Relativity 104c: Special Relativity - Time Dilation and Length Contraction Geometry

[@eigenchris]

Typo at the bottom of the slid eat 5:04. The last line should be x and y instead of t and x. Sorry for the copy+paste error.

[@alapandas6398]

This is the best channel for learning relativity and tensor calculus.

[@MasterHigure]

I agree that it is really good. And if you want both tensor calculus and special relativity in one place, yeah, the best one I've seen. But for just tensor calculus, the channel Lemma (previously Math the Beautiful) is in very close competition. And since Chris has done no general relativity yet, the best alternative I've found is the Hereaus International Winter School on Gravity and Light lecture series from 2015. It is phenomenal and of course available here on youtube.

[@mobilephil244]

The patience and clarity of your explanations is absolutely fantastic. Thanks for taking so much trouble and care with this.

[@81546mot]

CLEAR, STEP-BY-STEP EXPLANATION WITH SUPERB GRAPHICS!!! GREAT JOB.

[@JoeHynes284]

you have helped me immensely, i have was in the navy and have worked various engineering related jobs for the past 24 years but have never actually sat in a college classroom. Thank you for your work!

[@NicolasSchmidMusic]

I have just been watching 10 of your videos in a row. It is excellent! Everything is well structured, you understand the misconceptions a beginner could have. Thank you !

[@killer12543]

I really appreciate your efforts to explain everything geometrically. This will surely make the introduction to general relativity a lot smoother.

[@ultramadscientist]

I really like your approach of introducing galilean relativity in geometric, algebraic, and tensor formulations so that we not only keep using the same language all the way through but also have a very strong basis in translating between our three languages on the way. I know you said it'd make general relativity a little easier but it even made special relativity a little easier.

[@eigenchris]

Glad to hear it. That was my goal for these videos.

[@joeheafner2495]

Another excellent video. I will once again recommend this series to my physics students this semester.

[@paulbizard3493]

Thank you Chris. I overviewed briefly Special Relativity as an engineering student long ago. And now that I'm retired and watching your videos, you are answering all my questions. This Minkowski space time diagram is a so powerful tool. LOL. Hope you win the Nobel asap. 😁

[@leonardp8517]

12:20 this right here always confused the hell out of me. How can you scale with gamma back and forth and not stretch the coordinate system out infinitely in the process? Now I get it. I've been actively trying to understand relativity for some time now, and this series is better than anything I've seen both in other youtube videos or physics textbooks. Thank you comrade.

[@Oh4Chrissake]

"So if we have this spacetime vector *S* connecting a sneeze event and a clap event, time dilation is just measuring this vector with different basis vectors and we'll get a different time component in each basis." There is a beautiful economy to this 'spacetime separation vector' approach (an approach I have never seen done elsewhere).

[@SaberToothPortilla]

Yeah! Expressing space and time as one state space has a lot of utility.

[@eigenchris]

The more mainstream word for "spacetime separation vector" is "Four-position". I'll start using that term instead when I introduce four-vectors in a couple videos.

[@hazratprince956]

Sir, I really appreciate your struggle One of great teacher takes helping from ur channel,

[@ShadowZZZ]

As a physics student I'd like to show my apprechiation here for your well explanation of einsteins special relativity, the deratavion of the equations, explaining time dilation & lenght contration on vector components.

[@vitrums]

22:09 a great image. It finally made the idea click for me. As soon as something is being observed moving relative to us (in this case it's Einsein chilling with the clocks in his hands, who seems to be in motion from our perspective while we drive a car), all observed unit time interval events (such as a full tick cycle of his clocks) happen in our reference frame every γ≥1 time units. Therefore, despite the fact Einstein himself is moving away from us with a great speed, from our perspective all the motions of his body and of the objects surrounding him appear in slow-mo.

[@yongsikchu8258]

@26:59 I think the train in the car frame should be aligned along e_x tilda because it is in the moving frame at the simultanity .

