- Introduction to the Lorentz transformation
- Evaluating a Lorentz transformation
- Algebraically manipulating Lorentz transformation
- Lorentz transformation derivation part 1
- Deriving Lorentz transformation part 2
- Lorentz transformation derivation part 3
Lorentz transformation derivation part 1
Using symmetry of frames of reference and the absolute velocity of the speed of light (regardless of frame of reference) to begin to solve for the Lorentz factor.
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- At3:00what is the reason to suppose that the relationship between the Galilean transformation and the relativistic transformation is simply a scaling factor (gamma) rather than some more complex relationship?(11 votes)
- gamma is derived from the assumption that the speed of light has to be the same in all reference frames
For a derivation, go to youtube and search for "Sixty Symbols Gamma"(3 votes)
- At3:59-- aren't you assuming too early that gamma(-v) = gamma(v)? If gamma were equal to say, 1/(1-v/c), for instance, then the scaling factor would, in fact, be different.(4 votes)
- no it wouldn't because if you substitute (-v) or (v) you are gonna get the same result since v is squared and if you square the negative you get positive.(2 votes)
- Could someone explain to me about the Galilean transformations? Thanks.(3 votes)
- Newtonian physics codifies Galilean Transformations in its mechanics in velocities and implies that velocity depends on how you move even when you consider yourself inertial,for instance,if you're running at 1m/s and you were passed by a car whose velocity was 3m/s,then the way you view its movement is 2m/s because of its velocity 3m/s - 1m/s =2m/s. This is a brilliant analogy yet it's wrong,because it implies that at high speeds when you're stationary and you see a super vehicle moving at a half of speed of light and then he uses a laser pointing at the positive x direction,then it implies from there that the cosmic speed of light will be faster which would,in other words,imply that c is infinite and doesn't apply to e=mc^2 because it'd require an infinite amount of energy to support higgs mechanisms contribution to mass.
(I think I may make mistakes so uhm you can see few ideas on the internet that may help.)(3 votes)
- At 9.29 to 9.38 what does the line in the purple and later in the yellow mean and represent? I think it is the world line path of the photon shot out of Sallys flashlight relative to the S' system and it is not (but I am not sure about this) the path of the photon shot out of Sal flashlight relative to the S system . Is this true?(3 votes)
- It could be the world line path of light from both Sal and Sally's flashlight, in both of their reference frames. This yields 4 possible scenarios!
Which of them had the flashlight does not matter because it was shone at t = 0, when their positions overlap anyways.
The reference frame does not matter because the path of light is the angle bisector between the two axels. This might be a better definition than that the path of light is at 45 degrees.(1 vote)
- Why does the x-axis in the right graph not point in the second quadrant as well?(2 votes)
- I think it is for consistency. If we call the origin O, the ct label C, the ct' label C', the x label X, the x' label X', then the two congruent angles in the left graph would be COC' and XOX'. Similarly, the two congruent angles in the right graph are COC' and XOX' with the x' label pointing into quadrant 4.(1 vote)
- At2:51, why is it x' = (x-vt), as opposed to x'=(x+vt)?(2 votes)
- at8:46why we assumed that the event which they're observing is a beam of light why won't it be another thing?and if it is another thing will the Lorentz factor be different?(1 vote)
- Why gamma is used as scaling factor over there?(1 vote)
- Yeah but, from both of our frames of reference, we see each other going to the "right" of us at velocity v, so why are we saying -v??
I mean, imagine that if you see me travelling towards your right, then from my frame you are travelling to the right.... same speed...
So right is positive. And thus the positive itself shifts as frame of reference changes!
Is it like a convention or so?(1 vote)
- "right and left" are only defined relative to which way you are facing, which is ignored here. A better (although still flawed) analogy would be compass directions. Eg, if we think of positive x as East and negative x as West, then v and -v go in opposite directions. if I am travelling at v relative to you, then I am moving "East", positive along the x axis from your point of view, and you are moving at -v, which is "West", negative along the x axis from my point of view.(1 vote)
- I don't understand the difference between -v and v.
Wouldn't it only happen if we define one direction as positive and another as negative?
Shouldn't time dilation and length contraction be independent of the direction(we assume +ve)?
