Relativity 105d: Acceleration - Twin Paradox and Proper Time Along Curves (Rindler Metric)

[@eigenchris]

Error at 25:00, the components of ex-tilde should be cosh and then sinh, but I have written them as sinh and then cosh by accident.

[@eigenchris]

Re-uploaded from a couple days ago because someone caught a math error.

[@skyfall-t8p]

Did you re-record the entire audio?

[@eigenchris]

@紅樓鍮 No, just spliced in about 10-15 seconds where I messed up.

[@skyfall-t8p]

@@eigenchris Wow, your editing is really seamless

[@eigenchris]

@@skyfall-t8p I guess I got lucky that the sound came out the same. Usually you can tell when I've recorded lines on different days.

[@azeds]

I dont know if someone told you this but man you are made to explain math The tensor series i downloaded it . Its a tresor . Finally i wish from you yhat you publish a post recommending books for all math . Thanks

[@lrlrch8351]

Thank you for all of these videos. I know that a lot of people must say this, but you deserve more subscribers. Your videos inspired me to study more physics and maths :)

[@ProfeARios]

One of the best explanations I've watched about the Twin Paradox. Thank you so much for sharing. Regards from the Republic of Panama!

[@Anonymous95202]

Panama is a beautiful country, hope I can visit someday

[@oleg-avdeev]

Thank you so much for these videos! I found very enlightening the visual demonstration of how the zig-zag non-inertial frame of reference fails to "flatten" under Lorenz Transform. I think that this is the very source of confusion that people have with the Twin's Paradox-Galilean Transform is capable of straightening these kinks, and it is our default intuition for frames of reference, so we fail to imagine that this is not a fundamental property of any Relativity, but very specifically a consequence of Galilean Relativity assumptions. I also find it cute how Galilean spacetime sits in the middle of space and Minkowsky spacetime triangle inequalities, with ||Sb1 + Sb2|| being always equal to ||Sb1|| + ||Sb2||, and how this relates to Complex, Hypercomplex and Dual numbers. Makes one wonder what a "Complex"-based (completely un-physical) relativity could look like!

[@eigenchris]

A galilean transformations can be formulated using dual numbers (e^epsilon*a), and Lorentz transformations can be formulated using split-complex numbers (e^j*phi), similar to how circular rotations are formulated using complex numbers (e^i*theta).

[@vitrums]

31:59 wow, it took some time to make it click for me. But I finally figured, that proper time measures time of events on a world line. So by looking at a space-time diagram even if we notice a segmented world line turning different ways throughout its development in time, then it still measures spatial deviation = 0 along the whole way. All we have to do is to imagine ourselves in rest along such world line and we'll start to see how it's the whole world that starts to shift left and right instead. And the cool part about this thought experiment is that we essentially become non-inertial Bob observing inertial Alice. So even though it's Alice who's in motion not us, we still measure less time than her.

[@ericbischoff9444]

27:00 lambda final -> lambda initial As always, thanks for the great video

[@patriciacosson144]

Very nice video as usual félicitation for your great job

[@Archigamers]

Thank for ur work, you are really clear , I understand a lot more of relativity compare to any other media. (Jean Pierre Petit is also really good, but he is french)

[@Mysoi123]

Thank you! now I have a clear understanding of the spacetime interval. It is a 4-position vector in spacetime that measures the distance between two events. It is distinct from proper time, just like how the integral of dR/dlambda dlambda gives us the arc length L, not R.

[@narfwhals7843]

In special relativity it is essentially the same to consider the "length" of the displacement four vector as the spacetime interval between two events. In general relativity this doesn't make sense anymore, because we can't apply displacement vectors on manifolds. So we have to integrate along a curve. And we can't really talk about the spacetime interval between two distant events anymore because this depends on the curve we chose and you can't trivially minimize this arc length due to the metric signature.

[@narfwhals7843]

I think an interesting thing to do with this video would be combining the two topics you discuss. What does the twin paradox look like with a twin who is constantly accelerating towards earth? (aka gets thrown up and falls back down) In that case the acceleration _has_ to be the cause of the age difference since there is no inertial frame to consider. This could also be a nice tie in to General Relativity later.

[@eigenchris]

I haven't done the calculation, but I'm tempted to say the twin that gets thrown up and falls back down will be the older twin, since freely falling in a gravitational field is more or less equivalent to an inertial frame. Meanwhile the twin on the ground is in the non-inertial frame since the surface of the earth is pushing up against them, forcing them to be in a non-inertial frame, and so they'd age less. But maybe I'll come back to that in a later video.

[@narfwhals7843]

@@eigenchris In special relativity earth twin should be considered inertial while the thrown twin has a gravitational acceleration on them. So during the acceleration, which is the entire trip, you can use Rindler Coordinates for the traveling twin. In General Relativity we will know that the free fall path is the one with the longest proper time so the thrown twin has to be older.

[@dantoro648]

If I understand correctly, the twin paradox cannot be solved without involving an acceleration. If Alice and Bob have different speeds relative to each other, even if they are both in a frame inertial, they will never agree on their respective proper times unless they go through the Lorentz transformation. And to compare the two clocks side by side, one of the two must accelerate or decelerate to reach the other and thus pass through a non-inertial frame. However, if Alice and Bob have constant velocities and are converging on each other, at some point they will cross paths and their proper times will coincide without any acceleration. How do we represent this on the space-time diagram?

[@narfwhals7843]

You don't _need_ acceleration to resolve the twin paradox. If you treat Bob as two different inertial frames you get the same result. The physical explanation is then that the receding and the approaching frames do not see equal halves of Alice's worldline. There is a "simultaneity gap" between the two frames.

[@eigenchris]

That's a reasonable way of looking at it, but I think it still comes down to the fact that taking two inertial worldlines that point in different directions and "gluing" them together doesn't result in an inertial worldline. The "corner" of the zigzag is technically a point of acceleration since the 4-velocity vector changes.

[@tnagel88]

Was coming here to say this! Even Feynman used this incorrect explanation, so it's pretty prevalent. Another easy way to see that it is incorrect is to also allow the 'stationary' twin to accelerate out at exactly the same amount and then instantly accelerate back as soon as she gets to whatever travelling speed the 'travelling' twin will maintain. Then, when they've rejoined, they have accelerated in exactly the same way but the 'stationary' twin will still be older because of how 'long' the worldlines are in spacetime. We don't even need to calculate the accelerations of either twin to predict the effect, we only need to measure their respective worldlines. I'm a fan of the videos! Just wanted to make sure this point was made!

[@narfwhals7843]

@@eigenchris That's true, but using the acceleration to explain it makes it look like the acceleration causes the difference. But the difference is already there if you only compare the relevant vectors without the glue-point. The acceleration is what physically causes the change of reference frame, but the fact that there are two frames is what causes the time difference. The magnitude of the difference depends on how long Bob spends on his reference frames, not how hard he accelerates. That being said you never claimed the acceleration was that important, only that Bobs point of view is not equivalent to Alice's. And that remains true. Acceleration also takes away the simultaneity gap which only results from an instantaneous change of reference frames, so including it is definitely valuable either way.

[@eigenchris]

@@tnagel88 Sorry, I'm not sure I follow your point. In your example, when you say "Alice" should "accelerate out at the exact same amount", what do you mean, exactly?

[@tnagel88]

@@eigenchris Yeah, it's kind of difficult to explain without drawing the diagram. The point was that you could have had them accelerate in the same way but follow different trajectories and the time disagreement would have persisted even though they accelerated equally. Here's probably a better way to say what I'm trying to get at: you did the math correctly, but the fact that Bob accelerated doesn't show up in the math. What shows up is the ratio of their intervals through spacetime. The acceleration is needed *in special relativity* in order to produce such a trajectory (because Minkowski ST is 'flat'), but the acceleration isn't what explains the time disagreement. It's easier to see if we make the point from the point of view of GR: consider two inertial trajectories in a curved spacetime. The trajectories could start at the same point, end at the same point, but travel along different trajectories. In general, they will disagree on time intervals (one trajectory could have a longer path through spacetime to end up at the same point), but we cannot blame the disagreement on the 'non-inertialness' of the respective trajectories because *both* are inertial. I'm currently working on a PhD, and I don't claim to be an expert, so I might have it mistaken, but that's how I understand it. Hope that helps!

