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The Special Theory of Relativity

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Einstein published his Special Theory of Relativity in 1905. It ushered in a whole new way of thinking about space and time and led to some amazing and terrifying scientific developments. So what's so special about it?

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Relativity

Imagine you are cycling along at 15 miles per hour. You throw a ball forward at 30 mph. How fast is the ball travelling?

If you answered 45 mph, then well done! That is a good solid logical answer. But anything from 0 to 671,000,000 mph is still correct, and the best reply is "Relative to what?"

That's because all speed measurements are relative. Relative to the ground, the ball is going 45 mph. Relative to the bike, 30 mph. Relative to itself, it's not moving at all. Relative to Usain Bolt's fastest sprint, 18 mph. And relative to the Andromeda Galaxy, which is hurtling through space towards the Milky Way, approximately 250,000 mph.

In other words, there is absolutely no way of telling if you, the cyclist, are standing still and the ball is moving. Or if you are moving and the ball is standing still. Or if you are both moving, and it is the Earth which is standing still. There is no physics experiment or observation that can be made to tell there difference. And therefore there is no place anywhere in the Universe that is absolutely standing still.

The first postulate (statement) of relativity sums this up: the laws of physics are the same in all inertial frames of reference. "Inertial" just means "non-accelarating" or "moving at a constant speed". And a "frame of reference" is simply all the stuff that is moving together - you and your bike, or the ball, or Usain Bolt, or the whole Earth. Since the laws of physics are always the same, there's no way of telling the difference between frames.

The Speed of Light

The second postulate is even simpler: the speed of light is the same in all inertial frames of reference. The speed of light in a vacuum is always about 671 million mph, or 186,350 miles per second, or 299,792,458 meters per second. This vacuum does not refer to your hoover, but to the vacuum of space where there are no atoms for the light to bump into. In air and water and other materials light does slow down a bit.

This is a very weird statement. If you turn your headlight on while cycling, the light will shoot away at about 671 million mph. It will overtake the ball at the same speed. It will shine onto an oncoming tree at the same speed, and may eventually reach the Andromeda Galaxy at the same speed. Even if you are really really fast, cycling at 600 million mph relative to the ground, the tree will still measure the light at 671 million mph and not 671 + 600 million mph.

This speed was calculated by James Clerk Maxwell in the 1860s and verified by the famous Michelson-Morley experiment in 1887. It is not just the speed of light, but of all electro-magnetic waves, like microwaves, radio waves and Xrays. The speed of light is usually labelled as \(c\), and something travelling near the speed of light can be described as a fraction of c, such as \(\frac12c\) for half light speed.

Starting from only these two postulates, Albert Einstein turned physics on its head in the early 1900s. But first we'll go even further back.

Coordinates and Transformations

Isaac Newton laid the foundations for much of modern physics in 1687. He set his laws of motion and force against a back drop of absolute space and time. His laws still work very well for things travelling much less than light speed, such as cyclists and balls and trees. When physicists say that "the laws of physics are the same", they include the sorts of things Newton described: distance, speed, force, energy.

So imagine that you decide that everything in the Universe should be measured relative to you. you are at the beginning of space and time, and at the centre of your very own coordinate system. Any event in the Universe can be assigned coordinates \((x, y, z, t)\). So if a firework goes off 10 seconds from now, 8 meters to the right of you, 2 meters behind you and 40 meters above you, you can assign it the coordinates (8 meters, 2 meters, 40 meters, 10 seconds) or (8, 2, 40, 10).

And let's pretend your friend Jill is also a solipsist (thinking she is the centre of the Universe). She has her own coordinate system with coordinates \((x', y', z', t')\). The \('\) is pronounced "prime", so \(x'\) is x-prime. She starts her Universe at the same time as yours. She gives you a cuddle at (0, 0, 0, 0) and then starts walking slowly to the right at half a meter per second. After 1 second she is at \((x, y, z, t)\) = (0.5, 0, 0, 1) relative to you, and after 2 seconds (1, 0, 0, 2), etc. Relative to herself, she is always at the centre - after 1 second she is at \((x', y', z', t')\) = (0, 0, 0, 1).

We can swap between Your and Jill's coordinate systems with a Galilean transformation. This is a way of relating Jill's four coordinates to your four coordinates. In the equation below \(v\) is Jill's velocity relative to you, half a meter per second:

\(x' = x - vt\)

\(y' = y\)

\(z' = z\)

\(t' = t\)

So after 4 seconds Jill will be at \((x, y, z, t) = (2, 0, 0, 4)\) according to you. But from her point of view, she is still at the centre of her Universe. To transfrom this position from Your's to Jill's coordinates:

\((x', y', z', t') = (x - vt, y, z, t) = (2 - 0.5 * 4, 0, 0, 4) = (0, 0, 0, 4)\)

And from Jill's point of view, you will have transformed from \((x, y, z, t) = (0, 0, 0, 4)\) to:

\((x', y', z', t') = (x - vt, y, z, t) = (0 - 0.5 * 4, 0, 0, 4) = (-2, 0, 0, 4)\)

The distance between you and Jill is a law of physics, and should be the same in all coordinate systems, aka frames of references. We can see that it definitely is. The distance between (0,0,0,4) and (2,0,0,4) in your coordinates is 2. And in the Jill-centred Universe, the distance from (-2,0,0,4) to (0,0,0,4) is also 2. The laws of physics are preserved! The laws of physics do not vary during a Galilean transformation! Or more officially "the laws of physics are invariant under a Galilean tranformation"! Hurray! Let's celebrate with a couple graphs:

x meters The Universe according to You You Jill       x meters The Universe according to Jill You Jill

These graphs show You and Jill as fellow travellers through time. Time goes up from 0 to 4 seconds. The x position goes across. Each coloured arrow shows a person as they grow and mature between \(t=0\) and \(t=4\). For instance, in the second graph according to Jill, Jill stays stationary at \(x=0\) for the 4 seconds while you travel from \(x=0\) to \(x=-2\). Please take a moment to get familiar with this kind of graph, as there are a lot more of them coming up.

Messengers

That's all well and good. Let's complicate things. Jill is walking away from you at half a meter per second. You are very impatient and after 1 second at \(t=1\) you decide to send her a messenger to see how she is doing. Your messenger is a fairly brisk walker. His name is Gerald. He walks towards Jill at 1 meter per second. He arrives there a second later at \(t=2\), checks up on her very quickly and skips back at the same speed in time for tea at \(t=3\). Here is what that looks like according to you and Jill:

x meters The Universe according to You You Jill messenger       x meters The Universe according to Jill You Jill messenger

In Jill's Universe, you are moving to the left at 0.5 m/s (meters per second). At \(t=1\) you send Gerald her way. Relative to Jill, Gerald appears to be walking at only 0.5 m/s. On his way back, he appears to be hurrying away at 1.5 m/s. That is how it looks to Jill in her Universe. Gerald slowly catches up with her, and then quickly runs away.

