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>> This is the first of several
lectures on gravitational waves.
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[ Typing ]
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The exciting thing about gravitational
waves is that they allow us
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to make direct observations of
strongly gravitating systems.
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[ Typing ]
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As you probably know, we have a growing network
of ground-based gravitational wave detectors
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that have recently detected a number of
collisions among black holes and neutron stars.
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[ Typing ]
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The first gravitational wave detection
occurred in September of 2015.
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This event consisted of two black holes
orbiting each other, then spiraling together
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and merging to form a single black hole.
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Now this system originally consisted
of a 36 solar mass black hole,
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and a 29 solar mass black hole, and they
merged to form a 62 solar mass black hole plus
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about three solar masses
worth of gravitational waves.
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[ Typing ]
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So let me complete my picture here
by showing some gravitational waves.
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This first detection was made by the LIGO
detectors, which are in the United States.
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There are actually two detectors;
one in the state of Washington --
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-- which is in the Northwestern part of
the country, and the others in the state
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of Louisiana, which is in the
south central part of the country.
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Since this time, there have been
a number of other detections.
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One of the most notable occurred
in August of 2017.
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This was a 25 solar mass black hole --
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-- merging with a 30 solar mass black hole.
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And this was seen not only
by the two LIGO detectors,
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but also by the VIRGO detector,
which is located in Italy.
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Then just a few days later,
also in August of 2017 --
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-- the two LIGO detectors and the VIRGO
detector detected a binary neutron star merger.
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And this event was also seen in the
electromagnetic spectrum as a gamma-ray burst.
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In addition to the LIGO and VIRGO detectors,
there are a number of other detectors
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around the world that were not sensitive
enough to see any of these events,
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and they probably won't see
any future events either,
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but these instruments played an
important role in the development
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of the technology needed to
detect gravitational waves.
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Among these other detectors --
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-- the most notable are GEO in Germany --
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-- and TAMA in Japan.
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Currently plans are underway
for the construction
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of several new detectors whose sensitivities
will equal or exceed the sensitivities
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of the current LIGO and VIRGO detectors.
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These new detectors include --
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-- KAGRA, which is in Japan.
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And might be online is, early as 2020.
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And Indigo --
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-- which is in India.
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And construction is expected
to begin around 2020.
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There's also a lot of interest
and serious discussions
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about placing gravitational wave
detector in Australia in the near future.
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Now all of these detectors that I've
mentioned so far are ground-based instruments.
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They're large scale interferometers.
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They have arm lengths --
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-- in the range of three to four
kilometers and their sensitivities --
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-- are in the range of tens
to thousands of Hertz.
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So these instruments are all sensitive to the
gravitational waves from events like the merger
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of approximately equal mass
black holes or neutron stars.
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[ Typing ]
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One of the most exciting plans for
gravitational wave physics is the LISA mission,
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which will place the gravitational
wave detector in space.
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So this is the LISA mission.
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It's being led by the European space agency
with input also from NASA in the United States.
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Now the LISA detector will consist of satellites
working together as a large interferometer
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with arm lengths of millions of kilometers --
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-- and a sensitivity --
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-- in the range from 10 to the
minus 4 to 10 to the minus 1 Hertz.
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This is the appropriate frequency
range for an event
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such as the inspiral of a roughly solar mass --
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-- black hole or neutron star
into a supermassive black hole.
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Currently, the expected launch date for
the lease submission is around 2034.
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So what is a gravitational wave?
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I don't have a simple one
sentence answer to that question,
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but in a few sentences, it's a fluctuation --
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-- in the curvature of spacetime.
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It produces oscillatory tidal effects.
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So these are the world lines of
two freely falling particles.
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These particles are an accelerated, but
the space in between them is expanding
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and contracting due to the
passing gravitational wave.
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And finally, I would add to this
description that the fluctuations --
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-- propagate through space --
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-- at the speed of light.
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Now, gravitational waves are a bit
complicated to describe mathematically --
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-- and this is for two main reasons.
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The first reason is the Einstein
equations are nonlinear.
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As a result, we can't write down
exact solutions in most cases;
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we have to work with solutions
that are only approximate.
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And another consequence of the nonlinearity
is that waves can't be superimposed.
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But what we can do in spite of this nonlinearity
is develop the theory of a weak waves
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as small perturbations of flat spacetime.
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So weak gravitational waves as perturbations --
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-- on flat spacetime.
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And this should be sufficient
for applications in astrophysics
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where the gravitational waves we detect from
distant cosmological events are very, very weak.
