The following content is provided
under a Creative Commons license.
Your support will help MIT OpenCourseWare continue
to offer high quality educational resources for free.
To make a donation or to view additional materials
from hundreds of MIT courses, visit MIT OpenCourseWare
at ocw.mit.edu.
PROFESSOR: OK, in that case, let's go on.
At the end of last class, we were
beginning to talk about the Doppler shift.
And we defined our terms.
And I guess I'll repeat the definitions on the blackboard
here.
We are talking initially about the case where the observer is
stationary and the source is moving with velocity v. We
are talking initially about sound
waves, waves which have a fixed speed relative to some medium.
And the speed relative to the medium will be called u.
v will be the velocity of recession,
as shown on the diagram.
Delta t sub s, s for source, will be the time interval
between wave crests as measured at the source-- so period
of wave at source.
And delta t sub o will be the period
of the wave at the observer.
And what we're trying to calculate
is the relationship between delta t sub o
and delta t sub s.
OK, at this point, I'd like to go to the screen
and go through the different stages of what
happens as this process takes place.
We start in frame one with an observer
in some location, a source at some different location.
Source is moving to the right.
Source is emitting the first wave crest
in this slide number one.
Nothing too interesting so far.
Next picture, the source emits the second wave crest.
But meanwhile, the source has moved.
The time between wave crests as seen by the source
is delta t sub s.
So the distance that the source will move during that time
integral is v times delta t sub s.
And we'll call that delta l.
And this really is the important slide.
I think I have it highlighted here.
That picture is what really counts for the whole Doppler
shift.
It says that the second wave crest
has to travel a little further than the first wave crest
by this amount delta l.
So delta l will be the crucial quantity
that will control the answer.
Third slide, the waves have traveled.
Now the first crest in this third frame
has just hit the observer.
Next frame, the last crest, the second crest,
has hit the observer.
And to figure out the Doppler shift
from those images, all we have to do
is realize that if both objects were stationary,
there'd be no time difference between observation and source.
Each could occur later by some fixed amount, the amount
of time it would take the sound wave to travel.
But they will occur later by the same amount
if there was no motion.
So if there's no motion, delta t sub o
would be equal to delta t sub s, if there was no motion.
But because there is motion, we said
that the wave, the second crest, is
going to have to travel further by the amount delta l.
So it'll be delayed by the amount of time
it takes for the wave to travel delta l.
And that's just delta l divided by v--
so plus delta l divided by v. And we know what delta l is.
Delta l is just v times delta t sub s.
I'm sorry, this is divided by u.
u is the speed of the wave.
v is the speed of the source.
So here we have in the numerator v times delta t sub s,
and the denominator is u.
This is just what we said delta l is.
So this really is our result.
It tells us what delta o is in terms of delta t sub s.
And we can solve for the ratio.
It tells us that delta t sub o over delta t
sub s is equal to 1 plus v over u.
Solve that equation.
Now there's a standard definition
that's used to describe redshifts, which
is that this ratio, which is also the ratio of wavelengths,
which is how I usually think of it,
the wavelength at the observer divided by the wave
length at the source-- the wave length
being disproportional to the delta t.
The wave ones are just the wave speed times delta t.
This is defined to be 1 plus z where
z is what's called the redshift.
And the astronomers take out the one
so that if both objects are stationary, z is equal to 0.
That corresponds to know redshift.
That means the wave length is the same at the source
and at the observer.
So the ratio of the wavelength as its observed
to the wavelength of the source is what's called 1 plus z.
So in this case, it follows immediately
that the redshift for this case is just v over u.
So maybe I'll just write that again in a box and label it.
z is equal to v over u corresponds
to the nonrelativistic case, or the sound wave case,
with the source moving.
OK?
Everybody on board?
Any questions?
OK, straightforward enough, I think.
OK, now we will go on to do the alternative simple case, where
the observer is moving and the source is stationary.
So keep the source on the right and the observer on the left.
But this time it's the observer that's moving again with speed
v. So v is always for these two cases in my notation
the relative velocity between the source and the observer.
So now we have a new sequence of pictures.
So first picture again is fairly trivial.
The first wave crest is being emitted by the source.
Picture number two, the second wave crest
is emitted by the source.
Picture number three, the first wave crest
arrives at the source.
And picture number four now, the second wave crest
arrives at the source.
And for this case, it is this last frame
where all the action is.
The point is that between the time when the first crest hits
the source and the time when the second crest hits
the source, that is the time between the third and fourth
frame, the source has moved.
And it's moved by distance which is
v times the time between those images.
And the time between these images
is just the time that the source experiences
between the receipt of the two waves.
And it's therefore what we've been calling delta t sub
o, the time interval as measured by the observer.
So the distance traveled is just v times t sub o.
And it's that last frame inside that box where all the action
happens for determining the answer for this problem.
So we can put that into equations also.
This time it's slightly more complicated.
It starts basically with the same idea.
Delta t sub o is equal to delta t sub
s, which is what it would be if there was no motion.
But it's a little bit longer because of the extra distance
the second pulse travels.
And that extra distance is again called delta l.
So the time delay is again delta l divided by u, the wave speed.
But this time we have a different formula for delta l.
This time delta l is v times delta t sub 0, instead of
last time it was v times delta t sub s.
So this time the equation is ever so slightly more
complicated, because delta t sub 0
appears on both sides of the equation.
Last time we immediately wrote down
an equation for delta t sub o.
But nonetheless, this is one equation with one unknown,
so of course it's trivial to solve it for delta t sub o.
And when we do that, we get delta t sub o.
I'll maybe divide by delta t sub s.
By just doing a little bit of algebra on that,
you discover it's 1 minus u over v inverse.
And then z, the redshift, is delta t sub 0 over delta t sub
s minus 1 by definition.
AUDIENCE: Is that 1 minus v over u?
PROFESSOR: Oh, do I have it wrong here?
Yes, it's 1 minus v over u.
Thank you very much.
Right, coming from the v over u here,
obviously, as you noticed.
OK, so to write down a final equation for z,
it's delta t0 over delta t sub s minus 1.
