Relativity 104e: Special Relativity - Spacetime Interval and Minkowski Metric
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- เผยแพร่เมื่อ 21 ก.ค. 2024
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0:00 Intro (Spacetime Interval & Minwkoski Geometry)
1:55 Length vs Spacetime Interval
5:26 Invariance of s^2
8:50 Physical meaning of s^2
16:07 Euclidean vs Minkowski Metric
23:06 Hyperbolic Trigonometry
27:46 Hyperbolic Rotations
31:30 Summary
A number of people have asked why I'm not using the imaginary unit i to get i(et).i(et) = -1, or i(ex).i(ex) = -1. The short answer is that this "i" trick is a "hack" that works in some coordinate systems, but not others. In particular, this trick fails if our spacetime basis is non-orthogonal. Allowing dot products of basis vectors to be negative is a better solution because it works it absolutely all coordinate systems.
As an example, if we wanted we could take et and ex~ and use them as a non-orthogonal basis for 2D spacetime. This wouldn't correspond to a physical reference frame, but we can still do all the math I've talked about (calculate proper times, proper accelerations, etc.). This non-orthogonal basis would give us a metric with off-diagonal elements (where et.ex~ is not zero), and the "i" trick cannot handle these off-diagonal terms properly since we're stuck with a factor of "i" leftover. We could also take a vector pointing along a leftward lightbeam and another pointing along a rightward lightbeam and use it as basis for spacetime (in this case, the diagonal elements along the metric diagonal would be zero, and only off-diagonal elements are non-zero). This is called a "null basis" and is mentioned here: en.wikipedia.org/wiki/Minkowski_space#Causal_structure. Again, this doesn't correspond to a reference frame, but the mathematics of relativity still work. I have not studied it yet, but I know the Kerr metric from general relativity has off-diagonal elements in some coordinates: en.wikipedia.org/wiki/Kerr_metric
To sum up, the "ict" or "ix" trick is more of a "hack" that works nicely in some specific cases, but not other cases. But using a metric (defining dot products between all basis vectors) is the more "correct" way to handle things, and it will work in any coordinate system, no matter how crazy it is.
@@theta-rex In my view the "negative dot product" and "imaginary time" are both arbitrary. Either method is a possible choice for describing hyperbolas. A number of coordinate systems in relativity involve basis vectors that are not normalized to length 1 (Flat space metric in spherical coordinates, Rindler metric, Schwarzschild metric, Kerr metric). In particular the Kerr metric (for a rotating black hole) is often expressed in a non-orthogonal basis, so the metric has off-diagonal elements. So I think it's best to take a approach that can handle all coordinate systems.
You are spot on with this.
@@theta-rex I feel like you're conflating two uses of the word "imaginary". When talking about "imaginary numbers", the word "imaginary" is just a name. They could just as well have been called "red numbers" or "happy numbers" or "vertical numbers". Then there's the common English use of the word "imaginary" which means "existing in your head", and that's unrelated. All numbers exist inside your head, from "integers" to "real numbers" to "imaginary numbers" to "quaternions" and so on. I don't see any of this being related to how time is measured or experienced. At the end of the day, we have the postulate that the speed of light is the same in all inertial frames of reference, and this leads us to Lorentz transformations, which leads us to hyperbolas and the formula for the spacetime interval. The fact that the metric signature lets us calculate the spacetime interval seems like a perfectly valid justification for it. The choice of +--- or -+++ being arbitrary is also fine, just as the choice of saying electron charge is - and proton charge is + is arbitrary, or the choice of right being +x and left being -x is arbitrary (you can swap conventions in all of these cases and get the same physics). All I really see to this whole issue is that spacetime doesn't obey Euclidean geometry, it obeys Minkowski geometry. The distance between two separate points in Euclidean geometry being zero is impossible, but the distance between two separate points in Minkowski geometry being zero is fine and ordinary.
@@theta-rex I'm interested to know why you think the Kerr metric is wrong, since it's the standard accepted description of a rotating black hole, and has been around since 1965. I'd be curious to know how far you've studied into General Relativity. I think in GR, there's a much larger emphasis on the fact that coordinate systems don't matter too much--the underlying geometry is the more fundamental thing, in the same way that a flat table is flat, regardless of whether or not you put an orthonormal basis on it. Coordinate systems and basis vectors don't really play a true role in physics... it's the underlying geometry that determines the physics, so we always have the freedom to put new coordinate systems on our geometry, even if they are very strange once. All of this is true in Special Relativity as well, but maybe it's not emphasized as much in textbooks.
@@theta-rex I think the names "real" and "imaginary" are rather confusing names that could be improved, but they're just the names we got stuck with for historical reasons. But I'm not sure what other arguments I can make against your points other than just repeating what I said above, so I don't have much else to say.
