Introduction.

A lot of people are interested in understanding special relativity and how is it derived. Most of

the time, the offers are the ‘complicated’ Lorentz transform, or some very specific experiments

(light bouncing of a mirror in a rocket etc.), and some even make weird mistakes or contradict

themselves. After reading a wonderful book written by David Bohm about special relativity, in

that book there were a few chapters that approached special relativity in a geometric way,

namely the Minkowski diagram/space which I find very enlightening and interesting, therefore

I decided to share it around. Note that this is targeted to mainly high school students but anyway

is welcomed to give me comments. (P.S.:This is very long and I hope that you can bear with

me, I would be very glad if you read this to the end!)

So what is this theory of relativity about?

The special theory of relativity is about arguing that time and space are not absolute and

therefore not the same for all observers in different frames. To illustrate this concept of frames,

suppose your friend is on a train and you are observing outside. You might seem to be moving

from your friend, but from your point of view, your friend is moving. Now both of you see

each other moving, but see yourself at rest. This is because that you two are observing from

different inertial frames, so you cannot say your friend is moving, but relatively he or she is

moving from your view. This is pretty much relativity is about.

Special relativity vs. General relativity? So when we start off relativity, we first make two

postulates (assumptions). Physics is not something that can be proved, every theory as to start

off with certain assumptions, and these assumptions cannot really be absolutely proved. So

Einstein postulated that, first, the laws of physics are the same wherever you are, regardless of

your frame you observe things. Well this seems very obvious, but who knows if these laws act

differently? You really cannot prove this although it is just so obvious. Secondly, the velocity

of light in vacuum is constant in all inertial frames. But some people just think why must it be

the speed of light? Why not others like the speed of sound in a certain matter? Just because that

it seems to be the fastest thing? No. So again we cannot really prove the speed of light being a

constant, but to verify them. Experiments have shown light to have a constant speed, which

that is one reason. Apart from that, when we say light, we actually mean electromagnetic

waves, and electrodynamics has been described by Maxwell’s equations. In those equations,

there are 2 very important constants, which can be loosely described as a measure of resistance

of electric and magnetic fields in vacuum. Surprisingly, light’s speed just turns out to be a term

that is exactly a combination of these two constants, so people just take light’s velocity as this

constant which is absolute. So what makes special relativity special? Special relativity is just a

special case of relativity, loosely defined that it is assumed zero acceleration in it, which means

everything has a constant speed, so there is no gravity or other stuff. Therefore it is not very

practical with the existence of gravitational forces. But it is possible to investigate constant

acceleration in special relativity which we actually do not cover. Natural forces such as gravity

may not have a constant force therefore not necessarily constant acceleration, and we then need

general relativity to describe it. Anyway it is a very good concept to understand, and

unfortunately general relativity is just out of the scope in this article. Enough of these concepts

and stuff, let us move on.

Get to the basics and start working!

So in high school, you probably learn things like if a triangle has two equal angles, it has two

equal sides or maybe say Pythagoras’ theorem. But now we develop a special kind of geometry

that not everything you learnt can be used, for example in this Minkowski diagram, Pythagoras’

theorem is "false". Furthermore, one may see two segments that look identical but have

different lengths, and two segments that actually have the same distance but look very different.

Why is this so? Well high school geometry as we learn it was developed by Euclid, hence

named Euclidean geometry. He also needed to start with certain postulates, such as a unique

straight line can be drawn through two points and all right angles are congruent etc. But now,

this is a new demonstration of geometry, which we will see symmetry is a big theme. The first

thing is, for simplification, we deal with one time dimension and one space dimension, where

the space dimensions can be extended to three which is similar (which is named Minkowski

spacetime), but in this case we have a special case of special relativity, dealing with one

dimension.

