these lectures.
Ladies and Gentlemen, and variations thereof, It is time for us to receive the next lecture for our desire to learn about the cosmos. The mighty eigenchris is here today to teach us about the wonder that is General Relativity and its offspring - the Schwarzschild metric. I can only ever hope but to master GR like those who come before us, yet I trudge onwards towards my goals. There on the horizon, on the void horizon. It seems as though my journey here is synonymous to the journey of an unfortunate into the horizon of a black hole. Soon enough I shall crack this for I must - for my own sanity.
Hey @eigenchris, I'm currently following an introductory course on general relativity where we learned about the Schwarzschild metric and its properties. This video (and the previous one) have made this so much clearer and more understandable for me! Thank you for making these videos with such good explanations!
Dear @eigenchris, I covered your all lectures..and really your teaching capability is outstanding.. You are more than a professor for me. Firstly tensor for beginners,then tensor calculus and recently i finished the general relativity series.. I don't know where are you from,who you are I will say.. you are awesome. A true genius. Now my next plan is to cover your new series "spinors ".. A little request for you.. If you can then please make a detailed series on " kerr black hole".. Since I am doing a thesis on "Spacetime Singularly and black holes" , your whole lectures gave me an extra power.. I am from Bangladesh. Always love and respect for you my professor ❤
Thanks for posting content like this the way you expose it is just wonderfully clear i like a lot all the details and the modern presentation... we can see the huge amount of work behind such videos, i am very grateful
The switching of the timelike/ spacelike properties of the t/r coordinates corresponds, also, to the fact that the Schwarzschild geometry, although it is static from the perspective of the outside observers, it's dynamical ( time dependent) inside the Schwarzschild radius. This relates to the nature of the singularity ( r= 0) that is spacelike and also to the very interesting property that the maximal volume of the interior ( see the related papers by Christodoulou / Rovelli and others for details) is evergrowing. That means that two black holes with the same mass and Horizon area may have vastly different interior ( maximal) volumes if they differ in age, by many orders of magnitude! Of course this is a property of black holes from gravitational collapse. Eternal Schwarzschild black holes ( if existed) could have already infinite internal maximal volume...
I'm quite a calculus dummy, but according to the calculus textbook you were supposed to substitute (1-r/z^2)=t^2 which would turn the integrand into something like r^2*dt/(1-t^2)^2 if my algebra is right and then can be expanded into partial fractions. And thank you for your videos, I love them ^_^
Schaum's Outline Series Mathematical Handbook by Murray R. Spiegel, on page 68 under "Intervals involving sqrt(x^2 - a^2)", agrees with your boyfiend's solution at 13:34 😀 Lovely video again Chris. Thankyou JJ.
Two conclusions, maybe provocative for some, not mentioned explicitly in this great video: 1. Free fall in a gravitational field does NOT cause any physical clock change (while falling), compared to a clock completely outside of the gravitational field. Gravitation is not a force, in that sense. A physical clock-change only occurs if a mechanical force (for example a rocket engine) or a planet surface acts against the gravitation. So the freezing of the object falling towards the event horizon does not mean that a clock inside that object ticks slower and slower, the freezing effect is only due to the light propagating slower and slower from the object towards an outside observer. 2. The comparison between the two formulas (Gravitational time dilation, Kinematic time dilation) at 6.40 must mean that a physical clock change in flat Minkowski space (no gravitation) only happens during acceleration/deceleration phases of the clock. There can not be a physical clock change without a non-gravitational force affecting it. This also means that the twin "paradox" has a fully intuitive/logical/physical/ solution. Only the acceleration/deceleration phases of the outgoing twin affects his/her clock, cruising phases with constant speed do not have any physical effect on the clock. And there is no distance problem, the travelling twin can travel the double distance with the same acceleration/deceleration phases, only the time intervals for the acceleration/deceleration are valid in the kinematic time dilation formula. The squared velocity variable in that formula is only valid for acceleration/deceleration phases. Newton: v^2 = 2 • acceleration • distance (assuming starting with v = 0). You can not put cruising (non-accelerated) time intervals into that formula. Agree ?
I do not know if this has been said before but the integral at 13:30 can be solved by guessing the answer and checking for it very easily without needing to do any substitutions besides the first one
Just a small request if I can; I often watch your videos at night and the white background becomes really eye-searing after a while. Could you consider using a dark one please? Love your content, all the best
If we can remove the event horizon of a black hole with a coordinate change that means that if we go in real life near a black hole we are going to observe an event horizon or not?Thank you for yoyr time amazing video!
The "event horizon" is still a meaningful mathematical idea in all coordinate systems... but the event horizon is also a coordinate singularity in Schwarzschild coordinates. In other coordinates, the event horizon is still there, but it's no longer a singularity. You can google "Kruskal Szekeres coordinates" to learn more.