So it's 1 minus v over u inverse minus 1.
And then just putting things over our common denominator,
it ends up being v over u divided by 1 minus v over u.
And that then is our final answer.
So this now is the answer for the nonrelativistic case--
again, we haven't done relativity yet--
and the observer moving.
OK, now it's worth pointing out that when
velocity is small compared to the wave speed,
as is often the case if we're going
to be talking about light which we'll
be talking about in a minute, but could
be the case with respect to sound as well.
Then these two formulas are almost the same.
They're both proportional to v over u,
when I'm talking about the case where v over u
is a small quantity, where the motion is slow compared
to the sound speed or wave speed.
And the only difference is that denominator.
Here we have a denominator of 1 minus v over u.
And here the v over u is the whole show.
There's no denominator.
And if v over u is small, the denominator is close to 1.
So the two answers are going to be almost the same.
And one can describe that a little bit more succinctly
perhaps, by saying that z with the observer moving minus z
with the source moving is equal to v over u quantity squared
times 1 minus v over u in the denominator.
Just a little bit of algebra to get that result.
And what that shows explicitly is
that second order in v over u.
It's proportional to v over u squared, not to v over u.
So if v over u were a part in 1,000,
the difference would be a part in a million.
So for slow velocities, it doesn't
matter whether it's the source that's moving
or the observer that's moving.
But these answers can, of course,
be very different if the velocity v is comparable to u.
OK, any questions about that?
Yes?
AUDIENCE: Does this violate Galilean relativity?
PROFESSOR: Does it violate Galilean relativity?
Not really.
Although, one has to maybe be careful about how one defines
Galilean relativity.
What makes it legitimate as a classical mechanical
calculation is that unshown, but crucially
important on the blackboard, is the air
that the sound wave is moving through.
So in one of these pictures, the air was at rest.
The other picture, the air was moving.
Or rather, well, I said that wrong.
The air was at rest in both pictures.
But if you want to make a Galilean transformation
to relate one picture to the other,
then after you've made the Galilean transformation,
the air would be moving and it would not really
be the same picture.
So it is all consistent with Galilean relativity.
And one has to remember that the air is
playing a crucial role here.
When we say that the observer or the source is at rest,
the full sentence really is that it's
at rest relative to the medium that's transporting the wave.
Any other questions?
Yes?
AUDIENCE: This is not really a question, just
kind of an observation.
I just find it interesting that in the case
if v is greater than u that this one is always positive,
no problem.
But if v is greater than u in the other case,
then you have a negative number, which,
I don't know if that-- I just find that weird.
PROFESSOR: Right, let me think.
If v is greater than u, then the observer moving case
becomes negative.
And it presumably means the wave never reaches it.
If the observer is moving faster than the wave speed,
the wave never reaches them.
And that's why the answer is peculiar.
If the source is moving faster than the wave speed,
the wave still reaches the observer.
So there is a nontrivial and believable answer in that case.
OK.
What we're going to move on to now is the relativistic case.
And as I think I described in the course handouts, relativity
is has sort of an odd role in this class.
We need a little bit of relativity.
But at the same time, there are the courses in relativity.
So I don't want this to be a course in relativity.
In the old days, when there weren't so
many other courses in relativity,
we did in fact spend two weeks of this course doing
special relativity.
But I don't think that's worthwhile anymore.
Well, I'll maybe ask for a show of hands.
How many of you have had relativity
in some other course?
That's what I thought, most of you, perhaps not all.
Well, maybe I should ask the other question.
How many of you have not had relativity?
OK, a number.
So I do want to make the course completely
intelligible to the people in that second group.
Special relativity is not a prerequisite for this class.
So what my goal will be to tell you enough special
relativity so that you'll be able to follow what comes next.
But I will not be driving those results.
I'll just leave for other courses
for people who want to take them.
And if you don't want to take them, that's fine too.
But I want to make this course coherent.
So what we're going to do is discuss the consequences
of special relativity without trying
to relate those consequences directly
to the underlying ideas of special relativity.
I will, however, mention where special relativity comes from.
It arose in the mind of Albert Einstein,
because he realized that the physics that we knew well,
basically Newtonian physics at that time,
possess this property, Galilean relativity, which came up
in a question just a minute ago.
Galilean relativity says that if you
look at any given physical process in a frame that's
moving at a uniform velocity relative to the first frame
that you use to describe it, it should also
be consistent with the laws of physics in the second frame.
Incidentally, I-- maybe I'm more ignorant about history
than most-- only learned a few years ago what
Galileo had to do with this.
It did actually play a very important role
in the history of Galileo in the physics
that he was debating about.
A key issue in the time of Galileo
was whether the earth moved around the sun or the sun moved
around the earth.
And that was something Galileo was intensely involved in.
And one of the arguments that said
that it must be the sun that moves--
it can't be the earth that moves--
was that if it's the earth that moves,
it means we move at very high velocity.
The velocity of the earth around the sun
is high by ordinary standards.
And obviously, we'd feel that motion, people thought.
So it was a proof that it must be
that the earth is stationary and the sun moves.
Because otherwise, we'd feel the effect of this high velocity
motion.
So it was crucial to Galileo's point of view
that it was really the earth that
was moving that you don't detect motion.
If you're in uniform motion, the laws of physics
are exactly the same as they would be if you were at rest.
And that's basically what Galilean relativity
is all about.
And it was in fact very clearly enunciated by Galileo
in his writings.
So that was the case for mechanics.
But at the same time, in the 1860s,
Maxwell invented Maxwell's equations,
or completed them is maybe a more accurate description
for what Maxwell really did.
Most of those equations already existed.
And a prediction of Maxwell's equations
is that light travels at some fixed speed, which
could be calculated in terms of epsilon naught and mu
naught that appeared in those equations, a speed
that we call c.
Now if light travels at speed c, it
would mean that if you got into a spaceship
and chased a light beam, it'd say half the speed of light.
The implication of the physics that was known at the time
would have been that, from the point of view
of that spaceship traveling at half the speed of light,
the light pulse would only be receding
at half the speed of light.