Now imagine another scenario in which Jill is much much faster. We'll call her Super Jill. Instead of 0.5 m/s, she moves at half the speed of light 0.5c. In the graphs below, the vertical time axis is still in seconds, but the horizontal access is now in light-seconds rather than meters. A light-second is the distance light travels in one second. So Super Jill, travelling at half the speed of light, covers half a light-second every second. And instead of Gerald, you are now shining a beam of light on her. The light moves diagonally at one light second per second. The interesting thing is that light always moves diagonally in these graphs. In the situation above Gerald appeared to approach Jill at 0.5 m/s and leave at 1.5 m/s. Light can't do that, it always moves at light speed. I've also given the near-light-speed graphs a pinkish background, which has no basis in reality but makes the page look nicer. Notice the difference according to Super Jill:

x light-secs The Universe according to You You SuperJill light       x light-secs The Universe according to Jill You Super Jill light (not correct)

Because light insists on always arriving at light speed, Super Jill now receives the light messenger after 1.5 seconds instead of 2 seconds. Which seems very odd. Has time slowed down for her? Not really. This is one of the strange and wonderful properties of the Theory of Relativity. But even so the second graph is wrong. And not because of the elongated arrows (which will be explained later).

Time And Light

Benjamin Franklin was the first to claim that "time is money". That was long before Einstein. But in fact, time is more like light. If you look across the road, you're not seeing the other side as it is right now. you're seeing it as it was 0.00000002 seconds ago. Looking into the sky, you're seeing the sun as it was 8 minutes ago. At night in the northern hemisphere, you're seeing the brightest star Sirius as it was over 8 years ago. Unless it's cloudy. We know about the world through the light that enters our eyes. But that light takes time to get here. So on an astronomical scale at least, we measure time by light and the two concepts are thoroughly intertwined.

I could make a similar argument about sound, that the thunder which hits your ear drum may have happened several seconds ago. And even about smell. That the awful stink in your bathroom may have begun several minutes ago. Even if we always live in the moment, we sense the world as it was in the past. But sound and smell don't enter into the Theory of Relativity, so I'll drop this tangent.

Now think of yourself and Super Jill. If you are brushing your teeth at \(t=1\), Super Jill should see that at \(t=2\) according to the left graph above. If you rinse at \(t=10\), Super Jill sees it at \(t=20\). In other words, because of the time/light delay, all your actions appear twice as slow to Super Jill. How about if Super Jill brushes her teeth at \(t=2\)? According to the graph, you will witness that at \(t=3\). And if she rinses at \(t=20\), you'll see it at \(t=30\). In your view of the Universe, Super Jill appears to be acting 1.5 times slower than normal.

But this is not right! This means that the laws of physics, which definitely include brushing teeth, are not the same in all frames of reference! It could well be that Jill is loitering on Earth and Super You is the one moving away at half the speed of light. In which case, Jill would appear 2 times slower, and Super You would be only 1.5 times slower. Which situation is the real one? Who is really going slower? Surely speed is relative and you should both being seeing the same thing? This is terribly inconsistent, and is one of the fundamental problems that the Theory of Relativity resolves.

To restore order to the Universe and ensure the laws of physics work for everyone, Super Jill and Super You must witness the same rate of slowness. In this case, for people moving away from each other at half the speed of light, that rate of slowness is \(\sqrt3\) or about 1.732.

To explain, imagine Super Jill is holding a mirror. From your point of view, you send out a beam of light at \(t=1\) and it is reflected back at \(t=3\). The image of you rinsing at \(t=10\) comes back at \(t=30\). As you watch the mirror, you see yourself doing things 3 times slower. That combined delay is your delay multiplied by Super Jill's delay. The only number that will make you both see the same delay is the square root of it. So in fact, when you brush your teeth at \(t=1\), Super Jill sees it at \(t=1.732\) in her coordinate system. And when she sends the signal back, you receive it at \(t=1.732 * 1.732 = 3\). You both see a delay of \(\sqrt3\).

The corrected graph according to Super Jill is therefore:

x light-secs The Universe according to You You SuperJill light       x light-secs The Universe according to Jill You Super Jill light (correct)

The light beam reaches Super Jill at \(t=1.732\) seconds. The light still always moves diagonally, so it also leaves and arrives at different times (according to Super Jill). In the next sections, we'll learn how to calculate all this and more.

The k Factor

This delay factor is mysteriously known as \(k\), and this approach to explaining special relativity is known as the k-calculus, originated by Hermann Bondi in the 1960s. In the example above \(k=\sqrt3\). We'll now figure out how to calculate \(k\) for any velocity. This time velocity is given as a fraction of the speed of light. We'll call it \(\beta\) instead of \(v\), because that's the letter most physics textbook authors use. It is the second letter of the Greek alphabet: beta. In Super Jill's case \(\beta = 0.5\), half of light speed. Here's the same graph with a few additions:

x light-secs The Universe according to You You SuperJill R 1 t

You send out a light beam at \((x,t) = (0,1)\), it is reflected at the point R and arrives back at \((x,t) = (0,t)\). Because the light is reflected exactly halfway through its journey, point R's time coorindate is halfway between 1 and t or \(\frac12 (t + 1)\). Because the light travels diagonally, it travels the same across as up, and the x coordinate is half the distance from 1 to t or \(\frac12 (t - 1)\).

Super Jill's velocity is her x distance divided by the time. We can compute it at point R. And in the previous section we found that \(t\) was 3 seconds and the delay for each person \(k\) was \(\sqrt3\), so \(t = k^2\). Putting it together:

\(\beta = \frac{distance}{time} = \frac {\frac12 (t - 1)} {\frac12 (t + 1)} = \frac{t-1}{t+1} = \frac{k^2-1}{k^2+1}\)

Checking this equation for \(k=\sqrt3\):

\(\beta = \frac{k^2-1}{k^2+1} = \frac{\sqrt3^2-1}{\sqrt3^2+1} = \frac{3-1}{3+1} = \frac{2}{4} = 0.5\)

It works! With some algebra we can swap the equation around to derive \(k\) from \(\beta\):

\(\beta = \frac{t-1}{t+1}\)

\(\beta (t+1) = (t-1)\)

\(\beta t - t = - 1 - \beta\)

\(t(\beta-1) = - 1 - \beta\)

\(t = k^2 = \frac{-1-\beta}{\beta-1} = \frac{1+\beta}{1-\beta}\)

\(k = \sqrt{\frac{1+\beta}{1-\beta}}\)

Checking this for \(\beta = 0.5\):

\(k = \sqrt{\frac{1+\beta}{1-\beta}} = \sqrt{\frac{1+0.5}{1-0.5}} = \sqrt{\frac{1.5}{0.5}} = \sqrt{3}\)

Now we know the formulas for figuring out \(k\) from \(\beta\) and vica-versa.

Time Dilation

If you are watching someone like Super Jill fly away at half the speed of light, she will appear to slow down by \(\sqrt3\). If you brush your teeth at \((x,t)=(0,1)\), Super Jill will witness it at \((x',t')=(0,\sqrt3)\). Similarly, if Super Jill combs her hair at \((x',t')=(0,1)\), you'll see it at \((x,t)=(0,\sqrt3)\). That's how much she appears to slow down. But that appearance includes the time it takes light to travel between you both. Without that travelling time, how much does she actually slow down? We can compute this by referring to Super Jill's view again. (Note that in all of this you and Super Jill both experience time passing normally, it just looks slower.)

x light-secs The Universe according to Jill You Super Jill (0, k) E

You emit your light beam at point E (for event), after 1 second has passed for you at \((x_h,t_h)=(0,1)\). This is called the proper time or rest time. I've added the little h to remind you (well mainly me actually) that this is where the event actually happened. Looking at the graph, it looks like slightly more than 1 second has passed for Super Jill. To find out exactly how many seconds, we need to find out the coordinates of point E.

