- [Voiceover] Let's say this is me
and I am floating in space.
My coordinate system, my frame of reference,
we've seen it before,
we'll call it the S frame of reference.
Any point in spacetime,
we give it an x and y coordinates.
And let's say that we have my friend.
We've involved her in many other videos before.
She is traveling with a relative velocity
to me of v.
So v right over there.
Her frame of reference,
we call it the S prime frame of reference,
and you can denote any event
in spacetime by the coordinates
x prime and
ct prime.
Now you could also do it as t prime if you like,
but we've been doing it as ct prime
so that we have similar units.
Now let's say that we have a third character now.
This is going to get interesting.
Let's say this third character is traveling
with a velocity u in my frame of reference.
So u is equal to change in x
over change in time.
If we know all of this information,
let's see if we can come up, we can formulate a way
if we know what the change in x
and change in time is and we know what v is,
if we can figure out what is this velocity going to be
in the S prime frame of reference,
in this purple friend's frame of reference.
What we want to figure out is what is,
let me do it in that purple color,
we wanna figure out what is going to be the change
in x prime over change in t prime.
If we figure this out then we know
what will this velocity look like in the S prime
frame of reference.
Well let's just go back to the Lorentz Transformations.
Let's first think about what our...
Let's first think about
what change in x prime is going to be.
Well our change in x prime is going to be
the Lorentz factor times our change in x,
change, this in white, times our change in x
minus beta
times change in ct.
Let me close those parentheses.
Then we wanna divide that by our change in time.
We wanna divide it by our change in time.
So change in time
or change in t prime I should say.
Well let me just go back to the Lorentz Transformation.
Let me write it the way that I'm used to writing it.
I'm used to writing it as
change in ct prime, which is the same thing as
c times the change in t prime,
is equal to the Lorentz factor
times
this is going to be change in ct
or I could write it as c change in t
minus beta
times change in x.
Once again I could have written this as change in ct
or I could write this as c times change in t
because the c isn't changing.
So we know that all ready.
If we wanted to solve for just change in t prime,
we could just divide both sides by c.
So let's do that.
If we divide all of this by c,
what do we get?
This is a form, we've seen this in other videos
that you might recognize and you might see this
in some textbooks.
Our change in t prime is going to be equal to gamma
times, well c times delta t divided by c
is just going to be delta t.
And then if you take, so minus, beta divided by c.
Well let's just remind ourselves
that beta is equal to v over c.
So beta divided by c is
v over c squared.
I could write this as minus v over c squared
change in x.
Change in t prime, let's write that now.
That's going to be equal to
you have your gamma
times change in time
in my coordinate system.
So not the t prime just change in t
minus v over c squared change in x.
Immediately there's at least one simplification
we can do.
We can divide both the numerator
and the denominator by gamma.
I keep wanting to change...
So we can divide the numerator and denominator by gamma
that simplifies it a little bit.
We can write this as our change in x prime
over change in t prime
is equal to
in the numerator we have change in x
and actually let me just write out what beta is.
Beta, I'll do it over here,
minus v over c that's what beta is
And actually let me take the c out.
Change in ct is the same thing as c times change in t.
So times c delta t.
v divided by c times c is just going to be v.
Let me write it that way.
This part, these cancel out.
This is the same thing as change in x minus v change in t.
Let me write it that way, simplfy it.
Minus v change in t, that's our numerator.
And then our denominator is change in t
minus v over c squared change in x.
It looks like we're getting close here.
But we don't have the change in x change in t.
We have how much of a change in x
we have for a given change in t.
So what we could do is we could divide the numerator
and the denominator by delta t.
We could multiply the numerator by one over delta t.
And the denominator by one over delta t
which is equivalent to dividing
the numerator and the denominator each by delta t.
And why am I doing that?
Well, if I do that this first term right over here
is going to be delta x over delta t
which we know.
Then we have minus v, cuz delta t divided by delta t
is just going to be one.
And then our denominator
delta t divided by delta t is just one.
Then we're gonna have minus v
over c squared
and then times delta x divided by delta t.
This is cool.
We've just been able to figure out what our velocity
in the primed coordinate system,
in the S prime frame of reference is in terms of
the relative velocity between the S prime frame of reference
and my frame of reference.
Delta x delta t the velocity in my frame of reference.
And well we know all of this other stuff too,
this is just delta x delta t and this is v.
Or we can even write it like this.
We said delta x over delta t is going to be equal to u.
So it's equal to u minus v
over one minus,
delta x over delta t is just u.
So this is just u here.
It's u times v
or v times u,
all of that over c squared.
This is a really, really, really useful deriviation
which we're going to apply some numbers to
in the next video so we can appreciate
kinda how fun this is on some level.