I wrote and recorded this entire video, calling s^2 the "spacetime invariant" before realizing it's actually called the "spacetime interval" by pretty much everyone. Sorry if that confuses anyone.
...or at least it was before your video.
You did mention at the very beginning that it's also called the spacetime interval, and it is an invariant, so no harm done.
Invariant seems like a better term
Didn't Einstein prefer the term 'invariant'?
@@phugoidoscillations 😂 I'm gonna call it spacetime invariant too from now on
I've watched Leonardo Susskind, MIT Open Courseware, ICTP, and of course it's some random ass guy on TH-cam who has the clearest explanations of these topics.
Thank you, EigenChris
That's because Chris knows that we are dummies, while Leonardo doesn't have a clue about our profound ignorance. LOL.
@@paulbizard3493 this guy is smarter than Leonardo Susskind. To have such videos on Relativity you need to have a deep level of understanding that most university professors simply do not even at the highest level (I speak as someone that went through Physics at a top University only to learn Relativity and Quantum Mechanics by myself later on in life at a much deeper level). This "random" guy on TH-cam is a genius regardless of what his actual accomplishments in the field are.
This is better than the university lectures for which I've paid 9250 for. Just incredible. You're the living proof that teachers cannot explain things and blame it on students cause you've literally opened up an entire new level of understanding of special relativity to me. Thank you so much!
I made an attempt to teach myself tensor analysis years ago and never finished. Thanks to Chris I've pretty much nailed it now. (Over the last 2 weeks I watched the "Tensors for Beginners" and the "Tensor Calculus" playlists both twice!) as well as a few of his Relativity videos (although I'm pretty good w/ relativity). Chris is a unique individual. I used to get mad every time a paper brought up a "covector". Now I know EXACTLY what that thing is!
I have been searching this kind of amazing graphical interpretation of the mathematical underpinnings that encapsulates and correlates general and special theories of relativity so beautifully, for at least a decade.
I am not a student of physics or mathematics but a middle aged surgeon right now.I used love physics during my premedical days.
TH-cam had made possible for me to be introduced to the world of quantum mechanics and special theory of relativity,without getting into the depth of formalism.Some videos by Sean Caroll and Leonard suskind that I tried to follow did no good to me ,as I hadnot had the foundation of formal university education in physics.Same was true for Veritasium videos ,which dealt with complex stuffs without mathematical stuffs.In a nutshell,those videos got bogged down with hardcore math or did no math at all for lack of proper lucid approach.As all these were happening,I yearned to learn why geometry was indeed hyperbolic ,but found no answer, which I could grasp,up until today,when I had gone through this wonderful mathematical explanation by Eigenchris...thanks a lot for your labourious work
Hi. 'science clic English' complements this video nicely.. You're welcome. Old UK duffer here :)
This is fantastic, thank you. One of the clearest explanations I've seen,
Thrase lectures are pure gems. Thank you!
right on time! My professor just talked about invariant interval and I had no idea what was that lol
Interestingly, the mostly minus metric is called the “West Coast” metric and the mostly plus metric is called the “East Coast” metric. The names come from prominent physicists who used each sign convention (probably Feynman and Schwinger, respectively). The mostly minus convention is favored in particle physics. The mostly plus convention is favored in general relativity (e.g. in MTW) and I tend to like it because for students, it more closely resembles the Pythagorean Theorem. In the end, it’s just a sign convention though, and either is fine to use.
I'm hoping my choice doesn't come back to haunt me later. I chose mostly-minus because I like having positive results for timelike vectors (+proper time for 4-position, +c for 4-velocity, and +mass for 4-momentum). Do you know why MTW and other GR use mostly-plus?
eigenchris I don’t know what drives one choice or another besides appeal to authority. LOL Seriously though it may be the resemblance to the Pythagorean Theorem, but cursory googling shows a multitude of justifications for each. I think it’s great to see everything worked out both ways! Great for teaching.
@@eigenchris Something Alex Flournoy said in his GR course is that he likes the mostly plus metric because he does a lot of work in higher dimensions and the determinant is always the same. With the mostly minus metric it has the opposite sign for even/odd dimensions.
@@joshthompson6548 I've watched some of his videos. That's an interesting point. I don't see myself working in >4 dimensions anytime soon, though.
Thank you sir your videos are amazing . I understood all concepts easily
Encore une belle vidéo toutes mes félicitations
Merci!
That's great! Thank you so much!
i dont understand why such great video only have 600 thumbs up
ThankU for sharing and posting
Wow... this is amazing
17:44 It is also interesting how one can, from the "intuition" that a spacetime measure (as in self dot product) needs to be invariant WRT inertial frames, work out which characteristics (the values of the basis dot products) that such space must have in order to abide to such rule.
Just using the simple dot product definition, and the Lorentz transforms, lead to a system where the only allowed signatures are +--- or -+++ (both solve it).