The light beam reaches Super Jill at at \((x_s',t_s')=(0,k)\). The little s reminds us that this is where the event was seen to happen. The coordinates of E are when the green line and bottom red light beam cross. The green line is your position through time, moving away from Super Jill at velocity \(\beta\):

\(x_s' = -\beta t_s'\)

The bottom light beam is moving upwards diagonally and crosses the \(x\) axis at \(t_s' = k\). Its equation is:

\(t_s' = x_s' + k\)

We just need to find out when these lines intersect. Using some algebra:

\(x_s' = -\beta t_s' = -\beta (x_s' + k) = -\beta (x_s' + \sqrt{\frac{1+\beta}{1-\beta}}) = -\beta x_s' -\beta \sqrt{\frac{1+\beta}{1-\beta}}\)

\(x_s' + \beta x_s' = -\beta \sqrt{\frac{1+\beta}{1-\beta}}\)

\(x_s' (1 + \beta) = -\beta \frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}\)

\(x_s' = -\beta \frac{\sqrt{1+\beta}}{(1+\beta)\sqrt{1-\beta}} = \frac{-\beta}{\sqrt{1+\beta}\sqrt{1-\beta}} = \frac{-\beta}{\sqrt{(1+\beta)(1-\beta)}} = \frac{-\beta}{\sqrt{1-\beta^2}}\)

And from there we can compte \(t_s'\):

\(t_s' = \frac{x_s'}{-\beta} = \frac{-\beta}{-\beta\sqrt{1-\beta^2}} = \frac1{\sqrt{1-\beta^2}}\)

For a velocity of 0.5, the coordinates of E are:

\((x_s', t_s') = (\frac{-\beta}{\sqrt{1-\beta^2}}, \frac1{\sqrt{1-\beta^2}}) = (\frac{-\frac12}{\sqrt{1-\frac14}}, \frac1{\sqrt{1-\frac14}}) = (\frac{-\frac12}{\sqrt{\frac34}}, \frac1{\sqrt{\frac34}}) = (\frac{-\frac12}{\frac{\sqrt3}2}, \frac1{\frac{\sqrt3}2}) = (\frac{-1}{\sqrt3}, \frac2{\sqrt3}) = (-0.577, 1.154) \)

Which looks about right for the graph. In fact, I know it's right, because I used those calculations to create the graph in the first place. But what does it mean?

First of all, let's call a spade a spade. The fraction above comes up a lot in special relativity. Physicists affectionately call it \(\gamma\), the Greek letter gamma:

\(\gamma = \frac1{\sqrt{1-\beta^2}}\)

Or the same thing more compactly using fractional exponents:

\(\gamma = {(1-\beta^2)}^{-\frac12}\)

To understand what it means, remember that \(t_h\) is the proper time measured on your clock when an event happens. \(t_s'\) is the time measured on Super Jill's clock when she sees that event. So if you drink a cocktail at \(t_h=1\), Super Jill sees it in slow motion at \(t_s'=1.154\). Your time has dilated or stretched by 1.154. And not just for when \(t_h=1\). It works for any \(t_h\). This can be expressed as:

\(t_s' = \gamma t_h\)

We can drop the subscripts and give the time dilation equation as it appears in some textbooks. Just remember that stuff happens in your \((x,t)\) coorindates and is seen in Super Jill's \((x',t')\):

\(t' = \gamma t\)

We can also express the inverse. An event seen by Super Jill at \(t'\) must have happened in Your coordinate system at \(t\):

\(t = \frac{t'}{\gamma}\)

In another variation, the time in your Universe, the proper time, has a subscript 0 and everything is written in capitals:

\(T = \gamma T_0\)

By the postulates of relativity, you and Super Jill can swap Universes and all the same laws should still be true. I find this very confusing. It means that \(t\) and \(t'\) can essentially swap places. And in some textbooks, they do swap, and without warning! And the reader is just supposed to accept it all and keep up. The confusion is that if we swap \(t\) and \(t'\) in the equations above then \(t = \gamma t'\), which is the exact opposite. What they don't mention is that when the \(t\) and \(t'\) swap, the happening and the seen swap too. So really \(t_s = \gamma t_h'\) which can be true at the same time as \(t_s' = \gamma t_h\). In fact we've already seen an example of that with the beam of light being delayed when it gets to Super Jill, and delayed again on the way back to you.

Time Dilation Another Way

Maybe the time delay from the k-factor feels a bit vague to you? We justified it just by saying the laws of physics have to work for both you and Super Jill. Here's a more concrete explanation. The graphs below show two spatial dimensions, \(x\) and \(y\), instead of time and \(x\). In your Universe is a clock which measures time by bouncing light back and forth between two mirrors:

Your clock according to You       Your clock according to Jill 1 βt' t'

The mirrors are 1 light second apart, so in your coordinate system it takes the light 2 seconds to get there and back. You could build this into a grandfather clock which measures time very accurately. Of course, it would be a massive clock. The box would need to be nearly 300,000 kilometers high. It would cost a lot in delivery. When you opened it, the little styrofoam balls used as packaging material would flood a large part of the Earth's surface.

Super Jill, who is whizzing by at half the speed of light (as indicated by the thick blue arrow) sees your clock according to the graph on the left. Both your and Super Jill's Universe's begin at the same moment. But by the time the light reaches the far mirror a second later, Super Jill is long gone. In her view the light has travelled at an angle, and so has travelled farther, and has taken more than 1 second. (The light beam is not diagonal here as this is not a graph of distance vs time.)

How long does the light take? We can use Phythagoras' theorem (more on him later) to calculate the time \(t'\). In these calculations the speed of light \(c = 1\):

y distance between the mirrors in both coordinate systems: \(1\ light\ second\)

x distance to first mirror in Super Jill's coordinates: \(distance = velocity * time = \beta t'\)

distance light travels to first mirror in Super Jill's coordinates: \(distance = velocity\ of\ light * time = 1 * t' = t'\)

By Pythagaros' theorem on the triangle with dashed lines: \(t'^2 = 1^2 + (\beta t')^2\)

\(t'^2 - \beta^2 t'^2 = 1\)

\(t'^2 (1 - \beta^2) = 1\)

\(t' = \frac1{1 - \beta ^ 2} = \gamma \)

But the mirrors don't have to be 1 light-second apart. They can be any distance from 1cm to 12 light years and more. The light will be delayed proportionally:

\(t' = \gamma t\)

And in fact, because this happens with a light clock, it happens with any clock. Time itself is dilated, just as above, but a bit less wishy-washy.

Human Speeds

These effects are significant at \(0.5c\) but what about normal human velocities: cars, bikes and people? A cyclist going 15mph or 6.7 m/s is travelling about 0.000000022 times the speed of light relative to the ground. To a person standing on the side of the road, the cyclist will appear a bit slower:

\(k = \sqrt{\frac{1+\beta}{1-\beta}} = \sqrt{\frac{1+0.000000022}{1-0.000000022}} = \sqrt{\frac{1.000000022}{0.999999978}} = \sqrt{1.000000044} = 1.000000022\)

But at such low speeds, the time experienced by the cyclist's watch won't be nearly as dramaticly dilated:

\(\gamma = \frac1{\sqrt{1-\beta^2}} = \frac1{\sqrt{1-0.000000022^2}} = \frac1{\sqrt{1-0.00000000000000048}} = \frac1{\sqrt{0.9999999999999952}} = \frac1{0.999999999999998} = 1.000000000000002 \)

My calculator did not accept that many decimal places, so I've sort of estimated some of those calculations. At this approximate rate the cyclist will lose about 1 second in every 500 trillion, or 1 second for every 30 million years of riding. That would be very good exercise, but very hard to detect or measure.