@@namesurname1040 You can think it like this: the event horizon is a null/light-like "surface", the same type of "surfaces" which are given by the "trajectory" of light in a space-time diagram: since light is constant in every reference frame then any null surface (in this case the even horizon) is always present in any coordinate system there can be
13:08 I used y = sec(θ) as the trig sub. The integral was a bit messier, but there wasn't nearly as much algebra after. 2r∫y²/√(y²-1) dy y = sec(θ) dy = sec(θ)tan(θ) dθ 2r∫sec²(θ)/tan(θ) sec(θ)tan(θ) dθ 2r∫sec³(θ) dθ ∫sec²(θ) = tan(θ)+c (for integration by parts) 2r∫sec(θ)sec²(θ) dθ 2r(sec(θ)tan(θ) - ∫tan²(θ)sec(θ)dθ) 2r(sec(θ)tan(θ) - ∫(sec²(θ)-1)sec(θ)dθ) 2r(sec(θ)tan(θ) - ∫sec³(θ)dθ + ∫sec(θ)dθ) 2r∫sec³(θ) dθ = 2rsec(θ)tan(θ) - 2r∫sec³(θ)dθ + 2r∫sec(θ)dθ 4r∫sec³(θ) dθ = 2rsec(θ)tan(θ) + 2r∫sec(θ)dθ 2r∫sec³(θ) dθ = rsec(θ)tan(θ) + r∫sec(θ)dθ rsec(θ)tan(θ) + r*ln|sec(θ)+tan(θ)| + C ry√(y²-1) + r*ln|y+√(y²-1)| + C r(z√(r)/r)√((z√(r)/r)²-1) + r*ln|(z√(r)/r)+√((z√(r)/r)²-1)| + C z√(z²-r) + r*ln|z/√(r) + √(z²/r-1)| + C z√(z²-r) + r*ln|z + √(z²-r)| + C
Nice. Very admirable. I was really hopeless when it came to solving that integral. Wouldn't have been able to do it without help from humans and computers.
Welcome back! I think the "switching of roles" of time and space coordinates actually has a very pretty physical interpretation, telling us that the singularity is not a position in space but a point in the future. As it is unavoidable to reach it no matter what you do. About the time dilation at constant r without angular motion. Since these observers _have_ to be accelerating, does this mean there are additional Rindler time dilation effects from this acceleration or is that built into the metric?
I think it is "built into the metric" as you say. To maintain a constant r in schwarzschild coordinates, one has to accelerate outwards which, from the point of view of the accelerator (non inertial frame), looks locally like Rinddler metric.
About 12:00 ff: I once read another substitute in John Legat Martin's book _General Relativity_ (1988/1996) where, in the hints to problems he provided the reader with: √{1 − rₛ/r} =: q. The reader was to show that this substitution rationalises the square root and leads to a solution including the hyperbolic tangent. Maybe, this integral might be even easier to solve.
@eigenchris at 31:34 a outgoing light geodesic is locked in the event horizon, can I understand it as outgoing light can orbit arround the event horizon in the same fashion of photon sphere at 1.5 Swartzschild radius which is covered in the next video? Can light orbit arround both 1 and 1.5 Swartschild radius?
@eigenchris at 14:48, doesn't the L0 of schwarzschild spacetime should be the high point - low point of the complicated anti derivative instead of the anti derivative itself? And if the low point is the Schwartzschild radius rs, doesn't the L0 of flat spacetime should be r - rs? So the L0 axis position of the red line is the same as the blue line's when r = rs, and is always greater than the blue line's when r > rs, as it is the integration of the proper length from rs to r
The simple reason is that everything inside the event horizon must travel into the singularity. Up and down are spacelike directions inside the event horizon, so observers can freely travel up and down. But they are forced to travel left. This will be explained in Relativity 108d when we investigate alternative coordinate systems.
Fun fact: If you measure the fuel necessary to keep a rocket hovering at constant space-coordinates, using clocks onboard the rocket, and use that as a gravity measurement (or do any other local gravity measurement, for that matter, like dropping a ball onboard the rocket and measuring the distance to the floor and the time it takes to hit the floor), you get exactly Newtonian GM/r^2 gravity. At least that's what I recall from when I did the calculations on my GR exam a decade ago.
