Relativity 103b: Galilean Relativity - Spacetime Separation (Interval) Vector and Invariance

[@eigenchris]

I accidentally say "Emily Noether" instead of "Emmy Noether" in this video. Apologies.

[@Suav58]

Amelie, Emmy for short. Aren't you making it too easy for us, though? Two material points can not share place in event space, can they?

[@eigenchris]

@@Suav58 I guess you are correct. I was trying to use examples that are somewhat related to real life, instead of just saying abstract things like "point A" and "point B". But two different things can't exist at the same point, technically speaking.

[@mastershooter64]

@@Suav58 what about bosons? they can occupy the same states as one another, so two photons moving through space can technically share the same place in event space? or am I wrong?

[@mastershooter64]

@pyropulse seems like a semantic argument rather than a physical one, this would fall under ontology

[@IntelR]

But is it possible for two objects (Einstein and the car) moving relatively to each other to agree that an event (Feynman) happened at the same time for both of them? I don't think so, because the car will measure a different time due to time dilation. Or you can think that the temporal component of the velocity of Einstein can not be equal to the temporal component of the car, because the total spacetime velocity of a particle in spacetime must be the speed of light and since the car has spatial velocity in the Einstein reference frame, the time component has to be less than the speed of light

[@camorimd]

Eigencris I can't stress enough how valuable your videos are. I've watching all the series from the Tensor Algebra, Tensor Calculus and now I'm overexcited about this new series. Thank you.

[@Oh4Chrissake]

Note to self: (9:05) It happens 2.5 units to the right of the sneeze because in the car's reference frame it (the car) remains where the sneeze occurred and it is Einstein and Feynman moving to the left; and three time units after the sneeze, Feynman is 2.5 space units to the right of both the sneeze and the car.

[@Jabber_Wock]

These videos are great, thank you for posting. Hope you and family are keeping safe and in good health.

[@chavab8753]

This series is exactly what I've been looking for. I realized while trying to understand special relativity that I did not completely understand Galilean relativity, so decided to go back and get a strong grounding in that in order to see where GR and SR diverge. Thank you!

[@mwerensteijn]

Thank you so much for creating these videos!! Extremely helpful!

[@peterkokarchev6141]

I am so excited about the new Relativity Series! I hope you will cover and the General Relativity. I have never seen someone explaining that good Tensor Calculus and Differential geometry. Your videos are the best!!!!!!!!!!!

[@samymaziz8272]

What is the physical meaning of the spacetime separation vector and how it can be invariant neither if we change inertial frame of reference

[@m_bm_a7884]

This is an important step toward the general relativity..Thank you very much..waiting with an enthusiasm the next video!

[@AdrianBoyko]

Thanks for saying their names

[@laonza7036]

Suggestion please. At 8:10 the magnitude of “e_t tilde” looks to be a bit greater than 1, while at 9:10 the magnitude of “e_t tilde” looks to be about equal to 1. I ask as I am just coming to terms with the difference between basis vectors (which can be vary in magnitude) and unit vectors (which are always magnitude 1). At both 8:10 and 9:10 the magnitude of “e_t tilde” should be slightly greater than one, correct? Or, am I missing something here? BTW - thank you for theses videos.

[@eigenchris]

There is something you're missing: the "metric" used to measure the length of the time basis vectors is not the standard Euclidean metric you're used to (the result of the Pythagorean Theorem). The metric that measures time basis vectors is "measure the vertical component only". So when time basis vectors get "sheared" by the Galilean transformation, the vertical component doesn't change, so the time vector's length doesn't change. You'll have to be careful with this in special relativity too. The lengths of vectors is given by the minkowski metric, so your eyes won't be able to tell you which vectors are long amd which are short.

[@laonza7036]

Ok. I think I see now that the time basis vector gets sheared - I was thinking it would be rotated. Thank you very much. @@eigenchris

[@Physics_PI]

I am very interested to watch every lecture series.... nice lecture series sir

[@doaamohamedz9z9z9]

Amazing , God bless you

[@okoyoso]

Einstein, who just sneezed: thanks, bro

[@kubrickkubrick7045]

great work love your video so much waiting for next epidode!

[@СпасСтоилов-с2ю]

(9:12) What is the conversion factor of "t" to distance?

[@eigenchris]

In Special Relativity, the speed of light "c" is usually used to convert between distance and time. But in theory you can use any conversion factor you like.

[@przadka]

At 13:10, when you change the reference frame to the one of the car, and shift the coordinate lines, why is the vector S changing it’s geometrical shape as well? I’m struggling to understand that part. Didn’t we say that S is invariant under change of reference frames?