So Figure 1 just shows an observer being the dot, and is not moving. As he is stuck in this one

dimension, he can only move along the line. here represents the one-dimensional space. So we

now want to create a diagram that can represent space and time for all events. This observer

then stays at the same space but goes forward along time. We can have a vertical time axis, but

as space (displacement) does not have the same units with time, we dilate (scale) the time axis

by the speed of light, and make both axes be having units of length, although one represents

space and the other is time. We choose as it is a universal constant for everyone, and just also

for simplicity because will have to deal with light. So we have a vertical axis ct , with t being

time, and we just denote it as 𝜏(𝜏 = 𝑐𝑡). Keep in mind that although being multiplied by a

constant, it still represents time.

And now from Figure 2, we can see the stationary observer then ‘moves upwards’ along the

vertical axis, which we can think as continuous points representing continuous events that have

the same space but not the same time. With this, we now need to bring light into the diagram,

as it is the invariant in relativity. If the observer shoots two rays of lights directed to his left

and right, how should we represent that? Well as for light, by the formula distance equals time

times speed, we can see that ∆𝑥 = 𝑐∆𝑡 = ∆𝜏 ( ∆ means change in). Therefore the rays can be

represented as the two straight lines that form = |𝑥| . Moreover the light rays are then the angle

bisectors of the two axes as in Figure 3. Again these lines are continuous points that represent

continuous events, which these events are the travels of the light.

And that is it for this time, I will be posting the next part of the series soon and eventually add

all the links to all parts here. I hoped you did read this rather lengthy thing until here, as it is

very important to get the ideas and concepts right. In the next part, we will get some motion

coming, and hopefully have some nice arguments. Although the first part is sort of boring, but

I am sure in the future this would get interesting!

You will need that to understand that part's concept, so it is encouraged to read the previous

part first.

The "moving" observer. So in the previous part, I believe you have a good grasp of what the

axes in that Minkowski diagram represented and how it works. But now, we are introducing

more of that! Now consider a "travelling observer" that passes through O is moving at a

constant velocity v relative to the stationary observer. Of course the travelling observer will

observe that the stationary observer is moving at a constant velocity –v (the opposite direction)

and seeing he himself staying still (but we still call them the stationary, and the moving

observer to make it clearer). As we had seen the first observer’s line of events is the vertical

axis, but as this observer travels through space, his line of events will be different as seen from

the stationary observer. Let his line of events be ′ . We already know that this will be a straight

line passing through O since he is moving at a constant velocity, but we need to know the angle

it makes with the vertical axis. This is not hard to find out as one already knows the gradient

of the line is

∆𝜏

∆𝑥=

𝑐∆𝑡

∆𝑥=

𝑐

𝑣

since ∆𝑐

∆𝑡= 𝑣 . Therefore to find the equation for 𝜏′ is simple, which is

𝜏′ =𝑐𝑥

𝑣

Therefore it makes an angle of 𝜃 = arctan𝑣

𝑐 with the vertical axis (notice the fraction flip as

we are considering the vertical axis).

So as in Figure 4 we see that 𝜏′, being the line of events for our moving observer, and therefore

time, as compared between the two observers seems ‘distorted’. This suggests that space as

seen by the moving observer, represented by the line ′ , might be similarly 'distorted' too. In

fact, what does the horizontal axis x actually mean of the stationary observer? If two points are

connected by a line that is parallel to that, those two events represented are simultaneous. Also

obviously it is a measure of space too. So as we take the space for the moving observer as a

line, we can see that due to symmetry and that both observers have the same set of laws as

postulated, the light-line must be the "angle bisector" of 𝜏′ and 𝑥′. And therefore we have the

second line 𝑥′ as in Figure 5.

So it seems like the stationary observer’s lines are ‘correct’ and the moving observer’s lines

are being really distorted, like how people say about curving space-time, and you can actually

imagine these two being similar. But anyway, is the moving observer than ‘wrong’? We then

need to keep in mind that, everything is relative.