However in 1971 a physicist Hafele and an astronomer Keating sent cesium-beam atomic clocks on airplanes around the world. Very roughly, an airplane flies about 1000 km/hour or 0.00000093 times the speed of light, so the time delay is 1.00000000000086 or 1 second in every 1 trillion. The Earth is 40000 km around, so the flight would have taken about 40 hours and the flying clocks should have measured:

\(\frac{40\ hours}{1.00000000000086} = \frac{144000\ seconds}{1.00000000000086} = 143999.99999987\ seconds\)

This is a delay of about 0.00000013 seconds or 130 nanoseconds during the flight. The actual delay they measured depended on whether the plane flew with or against the Earth's rotation, and was from 60 to 270 nanoseconds, which broadly agrees with the quick esimate above. But not something you are likely to notice in your day to day life.

Twin Paradox

How about for speeds even faster than Super Jill? As \(\beta\) gets bigger, so does \(k\) and \(\gamma\). When Super Jill's velocity is 99% of the speed of light \(\beta = 0.99\):

\(k = \sqrt{\frac{1+0.99}{1-0.99}} = \sqrt{\frac{1.99}{0.01}} = \sqrt{199} = 14.1\)

\(\gamma = \frac1{\sqrt{1-\beta^2}} = \frac1{\sqrt{1-0.99^2}} = \frac1{\sqrt{1-0.9801}} = \frac1{\sqrt{0.0199}} = \frac1{0.141} = 7.1\)

An astronaut like Super Jill travelling at 99% of the speed of light experiences time passing as normal. To her family back on Earth, she seems to be moving 14.11 times slower. And her clock will measure time as passing 7.1 times slower. For every one year she ages, her mates back home will get about 7 years older.

What if she had a twin sister? Or a brother actually. The paradox doesn't specify if they are identical or fraternal twins. Since the laws of physics should be the same for everyone, shouldn't the twin also appear to be moving 14.11 times slower according to Super Jill? And shouldn't the twin also age seven times less? If Super Jill spends seven years flying to a distant star and back, who will be older when she returns? This conundrum is known as the twin paradox, and was one of the main criticisms of the Special Theory of Relativity in its first few decades.

The resolution is that the paradox breaks the postulates that special relativity relies on, because Super Jill is not in an inertial frame. For Super Jill to start on Earth and accelerate up to 99% of light speed requires an incredible amount of time and energy. Then she has to turn around and deccelerate on the way back. In fact, Super Jill's frame is incredibly non-inertial and she will be the younger one on return to Earth. It took another 10 years for Einstein to extend the Special Theory into the General Theory of Relativity and deal with accelerating frames of reference.

Light Speed Ahead

What happens at light speed, when \(\beta=1\)? Mathematically both \(k\) and \(\gamma\) will involve divisions by zero and therefore be infinite. This means that someone travelling at light speed will appear to be completely frozen, and so will their clock. But in theory, the traveller should experience time passing as normal. It's just the observers who notice the time dilations. How can this be?

Special Relativity's answer is that it is impossible to accelerate anything up to the speed of light. It would require infinite energy (we'll find out why later). Only light, which is already travelling at the speed of light, can travel at light speed. And for light, time is frozen.

Incidentally, there is nothing in special relativity which forbids something going faster than light speed. It's just that light speed is a barrier beyond which subluminal (slower than light) objects like humans can never pass. Superluminal (faster than light) particles may exist but have never been detected.

Lorentz Tranformations

A couple of sections ago we worked out how to transform the point \((x,t)=(0,1)\) from your coordinate system into Super Jill's. It ended up as:

\((x',t')=(\frac{-\beta}{\sqrt{1-\beta^2}}, \frac1{\sqrt{1-\beta^2}})\)

In this section, we'll figure out how to transform any point between the coordinate systems. These transformations were first formulated by Hendrik Lorentz in 1890s to explain unexpected experimental results. This was several years before Einstein announced Special Relativity. The tranformations fitted his theory perfectly.

First we'll define an arbitrary point in both your and Super Jill's coordinate systems. The light will be reflected from this point instead. Your point \((x,t)\) will transform to Super Jill's \((x', t')\):

x light-secs The Universe according to You time t+x (x, t) time t-x       x light-secs The Universe according to Jill time t'+x' (x', t') time t'-x'

I've also put in the times that the beam was emitted and received back in your Universe, and the times it passed Super Jill in her Universe. These times are easy to compute because light always travels diagonally so it covers the same distance up/down as left/right. So you sent the beam at your time \(t-x\). It reached Super Jill at her time \(t'-x'\). It was reflected and passed Super Jill on the way back at her time \(t'+x'\). It got back to you at your time \(t+x\).

Now according to the k-calculus, the beam first reached Super Jill after a delay of \(k\). In our example is was emitted at time 1 and reached her at time \(\sqrt3\). Therefore:

\(t'-x' = k(t-x)\)

Since the laws of physics are the same everywhere, the same logic also applies on the way back. There is also a delay of \(k\) from the time the beam passes Super Jill on the way back to when it is received by you:

\(t+x = k(t'+x')\)

We can solve these equations algebraically to get \(x'\) and \(t'\) in terms of \(x\) and \(t\):

\(t' = k(t-x) + x' = kt - kx + x'\)

\(t+x = k(kt - kx + x'+ x') = k^2t - k^2x + 2kx' \)

\(t+x - k^2t + k^2x = 2kx' \)

\(2kx' = t+x - k^2t + k^2x = t (1-k^2) + x(1+k^2)\)

\(x' (2\sqrt{\frac{1+\beta}{1-\beta}}) = t (1-\frac{1+\beta}{1-\beta}) + x(1+\frac{1+\beta}{1-\beta}) = t (\frac{1-\beta-1-\beta}{1-\beta}) + x (\frac{1-\beta+1+\beta}{1-\beta}) = t (\frac{-2\beta}{\beta-1}) + x (\frac{2}{\beta-1})\)

\(x' = \frac{-\beta t + x}{1-\beta} \sqrt{\frac{1-\beta}{1+\beta}} = \frac{x-\beta t}{\sqrt{1-\beta}\sqrt{1+\beta}} = \frac{x-\beta t}{\sqrt{1-\beta^2}} \)

We can do the same for \(t'\) and get:

\((x', t') = (\frac{x-\beta t}{\sqrt{1-\beta^2}}, \frac{t-\beta x}{\sqrt{1-\beta^2}})\)

Admire this transformation for a moment. It is one of the most important equations in the Special Theory of Relativity. And it should look familiar. The transformation of \(x\) is the same as the Galilean transformation times the time dilation factor \(\gamma\). The transformation of \(t\) looks similar as well, but unlike the Galilean transformation it also depends on the velocity and the \(x\) position. We can write it as:

\((x', t') = (\gamma (x-\beta t), \gamma (t-\beta x))\)

In fact \(\gamma\) is known as the Lorentz factor. We can check a couple of points. For half the speed of light:

\(\gamma = \frac1{\sqrt{1-\beta^2}} = \frac1{\sqrt{1-\frac12^2}} = \frac1{\sqrt{1-\frac14}} = \frac1{\sqrt{\frac34}} = \frac1{\frac{\sqrt3}2} = \frac2{\sqrt3} = 1.154 \)

I bet your are burning with curiosity by now, desparate to know why the green and blue arrows suddenly got longer when we switched to much faster speeds. The secret is about to be revealed. The point \((x,t) = (0,4)\) is where you end up after four seconds. In Super Jill's world view:

\((x', t') = (\gamma (0-\frac12*4), \gamma(4-\frac12*0)) = (\frac2{\sqrt3} * -2, \frac2{\sqrt3} * 4) = (\frac{-4}{\sqrt3}, \frac8{\sqrt3}) = (-2.308, 4.616)\)

Which is exactly where your green arrow ends in Super Jill's coordinate system. Because at the instant you tuck into your sandwich after 4 of your seconds, Super Jill measures that 4.616 have passed.