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So this is okay for applications
in astrophysics.
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Now, the second main reason why
gravitational wave theory is complicated is
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because general relativity
contains a large gauge freedom.
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Now, what I mean by gauge freedom
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in this context is the freedom
to change spacetime coordinates.
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So change of spacetime coordinates doesn't
actually change the spacetime geometry.
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The term gauge comes from
electromagnetism and it's also used
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in Yang-Mills theory's in particle physics.
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And in this context of gravitational wave
physics, we often refer to this freedom
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of changing spacetime coordinates
as a gauge freedom.
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In practice, this is an important issue because
we need to know how to interpret the metric
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to understand whether fluctuations
are coming from the spacetime geometry
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or from the choice of coordinates.
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[ Typing ]
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So fluctuations in the metric could
be coming from the geometry --
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-- or from the choice of coordinates --
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-- or of course a mixture of both.
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Let me give you an example of this.
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Let's take flat spacetime
in Minkowski coordinates.
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So let's -- let those coordinates
be t bar, x bar, y bar and c bar.
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So the metric is minus dt bar squared plus
the dx bar squared plus dy bar squared plus dz
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bar squared.
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And now let's make a change of coordinates.
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So the new coordinates are defined by t bar
equals t, x bar equals x, y bar equals y
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and z bar equals z minus a divided by Omega
times cosine of Omega times t minus z.
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In this expression, Omega is an angular
frequency and a is an amplitude,
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it's a dimensional amplitude, and we want
to think of this as being very small.
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So this coordinate transformation
is close to the identity.
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Now let's work out the metric and
these new unbarred coordinates.
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You have ds squared equals minus dt
bar squared, but t bar is equal to t,
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so this is just minus dt squared.
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Likewise, we have plus dx squared plus dy
squared plus dz bar is dz plus a time sine
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of Omega times t minus z times dt
minus dz, and all of that is squared.
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Now let's simplify this.
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We have minus dt squared plus
dx squared plus dy squared.
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Then squaring this term, we have dz squared,
we have the cross terms which are 2a times sine
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of Omega t minus z times dz times dt minus dz.
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And then we have the term that
comes from squaring this term
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and that term is proportional to a squared.
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And remember a is very small,
so we'll drop these terms.
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So putting this together we have minus dt
squared plus dx squared plus dy squared plus,
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now the dz squared terms, we have
1 times dz squared minus 2a sine
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of Omega t minus z times dz squared.
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And then we also have these terms dz times dt
that's plus 2a sine Omega t minus z times dtdz.
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And there are higher order
terms in a that we're dropping.
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Now because a is small, this metric
is close to the Minkowski metric.
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In fact, we can write the metric as a mu nu plus
a small perturbation, which we'll call h mu nu.
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So h is a perturbation of
the Minkowski metric --
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-- which is proportional to the small
amplitude a. Let me write down h explicitly.
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From the line element here we have h mu nu
as a matrix equals zero zero zero a sine
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of Omega t minus z. And the second row
was all zeros, third row is all zeros,
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and we have a times sine of Omega t minus
z zero zero and finally minus 2a times sine
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of Omega t minus z. So this metric
looks a lot like what one might think
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of as a gravitational wave, but
it's not a gravitational wave.
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It's not.
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The oscillations in the metric are just
coming from oscillations in the coordinates.
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Remember, this is just flat spacetime
written in a new set of coordinates.
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We can see what's happening if we recall the
coordinate transformation was z bar equals z
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minus a over Omega times cosine of Omega times t
minus z. And now remember t was equal to t bar,
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so we can also write this as t bar
minus z. And in this expression,
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z bar and t bar are Minkowski
coordinates for flat spacetime.
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So if we plot a Minkowski diagram, this is
the t bar axis and this is the z bar axis.
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Then the curves with constant z bar
are of course just straight lines.
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So this is constant z bar.
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Now what about the curves with constant z?
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For example, let's take z equals zero and
plot that curve on this t bar z bar diagram.
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So if z is equal to zero,
there's zero here, zero here,
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we have z bar at t bar equals
zero z bars minus a over Omega.
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So right here, and then as t bar increases --
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-- we get a sine or cosine wave like this.
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So this is the z equals zero curve.
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Now let's pick another value
of z, say pi over 2.
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So we start shifted to the right by pi over 2
and also shifted in phase upward by pi over 2.
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So that curve would look something like --
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-- this. So this is the pi over 2 curve.
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So the lesson from this example
is that we need to be careful.