Using what was presented in this video as well as in 107d, I tried to calculate R_rr (Ricci tensor in the r,r direction) for the Schwarzschild solution. I want to see if space gets more spacious as one approaches a black hole. After a lot of index tracing, I got, for the Schwarzschild solution, that R^t_rtr =\frac{r_s}{r^2 (r-r_s)} and R^{\theta}_{r \theta r} is negative half of that, as well as R^{\phi}_{r \phi r}. R^r_rrr = 0 by Bianchi Identity. So adding all of them up, I get R_rr = 0, meaning that volume spanned by all other basis vectors remains constant. I'm not sure what this means because the t component cancels out the other two. So I'm wondering: what kind of quantity would capture the notion that "space gets more spacious?" Should we take the 3d spatial subspace of spacetime (thus use the 3*3 spatial submatrix of g to calculate its Ricci tensor)? Can we just remove the R^t_rtr term? Thanks!
The Ricci tensor being zero was an assumption we used in the derivation of the Schwarzschild metric. You can loosely think of this as meaning a body will undergo tidal forces where the "pulling" in the vertical direction will cancel out with the "squashing" in the horizontal direction, so that there's no overall volume change up to a certain order. The issue with "space getting more spacious" in the case of the schwarschild metric is that it's a non-local phenomenon. You can't determine that it's true just by looking at a point or the tangent space at a point. I think it requires you do an integral over a path. I'm not sure what type of mathematical object would describe this.
Hi! Can you recommend me some free pdf materials for learning physics and math by myself? And some exercise books with solved problems? I have just finished highschool and I am a med student now, but I love physics and I'd like to continue learning in my free time. I don't know from where to start. And maybe a plan of studying?
If the mass energy exists on the SR the point singularity is a inverse spherical singularity of the Obverse Universe. There is now way you could distinguish which side you where on ?like a point on a hoft sphere ?.If the particle problem in GR is not answered can anything be answered ?.
Hi eigenchris, if a object can never actually fall on to the event horizon from a far away observer point of view, then now can a black hole actually “swallow” any thing and “grow” bigger? I am pretty confused with this bit.
This is a good question, and unfortunately I don't know the answer. A lot of people have asked this, so hopefully you can find the answer elsewhere. The math tells us that an outside observer will never actually see anything cross the event horizon--it merely gets closer and closer to the horizon without crossing it.
About 13:08 f: The arrow from the "(cosh θ)² − (sinh θ)² = 1" box is mispointed; it should point at the second box-free term where this identity is first used.
excuse me teacher! inside the event horizon , why we can change sign inside the natural logarithm from r-rs to rs-r? please give me some reason . Thank you!
Both versions are acceptable solutions to the differential equation. That's all you need. If you prefer you could change the "log(r-rs)" to "log|r-rs|", where |r-rs| is the absolute value, so the result is always positive. This will take both inner and outer solutions into account.
Yes! Could never have enough of physical pictures of advanced theories like this. Things that stuck in my mind tho, Considering analogies between special and general relativity, energy-momentum tensor can effect the rate of clocks ticking just like when we move in an inertial frame with respect to the object we're measuring. How can we sure that dimensions (lengths) of material objects aren't also affected by gravity? I was thinking that the 3 diagonal components of metric (the spatial ones squared) might have something to do with this. And another one, the whole "general" thing of general relativity is that no matter how we measure lengths, time intervals, and angles, no matter what convention we use (as long as its consistent) its not going to change the result right? But can we just get rid of coordinates entirely? Is having *any* coordinate system an inconvenient truth if we want to apply general relativity in the real world?
Oh so proper length *does* spatially change! I wonder what would happen if we have a pocket of gas with volume V0, free falling into a schwarzschild black hole... will the volume expands/contracts?
I think all the laws in SR/GR can be formulated in a coordinate independent way. But in order to do practical calculations, we need to choose a coordinate system. This is kind-of-sort-of similar to classical electromagnetism where the laws of E&M (Maxwell's Equations) don't depend on the choice of electromagnetic potential.., they only depend on the E and B fields. But sometimes to do practical calculations, we need to choose a specific potential to work with.
26:53 If gravitational time dilation causes faraway observers to see in falling objects to never cross the event horizon’s surface, then how do we explain the formation of supermassive black holes in the center of galaxies when they supposedly grew larger and larger by accreting nearby stars or by merging with another galaxy’s supermassive black hole during collisions between galaxies?
I'm not sure, and I'm not sure if there's a consensus in the scientific community to this question either. I know PBS Spacetime did a video on this a few years ago. You can try watching that.
Wonderful video! I really appreciate your work. By following your interpretation, I found that maybe it is not necessary to involve chain rule to derive time dilation. But I'm not sure if it is ok to write in this way. Just take $$(d\tau)^2=g_{\mu u}dx^\mu dx^ u,$$ and set the object we want to observe to stay still, which means $dx^i=0$. The result should be the same.