[@partakerofbread]

Now see your explanation on the two definitions of proper time tau. 😊 It's not an easy topic to explain, but I think you are doing a great job! It would help if you could explain how this is a leading into general relativity, but otherwise your explanations are very thorough. JS

[@j.k.sharma3669]

Very conceptual video Sir , but I have a question ❓ If there are two bulbs on two different walls of a room , now one bulb is made on and after some time second bulb is also made on . Time between these two events is measured 2 seconds by a watch on another wall of that room . Does this 2 seconds is proper time ? We should remember that these two events did not occur at same place ( which is the condition of proper time).

[@azeds]

As usual unique content

[@signorellil]

Re-loving this

[@pedrolopa2]

these videos are great! I just wish you went into a bit more detail in the twins paradox, especially because we've learned that sr can handle acceleration in flat spacetime. How to calculate alice's proper time from bob's perspective during his acceleration? Thank you

[@gilleseveloy]

At the end of the section on the twin paradox where the conclusion is that bob reference frame is not an inertial reference frame, why can't we split the problem into 2 distinct inertial reference frames, e.g first inertial reference frame includes Bob traveling at constant speed to the right and then a second inertial reference frame where he travels at the same constant speed to the left. Why can't we decompose the problem into these 2 inertial reference frames?

[@eigenchris]

You can do that for the calculations, but physically, two initial frames put together on a spacetime diagram is not an inertial frame. If it was, Every path would be inertial because it could be broken down into straight line segments.

[@slobodannedic1799]

Related to 20:00+, it should be noted that one cannot evaluate length of a ‘line’ which is not straight or circular unless one has such parameterization to ensure that the each an ver increment is of the same length, that is the tangential velocity is constant; good indication of this is that the surface of an ellipse cannot be integrated to ab times pi - by integrating r^2 times delta-angle, unless the angle’s time-parameter dependence is not determined by the Kepler’s Equation. That ver well might be the problem in still unsolved closed form expression for the perimeter of an ellipse … !?!

[@freydrik]

Velocity is not completely relative… it is limited by c. So it seems that nature can sort things out in terms of breaking the symmetry, knowing who is moving faster, as on such absolute velocity scale. One of the major successes of SR is the explanation of the muons detected on Earth and explained by fast moving muons time dilation or proper time increase. But this could be interpreted in the reference frame of the muon as if the Earth went into it at very fast speed, near c… not affecting the muon’s proper time, but instead dilating the time of Earth (and of the entire universe actually).

[@nityadas6846]

Once again thanks for this Excellent video. Thanks , dear Eigenchris. Regards - Dr. N Das

[@massimilianodellaguzzo8571]

The video is interesting, but I disagree with what you say from 11:40 to 12:55; Alice's frame and Bob's frame are both inertial. (at constant speed) I like to solve the twin paradox with another strategy. F: Alice's frame (frame of the Earth) F_1: Bob's frame (frame of the spaceship) We know that Bob's spaceship moves in the frame of the Earth with uniform rectilinear motion at speed v. (x = v * t) At the initial time (t = t_1 = 0), the nose of the spaceship (Bob) and Alice have the same position. (x = x_1 = t = t_1 = 0) Consider the point P(-d; 0) in the frame of the spaceship, point P represents the tail of the spaceship; imagine that the Bob's spaceship has a tail and let be the length of the tail d in the frame of the spaceship. (at the initial time t = t_1 = 0) Also consider point Q(d: 0) in the frame of the Earth, point Q represents a star at distance d in the frame of the Earth. (at the initial time t = t_1 = 0) d: Earth-star distance in Alice's frame d: tail length of the spaceship in Bob's frame Consider the point R (d / gamma, 0), and think now that the point R belongs to the frame of the Earth. In the frame of the spaceship, the two events A and B representing A: the Earth reaches point P, B: the point R (d / gamma) reaches the spaceship are not simultaneous! The two events A and B are simultaneous in the frame of the Earth, but they are not simultaneous in the frame of the spaceship! Let’s analyze event A: the Earth reaches point P. In the frame of the Earth, the point P reaches the Earth at time t = d / (gamma * v) and, in the frame of the spaceship, the Earth reaches point P at time t_1 = d / (gamma * v). In the frame of the Earth, if the point P reaches the Earth (for Alice, the length of the tail is contracted), the spaceship also reaches the point R at time t = d / (gamma * v). Point R reaches the spaceship at time t_1 = d / (gamma * gamma * v), t_1 < t. When in the spaceship frame the elapsed time is t_1 = d / (gamma * gamma * v), in the frame of the Earth the elapsed time is t = d / (gamma * v). If x = v * t, it makes no sense to imagine the tail of the spaceship moving with respect to the Earth. (t = t_1 and t_1 < t, ?) Now consider the point W (-d / gamma, 0) of the spaceship frame. If we know that x = v * t, at time t_1 = d /(gamma * gamma * v) the point R reaches the origin O_1, and the origin O reaches point W. In the frame of the spaceship the origin O (the Earth) reaches the point P at time t_1 = d / (gamma * v), exactly when the point Q (the star) reaches the spaceship. Point P of the Bob's frame is a second spaceship in motion with uniform rectilinear motion at speed v in the frame of the Earth (and not the tail of the spaceship as analyzed previously), THE EARTH'S FRAME IS AT REST. The Earth is in advance of the uniform rectilinear motion x_1 = - v * t_1, because the Earth reaches point P at time t_1 = d / (gamma * v), the motion of the Earth in the frame of the spaceship is so represented by: x_1 = - gamma * v * t_1. (x_1 is not - v * t_1) The uniform rectilinear motion of the spaceship x = v * t in the frame of the Earth "forces" the Earth to move with a faster speed. In the frame of the spaceship, the Earth reaches point P at time t_1 = d / (gamma * v) and, in the frame of the Earth, the spaceship reaches point Q (the star) at time t = d / v. (t_1 < t , FOR ANY DISTANCE d) BOB'S CLOCK SLOWS DOWN RESPECT ALICE'S CLOCK, EVEN IF THE SPACESHIP DOES NOT GO BACK. (AND THE SPACESHIP CONTINUES TO TRAVEL AT CONSTANT SPEED) The frame of the Earth is not a privileged frame, the spaceship is moving in Earth's frame with uniform rectilinear motion at speed v! The spaceship actually moves with uniform rectilinear motion between any two points of the Earth’s frame (if the acceleration of the spaceship is zero). The astronaut twin leaves from the Earth to reach a star, the Earth and the star belong to the frame of the Earth! (all the other points reached by the spaceship also belong to the frame of the Earth) We cannot choose x_1 = - v * t_1, because we know that x = v * t ! x = v * t and x_1 = - v * t_1 are mutually exclusive!

[@pedrolopa2]

I just want to make sure i understand something : for the accelerating obverser, clocks that are farther away tick faster than those that are close?

[@eigenchris]

Clocks "ahead" of an accelerating observer will tick more quickly. Clocks "behind" of an accelerating observer will tick more slowly. This is sort of similar to what someone hovering above a black hole with their rockets turned on would see... clocks near the black hole are slow and clocks far away ahead of them tick fast. Although it's not a perfect correspondence and the time dilation formulas are different.