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This metric with this perturbation on flat
spacetime is not a gravitational wave,
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the wave like elements in this perturbation
just come from a choice of wavy coordinates.
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In later lessons, I'll go through the analysis
of the gauge freedom for perturbations
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on flat spacetime and show how we can
restrict the choice of coordinates
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to eliminate solutions like this one.
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For now, I think it will be useful to put these
issues aside and show you a metric perturbation
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that does represent a gravitational wave.
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So here's the example that I want to use.
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[ Typing ]
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The metric is close to flat spacetime
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with Minkowski coordinates, and
the perturbation is h mu nu.
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And h mu nu written as a matrix is
zero zero zero zero zero times a sine
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of Omega t minus z zero zero
zero zero than minus a sine
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of Omega t minus z zero then
the last row is all zeros.
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This metric represents a plane wave --
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-- propagating in the z direction.
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Now this dimensional amplitude a, we're
assuming is small, much smaller than one.
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And you can check that this metric
satisfies the vacuum Einstein equations.
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So that would be g mu nu equals zero
up to linear through linear order in a.
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So there are 10 error terms of order a squared.
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So that just means this is
an approximate solution
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of Einstein's equations, not an exact solution.
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Now the first thing I'd like to do with this
metric is to compute the geodesics equations.
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For that purpose we need to Christoffel symbols.
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So gamma mu alpha beta is 1/2g upper mu nu
partial alpha g beta nu plus partial beta g
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alpha nu minus partial nu g alpha beta.
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So we're going to need the inverse metric.
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What is that?
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Well, let's recall the metric itself is
the flat metric plus a perturbation h.
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So we expect the inverse metric to
also be close to the flat metric.
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So g mu nu is a mu nu plus some small
perturbation, let's call it k upper mu nu.
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And now we want to define k such that g mu nu
g upper nu sigma is equal to delta sigma mu.
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So let's work that out.
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That's g mu nu plus h mu times ADA
upper nu sigma plus k nu sigma.
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And now ADA times ADA gives delta mu sigma.
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We have a term ADA times k
that's ADA mu nu k nu sigma.
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We have this term h times
ADA h mu nu ADA nu sigma.
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And then we have a term h times k, but
that's second order in small quantities.
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So we'll ignore that term.
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And now all of this must equal delta mu sigma.
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So that tells us --
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-- that ADA mu nu k nu sigma
equals minus h mu nu ADA nu sigma.
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And now if we multiply this equation by
the inverse of the ADA, say ADA Rho mu,
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so we're multiplying this
equation by ADA Rho mu,
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we find k Rho sigma equals minus
ADA Rho mu h mu nu ADA mu sigma.
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So this is k, the perturbation
of the inverse metric.
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And for notational convenience we're going
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to define this combination ADA
inverse times h times ADA inverse as,
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here's the minus sign, as h upper Rho sigma.
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So here I'm using for the first
time the convention that indices
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on h the metric perturbation
are raised and lowered
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with the flat metric, with ADA and it's inverse.
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So now we can write the inverse
metric as ADA mu nu minus h mu nu.
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And notice also when we insert
the metric in here,
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all the terms involving derivatives
with ADA vanish.
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So Christoffel will symbols become gamma mu
alpha beta equals 1/2 ADA mu nu minus h mu nu
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times partial alpha h beta nu plus partial
beta h alpha nu minus partial nu h alpha beta.
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For now, we're just carrying
out our calculations
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to linear order in small quantities.
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Remember h is proportional
to the small amplitude a
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and so also derivatives of
h are small quantities.
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So in this expression, h times derivatives of h
are second order, and we'll ignore those terms.
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So we'll write gamma equals 1/2 ADA mu nu
partial alpha h beta nu plus partial beta h
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alpha nu minus partial nu h alpha beta.
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Then there are second order
terms that we'll ignore.
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Now let's recall the mu nu zero components
of h are hxx, which was equal to a times sine
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of Omega t minus z. And that
was also equal to minus hyy,
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and all the other components of h are zero.
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So when we compute Christoffel symbols, the
only possible answers we can have or either zero
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or plus or minus 1/2 times the
derivative of this function.
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So let me just skip to the results.
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The Christoffel symbols that equal
plus 1/2 a Omega times cosine
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of Omega t minus z are gamma txx,
gamma zxx, gamma xtx, and gamma yzy,
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and of course also gamma xx, gamma yyz,
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since the Christoffel symbols are
symmetric in their lower indices.