22:27:
In 103d (specifically @ 10:33), You also used to put contravariant in in a row during Euclidean geometry, so I assume that using it again in a row in Minkowski Geometry shall have no surprise !.
Could you explain about twin paradox, because I have trayed looked many youtube to explain it but I still confuse, one say don't need acceleration to distinct frame of reference, in contrast the other say need acceleration. And how about in GR I think we should differ frame of reference because Gravity for sure need acceleration, but many explanation stay use relative of frame of reference?
I have that video planned for Relativity 105. The short explanation is that if you calculate the proper time for a stationary observer (one vector) and then calculate the proper time for the moving observer (2 vectors, there and back, with their proper times added together), the moving observer will have a shorter proper time. The moving observer is not inertial because their path cannot be made into a straight line using a Lorentz transform, so it is fundamentally different from the stationary observer (which is an inertial frame).
Thanks a lot 💙
I really loved your video! I had watched many videos about Relativity in the past and wanted to understand the math underlying it. I have a small doubt though, should'nt it be s^2=3^2+5^2 at 14:35 , line no.3. Thank you making this series😄
Yeah, my bad.
I do not understand how a space basis vector dotted by itself =-1. I mean it is obvious because it is the only way to make S invariant? If the "most pluses convention" applies, then that dot product can be either 1 or -1 ?
Any clue I missed in there?
It's just a definition. We define a unit spacelike vector dotted by itself to be -1. And yes, we do it so that s^2 can be invariant. You could also use the opposite definition and use +1 instead, if the time vector dotted with itself becomes -1 instead.
@@eigenchris Isn't defining such dot product similar to defining 2+2=5? Doesn't look compatible with linear algebra.
@@pedrokrause7553 The dot product is just a definition. There's nothing objectively incorrect or correct about it. In standard linear algebra, you would say that ex.ex = +1. But we can define a new dot product operation that gives a different answer, like ex.ex = -1. The dot product I use here with ex.ex = -1 is sometimes called the "Lorentz" dot product or the "Minkwoski" dot product. This is definitely different than the standard dot product and it will lead to geometry that behaves in non-obvious ways, but there's nothing inherently contradictory about it. In more advanced geometry classes, the dot product is called "the metric" because it helps us measure lengths and angles. A standard metric has the property that v.v >= 0. But we can also invent a metric where v.v gives negative results as well. The technical term for a metric like this is "pseudo-metric".
thank you!
Chris: Did you ever notice that the timelike interval looks just like the square of the magnitude of a quaternion?
Q = a + i*b + j*c + k*d .......and...... |Q|^2 = a^2 - b^2 - c^2 - d^2 en.wikipedia.org/wiki/Quaternion.
It looks as though the same form tells us about events in spacetime as well as 4D rotations
I never did notice that. There might be some interesting connections to investigate there. I've been learning about clifford algebras/geometric algebra recently and they can be used to do Lorentz transformations as well as to form quaternions. I'm not sure if those are related to the connection.
is the spacetime invariant event which both observers agrees upon located outside both reference frames?
Spacetime invariant is just the distance between events. All observers agree on that. Time and space separately don't mean anything.
Do hyperbolic angles still work in the full 3+1 dimensional setting? How do we add velocity vectors when they're not parallel?
For spatial rotations in 3D, you start with a 3x3 identity matrix and then pick 2 axes you want to rotate. Let's say you want to rotate in the xy-plane, then you will populate the xx, xy, yx, yy entries of the matrix with sin and cos as needed, and leave the rest as 0s and 1s from the identity matrix. If you want to combine several rotations, you just multiply several rotation matrices with different circular angles.
It's similar for hyperbolic rotations in 4D: you start with a 4x4 identity matrix and pick 2 axes you want to rotate (must be time and 1 spatial dimension). Then you fill the 4 entries (say tt, tx, xt, xx) with sinh and cosh as needed, leaving the other entries as 0s and 1s. If you want to combine several different rotations together (say tx, ty, tz), you just multiply the several hyperbolic rotation matrices together with different hyperbolic angles.
@@eigenchris Does that mean that the formula will be different depending on order of rotation?
@@kyrilo1993 Yeah, Lorentz transformation matrices don't commute, so the order of multiplication matters. The resulting matrix is still considered a Lorentz transformation because it will preserve the spacetime interval.
Hi Chris I have a question about hyperbolic functions. The angle (theta) between axis ct and ct prime some books say that tan (theta) = v/c , but also there are other books that say that tanh (theta) = v/c. The problem is that tan and tanh are not the same functions. In particular, if you use hyperbolic function the angle theta is compressed as you are close to 45 degrees that is expected as v is close to c. So, I think is wrong to use tan instead of tanh . What is your opinion about it?
