Georgi, Howard. Physics 16 - Mechanics and Special Relativity (англ.)

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different frame on the way back. It is the fact that twin 1 must switch from one inertial frame to

the other that makes his experience different.

Relativity may be strange, but it is consistent.

lecture 10

Topics:

Where are we?

Spacetime events

The space-time interval

The invariant interval

Lorentz transformations

Interlude on relativistic units

Varieties of space-time intervals

The tip of tomorrow

Where are we?

I hope that I have convinced you that relativity is strange. We have discussed time dilation and

the twin paradox in the last lecture.

You should have read about the relativity of simultaneity and

Lorentz contraction in David’s book. I am not going to talk about them here because I want you

to avoid using them whenever possible! As I mentioned last time, whenever possible, when doing

relativity problems, use time dilation. It is almost always easier to understand because it is related

to a very very obvious set of measurements - the readings on a clock or set of clocks. All you have

to do is ﬁgure out what times you need to measure and what clocks you need to measure them, and

you will be able to use time dilation successfully to solve problems. This is important!

By next week, I will be able to explain why I am so sure that relativity is true. For now, just

try to accept that as strange as this is, there is absolutely convincing evidence that the world really

works this way. Since we are stuck with it, we should ﬁgure out how to make sense of it, and how

to avoid getting confused by the strangeness of it all. To do this, you must train yourself to consider

space and time together, because what is time in one inertial frame is a combination of time and

space in another. There are a number of closely related ways of dealing with this. I will talk today

about a few things that you have already read about in David Morin’s book: spacetime events, the

space and time intervals and the invariant combination of the two, and Lorentz transformations.

Spacetime events

The concept of a spacetime event is very simple. Even in non-relativistic physics, to describe an

event completely you must specify not only where it is but also when it takes place. What’s the

big deal? The importance of the spacetime event in relativistic physics is really more of a negative

one. There are many things that don’t make absolute sense in the relativistic world because they

look different in different inertial frames. Time is relative. Distance is relative. Simultaneity is

relative. The spacetime event gives us something to hang on to. The coincidence of spacetime

events is NOT RELATIVE. If spacetime events coincide, that is if they things happen at the

I realized that I should have emphasized that there is nothing genuinely paradoxical about the twin paradox. It is

just strange. The resolution of the “paradox” is that the traveling twin really does come back younger.

same place and at the same time, they coincide in all inertial frames. Spacetime events are thus

the sensible things to talk about. Though the coordinates of a given spacetime event will change

depending on what inertial frame you’re in, the event itself does not change.

The right analogy to keep in mind is that different inertial frames in space and time are like

different coordinate systems in three-dimensional space. Time and distance change in going from

one inertial frame to another just as the x component of a point changes in going from one coordi-

nate system to another. But spacetime events in space and time are like points in three-dimensional

space. If points coincide in one coordinate system, they coincide in all coordinate systems, just as

spacetime events are coincident in all inertial frames if the coincide in any one.

One of the things that we will try to train you to do is to identify the important spacetime events

in a process, and to be very suspicious of any statement or question that cannot be described in

terms of spacetime events. Let me illustrate the difference in the twin paradox discussion we gave

in the last lecture, where twin 1 left twin 2 on earth and traveled to planet X and back. Here the

crucial spacetime events were:

Twin 1 leaves twin 2 — x = 0, t = 0 in Earth’s frame, with Earth at the origin;

t = 0

1 v →

∗ ∗ ∗

Twin 1 arrives at planet X — x = L, t = L/v in Earth’s frame;

t = L/v

←- v 1

∗ / ∗

The turn-around signal reaches twin 2 — x = 0, t = L/v + L in Earth’s frame;

t = t

∗ / ∗

← v 1

Twin 1 and twin 2 are reunited — x = 0, t = 2L/v in Earth’s frame.

t = 2L/v

∗ ∗ ∗

These events have an invariant meaning even though their coordinates, both their space and their

time coordinates, will change if they are viewed in a different coordinate system.

It is perfectly meaningful to put together events, so long as they occur at the same place and

at the same time. For example, again from the discussion of last lecture, “Twin 1 arrives at planet

X at t = L/v, turns around (quickly) and sends a turn-around signal to twin 2, indicating that

he has reached the planet.” This describes two separate events — reaching the planet and sending

the signal — but because they happen at the same place and at the same time, their space and

time coordinates will be the same in all reference frames. Thus the statement makes sense in all

reference frames

If we don’t stick to spacetime events, there are lots of questions that we can ask that don’t mean

anything at all unless we specify the reference frame. For example

Meaningless questions

How far does twin 1 travel?