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And the Christoffel symbols that are
equal to minus this same quantity --
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[ Typing ]
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-- are gamma tyy, gamma zyy,
gamma xzx, and gamma yty,
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and all the other Christoffel symbols vanish.
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And now for the geodesic equations.
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[ Typing ]
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Remember the geodesic equation is x mu double
dot plus gamma mu alpha beta x dot alpha x dot
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beta equals zero.
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So for the zero component for t
component, you have t double dot plus,
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you have the nonzero Christoffel symbols
with an upper t index, that's txx and tyy.
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So we insert those in here we
obtain 1/2a Omega times cosine
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of Omega t minus z times x dot squared minus
y dot squared and that's equal to zero.
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So the x component, we have x
double dot plus a Omega times cosine
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of Omega t minus z times the
[inaudible] Christoffel symbols are,
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have lower indices tx and zx.
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So this is times t dot x dot
minus z dot x dot equals zero.
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For the y component it's y
double dot plus a Omega cosine
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of Omega t minus z times z dot y
dot minus t dot y dot equals zero.
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And finally, for the z component, we have
z double dot plus 1/2a Omega times cosine
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of Omega t minus z times x dot squared
minus y dot squared equals zero.
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So these equations are fairly complicated,
but there's one family of solutions
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that is fairly easy to identify --
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[ Typing ]
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-- and it's the following.
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Remember, dot stands for derivative
with respect to proper time.
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So one solution is t equals Tao [phonetics]
proper time x equals x naught some constant,
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y equals y naught and z equals z
naught, where these are all constants.
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[ Typing ]
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And of course we could add a
constant to t to tell here as well.
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You can verify just by inspection that these
world lines satisfy the geodesic equations.
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So all of these terms with double dot vanish,
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and also all of these other terms are
proportional to at least one factor of x dot,
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y dot, or z dot, so they all vanish as well.
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Let me draw a picture of these world lines.
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So here's the t axis --
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[ Typing ]
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-- y axis, and the z axis.
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Try to suppress the x axis.
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Each of these world lines is
at a constant value of x, y, z,
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and just runs parallel to the t axis.
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So this is the family of geodesics.
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Each one of these geodesics has a four velocity
x dot mu, which is 1, zero, zero, zero.
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This might seem like a strange
result, after all they're supposed
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to be a gravitational wave propagating
through space along the z direction.
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[ Typing ]
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Yet these particles don't seem to be influenced
by the wave; they're just staying still
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at the same location, x naught, y
naught, z naught, as time passes.
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The key to understanding what's happening is to
recognize that as the gravitational wave passes,
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the particles remain at rest
in this coordinate system.
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[ Typing ]
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The emphasis here is on this coordinate system.
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[ Typing ]
00:31:27
So when we say that the particles
remain at rest,
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that statement has no intrinsic geometrical or
physical meaning, it's a statement that depends
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on the coordinate system that we've chosen.
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So what we'd really like to do is
compute a proper separation between two
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of these particles, say between
this world line and this world line,
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and monitor that separation in time.
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But how would we do that?
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For example, we can pick an event on this world
line, and another event on this world line,
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and then choose a path between those
events and compute that proper distance.
00:32:01
But what if we chose a different point
on this world line or a different path?
00:32:06
So you can see from here that
this notion of the proper distance
00:32:10
between world lines is ambiguous.
00:32:12
[ Typing ]
00:32:23
It's not well defined --
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[ Typing ]
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-- unless we do something to specify
which points and paths we're taking.
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So it's ambiguous without Further input
for choosing the paths and points.
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[ Typing ]
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Points or events on the world lines.
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I think the best way to supply this further
input is to ask, what would an observer see?
00:33:02
An observer who's riding
along one of these geodesics,
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what would they see happening
to the particles nearby?
00:33:08
So to answer that question we'll choose an
observer who's riding right along the t axis --
00:33:13
[ Typing ]
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-- and construct fermi normal coordinates
in the neighborhood, this observer,
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and use the fermi normal coordinates to
track the motion of the nearby particles.
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So we'll choose our observer --
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[ Typing ]
00:33:37
-- at x equal y equals z equals zero.
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This is one of the geodesics we computed before.
00:33:44
[ Typing ]
00:33:47
And then we'll construct fermi
normal coordinates for this observer.
00:33:51
[ Typing ]
00:34:05
I discussed fermi normal
coordinates in detail in Lecture 10.
00:34:08
So see Lecture 10.
00:34:11
[ Typing ]
00:34:14
And there I used a bar to
denote the coordinates.