[@eigenchris]

There are 2 types of transformations: "active" and "passive". "Passive" is when you leave everything on the spacetime diagram the same, except you just pick a different basis to measure things with. With a "passive" transformation, the S vector wouldn't move. With an "active" transformation, we actually move points on the spacetime diagram around. This is what we're doing at 13:10: we're doing an "active" transformation to force the time-basis-vector for the car to be vertical. Under this "active" transformation, the "snap" event is moving to a new location. The S vector is defined as connecting the sneeze-point and the snap-point. If those points move under an "active" transformation, the S vector must move with them. Does that clear things up?

[@przadka]

@@eigenchris thanks, I didn’t realize there were two things. I thought we were making the time axis vertical by rotating the whole plane. I need to think about it more but thanks for clarifying. I love your channel - huge fan since I found the tensor series :)

[@HowDoYouKnowThough]

@@eigenchris Thanks for the series, Chris. I respect your minimal examples and emphasis on invariants. I had the same question as Michal but am still a bit confused. My understanding is that when you have the "passive" perspective, you have only a change of basis, but the vector itself is still invariant. Take your pencil example with the position vector of the tip of the pencil relative to the eraser. If I have a basis fixed in my head, and then rotate it, then to me, the pencil appears to rotate. But with knowledge that I am rotating my head, I know that the position vector is still an invariant and the pencil was not rotated. My observation is an artifact of me making measurements along a co-rotating basis. In contrast, for an active transformation, I apply a torque to the pencil. Material points of the pencil undergo accelerations. A stress develops in the material during the process. But in the end, the original position vector becomes a new vector, which is delivered by a rotation tensor. To me, it is a very different physical process from the passive transformation. By analogy, if we consider the active transformation in the spacetime diagram, shouldn't the spacetime separation vector go to a new vector, which is delivered by a tensor operator?

[@eigenchris]

@@HowDoYouKnowThough The vectors aren't changing/moving here. Maybe "active transformation" isn't the best term for what's going on. What's happening is we are re-drawing the spacetime diagram so that the red/tilde basis vectors are at 90-degrees, representing a "stationary" reference frame. So we're re-drawing spacetime from the perspective of someone using the red/tilde basis. All the vector relationships on the diagram are still the same, they just look a little different because they are re-drawn from the perspective of another reference frame. Does this make it clearer?

[@HowDoYouKnowThough]

@@eigenchris It is making a little more sense. I think the root cause of my misunderstanding is that I don't understand the shearing of the spacetime manifold. Doesn't that require an extrinsic point of view, so that the manifold has something to shear "into?" For example, take a universe that has a definite start and end in space and time. Then the spacetime manifold is a square with boundaries. In my mind, I want to put down all elements (events) of the spacetime manifold into a square, fix the manifold, and then put different coordinate systems on top of that. But how I lay out the events in the first place is observer-dependent (Event A is directly to the "right" of Event B is only true for one type of observer). This is where my hang-up is. I think the answer to my question is something along the lines of: "Topologically, a sheared plane is still a plane."

[@mskEduTech]

Again Good Explanation. Waiting for next .....

[@alex-my8hp]

do physicists ever think of vectors as abstract elements of a vector space? how can a vector be invariant if it isn't an arrow in space but just an element of a set? if the standard basis is of no more important then any other basis then how does one denote the apparently invariant abstract element of the VS if not with a coordinate vector? these questions haunt me. it seems to me like the notions of vectors being invariant and all bases being equal are incompatible.

[@eigenchris]

I think everything I've said in this video is compatible with the "abstract vector space" approach. You'd denote an invariant vector just by giving its name (as I've done with "S" in this video). You can expand S as a linear combination of any basis you like (you can come up will a million different bases if you like), but you're still just talking about the vector "S". Does that answer your question?

[@alex-my8hp]

@@eigenchris but how do you know what and how many basis vectors it's made of if it's not being thought of as an arrow in space or even an ordered tuple?

[@jakeaus]