[@FakePhilospher]

good lord , you have crafted a masterpiece

[@hilmierdem5998]

Wonderful series. I have learned a lot. Regarding the length contraction, I wonder if a better explanation would be to indicate that the displacement vector M is the one that connects two events (beginning of the train and the end of the train on the two vertical worldlines) at the same time ( hence the simultaneity diagonal)

[@Vkonto]

This is incredible ... The best explanation on KZbin for time dilation ..can you please add a video on twin paradox

[@eigenchris]

That's my next video. Should be out this comong week.

[@Oh4Chrissake]

Note to self (17:25): Remember, from the car's point of view (i.e. frame) it (the car) remains in the same place as the sneeze, and it is Einstein who moves to the left before clapping (3.46 time units later, in the car's frame).

[@BruinChang]

Hello, thanks for the series. At 2:01, the S is decomposed into two components of the moving frame. I am confused about the one for moving x (the value in ex tilt ). Because it points to down left, does it mean the moving one goes backward on time axis of stationary frame?

[@eigenchris]

Sorry, I missed this comment from 3 months ago! You need to remember that the lines of simultaneity for the moving frame are tilted to be diagonal. So these "diagonal" ex-tilde vectors pointing "down-left" are actually pointing along the line of simultaneity for the car. So the ex-tilde vector is just pointing left, not back in time.

[@parthvarasani495]

28:29 @eigenchris but if we calculate the distance w.r.t. to mueon it turns out be d=vt , d=0.995c * 2mu = 597m. Did i make any mistake?

[@asmatminhas8644]

1:42 plz explain what do you mean by moving with the clock and the second reference frame watching moving clocks

[@laonza7036]

Question please: At the bottom of the slide at 5:04, the vector S is written in terms of basis vectors e_x and e_y (please excuse that I can’t write tildes here) on the left side of the equation while it is written in terms of e_t and e_x on the right side of the equation. Was that deliberate? I’m confused because I would expect e_x and e_y to be on both sides of the equation for this example - although we can write S in terms of any basis vectors so maybe that is the point you are trying to make? And thank you for this fantastic video series.

[@eigenchris]

That's a typo, unfortunately. Copy+paste error. IT should be x~ and y~.

[@laonza7036]

Thanks. @@eigenchris

[@comfuse]

@26:13, shouldn't the train be parallel to the cars time-axis instead of Einsteins in this situation (because otherwise we would be measuring the train at different moments in the cars frame of reference)? Great video series btw!

[@eigenchris]

Sorry, I misread your question and typed an incorrect response. The train is parallel to the car's time axis here, so I don't see the problem.

[@comfuse]

@@eigenchris Sorry, I meant to say parallel to the space-axis (of the car). Please allow me to clarify. In this diagram, (the green box representing) the train is drawn parallel to the space-axis of Einstein (e_x without the tilde, or in other words, is measured along a line of simultaneity in Einsteins FoR). If I understand correctly though (but I might be missing the point :D), you're trying to explain that: The length contraction for the train measured along a line of simultaneity in the cars FoR, contracts with the same factor. Then shouldn't the the green box should be drawn parallel to e_x with the tilde?

[@eigenchris]

@@comfuse Oh, yes. I see your point. The box should probably be slanted, if you were looking along lines of simultaneity from the car's/train's perspective. I didn't think too hard when making that diagram. The vector algebra and numerical result are correct, though.

[@jomermcjomer]

I just have a small question in 29:20, the 20 microseconds in total is the sum of all those time vectors? I'm just confused because there are horizontal time vectors... shouldn't they all be vertical in the case of the Earth's reference frame?

[@jomermcjomer]

I mean the sum of all those horizontal and vertical time vectors. I know the scale is off, but I got confused as to how you're using the horizontal (space direction) for a time measurement.

[@eigenchris]

Yes, the 20microseconds are the sum of the vertical time vectors. The horizontal vectors are spatial position vectors. They just indicate that the start and end of the muon's journey are not at the same location in this frame. In the muon's frame, the start and end of the journey are at the same position, so there are no spatial position vectors in this frame.

[@eigenchris]

Looks like I mis-labeled the blue time and space basis vectors... they should be switched. My bad.