[@stewiesaidthat]

​@eigenchris you do understand that clocks measure motion in space. Not motion in time. In fact, the caesium-133 atom in the atomic clock is chilled to absolute zero to prevent it from being accelerated in time when a force is applied. You will also notice that clocks in motion use the same amount of energy as stationary clocks. Both frames experience the same amount of time. One just experiences more space is all. Third. The observer is not in the same frame as the clock as biological processes are accelerated by different forces than atomic clocks. Plants are accelerated by energy from the sun. Animals are accelerated by the food they consume which is distributed to the individual cells (another frame of reference) by the circulatory system. The cesium-133 atom is accelerated by an electromagnetic wave that is emitted at a CONSTANT frequency which the redshifts with motion. Einstein’s Relativity is junk science. The properties of light make it an absolute reference frame. Newton's Laws of Motion make ACCELERATION the reference. This is why UV rays cause premature skin aging of the exposed skin and not the whole person's body. Quit putting out these garbage science videos and learn what a proper frame of reference is.

[@antontabjopko6434]

Your minkowski diagram is in error. The path on right is ok. But for the path on the left, the paths dont meet on ct axis. Rather the left path reaches ct axis lower than where right path reaches ct axis. Similarly the distance the left path on the x axiz is shorter than the x distance on right. I have a diagram to illustrate this but i dont know where to send it to you.

[@signorellil]

Hi Chris, as other people said, no need to include acceleration to solve the "paradox" (which is not really a paradox). In the context of SR saying that Bob will experience acceleration makes the "paradox" trivial, but you can build a scenario where there are three people (Alice standing on Earth, Bob and Chris passing by on two starships flying in opposite direction at constant speed) who gives the same result as the classic twin paradox. The point is, the "relativity principle" just says that for every inertial frame, you can find a set of coordinates for which the frame is at rest and inertia is homogenous and isotropic in every direction. Just this. This does not imply that "all inertial movement is relative", as the popular vision goes. It means that Newton's laws of motions don't allow you any way to discover if you're moving or not in an "absolute" sense. But beyond this SR is not a relational theory - it does not say "all movement is relative". In fact spacetime in SR is "absolute"; so, if like in the twin paradox you follow two different routes in spacetime to go back and forth from the same point, one will always provide the longest proper time. No way around it. And this doesn't contradict the "relativity principle" because taken individually all inertial frames can be seen at rest and inertia will be homogenous and isotropic in all directions. So SR is not a relational theory, and inertia is a preferred frame of reference - it’s “different” in a very specific way. This is a epistemological problem of SR (not a logical contradiction!) and disturbed Einstein enough to be one of the reasons he began his search for GR and a theory that could explan inertia. For some time he believed that Mach’s Principle could be right and so inertia (and the fact that moving faster through spacetime generates a shorter proper time) could be explained by the distribution of all the masses in the universe, and so GR could become a fully relational theory. This failed - GR is still not a relational theory, it doesn’t explain inertia, and spacetime are “absolute” in it. It is a fascinating tale - read Kevin Brown’s “Reflections on Relativity” more more insights on this - it’s an advanced text but it’s incredibly deep.

[@eigenchris]

I'm confused why people are saying acceleration is not required to solve the twin paradox. As far as your 2nd paragraph goes, you seem to be repeating what I said in the video, and so I agree with it. As far as I understand, the only way to determine if your own frame is inertial or not is to have an accelerometer with you. If you accelerometer reads "zero", then you're in an inertial frame. If it reads non-zero at any point, you're worldline is non-inertial. My point is that the reason Alice and Bob's worldlines cannot be considered equivalent is because Bob, at one point, reads a non-zero value on his accelerometer, and thus his worldline is not inertial. Could you explain your example in the first paragraph in more detail? I'm not able to visualize it.

[@signorellil]

@@eigenchris Here's a description of the scenario, taken from a paper available on ArXiv. "It is possible to invent a second scenario also compatible with the Special Relativity but without intervention of neither accelerations nor decelerations. The traveler gains his velocity in the region x < 0 and moves in the direction x > 0 with constant velocity V. He passes in front of the earth observer located at x = 0 with that velocity. At the coincidence of the two observers, they adjust their clocks to zero. The traveler goes on until he crosses another missile moving with velocity − V exactly at the point he wanted to reach. Now, he jumps into this new missile to come back to earth and the clock of the reference frame of this new missile is adjusted to that of the first missile. When he passes again in front of the earth observer, they compare their clocks. The traveler does not begin his motion at x = 0 nor acceleration

[@signorellil]

@@eigenchris There's also a graphical description of this scenario in this video (unfortunately the "explanation" that follows is very bad). The scenario is shown from 09:21 kzbin.info/www/bejne/pKDEeIGrhqiKbqM

[@signorellil]

@@eigenchris I think you give a simple and beautiful explanation at 13:21. It is not a matter of acceleration or not - it's just a matter of geometry. Alice is in a single inertial frame, and no transformation will magically change in two different frames with different directions. That's not how Poincaré symmetries work. So any attempt to transform Bob two different frames into Alice single frame wont work.

[@eigenchris]

@@signorellil I appreciate the links, however I feel like we are "splitting hairs" here somewhat. I don't believe I've contradicted that excerpt from that ArXiv paper, or the FermiLab video. I think we both agree, the twin paradox is basically just a more common way of saying "the reversed triangle inequality is obeyed in Monkwoski space for timelike vectors". I agree with that. The point I was trying to get across in the video is to say that Alice's frame is not equivalent to Bob's frame, because this is the point of confusion that I think most people get tripped up on. Bob's frame is two inertial frames "glued together", which when taken as a whole is a non-inertial frame. I feel like we agree on this point. Now, the reason his frame is non-inertial is because he experiences a non-zero accelerometer reading at the "bend" in his worldline. I feel like you can agree with me there as well. Now, the question is: "is the twin paradox caused by acceleration"? I guess it depends on how you define the twin paradox. If the younger twin is a single observer only, as in the case of my video, then I would say acceleration is definitely needed. If you allow the younger twin to really be two people in two different frames, then with Fermilab's reasoning you can come up with a way of expressing the "reversed triangle inequality" concept without appealing to acceleration. Does it sound like we're on the same page?

[@fkeyvan]

Here is another vexing problem. In the case of curved wordlines like Rindler's, you mentioned MCRF or instantaneous inertial frames. The whole concept of instantaneous velocities or instantaneous shift in wordlines tangent to the curve defy the axioms of SR. Therefore, one cannot assume that smooth curved wordlines have any physical meaning. They are only mathematical constructs.

[@eigenchris]

I'm not sure why the concept of instantaneous velocities would violate SR? It's just a tangent vector along a curve.

[@fkeyvan]

@@eigenchris A curved wordline requires instantaneous change in instantaneous velocity tangent to the curve. It also requires instantaneous acceleration. That is because curved lines are smooth, differentiable and continuous.

[@eigenchris]

Right. In the case of the twin paradox in this video, the worldline is not smooth, so it is not technically possible to traverse it in real life. Realistically, you would need to change your speed continuously. But this is just a simple example to show what the geometry of special relativity is like, and how the reversed triangle inequality works.

[@fkeyvan]

@@eigenchris so you agree that a curved word line is physically impossible in SR . But it is fine in Newtonian physics. I have no disagreement with the math. It’s the physics that is the problem

[@eigenchris]

I'm fine with curved worldlines in SR. I just was pointing point that "sharp angles", like the one in this video's example, are not allowed because they have infinite acceleration.

[@A-_--]

@eigenchris It's more of a theoretical argument than anything, I'm not a cosmologist but from my understanding the exact shape of the universe is still up for debate. In fact the same thing would hold for a 3-cylinder and various other shapes. As for a solution, It is pretty obvious in hindsight that you'll still need to accelerate when leaving and returning to the earth, but there would be one less acceleration than in the normal case, so it can't quite be it. If I were to do a loop around the universe and start time when I pass the earth, and someone else on earth did the same, the clocks would show that less time has passed for me when I return to earth again, even if I never stop when I pass by. How does this work? Well, in a toroidal universe there actually turns out to be a prefered refrence frame, namely one in wich shooting two light beams in opposite directions makes it so the beams return at the same time, that is if the universe is toroidal with the lorentz metric.

