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- [Voiceover] Let's say this is me

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and I am floating in space.

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My coordinate system,
my frame of reference,

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we've seen it before,

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we'll call it the S frame of reference.

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Any point in spacetime,

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we give it an x and y coordinates.

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And let's say that we have my friend.

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We've involved her in
many other videos before.

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She is traveling with a relative velocity

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to me of v.

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So v right over there.

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Her frame of reference,

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we call it the S prime frame of reference,

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and you can denote any event

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in spacetime by the coordinates

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x prime and

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ct prime.

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Now you could also do it
as t prime if you like,

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but we've been doing it as ct prime

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so that we have similar units.

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Now let's say that we have
a third character now.

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This is going to get interesting.

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Let's say this third
character is traveling

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with a velocity u in
my frame of reference.

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So u is equal to change in x

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over change in time.

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If we know all of this information,

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let's see if we can come
up, we can formulate a way

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if we know what the change in x

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and change in time is
and we know what v is,

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if we can figure out what
is this velocity going to be

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in the S prime frame of reference,

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in this purple friend's
frame of reference.

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What we want to figure out is what is,

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let me do it in that purple color,

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we wanna figure out what
is going to be the change

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in x prime over change in t prime.

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If we figure this out then we know

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what will this velocity
look like in the S prime

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frame of reference.

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Well let's just go back to
the Lorentz Transformations.

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Let's first think about what our...

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Let's first think about

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what change in x prime is going to be.

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Well our change in x prime is going to be

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the Lorentz factor times our change in x,

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change, this in white,
times our change in x

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minus beta

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times change in ct.

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Let me close those parentheses.

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Then we wanna divide that
by our change in time.

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We wanna divide it by our change in time.

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So change in time

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or change in t prime I should say.

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Well let me just go back to
the Lorentz Transformation.

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Let me write it the way
that I'm used to writing it.

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I'm used to writing it as

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change in ct prime, which
is the same thing as

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c times the change in t prime,

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is equal to the Lorentz factor

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times

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this is going to be change in ct

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or I could write it as c change in t

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minus beta

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times change in x.

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Once again I could have
written this as change in ct

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or I could write this
as c times change in t

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because the c isn't changing.

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So we know that all ready.

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If we wanted to solve for
just change in t prime,

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we could just divide both sides by c.

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So let's do that.

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If we divide all of this by c,

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what do we get?

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This is a form, we've
seen this in other videos

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that you might recognize
and you might see this

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in some textbooks.

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Our change in t prime is
going to be equal to gamma

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times, well c times delta t divided by c

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is just going to be delta t.

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And then if you take, so
minus, beta divided by c.

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Well let's just remind ourselves

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that beta is equal to v over c.

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So beta divided by c is

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v over c squared.

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I could write this as
minus v over c squared

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change in x.

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Change in t prime, let's write that now.

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That's going to be equal to

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you have your gamma

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times change in time

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in my coordinate system.

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So not the t prime just change in t

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minus v over c squared change in x.

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Immediately there's at
least one simplification

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we can do.

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We can divide both the numerator

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and the denominator by gamma.

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I keep wanting to change...

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So we can divide the numerator
and denominator by gamma

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that simplifies it a little bit.

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We can write this as our change in x prime

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over change in t prime

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is equal to

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in the numerator we have change in x

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and actually let me just
write out what beta is.

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Beta, I'll do it over here,

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minus v over c that's what beta is

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And actually let me take the c out.

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Change in ct is the same
thing as c times change in t.

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So times c delta t.

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v divided by c times c
is just going to be v.

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Let me write it that way.

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This part, these cancel out.

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This is the same thing as
change in x minus v change in t.

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Let me write it that way, simplfy it.

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Minus v change in t, that's our numerator.

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And then our denominator is change in t

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minus v over c squared change in x.

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It looks like we're getting close here.

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But we don't have the
change in x change in t.

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We have how much of a change in x

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we have for a given change in t.

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So what we could do is we
could divide the numerator

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and the denominator by delta t.

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We could multiply the
numerator by one over delta t.

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And the denominator by one over delta t

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which is equivalent to dividing

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the numerator and the
denominator each by delta t.

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And why am I doing that?

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Well, if I do that this
first term right over here

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is going to be delta x over delta t

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which we know.

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Then we have minus v, cuz
delta t divided by delta t

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is just going to be one.

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And then our denominator

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delta t divided by delta t is just one.

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Then we're gonna have minus v

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over c squared

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and then times delta x divided by delta t.

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This is cool.

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We've just been able to
figure out what our velocity

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in the primed coordinate system,

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in the S prime frame of
reference is in terms of

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the relative velocity between
the S prime frame of reference

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and my frame of reference.

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Delta x delta t the velocity
in my frame of reference.

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And well we know all of
this other stuff too,

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this is just delta x
delta t and this is v.

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Or we can even write it like this.

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We said delta x over delta
t is going to be equal to u.

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So it's equal to u minus v

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over one minus,

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delta x over delta t is just u.

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So this is just u here.

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It's u times v

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or v times u,

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all of that over c squared.

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This is a really, really,
really useful deriviation

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which we're going to apply some numbers to

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in the next video so we can appreciate

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kinda how fun this is on some level.