[@SzTz100]

Best explanations of the subject

[@AndreaPancia1]

Hi Chris thanks a lot for your work, just a question, how can we measure on earth the decay time of a stationary muon?

[@biblebot3947]

You said a few times in previous videos that a basis covector is contravariant and that a basis vector is covariant. Could you elaborate on how an invariant object follows a transformation rule?

[@eigenchris]

For basis vectors, they are the starting point for how we define what "covariant" means. When we do something to the basis vectors, that thing is "covariant" by definition. Every set of basis vectors also has a unique "dual basis" set of basis covevectors. And when we change the basis one way (the covariant way), to get the new set of dual basis covectors, we apply the contravariant transformation to the old dual basis. For example, if we make the basis vectors twice as big, we must double the spacing of the dual basis covectors (i.e. halving their density, so they are half as big) to get the correct new dual basis. So the reason basis covectors are contravariant is due to the way we define the dual basis.

[@MRich955]

As I understand it, contravariant doesn't mean invariant. Covariant = "varies with", contravariant = "varies against", but both involve changing values. Covariance (in this context) involves multiplying by a matrix, while contravariance involves multiplying by the inverse of that matrix. The broader definitions of co/contravariance (they involve multivariable derivatives) are described in Eigenchris's tensor calculus videos, but, nice as that series is, the concept is still hazy for me. As far as the co/contravariance definitions apply to this general relativity series, so far I've gotten by thinking of them in terms of matrix (and inverse matrix) multiplications. However, I have this sneaking suspicion that the general, multivariable-derivative definition will come in handy when we get to special relativity and curved spacetime!

[@eigenchris]

The only difference between this video and the "general" case is that for the general tensor calculus case, the pair of matrices used are the jacobian and inverse jacobian. You can try re-writing the matrix equations I show between 5:00 and 7:00 using the jacobian and inverse jacobian (with the row of basis vectors being partial derivative operators instead, as described in my tensor calc series, and the column of components being derivatives of coordinates with respect to a path parameter lambda).

[@fixed-point]

These videos continue to be fantastic. I'm curious: What takes the most time when producing these videos? Is it writing a coherent script? Is it the animation?

[@eigenchris]

I've never timed it. The graphics/writing *seem* to go faster because that's the part I like. The editing/fixing mistakes feels like it takes the longest because I hate it (and therefore also procrastinate on it). Sometimes there's a typo I need to fix, or I say the wrong variable name by accident, or I accidentally say something that's just scientifically false, so I need to re-record and splice it into the project. This video was particularly bad and had 15 or so errors of various kinds that needed fixing.

[@messapatingy]

At 2:03 you say the moving clock is ticking slowly, yet has a greater time. 3.46 is slower than 3. Looks like faster. Please explain my confusion. Normal intuition: 3 m/s is slower than 3.46 m/s.

[@eigenchris]

The clock's path in spacetime is given by the S vector. The clock is stationary in the frame (left) image, and moving in the red (right) frame. In the red frame, the clock itself ticks 3 seconds, even though observes in the red frame count 3.46 seconds going by. Therefore, the moving clock appears to tick more slowly than expected.

[@imaginingPhysics]

21:35 And time between events is even shorter for a non-inertial observer who makes a round trip. See the twin paradox.

[@denizkaplan4988]

You are a king

[@Oh4Chrissake]

Should the Earth's basis vectors be the other way around at 29:18 and again at 29:43?

[@eigenchris]

Yeah, that's a mistake.

[@Oh4Chrissake]

@@eigenchris Okay, no worries. Many thanks.

[@issraali7715]

@eigenchris just a quick question, why doesn't it make sense to measure the spacetime vector R at 23:03 in the et tilda ex tilda basis?