[@A-_--]

​ @eigenchris For some reason my replies aren't appearing (thanks youtube) so I'll post it here instead. A topological 3-torus can still be flat, but yes, a similar situation could arise if the universe has (roughly) constant positive curvature. Maybe you missunderstood my post, but I wasn't talking about any of the twins being in the vicinity of a massive body per se (well, I guess the twin on eath is), but rather the large scale structure of the universe. If the universe were a flat 3-torus, you could walk in a straight line and return to the place you started at.

[@eigenchris]

That seems like a pretty specific thing to believe. Do we have any reason to believe the universe is a 3-torus?

[@WildGamez]

Very interesting that frames of reference are mathematical...whether minkowski or rindler

[@apolloniuspergus9295]

What if both of them are inertial frames, travelling at different speeds and communicating at the speed of light? One would see the other as moving and oneself as being still symmetrically, yet they would disagree as to who has aged more, as much as they disagree about who is moving. Wouldn't they?

[@eigenchris]

They will each think they are stationary and the other is moving, yes. But the key to the twin paradox is that one of the twin's worldlines is non-inertial (they will experience a "proper acceleration" on their accelerometer at some point on the journey. This is an objective way of differentiating the twins. The twin with a non-zero accelerometer reading will always be the younger twin.

[@apolloniuspergus9295]

@@eigenchris Well, I am talking about a case where they never stop or change directions and just send informations to the other through radio waves or anything alike. It does not seem necessary that any of them experience proper acceleration if they never have to meet again, just communicate indirectly

[@eigenchris]

@@apolloniuspergus9295 Sorry, I misunderstood. So just two inertial frames being involved? Yes, each will think the other has aged less. This is possible because each observer "slices up" spacetime into planes of constant time differently.

[@apolloniuspergus9295]

@@eigenchris I am now confused, isn't that precisely the claim of the paradox?

[@eigenchris]

I think the "twin paradox" (at least as it's described on Wikipedia) involves one twin doing a there-and-back journey in space. So the twins start at the same point, separate, and then end at the same point, where they have the opportunity to compare their clocks. One twin will have aged less. The "paradox" comes from the idea that, since each twin sees the other as moving, both should see the other as having aged less. But this is a misunderstanding because (as I explained above), one twin must have a non-zero accelerometer reading in order for them to rejoin the other twin, which breaks the "symmetry" of the twin's journeys in spacetime. In the case of two twins moving away from each other in inertial frames, never to see each other again... they can't directly compare their clocks since they are separated from one another. They can use indirect methods of communication like light signals, and each twin will believe the other twin's clock is ticking more slowly. There's a perfectly sensible way to draw this out on a spacetime diagram, as seen on the "v-shaped diagram" on this section of the Wikipedia article on time dilation: en.wikipedia.org/wiki/Time_dilation#Reciprocity Does this answer your question?

[@slobodannedic1799]

Around 2:00 - the proper time pertains only clock with the observer at the beginning of the coordinate system, right!? Around 14:00 - shouldn’t be both reference systems/observers non-inertial, or at least to exchange the texts in the boxes on the right-hand side?!

[@eigenchris]

Any curve drawn on a spacetime diagram can be assigned a proper time. Proper time is basically like the "length" of a curve on a spacetime diagram, similar to how any curve you draw on a sheet of paper has a length. The main difference is that in spacetime we use a metric that has some negative diagonal components. For 14:00, I don't understand why you think both systems should be non-intertial (I'm neglecting gravity here). I'm saying Alice's frame is inertial because (ignoring gravity) she would measure zero on an accelerometer throughout her journey.

[@slobodannedic1799]

In that case you would proclaim the Alice’s inertial reference frame the absolute reference frame !? In such case RT becomes obsolete, for one thing, and for the other - based on my recent findings/insights even in “flat-space” geometry, when it comes to transformation of coordinates from say rectangular or polar ones, the so called non-inertial (two-components) radial and transverse accelerations are just the rotationally invariant forms of those in the rectangular coordinates … so, in general, what should be differentiated is between a position vector and the vector (velocity, acceleration, etc.) AT THE POINT, no matter if Christoffer symbols are used to attain the invariance … ok, this might lead far away from the topic here, but by all means I would appreciate if you would take a look at my transcripted presentation (ResearchGate or AcademiaEdu) titled “Problematizing of the Orbital Mechanics Two First Integrals”.

[@eigenchris]

I'm not saying Alice's frame is "absolute", but I do think there is a class of "special" reference frames called "inertial reference frames", where an accelerometer will measure zero. I don't quite follow the rest of your comment. Can you explain again what you mean by radial/transverse accelerations?

[@slobodannedic1799]

@@eigenchris Besides possibly inadequately ‘mixing’ kinematic sand dynamics, it should be aware that if even be functional accelerometers (as gyroscopes - for example) need a ‘substrate’ to interact with, and here besides being just impulsively excited with enormous energy for whatever “material point’s mass” needed energy) at the middle of journey, that should be the case (with the half of that amount) at both the staring and ending positions … How would it be with both Alice and Bob setting of for journeys symmetrically, a kind of analogously with Bell’s spaceships. You know, if a theory - either mathematical or physical - has one contradiction, than everything can be proved if one truly ‘wants’; the point should rather be to attempt refutation (not even wanting to lose job, as myself as an electronic and telecommunications engineer happened to experience in SiRF in California with having the GPS signals multipath measurements clash with the Einstein’s 2nd postulate:). Subsequently, to get back to your question, unexpectedly I was in situation to gradually arrive at refutation of the foundational basis of the Newtonian mechanics and physics, which are basis for all the contemporary theories, and thus automatically all being flawed: under the radial component of acceleration I meant the two-term expression r’’ - r times (Phy)^2, wherein the second part is termed the ‘fictitious’ centrifugal acceleration … In your very appreciable videos, reflecting - I would say/guess - the current state of affairs in physics, there is apparent ‘transitioning’ between the notion of two inertial referent frames of whatever space-time event, putting aside the unfortunate “primarity” of incomplete Maxwell’s Equations, to the notion of the so-called non-inertial referent systems, as in case of the Rindler’s coordinates for which it is not clear if they represent coordinates with respect to a common (not to insist - inertial/absolute) coordinate system, or the coordinates of “vectors at the pointiserve ” !? Anyway, if I have found out that the so-called non-inertial polar coordinate system does at all not ‘deserve’ such characterization, but simple the referent system for the straightforwardly rotationally invariant (without any reliance on the Christoffel’s symbols) components of - velocity, acceleration, etc. - vectors at the point, what then can potentially turn out to be in the domain of GRT and the covariant differentiation/integration. The influence of the (G)RT is such that Christopher-symbols are officially considered indispensable for technical domains with extremely non-relativistic velocities, and my main intention has been to a kind of ‘expose’ that mistaking/misconception, and if even that would be possible within the RTs. In that sense, I would propose to continue interacting through e-mails - at least for me to more efficiently communicate to you my comments/questions, answering those of yours, and - to possibly work together towards such a goal !? my e-mail address is [e-mail]

[@TheAdithya1991]

How can I support this channel?

[@eigenchris]

If you want, you can leave me a tip here: ko-fi.com/eigenchris Or you can just donate the money to a charity of your choice. Both are good with me.

[@warfyaa6143]

Very useful, thanks alot. @ 25:00 ،derivatives of vector components of et tilda are swapped.

[@eigenchris]

Yes, my bad.

[@mohdzikrya5396]

32:27 will an incoming traveller to earth experience slowly ageing if he comes to earth with constant speed unlike the one who goes out and then comes back. I am feeling it hard to convince myself about the traveller in this lecture being in a none inertial frame.

[@mikegale9757]

The traveller changes headings at the waypoint so the inbound and outbound segments of the journey occur in different inertial frames. But SR does indeed give ambiguous answers for one-way trips. That ambiguity leads many to believe that moving clocks run slower than one another. The discrepancy, they say, is due to the subjective nature of "now" over there. That's what Einstein was talking about with his stubbornly persistent illusions. But there are practical examples of one-way trips at near light speed. High-speed muons are created by gamma rays colliding with atmospheric particles at high altitudes for example. They can be detected on the ground even though their flight time exceeds their lifespan at slower speeds in the lab (hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html.) That's only possible if flight-time

[@ibrahimalsamak5268]

İs this a reupload?