It's definitely tanh (θ) = v/c = β. It's very quick so you may need to pause but I show this on the slide at 30:00. I suspect with those books it is either a typo, or by "tan" they are somehow implying "tanh", or possibly that they are just wrong. You might be interested in reading this wikipedia article: en.wikipedia.org/wiki/Rapidity
Chris, why not put all these great explanations into a book?
3:25 invariance of the expression (ct)^2-x^2-y^2-z^2 is another way of stating that the speed of an object moving at velocity c is the same in all coordinate system. Or am I mistaken?
Reasoning: if I demand the above expression must be unchanged in coordinate a linear transformation the only options are Lorentz transformations. And Lorentz transformations imply the speed of any object moving at speed c is the same in all systems. (so, since light moves at speed c, the speed of light is always the same)
Yes, you can think of it this way. The metric value along a beam of light is always zero, in all coordinate systems.
Thanks. Then it is a techical but equivalent way of stating the second postulate of special relativity. Funny.
14:40 I find it a bit confusing to bind the definition of "Proper Length" to a frame in space-like domain.
From this, it seems like the Proper Length cannot be defined for a time-like frame, which is not true, being it defined as the length measured in a frame at rest with the endpoints to be measured (symmetrical to Proper Time).
It is true that the real (real as in not-C) math breaks down with s^2 being positive (the square root to get x becomes one of a negative number) but maybe this requires a different reasoning on the relations of s WRT Proper Length?
What software do you use to create these? Its really smooth.
Microsoft Powerpoint.
Would love a full course on non euclidean geometry.
I briefly talk about some non-euclidean geometries in my cosmology videos. "Specifically Relativity 110b" (where I talk about Spherical and Hyperbolic universes) and "Relativity 110f" (where I have an 8-minute segment where I talk about de Sitter and anti-de Sitter spaces). But I don't plan on making more videos on those. Hopefully you can find the videos you want somewhere else.
I don’t understand why ex dot ex = -1 in Minkowski metric. We are just deciding to be like that and see if things work correctly?
Yes, we define it that way because we want a value (spacetime interval) which doesn't change under Lorentz transformations.
How to derive the Minkowski metric? or is it just the definition
How Minkowski himself had got the metric form?
It's the metric held constant by the Lorentz transformation. In 2d spacetime, you can look for a 2x2 metric matrix that is constant before and after a Loretnz transformation. The result is the Minkowski metric.
If there is a notion of negative distance in the minkowski metric, does it mean it is a pseudo metric space?
Yes, that's correct.
Hello,
In minute 29:58 it was displayed that coshφ=γ. I would like to ask why is that? I mean this equality is arbitrary or comes from something else?
Thank you very much for your time
That equality is arbitrary (if you like, you can think of it as a definition for φ), but the important thing is that we can derive that tanhφ = sinhφ/coshφ = β (and therefore sinhφ = γβ) without any further assumptions.
@@eigenchris thank you for your quick reply!
Hello. I have a question about tensors for which I would like to attach an image. What's the best way to contact you?
Maybe upload it to an image-sharing site like imgur and paste the link in a comment?
@@eigenchris very well. I have written it in the following google doc: docs.google.com/document/d/12nwhTvCLx11sW-aRIjM_UXwi-u8GAEg-qH0Fr6sbhDk/edit?usp=sharing
I'm a bit unsure why you would create a 6x6 matrix here. Usually a vector U and its opposite -U are not taken to be independent directions. Your 6x6 matrix will be composed of four 3x3 sections that are all identical...except two will be positive and two will be negative... again because we don't consider U and -U to be different directions. It seems you are trying to take the "outer product" / "kronecker product" of the vector A and the vector V to get a rank-2 tensor, aka matrix (probably a 3x3 matrix). While you can do this mathematically, I'm not sure what this matrix would represent physically. Is there a certain physical situation or problem that made you consider doing this?
@@eigenchris thank you! The motivation for this just came to me as an idea for trying to comprehend tensors. Although this has no physical meaning, I thought this would be an analogy to try to understand the concept. I agree that the negative dimensions of the vector would be neglected so it would actually be represented by a 3x3 matrix. Although this isn't a "real" physical tensor, this exercise definitely helped me to understand Tensors a bit more since I "invented" one. Thank you!
13:51 Beta represents the slope (in x/ct terms) of the ct-tilde axis, and 1/Beta (or ct/x) the slope of the x-tilde axis, shouldn't x-tilde and ct-tilde be swapped when Beta>1 ?
beta>1 would imply we're going faster than light. Lorentz transformations cannot handle this because they break at beta=1. You could maybe invent a version of it that swaps the x and ct axes but I'm notnsure wha this would mean. I talk a little bit about this in thebprevious video 104d.
@@eigenchrisYeah, what I meant is that when slope is greater than one, ct-tilde crosses over (over the x=ct boundary line) outside the light cone, while x-tilde crosses over inside the light cone, if one follows the geometric constructions laid by the previous videos.