[@eigenchris]

You can technically measure any vector using any basis, but the numbers you get will have different meaning depending on the basis you use. In this part of the video, we want to measure the physical length of something. To measure length, we have to pick points on our spacetime diagram that all correspond to the same moment in time. If you took note of where the front of the train was at noon, and then looked at where the back of the train was an hour later, you haven't measured the train's length, because the train has traveled forward in that hour. If we measured the R vector, this is exactly what we would be doing. We need to get the "back" and "front" locations of the train at the same time to measure it's length correctly. This means the "back" and "front" locations of the train need to be on the same line of simultaneity on our spacetime diagram. This is what we do when we measure the M vector.

[@issraali7715]

@@eigenchris I see, thank you so much. Also, another question, regarding the muon and Earth example at the end of the video: would it be accurate to say that from the muon's perspective (whereby it thinks itself to decaying in 2 microseconds because it believes that it is stationary), the Earth is moving towards it at 0.995c?

[@eigenchris]

@@issraali7715 Yes, the earth is moving at 0.995c and so it is length contracted such that the atmosphere is much thinner.

[@issraali7715]

@@eigenchris Thank you sir

[@MrYahya0101]

Hello, at 10:33 you say to reverse the transformation equations, you swap the basis vectors and change sign of B. Why do you do that? Why can't you just solve for the other basis vector?

[@eigenchris]

You can also do that. But in this case, since the transformation represents a velocity change, the easiest way to reverse it is to just flip the sign of the velocity.

[@MrYahya0101]

@@eigenchris I see, thanks. One thing I'm not getting about basis vectors is that time component has a basis vector, so does that mean time is a vector? Wouldn't that mess with the equation x_vector=v*t_vector, making velocity a scalar, instead of a vector? Other equations with time would also be affected?

[@eigenchris]

@@MrYahya0101 position (x,y,z) can be thought of as a 3D vector, and velocity (vx,vy,vz) can be thought of as a 3D vector, and time can be thought of as a scalar. Then we still have (x,y,z) = (vx,vy,vz)*t. But as this video series goes on, you'll see in relativity that it's better to treat space and time as part of the same vector.

[@MrYahya0101]

​@@eigenchris So you'd say it's more appropriate to treat position and time as vector components (and spacetime as vector)? If so that would mean x=v*t is only a relationship for the magnitude of the time not time vector, right?

[@redbel2624]

This is very easy, but a lot of information for me to save all of this

[@ahmedkhalifa5190]

@eigenchris - do I understand correctly if i say, given an observer is associated with the muon .. once the muon's clock ticks 2 seconds (to the muon observer) .. despite that the world's clock (from the muon's observer perspective) is smaller than 2 .. but still the world will vanish to muon .. ? Also this means the the twins in the famous Twin Paradox will end up when they meet up having the same age?

[@ericterry4544]

17:16. Forgive me, but shouldn't the car's clock be slower than Einstein's? (After all, the car is moving at .5c while Einstein is stationary) But yet the car thinks that ~3.5 time units have passed while Einstein thinks 3 time units have passed. Not sure what I'm misunderstanding here.

[@eigenchris]

Einstein will see the car's clock running slower, and the car will see Einstein's clock running slower. In this case we're looking at Einstein's clock (since the vector S is pointing along Einstein's worldline). When Einstein measures S with his basis, and the car measures S with its basis, the car will measure more time. This means that the car sees Einstein's clock tick more slowly. If we repeated this experiment with a vector T that points along the car's worldline, Einstein would measure more time than the car.

[@anmolmehrotra923]

24:22 Why does vector M has to join the ends of the train?

[@eigenchris]

The vector M represents the length of the train in the movie frame of the car.

[@anmolmehrotra923]

@@eigenchris yeah I was confused by why should It exactly be such that it lies between the worldlines of ends of the train, like it could be smaller or longer too. But then I realized that since the x' represents line of simultanaity, it must be so that M is exactly what you drew. BTW where did you learnt this material yourself from? It's so original and exactly what I wanted for years, literally. Thank you very much for providing them

[@eigenchris]

@@anmolmehrotra923 I took a special relativity course in undergrad, but the particular explanations in this video I largely came up with myself. After completing my "Tensors for Beginners Playlist", I realized it makes a lot of sense to write basis vectors as rows, and vector components as columns. I wish I had realized it beforehand, but oh well. I found most "derivations" of length contractions to be impossible to follow and very confusing, so I tried to come up with a vector-based approach for this video. Personally I like it a lot better than the standard way.