You would have half caught up with it.
But that would mean that from the frame of this rapidly
moving spaceship, the laws of physics
must somehow be different.
Maxwell's equations must not hold in their standard form.
So there was this tension between Maxwell and Newton,
if you like.
The tension was not a contradiction.
It would be perfectly possible for there
to be a fixed frame in which Maxwell's equations had
their simple form.
But Newton's equations could perhaps
have the same form in all frames.
And that in fact was what people thought at the time.
To account for this situation, physicists invented the idea
of an ether which was a medium through which light waves
travelled, similar to the air in which sound waves traveled.
And the frame in which Maxwell's equations had their simple form
was the rest frame of the ether.
And if you moved relative ether, the equations
would be different.
And that's what people thought in 1904.
And it was a consistent point of view,
but it meant that there was this dichotomy
between electromagnetism and mechanics.
So Einstein thought maybe physics is not so sloppy.
Maybe there's a more elegant way which all this plays out.
And he realized that if you modified the equations that
are used to transform between one frame and another,
you could make Maxwell's equations frame invariant.
You could make it so that Maxwell's equations
are valid in all frames.
And if we go back to our example of the spaceship chasing
the light beam, with these new transformation equations
that Einstein suggested, it would turn out,
even though it's very contrary to intuition,
that when the spaceship measures the speed of the light pulse,
it would still measure that the light pulse was moving away
at speed c, even though it had moved at half c trying
to catch up with the light pulse.
So it's not obvious how that can happen.
But it turns out it can happen.
We'll be talking a little bit more about how it happens.
And that was basically Einstein's proposal.
It was a proposal that there is no ether,
that the laws of physics both electromagnetism and mechanics
are the same in all frames.
And in order to do that, he had to say
that the equations of transformation
between one frame and another are
different from what Galileo believed.
So these are what we call the Lorentz transformations.
We might write them down later in the course,
but we're not going to write them down today.
But what goes into them are three physical effects,
which we will talk about.
If we're talking about the time dilation,
we only need one of those three effects.
So I'm going to start by just discussing that for a minute,
and then we'll come back at the end of the class, or perhaps
next class, depending on how timing works out,
to discuss the other two primary effects that
are needed to make up the theory of special relativity
and explain how it could be that the speed of light
could look the same to all observers,
even if we're talking about observers that might be moving.
OK, so time dilation is the simple statement
that if I were to watch a moving clock,
the moving clock would appear to me to be running slower.
I'll just mention for a moment now
that I put the word appear in quotation marks.
That means we're going to come back and discuss in detail
what is meant by the word appear.
But to finish the sentence first,
the moving clock would appear to me in my reference frame
to always be running slower.
And so by a very predictable amount, a famous expression
in special relativity, gamma, where
gamma is 1 over the square root of 1
minus beta squared, where beta is just an abbreviation for v
over c, the velocity of the clock
divided by the speed of light.
So as long as v over c is small, this is a small effect.
Gamma is near 1.
And running slowly by a factor of 1
means not running slowly at all.
So the fact it was near 1 means it's a very small effect.
But moving clocks will always appear to be running slower.
That's one of these three effects of special relativity
that we'll be discussing in the course of lecture notes one.
Now let me come back now to talk about this word
"appear," because that's a little bit subtle.
Might just add that there was a series-- this
is just an aside-- but I guess broadcast last year
there was a four-part series of Brian Greene's
Fabric of the Cosmos that was broadcast on PBS.
And the interesting thing about that, which is relevant here,
is that he tried to illustrate time dilation.
And he did it by having sort of a parable of a person
sitting in a chair and somebody else
carrying a clock over his head, walking
towards the person sitting in the chair.
And the camera showed what the person sitting in the chair
would see and showed the clock running slowly.
That's wrong.
It's not what he would actually see.
And that's the crucial issue of the word "appear" here.
When we say that the clock appears to run slowly,
we're not talking about what an observer would actually see.
The complication of literally seeing
is that when you see something, what you're doing
is you're measuring the light pulses as they arrive
at your eyes at a given time.
And since light has a finite travel time,
it means that you're seeing different things
at different times.
In particular, if there's a large object coming
towards you, say this laser pointer coming towards me,
I would be seeing the front of it
at an earlier time than I would be seeing the back of it.
The other way around, actually.
I'd be seeing the back at an earlier time,
because the light that leaves here at an earlier time
would take longer to reach my eye, reach my eye
the same time as light which leaves the front of the object
later.
So the point is that as this is coming toward me,
I'm seeing different pieces of it
at different times in terms of the actual existence
of this laser pointer.
And that makes things complicated.
So what you actually see when you take into account
special relativity is fairly complicated.
You can calculate it, but there's
no simple expression for it.
You really just have to calculate point
by point what you'll be seeing for every part of the object
at a given time, nothing very simple.
So the simple expression, which just says,
clocks run slowly by a factor of gamma,
and we'll learn later e expressions
about how things contract and how
simultaneity changes, those simple expressions
are not based on what any observer would actually see.
But rather, they're based on what
ends up giving a simpler picture, a picture in which you
imagine that what we're discussing
is not what an individual sees but rather what
a frame of reference sees.
So we're talking not about what the observer sees
but rather what is seen in the observer's frame of reference.
And a frame of reference, I think
you could think of it pretty concretely, as kind
of a structure of rulers connected together
to each other to form a grid of rulers
and with clocks located everywhere along this grid.
So all the observations are made locally.
That is, if you want to measure a time in a given reference
frame, you don't use a central clock, waiting
for the light pulse to reach that central clock.
Rather, you measure the reference frame
is filled with clocks, all of which
have been synchronized to start with.
And if you want to know what time an event happens,
you look at the clock that's next to it.
And that clock tells you what time that event happened.
So that's typically what we draw when
we draw a coordinate systems and so on.
It's really the way we normally think.
The point is, though, if you really
want to think what one observer would see,
it's more complicated.
Then you have to take into account the light travel time.