How long does it take twin 1 to get to Planet X?

Where is twin 1 when the turnaround signal reaches Earth?

How old is twin 2 when twin 1 reaches planet X?

The last two are a classic kind of pitfall, because they attempt to compare two events at different

places. Think about them carefully.

The space-time interval

Let me illustrate some of these issues by talking about a particularly useful sort of spacetime event

— the tick of a clock. If I have a clock sitting at the origin of the reference frame of the lecture hall,

ticking every T seconds, the ticks of the clock deﬁne a series of spacetime events. In the frame of

reference of the lecture hall, all of these events have space coordinates 0, because they are all at

the origin, and their time coordinates are just 0, T , 2T , etc. Thus the time and space coordinates of

the series of ticks would look like

event E

···

(t, ~r ) (0, 0) (T, 0) (2T, 0) ···

(1)

In general, if we hadn’t put the ﬁrst tick at the origin, the ~r and t coordinates of the later ticks

would all just have the initial ~r and t positions added on to them, so the sequence would look like

event E

···

(t, ~r ) (t

, ~r

) (t

+ T, ~r

) (t

+ 2T, ~r

) ···

(2)

We know that there is nothing very interesting about the ~r

and the t

, which can be changed by

moving to a different coordinate system by translations in space and time. Thus it is useful to

The word “meaningless” in the table is a little strong. It might be fairer to say “questions whose answers depend

on the reference frame,” but we are trying to get your attention!

get rid of this dependence. We can do this in a way that is completely independent of our choice

of origin of coordinates by considering the space-time intervals, which are just the differences

between the coordinates of space-time events

∆E

= E

− E

= (∆t

, ∆~r

) = (t

− t

, ~r

−~r

) (3)

intervals ∆E

∆E

···

(∆t, ∆~r ) (T, 0) (2T, 0) ···

(4)

Events are the important underlying things, but have all this extraneous dependence on the origin

of the coordinate system in space and time, so it is almost always useful to think about space-time

intervals when you want to calculate something.

So far there is nothing relativistic about this, but here it comes. We also know from our discus-

sion of time dilation what these same events look like in a reference frame in which the clock is

moving. Suppose that the clock is ﬁxed in the lecture hall, but describe it in an inertial reference

frame is moving in the −x direction with respect to the lecture hall. According to the series of

clocks that move by in the moving reference frame, the single clock we have been considering

ticks slowly. This is time dilation, the ticks are spread out by a factor of γ, so that the time co-

ordinates of the ticks in the moving frame are 0, γT , 2γT , etc. Let us also assume that the space

coordinates are 0 for the ﬁrst tick (at t = 0). Then because we know the velocity of the clock, we

can easily work out the other ~r coordinates –

event E

···

, ~r

) (0, 0) (γT, ˆxvγT ) (2γT, 2ˆxvγT ) ···

(5)

Again, if we hadn’t put the ﬁrst tick at the origin in space and time, the x and t coordinates of the

later ticks would all just have the initial ~r and t positions added on to them, so the sequence would

look like

event E

···

, ~r

) (t

, ~r

) (t

+ γT, ~r

+ ˆxvγT ) (t

+ 2γT, ~r

+ 2ˆxvγT ) ···

(6)

And again, we can get rid of the dependence on the origins in space and time by looking at the

intervals:

intervals ∆E

∆E

···

(∆t

, ∆~r

) (γT, ˆxvγT ) (2γT, 2ˆxvγT ) ···

(7)

Finally, there is nothing special about the x direction. For a different choice of coordinate system,

the motion might be rotated into a different direction, ˆx → ˆv

intervals ∆E

∆E

···

(∆t

, ∆~r

) (γT, ˆvvγT ) (2γT, 2ˆvvγT ) ···

(8)

While you are used to thinking of the difference between (1) and (2) or between (7) and (8) as a

change of coordinate system, you are probably not used to thinking about the difference between

(1) and (6) as a change of coordinate system. But what you see here is that these spacetime

coordinates really can describe the same set of events, the ticks of a clock, in different reference

frames. Going from one reference frame to another IS like going from one coordinate system

to another — it is just that the time “coordinate” as well as the space coordinates are involved.

The events don’t change, but both the time and space coordinates change when we go from one

inertial frame to another. What this means is that we have gone from a 3-dimensional space to a

4-dimensional spacetime.

