WEBVTT
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MARKUS KLUTE: Welcome back
to special relativity, 8.20.
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After discussing energy
and momentum and examples
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with collisions, we now
want to talk about forces.
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And we get back to
the example of Alice
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traveling to the center
of the galaxy and asking
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what does it mean in
terms of acceleration.
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So we start from
Newton's second law.
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We know that a force is
a change in momentum.
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We can write this
down as d dt m0 times
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u over the square root of 1
minus u squared over c squared,
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or just with a gamma factor.
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And the thing to
consider here is
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that now the gamma
factor and the velocity
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are actually time dependent.
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So there's two
components to this.
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We'll come back to this.
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The kinetic energy is the work
done by an external force.
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And you can get to the kinetic
energy by just integrating,
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let's say, for a particle
which is accelerated
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by an external force
from the velocity 0
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to some velocity v.
That's the integral
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over the path of the particle,
over the path of this object,
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times the force.
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If you just assume here
uniform motion in x-direction,
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then this simplifies
to just an f dx.
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So, as the first
activity, I want
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you to find the kinetic energy
of an object with velocity v
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and the mass and mass m0.
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And, as the second
part, I want you
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to test this result
for velocities much,
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much smaller than
the speed of light
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where you're used to doing
this kind of calculation,
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and you're familiar
with the outcome.
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So we just have to integrate,
just have to integrate.
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So this is a little
bit involved here.
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So we have to integrate
from 0 to v m0--
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that's our constant;
we can take that out--
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d dt of this u times gamma dx.
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OK, so you find there's
two components here.
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And then we do a trick where
we introduce this du dx dx.
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And then the integral becomes
an m times u du over 1
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minus u squared over c
squared to the third,
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to the third power half.
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OK, and then you can just look
up the integral or work it out.
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It's not that
difficult actually,
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but you find that this
is equal to m0 times
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c squared over square root 1
minus u squared over c squared,
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which you have to evaluate
for velocities v and 0.
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And, when you do that, you
find those two components here.
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OK, the first one is m0
c squared times gamma.
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And the second one
is m0 c squared.
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So that result is actually
not too surprising,
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as we saw that we can write the
energy equal to m0 c squared
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plus k.
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And what we just calculated
here from this example
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is k is equal to energy
minus m0 c squared, OK?
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So that result already
makes sense with respect
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to the discussion we
had to this point.
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Or you can simplify
this by saying,
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the kinetic energy is gamma
minus 1 times m0 c squared, OK?
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So, if we now evaluate
this for small values of v,
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as we did before, we find
that the kinetic energy
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is 1/2 m0 v squared.
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And I find it
interesting, illustrative,
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to plot what this means now.
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So, if we plot
the kinetic energy
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of a particle as a
function of its velocity,
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we find that, for small values,
those two curves basically
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overlap.
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For small values, m0 c
squared times gamma minus 1
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is basically the same as 1/2 mv
squared, which we just derived
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here from the Taylor expansion.
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But, for larger
values, this diverges
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and especially when you get
closer to the speed of light.
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Just to get a
quantitative example,
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I asked you to do
another calculation here.
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I want you to reinvestigate
Alice's journey
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to the center of
the galaxy where
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she has a spacecraft,
which moves with a gamma
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factor of 15,000, an
acceleration of 10
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meters per second squared, and
the mass of the spacecraft,
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let's say, is 10,000
metric tons or 10,000--
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sorry, 100 metric tons
or 100,000 kilograms.
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So compare the kinetic energy
using Newtonian mechanics
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or special relativity.
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And you find that the difference
is astonishingly large.
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So, if you just work this
out, 1/2 mc squared--
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we can just use c squared
here because the velocity is
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basically c--
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5 times 10 to the 22 kilograms
meter squared over seconds
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squared.
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And, in relativistic
terms, the answer
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is 30,000 times larger,
so 30,000 times larger
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than the classical case.
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So there's a huge
difference between
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the classical evaluation
and the evaluation
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with special relativity.
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One more word on F
equal ma, the question
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is, how does this transform
under Lorentz transformation?
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It's something we already
halfway figured out.
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So here you basically want
to see how a transforms
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under Lorentz transformation.
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We have started the
discussion by saying,
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you know, in Galilean
transformation,
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the acceleration is
invariant, while, in Lorentz
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transformation,
that's not the case.
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But, if you investigate again
the second law of physics,
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the force as a
change in momentum,
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you find that you get
those two components here.
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One is parallel to the
acceleration, so m times gamma
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a.
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But the second one is not.
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The second one is m0
times u times the change
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in time of the gamma factor.
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And that's not
parallel to F or to a.
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And so you find
that there's two--
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the new vector or the
new force of a particle
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is not parallel to the
acceleration anymore.
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That's kind of counterintuitive.