And for length contraction, the integral is really performed geniusly. I use a way trying to avoid doing the integral: set the centre between r_H and r_L to be r_C. Then can write the integral as \begin{align*}\int^H_L f(r)dr&=[F(r)]^H_L\\&=F(H)-F(L)+F'(H)H-F'(L)L+o(H^2)+o(L^2)\\&\approx F(C)-F(C)+F'(C)H-F'(C)L\\&=f(C)(H-L)\end{align*} And again I'm not sure if it is appropriate to write it in this way. Sure it won't get the precise result but it can somehow prove that the length is contracted. Thank you for your excellent work again!
For the Schwarzschild radius of the Earth, if we are close to it would we experience gravitational time dilation and length extension? Is there a singularity at the centre of the Earth? Is there a singularity everywhere??
Singularities only happen when the Schwarzschild radius is larger than the physical radius, so ordinary stars and planets don't have singularities. Ordinary stars and planets do cause gravitational time dilation, though.
Note that Schwarzschild radius assuming mass be points. It is no longer work accurately when the situation is inside the physical body, its most accurate for physical compact objects like neutron star.
Eigenchris, your video's are really great!!! I watched this one the other night and I was left pondering a question. For timelike trajectories, the concept of proper time is pretty straightforward, but how do we calculate the distance traveled along a timelike trajectory from the perspective of a comoving observer. Recognizing, of course, that the comoving observer could consider himself stationary, in which case we are interested in the total distance he measures the center of mass of the massive object traveling. I haven't put effort into working out the details. Have you seen that calculation before? I'm wondering if it is as simple as L0*tau/delta t? It feels like it needs to be simple like that to make c invariant.
Every observer in relativity takes themselves to be stationary. When calculating proper lengths, you have to pick two points in spacetime, draw a spacelike curve between them, and integrate tangent vectors over the curve to calculate the proper length L0. This is what I've done when calculating radial distances near the Schwarzschild radius. It's the same strategy for any region of spacetime. Does this answer your question?
@@eigenchris Thanks for answering so quickly. My question was a little different. Let me give an example. A muon sees the distance from the upper atmosphere to the ground as being short while ground based observers see its life prolonged. In GR what is the integral used to determine the equivalent shortened path length for some time like trajectory? Thanks
@@justalittlestretch9404 Since any timelike worldline can be made "straight"/stationary by an appropriate choice of coordinates, I'm not sure how to answer this. The muon would see a length-contracted earth rushing towards it. To measure the length of the earth (or its atmosphere), you'd need to pick two points that lie on the muon's "plane" of simultaneity and integrate along the path between them. I feel like I haven't answered your question. Can you be more specific about your example?
I’m betting I’m having some fundamental misunderstanding that is causing the problem. Rather than wasting your time with another short attempt at the question, let me formalize my thinking and ask again maybe in a day. Sorry and thanks again.
@@eigenchris Hi I created a video, which has a diagram and hopefully explains my question a little better. Thanks again ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-WFXdyLyzqFI.html
Did I understand correctly that light beams which are exactly on the event horizon travel should in a circle exactly on the event horizon? Or do they get stuck in space coordinates and just sit there in one point of space? I thought that around black hole there's an orbit at 2r_s or 3r_s (I don't remember exactly), where light can orbit (except it's unstable equillibrium point)? What does it mean when you say that we can travel along e_t because it's space-like now? It means that we can travel in time behind the event horizon? It doesn't make sense to me.
1. If you are exactly on the event horizon r_s and aim a flashlight OUTWARD (in the direction of the r-coordinate), the light beam will get stuck in place 2. If you are at 1.5*r_s and aim a flashlight SIDEWAYS (in the direction of the phi-coordinate), the light beam will orbit the black hole in a circle. This is called the "photon sphere", and it is an unstable equilibrium, as you say. This will be included in my next video (108c) on Schwarzschild geodesics 3. I will try to clear this up in video 108d. Inside the event horizon, the schwarzschild coordinate names "t" and "r" do more harm than good for our understanding. You need to think of spacetime as a curved surface that exists independently of coordinates, the same way the earth exists without the concepts of longitude and latitude. Coordinates are human inventions that we place on top of curved surfaces. Just about any coordinate system for earth or spacetime is allowable, as long as coordinate curves don't self-intersect. On earth we could invent coordinate curves that goes through Canada, then goes to Europe, then zigs backs to the United States. In spacetime, we could invent a coordinate curves that go forward in time, then move sideways in space, then go back in time. These coordinate systems would be weird and confusing, but there is no mathematical reason preventing us from using them. Inside the event horizon, the Schwarzschild coordinates fall into the "weird and confusing" category where the "t" coordinate points along space and "r" coordinate points along time. Since "t" measures space, we can move along it in either direction, just like you can move left and right in real life. The "r" coordinate points towards the future, so it's impossible to avoid going into the singularity.