@@alex-my8hp I dare say the truest answer to your question involves understanding how a manifold is formally constructed in a mathematical sense. I don't have Chris's gift of simple-yet-concise explanation so what follows may be too much info. Also I'm just a self taught hack so take it with a grain of salt :) In one way of thinking, the vectors we casually view as arrows joining two events in spacetime are most fundamentally just abstract vectors, belonging to the abstract vector space called the tangent space, which is an object that lives on a manifold. Now, talking about manifolds (curved or flat) and the tangent spaces at points in those manifolds is a game we can play with total ignorance of what the real world is like. But when we want to do physics we need to make a few leaps of intuition, and claim that the objects we have formally built using our mathematics are the right kind of objects to describe things in the real world (Riemann created much of the mathematical machinery that Einstein required to formulate general relativity, but Riemann did it in the 1850s, as an exercise in pure maths, without ever being motivated by the need to formalise a physical theory). So, to answer one of your questions explicitly, we simply claim that spacetime is a 4 dimensional manifold which is equipped with a 4 dimensional tangent space at each of its points, and since vectors live in the tangent space they must be 4 dimensional objects that require 4 basis vectors to describe in component form. The dimension of the vector is an intrinsic property of that vector, which emerges from the dimensionality of the underlying spacetime, which we have assumed a priori. Whether or not it is true that spacetime has 4 dimensions, or even whether there is truly a thing in the world that has the properties of a (purely abstract mathematical) spacetime manifold requires us to test the theory through experiment. If you're interested to see how exactly you can get from "element of a set" to "arrow in space" I recommend Frederic Schuller's lectures. Basically you can start with literally a set, then keep giving it more and more structure (a topology, a maximal atlas, a tangent bundle, a metric connection, etc) until its properties are so constrained that the thing that you have been calling an abstract vector is in every way exactly the same as the cartoon arrow version of a vector.

[@imaginingPhysics]

Is this your dilemma? : A vector is supposed to be a coordinate independent object, abstract (like velocity) or concrete (like the pencil). But to describe a vector we often to do it by using some coordinate system, which is obviously coordinate dependent, so not invariant. Don't these contradict each other?! The answer is to keep track of the coordinate system by coordinate vectors e_i. Details of the proces are demonstrated in these videos. If we forget them and only keep track of the coordinates/coefficients v^i we lose invariance. Invariance is a profound idea. Makes me positively confused too. Don't worry.

[@hariszachariades8299]

If the "spacetime separation vector" has components (Δx,Δt), there is no combination a Δx^2 + b Δx Δt + c Δt^2 that remains invariant under the Galilean transformation Δx' = Δx - v Δt, Δt' = Δt. Maybe this is the reason that we do not usually see discussions of Galilean spacetime.

[@eigenchris]

Yeah. The only options are to set either "a" to zero or "c" to zero. So you end up with a metric for only time or only space.

[@SzTz100]

Brilliant video

[@rupabasu4261]

Great explanation👍👍👍👍👍😍😍😍

[@colinn4239]

Great series thank you so much.

[@-_Nuke_-]

Is it true that in Galilean Relativity, the speed of each object in time is always V=1s/s? And that this speed is a constant and invariant speed? Constant because it can't be changed, so you can't move faster or slower in time... And invariant because it will always be the same for all observers. Correct? (ofcourse that is not true for Special or General Relativity...)

[@eigenchris]

You can think of it like that, yes.

[@-_Nuke_-]

@@eigenchris nice!

[@jigold22571]

ThankU for posting and sharing.

[@aBigBadWolf]

It felt like 90% of this video was already covered previously but maybe that's just because I've seen this stuff too many times.

[@eigenchris]

I'm trying to make this series accessible for people who haven't watched my other tensor videos. I realize I'm repeating myself.

[@zulqarnainchaughtai]

Really awesome❤❤

[@matthewbradley4644]

im gonna guess that special relativity doesnt shear time and space, but rotates it?

[@eigenchris]

Sort of. It's not a circular rotation, it's something called a "hyperbolic rotation". (Also called "Squeeze transformation" in math, or "Lorentz transformation" in physics).

[@astrolillo]

Wait: This video is still about Galilean relativity, so you CANNOT rotate the spacetime diagram when moving from one observer to the other since time is absolute so the same for all.

[@eigenchris]

In Galilean relativity, it's acceptable to do Galilean transformations, which is what I'm doing in this video. Each "time slice" stays at the same vertical location, so absolute time is still a defined concept.

[@LamontGranquist]

Pedagogically I wonder if planes and cars are the best use cases to teach relativity? People so instinctively mentally "look out the window" and do physics in the reference frame outside of them that they think is the correct one to be at rest. It might be better to do the examples in the back of a semi trailer which has no windows, or in the cargo hold of a plane. Considering something like a ping-pong game played under all three circumstances might world.

[@quantumsoul3495]

I know it's not the subject here m, but will you someday make a video about spinors and twistors ? There are many different definitions as with covariant derivative which you untangled, so it would be interesting to understand them more

[@eigenchris]

I don't understand spinors very well and "twistor" is a word I only know by name, so it's unlikely I'll be talking about them.

[@quantumsoul3495]

@@eigenchris Oh Ok

[@Sharikkhursheed]

This was brilliant bro.. but upload every day one video

[@arisperakis8550]

nice. keep going.

[@DrChrisB]

I guess you mean Maria Skłodowska-Curie…

[@averagecornenjoyer6348]

Gesundheit