1 Special Relativity

1.1 Galiean Relativity

1.1.1 Spacetime Diagrams

We keep seeing the word “relativity” appear in our discussion. To the person on the street, relativity (nor-mally associated with Einstein) elicits the statement “everything is relative,” implying that we cannot knowanything objectively. In fact, both Einstein and Galileo (who we will also discuss) would have emphaticallydisagreed with this assessment. We can say things objectively – it’s just that part of our objective statementneeds to include information about how to translate what I see, to what you see.

Our standard piece of equipment for this experiment will be a train moving to the right at some knownspeed, v. You will be on that train, while I will be on the platform.

Note that the train is moving at all times. The “Special” in special relativity comes from the fact that thereare no accelerations.

Of course, you also know from your own experience that from your vantage, it appears that I am movingbackwards at speed, v.

In general, when we talk about different perspectives, we’ll refer to the “unprimed” (meaning, in this case,your) or “primed” (meaning my) frame. For example, if you saw a puppy roll over 10 meters in front ofyou, you’d say, x = 10m. However, if I also observed the same “event,” but it was behind me 5m, I’d say,x′ = −5m, which is just a shorthand for saying that the position of the puppy rolling over is -5m from theperspective of the person outside the train.

I used the word “event” in the previous paragraph. For us, there are only events, and events have a time,and a position. Thus, “The party begins at 9:00pm at the funky disco” indicates an event.

We can imagine a particularly simple 1-d universe full of events, and plot it up on a “space-time” diagram.

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In general we tend to scale these things so that 1 unit in the x-direction corresponds to, say, 1 light-second,and 1 unit in the y-direction corresponds to a second, and thus, light travels at 45 degree angles (or -45 deg.if it’s traveling to the left). Someone sitting still will move vertically through time (but not through space).And, of course, since no-one can travel faster than light, your “world line” can’t ever be shallower than 45degrees.

On the space-time diagram, I’ve drawn a light-beam fired from a ship moving half the speed of light towardan observer at rest (Event 1). The beam is detected by the stationary observer (event 2), and re-deflected.Finally, the the lightbeam is detected by the moving observer (event 3).

For each observer, each event has a definite time and position. I have drawn things from the perspective ofthe stationary observer, since from your own perspective it always seems as though you are sitting still.

1.1.2 Galilean Relativity

All of this is just notation leading us up to relativity. And when you think relativity, you no doubt think ofGalileo. Well, probably not. But in reality, Galileo’s relativity is the one in our everyday experience.

Time: For Galileo, time was an absolute, and thus, all clocks ran at the same rate. So, even though it islegitimate (say if you and I are in different time-zones) for us to measure an event at different “times” – sayt1 = 0, and t′1 = 3600s (if you measure a particular event as being one hour later than may because younever reset your watch from daylight savings), it is true that our watches run at the same rate. Thus, if wemeasure a second event, and I measure it at t2 = 10s, you’ll measure it as t′ = 3610s.

According to Galileo, For both of us:

Galileo : ∆t = t2 − t1 = ∆t′ = t′2 − t′1

Space: With space it’s a different matter. Consider that for a moving observer, objects, and hence events,will look like they’re in different positions at different times. Thus, if a particular event appears to me atsome position, x, you’ll observe it as

x′ = x+ vt

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which should accord with your intuition. If you don’t believe me, think about how it looks when someonewalks forward in a moving train.

Finally, we can relate the velocity of an observed object. Consider that velocity is simply defined as:

u ≡ ∆x

∆t(1)

where we simply observe a moving object at two different times. Using the relations:

∆x′ = ∆x+ v∆t (2)

∆t′ = ∆t (3)

simple algebra yields:u′ = u+ v (4)

In other words, if you were on a train going 50mph, and sprint forward at 10 mph, outside the train it wouldseem like you’re going 60 mph.

Finally, note that these equations are completely symmetric. We could just as easily say:

∆x = ∆x′ − v∆t′

∆t = ∆t′

u = u′ − v

And thus, there’s no way from Newton physics to tell who is moving and who is staying still.

According to Newton, the only thing that matters are accelerations, and consider that a system which obeysNewton’s laws in one frame will also satisfy them in any other inertial frame.

No problem, right? Wrong.

You’ll get a hint of the problem when you consider the following riddle: What happens if you’re traveling90% the speed of light and you turn on your headlights?