00:34:17
[ Typing ]
00:34:20
So the fermi normal coordinates were
called x bar mu, and they're often used
00:34:25
for x bar zero, I'll often use Tao.
00:34:29
So Tao is also x bar zero tabbing proper
time along the observers' world line.
00:34:36
And then the spatial coordinates
were just to know x bar a,
00:34:40
where a ranges over one, two and three.
00:34:44
The key result that I showed in this
lecture 10, was that the components
00:34:48
of the metric in fermi normal coordinates --
00:34:50
[ Typing ]
00:34:59
-- call that g mu nu bar,
the components of the metric
00:35:03
in those coordinates are just ADA
mu nu plus terms that are quadratic
00:35:08
and higher order in the x bar a's.
00:35:10
[ Typing ]
00:35:21
This result tells us that close to
the world lines of the observer,
00:35:25
let's call the observer o,
so close to o's world line.
00:35:32
[ Typing ]
00:35:35
In other words, where x a bar is small,
and we can ignore these quadratic terms,
00:35:44
then the observer o can use special relativity.
00:35:47
[ Typing ]
00:35:55
They can use everything they know about
special relativity to interpret all
00:35:58
of the results and all of the physics.
00:36:00
And in particular, x bar as' are
just spatial Cartesian coordinates.
00:36:07
[ Typing ]
00:36:14
And x bar zero of course is Tao is
just proper time for the observer.
00:36:21
[ Typing ]
00:36:28
Now let's recall how fermi normal
coordinates are constructed.
00:36:32
And just for notational simplicity
and consistency,
00:36:35
I think I'll denote the x bar zero coordinate,
00:36:38
the proper time along the
observers world line is t bar.
00:36:42
So the fermi normal coordinates will be t bar
and x bar a, and sometimes I'll use x bar,
00:36:48
y bar, z bar, for these spatial
fermi normal coordinates.
00:36:52
So here's the observers world line, the
observer's o, and the worldwide is defined
00:36:58
in terms of our original spacetime
coordinates as x mu as a function
00:37:03
of proper time along the world line.
00:37:05
Proper time I'll denote by tab -- t bar.
00:37:08
And just to remind ourselves that
this is the observer's world line,
00:37:12
I'll put a little subscript
script o here for observer.
00:37:16
So the four velocity of the observer
is u sub o equals u sub o dot.
00:37:24
The observer constructs fermi normal
coordinates by first picking a time,
00:37:29
let's say t bar equals zero, and
here's the four velocity at that time.
00:37:38
The observer then chooses three
spatial basis vectors, e1, e2, and e3,
00:37:42
such that these basis vectors are
orthogonal to the four velocity u
00:37:54
and also orthogonal to each
other and normalized.
00:37:58
So if we refer to these three basis factors
collectively as e sub a, and we have e sub a
00:38:04
as orthogonal to the observers world
line, and also these are orthonormal.
00:38:10
So ea dot eb is delta ab.
00:38:14
And now the observer defines basis
vectors all along the world line
00:38:21
by parallel transporting these basis vectors.
00:38:24
[ Typing ]
00:38:27
So define the ea's along the world line --
00:38:30
[ Typing ]
00:38:36
-- by parallel transport.
00:38:37
[ Typing ]
00:38:43
So that's mu alpha dell alpha
e mu a equals zero.
00:38:49
As I showed back in Lecture 10, if
these orthogonality conditions hold
00:38:53
at the initial time at t bar equals zero,
then they'll continue to hold into the future,
00:38:59
as these basis vectors are parallel transported.
00:39:02
Physically, the way the observer defines these
basis vectors is with a set of gyroscopes.
00:39:08
The access to the gyroscope
defines a spatial direction,
00:39:12
and that direction is parallel
transported into the future as long
00:39:15
as there are no torques acting on the gyroscope.
00:39:22
The main result of this construction
is a coordinate transformation
00:39:25
between the original spacetime coordinates
and the fermi normal coordinates.
00:39:29
[ Typing ]
00:39:37
So the original coordinates x mu has functions
00:39:41
of the fermi normal coordinates t bar x bar
a were forgiven by x mu of o's worldwide
00:39:50
and that's function of t bar
plus the basis vector ea mu.
00:39:56
And that's a function of t bar, since the base
vectors have been parallel transported along the
00:40:01
world line, times x bar a minus 1/2 the
Christoffel symbols, gamma mu alpha beta.
00:40:09
And these are evaluated along the world lines
so there are functions of t bar times e alpha a,
00:40:16
which is a function of t bar times e beta b,
which is function t bar times x a bar x b bar.