The invariant interval

We can extend the analogy between going from one coordinate system to another and going from

one reference frame to another in the following way. When you make a change of coordinate

system in space, the distance between points doesn’t change. You can easily compute the square

of the distance between two points labeled by vectors ~r

and ~r

in terms of their coordinates,

≡

−~r

j=x,y,z

− r

(9)

Two things have happened here. First we have subtracted the coordinates of the two points. This

eliminates dependence on the origin of the coordinate system, which just cancels when we subtract.

Another way of putting this is that we are not interested in the length of the vectors themselves,

because this depends on the arbitrary position of the origin. But if we make the vector from ~r

by forming the combination ~r

−~r

, it doesn’t depend on the origin. Then we sum the squares

of the coordinates to get something that doesn’t change when we make a rotation.

When you make a change of coordinate system in spacetime, by going to a new reference

frame, there is a similar thing that doesn’t change. If I have two events, event 1 at time t

and

position ~r

and event 2 at time t

and position ~r

, the following combination doesn’t change when

we go from one reference frame to another:

≡ c

− t

−

−~r

(10)

doesn’t change when we go from one reference frame to another. Again we have done two things.

The obvious one is to subtract the space and time coordinates of the two events, so that s

doesn’t

depend on the origin of coordinates. The other, much less obvious one, is to combine the difference

in time and the difference in space in a very particular way. You can see that this is the right thing

to do by looking at two successive clock ticks in any of (4), (7) or (8), we have

= c

− 0

= c

(11)

In (6), we have

= c

γT

−

vγT

= (c

− v

/(1 − v

) = c

(12)

This is independent of v, so it looks the same in any frame of reference.

The quantity s

in (10) looks a lot like the quantity `

in (9). There are three differences. One

is that time is involved. The second is the minus sign in front of the second term. The third is the

factor of c

in the ﬁrst term.

Let’s deal with the easy one ﬁrst. The factor of c

is nature’s way of telling us that we are using

a really stupid system of units. Because time and space get mixed up in relativity, and because the

ratio v/c appears in many many of the relations of relativity, it makes sense to use units in which

c = 1. One way of doing this is to use seconds as your unit of time and light-seconds, that is the

distance light travels in a second, 299,792,458 meters. Of course, because we are so slow, this is

an inconveniently long distance (which is why we don’t adopt this convention as a standard part of

our metric system) but you will get used to it. Anyway, 1 is easier to remember than 299,792,458.

So from now on, we will follow nature’s advice and set c = 1, which we will call “relativistic

units.” This will make our formulas look simpler. We also lose something by doing this, but it is

not very important, and we will come back and discuss it at the end of the lecture today.

The other two differences between s

and `

express the crucial strangeness of relativity. Time

is not quite the same as space — that is obvious from the minus sign. But it is not completely

different either, because it must be included to get something that looks the same in different

reference frames.

Notice that because of the minus sign, there is no reason why s

has to be positive. It is positive

in our example of the interval between two ticks of a clock, but for other kinds of intervals, such

as the interval between two events that are in different places but at the same time in some frame,

can be negative as well. The quantity s

goes by various names. I will call it the “invariant

interval.”

Let’s illustrate this with the twin “paradox.”

Twin 1 leaves twin 2 — Event E

with (t, x) = (0, 0) in Earth’s frame, with Earth at the origin;

t = 0

1 v →

∗ ∗ ∗

Twin 1 arrives at planet X — Event E

with (t, x) = (L/v, L ) in Earth’s frame;

t = L/v

←- v 1

∗ / ∗

The turn-around signal reaches twin 2 — Event E

with (t, x) = (L/v + L, 0) in Earth’s frame;

t = t

∗ / ∗

← v 1

Twin 1 and twin 2 are reunited — Event E

with (t, x) = (2L/v, 0) in Earth’s frame.

t = 2L/v

∗ ∗ ∗

We can easily write down the coordinates of events E

and E

in a frame of reference moving

in the +x direction with velocity v, because in this frame, the spaceship is at rest for this leg of the

trip. Thus if we keep E

as the origin of space-time, the coordinates are

: (t

, x

) = (0, 0) and E

: (t

, x

) = (L/(γv), 0) (13)

incorporating the fact of time dilation. But now you can see that s

is the same in either frame. In

the earth’s frame

= c

− x

= c

− L

(14)

In the ship’s frame it is

− x

= c

L/(γv)

/γ

(1 − v

)

= c

− L

(15)

Note that it would NOT be trivial to write down the coordinates of events E

and E

in the

primed frame. But that brings us to the next topic.