But if s^2 < 0 then Gamma turns complex, and although the math remains valid, things gets weird from a physics POV 🙂
Hi Chris, I'm trying to find a maths proof that the spacetime interval (ct)^2-x^2 remains unchanged under Lorentz transformation. So far I can't find it in your excellent videos.
Is the explanation at 5:26 not good enough?
@@eigenchris I was looking for a derivation for s^2=(ct)^2-x^2. I agree though that the proof at 5:26 is sufficient. Thanks very much for taking the time to reply and for your excellent videos.
@@EventHorizon618 As in a derivation for where the formula for s^2 comes from? I think it just comes from noticing that the Lorentz transformation leaves it unchanged. It's similar to how you could get the formula r^2 = x^2 + y^2 by noticing that it remains unchanged after a circular rotation... although the Lorentz-version is probably less obvious since circles are much more familiar to the average person compared to hyperbolas.
You said that we can find the dot product even for non orthonormal bases. But what even is the definition of dot product in that case? For orthonormal bases we have defined dot product such that et.et=1, and so on, but I dont see how this can be extended to non orthonormal bases.
If you accept the dot product definition for orthogonal vectors, and also add definition that the dot product is bilinear (you can distribute over it and pull scalar multipliers out) you can get the dot product for any pair of vectors just by writing them as a linear combination of orthonormal vectors. Alternatively, you could make the definition that a.b = cos(θ)/|a||b| when both vectors are spacelike, and a.b = cosh(θ)/|a||b| when one vector is timelike and the other is spacelike.
Hey Chris, at 3:08, what if you choose an orthonormal basis whose magnitude wasn't the same as the original basis? For example what if R = 2 e tilda instead of 5 e tilda? Wouldn't the magnitude be 4 instead of 25? Thus not equal?
If the basis vectors don't have length 1 (they are orthogonal, but NOT orthonormal), then the entries in the metric tensor matrix would change in such a way that the vector length would stay the same. You can't use the simple formula x^2 + y^2 in this case... you'd need to take the basis vector dot products into account and do (x^2)(e_x · e_x) + (y^2)(e_y · e_y). If R = 2 e_x~, this would mean (e_x~ · e_x~) = 6.25
@@eigenchris I see, thanks a lot. Your videos are the best on TH-cam!
I dont understand one thing, please help: why the dot product of ex.ex is -1? Whats the definition of the dot product for basis vectors? Or do you start by saying (ct)2^2 - x^2 is the general definition of the dot product and the dot products of basis vectors are calculated always with this definition and this gives the metric tensor?
I started with (ct)^2 - x^2 and defined the vector dot products from there.
at 11:00
i derived F=d(mv)/dt and E=mc^2, with my time( dt) from the stationary frame. i had no problems. but the proper time concept is so confusing. it looks like time contraction, completely opposite of time dilation. a person moving with the clock measures his time is the stationary time, just like i measure my own time when i am NOT moving. then both people think the other's clock is ticking slower, then the other who sees my clock and thinks his universe as a whole must be ticking faster. can you fix my brain? by adding some more words and details for this proper time thing? i hate to do this way F= d(mv)/d tau
Remember the equation Δt = gamma (Δt'+v/c^2 Δx')
Proper time is important in relativity because two observers disagree on simultaneous events.
if two events are separated by Δx' , then the time between two events in its frame is Δt' = 0 if they were simultaneous.
in another frame, Δt > 0, and therefore in an unprimed frame, the time between two events is greater than zero.
t' becomes Tau as the distance between two events in x' coordinate approaches zero, and hence
Δt = gamma (Δt'+v/c^2 x') is simplified to Δt = gamma ΔTau
F= dgamma(mv)/dt = gamma^3 m (dv/dt), where t is the time in your own reference frame, you can also call it Tau because it records your path through spacetime, but Tau is usually describe another moving body in space that not in your frame.
hence d Tau is not needed.
proper time is just the time between two events that happened at the same location in its own reference frame.
what you're describing dp/dTau is called four-force.
the change in energy and momentum in all three spatial dimensions over proper time.
Maybe this is a silly question, but when we hear about the equations for hyperbolas we always hear about them equaling one. So, why in this case can we have s in the equation for a hyperbola this value may be larger than one?
Just as you can have a circle of and radius, you can have a hyperbola of any "s" value. A circle with radius 1 is a special circle called the "unit circle". A hyperbola with s=1 is a special hyperbola called the "unit hyperbola".
@@eigenchris
Thank you for clarifying!
Don't you think that proving the invariance of space-time interval s^2, by using Lorentz Transformation is a circular argument and shows nothing? Lorentz Transformation can only be derived if we assume that S^2 is invariant. What is the point of this argument?