[@joshuapasa4229]

Yh

[@se7964]

SR certainly doesn’t treat inertial and non-inertial frames the same way, but it also doesn’t offer any real insight into what distinguishes the inertial from the non-inertial frame, so on some level it’s fair to argue that there’s still a paradox. Also fun fact: in certain curved spacetimes, the “straight line” geodesic is no longer necessarily the longest path between two spacetime points, and so you can have an inertial twin age less than a non-inertial one.

[@eigenchris]

Non-inertial frames will see their accelerometer measure a non-zero value at some point, and for inertial observers the accelerometer will always measure zero. Can yoy give me an example in GR for when the inertial twin will age less?

[@se7964]

@@eigenchris If you place a twin in orbital free-fall around a mass in a Schwarzschild-like metric, then they’ll age less than a twin who accelerates continually in order to remain stationary relative to that mass. An accelerometer is an import from classical Newtonian thinking. I’d say its reliance on the idea of absolute acceleration is problematic, especially since it only measures the acceleration of the instrument relative to the observer, who then interprets this acceleration to be invariant for all other observers via logical inferences based on their familiarity with a global class of systems. This is why, if you restrict your view to only a few simple systems, you’re likely to conclude, as Einstein did in the preface to his 1916 GR paper, that defining acceleration seems to require arbitrarily privileging one system over the other. Of course the GR twin scenarios seem to indicate that acceleration isn’t relevant to the paradox at all, other than being a requirement for making a round-trip in flat spacetime. But still one can wonder where in the formalism of SR one can derive a notion of acceleration if they only have a few simple frames to work with from the outset.

[@eigenchris]

I admit I haven't done that calculation in the Schwartzschild metric, but my guess is that free-fall geodesics always have the longest proper time in relatively small regions of spacetime, but this rule might not apply globally.... similar to how a spatial geodesic on a sphere will give the shortest two distance between any two points in a small enough portion of the sphere, but geodesics can also take "the long way around" and not give you the shortest distance. There's several different meanings that "acceleration" can have. There's newtonian 3-acceleration, the 4-accleration vector from SR and GR, and also the proper acceleration (which is a scalar, and is the "length" of the 4-acceleration vector). Trying to find an absolute meaning in 3-acceleration is hopeless, I agree, but my understanding is that 4-acceleration (and it's length, the proper acceleration) are absolute quantities. The proper acceleration is just what you measure on an accelerometer you're carrying with you, and it's an absolute/invariant/objective quantity, just like proper time. So I feel accelerometers are still very important in SR/GR, not just Newtonian physics. I agree the fundamentals of twin paradoxes is not acceleration necessarily, but the geometry of spacetime and the amount of proper time you accumulate while travelling along a worldline. But as you say acceleration may play an incidental role in the shape of the worldlines involved.

[@se7964]

@@eigenchris I’m not positive on this point, but I think that the maximal spacetime path between any two events will always be an inertial geodesic, even from a global standpoint - this doesn’t mean however, like you pointed out with the globe example, that there won’t be other shorter geodesics also connecting the two spacetime events. 4-acceleration is certainly postulated to be, and treated as, invariant by SR/GR, but the fact is that its definition is still ultimately based in 3-acceleration. This is because, since 4-acceleration is an acceleration due to any force which is not gravity, it still requires, in order to be identified, a 3-acceleration measurement. Breaking this down further: calculating 4-acceleration requires first calculating the amount of proper time elapsed (whether over an infinitesimal or approximate interval). But how do we determine the amount of proper time that has elapsed on a clock? By, as your video explains, knowing the length of the spacetime path which that clock travels. So then we next have to ask: how do we determine the length of the spacetime path which the clock has traveled? In flat spacetime, as again demonstrated in your video, we add up all the lengths of the straight segments - approximating curves with small segments if need-be, in order to get the total length. But then the “notches” or cusps in the path ultimately determine its make-up and total length. So then we ask, what determines these notches? This is where the formalism of SR breaks down and gives no real answer - like you assert, an accelerometer has to be empirically invoked to determine that a “notch” in a spacetime path has occurred. But again, an accelerometer only takes a 3-acceleration measurement of an instrument, and then from that measurement extrapolates to impose the idea of “absolute” movement. Thus, the concept of 4-acceleration is still innately derived from 3-acceleration.

[@krzysztofciuba271]

@@se7964 ??? acceleration - an absolute property??? Do you know the definition of acceleration? An accelerometer introduced into the system (of two bodies twins) for what? C. Moeller almost openly admitted his "trick" (only on one twin acts....a force- from a blue sky): let's assume (!) that only one gay/twin is under gravitational field and not the other! "Assume"- on what ground? The equivalence principle- have you heard about it? One can speak in metaphor only that if a closed system of bodies has gravitation and potential then we can make it equivalent to an acceleration effect. No wonder A./eintein finally rejected his own 1918 argument, but his 1905 with "peculiar" (!) consequences" is still ...fantasy. Do not forget that u are travelling with the speed of almost light in reference to the observer at the Frontier of Universe and your mass is... growing into ..infinity -according to your type of interpretation of Lorentz formula (and you are squeezed to tiny sheet in the direction of motion. Amazing - a proof of global idiocy in now 5 generations since AD 1911 (even A.Eisntein,1905)

[@mohdzikrya5396]

Can proper time be considered as invariant?

[@eigenchris]

Yes, proper time is a scalar/tensor/invariant quantity. Everyone in all reference frames will agree on the proper time for a given worldline segment.

[@VortekStarling]

If what you described were actually correct, it would invalidate the entire premise of relativity, that there's no preferred frame. Einstein didn't actually say it depended on the path you take through spacetime though. He said that going from point A to Point B in a straight line is the same as going on a polygonal path, or even a circular path, with the start and end points being the same. See for yourself, from his 1905 article: "From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2tv2/c2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B. It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide. If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be 1/2 tv2/c2 second slow." We now know that a circular path could involve a Sagnac effect, and/or gravity-like effects (centripetal acceleration), but Einstein didn't involve those factors so neither will we for this particular thought experiment. So what's the difference between the rocket going in a circular path, say clockwise, and the earth going in a circular path counterclockwise? Either way, they both start and stop at the same place.

[@Mysoi123]

Two different frames have two different planes of simultaneity, breaking the relative motion symmetry. They don't agree on the order of events; Lorentz transformations only apply to a single frame with a single plane of simultaneity. Rindler coordinates can handle an infinite number of planes of simultaneity at different moments for constant proper acceleration, and you can be considered stationary in that coordinate. However, your proper acceleration indicates that your frame is non-inertial. In General Relativity, although you may perceive yourself to be always standing still and believe there is a fictitious gravitational field when you accelerated, the mathematical results are still completely equivalent to Rindler coordinates."

[@VortekStarling]

@@Mysoi123 The speed of light is not constant though, only gravity is constant, light is a slave to gravity, not the other way around. Why would anyone want to use light as a time keeper when it's clearly not at all stable? A cesium clock is timed by microwave frequency and it's easy enough to find experiments showing that the time shown is different at the top of a mountain than at the bottom, therefore cesium clocks have been conclusively proven unreliable because they're not based on the real constant of time, gravity, meaning the amount of time it takes one mass to orbit another while covering a particular area. Mass is constant, therefore gravity is constant and the time for orbits are constant, the speed of light is not constant.

[@Mysoi123]

@@VortekStarling in an inertial reference frame, it is constant.

[@VortekStarling]

@@Mysoi123 Yeah but how often are you in an inertial frame with no gravity source around?

[@Mysoi123]

@@VortekStarling In the realm of mathematics, simply remove all the mass, and you will revert to the Minkowski metric. This metric describes an empty spacetime of special relativity, wherein the speed of light remains constant. Although in general relativity, the speed of light is not constant, the formulas for the speed of light at each point in space still incorporate the factor "c." That is precisely why "c" continues to be a fundamental part of the universe.