Remember to step in other people's shoes! The answer is, in relativity, no one is absolutely

correct, from the moving observer’s point of view, he will see the stationary observer relatively

moving at a velocity of -v , while seeing himself at rest. Therefore he sees his axes of time and

space as if they are horizontal and vertical, and since the velocity of the stationary observer to

him is the opposite of the velocity of the moving observer to the stationary observer, 𝜏 in Figure

6 will be the reflection 𝜏′ of in Figure 5. As for x, since the green line would be the "angle

bisector" of x and , try to figure out where the line should be.

Observe that the angle between 𝑥′ to x and also 𝜏 𝑡𝑜 𝜏′ are both clockwise directions in both

diagrams, and in the same magnitude. It is then clear to us the symmetry that is preserved

between what is observed by the two observers from Figure 5 and Figure 6. The theme of

symmetry would reoccur , and this actually concludes part 2! I hope enjoyed this exploring,

and in the next part, we will be showing an idea of nothing can travel faster than the speed of

light. Comments and any suggestions are very welcome!

Simultaneity.

With this time and space being different for different inertial frames, it is interesting to

investigate whether two events can be simultaneous, that is, two events being at the same time.

So will the two observers agree a pair of events to be simultaneous for both of them?

Apparently, most of the time they will not. But what does it mean by simultaneous? So if two

events are simultaneous, they are in the same time, that is, parallel to the space-dimension line.

See Figure 7 as an example.

As the line AE is parallel to 𝑥′ (in fact they concur), the two events situtaed O and E have

zero time difference for the moving observer, who takes them as simultaneous. But for the

stationary observer, the line OE it is not parallel to x . So to find the difference of time, we

draw a line parallel to his time-line, 𝜏 , intersect it with the line parallel to his space-line, x that

passes through O, that is, x itself, and mark it as EE’ . The time lapse between the two events

is therefore 𝐸𝐸′̅̅ ̅̅ ̅

𝐶 for the stationary observer.

Suppose E' itself is also an event. It is clear that O and E’ are simultaneous to the stationary

observer, but similarly since line OE’ is not parallel to x’ , they are not simultaneous to the

moving observer.

Using the same method, we find the distance between E’ and the line passing through O that is

parallel to x’ which is x’ itself. Note that we do not draw a line perpendicular to x’ as we will

do it in Euclidean geometry to find the distance between a point and a line 𝜏′, but draw the line

parallel to . The time lapse is then 𝐸′𝐸′′̅̅ ̅̅ ̅̅ ̅̅

𝑐 as in Figure 8.

With this we see that two events cannot be both simultaneous to the stationary observer and

the moving observer, unless the two events happen at the exact same time and space.

Faster than light?

Having understanding simultaneity, we actually have covered a lot of concepts in relativity.

We now demonstrate a contradiction if one assumes that there exist some communication that

travels faster than the speed of light, which for the sake of simplification, we assume this

communication has infinite velocity, which is equivalent to moving through space instantly.

The reason we can make this assumption is that we can first just observe the above figure. Note

that light is represented throughout as green lines. Anything that travels slower than light is

between 𝜏 and the green line, such as the moving observer represented by 𝜏′ , and anything

faster should be between x and the green line, with something travelling at infinite speed having

a line that is parallel to the space-line.

With this in mind we know introduce even more observers that join the party. Let A be the

stationary observer and B the moving observer that we have discussed until now.C is another

stationary observer at point P, D is a moving observer that moves at a velocity v relative to C,

which is the same B velocity is relative to A , but passes through point Q instead, as shown in

Figure 9.

We can see that A and C would see two events connected by a line that is parallel to x being

simultaneous, whereas for B and D , two events connected by a line that is parallel to x’ being

simultaneous. Now suppose at a particular instant A and B are at O,C and D are at Q . C first

transfers a signal to D without any time as they are at the same point, and D transfers it to B

simultaneously as in our assumption, B passes it to A at the very same time instant, and then

since the laws of physics are same in all frames as postulated, A has the ability to pass the

signal simultaneously to P. This is equivalent to D, being at Q, has contacted himself in the

past at P. Similarly, D at P can contact himself in the future at Q .