@@eigenchris 2. Like, photons will be like stationary waves/particles in space or something? Is this even possible to c == 0? Would someone falling into black hole see a flash of light because he would hit this "wall of stationary light"? (Technicaly I doubt it, because of accretion disc and probabillistic nature of quantum things, etc., but in this theoretical model, could he?) 3. Thank you very much for elaboratin on this! Previously I saw different explanations of Schwarzschild coordinates in a different way, like people really think that it have physical meaning, making this topic really confusing. Can't wait for next videos! p.s. thanks for tensor for begginers series, it's great, I finally covering up stuff I didn't get in university!
2. So, I've said a few times in my videos that "c" is defined as "c = dL0 / dtau", and should not be defined using coordinates, as in "c = dr/dt". In schwarzschild coordinates, the r-coordinate is not changing for the light beam on the event horizon, so "dr/dt" is zero, but that doesn't mean "dL0/dtau" is zero. I admit I haven't done the calculations in alternative coordinate systems yet, but since "dL0/tau" is a quantity that doesn't depend on coordinates, its value should be the same in all coordinate systems: "c". I honestly don't know if someone would see a flash of light when falling into a black hole. It seems reasonable to think so, but I haven't thought about it much.
Do you know of any metric that solves the einstein field equation for a black hole, but doesnt assume cosm. const. = 0 ? Since the volume goes infinite, could it have a significant influence on the solution?
The "Friedman-Lemaitre-Robertson-Walker" or "FLRW" metric is the standard metric used for an expanding universe, and it uses a non-zero cosmological constant. I'm not aware of any black hole solutions that use a non-zero cosmological constant. I think black holes are treated to be small enough that cosmological scales don't matter.
@@eigenchris Thank you. I think I am not deep enough into the matter to understand why an evergrowing volume of space in a black hole, as r→0 can be treated as "non cosmological" in size therefore irrelevant in respect to lambda. Am I missing something? Doesnt Lambda scale with the non proper volume given by the Schwarzschild metric (where V→∞ as r→0), therefore exerting infinite negative pressure inside a black hole, wich would somehow counteract gravity? Do you know of any sources that deal with Lambda in extremely curved space?
Hey Chris, what is this L0 = rΦ trying to measure here? 16:00 Because r is not allowed to vary, I thought it means a person hovering at a constant radius from mass will experience larger proper distance the more he travels angularly. But it seems the correct answer is otherwise?
@@eigenchris Got it! So closer to singularity the L0 is smaller due to downward curvature; meanwhile L0 at 17:29 appears to be length between concentric circles?🤔😬
@@greenguo1424 you can take it that way but remember spacetime is intrinsic. The math doesn’t require an outside dimension so there is actually no downwards.
Yeah. I looked at tables of common integrals and found a bunch that looked *almost* like it, but not quite. I was prepared to just tell everyone I couldn't do the integral, but my boyfriend saw I was working on it and found the right trig substitution in maybe 20 minutes. I put together the rest using wolfram alpha and a few other websites. I'm actually not much good at math problem solving, despite the fact that I'm teaching GR. I haven't seen an explicit solution anywhere else online or in any textbooks, though, so it's nice to have the solution somewhere.
@@eigenchris congrats to your boyfriend, the two first substitutions were far from being obvious, and congrats to wolfram :-) , the dirty computations that follow seem to be really a pain.
Something that always confused me, but I always brushed off, was the role of coordinates in relativity. Doesn’t a change of coordinates transform between frames of reference? Like we saw in the rindler spacetime video, it is a coordinate transformation that changes our frame from inertial to rindler. So when we say “we can remove this singularity with a change of coordinates”, does it not mean that a different frame of reference sees no singularity? But one frame of reference will still see a singularity. Do coordinates not represent actual physical reference frames?
Coordinates can sometimes line up with the proper times and proper lengths that an observer in a given reference frame would measure. In cartesian coordinates, ct lines up with cτ for constant position, and x lines up with L0 for constant time. In rindler coordaintes, x lines up with L0, but ct only lines up with cτ for the rindler observer at a constant position of c^2/α and nowhere else. In general, there is no guarantee that coordinates need to match up with either proper time or proper lengths. Coordinates are just mathematical curves we put on a surface, so they don't need to have any physical meaning. If you took the 2D surface of a table, you could draw any curves you like on it to make a coordinate system (cartesian, polar, or anything else). The same is true for spacetime. The coordinate lines don't have to mean anything. They're just arbitrary ways of identifying points in spacetime.
