# Theory of relativity/Special relativity/space, time, and the Lorentz transform

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Special relativity is a physical theory that states that the laws of physics are the same for all observers moving at a constant speed. This seemingly simple assertion has several non-intuitive consequences, as we will see next. It was developed by Albert Einstein, and others, in 1905.

## Frames of Reference

Suppose that a person is standing in a plane in the middle of a flight. Is he moving, or is he at rest? He is moving with respect to the Earth, and he is at rest with respect to the plane. Hence, to describe the motion, it is necessary to state a reference frame with respect to which the object or person is moving. In our example, both the Earth and the person are reference frames. Not all reference frames are created equal. If you put a ball on level ground, it will stay there. The same happens if you put the ball on an airplane moving at a constant speed. But if you do the same on an accelerating car, the ball will move even if nobody is exerting a force on it. An inertial reference frame is a reference frame in which the First Newton Law is true, that is, a reference frame in which an object will remain at rest if it is no subject to any force. It is easy to show that if S is an inertial reference frame, so is any frame S' that moves with a constant speed with respect to S:

Let frame S' move at a speed v with respect to S to the right. S' can be for example a plane moving at a speed v with respect to the Earth. Let u be the velocity of an object in S and u' the velocity of the object in S'. Then u = u' + v. This is easy to see since if a person is walking on a plane at a velocity u', then its velocity with respect to the Earth is u' plus v, the velocity of the plane itself. If we assume S' is inertial, then a free object (that is, no subject to any force) will move at a constant speed u', according to Newton first law. But since v is constant, then u = u' + v is also constant, so Newton first law is also true in S and therefore it is also inertial. As we will see later, the equation u = u' + v is not true in the framework of relativity, but the main conclusion stands.

Galileo recognized that the laws of classical mechanics are the same in all inertial reference frames, which means that it is impossible to determine whether we are in a reference frame "at rest", or in a reference frame moving at a constant speed. That is, there is no such thing as an absolute rest. It is important to remark that this notion of the absence of an absolute rest is a consequence of classical mechanics, and there is on need to introduce the theory of relativity to arrive at such conclusion.

## Principles of Special relativity

Maxwell's Equations however brought a difficulty. These equations describe the electric and magnetic fields, and combining them it is possible to derive a wave equation whose solutions are electromagnetic waves. Maxwell's equations predict that these waves will travel at a constant speed, which was quickly identified to be the speed of light. But, the speed of light in which frame? Remember that a person on a plane is either at rest with respect to the plane, or moving at 500 miles per hour with respect to the Earth, so when we say a speed, we need to be aware with respect to which frame is measured. But Maxwell's equations do not mention any specific frame, and using the Earth as a frame as it is done in everyday life does not make sense in this context. It was then postulated that there exist an absolute reference frame after all, which was called the aether (or ether), against which the speed of light is the one predicted by the Maxwell equations. That implies that since the Earth is moving with respect to the aether, the speed of light measured on Earth will depend on the relative speed between the Earth and the aether. In particular, the speed of light in a direction parallel to the Earth's motion should be different from the speed of light in a direction perpendicular to it. However, careful experiments by Albert Michelson and Edward Morley showed that there is no such a difference.

Albert Einstein postulated the two principles on which special relativity is based:

1. The laws of physics are the same in all inertial frames of reference.
2. The speed of light in vacuum is the same for all inertial frames of reference.

The first principle is just a re-affirmation of the one that Galileo had postulated 400 years before. The second explains the results of the Michelson-Morley experiment, although it is probable that Einstein postulated is based purely in theoretical considerations.

## How it works

Figure 1—Mike's experiment. A laser beam goes from event A to event B and back to event C, which coincides with A.
Figure 2—Leslie's view of the experiment. Mike's apparatus appears to move to the right, so the light must take a longer path.

Suppose that there is a laser beam sent from point A to a mirror B at a distance of 4 light-seconds (1,200,000 kilometers) and reflected back, as shown in Figure 1. Two observers, Mike and Leslie, are going to record the time it takes for the laser light to go from point A to mirror B and back again. Mike is at rest with respect to the apparatus, while Leslie moves at a speed v with respect to it, in a direction perpendicular to the light propagation. Calculating the time for Mike is easy: the time is simply the distance traveled divided by the velocity. This is 4 seconds each way, or 8 seconds total.

## Time Dilation

Now Leslie claims that she is at rest, since all inertial frames are equivalent, and that Mike's experimental apparatus is moving with respect to her. Leslie is moving to the left relative to Mike. She therefore thinks that Mike and his apparatus are moving to the right. She will see the light travel a diagonal path, as shown in figure 2. They both agree that the vertical distance is 4 light-seconds—the effects of relativity only show up in the direction of motion, not in perpendicular directions. So Leslie sees the light traveling a longer distance from her point of view. In classical physics, the light would have to be moving faster in order for her to see it travel the longer distance in the same 8 second period, and there would be nothing wrong with this. But it would violate the second principle of relativity. Leslie thinks that each diagonal segment was 5 light-seconds long, so she must think it took longer—10 seconds instead of 8—for the light to make the trip over and back. By Pythagoras' theorem, Mike's apparatus moved 3 light-seconds while the beam was traveling from A to B, and another 3 light-seconds during the return trip to C. Since it moved 3 light-seconds in 5 seconds, it was moving (to the right) at speed 0.6c. Mike of course thinks Leslie is moving to the left at speed 0.6c.