00:40:26
And then there are higher
order terms in the x bar a's.
00:40:31
Now let's transform our gravitational
wave metric into fermi normal coordinates.
00:40:35
[ Typing ]
00:40:42
So remember the metric has the
form of the flat spacetime metric,
00:40:46
and Minkowski coordinates
plus a small perturbation.
00:40:50
And the only nonzero components of the
perturbation are the xx and yy components.
00:40:55
And those are equal to plus or minus a
times sine of Omega t minus z. So if we put
00:41:00
that together the metric is diagonal.
00:41:03
So let me write it this way,
diagonal minus 1, then 1 plus a sine
00:41:11
of Omega t minus z. The next entry, the yy entry
is 1 minus a times sine of Omega t minus t.
00:41:25
And the zz component is just plus 1.
00:41:27
And remember this amplitude a's small.
00:41:31
Now our observer o just sits at
x equal y equals z equals zero
00:41:37
at the origin and these coordinates.
00:41:41
And so the world line of the
observer x mu sub o is a function
00:41:47
of t bar is just t bar, zero, zero, zero.
00:41:54
So the four velocity of the
observer is just 1, zero, zero, zero.
00:42:02
And now we need to construct
these basis vectors, the ea mu.
00:42:05
Because the metric is diagonal,
it's pretty easy to guess.
00:42:10
We'll take e1 mu to be zero, 1 divided by the
square root of 1 plus a sine Omega t minus z --
00:42:24
[ Typing ]
00:42:27
-- zero, zero, e2 mu is zero,
zero, 1 divided by the square root
00:42:35
of 1 minus a sine Omega t minus c, zero,
and e3 mu is equal to zero zero zero 1.
00:42:50
Now because the metric is diagonal, you
can tell just by inspection that each
00:42:54
of these vectors is orthogonal
to u, and they're also orthogonal
00:42:59
to each other, and they're also normalized.
00:43:02
But let's remember, we're constructing a set of
basis vectors along the world line where x, y,
00:43:08
and z are equal to zero and t is equal to t bar.
00:43:12
So for the basis of vectors on the world line,
we can replace this combination t minus z
00:43:18
with t bar, and this t minus z is t bar.
00:43:25
Let's also remember that this parameter a
is very small and we're carrying out all
00:43:29
of our calculations just to linear order
in a. So let's write these basis vectors
00:43:35
to linear order in a, e mu 1 is zero, 1 minus
a over 2 times sine of Omega t bar, zero, zero,
00:43:49
e mu 2 zero, zero, 1 plus a over
2 times sine of Omega t bar,
00:43:58
zero, and e mu 3 is zero zero zero 1.
00:44:04
Now there's one more condition
that we need to check.
00:44:07
We need to know that these basis vectors are
parallel transported along the world line.
00:44:12
[ Typing ]
00:44:28
So it's possible to have a set of basis
vectors that are orthogonal and normalized
00:44:32
in space like, they're space like in respect
to the world line, but they rotate relative
00:44:37
to the basis vectors that
are parallel transported.
00:44:41
In other words, they rotate relative to the
directions defined by the inertial gyroscopes.
00:44:46
So we need to know that our set
of basis vectors is not rotating.
00:44:51
So we need to check that u alpha
del alpha e mu a is equal to zero.
00:45:01
Now this can be expanded in terms of
ordinary derivatives and Christoffel symbols
00:45:05
to give e mu a dot, the dot is
differentiation with respect
00:45:09
to proper time t bar plus gamma
mu alpha beta e alpha a u beta.
00:45:17
But u beta is just 1 zero zero zero.
00:45:19
So this is e mu a dot plus
gamma mu alpha zero e alpha a.
00:45:27
So let's first consider this basis vector.
00:45:30
Let's let a equal 1 so we can
immediately compute e1 dot is zero,
00:45:38
the derivative of this is minus
a over 2 times Omega times cosine
00:45:45
of Omega times t bar, zero, zero.
00:45:49
That takes care of this term,
now let's compute this term.
00:45:52
We have gamma nu alpha zero e alpha
a. Knowing the sum over alpha,
00:46:00
the only nonzero term is
when alpha is equal to 1.
00:46:03
So what this is gamma nu 1 zero times
1 minus a over 2 sine of Omega t bar.
00:46:12
So we need to compute these Christoffel symbols.