[@redbel2624]

I watched your entire series on tensors, and now I will start with general relativity, but I think that videos are not enough. I need a lot of exercises to understand more. Can you give us links to exercises

[@MudahnyaFizik]

You should try hartle gravity (easier for beginners) or caroll's GR (a little harder)

[@fkeyvan]

Hello. Here is another issue. In this same video you negate yourself. You mention that an accelerating Rindler in a curved world line have MCRF or instantaneous inertial frames than can undergo LT to Einstein's stationary frame. Correct? Then why did you say that this same LT from the curved path cannot be applied in case of the twin paradox to the worldline of the rocket?

[@eigenchris]

You can apply an LT to any instantaneous point on the non-inertial observer's worldline, but you can't find a global LT that matches the entire worldline (since it is broken up into 2 segments).

[@fkeyvan]

@@eigenchris that's exactly my point. why can't you apply LT on each leg of the rocket's inertial frames? Then you will have that in the rest framd of the rocket, it's the earthbound twin that has 2 inertial frames and therefore a longer wordline.

[@eigenchris]

@@fkeyvan Each "leg" of the path is individually inertial, but the complete path is non-inertial. You could think of the Rindler path as being a long sequence of extremely short inertial paths, but the overall path is non-inertial. You can't find a global Lorentz transformation that takes you from an inertial path to the "bent" path, so the bent path is non-inertial.

[@fkeyvan]

@@eigenchris , yes i understand that. my point is you can apply LT to each leg of the rocket frame which are inertial. Then in the stationary frame of the rocket you will have the earth going away and returning in 2 inertial frames. Is that correct? If so then in the frame of the rocket you measure less time has passed for the earth bound twin. The paradox cannot be resolved as long as the twins are in relative motion.

[@onebronx]

@@fkeyvan You do not apply LT to individual legs, it is nonsence. You apply LT to the whole spacetime, the whole world. And there is no combination of LTs that transform the spacetime so that two legs become a single straight line. As soon you apply an LT to move the second leg onto the time axis, the first leg will move away from it, and the Bob's worldline will remain consisting of two sections with an angle between them.

[@sleepy314]

10:00 remember the sign of the coefficient for time dimension is opposite of the coefficients for the space dimensions. that is why proper time is shorter.

[@-_Nuke_-]

SURE if we beforehand agree that the Earth is inertial then the other is non-inertial... But what about 2 objects that we don't know if they are inertial or not!? That is the essence of the paradox and it can't be resolved.

[@eigenchris]

If you are carrying an accelerometer with you (you can think of it as a box containing a ball, connected by springs along the x/y/z-axes) you will be able to tell which direction you are accelerating in by looking at the compression of the springs.

[@-_Nuke_-]

@@eigenchris So that means that we can always tell if we are accelerating or not... Which would also mean that we can always tell what our exact speed through spacetime is... Something that I asked in physics stack exchange and they said that we can't have any instrument on board that can tell us what our speed through spacetime is... I can't seem to be able to link the particular post from stack exchange but if you want to read it its called "Spaceships with relativistic instruments" Anyway! If I can always tell that I accelerated, then I can then measure how much I accelerated. And thus, when I cut off the acceleration and go back to being inertial, I can always know that my speed through spacetime is diffrent to what it was before I accelerated... If for example I accelerated at 60% C. Then my speed through time is 20% less than before - or in other words, I know that I experience time dilation of the factor of 20%... But the people in my post said that I can't do that! I'm confused :D

[@-_Nuke_-]

@@eigenchris Imagine that you have an empty Universe and 2 objects in in. The rest of the space is absolutely void of everything else. So no planets, no backrounds stars no nothing, just those 2 objects. Then I ask: If A sees B as moving from his refference frame close to the speed of light. And if B sees A as moving from his refference frame close to the speed of light - then who will ages slower? The answer might be something like - depends on who will turn around to meet the other. But that doesn't reveal the crux of the matter. And the crux is... Lets assume that they can communicate instantly with eachother. Who is aging less then? Because aging isn't something that depends (or should depend) on communication. Whether or not I'm able to communicate with someone is irelevant to if I can age more or less than him. And here we find 2 answers that I don't think that anyone alive today knows which one is the correct one: Possible Answer #1) If they can communicate instantly then they will find that they have the exact same age. Possible Answer #2) Or, one of them is definatelly older. The instant communication will reveal it, but withought it we can never know. Who is older will depend on who accelerated to reach the speed of light in the 1st place. We might have forgotten who that was (or we don't know) but DEFINATELLY someone did. You can't just reach the speed of light withought burning energy. Sure either A or B can be thought as moving at close to light speed due to being stationary and the other one moving at close to light speed. But that's just our ignorance - and our ignorance doesn't negate the fact that SOMEONE AT SOME POINT IN THE PAST REALLY DID BURN ENERGY TO ACCELERATE. (We are in flat spacetime so no gravity involved to make this even more complicated)... So the Twin Paradox can only be solved if we can measure the worldlines of the 2 objects. But there are cases that we won't be able to do that, because of how little information we have about the system. Especially if none of them turns around (burns energy) to meet the other. When our memory is erased and we don't know who accelerated to reach the speed of light and thus create such a big gap in A and Bs relative velocities - then either their age is still the same - or it isn't and we just can't know which one was the accelerating one. I personally believe the second answer, but it might just be the 1st one, I don't know. But the paradox definatelly can't be resolved under all circumstances.

[@freyc1]

​@@-_Nuke_- What do you mean by "my speed through spacetime"?

[@-_Nuke_-]

@@freyc1 Our speed through spacetime is the speed of light. I have taken this idea by science click, as he himself has said, when you accelerate you just change how you are moving through spacetime. You can allocate more or less of your speed in time to make it speed in space and visa versa... If that is true, then light is traveling at 100%C in space and 0%C in time... And things that are in freefall are traveling 0%C in space and 100%C in time. Anything else will travel at some percentage of C in time and some other percentage of C in space. If you are moving in space with a speed of, for example... 30%C... Then you won't be ageing at the speed of 100%C anymore... You will be ageing a little slower, and we can see how much slower if we solve the Lorentz factor γ = 1/sqr(1-(v^2/c^2)) where v is your speed in space as measured by a far away stationary observer, and c the speed of light... It seems like the only way to be truly static in space (if we forget about the expansion of the universe) and at the same time ageing at 100%C in time is to be in freefall. If that's true, then we definitely do have many absolute reference frames and that's honestly the true solution to the twin paradox... Because if I phrase the twin paradox differently, it might not have a solution! "Imagine an empty universe with only 2 spheres in it. Sphere A and sphere B such as, that from the perspective of A, B is accelerating towards light speed, turns around and comes back, while from the perspective of B, A accelerates towards light speed, turns around and comes back. Which ball at the end will have aged less A or B?" This, modified twin paradox, has NO solution. We somehow need to know which one is in freefall on some planet or something, in order to know which one will age more or less than the other;

[@seanmchugh840]

So what of two objects travelling and accelerating equally, making them entirely symmetrical, as with the galaxies in the large scale universe? There is a serious problem here.

[@eigenchris]

As in, if the worldlines of alice and bob form a "diamond" shape? Then they will age equally. You can measure the proper time of their worldlines using the method in this video.

[@seanmchugh840]

@@eigenchris I understand but here we're beginning to stray away from the Lorentz transformations of mass-energy and spacetime to account for light speed and laws of physics staying the same, this being the heart of relativity theory. Neither of the objects need to be accelerating, decelerating or returning, just that who it was who is 'travelling' is not determinable. You're trying to say that two galaxies moving at 50%c, each in opposite directions from us, or just our galaxy and each of those galaxies, don't really have the regular transformations that are directly observed by local high speed objects, such as in particle accelerators.

[@eigenchris]

@@seanmchugh840 I'm sorry, but I don't understand the argument you're making. Can you describe your example with the two galaxies in more detail? What would the spacetime diagram look like, and why do you see it as a problem?