How do we explain this? Suppose Mike holds out his wrist-watch where Leslie can see it. He says "Look. The light started from point A at time 12:00:00, and arrived back at time 12:00:08." Leslie says "Yes, I can see that. But, according to my watch, the two events were separated by 10 seconds. Since the hands of your watch only advanced by 8 seconds, your watch is 20% slow."

Leslie can do her own version of the experiment, with Mike seeing her apparatus move to the left and the light taking a diagonal path, so he will conclude that Leslie's watch is 20% slow. The effects are completely symmetrical. Each of them thinks the other's watch is slow.

Neither of their watches is defective. Relativity does not cause clocks to malfunction and register something other than the correct time. Relativity causes time itself to be distorted between two observers in inertial frames moving relative to each other. If we crunch the numbers for the general problem, using Pythagoras' theorem, we see that each party observes the other as having a clock appearing to go at ${\displaystyle {\sqrt {1-v^{2}/c^{2}}}}$ as fast as their own clock. So each observer sees an action in the other frame take ${\displaystyle {\frac {1}{\sqrt {1-v^{2}/c^{2}}}}}$ times as long as it should. In the above example, ${\displaystyle v=0.6c\,}$, so the time dilation was ${\displaystyle 5/4\,}$.

The clocks don't malfunction—time itself dilates. This was Einstein's insight. Whereas others (Morley, Lorentz, etc.) postulated that something made clocks tick more slowly than "real" time, Einstein said that time itself stretched out.

## Beginnings of the Lorentz Transformation

Mike and Leslie start to work out the transformation equations. They agree to write Mike's coordinates with a prime, and Leslie's without a prime. They also agree that event A, the instant the light beam starts, is the origin of their space and time coordinate systems.

At event A:        ${\displaystyle x'=0\,}$      ${\displaystyle t'=0\,}$   (Mike's frame)
${\displaystyle x=0\,}$      ${\displaystyle t=0\,}$   (Leslie's frame)

The Galilean transformation would be:

${\displaystyle t=t'\,}$     (works both ways)
${\displaystyle x=x'+vt'\,}$            ${\displaystyle x'=x-vt\,}$

But they have to scale the passage of time to deal with the fact that Leslie observes event C at time t=10.

${\displaystyle t={\frac {t'}{\sqrt {1-v^{2}/c^{2}}}}\,}$ ???            (draft of equation 1)

This gives ${\displaystyle t=10\,}$ when ${\displaystyle t'=8\,}$, as required.

But the transformation formula has to be symmetrical between Mike and Leslie, since neither frame of reference is the "true" frame. So, since Mike and Leslie both think the other's watch is slow, the formula in the other direction has to look similar:

${\displaystyle t'={\frac {t}{\sqrt {1-v^{2}/c^{2}}}}\,}$ ???            (draft of equation 2)

These two formulas can't both be correct. Mike and Leslie can't compare their two wristwatches and come to opposite conclusions.

So they do the experiment again, and Mike notices something.

"When we supposedly looked at each other's watches at event C, and my watch showed 12:00:08 and yours showed 12:00:10, that wasn't your watch I was looking at. It was your assistant Kate's watch. You positioned Kate 6 light-seconds to your right, so that she would be there when the light beam arrived at event C. You and Kate don't know how to synchronize watches."
"Not so. Kate and I synchronized our watches perfectly."
 How to Synchronize Clocks at Different Locations in an Inertial Frame Since the notions of space and time are starting to get slippery, and the familiar notion of simultaneity is about to disappear, we need to be clear on the fact that within any one reference frame, simultaneity works correctly. It is possible to synchronize clocks within one frame. To synchronize clocks between X and Y, send a radio or light signal from X to Y and back again. Measure the delay and divide by 2. That is the light-speed delay between the points, say 10 seconds. Then, at 12:00:00 at point X, send a message to Y "It is now 12:00:00, and the delay is 10 seconds. When you receive this, set your clock to 12:00:10." As we will soon see, synchronization between points in different reference frames is not so simple.

So they realize that there must be some effect causing Mike to think that Leslie and Kate's watches aren't synchronized properly, even though they really are. It must be proportional to the separation distance—the farther Leslie and Kate are separated from each other, the farther off Mike thinks their synchronization is. It must also depend on the speed.