Now suppose D is at P , receiving the signal from himself at Q does something that makes

himself at Q impossible to send the signal, we then reach a contradiction. Indeed, this also

applies to any transferring that is faster than light, as the light-line is the angle bisector of the

two space and time lines. The details are left to the reader. Therefore nothing can be travel

faster than the speed of light, or if you are not that convinced (or if you believe that tachyons

exist), travelling faster than light does result in travelling backwards in time.

And that is it for this time, which is again one of the shorter notes in the series. In the next part

we will be investigating even more about time in special relativity, where the maths starts to

really kick in, so, keep an eye on the series! Tell me what you think of this.

You should have a good understanding of the Minkowski diagram now, and we now attempt

the so called time dilation.

Strange geometry.

This is the point where we will use the diagram to its full extent that is within our knowledge

to derive some important results of special relativity. We return back to the case where there

are only the first two observers, but now they first agree that for both of them, the time

coordinate for O is 0. Refer to Figure 10.

Now suppose that the stationary observer sends a ray of light from point E to the moving

observer, who receives it at point F , and at the same instant, sends another ray of light from

point F to point G’. We use light again as the velocity of light is constant to both observers.

Now draw a line parallel to x that passes through F is drawn, which intersects 𝜏 at H . As in

Figure 3 we saw that light travels the same "distance" in x and , we can have 𝐸𝐻̅̅ ̅̅ = 𝐹𝐻̅̅ ̅̅ = 𝐺𝐻̅̅ ̅̅

. Note that these equations are derived not by geometry but by the invariance of the speed of

light. For simplicity we name some of these lengths, 𝑂𝐸̅̅ ̅̅ = 𝑇, 𝑂𝐹̅̅ ̅̅ = 𝑡, 𝑂𝐺̅̅ ̅̅ = 𝑇′, 𝑂𝐻̅̅ ̅̅ = 𝑡′ .

So if we want to know how time differs between frames, we then have to consider 𝑡′

𝑡 . An

average high school student would say by Pythagoras’ theorem we have

But unfortunately we did not assume Euclidean geometry here, so unfortunately we cannot

just simply use Pytaghoras’ theorem, although the answer is very close, which we will see soon,

it is a matter of changing that plus sign to a minus sign (for those of you who are eager to

know).

Now we consider something ‘easier’ first, that is to find 𝑡

𝑇. I say that this is easier because due

to the first postulate, the rules in all inertial frames are the same, thus we have

where is k just a shorthand of the ratio (now we can see that the Minkowski diagram does not

follow Euclidean geometry, since if we assume so, applying the Power of a Point Theorem on

𝑡2 = 𝑡𝑇 ↔ 𝑂𝐹2 = 𝑂𝐸. 𝑂𝐺 , which implies that OF is parallel to OE , due to OF being a

tangent of the circle EFG at point F , and this is obviously false).

We know that 𝑇′ = 𝑘2𝑇 , and it follows that

Note that by this we also have

And we get another expression for 𝑂𝐻̅̅ ̅̅ , which is

And we want to solve for this positive k is therefore possible, by equating both expressions

for 𝑂𝐻̅̅ ̅̅ and T the on both sides cancel out each other. We then have

And now after obtaining a formula for k , we then can find the original ratio we wanted to find,

namely 𝑡′

𝑡 . We have

(I may have skip some steps but it should be easy to follow). And there you have it, the time

dilation equation in special relativity, by a geometric diagram and some algebraic

manipulation. To sum it up, if the stationary observer observes a time lapse of t’ , for the

moving observer who moves at a velocity of v relative to the stationary observer, if he observes

a time lapse of t , it is actually related by the equation 𝑡 = 𝑡′√1 − (𝑣2

𝑐2) . Again I would like

to say that although in the diagram t’ appears to be longer than t, it is not, because this is really

some strange geometry. Anyway one more thing to note that is that if I put the velocity v>c ,

the expression under the square root will be negative which somehow “disallows” velocities

faster than light.