To follow up, the Carroll book explains the difference between the singularity at r=0 ('actual geometric blabla singularity') and the one at r=R_s (just a coordinate singularity) as follows: a scalar value can be derived from the Riemann curvature tensor (I don't think it was the Ricci scalar, but a higher order one) that goes to infinity at r=0. And as we know, scalars are independent of coordinates, so that singularity of the manifold is intrinsic to the spacetime manifold, as opposed to the one at r=R_s where it just happens that some coordinate-dependent expression diverges. I hope this is a decent explanation, I'm just learning myself. See Carroll's book for a better one:)
29:51 Why is et "spacelike" when gtt0? 26:13 The incoming geodesics are traced by the paths of light beams. How are geodesics different than worldlines since they both use spacetime diagrams?
For 29:51, this is the definition of a "spacelike" vectors (square length is negative) and "timelike" vectors (squared length is negative). I originally introduced this idea in the 104f video, I think. I'm just applying it to the e_t vector in this video. For 26:13, for massive particles, a "worldline" is any path the particle follows in spacetime. A "geodesic" is a special worldline a particle follows when there are no 4-force vectors applied to the particle (and so it follows the curvature of spacetime only, and is only affected by gravity). Light always follows light-like geodesics in all cases.
Hi, great video. I just don’t quite understand that what exactly it means by coordinate time and coordinate length anymore, can somebody explain a bit? Thanks
One their own coordinates don't mean anything. The physically meaningful quantities are proper time and proper length. You calculate those bybdo8ng integrals along a path.
Hi @@eigenchris , thanks so much for your reply. I did see you mention this in the beginning of your video but I am still having issue understanding it. For example if we really need to use the Schwarzschild Metric to calculate satellite orbiting the earth in circular motion, what is the coordinate value of r for the satellite? would that be the same thing as what we usually use in newtonian physics (km from the satellite to the the earth center)?
@@vincentxu2637 It's not exactly the same. The insane integral I do in this video shows there is a difference between the r coordinate and physical radial distance. But in earth's case there's basically no difference. It's maybe a fraction of a centimeter of a change.
Is it possible to show Spaghettification through the metric? It’s easily derived from Newtonian mechanics, but I am curious to see if maybe rather than the difference in force, the equations of motion of a body separated by of height r at time t1 can be shown after time t2. Great video really enjoyed!!
You could take two points that are falling to the center along the same axis, and calculate their geodesics. They will spread apart along the axis as time goes on (this is the typically result for tidal forces).
@@eigenchris I guess what I'm trying to understand is that since geodesics are paths where a test object doesn't feel any forces, and given a test object falling towards a massive object starts experiencing an increasing magnitude of time dilation. Then what name, other than a free fall geodesics, would you call those paths?
@@aneikei "Free-fall" and "geodesic" are basically synonyms in GR. It's often said massive objects i free-fall folllw "time-like geodesics" because their tangent vectors are time-like, and light follows light-like or "null" geodesics because their tangent vectors are light-like/null.
If proper length goes to infinity at the "coordinate singularity" rs, how does anything fall past the event horizon of a black hole? Wouldn't it have to traverse an infinite amount of space first?
Proper length doesn't go to infinity at rs. If you look at the proper length formula at 13:30 and the calculations on the following slides, you can see the space between r=rs and r=2rs has a proper length of L0 = 2.3rs, which is a finite number.
@@eigenchris Ok right. I guess I’m confused about how the radial component of the metric tensor can blow up to infinity at rs but the proper length can still be finite. Is there some calculus converge stuff I’m overlooking or something?
@@gman8563 I don't know all the details behind this, but as you mention, in calculus, when things go to infinity, they can sometimes go to infinity in a way that sums to a finite value (e.g. 1/2 + 1/4 + 1/8 + 1/16 + ...) aka "converge", or they might not sum to a finite value (e.g. 1/2 + 1/3 + 1/4 + 1/5 + ...), aka "diverge". In this case the g_rr value goes to infinity at a rate where integrals to the event horizon still converge to a finite value. I haven't studied convergence/divergence very deeply so that's all I can say.
@@eigenchris Appreciate the answer thank you, that at least clarifies where my confusion lies. EDIT: yup, just ran it through a Taylor series expansion and it does converge when r = rs.
I always thought that time and space have to dilate in the same way, so that v=s/t=c holds true. So time dilation goes to infinity at rs, and length too, but proper time is just a function of r, whereas proper length is an integration along the curved space between two points. But at the event horizon, time is frozen from outside as proper time is zero. Does the universe exist at all if one crosses the event horizon? We see that black holes get bigger in finite time, so yes. These might be philosophical questions, but as mass cannot reach the speed of light, no one can ever observe photons of the universe near the event horizon for a longer time except photons itsself (but for photons time does not exist as the don't have a resting mass). But mass cannot return from the black hole to the universe, because an infinitely long time would have to have passed. Simply fascinating how nature censors itself. And still a Hawking radiation/information comes out of the hole.
