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MARKUS KLUTE: Welcome back
to 8.20, special relativity.

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We're going to continue our
discussion of galactic space

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travel.

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Here the situation is slightly
modified from the previous one.

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We still have Alice being
our ground control and Bob

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riding on a spacecraft in order
to explore planetary systems

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and solar systems.

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This situation is
different in the sense

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that the spacecraft
has an escape rod.

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So it's able to send probes to
planets in order to study them.

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And so, in this specific case,
the velocity of this escape rod

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is uB delta xB over delta tB,
as measured in Bob's reference

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frame.

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The direction of this
velocity is the same

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as the direction
of Bob's spacecraft

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when looking in
longitudinal direction.

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Again, as a reminder, velocity
is distance over time or delta

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x over delta t.

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So the question now is,
what does Alice observe?

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Because this is an activity
you should try to work out

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yourself, stop the video.

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I'll just continue here.

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So, if you calculate now the
velocity as seen by Alice,

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delta xA over delta tA, that's
given by gamma times delta.

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We just use Lorentz
transformation-- gamma times

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delta xB plus v times delta
tB over gamma times delta tB

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plus v over c squared delta xB.

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Now, we can cancel the
gammas and take out

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delta tB out of the brackets.

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And then we find uB plus v over
1 plus v over c squared uB.

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So this looks like an
addition of velocities

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with a correction factor of
1 plus v over c squared uB.

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Good.

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So does this make sense?

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So, whenever we have a
calculation like this,

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we should check that
it actually works out,

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that extreme cases
are preserved,

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and that units work out.

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So, again, the
units work out here.

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On both sides, you
find meters per second,

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the unit of velocity.

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If you check now what happens
if we set uB equal to 0,

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we know that the
escape rod is at rest.

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There's no velocity with respect
to Bob's reference frame.

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In that case, we
find that uA is equal

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to v, exactly the
velocity difference,

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the relative velocity between
those two reference frames.

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If that velocity is 0,
we find uA equal to uB.

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Again, that's expected.

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If Bob and Alice are in
the same reference frame

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and they observe
the same escape rod,

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they better measure
the same velocity.

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And, lastly, if we now, instead
of having an escape rod,

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we send a beam of light
out, which has a speed--

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a beam of light has the speed
of light, uB equal to c,

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we find that the velocity
observed by Alice

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is also c, which brings us
to an interesting point here.

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Yes, we still add
velocities with a little bit

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of a relativistic
correction, but we will never

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get larger velocities
of the speed of light.

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So the speed of light is
an absolute speed limit.

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Let's analyze this
a little bit more

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in the context of
our light clocks.

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So what now happens if
the velocity is equal to c

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is that gamma goes to infinite?

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And, in the context
of the light clock,

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you can notice that the upper
mirror can never be reached.

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It's moving with
the speed of light,

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the same velocity
as the light itself.

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So light is never
able to reach this.

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The clock will stop.

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All right, so there's an
absolute limit of velocity

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at the speed of light.

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OK, so now, so far, we
discussed only velocities

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in the direction in which
the two reference frames move

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or the second
reference frame moves

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with respect to the first one.

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What now happens if we consider
perpendicular velocities?

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So, in this case,
Bob's spacecraft,

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this escape rod goes up.

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Maybe he's circling the
planet, or he's just

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approaching the planet.

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And [INAUDIBLE] when
it [? pops at ?]

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that planet specifically.

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So here we want to work
out the example in which

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the perpendicular
velocity is not 0,

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but the longitudinal
velocity is 0.

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So what does Alice observe?

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So we do this as a
concept question.

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Which of the four
answers is correct?

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Is the velocity
unchanged because we

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are studying
perpendicular velocity?

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Is the velocity
smaller, larger, or you

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don't know because
you actually have

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to figure it out, work it out?

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OK, so the velocity,
as observed by Alice,

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is actually the absolute
value is smaller

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than the one observed by Bob.

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We can do the very
same calculation.

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So we have uB y is
delta yB over delta tB.

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And then, for Alice, this is
uA y delta yA over delta tB.

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So the y-component,
the length measured

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in y-direction
between Bob and Alice,

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is invariant, as we saw
in the previous section,

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but the time is not.

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So we do have to do the Lorentz
transformation of delta tA

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and find that, in the case
where uB x is equal to 0,

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that we just have to
divide uB y over gamma.

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Situation is a little
bit more involved

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when there's also a
longitudinal velocity,

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but you see here how
this would unfold.

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So 2 was the
correct answer here.

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So, while the length in
longitudinal directions

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are invariant, the
velocities are not.

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And that's because
time is suspect.

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Time needs to be corrected
in the two reference frames.