[@anmolmehrotra923]

@@eigenchris true. This approach makes everything clear at once and is way more powerful than anything I have seen before. For example the fact that length only contracts in the direction of motion is next to obvious in this approach, in contrast to standard approaches where one needs help of certain "paradoxes" to state this fact (see for example, Griffiths electrodynamics)

[@tommy51s5]

For video time-point 2:19, isn't it so that for the person who is carrying a rod in the moving (car) frame, they can measure each end at their leisure, and the rod's length will always be the 3.46-unit-long red diagonal shown at 2:19. OTHO, the person in the stationary (Einstein) frame must measure the ends simultaneously, as the rod whooshes by, to get the horizontal purple vector. Einstein thus concludes that the length of the rod moving by him is only 3.00 units long, i.e., contracted by 1/gamma. The car-frame person measures the purple vector's length (at two different times, as shown in the fig) and it's still 3.46 units. BTW, I love this channel.

[@eigenchris]

The problem with 2:19 is that when the red et~ and ex~ basis vectors are measuring the rule, they are not measuring it in their rest frame. They are looking at the left-most point in their past and the right-most point in their future. To see this, you can draw the same vector on paper and "extend" the et~ and ex~ vectors into a grid of parallelograms. Each end of the purple S vector will be located at different points in time. This would be like trying to measure the length of a moving car by noting the position of the car's back and the position of the cars front 5 minutes apart. The car would be measured to be several kilometers long using this method. We always need to measure an object's length according to a specific rest frame, as done at 2:30.

[@tommy51s5]

@@eigenchris Yes. Sorry, I mis-attached the names of each frame ("stationary" vs "moving"). I should have said the rod is At Rest/Stationary in the "car" frame, so that "Einstein" would be the one to see the rod as Moving. This should allow the S vector at 2:19 to be the interval of focus for both observers. So, how about this: Assuming the et~/ex~ person (in the car) is the one carrying the rod, then they know its proper length (3.46 units), regardless of when they measure the ends. They also know that the ends of the S vector are the events associated with that rod being measured at different times (in their frame), so that the space-time interval-length of S vector is not the length of their rod, which is still 3.46 units (per the red component vectors drawn in the 2:19 figure). However, the et/ex person (Einstein) measures the ends of the rod simultaneously (for him), as it Flies By. For him, the interval-length of the S vector IS the length of the rod, since that interval has no time component in his frame. He concludes that the S vector is 3.00 units long, and thus the rod is 3.00 units long. To et~/ex~ person, the et/ex person has a problem, since he's not measuring the ends simultaneously at all.

[@tommy51s5]

This looks quite like 26:23. There the length of vector R (proper length of the object) will always be the length of the object for the "car" observer; its ends don't have to be measured simultaneously by him, since he's "carrying" R along with him. If the ends of a new, non-simultaneous interval are measured by the car observer at different times (e.g., vector M), he'll just decompose the interval into space and time components, and the space component will be R, just as if he had measured the ends simultaneously.

[@asmatminhas8644]

Nice very informative 😘😘

[@j.k.sharma3669]

Great video but plz clarify one thing. Can we consider time dilation and length contraction simultaneously ? In example of muon , for ground observer you only considered time dilation but why not length contraction and vice versa??????

[@eigenchris]

You can. For the ground observer, you could consider the muon itself length-contracting, but this doesn't lead to any noticeable effects because we only care about how far the muon travels, not how long it is. The muon would also see the clocks on earth time-dilate. But this doesn't matter since the laboratory on earth isn't decaying like the muon is.

[@WildGamez]

Bloody brilliant. Do u have a subscription website where i could donate to for such an amazing-mind-bending knowledge? It's surprising that u just provide all this knowledge for free in a systematic manner!