The "derivation" I used in the 104a and 104b be videos didn't make explicit use of s^2. I instead used a geometrical argument to show that a constant speed of light in inertial frames required a transformation that lead to the time and space axes being at equal angles with a beam of light, and also argued the determinant of the transformation matrix should be 1 to ensure there was no preferred direction in space. Maybe you could argue the concept of s^2 buried in there implicitly, but I wanted to show everyone explicitly that s^2 is unchanged after a Lorentz transformation.
Hi Chris! In video 14:52, in the vector's own reference frame it has length 4 and it called "longest possible length" right? But why in previous reference frame it has length 5? Shouldn't 5 is longer than 4?
If you watch the 104c video, around 22:25 of that video I talk about length contraction, and how you can't just measure a vector using 2 different sets of basis vectors to get length contraction. Instead, whichever reference frame you pick, you need to must measure an object's length along lines of simultaneity. It's a bit hard to explain in words, but the pictures in the 104c video make it clear, I think. Measuring the length-contracted distance using the blue basis will give you a shorter length.
@@eigenchris Thank you so much! I love your courses!
Chris, I did your tensors calculus series and generally I'm quite comfortable with that stuff, including arbitrary metric tensors. What I'm struggling to understand here is why we only limit ourselves to a very specific subset of coordinate changes (that are consistent with Lorenz transformations)? In tensor calculus, coordinate changes are very general and invariant objects tell us something fundamental about underlying geometry. Here it seems we just postulate a set of allowed coordinate changes and work from there. Is this starting point as fundamental as, let's say, axioms of euclidean geometry? I feel this must be the case but it seems too narrow as a starting point... Do my questions make sense?
You're right that a Lorentz transformation is a very specific transformation: it is the transformation used to move from one inertial frame to another. In inertial frames with Minkwoski-orthonormal basis vectors, the time and x coordinates match up with the real time and space measurements people in that frame would actually measure. In arbitrary coordinate changes, the coordinate variables might not mean anything physical at all... they would just be some strange number we use to mark a point on the spacetime plane. I do an example of a curvilinear coordinate change in the 105 videos when I talk about accelerating frames. The problem with those is that the coordinates you get don't necessarily correspond to physically meaningful time and space measurements.
@@eigenchris Thanks! Let me ask another question which in my mind is related to my first comment. What is the justification for treating the spacetime invariant as a length-like object? After all, we discovered it’s invariance in a very specific subset of transformations. The leap that confuses me is that after we calculate the invariant under Lorentz transformations, we then postulate that this will be invariant under all transformations and build our geometry around that assumption (with a specific metric tensor that is compatible with that assumption). At least this is what I think is going on. What if the spacetime interval that we discovered while studying the Lorentz transformations is postulated to be a different, not a length-like object? To me this seems as possible alternative. After all length is not the only invariant property there is. This question is puzzling me because I don’t follow this leap in your lecture - everything else is crystal clear, I just don’t understand why we are so quick to assume that the invariant t^2-x^2 will be “used as” length.
@@przadka It's not much different than the cast of spatial rotations and the "invariant" x^2 + y^2 (which is just the radius squared). Rotations will keep this quantity constant, but we can still calculate the length of a curve in any arbitrary coordinate system. The length of a curve is computed using 2 tangent vectors (2 contravariant things) and a metric tensor (a twice-covariant thing) so the resulting length will be a scalar that is the same in all coordinate systems. The spacetime interval is the same idea applied to a pythagoras formula with a minus sign: (ct)^2 - x^2. Lorentz transformations (hyperbolic rotations) keep it constant, but we can still calculate the "length (proper time)" of a curve in any coordinate system. Again, 2 vectors are 2 contravariant things, and the metric is twice-covariant, so proper time is a scalar that is agreed on in all coordinate systems.
@@eigenchris Thanks Chris. I think I understand it now. This is how I see it in simple steps:
1. We start with 2 things: 2 dimensional spacetime and a set of allowed coordinate transformations that preserve speed to light - Lorenz transformations.
2. We discover that under these transformations there exists an invariant - OP(S,S) defined as t^2-x^2.
3. Expressing this in tensor language: An operator that takes a vector S twice and produces an invariant has to be something similar to a dot product. Moreover, since S is expressed as a set set of contravariant coordinates we need to “process it” via a double-covariant tensor to produce an invariant with OP(S,S). There is no other way to produce an invariant from two contravariant vectors but to contract them with a double-covariant tensor.
4. This double-covariant tensor is a metric tensor that has a form of diagonal matrix with 1 and -1s -> Minkowski metric tensor.
5. So we arrive at a conclusion that a spacetime with Minkowski metric keeps the t^2-x^2 invariant.
I want to follow up with something that I understood only recently. The spacetime invariant allows us to calculate the length for any vector but also it allows us to calculate the dot product between any two vectors. This means that once we have the formula for the spacetime interval (which immediately gives the diagonal entries for the Minkowski metric), we also have all the dot products and all other entries in the Minkowski metric matrix. The dot-product can be calculated from lengths using the formula: |u+v|^2=|u|^2+2u.v+|v|^2. This formula can be used to derive that dot product e_t*e_x is equal to zero for the Minkowski orthonormal basis.