Inverse Lorentz Transformation

With a bit more algebra, we can also figure out the inverse Lorentz Transformations. We start with the normal Lorentz Transformation:

\(x' = \gamma (x - \beta t)\)

\(t' = \gamma (t - \beta x)\)

And then solve for for \(x\):

\(t = \frac{t'}{\gamma} + \beta x\)

\(x = \frac{x'}{\gamma} + \beta t = \frac{x'}{\gamma} + \beta (\frac{t'}{\gamma} + \beta x) = \frac{x'}{\gamma} + \frac{\beta t'}{\gamma} + \beta^2 x\)

\(x - x \beta^2 = x (1 - \beta ^ 2) = \frac {x' + \beta t'}{\gamma} = (x' + \beta t') \sqrt {1 - \beta^2}\)

\(x = \frac {(x' + \beta t') \sqrt {1 - \beta^2}} {1 - \beta ^ 2} = \frac {x' + \beta t'}{\sqrt {1 - \beta^2}} = \gamma (x' + \beta t')\)

So the inverse Lorentz Transformation has a similar form but it adds rather than subtracts the second term. We can compute \(t\) in the same way, and get:

\(x = \gamma (x' + \beta t')\)

\(t = \gamma (t' + \beta x')\)

Euclid And Pythagoras

Before 1905, we all thought we lived in a Euclidean Universe. Euclid was a Greek mathematician in the 200s BC and is known as the "father of geometry" because he laid down the rules of geometry so well. In Euclidean geometry the world and its objects have three dimensions \((x, y, z)\) or (width, depth, height) or (left/right, forward/back, up/down).

Two centuries earlier, Pythagoras came up with his famous theorem to work out the length of the sloping side of a right-angled triangle: \((a^2 + b^2 = c^2)\). This was extended to work out the distance to any point in three dimensions: \((x^2 + y^2 + z^2 = distance^2\)). Using this formula we can calculate the distance between you and the firework from above:

\(distance = \sqrt {x^2 + y^2 + z^2} = \sqrt {8^2 + 2^2 + 40^2} = \sqrt {1668} = 40.84\ meters\)

It is this distance which is "invariant under a Galilean transformation". In the example at the start, we did a very simple transformation, with Jill drifting to the right at a constant speed. It's possible for transformations to be much more complex involving displacements and rotations, but the distance between two points never changes.

In more mathematical notation, the difference between coordinates is denoted with a \(d\). So \(dx\) is the difference in x-coordinates, \(dy\) in y-coordinates and so on. To avoid overusage of the letter "d", many textbooks use \(ds\) for the overall distance. You can think of it as the separation of the two points. So to put it another way:

\(ds^2 = dx^2 + dy^2 + dz^2\)

We calculated the distance between you and Jill in both your and Jill's coordinate systems:

In your coordinates: \(ds^2 = dx^2 + dy^2 + dz^2 = (0-2)^2 + (0-0)^2 + (0-0)^2 = (-2)^2 = 4\)

In Jill's coordinates: \(ds'^2 = dx'^2 + dy'^2 + dz'^2 = (-2-0)^2 + (0-0)^2 + (0-0)^2 = (-2)^2 = 4\)

In both cases \(ds^2\) = the distance squared = 4. So the distance = 2 and is invariant. Now we'll examine this distance under a Lorentz transformation.

Invariants Under A Lorentz Transformation

Einstein showed us that we no longer live in a Euclidean world. The three spatial dimensions, \((x, y, z)\) upon which Newton's laws are built, are not alone. They are inextricably intertwined with a fourth dimension, time. And consequently, the centuries-old Galilean transformation has been superceded by the more intricate Lorentz transformation.

So is the distance \(ds^2 = dx^2 + dy^2 + dz^2\) also invariant under a Lorentz transformation? In case you haven't guessed it, the answer is an emphatic NO! To prove it, we will need to introduce a new person to your and Super Jill's Universes. Meet Super Jack. He clumsily involves himself in the group hug at the beginning of time, but escapes at half the speed of light in the other direction.

Super Jack's timeline is also elongated relative to you. According to you his path is the mirror image of Super Jill's and he ends up at:

\((x, t) = (\frac{-4}{\sqrt3}, \frac8{\sqrt3}) = (-2.308, 4.616)\)

Using a Lorentz transformation, we can transform Super Jack's ending position into Super Jill's Universe:

\((x', t') = (\gamma (x-\beta t), \gamma (t-\beta x)) = (\frac2{\sqrt3} * (\frac{-4}{\sqrt3} - \frac12 * \frac8{\sqrt3}), \frac2{\sqrt3} * (\frac8{\sqrt3} - \frac12 * \frac{-4}{\sqrt3})) = (\frac{-16}3, \frac{20}3) = (-5.333, 6.667) \)

Here's what that looks like graphically:

x light-secs The Universe according to You You SuperJill SuperJack       x light-secs The Universe according to Jill You Super Jill Super Jack

I bet you weren't expecting that! Out of the blue, the graphs have grown. From Super Jill's point of view, Super Jack is way out there. His clock is moving very slowly, taking 6.667 seconds (according to Super Jill) to tick a paltry 4 times. And clearly the distance between you and Super Jack, and between Super Jack and Super Jill, are not invariant. The distances are way bigger in Super Jill's coordinate system.

But we can fix the invariance, by subtracting the time:

In your coordinates: \(ds^2 = dx^2 + dy^2 + dz^2 - dt^2 = (-2.308 - 0)^2 + (0-0)^2 + (0-0)^2 - (4.616-4)^2 =\) \( 5.333 + 0 + 0 - 0.38 = 4.94\)

In Super Jill's coordinates: \(ds'^2 = dx'^2 + dy'^2 + dz'^2 - dt^2 = (-5.333 - -2.308)^2 + (0-0)^2 + (0-0)^2 - (6.667 - 4.616)^2 = \) \(9.15 + 0 + 0 - 4.21 = 4.94\)

Voila! The new 4 dimensional distance \(ds^2 = dx^2 + dy^2 + dz^2 - dt^2\) is invariant under a Lorentz transformation. And not just for the simple translation we have done here, but for all Lorentz transformations no matter how skewiff. It is vaguely intuitive as well. Because light takes time to travel, it makes things seem further away, so time has to be subtracted from the Euclidean distance \(dx^2 + dy^2 + dz^2\)to make the Universe work again.

We can use some algebra to check this result. Note that the \(y\) and \(z\) drop out because our transformation only involved \(x\) and time. I'll also drop the "d" as we are computing the distance to a point from the origin of the Universe, rather than the difference between two points. We can find:

\(s'^2 = x'^2 + y'^2 + z'^2 - t'^2 = (\frac{x-\beta t}{\sqrt{1-\beta^2}})^2 + 0^2 + 0^2 - (\frac{t-\beta x}{\sqrt{1-\beta^2}})^2 = \frac{(x-\beta t)^2 - (t-\beta x)^2}{1-\beta^2} =\) \(\frac{x^2-2\beta tx+\beta^2t^2 - t^2+2t\beta x - \beta^2x^2}{1-\beta^2} = \frac{x^2 - \beta^2x^2 - t^2 + \beta^2t^2}{1-\beta^2} = \frac{x^2 (1-\beta^2)- t^2 (1 - \beta^2)}{1-\beta^2} = x^2 - t^2 \)

Which is fancier way of saying that the distance is invariant under a Lorentz transformation.