Lorentz transformations

You have read about a nice derivation of Lorentz transformation in David Morin’s text. I will write

them down and talk about them. They look much simpler with c = 1. So suppose that we have

two events

event 1 at t

, ~r

and

event 2 at t

, ~r

(16)

The interesting thing is the interval in time and space between these two events:

∆t = t

− t

, ∆~r = ~r

−~r

. (17)

Now if we look at these same two events from a reference frame moving in the +x direction

with speed v, the coordinates of the events will change. We will get a new description of the events

in terms of new coordinates,

event 1 at t

, ~r

and

event 2 at t

, ~r

(18)

Again, the interesting thing is the interval in time and space between these two events:

∆t

= t

− t

∆~r

= ~r

−~r

(19)

Now the statement of the Lorentz transformation is that we ﬁnd that we can choose a coordinate

system in which

∆x

= γ(∆x − v∆t) ,

∆t

= γ(∆t − v∆x) ,

∆y

= ∆y , ∆ z

= ∆z .

(20)

If we set ∆~r = 0 and look at ∆x

and ∆t

as a function of ∆t, we can recognize the moving

clock. The condition ∆~r = 0 means that we are looking at a series of events that are ﬁxed in the

original frame. A pair of events with ∆~r = 0 and time difference ∆t could be two ticks of a ﬁxed

clock a time ∆t apart.

∆~r = 0

∆x

= γ(−v∆t) ,

∆t

= γ(∆t) ,

∆y

= 0 , ∆z

= 0 .

(21)

The γ in

∆t

= γ (∆t)

expresses time dilation. The −v in

∆x

= γ( −v ∆t)

describes the clock moving with x velocity −v in the new frame.

The other terms in the Lorentz transformation can be understood in a similar way.

Notice that the Lorentz transformations are linear transformations. This is important for many

reasons, but one of the most obvious is that the time evolution of a clock in some reference frame

can be described by a straight line in space and time. The trasnformation from one frame to another

must take any such straight path into another straight path.

It very important to understand why Lorentz transformations exist. They are a kind of space-

time analog of rotations, but they look different because of the minus sign in (10). I recommend

that you learn about Lorentz transformations, that you learn how to use them, but that you actually

use them in a problem only if forced to do so (that is if the problem says “Use Lorentz trans-

formations to ...”). There are usually easier ways of understanding any given problem. Lorentz

transformation will get you there in the end if you don’t make any mistakes, but you have to be

very careful and systematic to use them properly.

Interlude on relativistic units

Let’s talk about the addition of velocity formula you saw in Morin

+ v

1 + v

. (22)

In sensible units in which c = 1, this looks like

+ v

1 + v

. (23)

One can ask, suppose, in sensible units, you get a result like (23). How would you know how to

put the factors of c back into (23) to get (22)? This is a crucial step because it allows you to do

calculations with c = 1, to make things simple, but still get the full answer at the end of the day,

so you can talk to people who insist on using dumb units.

If you have done a calculation in relativistic units and you want to put the factors of c back in

to translate the result to conventional units, you have to know the conventional units of the objects

you have calculated. Then you put in factors of c to get the units right.

In the case of (23), we know that we are calculating a velocity, and the numerator is already a

velocity. So we can leave that alone and make the denominator dimensionless. The 1 term is OK,

and we can make the uv term dimensionless by dividing by two factors of c — and that is where

(22) comes from.

Another example: Suppose that you are calculating an energy in terms of a mass m and a

velocity v and you get the result

6π m v

(24)

The 6π is a dimensionless number — we can forget about that. But the rest would not have units

of energy in conventional units (remember, for example, that Newtonian kinetic energy is

We have two too many factors of v , so we must divide by c

— thus the result in normal units is

6π m v

(25)

Here’s another one. For a system of two particles, with masses m

and m

and speeds v

and

, suppose you calculate an object that is a mass, and the result is

+ m

+ v

(26)

Let us start with the denominator. We don’t know what the units should be of the denominator

alone (because we can always multiply numerator and denominator by a power of c), but it is clear

that the two terms in the denominator must have the same dimension. So for example, we can

divide the second term by c

. That makes the whole denominator dimensionless, so the numerator

must then have the dimensions of mass. We can arrange this by multiplying the ﬁrst term by c and

dividing the second by c — thus the result in normal units is

c/v

+ m

+ v

(27)

Varieties of space-time intervals

We spent a lot of time talking about the space-time interval (∆t, ∆~r ) between two ticks of a

moving clock. This is called a “time-like” interval, because there is a frame in which it has a time