[@eigenchris]

I have a ko-fi tip jar if you want to leave a tip. Thanks! ko-fi.com/eigenchris

[@canyadigit6274]

Amazing!

What actual measurement has already been made of the length contraction?

[@eigenchris]

I'm sorry, I don't understand your question. Je peux parler en francais si tu veux.

@@eigenchris _(Merci. A t'on déjà fait _*_une véritable mesure_*_ de la contraction des longueurs ?)_ Has *a real measurement* of length contraction ever been taken?

[@eigenchris]

@ Francais: Je crois qu'il y a de l'évidence pour l'effect de "time dilation", mais l'évidence directe pur l'effect de "length contraction" est plus difficile parce qu'on a besoin d'un object physical et pas seulement un horloge. On a observé les muons atmosphériques qui bougent à >99% le vitesse de la lumière qui arrivent au Terre. Dans notre cadre, ceci est à cause du "time dilation", mais dans le cadre des muons, c'est à cause du "length contraction" de l'atomosphère. (English: I think time dilation has experimental evidence, but direct evidence of length contraction is more difficult since it requires physical objects instead of clocks. We have observed atmospheric muons moving at >99% the speed of light reaching the earth. In our frame, this is because of time dilation, but in the muon's frame, it would be because the atmosphere of the earth is length contracted to be shorter.)

@@eigenchris _Ma question portait sur une mesure "dans notre référentiel" bien sûr._ My question was about a measure "in our referential frame" of course.

[@fsaldan1]

Good question. Which perhaps could be rephrased as asking what is the meaning of the word "appear" as used in the video.

[@drsonaligupta75]

Great content!!

[@manog8713]

Length contraction is a gemoetric or better to say kinematic or perspective effect. if a moving length stops and then examined, surely there is no change in the length (no physical law to expalin why there should be any change at all). Similiarly time dilation is a perspective effect. How then time dilation for Mueon could be so real that it actually lives 20 micro seconds? There is something not quiet right in this reasoning and similar ones elsewhere. The Mueon experience is heavily shouded in statictial issues.

[@Unidentifying]

needlessly complicate things with the e, why use that??

[@eigenchris]

It's a somewhat standard notation for basis vectors. It also makes it pretty easy to tell when something is a basis vector as opposed to something else. Which notation would you prefer?

[@Unidentifying]

@@eigenchris I think my issue is that my professor didnt go into vectors at all in the SR course, we mostly do all these things, velocity additions, contractions and dilation and other SR problems without vectorial things, maybe the stuff is all implied. She is an expert in Special and General relativity so I guess there is a good reason. Did you include them to be more encompassing to Relativity in whole?

[@eigenchris]

@@Unidentifying When I learned galilean relativity and special relativity, I probably learned it similar to the way you are learning it now (without the vectors). But when I got to general relativity, the math suddenly got extremely complicated. I had no idea what was going on and I got destroyed. In my relativity videos, I'm trying to teach in a way that will avoid this problem of "getting destroyed when you reach general relativity". You'll see I'm teaching galilean relativity and special relativity in ways that seem overly-complicated, with vectors and covariace and contravariance. But my hope is that seeing all this "complicated" stuff now, general relativity becomes easier when you get to it. It's impossible to learn general relativity without understanding vectors, covariance and contravairance. I want my viewers to learn this stuff now so they can learn general relativity without "getting destroyed".

[@Unidentifying]

​@@eigenchris Sir thank you for the reply, it makes a lot of sense now yes. GR is a totally different cookie yes, mostly the math is incredibly obscure , I'm only in the first year now but GR is for the third year! That says a lot :P Although SR is famous for its conceptual difficulty.. I still find myself struggling sometimes when trying to solve problems with velocity addition for instance.. and get very confused with reference frames. Final exam coming very soon, do you have some tips?

[@eigenchris]

I'm not really sure what to suggest, other than doing lots of practice problems and maybe trying to find other special relativity exams online to practice from. I find drawing a picture with basis vectors (et, ex or et~ and ex~) usually helps me get unconfused with reference frames. I try to show how it's done in this video at the end with the muon, where you can look at the same physical situation with two different coordinate systems. With velocity addition, it's hard to draw that, so I guess all I can suggest is do practice problems.