So it's only after you take out the way travel time
and calculate how local clocks would compare
that you see this time dilation in the simple form,
that the clock always runs slower.
So in particular, for this example
of the person sitting in a chair and the clock
coming towards them, that's exactly what
we're going to be talking about.
That is the Doppler shift.
And what we'll find is when the clock comes towards them,
he will see the blueshift not a redshift.
He'll see the clock running faster not slower,
the opposite of what was shown in the TV program.
But the difference is that what causes
it to look like it's going faster
is the fact that each pulse travels a shorter distance
if the clock is coming towards the observer.
And that becomes a bigger effect than the fact
the clock itself, if it were measured relative to clocks
that it passes, would be running slowly.
Yes?
AUDIENCE: So in that case, if the clock, let's say,
was moving toward you really fast,
could you measure it when it was directly perpendicular to you?
And then that would be the special relativity time
dilation?
PROFESSOR: Well, if it was moving alongside you.
If it was coming right at you, it would just hit you.
And that was the case, Sonya, more or less.
But yes, you're exactly right.
If the clock were moving at right angles to the observer
so that the observer saw it, what you really
want for this to be the pure effect
is you want to have the velocity of the clock
to be perpendicular to the velocity of the photon
that the observer is seeing, as measured in the observer's
reference frame.
Actually that matters.
Then you see would see pure time dilation effect.
That's exactly right.
OK, so might just add that Brian Green and a bunch of people
here at MIT, actually-- I was involved
and so were some others.
There's a number of MIT people on this program.
So we ended up having a long conversation
with Brian Green about it by email.
And everybody at MIT thought it was just plain wrong.
Brian Green actually took the position
that it was very intentional on his part,
and he was just trying to illustrate the time dilation
effect, and he didn't want to talk about the Doppler shift.
And since he didn't want to talk about it,
he could ignore the fact that it was there.
We all thought that was bad pedagogy.
But we never convinced Brian, I should say.
OK, now what we want to do is go through these Doppler shift
calculations again, this time recognizing
that moving clocks run slowly by a factor of gamma.
So we're now going to be doing the relativistic case where
the wave is sound waves-- excuse me,
where the wave is light waves.
I'll get this straight.
And the velocities might be comparable to the speed
of light.
So this time dilation effect is large
enough so that we want to take it into account.
Now in this case, what we are hoping,
and everything would be wrong, inconsistent, if we can find
this, we're hoping that in this case,
the two answers should be the same.
It shouldn't matter whether the source is moving
or the observer is moving.
Earlier it did matter, and we said that was explicable,
because we knew that air was involved.
And if we made a velocity transformation
to go from one picture to the other,
from the picture where the source is moving to the picture
where the observer was moving, the air
would have a different velocity in the two pictures.
It would be stationary in one and moving in the other.
So we would not expect to get the same answer
as we would have gotten assuming the air was stationary.
But in this case, if anything has a different velocity when
you go from the picture where the source is moving
to the picture where the observer is moving,
it would have to be the ether that has a different velocity.
But the basic axiom of special relativity
is that there is no ether, at least
there's no physical effects coming from the ether.
So you might as well pretend it does not exist.
You can't really prove that it does not
exist if it has no properties.
It still exists.
But the basic axiom of special relativity
is that there are no physical effects
coming from this either.
So for special relativity, we should get the same answer,
whether it's the source that's moving
or the observer that's moving.
It's really the same situation, just viewed
in different reference frames.
And special relativity says it cannot matter what reference
frame we're doing the calculations in.
So it's the same pictures, but this time we
want to take into account the fact that moving clocks run
slowly by a factor of gamma.
So looking at these pictures, we can first
start just glancing at them and saying,
where's there a clock that's moving?
That might be something we need to think about.
And maybe I'll ask you-- which of these four frames
shows a moving clock?
AUDIENCE: All of them?
PROFESSOR: Sorry?
AUDIENCE: All of them?
PROFESSOR: All of them, I guess so.
But the time interval measured on the clock
is only actually relevant to one of them.
AUDIENCE: Two.
PROFESSOR: Two, exactly, two.
Here we said that the source is moving,
and we said the time measured on the source's clock
between the emission of these two wave crests.
Incidentally, we're usually talking
about continuous waves like light waves.
And then we're talking about the time between successive wave
crests.
Well, if we just as well imagine that the source is emitting
a series of pulses where each pulse represents a wave
crest, and somehow to me that sounds a little simpler
to describe, because you don't have
to think about the sine wave associated with the signal
that the source is actually creating.
So in any case, the time between these pulses,
I'll call them, as measured by the source clock,
is what we call delta t sub s, is
the time that the source would actually measure.
And the source is moving in this picture.
So relative to our frame, we want
to think of this entire series of pictures
as all being consistent in our frame.
It's very important, since transformations between frames
are a little complicated in special relativity.
It's very important when you're doing any problem
to pick what frame you're going to use for your description
and be sure to stick to it.
If anything is initially described in another frame,
you have to figure out what it looks like in your frame
in order to fit it together with the other events
that you're describing in your own particular reference frame.
So for this problem, our frame will
be the frame of the slide, the frame, which
is at rest relative to the observer.
So we could also call it the observer's frame.
And relative to that frame, the source is moving.
And therefore, the clock-- a source
is emitting a series of pulses.
That's a clock.
Anything that does anything at regular intervals is a clock.
So the source represents a moving clock.
And we need to take into account the fact that the source will
be running slowly by a factor of gamma.
And otherwise, nothing changes.
The observer has a clock also, which the observer
is going to use to measure the time between crests.
But the observer's clock is at rest in our frame.
So there's no time dilation associated with the observer's
clock, only a time dilation associated
with the clock on the source.
So again, the important issue is all inside that yellow box.
And what do I do now is look at the equations
and see how the equations are modified.
I guess I should start back at the beginning
of the blackboards.
Maybe I should turn the blackboard lights
on when I work on the blackboard.
So the time integral as seen by the observer
will be equal to-- last time we just
had delta t sub s as our first term, which would be what it
would be if there was no velocity.