Minkowski Spacetime

So if we are not living in a Euclidean world, what type of world is this? It's a Minkowski one, named after Hermann Minkowski who ellucidated the maths in 1908. The special property of Minkowski spacetime is the minus sign in front of the time coordinate above. It gives Minkowski spacetime the signature \((-, +, +, +)\), with the minus time coming first. Sometimes all the signs are reversed and the signature is \((+, -, -, -)\). The maths still works out the same, but many of the details are opposite.

Minkowski spacetime lends some support to the notion that time is imaginary. There is a number out there that doesn't really exist. It is \(\sqrt{-1}\), also known as \(i\). It doesn't exist because the square of any number is always positive. It is therefore possible to write the distance calculation above with only + signs, but it does unfortunately mean that time is imaginary, which some philosophers have suspected for eons:

\(ds^2 = dx^2 + dy^2 + dz^2 + (idt)^2\)

What does Minkowski spacetime look like graphically? I'm glad I asked. Below is another view of your and Super Jill's Universes, but instead of changing the arrows, I have skewed the whole coordinate system. I've essentially applied the Lorentz transformation to everything. In the first graph, notice that your green arrow is still on the time axis. It still goes from \((x, t) = (0, 0)\) up to \((x, t) = (0, 4)\). You experience 4 seconds - your green arrow crosses 4 time lines. And Super Jill, even though her arrow looks shorter, actually crosses more time lines up to 4.616 seconds. Super Jill's Universe is a mirror image.

The Universe according to You You Super Jill       The Universe according to Jill You SuperJill

The graphs above are only for when the velocity is half of light speed. At day-to-day speeds like 30mph, the coordinates would be horizontal and vertical, and Minkowski spacetime would be just like the Euclidean world which Newton and you and me are more used to. Look what happens as speeds get even faster:

You Velocity   =   0.75 times speed of light

The axes get more and more skewed and Super Super Jill's arrow gets longer and longer as her clock appears to slow down more and more. As the speed increases, the axes will continue to get closer and closer, and the arrows longer, and the clocks slower. This is another demonstration of how things get really weird near light speed and why light speed is an unachievable barrier for Earthbound objects like us. At light speed, the axes would meet, time would stop and the arrows would be infinitely long.

The properties of Minkowski spacetime become especially important when considering the General Theory of Relativity. The minus sign in front of the time has all sorts of implications.

Length Contraction

If you've studied Special Relativity before, you may have heard that time dilates and lengths contract. While writing this, I found this incredibly confusing. Because the Lorentz Transformations are very similar mathematically for time \(t\) and distance \(x\). When you send off a ray of light at \((x,t) = (0,1)\), it appears to Super Jill to have come from \((x',t') = ( -0.577, 1.154)\) rather than the \((-0.5,1)\) which would have been predicted by a slow Galilean Transformation. We used that very result to demonstrate time dilation. Doesn't that surely show that lengths expand as well as time?

Of course, it doesn't make sense that lenghts would expand. Because then, a ray of light travelling at \(\beta=1\) would be infinitely long. The Universe would be full of these tiny, infinitely long photons. Doing the crossword would be next to impossible, and forget about archery and 3D video games. The resolution to this conundrum is subtle and not very well explained in the textbook I have been using. It took a lot of scribbling to finally understand it. First, feast your eyes on this:

x light-secs The Universe according to You String A B C SuperJill       x light-secs The Universe according to Jill A B C String Super Jill

That big green rectangle is your very very long piece of string, extending through time. It is 299,782km long. It would stretch three-quarters of the way to the Moon, and over halfway around Jupiter. In fact it is one light-second long. In your Universe, it stretches from point A to point B in the graph above. Since the string and you are in the same inertial frame, at rest to each other, the string's rest length or proper length is:

\(l = x_B - x_A\)

Now look at the same string from Super Jill's point of view. The end of the string at point B is actually in the past! It is also further to the right, it's x-coordinate \(x_B'\) has in fact expanded, just as I thought it would above. But at time \(t'=0\), which is the instant Super Jill's Universe supposedly began, the string covers the dashed green line and the end of it is at point C, which is indeed contracted. So we need to figure out this length according to Super Jill:

\(l' = x_C' - x_A'\)

And now we can figure out the length contraction. We'll start with \(l\) and use the inverse Lorentz Transformation. We'll also use the fact that \(x_B = x_C\) and \(t_A' = t_C'\):

\(l = x_B - x_A = x_C - x_A = \gamma (x_C' + \beta t_C') - \gamma (x_A' + \beta t_A') = \gamma (x_C' + \beta t_C' - x_A' - \beta t_C') = \gamma (x_C' - x_A') = \gamma l'\)

The grand result is therefore that:

\(l' = \frac {l}{\gamma}\)

As with time dilation, the proper length is sometimes written in capitals with a subscript like \(L_0\). But unlike time dilation, we are dividing by \(\gamma\) so the apparent length is smaller than the proper length:

\(L = \frac {L_0}{\gamma}\)

So that is why Special Relativity involves time dilation but length contraction. According to the Lorentz Transformations, the lengths do actually expand, but they expand into the past, and by the time they get to the present, they have contracted. If you see what I mean. I think I do.

Adding Velocities

Super Jack's entrance into this discussion has raised another very interesting point. Both he and Super Jill are travelling away from you in opposite directions at half the speed of light. I know they've had some trouble in the past with a pail of water, but how does Super Jill see Super Jack now? What speed does he appear to be going from her point of view?

First let's look at the situation at normal speeds. In the graphs below Jack and Jill are both moseying away from you at a tranquil 0.5 meters/second. From Jill's point of view however, Jack is moving away at double that speed, 1 meter/second. Jack's speed is the 0.5 + 0.5 m/s added together.

x meters The Universe according to You Jack You Jill       meters The Universe according to Jill Jack You Jill

But at light speeds, \(0.5 + 0.5 \ne 1.0\). You can see that with the big Super Jack graph above. Super Jack's arrow is nearly diagonal but not quite. This is because mortals like Super Jack can never reach light speed, even relative to super fast people like Super Jill. We can use the k-calculus to calculate his exact speed relative to Super Jill. If we label Super Jack's k-factor relative to you as \(k_A\) and Super Jill's relative to you \(k_B\), then the combined k-factor, Super Jack's k-factor relative to Super Jill will be the two k-factors multiplied together, \(k_{AB} = k_A k_B\). We can use that and some algebra to compute the combined speed \(\beta_{AB}\):

\(\beta_{AB} = \frac{{k_{AB}}^2-1}{{k_{AB}}^2+1} = \frac{{k_A}^2 {k_B}^2 -1}{{k_A}^2 {k_B}^2+1} = \frac{ \frac{1+\beta_A}{1-\beta_A} \frac{1+\beta_B}{1-\beta_B} - 1}{ \frac{1+\beta_A}{1-\beta_A} \frac{1+\beta_B}{1-\beta_B} + 1} = \frac{ (1+\beta_A)(1+\beta_B) - (1-\beta_A)(1-\beta_B) }{ (1+\beta_A)(1+\beta_B) + (1-\beta_A)(1-\beta_B) } =\) \(\frac{ 1+\beta_A+\beta_B+\beta_A\beta_B - 1+\beta_A+\beta_B-\beta_A\beta_B }{ 1+\beta_A+\beta_B+\beta_A\beta_B + 1-\beta_A-\beta_B+\beta_A\beta_B } = \frac{ 2\beta_A+2\beta_B }{ 2+2\beta_A\beta_B } = \frac {\beta_A + \beta_B}{1 + \beta_A\beta_B} \)

We can use this to compute Super Jack's speed relative to Super Jill:

\(\beta = \frac {\beta_A + \beta_B}{1 + \beta_A\beta_B} = \frac {\beta_{Jack} + \beta_{Jill}}{1 + \beta_{Jack}\beta_{Jill}} = \frac {0.5 + 0.5} {1 + 0.5 * 0.5} = \frac1{1.25} = 0.8\)

Relative to Super Jill, Super Jack appears to be moving away at 0.8 times the speed of light. This corresponds perfectly with the big graph above. Every second Super Jack moves 0.8 light-seconds to the left.