[@gman8563]

Your statement about time dilation vs. length contraction is not quite correct here. Measuring an instance of time dilation requires only looking at the component of a vector. Determining the symmetry (two observers each claiming the others clock runs slow) however requires measuring two separate vectors in two separate frames, same as length contraction. Edit: having reached 21:30, you clearly state that two separate vectors are involved for time dilation. So I’m not sure why you state otherwise at the beginning of the video

[@asmatminhas8644]

Plz make a video series on lorentz transformation as hyperbolic rotation

[@eigenchris]

That's video 104e. Slides are already done. Hopefully will be out by the end of the month.

[@asmatminhas8644]

@@eigenchris thanks very much

[@canyadigit6274]

After you’re done with general relativity and special relativity, what’s next?

[@eigenchris]

No idea. I have various other topics that interest me that I might do some videos on. Clifford algebras seem neat and have applications in physics.

[@RizkyMaulanaNugraha]

@@eigenchris Do you have plan to explain quantum physics next?

[@eigenchris]

I don't think I have much insight into quantum, so probably not.

[@se7964]

Rewatching this (still a masterpiece!) but think there’s a bit of an inaccuracy in your statement about measuring time vs length contraction. You claim we’re measuring two different vectors when computing length contraction, and only one for time dilation; but this seems incorrect because (as you clearly show) the fact that each observer perceives the other’s clock running slow is also the result of measuring two different spacetime vectors. If you study the graphs, you’ll see the same technique for judging time dilation by extending multiple lines of simultaneity in a given frame can equally be applied by extending multiple lines of “simul-spatialness” (lines which represent the same coordinate position in a frame) in a given frame.

[@magicglide3928]

I thought so too but it is not a symmetric situation since the system does not evolve along x. There is no horizontal worldline orthogonal to the time basis, even though you can draw such simul-space-neity line it’s not physical, otherwise you would have time contraction when t~ crosses the horizontal simul-space-neity line of Einstein, like spatial contraction by the same math. Maybe I am wrong but i struggle with this also and conclude as above.

[@rupabasu4261]

What is your major??

[@eigenchris]

My undergrad was in "engineering physics".

[@new-knowledge8040]

If you create a simple Space-Time geometric representation of motion, and do so by using both motion vectors and length scalars stacked on top of each other, you can use this geometry derive the SR equations and the Lorentz transformation equations, and complete this task in mere minutes. If you handed this specific geometric representation of motion to some Joe blow out on the street, a fellow that knows nothing about physics at all, the chances are still high that the person will be able to derive the SR mathematical equations.

[@eigenchris]

I don't understand why you think you can put a line on the end of a velocity vector and then rotate the two, as if they are connected somehow. That seems kind of baseless to me.

[@new-knowledge8040]

@@eigenchris Well that's odd ?

[@VortekStarling]

It's very easy to prove that the entire basis of Einstein's time dilation theory, his moving light clock thought experiment, was wrong. He suggested that the path taken by a light pulse moving between the top and bottom of a vertical light clock would actually be a diagonal, rather than vertical, and therefore take a longer time to complete the trip at light speed. Let's see what actually happens though. To make things simple, we'll suppose that the light clock is 1 meter high and traveling horizontally at a velocity of 0.866 c, which has a Lorentz factor of 2. We'll also suppose for simplicity that light speed is 1 meter per second, so that it takes 1 second for light to go from one end of the clock to the other when we're in the same frame with the clock. Now what happens if we say that the clock traveled 0.866 m at 0.866 m/s? We would draw a diagonal line from the top of the clock to the 0.866 m point along the x axis, meaning the horizontal path of travel of the moving clock. That diagonal line, the perceived path of the light, would not be 2 meters long but instead 1.322. Light should only take 1.322 seconds to travel that path but the time dilation factor is 2, what went wrong? The problem is that the thought experiment is flawed, the distance the clock would need to travel for the diagonal light path to be 2 m long instead of 1.322 is 1.732 m. As you can see, the distance the clock travels, to make the geometry work, has to be the velocity, 0.866 m/s, multiplied by the Lorentz factor, which in this case is 2. You can't use the velocity directly for the distance the clock travels to get the required diagonal light path length, you need to already know the Lorentz factor for the velocity and then multiply that velocity by it to get the length of the long side of the right triangle along the x axis. That's what makes it impossible to work, you would already need to know an unknown, the Lorentz factor, for the diagonal to be the right length for the light to take that much time to traverse. You could call it the "Einstein Impossible Moving Light Clock Thought Experiment", in which you already need to know the answer in order to get the answer.

