Main content

### Course: Physics library > Unit 16

Lesson 3: Lorentz transformation- 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

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Introduction to the Lorentz transformation

So we've got two coordinate systems from the perspectives of two observers. How can we convert spacetime coordinates between these? Enter the Lorentz transformation!

## Want to join the conversation?

- Did I miss something? I watched all the previous relativity videos, and I saw in "Angle of x-axis in Minkowski spacetime" that the two angles α must be identical, but where did we show the inverse tangent relationship (as Sal states at2:15)?(70 votes)
- You're right, the video must have been left out somewhere, but we can get to the angle pretty easily by thinking about how the relative velocity of Sal's friend relates to the slope of ct'. Since her velocity is .5c, the slope of ct' is 2. We could graph it out and see that if her velocity was .25c, the slope of ct' would be 4; and if her velocity was .75c, the slope of ct' would be 1.333. Which makes sense because as she goes relatively faster and faster, the slope gets closer and closer to 1 (the slope of light).

So we can see that the slope of any ct' would be**c/v**.

Now because we know that slope is**change in y/change in x**we can use a little trigonometry to find the angle between ct and ct' (Draw a right triangle with with ct' as the hypotenuse and see for yourself!)

The tangent of an angle equals opposite side over adjacent side which for this triangle is**change in x/change in y**or**v/c**.

So**tan α = v/c**

Or α = tan^-1 (v/c)(70 votes)

- I think I find this really hard.

I started relativity by watching pop-sci videos and pop-sci books, which now makes it even more counter-intuitive for me.

Especially those thought experiments of the train and so, I seem to have the wrong information. I solved some of the problems with the thought experiments (the problems I had) and now it still hurts my brain

I wanna learn this interesting stuff but this is counter-intuitive to the fullest like so much crazy. I have no problem in accepting speed of light will be constant, since the electromagnetic field is stationary with respect to any observer and so on. But I find other stuff crazy and confusing.... Will I be able to do it?

Sorry I'm a bit demotivated cause of sleep deprivation but ah....(8 votes)- Unfortunately this is counter intuitive and there is no way around it. Our intuition is mainly based on our every day experiences and they do not include any of these Relativistic effects. This is also a big issue with understanding quantum mechanics as well.

The only way to get past this is to keep at it.(7 votes)

- Can someone provide a proof or the derivation of the Lorentz factor? How was this factor discovered?(3 votes)
- Or study these

https://en.m.wikipedia.org/wiki/Lorentz_factor

https://physics.stackexchange.com/questions/173268/deriving-the-lorentz-factor-gamma

https://en.m.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations

All of these explain (hopefully) what you asked(2 votes)

- can you guys do a series using the alternate coordinate system?(3 votes)
- Why are the coordinate transformations of space and time between uniformly moving ststems, -(where the grid lines are set up for each system such that the transformation equations of Lorentz Result, where the space and time mix inextricably and intimately in the new spacetime concept)- able to be,(and why are they), expressed in terms of the factor 1/{sqrt(v^2/c^2)}?

What does this factor represent which makes the transformation equations be what relations they are between the relative velocity of the reference frames , the speed of light c which is the same value in both frames and the space and time coordinates used in its calculation for each of the uniformly moving reference frames?(1 vote)- I do not appreciate that you decided to track down my personal email and tried to contact me directly and to get Sal to get me to respond directly to you via email.

Did you watch the videos I gave the links to? Here they are again: I have updated the first 3 so that they go to the TedED education site where the original videos came from and the 4th is from the PBS Spacetime YouTube channel:

http://ed.ted.com/lessons/the-fundamentals-of-space-time-part-1-andrew-pontzen-and-tom-whyntie

http://ed.ted.com/lessons/the-fundamentals-of-space-time-part-2-andrew-pontzen-and-tom-whyntie

http://ed.ted.com/lessons/the-fundamentals-of-space-time-part-3-andrew-pontzen-and-tom-whyntie

https://www.youtube.com/watch?v=YycAzdtUIko

The first three links to the videos/lessons go through the reasoning behind the use of the Lorentz transformation. This stems from the fact that the space-time interval is defined by Δs^2 = (c * Δt)^2 - Δx^2 - Δy^2 - Δz^2 and that the space-time interval for light traveling in a vacuum is 0. Here is a link the the derivation of the Lorentz transform by Einstein from the appendix of his book on general relativity: http://www.bartleby.com/173/a1.html(3 votes)

- i have a question if lets say i am travelling at the speed of light in a rocket and ahead of me there is a wall since according to me time stops (by lorentz transformation) and so i will never hit the wall but according to some other rest frame who is watching both the wall and me will say that i collapsed into the wall and since both frames of reference are equally valid what will happen will i collapse or not please define in brief(1 vote)
- First of all, you can't go the speed of light.

Second of all, in your reference frame, time is normal. You will see a wall coming toward you and then - wham.(2 votes)

- Isn't the Lorentz factor time dilation?(1 vote)
- So can we say in other words that, conceptually, Minkowski graphs represent
*perceptual*differences of events (based on frames of reference) of space and time due to the fixed speed of light? Where I am in my mind conceptually is that the "bending" of spacetime is viewed from an arbitrary frame of reference that takes into account all given frames. If this is correct, please help me understand the implications of the concepts.(1 vote) - can someone tell me how energy of light when noticed from a moving frame of reference with velocity v is E(1+v^2/2c^2)(1 vote)
- So since the new pair of coordinate axes are not orthogonal to each other, this means that you can use express ct' in terms of x' and vice versa and hence there is a continuum of spacetime, they are not distinct of each other. Is this a mathematical way of realising the existence of spacetime?