00:46:15
[ Typing ]
00:46:18
We have gamma mu 1 zero is 1/2 g mu times
partial 1 g zero nu plus partial zero g 1 nu
00:46:29
minus partial nu g 1 zero.
00:46:32
Now remember the metric is diagonal,
so this term is equal to zero.
00:46:36
And also the only nonzero
term here is when nu is equal
00:46:40
to zero but g zero zero is just minus 1.
00:46:43
So the derivative of minus 1 is zero.
00:46:46
And now the only nonzero term here in
the sum over nu is when nu is equal to 1.
00:46:51
So this is equal to 1/2 g mu 1
times the derivative of g 1 1,
00:46:58
but g 1 1 is just 1 plus
a sine of Omega t minus z.
00:47:06
[ Typing ]
00:47:09
So this simplifies to 1/2 g mu 1 times a Omega
cosine Omega t minus z. Now let's remember,
00:47:21
we're evaluating this along the world line where
z is equal to zero and t is equal to t bar.
00:47:28
So gamma mu 1 zero on the world line is equal to
1/2 g mu 1 times a Omega cosine of Omega t bar.
00:47:40
And now we can insert this result in here,
and this index a should actually be a 1.
00:47:48
And then gamma mu alpha zero e alpha 1 is
equal to 1/2 g mu 1 times a Omega cosine
00:48:00
of Omega t bar times this factor 1 minus
a over 2 times sine of Omega t bar.
00:48:10
Now let's remember that we're carrying out
all of our calculations to linear order
00:48:14
in this small amplitude a. And
we already have in fact, a here.
00:48:18
So this term can be dropped because it just
contributes quadratic terms to our answer.
00:48:24
And also g mu nu, to leading orders, is
just a flattened Cartesian metric a mu nu.
00:48:29
So this is approximately 1/2 ADA
mu 1 a Omega cosine of Omega t bar
00:48:37
to linear order in the small parameter --
00:48:41
[ Typing ]
00:48:44
-- a.
00:48:45
[ Typing ]
00:48:49
Now of course, ADA mu 1 is only
nonzero when mu is equal to 1.
00:48:53
So let's write our answer this way, zero,
1/2 a Omega cosine Omega t bar, zero, zero.
00:49:03
Now we can compare this result to
our previous result for e1 dot,
00:49:10
and we see that they're just the
same but with opposite signs.
00:49:14
So the sum of these two terms equals zero.
00:49:17
So what we've just shown is
that for the basis factor e1,
00:49:21
it's parallel transported along the world line.
00:49:24
And by similar calculation, you can show
that e2 and e3 are also parallel transported.
00:49:30
So we've now confirmed that the basis vectors
are parallel transported along the world line.
00:49:35
So the ea's are parallel transported.
00:49:42
[ Typing ]
00:49:45
They satisfy u alpha del
alpha e mu a equals zero.
00:49:52
Now we can put all of this together to
construct the coordinate transformation
00:49:56
from our original spacetime
coordinates to fairly normal coordinates.
00:50:02
So here are the basis vectors that we need --
00:50:12
-- and this is the coordinate transformation.
00:50:20
This transformation is expressed
in terms of the series expansion
00:50:23
in the spatial Fermi normal coordinates,
the x bar a's, and for present purposes,
00:50:29
we only need this expansion out to linear order,
00:50:31
the x bar a. So we only need
this part of the expression.
00:50:35
So explicitly this coordinate
transformation is given by the following.
00:50:41
[ Typing ]
00:50:51
We have t equals t bar, x equals 1 minus
1/2a times sine of Omega t bar times x bar,
00:51:07
y equals 1 plus 1/2a sine of Omega t
bar times y bar, and z equals z bar.
00:51:18
And I'll remind you that there are higher order
terms in the spatial fermi normal coordinates.
00:51:23
So this is accurate through linear order --
00:51:25
[ Typing ]
00:51:32
-- in the x bar a's.
00:51:35
And as always, it's accurate
through linear order --
00:51:37
[ Typing ]
00:51:42
-- in small amplitude a. Now
let's invert these relations.
00:51:51
You have t bar equals t, and x bar equals
the inverse of this factor to linear order a,
00:51:59
is just 1 plus 1/2a times sine of Omega times t
bar, but t bar is equal to t plus corrections,
00:52:10
but this is already a small term because of
the factor of a so we can just replace t bar
00:52:14
with t here, then times x. And likewise
y bar is equal to 1 minus 1/2a time sine
00:52:23
of Omega t times y, and z bar equals z.