And that's still true if there was no velocity.
But if the clock is going to be running slowly
by a factor of gamma, that would mean that the time interval
that we would measure, even if there
was no change in path length, which will be the next term,
if there was no change in path length,
the time that we would measure as the observer would
be different from the time interval
as measured by the source by a factor of gamma.
But one has to figure out whether the gamma goes
in the numerator or the denominator.
And sometimes it's a little tricky.
It helps a lot, I think, to just sort of imagine an example.
Any example is clear, it turns out.
But when you try to write down the answer in general,
you sometimes get it wrong.
That's my experience with myself and with other people.
So this clock is wanting slower.
And say we're talking about a second time interval.
If this clock is running slower, it means it takes longer for it
to take off a second.
We would see it as maybe running slower by a factor of two.
It would mean that it would only take off
a second every two seconds.
So that means that the time interval that we would see--
and that's what we're trying to calculate,
the time interval as the observer--
is going to be longer than delta t sub s by a factor of gamma.
So the first term changes from what it was before.
And what we're doing is we're rewriting this equation.
So the first term changes from what
it was before by putting in a factor of gamma in front of it.
Because the source clock is running slowly.
And then the second term is still delta l divided by u.
But the equation for delta l changes also,
because delta l is the time interval
that it takes for the light to travel the extra distance,
where the extra distance was because
of the time between the clicks.
And that is now changed because of the time
dilation of the source clock.
So the second term also changes by a factor of gamma.
So it's gamma times delta t sub s
plus v times gamma times delta t sub s divided by u.
So the whole answer just changes by a factor of gamma.
So it's gamma times 1 plus v over c times delta t sub s.
And now if you do a little bit of arithmetic
here-- gamma, remember, is 1 over the square root of 1
minus v squared over c squared.
And a squared minus b squared-- I'm now
thinking of the denominator 1 minus v squared over c squared.
Maybe I'll write this on the side.
1 minus v squared over c squared--
it's worth remembering here-- can be written as 1 plus v over
c times 1 minus v over c.
And we have the square root of that appearing here
in the denominator.
Gamma is 1 over the square root of this.
So this becomes the square root in the denominator,
resulting the square root of 1 plus v
over c appearing in the answer.
And this remains in the denominator.
And what we get simplifies to simply 1 plus beta over 1
minus beta inside the square root times delta t sub s.
So this is the special relativity answer,
the relativistic answer, and this is for the source moving.
Now we expect that the answer won't depend on
whether the source is moving or not,
but certainly the calculations do.
So this is what we got from the calculation, which we assumed
that it was a source that was moving,
and the observer was stationary.
It's just corrected from the previous answer
by this factor of gamma.
Any questions about that?
OK, so to summarize on the slide what's changed
is that this time interval, this extra distance delta l
for the relativistic problem, is not
v times delta t sub s as it was.
But rather what we said is that it's gamma times v times
delta s, because the clock on the source
is running slowly by a factor of gamma.
And it was this difference that we just
used on the calculation on the blackboard
to get the new answer.
OK now the next calculation, now we're
going to do the same calculation again,
which we already did for the nonrelativistic case.
But this time it will be the observer moving.
And we're going to try to do the relativistic case.
So this time it's the clock carried
by the observer that's running slowly.
And remember, this is running slowly
relative to us, relative to our frame of reference, where
our frame of reference is by definition the frame
of reference of the slide that we're looking at.
Source is stationary, so delta t sub
s is just an honest time as we would measure it.
But times as measured by the observer, delta t sub o,
are going to be different.
So to correct the calculation for relativity,
that's the crucial box as it was even
in the nonrelativistic case.
And we're going to have to change
that equation by replacing delta l.
Instead of v times delta t sub 0 is v times delta t prime.
Now delta t prime is not exactly delta t sub 0
or gamma times delta t sub 0.
It's a little trickier.
Let's see.
Hold on.
The definition I want to use for delta t prime
is that it's the time as measured
between the third frame and the fourth frame.
But it's the time as we measure it.
I still want to describe everything
in terms of the way we would measure it.
So delta t sub prime is not measured either
by the source or the observer.
It's measured by us.
And it's related to the time as the observer would measure it
by a factor of gamma, because relative to us,
the observer's clock is running slowly by a factor of gamma.
AUDIENCE: So we're the source?
PROFESSOR: Well, we're stationary
relative to the source, because the source is stationary.
But we're the frame of reference of the slide,
is the way I like to think about it.
But it is the frame of reference,
same as the frame of reference as the source.
That's right.
So delta t sub 0 is related to delta t
sub prime by a factor of gamma.
And again, one has to think a little bit
to make sure one gets the gamma in the right place,
the numerator or the denominator.
We're saying that the observer clock appears
to be running slowly relative to us.
And that means that during the amount of time
it would take for it to say take off one second,
since it's running slowly, should take more than a
second relative to us.
And that's what this formula would give,
if we multiply by gamma.
It would say that delta t prime is equal to gamma times
delta t sub 0.
So we would say that the time that the observer clock ticks
off one second we might measure two seconds.
That's the direction of the time shift,
all given by that formula.
And those are all the pictures.
Now we just have to write down the equations that
go with those pictures.
And the hope is that we'll get the same answer
as we got last time.
I think I'll leave that on the board.
Let's see.
What do I do?
No, I won't leave it on the board.
OK, this time the calculation we're mimicking
is the calculation up here.
This was the calculation we had for the nonrelativistic case
where the observer was moving.
Now we want to put in the time dilation that
would correct that calculation.
And the key equation is that the time interval
between the receipt of the two wave crests
as we would measure it, which is why
I call it delta t sub prime, so prime
is not the source or the observer.
It's us.
It's the same frame of reference as the source,
but the source isn't located there.
So we think of it as the time interval
as measured in the frame of reference.
That's the frame of the slide, which is our frame.
And it is equal to gamma times delta t sub
0, where delta t sub 0 is the time as it would actually
be measured on the observer's clock.
So, we're keeping their labels s and 0
to mean what would actually be measured
on the clocks at the source and at the observer.