[@seanmchugh840]

@@eigenchris Hi. You’re trying to answer the Twin paradox in saying the traveller’s perspective is a kind of illusion due to their direction changes or acceleration that the stationary observer doesn’t have. Okay but now we need to explain about two observers who are entirely symmetrical, in relative motion with or without acceleration, and being in their frames since the beginning of time, like the galaxies or galaxy clusters- the entire universe is in this state. If you’re saying that their relative motion does not cause the Lorentz transformations, that are essential to keep light speed and physics and reality intact, do not take place and that instead both observers are now under illusion, and with neither object’s clocks being slow or fast, then relativity theory is incorrect in some important non-local way. The ladder paradox as looking at space rather than time transformations shows the same issue. There’s something very wrong here. I’m also reminded of the famous experiment I think in the 1980s that looked to the need for a cosmic ether.

[@eigenchris]

I don't think I'm saying anything is an illusion. My main point is that different paths through spacetime will have different proper times... and you measure these proper times according to the metric for whichever coordinate system that you're in. The same applies for galaxies. Just take their worldline, measure its proper times using the metric, and you are done. Saying "this person's clock runs slow" or "that person's clock runs fast" means you are making a comparison between two clocks. Bob's clock might run slow compared to Alice's, but it might run fast compared to someone else's. I would focus less on the question "does a clock run slow?" and focus more on the question of "what is this observer's proper time?". If you have 20 observers, just calculate all their proper times using the method shown in this video. After that you can worry about whose clock is slow and whose clock is fast.

[@-_Nuke_-]

At 14:00 the paradox is not resolved! Sure you can't perform a Lorentz transformation ok! But who is to say that Bob is the one that accelerate!? Did we knew that beforehand? Do we infere it because his vehicle has an engine on it while the Earth is just a spherical ball!? How do we know that Bob is the one that's moving!? Here is a twin paradox: 1) 2 objects A and B, are close by. 2) From the perspective of A, B is moving away from A, and then coming back while having traveled close to the speed of light. 3) From the perspective of B, A is moving away from B, and then coming back while having traveled close to the speed of light. Which one will age more or less? That paradox DOESN'T seem to have a resolution. The resolution to the paradox is ALWAYS solved beforehand, because we already know who is the planet and who is the spaceship. If we can't know that - we can't solve the paradox, because it has no solution. Relativity can't answer that question. The only thing that can answer that question, is we had some way of knowing more about the systems energy and entropy.

[@Torbu6286]

So it's not about refrence frames but about the fact that one travels near the speed of light which slows down the time for them.... I can make sense of how that happens with a light clock but I can't comprehend how it can happen with a biological body or a mechanical geared clock. That's what I wanna know to understand it.

[@tea3er250]

Thank you for your excellent video. I'm learning Relativity taking a note. I notice some errors: 22:15 In the figure both lambdas have "final" subscript. 29:45 "et" and "ex" vectors in the figure should have tilde, I think.

[@fkeyvan]

You explanation of the Twin Paradox seems incorrect. One of your reasons is that Bob’s zig zag world line cannot be straightened by a Lorentz transform. Same thing can be said about Alice’s path which would be a zig zag from Bob’s viewpoint. Another false reason is that you state Alice is an an inertial frame but Bob is in a non-inertial frame. That’s only from Alice’s viewpoint. From Bob’s frame it is Alice that decelerates and turns around. So Bob could say just as accurately that Alice is in an accelerating non inertial frame.

[@eigenchris]

Only one of them will have an accelerometer that measures a non-zero number at some point on the trip. The twin with the non-zero accelerometer reading is the one in the non-inertial frame, and the twin whose accelerometer is zero for the whole trip is in an inertial frame.

[@fkeyvan]

@@eigenchris what if all there is in the A frame is a clock and B also only contains a clock. This is a more general case than frames with accelerometers. Then there will be no way to know which frame experienced a force. A sees B accelerate relative to it, and B sees A accelerate relative to it. no asymmetry here.

[@fkeyvan]

a different argument. Considering frame B (rocket) move on a smooth curve without breaks, decelerate and then accelerate and return back to earth. In this case you can still use the SR axiom about all inertial frames being the same. That's because at each spacetime point on the trajectory of frame B you can do a LT to the inertial frame A. This means SR is satisfied locally. In your explanation you skip this argument. You do not mention that the 2 legs of B's travel can each be LT to A and thus satisfy SR axiom.

[@eigenchris]

@@fkeyvan I think when you're doing physical experiments, you need to assume that the proper equipment is provided. The lack of accelerometer doesn't mean there is no acceleration, just as lack of ruler doesn't mean there's no distance, or lack of clock means there is no time.

[@fkeyvan]

@@eigenchris during the relative travel observers A and B will measure the same flight times for the other frame no matter inertial or not. That’s because non inertial frames are locally inertial. Therefore during the travel period there is no asymmetry between the frames of reference regarding their own clocks or clocks measured in the other moving frames . The only time the asymmetry appears is when they meet back on earth. How do you explain that? The existence of an accelerometer makes no difference.

[@HighWycombe]

This doesn't really explain the TWIN PARADOX, it just tells us that the Lorentz Transformation isn't applicable in this situation and so isn't discredited. Looking elsewhere on KZbin, I can't see that ANYBODY has provided a convincing explanation for the Twin Paradox.

[@eigenchris]

I think the key idea is that the geometry of spacetime obeys the "reverse triangle inequality", so that doing a non-inertial "zig-zag" gives you LESS proper time than just following a "direct" inertial path between the start and end points. Do you feel this is not a sufficient explanation?

[@HighWycombe]

​@@eigenchris The "Reversed Triangle Inequality" is based on the stationary observer's reference frame and explains why the moving observer will always experience less proper time than the stationary observer. The problem comes if we want to see things from the (non-inertial) reference frame of the moving observer and hope to get the same result. You've explained that it is not valid to use the Lorentz transformation to switch bases, but I can't see what we should use in its place. We shouldn't need to use GR to handle accelerating reference frames should we?

[@eigenchris]

@@HighWycombe We can use Lorentz transformations for each individual leg of the non-inertial observer's journey. We just can't use a Lorentz transformation for the entire thing. There is a set of coordinates that can be obtained for the non-inertial observer, it's just not a coordinate system we get from a Lorentz transformation... and I suspect the coordinate system will "break" (with various coordinate singularities) because the observer experiences an infinite acceleration when they change direction.

[@HighWycombe]

@@eigenchris Many thanks, yes that makes sense. The Twin's Paradox is more involved than it first appeared to me. You have a reference frame for the stationary observer, one for the moving observer going out, and one for the moving observer coming back. They all need to work together to agree on the age difference at the end of the journey. ...Maybe we could imagine Bob as a photon, being reflected in a mirror at the half-way point? The infinite acceleration that happens when he changes direction would then be an "everyday" singularity, and possibly OK? 🙂

[@Torbu6286]

​@@eigenchrisyeah, this doesn't make sense

[@nicu_danciu]

You realy made my lough. Why should Bob go în a zig-zag way? He can go straight, he does not have to come back to the Eartht. Than just see what happens after 10 years. Who îs younger, who îs older and why? Moreover, you do not have to imply persons and accelerations, just use clocks. You do not have to accelerate, just syncronize the clocks while one is moving and see what happens after 10 years and why, since there îs no prefferance for one or the other frame. Well... at least one thing îs true: the special and general relativity are valid, în spite of all these paradoxes.

[@klevisimeri607]

Respect for you👏👏❗

[@mikegale9757]

Not so fast. The naysayers have a point. Forget about round trips. Let the traveller do a flyby of the waypoint and send a selfie to the bystander. There are at least 2 possibilities: 1. The bystander and waypoint are moving and the traveller is standing still. 2. The traveller is moving and the bystander and waypoint are standing still. The traveller will be younger in snapshot #2, but how can you predict which way it's going to go? We're missing an initial condition, which is who felt the force on the launchpad and the answer could well be both. Indeed, twins who go walkabout with equal and opposite velocities age at the same rate despite their relative velocity. More generally, one twin could have felt a greater force than the other. In all casea, the reference frame in which the time axis is parallel to the centre of momentum has the fastest clock. Not a lot of people know that, but it's high time to let that cat out of the bag so to speak. It's a stubbornly persistent delusion that moving clocks run slower than one another. Here's another thought experiment to drive this point home. Throw a clock at a wall and then throw the wall at the clock from the same distance. The elapsed time on the clock on impact is different in each case even if the relative velocity is the same. See the link on the About tab of my profile page for more details.