Besides ,assume if both Sal and his friend are looking towards and moving away from each other at C/2(in respective frames) ,shouldn't the relation between Sal's t and t' and his friends t and t' ,be the same due to symmetry?(1 vote)
- [Voiceover] So in all of our videos on special relativity so far, we've had this little thought experiment, where I'm floating in space, and right at time equals zero, a friend passes by in her spaceship. She's traveling in the positive x direction, velocity is equal to v, and we draw space-time diagrams for both of us. First, I draw my space-time diagram in white, and then I overlay her space-time diagram. And the angle that is formed between the time axes and the position axes, right over there, that's going to be dictated by how fast v is, how fast she is actually traveling. And we give the space-time diagram from her frame of reference, we see with the little primes right over there. Now, one thing that you might have been thinking about throughout this entire series, is well, if I perceive her traveling with a velocity of v in the positive x direction, if we took her point of view, and that's what I have right over here. If she views herself as just floating in space, what she will see me as right at time equals zero, actually I'd say right at t prime is equal to zero. We're saying that t prime and t equals zero are coinciding. Right at that moment, she will see me fly by at negative v, going in the negative x direction. Once again, there is no absolute frame of reference. These frames of reference are all absolute... I'm sorry, these frames of reference are all relative. And so you could imagine what it would look like, 'cause if we drew her space-time diagram where her ct prime axis and x prime axes that they are perpendicular to each other and then based on that, my space-time diagram would be at an angle. It's at an angle like this. You can kind of see the positive ct axis is in the second quadrant here because I'm traveling with the velocity of negative v, but these angles are going to be the same. This is going to be alpha and that is going to be, let me write this, is going to be alpha and this is going to be, and this right over here is going to be alpha. Now what I want to do in this video is use this symmetry, use these two ideas to give us a derivation of the Lorentz Transformation or the Lorentz Transformations. And the way we might start, and this is actually a reasonable way that the Lorentz Transformations were stumbled upon, is to say, all right, we could start with the Galilean Transformation, where we could say, all right, the Galilean Transformation would be x prime is equal to, is going to be equal to x minus v times t. V times t. Now we already know that if you just use the Galilean Transformation, then the speed of light would not be absolute, it would not be the same in every frame of reference. And so we had to let go of the constraints that time and space are absolute, and so there's going to be some type of scaling factor involved. And so we can call that scaling factor, gamma. So, we could say all right, let's just postulate that x prime, if we assume the speed of light is absolute, is going to be some scaling factor, gamma times x minus vt. Well, you could make the same argument the other way around. If you view it from her frame of reference, and you're trying to translate it to my coordinates, you could say, well, x, instead of just using the Galilean Transformation that x is going to be equal to, x is going to be equal to x prime, and now instead of a v, we have a negative v, right? So if you subtract a negative v, in fact, let me just write it that way. X minus negative v times t prime, that would be the Galilean Transformation, but whatever scaling factor we used here, there's a symmetry here. I shouldn't have to use a different scaling factor if I assume a different, kind of, if I'm in a different frame of reference. So, if we assume the absoluteness of the speed of light, we're going to have some other scaling factor just like that or we could rewrite this as x, let me do that same color, we could rewrite it as x is equal to this scaling factor. I'm really having trouble changing colors today. It's gonna be equal to that scaling factor times x prime, subtract the negative plus vt prime. And if you ignore the scaling factor right over here, this is the Galilean Transformation from the primed frame of reference to the non-primed frame of reference. So an interesting thing is what is this scaling factor? How do we figure out what that scaling factor is going to be? And so we can do a little bit of interesting algebra here. What we could do is, Actually, let me just write what I just wrote, let me write it right below here. So we could say that x... Once again changing colors is difficult. We could write that x is equal to our scaling factor, gamma times x prime, times x prime plus vt prime. And now what I'm going to do in order to have myself an equation that involves all of the interesting variables, I'm gonna multiply both sides of this equation by, one way to think about it, I'm going to multiply both sides of this top equation by x. So if I multiply the left-hand side by x, I'm going to have x times x prime, x times x prime. And then the right-hand side of the equation, I could multiply by x, but x is the same thing. I'm saying it's the same thing as gamma times all of this business. So, I'm just going to multiply the left-hand sides of the equation and I'm going to multiply the right-hand sides of the equation. So if I multiply the right-hand sides of the equation, I am going to get gamma squared times, and I'm gonna have a big expression here, and so just really applying the distributive property twice, x times x prime, x times x prime, and then x times positive vt, so, that prime doesn't look like a prime, x times x prime plus x times positive vt plus x times, actually positive vt prime I should say, gotta be careful here. X times positive vt prime and then I have negative vt times x prime. So it's gonna be negative vt times x prime, times x prime. And then finally, I'll have negative vt times positive vt prime. So that's going to be, I could write that as a negative. Let's write that as vt squared, I'm sorry, negative v squared. And actually, let me delete this parentheses. I don't wanna force myself to squeeze for no reason. So I'm gonna have negative v times v, so that's negative v squared times t, times t prime. Times t prime. And now let me place my parentheses. So how can I use all of this craziness here to actually solve for gamma? And here we're going to go back to one of the fundamental postulates, one of the assumptions of special relativity and that's the speed of light is absolute. You're going to measure it to be the same in any frame of reference. And to think about that, let's imagine an event that is connected with the origin with a light beam. So let's say right at time and t prime is equal to zero, I were to shoot my flashlight and let's say it hits something at some point. We look at some event right over there and they're connected by a light beam, by photons. So let me connect them. So let me connect them. And so if you say, and once again this could be me turning on my flashlight and the photon at some future, at some forward distance and some forward time, it could just be at some position or maybe it hits something, it triggers some type of reaction. Who know what it does, but we're going to talk about this event right over there. That event in my frame of reference, its coordinates are going to be x and ct. And since we know the speed of light is absolute and the way that we've set up these diagrams, any path of light is always going to be at a 45 degree or a negative 45 degree angle, we know that x is going to be equal to ct. X is going to be equal to ct for this particular case. I could draw it on this diagram as well, if I'd like. Just to show that I can, so let me draw that. So it would look like this. It would look like this and we would once again have x equaling ct. How would you read that? Well, to get the x coordinate, you go parallel to the ct axis. So that would be the x coordinate on this diagram. And then the ct coordinate, you go parallel to the x axis. So, just like that. But, once again, x is going to be equal to ct, and, similarly, because the speed of light is going to be absolute in any frame of reference, if we look at x prime, x prime is going to be the same, is going to need to be equal to ct prime, ct prime. If we look at over here, x prime is going to be equal to ct prime. Once again, because light, this is going to be at a 45 degree angle. So x prime is equal to ct prime. They're connected by light events. So if you take your change in x divided by your change of time is going to be the speed of light. So what we can do is use this information, for this particular event, if gamma's going to be true for all transformations, it definitely should be true for this particular event. I can use this information to substitute back in and then solve for gamma. And that's exactly what I'm gonna do in the next video, although, I encourage you to try it on your own before you watch the next video.