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### 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

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# Deriving Lorentz transformation part 2

Continuing the algebra to solve for the Lorentz factor.

## Want to join the conversation?

- But what if you take a point that isnt on the diagonal. (light). doesnt this assume that we are talking about light or something like that?(6 votes)
- The points on the diagonal world line of light is used to calculate the coordinate transformations required to keep the world line of light at a 45 degree angle.

Once the transform is derived you can use it for transforming the coordinates that are not one the diagonal world line of light.(4 votes)

- Aren't x=ct and x"=ct" only true when v=c?(4 votes)
- I'm confused; if we assumed that the event we were observing is a beam of light (or at least connected with light) shouldn't the transformations be only true in the case of light observation? And can we prove the Lorentz transformation assuming the observed object isn't connected with a light beam?(3 votes)
- Why do we multiply x and a prime? I know that it gives us the scaling factor but what is the significance of multiplying x and x prime? On what basis we multiply it?(2 votes)
- Because we say light has absolute speed and going with it everytime. This means light always has 45degree in minkowski diagrams. You can check pictures about this.

Also think like that, you shooted light and it moved for a sec, means 3x10^8meter, you check it on x side. Now look for ct side, t is 1 and c is 3x10^8 as you know. They same for anytime.(1 vote)

- I still don't understand why the light path is where it is (always at a 45 degree angle). Please explain if you understand. Thank you.(1 vote)
- That's how the axes of the graph are set up.. The unit on the axis is t and on the y axis it is ct. That means the slope of the graph is c, and if you draw equal-sized units, the line will be at 45 degrees because for every extra unit of time, light travels one extra unit of ct in distance.e(1 vote)

- Why would we not substitute in c for v as well? Isn't v just the velocity of whatever is moving, and since it is light in this case, wouldn't that make v equal to c?(0 votes)
- Why does the scaling of the Galilean transformations by the factor Y=[1/sqrt(1-v^2/c^2)]give us the Lorentz transformations which will keep the speed of light c( and the forms of all laws) the same in all uniformly moving frames when the coordinates are used to calculate the ratio Dx/Dt or Dx' /Dt' which is the speed of light when x and t or x' and t' lie along the photon world line path at 45 deg to the x axis where x' and t' are skewed symmetrically relative to x and t?? Why does the scaling of the Galilean transformations result in the Lorentz transformations x'=Y(x-vt), t'=Y(t-(v/c^2)x} which keep the ratio Dx/Dt=Dx'/Dt'=c=the same in all such systems?(0 votes)
- I can see that scaling the Galilean transformations by Y means that bi0th x and t coordinates are multiplied by Y so we get

x' =Y(x-vt)=Y.x-Yv.t which is in analogy with the rotation of coordinate systemswhich has its coordinates transforming as x'=(cos Q).x+(sin Q).y where (cos Q) is the projection function of x onto the x

axis so that by analogy this would seem to correspond to Y which scales x onto the x'axis of the squashed system and where Yv seems to be analogous to sin Q which projects the y=ct coordinate onto the y'=ct' axis.,now I see how the circular functions can perform the orthogonal perpendicular projection BUT I do not see how Y or Yv can perform the parallel tilting(squashing) projections for the Lorentz transformations of the x coordinate onto the x'axis(Y is the projector function), or the ct corrdinate onto the ct' axis(projectorYv)(0 votes) - How does the "stipulated condition" that the speed of light (as calculated from the coordinates ) be the same value of the ratios Dx/Dt and Dx/Dt' both be equal to c in all uniformly moving frames imply the particular way the coordinates transform between such relatively moving frames ,which transforrmations are called the Lorentz transformations and which are expressed in terms of the Lorentz factor

{1/sqrt(1-v^2/c^2)] which results from the invariance of the spacetime interval this invariance being equivalent to the condition of constant speed of light in all inertial frames.

The condition tha(t Dx/Dt=dx'/Dt' be a constant when calculated from the transformed coordinates of the frames tells how the coordinates have to be skewed(tilted) so that this condition Dx/Dt=Dx'/Dt'=c holds when the grid lines(coincidence and simultaneity lines) of the systems are tiled due to their relative uniform motion,

That is the invariance of dI^2=c^2dt^2-dx^2 shows how the coordinates have to be skewed(squashed) and tilted in order for this condition on the speed to hold .Then these transformations are gib=ven in terms of the Lorent factor Y.