More Mass

As velocities increase time dilates and length contracts. What happens to mass?

Mass is very similar to weight. It is the thing that gives you weight. But weight is not constant. If you weigh 60kg on Earth, you will weigh only 22kg on Mars because of the lesser gravity. You'll weigh 10kg on the Moon and 142kg on Jupiter! Your mass is the sum total of the mass of all the atoms in your body. It is your substance. It is about 6kg, approximately a tenth of your Earth-weight.

Mass increases as it gets faster. Not so coincidentally, it increases by the factor \(\gamma\). So a 1kg bag of sugar whizzing by at half the speed of light, will appear to you to weigh 1.154kg.

To compute this, consider two objects of equal mass colliding and combining. Imagine there is a 1kg blob of blue-tack hanging around with you at the centre of the Universe. Another 1kg blob of blue-tack comes flying in at 0.8 times the speed of light, whacks into it at this collosal speed, fully combines with it, and sends it hurtling off away from you at 0.5 times the speed of light. Super Jill sees it too, but according to her, two blobs of blue-tack come raging in at equal speeds from opposite directions, and smash together just over a light-second to her left. It looks like this:

x light-secs mass mcomb. at speed u mass mrest at speed 0 mass mfast at speed ufast The Universe according to You       x light-secs speed u The Universe according to Jill

From these diagrams, we can extract three useful equations. First of all, that velocity of 0.8 should sound familiar. It's how Super Jack looked to Super Jill. It's the relativistic sum of \(0.5c\) and \(0.5c\). Super Jill sees two equal blobs of blue-tack approaching at half the speed of light, just as you saw Super Jill and Super Jack zooming away from you. From your point of view however, the speeds are added together, and you see one blue blob at rest (\(m_{rest}\)) and the other (\(m_{fast}\)) coming in really really fast at \(0.8c\). After they collide, the combined über-blob of blue-tack (\(m_{combined}\)) is at rest in Super Jill's Universe and so moving away at speed \(\beta=0.5\) in yours. The speed of the fast one is therefore the addition of the two \(\beta\)s:

\(\beta_{fast} = \frac{2\beta}{1+\beta^2}\)

Isaac Newton had a productive life. He also explained the conversation of mass and momentum. In other words, the total mass and momentum must be the same before and after a collision like this one. The mass of blue-tack hanging about next to you is \(m_{rest}\). This is the rest mass, 1kg in this example. The mass of the blue-tack on the left is \(m_{fast}\). This is the one flying in at 0.8. Its rest mass is also 1kg, but we don't know what its mass appears to be when it's moving so fast. The two blue-tacks combine together and fly away with a mass of \(m_{combined}\). So looking at your Universe, the conversation of mass means that:

\(m_{rest} + m_{fast} = m_{combined}\)

An object's momentum is its mass times its velocity. It's the amount of ooomph it packs. Would you rather be hit in the nose by a bowling ball coming towards you at a snail's pace or a tennis ball at human throwing speed? A bowling ball weighs about 6kg and a snail slithers at up to 1 meter per hour, so the bowling ball will have momentum \(6kg * 0.0003m/s = 0.0018 kgm/s\). The tennis ball weighs about 60g and was thrown by our athletic cyclist at 30mph, so it has momentum of \(0.06kg * 15m/s = 0.9kgm/s\). The tennis ball has 500 times more oomph! If you ever have to make this difficult choice and you value your nose, choose the bowling ball. Again looking at your Universe, the conversation of momentum means that:

\(m_{rest} * 0 + m_{fast}\beta_{fast} = m_{combined}\beta\)

Now we want to work out the percieved mass of \(m_{fast}\) in relation to \(m_{rest}\) when it's flying in at speed \(\beta_{fast}\). Algebra to the rescue!

\(m_{fast}\beta_{fast} = m_{combined}\beta = (m_{rest} + m_{fast})\beta\)

\(m_{fast}(\beta_{fast} - \beta) = m_{rest}\beta\)

\(m_{fast} = m_{rest}\frac{\beta}{\beta_{fast}-\beta}\)

We just need to replace \(\beta\) with \(\beta_{fast}\). Unfortunately that is not so easy because we first need to solve the velocity adding equation in terms of \(\beta\):

\(\beta_{fast} = \frac{2\beta}{1+\beta^2}\)

\(\beta_{fast}(1+\beta^2) = 2\beta\)

\(\beta_{fast}\beta^2 - 2\beta + \beta_{fast} = 0\)

This is a quadratic equation and the solutions are given by the quadratic formula which you may remember from school algebra classes. The formula is:

\(x = \frac{-b \pm \sqrt{b^2 - 4 ac}}{2a}\)

In this case \(a = u_{fast}\), \(b = -2\) and \(c = u_{fast}\), so the solutions are:

\(\beta = \frac{2 \pm \sqrt{4 - 4 {\beta_{fast}}^2}}{2\beta_{fast}} = \frac{1 \pm \sqrt{1-{\beta_{fast}}^2}}{\beta_{fast}} \)

Since \(\beta\) cannot be greater than 1, the speed of light, only the negative solution is possible. We can plug this back into the mass equation above:

\(m_{fast} = m_{rest}\frac{\beta}{\beta_{fast}-\beta} = m_{rest}\frac{ \frac{1 - \sqrt{1-{\beta_{fast}}^2}}{\beta_{fast}}} {\beta_{fast}- \frac{1 - \sqrt{1-{\beta_{fast}}^2}}{\beta_{fast}}} = m_{rest} \frac{1 - \sqrt{1-{\beta_{fast}}^2}} {{\beta_{fast}}^2 - 1 + \sqrt{1-{\beta_{fast}}^2}} = m_{rest} \frac{1 - \sqrt{1-{\beta_{fast}}^2}} {\sqrt{1-{\beta_{fast}}^2} -(1 - {\beta_{fast}}^2)} = m_{rest} \frac{1 - \sqrt{1-{\beta_{fast}}^2}} {\sqrt{1-{\beta_{fast}}^2} (1 - \sqrt{1-{\beta_{fast}}^2})} = m_{rest} \frac1{\sqrt{1-{\beta_{fast}}^2}} = \gamma m_{rest}\)

Phew! That's the longest derivation yet. It is tempting to write this as \(m' = \gamma m\) but there is no \(m'\) because only \(t\) and \(x\) are the coordinates. Instead we'll use the convention of labelling the rest mass with a subscript 0:

\(M = \gamma M_0\)

Energy

This is all leading up to probably the most famous equation in all of physics. Unfortunately, it's quite difficult to derive so the approach here is an approximation of other approaches. First of all we'll start with another of Newton's gems, another top-ten physics equation: \(F=ma\). \(F\) is the force on an object. \(m\) is for mass and we've seen already. \(a\) stands for acceleration, which is a change in speed. The equation says that the force on an object equals its mass times its acceleration.