In your previous video 108a, you substitued C(r)*r = r~, from then you never convert r~ to C(r) & r, that means r, rs in this video are not really a coordinate radius of the sphere, becase of C(r) factor? as you mention RS of earth is 8mm is also not a true coordiate radius? why don't you solve for C(r) and get the original r?
That's correct. The true radius is giving by the integral I solve for at 13:30. This is basically equivalent to figuring out what the C(r) factor would do. The r-coordinate is schwarzschild coordinates is called a "reduced circumference coordinate" because it is equal to the circumference of a circle divided by ("reduced by") 2*pi. I talk a bit more about reduced circumference coordinates at the end of the 110b video.
17:31 Question about the observation of the the Lo value from an observer at 'infinity'. Does an observer at infinity see the Lo lengths as 'squished' ie Length Contracted? (in the r direction only).
I was thinking about this and I'm not sure. The answer is more obvious to me in flat space because basis vectors ("rulers") are the same everywhere in spacetime for an inertial observer. But you saw a spaceship approaching a black hole from the side, I'm not sure exactly what you'd see in terms of length effects. The way light beams from the spaceship reach your eyes would also play a role.
@@eigenchris from your description 'There's more space on the inside' I'm sorta assuming we take the length Lo and squish it into the length r , and, as an observer is in the frame of r, the observer sees the squish. But this begs the question of observed density. If the space all around a blackhole was filled with matter (like dust), would the observer see a higher density near the blackhole? Furthermore, similar to the electromagnetic special relativity effect, where electrons length contract and thus create a charge difference that makes an electromagnetic field, ...does matter around a blackhole create an observed gravitational field that is larger than the matters local field?
@@DanSternofBeyer When I was saying "larger on the inside", I was mostly referring to the fact that the "circumference/radius" ratio doesn't give the expected value of pi=3.14 that we're used to in flat space; the radius is larger than we expect compared to flat space. I don't think the "r" coordinate should be used for any physical reasoning. It doesn't correspond to an orthonormal coordinate system so I don't think anyone can ever be in the frame of "r". Again, I'm not really sure what the right answer is. In SR, "length contraction" involves two objects that are immediately next to each other, so lengths can be directly compared. But in this case, if we're talking about an observer close to the event horizon and another observer far away, they are not immediately next to each other, so we'd have to think about what exactly "length contraction" means.
@@eigenchris okay. That makes... some sense. I guess I'm assuming, since r and Lo become equal as they approach infinity, and infinity is so far away that it's basically Minkowski flat space, that r is basically a measurement in Minkowski space. But, correct me if I'm wrong, you are saying that this is not a relation between Flat Minkowski space (grey circles of r) and curved space (green circles of Lo).
@@DanSternofBeyer I'm not sure if that reasoning makes sense or not. Since I've started learning relativity, I've learned that "trusting coordinates" can very often lead to problems with reasoning, so I always to make claims using only tensors (including "true" scalars like proper time and proper length). I feel like reasoning with coordinates often becomes dicey. I think at this point I just have to admit I don't know how curvature of space looks to a distant observer.
The only reason i have a hard time following this video, its because i can't stop thinking and laughing about "Space Balls" Movie...TRUST THE SCHWARTZ LONESTARR 😂⭕
at 18:22, isn't important on this slide to recall that to stay of the Flamm's surface r>rs according to geometry of the Flamm's paraboloid? Using these algebraic expressions out of the domain of validity (governed by the geometry exterior to the body in this case) will looks like introducing an imaginary region in spacetime.
Yeah, that should only be valid for r>r_s. I guess I was implicitly assuming that was clear since the surface stops at r=r_s. The 108d video explores a more theoretical possibility where 2 Flamm's paraboloids join together to make a wormhole.
@@eigenchris Thank you, it's very interesting, I will follow-up with 108d. Otherwise, I'm still confused from a physical point of view because in 108a at 30:15, you specifically underlined that the Schwarzschild metric solution is only valid in the vacuum region outside the mass, but later in this video at 18:40 our are using it, the exact same Schwarzschild (external) metric, in the inside region (r
@@ohault My intention was to say that the "proper length" formula shown at 13:35 is only valid for r>r_s, because I only computed the integral for that region. I wasn't trying to say that the metric as a whole is invalid inside r
@@eigenchris Somehow, do you agree that there are two distinct topics here interleaved, with in one side the concrete case of a classical massive body and on the other side the special case of blackholes with physical singularities (not inherent to a coordinate system)?