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1.2 Special Relativity: Space and time

1.2.1 Einstein’s Postulates

# 1:

The laws by which the states of physical systems undergo change are not affected, whetherthese changes of state be referred to the one or the other of two systems of coordinates in uniformtranslatory motion. Or: the laws of physics are the same in all inertial frames of reference.

# 2:

As measured in any inertial frame of reference, light is always propagated in empty spacewith a definite velocity c that is independent of the state of motion of the emitting body. Or:the speed of light in free space has the same value c in all inertial frames of reference.

1.2.2 A Simple Experiment with Light

As I’ve mentioned before, you are in the primed frame (the one inside the train). This is a very special train,one set up with mirrors and lasers, and sophisticated chronometers and the like.

Light (in the form of lasers) is particularly important because of an experiment done in 1887 by Michelsonand Morely, which showed that light travels at a constant speed for all observers. Using that fact, and thefact that at low speeds our picture of the universe should be the same as that suggested by Galileo, will beour basis of special relativity.

So, consider the following apparatus within your train.

In this “experiment” there are 3 events:

1) A pulse is fired (at the speed of light) from the bottom of the train.

2) The pulse is reflected from the top of the train.

3) The pulse is detected at the bottom of the train again.

View from the unprimed frame:

Despite the small offset in the picture, events “1” and “3” take place at the same position (according to

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you), and thus:∆x = 0

However, since the height of the train is h, the light must have traveled a distance of 2h between the beginningand end of the experiment, and thus:

∆t =2h

cAll we’ve used here is the fact that light travels at the speed of light.

View from the primed frame:

Now, from the primed frame, the situation looks somewhat different. The detector has moved from whenthe laser was fired, and thus, from my perspective, the experiment looks like:

The solid train represents the position of the train at the beginning of the experiment, and the dotted linerepresents the position of the train at the end of the experiment. The detector must have moved a distance:

∆x′ = v∆t′

from event “1” to “3”. This is the same result we would have gotten from Galilean relativity. Well, exceptfor one thing. We are no longer assuming that ∆t′ = ∆t.

In this case, we note that the total trip length is:

d′tot = 2

√(v∆t′

2

)2

+ h2

In case it’s not obvious where I got that, each light beam can be written as an x-component (v∆t′/2, where∆t′ is the time observed by me), and a y-component, h. I’m just using the Pythagorean theorem to addthem.

We also note that the light travel distance can be related to the full time interval between “1” and “3” viathe relation:

∆t′ =d′totc

= 2

√(v∆t′

2c

)2

+h2

c2

=

√(v∆t′

c

)2

+

(2h

c

)2

=

√(v∆t′

c

)2

+ (∆t)2

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where I’ve used ∆t = 2h/c explicitly.

Interesting... I now have ∆t′ in terms of ∆t. Squaring the expression, I get:

∆t′2 =

(v∆t′

c

)2

+ (∆t)2(1− v2

c2

)∆t′2 = (∆t)2

∆t′2 =(∆t)2

1− v2/c2

∆t′ =∆t√

1− v2/c2

Thus, we’ve found an interesting relation:

∆t′ = γ∆t+ f(v)∆x (5)

where

γ ≡ 1√1− v2/c2

(6)

Note that I put a function f(v) in there. Why? Because we don’t know what happens if ∆x 6= 0. We’llfigure it out eventually.

Also, by the same argument, since we found:

∆x′ = v∆t′

then∆x′ = vγ∆t+ g(v)∆x (7)

These equations need to be symmetric. In other words, there shouldn’t be anything which can distinguishbetween the “moving” (unprimed) frame and the “stationary” (primed frame). To you, I should just appearto be moving to the left. And thus, we should have the relations:

∆x = −vγ∆t′ + g(−v)∆x′ (8)

and∆t = γ∆t′ + f(−v)∆x′ (9)

To figure out what f(v) and g(v) are, we first note, that by definition, ∆x = ∆x. Thus, plugging equations8 and 9 into 7 we get:

∆x = vγ(γ∆t+ f(v)∆x) + g(−v)(vγ∆t+ g(v)∆x)

= (−vγf(v) + g(v)g(−v)) ∆x+(−vγ2 + vγg(−v)

)∆t

Since the left has to equal the right, the terms in front of ∆x have to add to 1, and the terms in front of ∆thave to add to 0. Thus:

vγ2 = vγg(v)

vγf(−v) = 1− g(v)g(−v)

The first of these yield:g(v) = g(−v) = γ

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while plugging in gamma (after a bit of algebra) yields:

f(v) = γv/c2

We (finally!) have our Lorentz transforms:

∆x′ = γ∆x+ vγ∆t (10)

∆t′ = γ∆t+v

c2γ∆x (11)

∆x = γ∆x′ − vγ∆t′ (12)

∆t = γ∆t′ − v

c2γ∆x′ (13)

Note that if γ ' 1, then these are simply ∆x′ = ∆x + v∆t, and ∆t′ = ∆t, just as with the Galileantransforms.