For example, the Earth's gravity accelerates objects at about \(a = 9.8 m/s^2\). So if a skydiver jumps from an airplane, her velocity will increase by 9.8m/s every second. So after 0 seconds, at the airplane door, she'll be going 0 m/s. After 1 second 9.8 m/s. After 2 seconds 19.6 m/s, etc. In reality, there will be a lot of air resistance and her acceleration and velocity will be a lot less. If she weighs 70kg, then the Earth is exterting a force \(F=ma = 70kg * 9.8m/s^2 = 686 kgm/s^2\), also known as 686 Newtons in honour of Isaac, or 686 N for short.

In physics, energy is a force exerted over a distance. And kinetic energy is the energy a moving object possesses. So if our skydiver falls for 10 meters, she possesses a kinetic energy of \(Fd = mad = 70kg * 9.8m/s^2 * 10m = 6860 kgm^2/s^2\). The unit of energy is a Joule, so she'll have a kinetic energy of 6860 J. For comparison, an adult woman needs to consume about 2000 kCals (kilo-calories) of energy per day. A calorie is 4.184 Joules, so a woman ought to have 8,000,000 Joules per day. Roughly the energy the skydiver would achieve after 1166 meters (and without air resistance). She would be going extremely fast by then.

How fast? Her velocity is her acceleration times the number of seconds she has been falling \(v = at\). The distance she covers is her average speed during the dive times the number of seconds she has been falling. Her average speed is half of her final speed, so \(d = v/2\ t\). Therefore:

\(KE = Fd = mad = m\frac{v}{t}\frac{vt}2 = \frac12mv^2\)

This is the standard Newtonian expression for kinetic expression. Our skydiver would be going about 500m/s to equal the Recommended Daily Allowance of energy for a woman. The \(v\) here is the actual velocity, not \(\beta\) which is a fraction of the speed of light. They are related by \(v = \beta c\) and \(\beta = \frac{v}{c}\). The equation above only applies when \(v\) is much smaller than \(c\), which is when \(\beta\) is a very small fraction.

Now I'm going to diverge a bit from the textbook I've been reading. Since energy is related to mass, and objects have a rest mass, let's assume they have a rest energy too which obeys a similar relationship as mass and time (which also means it would take an object infinite energy to reach the speed of light):

\(E = \gamma E_0\)

Super Jill is at rest in relation to herself. So she does not feel herself to have any velocity or kinetic energy. But you, watching her zoom by, do perceive a velocity and a kinetic energy. The total energy you perceive is her rest energy plus her kinetic energy:

\(E = E_0 + KE\)

Putting these two together:

\(\gamma E_0 = E_0 + \frac12 m v^2\)

\(E_0 (\gamma - 1) = \frac12 m v^2\)

\(E_0 = \frac12 \frac{mv^2}{\gamma - 1} \)

Now we'll use two mathematical approximations which work for very small numbers:

First approximation: \(\frac1{1-x} \approx \frac{1+x}1\)

For example, if \(x=0.0001\), then:

\(\frac1{0.9999} = 1.000100010001 \approx 1.0001\)

Second approximation: \(\sqrt{1+x} = \sqrt{1+\frac12 x + \frac12 x} \approx \sqrt{1+\frac12 x + \frac12 x + \frac14 x^2} = \sqrt {(1 + \frac12 x)^2} = 1 + \frac12 x\)

This basically creates and then ignores the \(\frac14 x^2\) term. For example:

\(\sqrt{1.0001} = 1.00004999 \approx 1 + \frac12 * 0.0001 = 1.00005\)

Using these two approximations, \(\gamma\) becomes:

\(\gamma = \frac1{\sqrt{1-\beta^2}} \approx \sqrt{1+\beta^2} \approx 1 + \frac12 \beta = 1 + \frac12 \frac{v^2}{c^2}\)

And plugging that into the equation above:

\(E_0 = \frac12 \frac{mv^2}{\gamma - 1} = \frac12 \frac{mv^2}{1 + \frac12 \frac{v^2}{c^2} - 1} = \frac12 \frac{mv^2}{\frac12 \frac{v^2}{c^2}} = mc^2 \)

So at low velocities, the rest energy of an object of mass \(m\) can be written as:

\(E = mc^2\)

Which is probably the most famous physics equation ever. At higher velocities approaching the speed of light, the approximations no longer hold, and the maths gets much more complicated, but the same equation pops out.

The Equations

We have worked out almost all of the equations above using \(\beta\) which is a fraction of light speed. We'll now relate all the equations back to real velocities because \(\beta = \frac{v}{c}\). So someone moving at 0.5 times the speed of light is actually travelling at 0.5 * 299,792,458m/s = 149,896,299m/s. And a cyclist riding 15mph is pedalling along at 15mph = 6.7ms = 6.7/299,792,458 = 0.000000022 times the speed of light. We can now express the equations in actual velocities with \(v\) rather than fractional ones with \(\beta\).

First of all, the letter which has helped us throughout this whole experience:

\(\beta = \frac{v}{c}\)

And now please welcome the k-factor!

\(k = \sqrt {\frac {1+\frac{v}{c}} {1-\frac{v}{c}}}\)

And the stretching and contracting Lorentz factor which seems to pop up everywhere!

\(\gamma = \frac1{\sqrt{1-\frac{v^2}{c^2}}}\)

The Lorentz Transformation for someone moving off along the x-axis!

\((x', t') = (\gamma (x-t\frac{v}{c}), \gamma (t-x\frac{v}{c}))\)

And who could forget its inverse!

\((x, t) = (\gamma (x'+t'\frac{v}{c}), \gamma (t'+x'\frac{v}{c}))\)

Time dilation!

\(T = \gamma T_0\)

And now for the one which gave me so much trouble. Let's have a round of applause for the counter-intuitive length contraction!

\(L = \frac {L_0} {\gamma} \)

The heavy one about mass increase!

\(M = \gamma M_0\)

When velocities are added together!

\(v_{AB} = \frac {v_A + v_B}{1 + \frac{v_Av_B}{c^2}} \)

Computing distances in Minkowski space-time. Aren't you proud you now know what that means?

\(ds^2=dx^2+dy^2+dz^2−dt^2\)

And finally, the queen of them all... energy!

\(E = mc^2\)

Back To The Beginning

Seeing the equations this way should give an even clearer idea of why the effects of relativity aren't noticeable at low speeds. Returning to the example at the very beginning, when a cyclist going 15mph throws a ball 30mph, the combined velocity relative to the ground is:

\(v_{AB} = \frac {v_A + v_B}{1 + \frac{v_Av_B}{c^2}} = \frac{15 + 30}{1 + \frac{15 * 30}{450241000000000000}} = \frac{45}{1.000000000000001} = 44.9999999999999mph \)

Not very noticeable. But very interesting! In my opinion at least.

Acknowledgements

I used several sources to understand special relativity myself: chapters 2-4 of a textbook Introducing Einstein's Relativity Ray D'Inverno and lots of quick online searches. Just before I finished this article, I found a thorough book called The Wonderful World of Relativity by Andrew Steane. It explains things from the ground up, somewhere between popular science and a textbook, and about the level I was going for in here. I hope you find this article helpful. Please contact me if you have any comments or questions.