[@eigenchris]

Your error is assuming the top of the triangle is sqrt(3)/2 = 0.866 m long. The speed of the train is 0.866 m/s, but to calculate the distance traveled, you must use the clock in the frame of observer who watches the train move, which is t~. If the observer inside the train observes a duration of t=1s, then the t~ time will be dilated to t~ = 2*t = 2s. So the distance traveled by the train is actually vt~ = vγt = sqrt(3)/2 * 2 * 1 = sqrt(3) = 1.73. This leads to a sensible right triangle with sides 1, sqrt(3), and 2.

[@VortekStarling]

@@eigenchris Yeah, but if the observer on the ground saw the train move 1.732 m in 2 seconds on his clock, how far does he see it move in 1 second? Wouldn't you say 0.866 m? That doesn't work though, because at 0.866 m the diagonal light path would be 1.322 m, not 1 m.

[@eigenchris]

​@@VortekStarling If you are taking t~ to be 1s, then the "undilated" time for t would be t=0.5s. This means the sides of the triangle would obey (1/2)^2 + (sqrt(3)/2)^2 = (1)^2. You are getting a weird result because you are assuming t=1 and t~=1, which causes the sides of the triangle to obey (1)^2 = (sqrt(3)/2)^2 = (sqrt(7)/2)^2, where the sqrt(7)/2 = 1.322, but that isn't what relativity is claiming.

[@VortekStarling]

@@eigenchris Observer sees train move 1.732 m in 2 seconds. Only 1 second passes on the train. On the train, it would still be 1 second in their frame, they saw the light traverse the clock once. How far did they see the ground move in that time? You'll probably say that they would see the ground frame contracted to 0.866 m so they would perceive the relative velocity to be 0.866 m/s. The problem is that the people on the train weren't using the ground as their measuring gauge, they were using their own frame as the gauge, and their own frame was not subject to length contraction. They simply noted when a certain point, say a pole beside the track, coincided with markings in the front end of the train and then timed how long it took until the pole reached the back end markings. The distance would be same as when the ground based observers did the same thing, used their own frame as the measuring stick in the relative velocity determination. The ground observers didn't use a measuring stick on the moving train for that, they used the train track as the measuring stick so we have to do the same when we switch to the train frame, we have to use the train as the measuring stick. When the only thing you're using in the other frame as a relative velocity reference is a point, it makes no difference whether the frame is length contracted or not, because a point has no length. The only length being employed is in the inertial observer's frame and that never changes. The train people would see the same distance traveled but in 1 second instead of 2. They would calculate the relative velocity to be 1.732 c, an impossibility.

[@ashwinvishwakarma2531]

24:43 that was on purpose

[@eigenchris]

I don't actually know what you're talking about, so I'm guessing it wasn't.

[@ashwinvishwakarma2531]

@@eigenchris big d,

[@yq9hy6moon55]

Wow !!!

[@brendawilliams8062]

This is 3/5. 4/5.

[@jeremylakey680]

I dont know how I got here and whether or not yo take this seriously

[@eigenchris]

These are pretty standard results in special relativity: en.wikipedia.org/wiki/Time_dilation en.wikipedia.org/wiki/Length_contraction Full playlist is here if you want to start from the beginning: kzbin.info/aero/PLJHszsWbB6hqlw73QjgZcFh4DrkQLSCQa