That's what we're trying to establish
a relationship between.
So the basic formula that we have up
top there would first become delta t sub 0
is equal to-- hold on.
Now actually, we would not start by writing down
an equation for delta t sub 0.
Rather what we want to do is start
by writing down the relationship as we would see it
in our frame, which is going involve delta t sub prime.
We'll worry about delta t sub 0 later.
So our equation is going to become
delta t prime is equal to delta t sub s plus v times
delta t sub prime divided by the wave speed,
which now we'll call c, since we're
doing a relativistic problem.
So this is the basic equation for the time delay
as we would see it in our frame, where delta t sub prime is
the time between the receipt of the first and second crests,
the time between frames three and four
as we would measure it.
And now we can do the same thing that we did over there,
first for delta t sub prime.
And we discover that delta t sub prime
can be written as 1 minus v over c inverse,
just doing algebra on this equation, times delta t sub s.
So that equation relates delta t prime to delta t sub s.
And now we use the equation up top here
to see what the observer himself would actually measure.
And that becomes just 1 over gamma times delta t sub prime.
So delta t sub observed is equal to 1 over gamma times 1
minus v over c inverse times delta t sub source.
And now since it's important we get the same answer,
I'm going to write in some intermediate algebra
here just so we can all really see that it works out.
The 1 over gamma, I am going to write as-- remember gamma is
1 over the square root of 1 minus beta squared--
so 1 over gamma is the square root of 1 minus beta squared.
And I'm going to write that as the square root of 1
plus beta times 1 minus beta.
So this is the factor 1 over gamma.
Now we have explicitly here a factor of 1 over 1 minus beta.
The inverse makes it 1 over.
Did I write something wrong?
AUDIENCE: From the second until the third time?
PROFESSOR: Second to third, you're there?
AUDIENCE: I thought it wasn't--
PROFESSOR: Yeah, this goes to the other side.
It is minus.
OK, finally, the 1 minus beta, you see
occurs to the first power in the denominator.
And the 1/2 power to the numerator.
So we do indeed get exactly what we wanted.
Delta t sub 0 is equal to the square root of 1
plus beta divided by 1 minus beta times delta t sub s.
And now we know that this is the relativistic answer
for either source or observer moving.
And then if we want to write down what z is,
this is delta t observer divided by delta t source minus 1.
So that becomes just the square root of 1
plus beta over 1 minus beta minus 1.
So we found what we expected, what
we wanted to find to be consistent with the basic ideas
of relativity.
It's the same answer no matter which
one is moving, because it doesn't matter
what frame of reference we do the calculation in.
OK, questions about either the calculations or the ideas
behind them?
OK in that case, let's move on.
I do want to come back and talk about the other two
kinematic effects of special relativity, which
are Lorentz contraction and a change
in the notion of simultaneity.
But before I get to that, there's
one other issue I want to discuss first,
mainly because it's something you need to understand
to do the problem set that's due tomorrow.
And that is the situation that's needed
to describe clocks which might be accelerating.
Special relativity really only describes inertial reference
frames and how things change from one inertial frame
to another.
So if you know how a clock behaves
if it's at rest in one reference frame,
special relativity completely dictates without any ambiguity
what it would look like at a frame that
was moving at a uniform velocity,
relative to the original frame, which means it completely
dictates how that clock is going to behave if it is moving
at a constant uniform velocity.
But nonetheless, in the real world,
we have very few clocks around that are completely inertial.
Every clock that we see around us
from the clock on the wall, which
is moving with the earth or my wrist watch which moves more,
is constantly undergoing accelerations.
So we want to be able to talk about clocks
which are accelerating.
So we need to say a little bit about how we would do that
and how we do it if the clocks were moving at relativistic
speeds , which also happens with satellites, for example.
The GPS system as you've probably been told
wouldn't actually work unless the calculations were
done in such a sophisticated way that they even
take into account the effects of general relativity
as well special relativity.
So moving clocks and how they behave
is a crucially important topic technologically.
So what do we say that a clock that's accelerating?
There's I think a common myth that to describe acceleration
you need general relativity, and therefore we
have to put off talking about accelerating clocks
until we take a course in general relativity.
That's actually totally false.
What general relativity does is provide a theory of gravity
which basically says the gravity and acceleration are intimately
linked.
And that's where acceleration gets
pushed into general relativity.
But special relativity alone is enough for us
to describe any system that could be described by equations
that are consistent with special relativity.
Special relativity does not describe gravity.
So in any situation where gravity is important,
special relativity is at a loss to make real predictions
about what should happen.
But as long as gravity is absent, as long as we're only
dealing with electromagnetic forces
that we think we understand, there's
nothing that prevents us from using
the equations of special relativity
to describe what happens.
We have to use the dynamical equations of special relativity
that talk about how things respond to forces.
And whenever there's a force, there's an acceleration.
But there really are such equations.
We can combine, for example, electromagnetism
with relativistic mechanics to describe a system of particles
that are interacting electromagnetically, completely
consistent with special relativity.
And even those particles are accelerating,
we could say everything we want to say about them.
So in particular, if there's a physical clock,
to the extent that we could describe that clock as made out
of particles whose physics we understand,
special relatively will still tell us
what that clock will do even when that clock accelerates.
The answer, however, from that calculation, you might imagine,
is very, very complicated.
Because the physics of any actual clock,
my wrist watch as an example, is pretty damn complicated.
And we're not really going to write down
the equations that describe my wrist watch to figure out
how it's going to behave when it accelerates.
So what are we going to do?
Let me point out that you already
have pretty much experience with accelerating clocks,
because all of you-- well, many of you--
wear wrist watches like I do that are accelerating
all the time.
And they tend to work.
You basically assume that even though they're accelerating
they've been designed well enough so they can withstand
the accelerations that your wrist gives them
and still read the right time.
On the other hand, one could imagine contrary situations.
Probably my wrist watch would survive this,
but if you take a mechanical clock, a windup clock,
and heave it against the wall and let
it smash against the wall and come to a stop,
as it smashes against the wall, it
wanted to go a very large acceleration.