x'=Y(x-B.ct)=Y(x-vt)=Y.x -Yv .t =Y.x--YB.ct

t'=Y(ct-B.x)=Y(ct-Bx)=Y.t-Y(v/c^2).x=Y.t-Y(v/c^2).x

B=v/c, Y=1/sqrt(1-v^2x/c^2)(0 votes) - How does the factor [1/sqrt(1-v^2/c^2)] reflect the invariance of the interval dI^2=c^2dt^2-dx^2=c^2dt'^2-dx'^2=dI'^2? that is the constancy of the speed of light in all uniformly moving systems(inertial systems) ,systems that have no curvature but are flat and have intervals defined as c^2dt^2-dx^2 that is interms of corrdinates x,ct or x',ct'(0 votes)

## Video transcript

- [Voiceover] We left
off in the last video trying to solve for gamma. We set up this equation
and then we had the insight that look, we could
particular, we could pick a particular event that is
connected by light signal, and in that case x would be equal to c t, but also x prime would
be equal to c t prime. And if gamma's gonna hold
for any transformations between events, between x and
x prime and t and t prime, it should definitely hold
for this particular event, and so maybe we can use
this to substitute back in and solve for gamma, so
that's exactly what we're gonna do right now. So for all the xes I'm gonna
substitute it, with the c ts, so I'm gonna substitute with
a c t, so x, substitute ct. x, I'm gonna substitute c t. x, substitute c t, and that's it. And then all the x primes
I'm gonna substitute with a c t prime, with a c t prime. So x prime, c t prime. x prime, c t prime. And then I have an x prime here,
so it's gonna be c t prime. So let's simplify, and
now I'm gonna switch to a neutral color. So I'm gonna now have, I'm now gonna have c times t times t times t prime. So that's gonna be c
squared, actually let me just keep using the t, t primes, t, t primes. OK, I'll still do a little
bit of color coding. T prime is equal to gamma squared, gamma squared times, so
it's gonna be c squared c times c is c, let me do
that in the yellow color, so c squared, t times t prime, so t times t prime. and then we have plus c times v times t times t prime. plus c times v, I'll just write it this way. C times t, times v, times t prime. And then we have minus, minus, let's see, we're gonna have a c here. So minus c, minus c times t t times v, times t prime times v times t prime. I wrote this v in blue just
so it matches up with this. And we see something
interesting is about to happen. And then finally we have minus v squared. Minus v squared, times t times t prime. Times t times t prime. It doesn't look that much
simpler, but we're about to simplify it a good bit. And so we're gonna get
these two middle terms to cancel out. So plus c t v t prime,
minus c t v t prime. So those are going to cancel out. And then every other term has a t, t prime in it. So let's divide both sides of
this equation by t t prime. And so we're gonna get, if
we divide the left-hand side by t t prime, we're just
gonna be left with c squared, and then we're just gonna
divide everything by t t prime, and there our whole thing
has simplified quite nicely. Our equation is now, I'll
continue it over here. Our equation now is, c squared is equal to gamma squared,
is equal to gamma squared, times c squared, times c squared minus v squared. Minus v squared. Minus v squared. Close the parentheses.
And now we can divide both sides by c squared minus v squared, and we would get gamma squared, and I'm gonna swap the sides, too. So gamma squared is equal to c squared, c squared over, c squared, I'll write
it all in one color now. c squared minus v squared. Now if we like we can divide the numerator and the denominator by
c squared, in which case this will be equal to one over c squared divided by c squared is one, and then, v squared divided by c squared, and we are in the home stretch now, we can just take the
square root of both sides, and we get, we deserve a
little bit of a drum roll. Actually let me continue it up here where I have some real estate. We get gamma is equal to
the square root of this, well the square root of one is just one, over the square root of the denominator. square root of one minus v squared over c squared. So hopefully you found that as satisfying as I did, because all we
did, we just thought about the symmetry of x prime is
gonna be some scaling factor times the traditional
Galilean transformation, and x is going to be
some scaling factor times the traditional Galilean
transformation from the prime coordinates. We use that and it's important that we use one of the fundamental assumptions of special relativity,
that the speed of light is absolute in either frame of reference, that x divided by t is c, that
x prime divided by t prime is going to be equal to c, for some event that's associated with a, a light beam. We use that to substitute back in, and we were able to solve for gamma. So this looks pretty neat. And so some of y'all might be saying, well what about, what about, so we've been able to do the derivation for the x coordinates, but what about the, the
Lorentz transformation for the t and t prime coordinates? And I'll let you think
about how we do that. And I'll give you a clue,
it's just going to be a little bit more algebra,
and we're going to do that in the next video.