The two observers get on a rocket for the party!

Well actually there is not really a party, sorry. But we do have a rocket now.

So after time dilation, we look into the effects in space, that is, length contraction. Suppose C

and D are on a purple rocket travelling at the velocity v relative to the stationary observer. They

are on the endpoints of the rocket. Note that they are both on the same line x’ because for them

they are simultaneously on the rocket together, separated by a certain distance. This is because

C and D see each other at rest. Suppose at a particular instant C is at O. Let D shoot two rays

of light directed to his left and right at the same instant, but at a different place.

It should not take you long to realize Figure 10b is exactly Figure 10 from the previous post

reflected across the light-line (the ‘b’ in Figure 10b should have reminded you this). So the

moving observer observes that the rocket is at rest relative to him, while the stationary observer

sees it moving at a velocity of v. The length of the rocket to them would be l’ = OD and l =

OD’ to the moving observer and the stationary observer respectively. One can then relate OD

to OF in Figure 10, OD’ to OH in Figure 10.

Essentially, we are just explaining that

Where l’ is the length of the rocket observed on the rocket, and l is the length of the rocket

observed by the stationary observer. For simplicity, from now onwards we use the function

𝛾(𝑣) to represent

To summarize, if a time of t is observed in a moving frame of velocity v relative to a second

frame, the time t’ observed in that second frame would be 𝑡′ = 𝛾(𝑣)𝑡 . If a length is l’ observed

in the former frame, the length l that appears to the second frame would be 𝑙 =𝑙′

𝛾(𝑣).

Adding is not merely adding.

So we have done time and space. Let us combine them! What do you get when you combine

time and space? Yes, velocity, and we will see what relativity wants to do with velocity.

If a ball is rolling at a velocity u relative to a moving vehicle which is moving at a velocity v

relative to you observing outside the vehicle, does the ball roll at a velocity u+v relative to you?

This is a natural response, but what if the ball is replaced with light? Do you see the light

travelling at c+v? Not quite, since it must be c . So you cannot just add velocities like that. We

get back Figure 9 and introduce another moving observer who moves at a velocity of w relative

to the stationary observer at I , and a velocity of u to the observer who is moving at a velocity

of v relative to the stationary observer. Our aim is to link u and v to get w, and we use the

following diagram.

Now suppose that at time T the stationary observer shoots a ray of light and reaches the

observer who travels on OF at F at time t , and then reaches the observer who is travelling at

velocity relative to the stationary observer at I at time S . Declare the function K(v)

We then have 𝑆 = 𝑇. 𝐾(𝑤) , and also we have 𝑡 = 𝑇𝐾(𝑣), 𝑆 = 𝑡. 𝐾(𝑢) = 𝑇. 𝐾(𝑢). 𝐾(𝑣) .

This then gives us 𝐾(𝑤) = 𝐾(𝑢). 𝐾(𝑣) . Rearranging the algebra gives

And this is therefore called the formula for addition of velocities. Notice that if u and v are

much smaller than c , w is just approximately u+v . The same goes for time dilation and length

contraction. If we set u = c, we get 𝑤 =𝑐+𝑣

1+𝑣

𝑐

= 𝑐. This coincides with our assumption that the

velocity of light is invariant. Now as we can see if we increase u or v ,w will also increase. So

assume both u and v are at most c , then w is at most c too. And now we can show that nothing

can accelerate from a velocity slower than light to a velocity faster than light, as it has to

undergo continuous addition of velocities that are at most c , and therefore it cannot exceed c .

Warning: Math ahead!