And if the acceleration is large enough,
we can predict the effect it will have the clock, even
though it's a complicated interaction.
If the acceleration is large enough,
it'll simply break the clock and it will stop.
And that's one possible effect that acceleration
can have a clock.
And other effects are similar in nature.
If there's any effect that the motion of my hand
has on the wrist watch, it would be a mechanical effect
that you'd calculate by understanding
the mechanics of how the watch works,
not by understanding any principles
of general relativity.
What's at stake underlying this--
you might wonder what the real difference is--
special relativity can make precise predictions about how
a clock will behave if it moves at a uniform velocity,
even without knowing anything about the details
of that clock.
Special relativity could make that prediction
because there's a symmetry, Lorentz symmetry, which
relates those two situations.
And that's an exact symmetry of nature.
So no matter what the clock is made out of,
if it's moving at a uniform velocity,
special relativity tells you there's
no doubt it would run slowly by a factor of gamma.
On the other hand, there's no such principle
of any kind, either in special relativity
or general relativity, about acceleration.
So if you want to know the effect of acceleration
on a clock, it really depends in detail on how large
the acceleration is of course and the detailed
physics of the clock and how much acceleration it takes
to effect it and in what way it affects, precisely.
So what's the bottom line?
The bottom line is that when we want
to talk about an accelerated clock, which we really
do all the time, what we always do is simply
assume that the clock is built well enough
so that the acceleration does not affect its speed.
And that really can be said very precisely.
The assumption that we're going to be making
is that these are ideal clocks, meaning
that they're built well.
And when we say the acceleration does not
affect the speed of the clock, what we're saying
is that the clock will run at precisely the same speed
as another clock that's, say, moving instantaneously
alongside it with the same velocity
but with no acceleration.
So at any point in the motion of my arm here,
my wrist watch will have some specific velocity.
The velocity will affect in some tiny way
the speed of the clock by this factor gamma, which
will be very close to 1 for that case.
But we're going to assume, if we call my watch an ideal clock,
that in the even time, it will be running
at exactly the same speed as a clock which is not
accelerating, but which is moving
with the same velocity as the wrist watch.
Therefore, the factor of gamma will be there,
but there'll be no effect of acceleration.
The speed of the clock will be determined only
by its velocity relative to our reference frame.
OK, is that clear enough?
And that's what you need to assume about some accelerating
clocks that show up on the problem set.
That's why I wanted to get it in today.
OK, if there are no other questions about that,
I'd like to come back and talk a little bit more
about special relativity and its consequences.
Sometime later in the course, we'll talk a little bit more
about what I would call the dynamical consequences
of special relativity, which include
well known equations like e equals mc squared, for example.
But before one talks about energy and momentum,
which are quantities which I will dub dynamical,
there kinematic effects of special relativity
of which this time dilation is one.
And by kinematic I really just mean
the consequences of special relativity
for the measurements of times and distances.
And if one limits oneself to discussing consequences
for times and distances, kinematic consequences,
there really are precisely three and no more consequences
of special relativity.
And really all special relativity in some sense
is embodied by these three statements
that we're going to be talking about, the first of which
was time dilation, which we've already seen.
I'll just remind you.
Time dilation says that any clock which
is moving at speed d relative to a given reference frame
will appear in quotation marks to an observer using
that reference frame to run slower than normal by a factor
denoted by the Greek letter gamma.
Turn out the board lights in case they're distracting.
And again, "appear" refers not to how it would actually
look to any particular observer, because any particular observer
in a particular location is going
to be waiting for light rays to reach that observer.
And they'll take different amounts of time
depending on where they start.
"Appear" refers to measurements made in the reference
frame of the observer, where we assume
that all the actual measurements are made by on the spot clocks
and rulers who measure where the object actually
were at the time these events happened and not
what it looks like at sometime later when
light rays reach some observer.
OK, the second consequence, and again all these
will involve the word "appear."
And I'll always write it in quotation marks
to remind you that it's not exactly what a person would
see.
The second one is another famous effect of special relativity,
Lorentz contraction, or sometimes called
Lorentz-Fitzgerald contraction.
Any rod which is moving at a speed v along its length
relative to a given reference frame
will appear-- and again appear to an observer
using that reference frame-- to be shorter
than its normal length by the same factor, gamma.
A rod which is moving perpendicular to its length
does not undo that change in apparent length.
So these pictures kind of show it all.
A rod, that bar is the rod, a rod which
is moving at speed v will look contracted,
will appear to be contracted, by a factor of gamma.
And a rod which is moving perpendicular
to its length, a rod which is like this moving this way,
has no such effect, we'll appear to have its natural length.
And this is a very famous consequence
of special relativity.
It means that moving rocket ships shorter
and shorter and faster they go, and so on.
And again, you should remember, it's
not what you'd actually see.
It's what you'd measure if you had on the spot local observers
making these measurements, which you then later compile.
OK, that's actually all I want to say about the contraction.
Any questions?
OK, next and last is an effect we just talked about less
because it's a little bit more intricate to describe.
But the other crucially important effect-- these
would not be consistent if you didn't have all three.
The other crucially important effect
is the changing simultaneity, the relativity of simultaneity.
And it takes more words describe it,
so there's more words on the slide than on the other slides.
And the pictures are a little bit more complicated too.
But the pictures do say it all, really.
The point is that if you have a system consisting of two clocks
which have been synchronized in their own rest frame--
they're both at rest with respect to each other
and they've been synchronized in their own rest frame--
and they're connected by say a rod,
which has some length which we'll call l sub 0 in the rest
frame of the two clocks .
If that whole system moves relative to us
by speed v along the length, those
clocks, even though they were synchronized in the rest frame,
would not look synchronized to us
but rather would look like they're out of synchronization.
And in particular, it will look like the trailing clock,
the one on the back of this combination of clocks,
will look a little bit like it reads a little bit
later in the day by a factor which
is beta times l sub 0 times c.
Beta remember is v over c. l sub 0
is the distance between the clocks
as measured in the rest frame of the clocks.