Moreover, as with our Galilean transforms, we can combine these to figure out how velocity transforms:

u′ =u+ v

1 + uvc2

(14)

u =u′ − v1− u′v

c2

(15)

1.2.3 Some Examples

A Lightbeam:

Consider what happens if you shine a flashlight (u = c) on your train which is moving, say, at half the speedof light (v = c/2). The speed I observe the lightbeam to be traveling is:

u′ =c+ c/2

1 + c/2 cc2

=1.5c

1.5= c

A Meter Stick:

Now, what happens if you have a meter stick in your ship (which is lying on the floor in the i direction)?You observe it as having a length, of course, of 1 m.

I, however, notice something different. Since I’m measuring the length at a particular moment, I measure∆t′ = 0 (the front and back are measured at the same time). Thus, according to equation 11, ∆x = γ∆x′.Or, if I am measuring the length of the rod in my frame:

∆x′ =∆x

γ=L

γ

Since γ is always greater than or equal to 1, I always measure moving meter sticks to be shorter thanstationary ones. Not just meter sticks – everything. You, your ship, your control panels, and so on.

This is known as length contraction. Now here’s the wacky thing – because of the symmetry of theequations, you will find that a meter stick in my frame (again, oriented along the x-axis) looks short to you.

A Clock:

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What about a clock? Well, imagine that a clock ticks on your ship ∆t = 1s. For you, the two ticks of theclock happen in the same place ∆x = 0. Thus, according to equation 13, we have ∆t′ = γ∆t. In otherwords your clock (and your heartbeat, and the oscillations of your molecules and everything else, indeedtime itself) seems to run slow on your ship.

But again – to you, it appears that my clock (and heart, etc.) are running slow!

This is known as time dilation.

1.2.4 A Spacetime Example

Here’s a snapshot of a stationary lab in which a light beam is bouncing back and forth as seen from withinthe lab, and also as seen from an observer moving at 1/2 the speed of light:

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1.3 The Relation of Energy to Momentum

Thus far, we’ve described how to translate between space and time intervals from one frame to another.And beyond a constant speed of light, we’ve introduced very little actual physics into the mix. Where’smomentum? Where’s energy?

I want to begin by reminding you of the fundamental definition of energy:

∆E = v∆p (16)

Remember, we derived this from the Work-Kinetic Energy theorem. Turning this into a derivative, we get:

dE

dp= v

Remind you of anything?

How about:dx

dt= v

In other words, if we switch frames, momentum and energy chance. In the transforms, momentum plays thesame role as time, and energy plays the same role as position. Thus:

E′ = γE + vγp (17)

p′ = γp+ γv

c2E (18)

But wait! There’s more! Consider a particle to be stationary in the primed frame. There sits a particle withsome (unknown) energy,E = E0, and p = 0. Now, do a Lorentz transform into the unprimed frame with avelocity v << c. In this case, γ ' 1, and p′ = mv, since after all, we’re in the Newtonian limit.

Thus, according to equation 18:

mv = 0 +v

c2E0

or, rearranging:E0 = mc2 (19)

Even the most stoic among you must admit that this is freakin’ awesome!

But now that we know the rest energy, we can solve for momentum in general. Plugging in the energy toequation 18:

p′ = 0 + γv

c2mc2

= mvγ (20)

(21)

Exactly as we asserted in the first place!

Or, using equation 17, we find:E′ = γmc2 (22)

1.3.1 An example using an explosion

Okay, so perhaps you have a little trouble buying into my math tricks. Who could blame you? To demonstratehow all of this fits together, we’ll do a little example.

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Now, the important thing to remember here is that we physicists love conserved quantities.Momentum and Energy should be conserved no matter who is looking at the experiment. Ifthey’re not conserved, we’ve done something wrong.

This is what we call a nuclear bomb. We start with a big mass (at rest), and it explodes into two equal parts(each of mass, m).

Now, before getting into detail, the nice thing about this explosion is that it is symmetric. No matterwhether you use the relativistic or non-relativistic version of momentum, the explosion seems reasonablesince as much of the material flies to the left as to the right. Thus, momentum is conserved.