I'm actually having difficulty understanding a non-orthogonal axes system....(1 vote)

## Video transcript

- [Voiceover] So we've
already been able to explore a lot with our little thought experiment where I am floating in space. I'm at the center of
my frame of reference, and right at time equals zero in my frame of reference, a friend comes in a
spaceship passing me by with a velocity v, I'll say the magnitude is v, and it's going in the
positive x direction. We're just going to focus, and we have been focusing, just on the x dimension for simplicity. And we've thought about reconciling space and time in my frame of reference relative to her frame of reference. The conundrum that we've faced in previous videos is how do we reconcile that the speed of light is always going to be the same in
every frame of reference? And to reconcile them, we
had to essentially come up with the idea of spacetime. I should say it even faster, spacetime. Spacetime, let me write it out, spacetime. And the first time I
heard about spacetime, I assumed that people were just talking about space and time as independent things and just plotting your
point in space and time. But when people talk about spacetime, they're really talking about
this continuum of one thing, and we're just talking
about different directions in spacetime. They could have called
this something else. They could have called this spime, or tace, or stace, or a lot of different things. But this is spacetime, and it's this idea that
it's this continuum. And when we started to
make spacetime diagrams, we realized in order
for the speed of light to be absolute, that time and space weren't as independent of each other as we thought, and they
weren't as absolute as we thought. And we constructed these Minkowsky spacetime diagrams for each of our frames of reference. So my frame of reference, the spacetime diagram, is here in white, and for my friend's frame of reference, her spacetime diagram is here in this blue color. The angle formed between these axes, between the x and the x prime axis and the ct and the ct prime axis, this angle alpha here, this is going to be dependent on her relative velocity in
my frame of reference. So if her velocity is v, or the magnitude of her velocity is v in my frame of reference, this angle we've already seen is going to be the inverse tangent or the arctangent of the ratio between her relative velocity and the speed of light. So this is going to be
equal to the inverse tangent of v over c. So the faster she goes, these two things are going
to start squeezing together, and if somehow she were
to approach the speed of light, they would both
approach a 45 degree angle and actually start to coincide if they were actually able to approach the speed of light. And that's all interesting already, this idea that space and time are not as independent, that it's all a continuum called spacetime, but some
of you have probably said, well, I want to deal with some more tangible numbers here. For example, if this event right over here in spacetime, we can think about it from my frame of reference, and we can think about it from her frame of reference. In my frame of reference, I would view the coordinates here. This coordinate would be x, and this coordinate right over here would be ct. We had a whole video on why we think of time in terms of ct. The units here literally would be meters. We could think of it as
light-meters if we like. So that would be the coordinates in my frame of reference. Well, what would be the coordinates in her frame of reference? Well, we've already
thought about how to read these Minkowski spacetime diagrams. To find her x prime coordinate, we would just go parallel
to the ct prime axis. So that would be the x prime coordinate in her frame of reference. And to figure out the ct prime coordinate, we would just go parallel
to the x prime axis. So this would be the ct prime coordinate. Now how do you actually go in between, transform, from x to x prime and from ct to ct prime? And to do that, we're going to introduce in this video the Lorentz transformations. And what they do is they allow us to do exactly what we just needed to do. They'll allow us to go from x, ct to x prime and ct prime. And to help us think about it, I'm going to introduce some variables, and hopefully it will show the symmetry of the Lorentz transformations. You might see them written in other ways in other sources, and we'll reconcile all of those in the future. But the Lorentz transformations, we'll start with what we
call the Lorentz factor because this shows up a lot in the transformation. So I'll just define this ahead of time. So the Lorentz factor,
denoted by the Greek letter gamma, lowercase gamma, it is equal to one over the square root of one minus v squared over c squared. Now sometimes you might
even see it written like, well, I'll write it another way. Sometimes you might see
it written as gamma, let me do it in that same color, same reddish color, gamma is equal to one over the square root of one minus beta squared. You might say, well, what is beta? Well, beta is another
variable that shows up a lot when we're thinking
about special relativity. And beta is just the ratio between the relative velocity, her relative velocity in
my frame of reference, and the speed of light. It shows up a lot, even this angle alpha here, we could have said this
is the inverse tangent of beta. This becomes a little bit
simpler when you write the Lorentz factor. And when we look at the
actual transformation between the coordinates,
we'll see that beta becomes useful again, at least the way I like to write it. So if we want to think about what x prime is going to be, so we can write x prime is going to be equal to the Lorentz factor, let me do it in that red color, is going to be equal to the Lorentz factor times x minus, and now we're going to say beta times ct, and ct prime is going to be equal to
the Lorentz factor gamma, let me do that same color again, switching colors is sometimes difficult, gamma times, and we'll see it's just the other way around. It is going to be c, let me do it in that green color, ct minus, you might even guess what I'm about to write based on the symmetry that we see here, ct minus beta times x. And I really want you to appreciate this because it really shows
that space and time are really just different directions in the spacetime and
there's a nice symmetry to them right over here. Notice we have an x and we have an x. We have a ct and we have a ct. So when we're thinking about x prime, it's x minus beta times ct, and we're scaling it
by the Lorentz factor, and then when we're thinking about time, well we do it the other way around. We're still scaled by the Lorentz factor, but now it is ct minus beta times x. Now this might all seem like Greek to you, and we actually are using Greek letters, but in the next video, I'll actually use some
sample numbers here, and you'll see that evaluating these is just a little bit of
straightforward algebra.