And yes I do mean a bunch of math ahead, and also a bit of calculus, since we are going to

prove the so called most famous physics equation, = 𝑚𝑐2 . Firstly let us discuss some stuff

about momentum.The measure of momentum can be interpreted as how difficult is it to make

an object rest. Obviously mass and velocity will contribute to this, and indeed it is defined that

momentum is the product of mass and velocity. In a closed system with no external effects,

momentum is conserved. To show this we will need Newton’s laws, namely the second and

third law The second law gives the definition of force, =𝑑𝑝

𝑑𝑡 , where p is momentum and t is

time. The third law says that for every force there is a reaction force in the opposite direction

with the same magnitude. Therefore assume we have two particles in a closed system

interacting with each other. Let the first particle have momentum 𝑝1 and experiences a force

𝐹1 , the second with momentum 𝑝2 and force 𝐹2. By the third law, which means that the sum

of momentum does not change. Therefore momentum is conserved in a closed system.

Now suppose these two particles have velocity 𝑣1and 𝑣2 , with zero total momentum. We write

the sum of their mass as = 𝑚1 + 𝑚2 , sum of their momentum as 𝑃 = 𝑝1 + 𝑝2 = 0. This leads

to

Suppose in a different frame that sees the system relatively moving at a velocity of V, we

observed a different set of 𝑚1, 𝑚2, 𝑝1, 𝑝2, 𝑣1, 𝑣2, 𝑀, 𝑃 𝑎𝑠 𝑚1′ , 𝑚2

′ , 𝑝2′ , 𝑝1

′ , 𝑣1′ , 𝑣2

′ , 𝑀′, 𝑃′ instead.

We then obtain

After a slight arranging, we have

And also by the addition of velocities, we get

Now we substitute this back and use Componendo et Dividendo to simplify,

From here, we would now like to express the term 1+

𝑉𝑣1𝑐2

1+𝑉𝑣2𝑐2

into something that can be associated

with . Note that

Similarly we can obtain this kind of result for 𝑣2′ , and therefore

Note that this relation is independent of V . Next we introduce the factor R, where

If we consider the special case 𝑣1 = 0 , which means that the particle is at rest, at we let that

particle have its rest mass (mass of object when it is at rest) 𝑚0 . Since (0) = 1 ,

If we consider another special case, which is 𝑣1′ = 0 , since now both frames are not moving

to each other yet R must be a constant, we have 𝑚1′ = 𝑚0 which implies that R = 1and we

conclude that the relativistic mass of an object, m , is

for an object with rest mass 𝑚0, moving at a velocity v of with respect to the reference frame.

One may multiply this by v and transform this formula of relativistic mass to relativistic

momentum, that is,

We are now ready to tackle the equation 𝐸 = 𝑚𝑐2 Recall that the change in kinetic energy is

mechanical work, and work is,

Note that we use the product rule in the last equation. Note that in Newtonian mechanics where

m is taken as a constant p=mv, we would have obtain instead

which is the formula for kinetic energy in classical physics. Now observe that

by the chain rule. Note that since 𝑣2

𝑐2 = 1 −1

(𝛾(𝑣))2, we can have

This means that

or just

Take v=0 and thus obtain the well-known equation,

Of course in the general case we can also have 𝑚 = 𝑚0𝛾(𝑣) to get

which is the full form of the equation. Note that in many cases we also write it as

or even

So most of the time when people say ‘E is equal to m c squared’, it actually means an object

has energy due to its mass even at rest, that is, its mass at rest multiplied by 𝑐2. This gives

equivalence between mass and energy, creating the conservation and interchanging of mass

and energy. Furthermore mass is then not how much matter an object contains, but a measure

of its energy content. So if you add mass to an object, it has more potential to release more

energy. Similarly, if you add energy to an object, no matter kinetic, heat, sound or any other

form, then you actually be adding its mass! And again we see that if something is travelling

faster than the speed light, we again need to take the square root of negative number, and its

energy is going to be something very nasty.

And we are done! Now you might expect this as the finale, but actually I still have one last

long post that would be coming up. So I hoped you have enjoyed this post and thanks again for

reading!