However, this is not necessarily true if we look at the same explosion in a frame moving to the left at speedv:

Now, if Newton came along and looked at the explosion from this point of view, he’d be baffled. The massof each of the pieces of debris should just be half that of the original, and thus, the total momentum beforethe explosion is:

p(Newton)i = Mv

while the total momentum afterwards is:

p(Newton)f =

M

2

2v

1 + v2/c2

which is not the same thing at all! In other words, our traditional interpretation of momentum does notwork.

We can get around this if we use the form of the momentum:

p = mvγ (23)

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Well, with one more complication. In Einstein’s view, the energy of a particle (kinetic+mass energy) is givenby:

E = mc2γ (24)

where γ is based on the velocity that the particle is moving in a particular frame. Unsuprisingly, a particlewhich is at rest according to one observer, and moving according to another, will appear to have a higherenergy according to the moving observer.

Because the mass energy is so big, it is a huge reservoir for kinetic energy, and thus is no longer conserved.How much mass is lost? Looking at it in the unprimed frame, the initial energy is:

Ei = Mc2

(since the bomb is initially at rest γ = 1).

The final energy is:Ef = 2(mc2γ)

where the factor of “2” comes from the fact that there are two pieces of debris and the mc2γ includes all ofthe energy (mass energy+kinetic energy). Thus, conservation of energy gives us:

2mc2γ = Mc2

2m

M=

1

γ(25)

(26)

In other words, if the explosion is very mild (small v), then γ will be close to 1, and it looks like mass willbe more or less conserved. However, if the pieces fly off at, say, v = c/2, then γ = 1.15, and thus, about13.4% of the mass will be converted into energy!

1.3.2 Putting some numbers in.

The unprimed frame:

Let’s do a concrete example to show that our approach does work. Consider a bomb which has an initialmass of 1kg. Its rest energy is thus:

Ei = 1kg × c2 = 9× 1016J

It then explodes into two equal masses, each with a speed of 0.8c, giving us γ = 1.67 (plug into the γ equationand verify for yourself!). Thus, according to equation 26, we have:

2m

1kg=

1

1.666= 0.6

In other words, each of the two particles will have a mass of 0.3kg. However, ∆M = 0.4kg were convertedinto energy.

Using this, energy (because that’s how we derived the mass loss relation) and momentum (because it’s zerobefore and after) are conserved in the rest (unprimed) frame.

The primed frame:

The primed frame is more complicated. Imagine that we’re running to the left at a speed 0.8c. Before theexplosion, the big mass is moving within that frame at a speed u′M = 0.8c (γ = 1.67). One of the pieces ofdebris is at rest, while the other is moving at a speed:

u′m =0.8c+ 0.8c

1 + 0.8c× 0.8c/c2= 0.976c

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which puts γ′ = 4.55.

So, before the explosion, the big mass has a momentum of:

p′M = mu′Mγ

= 1kg × (0.8× 3× 108m/s)× 1.67

= 4× 108kg m/s

Likewise, the energy is:

E′M = Mc2γ

= 1kg × (3× 108m/s)2 × 1.67

= 1.5× 1017J

Now, after the explosion, the stationary mass obviously has zero momentum, while the moving piece has amomentum of:

p′m = 0.3kg × (0.976× 3× 108m/s)× 4.55

= 4× 108kg m/s

And momentum is conserved! (As it must be)

Likewise, the energy afterwards from both the stationary and the moving pieces are:

E′fin = mc2 +mc2γ′

= 0.3kg × (3× 108m/s)2 + 0.3kg × (3× 108m/s)2 × 4.55

= 1.5× 1017J

Tada!

1.3.3 A real world example

You’ll notice that the number of Joules, above, tend to be enormous, and thus particle physicists tend towork in units of energy called MeV (mega-electronvolts), and units of mass of MeV/c2. It cuts down on theexponents.

So, consider the following relation (the basic nuclear fusion relation we’ve discussed earlier):

4H → +2e− + 4He+ 2ν + γ (27)

In this case, we find that the total masses on the left hand side of the equation are considerably less thanthe masses on the right.

mp = 938.3MeV/c2, mHe = 3728.1MeV/c2, and me = 0.511MeV/c2.

So, Ei = 3754.2, Ef = 3728.1, so, ∆E = 26.1MeV .

Where does it go? It goes into the kinetic energy of the resulting helium atom, and the energy of theneutrinos and photons.

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