Dear Eigenchris
at 104e 18:57 after intoducinng the Minkowski metric you show that, using the Minkowski metric, S*S is invariant under Lorentz transform.
What I could not find - or may be I simply missed - in you lessons is weater the reverse is true; I mean:
If a tilda space is transformet into smile space through a Lorentz transform the ( only ?) invariant metric need be the Minkowski's?
I have checked the net but could not find much so I devised a simple proof of sort that you might be interested in.
I am deeply gratefull for you incredible work.
Best Regards
Ezio Oreste Rivetti
The answer is "yes", but I'm not sure how to go about it algebraically. If you take a point in spacetime and continuously apply a Lorentz transformation to it, it will trace out a hyperbola. The Minkowski metric is the metric that keeps "distances" along hyperbolas constant, similar to how the Euclidean metric keeps distances constant along a circle.
@@eigenchris how am i to send you a .pdf
Hi Chris I'm a bit confused here. First, when S < 0 does it imply the object moves at a speed bigger than C? About Lnot "proper lenght" I assume we can calculate it even when S>0 just changing the inertial reference frame in order to get the Ct component = 0.... Why do you refere to Lnot just when S
If an object were to travel along a path with "S^2 < 0", then it would be travelling faster than light. But that can't physically happen. All massive particles must travel along paths with "S^2 > 0", and massless particles must travel on paths with "S^2 = 0".
Intervals with "S^2 < 0" should instead be interpreted as physical lengths/distances. We can always find a frame where the two ends of an interval with "S^2 < 0" are at the same time, so it can be thought of as two ends of a path in space.
Are you working in a system where c=1? It is much more confusing cause in this system you can't define c from other power of c (c^2 from c for example). SI system is better in this case on my opinion.
I'd rather not work in a system where c=1, because then I have no idea where the c's belong in the formulas. So I'm just letting c be itself.
@@eigenchris I wrote that cause you wrote S^2=(ct)^2 - x^2 = 5^2 - 3^2 = 16 not taking into account c=c (299 792 458 m/s in SI system). The correct equation in SI is S^2 = (c^2)*(t^2) - x^2 which is equal to 25*c^2 - 9 which is not equal to 16. It is equal to 16 only in the system where c=1.
@@DmAlmazov I'm assuming ct=5
Chris! I just found you on an Alan Macdonald video. Do you think you'll cover some GA/GC topics applied to physics on the future?
I'm sorry, but what are "GA" and "GC"?
@@eigenchris Geometric algebra and geometric calculus, sorry.
@@Ottmar555 Oh, gotcha. If I ever do that, it would probably be at least a year from now. I'm not sure GA/GC has really "clicked" for me yet. I wanted to use it to understand spinors in quantum mechanics, but my understanding there is still foggy.
@@eigenchris I know what you mean. I'm a chemical engineer, so I mostly work in continuum mechanics/transport phenomena. I must say that your videos on tensor algebra and calculus really helped tensors "clicking" for me. However, now I'm trying to translate the tensor formalism into a geometric calculus one, and I must say although hard, it seems to really pay off in many areas.
I'm curious where it's "paying off"? To me GA/GC seems very pleasing in theory, but I haven't seen any examples of problems that it actually makes easier to solve.
But why assume that gamma = cosh to begin with?
When the spacetime interval is constant, it traces out the shape of a hyperbola, so it makes sense that the transformation that preserves it is a hyperbolic rotation (similar to how constant Euclidean distance traces out a circle, and so the transformation that preserves it is a circular rotation).
Thank you.😍😍❤❤
Please start General Relativity as soon as possible.
At time 14:37 the third line reads (9)2. It should read (3)2
Yeah, 3^2 = 9. My bad. :)
How Minkowski metric is derived?
The definition comes from the Lorentz transformation. The Minkowski metric is the quantity that the Lorentz transformation leaves constant.
But why ex dot ex is equal -1, is it a definition or what
It's a definition for spacelike vectors of length 1.
@@eigenchris thx
At 19:11 shouldn't it be ë_t=gamma(e_t-beta*e_x) and ë_x=gamma(e_x-beta*e_t). and not plus. I've tried and it seem to give me the Minkowski metric tensor. Great videos by the way.
For the basis vectors +beta will Lorentz transform you to a frame moving right and -beta will Lorentz transform you to a frame moving left. beta is just a velocity, so it can be positive or negative.
@@eigenchris Okay, thanks.
14:34 You made a slight mistake 9^2 should be just 9, or else 3^2 on the third line.
Great series, though.