[@eigenchris]

I don't follow your first example about the "flyby". Can you describe what the spacetime diagram would look like in this situation? For your example of the clock/wall, I don't see the contradiction. The clock will only every experience one time: its proper time. This is invariant and agreed on by everyone in the universe. Other observers might measure different times on their own clocks for how long it takes the thrown clock to meet the wall.

[@mikegale9757]

@@eigenchris The spacetime diagram for the flyby scenario is the 1st half of the conventional twin paradox except everyone maintains their speed and heading after the traveller reaches the waypoint (case #2) or the waypoint reaches the "traveller" (case #1.) The clock/wall experiment is the 2nd half of the conventional twin paradox. Just the return trip. No outbound leg. The issue there is whether the clock is the traveller or the bystander.

[@eigenchris]

@@mikegale9757 I don't think it's a true twin paradox unless there's a twin that has a "turnaround" path. The idea is that the twins start at the same location in space, then one twin goes off and comes back, then they compare clocks at the same location in space again. They can't truly compare the amount of time that has passed for each of them unless they are standing next to each other. If you just have "half" the twin paradox (only incoming or only outgoing), you can just use the standard time dilation formula to. Each frame will see the other frame's clock ticking more slowly. But given a path in spacetime, everyone will agree on how much proper time has elapsed on that path; there's no disagreement for proper times.

[@mikegale9757]

@@eigenchris You certainly need a return trip if you want to compare ages at the family reunion (because everyone has different notions of "now" over there in the mean time), but your age at the waypoint has nothing to do with your antics on arrival. It's all about the forces you experienced on the launchpad. i.e. How far you deviated from the centre of momentum heading. The equal and opposite velocities case epitomizes that point. Note also that we don't need to be co-located to compare elapsed time measurements. Co-moving will suffice because, in that case, everyone has the same notion of "now" over there. The elapsed time measured by an observer who remains at the waypoint is the same as that measured by an observer who remains on the launchpad. It's just a question of whether the traveller measures more time or less after traversing the intervening space. SR can accommodate either case (and anything in between), but it's up to you to figure out which answer is correct if there's no return trip (as is the case for muon decay.) In his WSU video on SR, Brian Greene admits that he uses intuition to resolve that ambiguity. My point is, we can do better than that by keeping track of the centre of momentum.

[@mikegale9757]

@@eigenchris I should also point out that each observer sees the other in fast-fwd on approach and slow-mo in recession, just as you would expect from a Doppler analysis of light waves in an aether. Time dilation is what's left over after you correct for that effect and it does not depend on the direction of motion.

[@Physics_PI]

As usual 🙂👍

[@A-_--]

This is only a satisfactory explanation for a universe that has a certian topological structure. Particularly, what if the universe were a 3-torus, you could travel along a geodesic in spacetime from the earth and back again, meaning you were always in an inertial frame of refrence.

[@eigenchris]

I'm assuming flat spacetime here. In general relativity it's possible for a twin orbiting a planet ("inertial frame") to age less than a twin firing its rockets above a planet at the same radius ("non-inertial frame"). I might make a video about this later.

[@gustavitch8518]

very interesting but this voice fry is pretty horrible. I ended up looking for another video :/

[@redbel2624]

Haha, I forgot what an alpha is. Remind me what alpha is

[@-_Nuke_-]

I wish you would take google translate, put τ in there and hear how we Greeks correctly pronounce it... Its not tau, its taf. To anyone non Greek that would sound weird. But to us, hearing "taf" being pronounced like "tau" is beyond cringe. "Tau" doesn't even sound Greek! Its sounds Korean, Japanese or Chinese like... How on EARTH did you guys ever thought that a word like "Tau" will ever EVER, have anything to do with Greece!? Its TAF, not TAU... The "u" at the end of every Greek word, gets an "f" sound, not a "u" sound. For example: Tautochrona - (simultaneously) is pronounced "taftochrona" Please if you are going to use our Greek letters, LEARN how to correctly pronounce them...

[@eigenchris]

Sorry, I'm just pronouncing them how they are pronounced in Canadian schools. I imagine we pronounce most Greek letters incorrectly. My understanding is that π is correctly pronounced "pee", but every English speaker I've met calls it "pie"/"pye".

[@-_Nuke_-]

@@eigenchris Yep! π is pronounced like p is pronounced! Its the same letter after all! So its indeed "pee" but it doesn't refere to peeing :P I wish one day people will start calling τ taf as its meant to be pronounced! Then more people will find it weird and learn the truth about it :D Your videos are the best in the entire KZbin btw! Sorry for my tone of voice if it sounded harsh! :)

[@freyc1]

@@-_Nuke_- We use latin letters too, but we don't pronounce them the latin way either. If I spell an English word to a French person, I'll use the name of the letters in French. Those letters as they are used in maths don't really have anything to do with modern Greek. Their pronounciation in English is codified and I think you should just accept it.

[@-_Nuke_-]

@@freyc1 I do accept it, I'm just proposing something that is both better sounding (taf sounds better than tau) and it's also more original to ancient Greek. You guys can take from this what you want!

[@stewiesaidthat]

Special relativity is junk science. If you want to learn physics, stick with Newton's law of motion F=ma, it is the only one that works in ALL frames of reference. Broken down, it becomes Force Equals Acceleration. The frame of reference is centered on the acceleration event. The point in space with the least amount of acceleration for that event. The earth rotating on its axis, the axis the frame of reference. The earth orbiting the planet, the sun is ground zero for the Earth's acceleration frame. The two frames combined give you the daily tides and the yearly tide as laid out by Kepler's law of Motion. The proper frame of reference is the acceleration event, not the mass. You are wasting your time with Einstein’s relativity nonsense as it is using mass as the frame of reference which creates a mirror image of reality. Exactly 180 degrees in fact. Lastly. Space and time are TWO separate frames of reference. Boiling water is acceleration in Time. Throwing an ice cube is acceleration in Space. How does accelerating an ice cube into the sun cause it to exist longer in its frozen state?

[@krzysztofciuba271]

Sorry, it is ...scandalous as in ..textbooks. You did not resolve Paradox! You wrongly claim that "some" claims..inconsistency: no, only claim that you, the majority do not understand..the time parameter in Theory. More; you do not understand the relationship between a mathematical model and the physical (real, data ). Hm: what is your physical "unit proper time", again "unit time" in the term "proper time"? Ok: why do you introduce an outside the "system", a frictional force that acts only on one object/twin and uses it into the equation, dictionary one in this case? You have only two coordinate systems: from A-twin, or B-twin; you are free to make one at rest or the other moving but you make (and your BS authors in textbooks) only one "turning and accelerating and deaccelerating".Why? From the point of view of "accelerating" the other twin does the same! AGAIN: WHAT IS the "unit of proper time" both in Special and General Relativity? If you measure something you. must have a physical unit time (and "unit length" though ....not necessary as some pointed if having already the "unit time"!).ps. don't be upset, also A. Einstein could not resolve it unit his...death though he rejected your argument in this video and opted for a "snake" case with "3 inertial twins;A,B, and C where A-would be at "rest". At the foundation of the theory is the assumption of...Symmetry and that is included in math's equation (just switching the apostrophe for the designation of coordinates, time, and distance. More,id you claim "less aging" of the "traveller" you must also claim the Traveller is shortened in the direction of ...travelling,his mass ...increased,....and even his temperature (if treated as a gas system) decreased to...Zero if his speed close to that of light!