And c is of course the speed of light.
On the other hand, if these two clocks
were moving in a direction which is
perpendicular to the line that joins them,
then there's no change in the synchronization.
Now I should mention that this really is crucially
important to the consistency of the whole picture.
And actually showing that the picture is consistent
is more than we're going to do.
It's not impossible to do at this level by any means.
But it's more than we're going to do in this class,
since we're not focusing on special relativity.
But when you hear about special relativity
and look at these postulates, you
might realize that there seems to be a pretty obvious tension
between the idea that moving clocks run slowly
and that all observers experience
the same laws of physics.
Because it means that if you and I are moving with respect
to each other, I would claim that your clock
was running slowly.
But at the same time, you would claim
that my clock was running slowly.
Because from your point of view, you're
at some fixed velocity and therefore an inertial frame.
And I'm moving relative to you.
So as you would describe it, I would be a moving clock.
And my clock would be running slowly.
So I think your clock is running slowly.
You think my clock is running slowly.
That seems to contradict.
What happens if we hold the clocks next to each other
and really just watch how they compare?
Which one gets ahead?
How can we disagree on that?
Well of course, we can't hold the clock next to each other
and also have them moving relative to each other.
That's part of how you get out of this conflict.
But let's think in a little bit more detail
about what we're really saying when
I say that your clock is running slowly.
Remember, I want to make all my observations
not by watching you, because then there's this time delay
effect which complicates things.
I make all of my observations by having
a family of local observers surrounding me all at rest
relative to me.
And they report back to me.
And only when I receive those reports and piece
them together do I get the simple picture
of what happened where when, which
is the simple picture that is described
by these "appear" relationships.
So when I say your clock is running slowly, what I mean
is that if I have a network of clocks all at rest relative
to me, and your clock comes shooting
by, I would measure what time it reads as it goes past all
my local clocks.
Rather, they would measure what time your clock reads
as it goes past each local clock.
And then they would all report back to me.
So if your clock is running slowly, for example,
let's say by a factor of two, it would
mean that when you're clock passes my clock
and my clock reads one second, your clock would only
read half a second, because it's running slowly.
It hasn't ticked off as much time.
When it passes some later clock of my sequence of clocks, where
my clock reads two seconds, your clock
will read one second, and so on.
So in that sense, I would say your clock was running slowly.
Now that has to be consistent with you thinking
that my clocks are all running slowly as well.
So if you agree that my clocks were all synchronized,
then you would conclude that my clocks must be running fast.
Because when your clock reads a half second,
my clock reads one second.
When your clock reads one second,
my clock reads two seconds.
You would say that my clocks were reading fast
if we just made that direct comparison.
But at the same time, we know that's not the right answer.
You should see the same physics and I see.
If you and I are moving with respect to each other,
you should see my clocks running slowly.
So the way out of that is this question of simultaneity.
From the point of view of your clocks
going past all my clocks, if you just
looked at the time on my clocks as you passed them,
you would actually think that they
were running fast relative to your clock.
But you would also, however, not think
of those clocks were synchronized with each other.
So you don't determine what speed
they're running by looking at two different clocks.
If you want to figure out whether my clocks
running fast or slow, you want to look at one of my clocks
and see how it changes with time,
not comparing different clocks.
Because the different clocks would just
be out of synchronism with respect to each other as you
would see it.
And we're not going to go through the details.
But if you do look at my clocks consistently
using a family of your clocks that
are stationary relative to you, just
as I thought about a network of clocks
when I was trying to measure the speed of your clock,
then everything's consistent.
You would see all my clocks running slowly.
I would see all of your clocks running slowly.
And because we disagree on what's simultaneous,
there are no contradictions.
So simultaneity is absolutely crucial
to get out of what would otherwise
be a glaring contradiction in the whole system.
OK, that's about all I really wanted to say today.
But let me just give a preview of things
that we'll talk about later in the course concerning
relativity.
So far I think we've said all we are going to say,
unless the questions.
I think we've said all we're going
to say about the kinematic consequences
of special relativity.
And we're not going to be trying to derive them, as I said.
If you're interested, by the way,
the notes recommend several references,
including the lecture notes from eight to 86
from earlier years when special relativity was
included as a real topic.
So certainly if you're interested in learning
about this and you haven't already seen it,
I'm happy to help you.
But otherwise, it will not be part of this course
to discuss how these three consequences
of special relativity arise from the basic postulates
of special relativity.
But later we will be saying things that follow further
along the line by pursuing the consequences
of special relativity for momentum and energy,
which will be important to us.
The connection is the important connection
and simple connection that energy and momentum
are only interesting to us if they're
defined in a way which makes them conserved quantities.
That's why energy and momentum are important in physics.
Because for a closed system, the total energy
and the total momentum do not change.
Energy and momentum can be transferred
from one part of a system to another.
But energy and momentum cannot be either made nor destroyed.
Now if we took Newton's definitions of energy
and momentum, and used relativistic kinematics, what
we would find is that if we looked
at, say, a collision of two particles,
the Newtonian definitions of energy and momentum,
if we took those seriously, would tell us
what might happen in a collision.
Usually there's an angle that's undetermined.
But given an angle, it determines everything else.
If we used special relativity to then describe
what that same collision would look like in a different frame,
we would find that these Newtonian definitions of energy
and momentum would not be conserved in the other frame,
if they were conserved in the first frame.
The conservation laws are dependent on what
reference frame you're using.
So what Einstein amended was slightly
modified definitions of energy and momentum, which
are determined by the criterion that these slightly modified
definitions of energy and momentum should,
if they are conserved in one frame
be also concerned in any other frame, which
are related by the relationships to special relativity.
So that's why it was essential once one changed
the kinematics of going from one frame to another
to also change the definitions of energy and momentum
so that the conservation laws would hold in all frames using
the new transformation equations to get from frame to frame.
And that's why later in the course
we will be introducing a slightly modified, slightly
non-Newtonian, definitions of energy and momentum
of moving particles.
OK, that's all for today.
I will see folks next Tuesday.