Now recall from earlier in the lecture,
00:52:34
we found a family of geodesics --
00:52:39
[ Typing ]
00:52:44
-- which were given by x equals a constant
x naught, y equals a constant y naught,
00:52:51
z equals a constant z naught, and t just
equals the proper time along the geodesic Tao.
00:52:58
Using this coordinate transformation, we can
write down these world lines, these geodesics,
00:53:03
in terms of fermi normal coordinates.
00:53:05
You have t bar equals Tao, x bar equals 1 plus
a over 2 times sine of Omega Tao times x naught,
00:53:20
y bar equals 1 minus a over 2 times sine of
Omega Tao times y naught, z bar equals z naught.
00:53:32
Keep in mind that the observer who's sitting
at the origin can interpret these coordinates
00:53:37
to spacial coordinates x bar, y bar,
and z bar as Cartesian coordinates.
00:53:42
[ Typing ]
00:53:49
So the values of these coordinates give the
actual proper distances along orthogonal spatial
00:53:54
directions, and at the same
time is defined by the observer.
00:53:58
So what the observer sees is that any
particles that are displaced from your origin
00:54:02
in the x bar y bar plane, will undergo
oscillations with angular frequency Omega.
00:54:08
However, notice that the oscillations in the
x bar and y bar directions are out of phase
00:54:13
by 180 degrees due to the sine difference.
00:54:17
So for example, let's take a
look in the x bar y bar plane.
00:54:27
So this is the x bar axis,
this is the y bar axis.
00:54:32
Now here's a test particle on the
x bar axis at Tao equals zero,
00:54:37
so it's at location x bar equals x naught.
00:54:41
Now as Tao increases the bar code moves outward.
00:54:47
So after a fourth of the period, it moved
to location 1 plus a over 2 times x naught.
00:54:52
And similarly, if we have a particle on the
y axis during the first 1/4 of a period,
00:54:58
it will move towards the origin by an
amount minus a over 2 times y naught.
00:55:05
We can in fact consider a
whole ring of test particles.
00:55:10
So initially these particles are spread out in
a circle and during the first 1/4 of the period,
00:55:17
this particle will move outward, find
them out, they over 2 times x naught.
00:55:23
This particle will move inward.
00:55:25
So the whole pattern, the whole
ring distorts into a ellipse.
00:55:34
So this is the ring at Tao equals zero,
and this is the ring at Tao equals pi
00:55:43
over 2 Omega, which is 1/4 of the period.
00:55:47
Then after another 1/4 of a period,
the ring returns to its original shape.
00:55:53
So this is also the shape at Tao equals
pi over Omega after half a period.
00:55:59
Then during the next fourth of a period,
00:56:01
these particles all shift in
the opposite directions --
00:56:04
[ Typing ]
00:56:07
-- so the ring distorts into an ellipse
that's elongated in the y bar direction.
00:56:17
So this is the ring at Tao
equals 3 pi over 2 Omega.
00:56:23
And finally, after a full period, 2 pi over
Omega, the ring returns to the original shape.
00:56:31
So this picture shows the physical measurable
effect that a gravitational wave has on a set
00:56:36
of freely falling on accelerated test masses.
00:56:40
So these are freely falling and
therefore are accelerating --
00:56:45
[ Typing ]
00:56:49
-- test masses.
00:56:51
[ Typing ]
00:56:56
Let me summarize.
00:56:59
What I've shown is an example
of the gravitational wave.
00:57:02
[ Typing ]
00:57:11
And it was defined by the metric probation
h mu nu which was diagonal zero a sine
00:57:20
of Omega t minus z, a minus a
sine of Omega t minus z, zero.
00:57:32
And this is a plane wave
propagating in the z direction --
00:57:39
[ Typing ]
00:57:49
-- and with what is called
the plus polarization.
00:57:53
[ Typing ]
00:58:02
The term plus for the polarization comes from
the effect that it has on a ring of test masses.
00:58:07
They distort into a sort of plus pattern.
00:58:11
And another thing that you
should notice from this example,
00:58:13
that's true of all gravitational
waves, is that the wave is transverse.
00:58:19
[ Typing ]
00:58:25
In other words, the wave only affects
the separation between test masses
00:58:29
in the plane that's transverse
to the direction of propagation.
00:58:33
In the next lecture, we'll show how
this example is derived and also show
00:58:38
that there's another polarization
state called cross polarization.
00:58:42
So next lecture --
00:58:44
[ Typing ]
00:58:48
-- we'll derive the plus and the cross
polarization gravitational waves.
00:58:57
[ Typing ]