@@ohault I guess the "classical" case you refer to is when the physical radius is larger than r_s, and the singularity cases is the case where the physical radius is smaller than r_s. Is that what you mean?
The "proper length" formula I derived at 13:45 is only valid for r>rs because (if I remember correctly) I used the assumption r>rs while computing the integral to get a result. The Schwarzschild metric is, theoretically, valid for all values of r (although the r=rs is a coordinate singluarity in the coordinate system here, but we can change coordinates to fix that problem). Granted, we can't actually examine the inside of a black hole in real life, so we're just letting the math guide us here. The r
@@eigenchris I'm not a math wizard but the proper length to the event horizon is infinite. Since proper length is an invariant, changing to another coordinate system shouldn't change the proper length. You can make the singularity go away but it won't change the proper length.
@@rajeshraut6447 When you say "the proper length to the event horizon is infinite", are you talking from the perspective of the origin of the coordinate system?
@@eigenchris From any point outside the event horizon, the proper length to the event horizon is infinite, because as close as you get to the event horizon, the remaining distance is still infinite, that last little stretch is bigger than the distance you have already traveled. No series of finite numbers adds up to infinity.
@@rajeshraut6447 Note sure I agree with that? At 14:22 I use the result of the integral to compute the distance between r=rs and r=rs+1. The result is a finite number (2.6*rs), which doesn't depend on the coordinates.
How can ( e^x ) * (ln(x+1)) = e^(x*ln(x+1)) is possible ? ln() is multiplied with e. How can ln() be power of e ? It is in your boyfriend's calculation. Please answer. I am doing step by step calculation. You may have skipped some steps.
Very good video.Why is a photon not describe as having a Schwarzschild radius(all energy must ?) so surely it must cross and you now have a hole-in-a-hole ?
How I wish to understand these things: i am an engineer. i have a good mathematical background, but I find it very hard to follow the reasoning well, since I was in high school, i was fascinated with the relativity and Einstein.
Have you been following the entire series so far? The entire playlist is intended to be watched in order. Although it's about a dozen hours of content before now, so it's a big commitment. Is there something in particular you don't understand.
@@eigenchris Yes, I realize that. They must be watched sequentially and slowly, and I do not do that even though I intend to. All in all, thank you so much for making this high-quality content accessible to everyone. 🙏🙏🙏
Bro that integral for the length it's very similar from the one you get when solving for the time needed for two masses to get together by gravitational attraction in Newtonian mechanics. Weird!
At 14:44 you present a function for the distance to the horizon at constant t. My calc84 app on my iphone arrives at a different result. This app can do integrals too. I checked and the results are truly different. This is what it finds: UndefinedIntegrate(1/Sqrt(1-a/x),x) = (a*Sqrt(-a/x+1))*(((1/Sqrt(-a/x+1))*ArcTanh(Sqrt(-a/x+1)))+((1/a)*x)) Where “a” stands for the Schwarzschild radius and “x” for r. Edit: I conclude your result has to be correct Edit: both results are correct
Okay as a gay in STEM, I was really touched to hear you've got a boyfriend (and thus, I'm assuming, are part of the queer community). Makes me feel very cool to know there are more of us out there.
When you want to express that two things have different sign, although it can be any sign otherwise, this is usually expressed using the symbols ± and ∓
Beware of the cranks in the comment section (it is infested with them!!). Stick to the material presented in the video and the standard texbooks (Wald, Hartle,Carroll, MWT, Weinberg). Run away very fast from any "interpretation" presented in the comments.
This is like a gravitational version of length contraction. So, your physical space, denoted as L_0, remains the same everywhere in the universe, but you might appear shorter or longer than that in some reference frames due to the distances between objects and the gravitational field.
dear fellow, I dont mean or intend to be a fun spoiler or mean to you but although the lecture is interesting with respect to the subject, your voice is extremely monotonous (and thus boring) to listen to, it is close to robotic. Could you please please please introduce more enthusiasm in to your eduction & voice that would help a lot and makes it more fun to listen to. Again I am not saying this to be rude or abbrasive or insulting but it could help.....
his voice is fine, yeah i did find myself a bit tired after watching various videos in a row, but he explains really clearly and straightforward, and i find it hard to be excited about a bunch of computational steps so idk how he'll be able to put that off. And i don't think monotonous voice is bad/boring, e.g. CGPgrey & 3b1b, but that may be just me.
@@abrahamx910 well every one is entitled to his/her own opinion. and yes he explains it very well... that is not what I commented about. unfortunately to me it is a bit monotonous sorry.. and I did NOT meant it in an offencive way but more like a constructive meant remark so he can improve this.... I dont mean it to put him down because he does a great job in explaining it. On that I agree...