I watched the earlier videos in this series because I was learning for my entrance exam. Now I am watching this because I am bored
Why someone would use such math to deacribe all these things? Where is the step where euclidean geometry fails? What does it describe better?
The formula (ct)^2 + x^2 doesn't have much meaning in special relativity, since it is not a number that different reference frames agree on. The definition of proper time (tau) is very important because it gives us a way to take derivatives with respect to time in a way that all reference frames can agree on (this will be shown in the next video).
In euclidean geometry the physical length that you measure is invariant (in O.N basis).
But, in reality you can't use euclidean metric anymore because the physical length that you measure is not invariant. It actually depends if the length is moving or not. That is why we use minkowski metric.
In Euclidean geometry, what is shown in the video is the space diagram (it doesn't take into account the time component because it's assumed to be invariant).
In Minkowsky geometry that is shown in the video, the diagram is spacetime and it takes into account the time component instead of just space.
Euclidean geometry fails because it assumes time is invariant while in fact it is not. It was the spacetime interval that is invariant.
To put it simply, the area that he is talking about in Euclidean is a circle as you see it in real life.
But in the minkowski diagram it's not "rectangle" as you see it in real life, because one component is length, the other component is time.
hoooo! If I watch n videos on special relativity, I'll see n different theories of special relativity.
Sir, tensor calculus is not completed yet. Plz do something.
What else did you want to see covered in Tensor Calculus?
1st comment 😊😁
There is a type of Space-Time diagram in which the scale for both systems is the same. They are called "Loedel Palumbo Diagrams" and with them any analysis of special relativity is significantly simpler. They were developed in the mid-20th century by the Uruguayan physicist Enrique Loedel Palumbo from the simple, but brilliant idea, of considering in a diagram of Minkowski not one, but two "mobile" systems with the same speed, but in opposite directions and then remove the "fixed" system from the middle and... voila! you have two systems with the same scale! .The relative speed between these two systems is now given by the sine of the angle between the axes, not by the tangetic and trigonometry is that of all life. It is a shame that they are not very widespread.The deduction is very simple and can be found in the following link
th-cam.com/video/o4kKeG8PyyM/w-d-xo.html
Hello, what is your opinion about the following transformations: X'=X - kVt , and t'=kt , (uppercase= vector), k = [(vcos(n)/c +sqrt(1-v²sin²(n)/c²)]^(-1)? THANK YOU.
I'm not sure what to think about it, just looking at the equations. Where did those equations come from?
Mmmm 😊
dude u know that ur crazy? Like I understand the stuff you are talking about?!? My ego is getting so big, I can feel it's gravitational pull on me...
Glad to hear it. General relativity doesn't start until the 107 videos, though.
Trying to figure out how to use this math to make a lasagna
Please let me know if you figure that out. I could probably convince way more people to learn it.
great example of making total nonsense seem half way rational.
Bla Bla Bla. My videos show how any bone head can derive the SR mathematical equations without any knowledge of physics.
Are those videos taken from this channel?
th-cam.com/users/NEWKNOWLEDGEvideos
@@eigenchris Yes. They cut me off from my own channel. I had decided to add two touch safety verification keys, and did so, but they seemed not to be working. Both were verified as being active, as was my verification email, but at the same time it was said that my two step verification was not enabled, so I enabled it. That simple action then wiped out both the two safety keys, along with my verification email. After that, there was no way what so ever that I could open my own channel account. So here it was me trying to make things safer, but I got tossed out instead. So I created three new channels, and loaded them with my videos once again. HELIFY NOE, HeliFy NOE, and Helify Noe. Why they did this to me, all I can say is, hell if I know. Sorry if I sound pissed off. Anyhow, I still did manage to derive the SR equations despite me being a high school dropout with no physics education at all. I was classified as being the dumbest person in the class due to my way of thinking being NOT the norm. My report card was full of F's and D's etc. I suffered from a nasty head injury at the age of 10. From then onwards, I could no longer hold on to knowledge memories like every one else, which is why I dropped out of school, but that did not make me stupid. I created those videos to prove it. 55 YEARS NOW WITH A CERAMIC PLATE ON THE LEFT SIDE OF MY HEAD, but I am still ticking.
this is interesting but the monotone voice has made me stop watching
why do the tilda vectors examples have 0 for their x and y components. The picture shows the unit vector of length 1.
What part of the video is this? Did you mean why are thr y and z components zero?
@@eigenchris 3:20 you rotated and the y component became 0 and at 4:46 x component became 0. How is this possible their is still a length in your picture or are we just imagining?
@@BrianRiendeau At 3:20 I'm measuring the length of the R vector using the ex and ey basis vectors. In the 2nd example with ex-tilde and ey-tilde, the vector R is parallel to ex, and so it requires no ey vectors to build it. Same goes for 4:46: in the 2nd example the vector S is parallel to the et basis vector and so it requires no ex basis vectors to build it.
Thank you