At event C:        ${\displaystyle x'=0\,}$      ${\displaystyle t'=8\,}$   (Mike's frame)
${\displaystyle x=6\,}$      ${\displaystyle t=10\,}$   (Leslie and Kate's frame)

The correct transformation, taking into account that Kate is at ${\displaystyle x=6\,}$ instead of ${\displaystyle x=0\,}$, has to be:

${\displaystyle t'={\frac {t-vx/c^{2}}{\sqrt {1-v^{2}/c^{2}}}}\,}$            (real equation 2)

Going the other way, equivalence of the two frames requires that the formula be the same except for the change in the sign of ${\displaystyle v\,}$:

${\displaystyle t={\frac {t'+vx'/c^{2}}{\sqrt {1-v^{2}/c^{2}}}}\,}$            (real equation 1)

(Since ${\displaystyle x'=0\,}$ at event C, this is still correct.)

## Length contraction

Figure 3—Layout of Mike's meter stick. It doesn't move (t'=anything.) Light starts event P at time t'=0, and arrives at event Q at time t'=10.
Figure 4—Leslie's view of the meter stick at t=0. Light is starting at event P.
Figure 5—Leslie's view of the meter stick at t=20. Light is arriving at event Q.

Now try another experiment. Mike shines a laser beam directly to the right, from point P (x' = 0) to point Q (x' = 10), a distance of 10 light-seconds. He has a 10 light-second (3,000,000 kilometers) long calibrated "meter stick", running from P to Q. Mike and Leslie measure a few things.

At event P:        ${\displaystyle x'=0\,}$      ${\displaystyle t'=0\,}$   (Mike's frame)
${\displaystyle x=0\,}$      ${\displaystyle t=0\,}$   (Leslie'sframe)
At event Q:        ${\displaystyle x'=10\,}$      ${\displaystyle t'=10\,}$   (Mike's frame)
${\displaystyle t=20\,}$   (Leslie's frame, using equation 1)
${\displaystyle x=20\,}$   (Leslie's frame, using the speed of light)

They don't yet have a formula for transforming ${\displaystyle x\,}$ values. But since they are tracking a light beam, and, at event Q, ${\displaystyle t=20\,}$, Leslie knows that ${\displaystyle x=20\,}$. The right end of Mike's meter stick was at event Q. Did the meter stick somehow grow to be 20 light-seconds long? Of course not. Since it, and Mike, were moving at a speed of ${\displaystyle 0.6c\,}$, the left end is now at ${\displaystyle x=12\,}$. And the right end is at ${\displaystyle x=20\,}$. Mike's meter stick has shrunk by 20%, to 8 light-seconds in length. At least, that's what Leslie observes. This is the famous length contraction phenomenon of relativity. Of course, if Leslie displayed a meter stick, Mike would think that it had shrunk.

Working out the algebra, the ratio is ${\displaystyle {\sqrt {1-v^{2}/c^{2}}}\,}$, the same as in the time transformation formulas. So the space transformation formula, instead of being

${\displaystyle x=x'+vt'\,}$            (Galilean)

becomes

${\displaystyle x={\frac {x'+vt'}{\sqrt {1-v^{2}/c^{2}}}}\,}$            (equation 3)

Going the other way, the formula has to be the same except for a change in the sign of ${\displaystyle v\,}$:

${\displaystyle x'={\frac {x-vt}{\sqrt {1-v^{2}/c^{2}}}}\,}$            (equation 4)

The formulas 1-4 that we have derived are satisfied by all of the data above.

They are the celebrated Lorentz transform. The reader can check that they are inverses of each other; that is, a conversion from (t', x') to (t, x) and back again gets the same numbers.

 The Lorentz transform ${\displaystyle t={\frac {t'+vx'/c^{2}}{\sqrt {1-v^{2}/c^{2}}}}\qquad \qquad t'={\frac {t-vx/c^{2}}{\sqrt {1-v^{2}/c^{2}}}}}$ ${\displaystyle x={\frac {x'+vt'}{\sqrt {1-v^{2}/c^{2}}}}\qquad \qquad x'={\frac {x-vt}{\sqrt {1-v^{2}/c^{2}}}}}$ This is needed only in the direction of motion. In other directions, there is no change. So, for completeness, we have: ${\displaystyle y=y'\qquad \qquad \qquad \qquad y'=y}$ ${\displaystyle z=z'\qquad \qquad \qquad \qquad z'=z}$ When it is clear what velocity is being referred to, it is common to define ${\displaystyle \gamma ={\frac {1}{\sqrt {1-v^{2}/c^{2}}}}}$ This is called the Lorentz factor. It allows the transform to be written more conveniently: ${\displaystyle t=\gamma \ (t'+vx'/c^{2})\qquad \qquad \quad t'=\gamma \ (t-vx/c^{2})}$ ${\displaystyle x=\gamma \ (x'+vt')\qquad \qquad \qquad x'=\gamma \ (x-vt)}$ ${\displaystyle y=y'\qquad \qquad \qquad \qquad \qquad y'=y}$ ${\displaystyle z=z'\qquad \qquad \qquad \qquad \qquad z'=z}$ Now the Lorentz transform is not the only physical transformation of coordinate systems that can be used. Any ordinary coordinate transformation that doesn't involve relative motion can also be applied—translation, rotation, or time shift. Also, any composition of legal transformations is legal. So, if a problem involves motion in some diagonal direction, apply a Euclidean rotation so that the motion is along the x axis, apply the Lorentz transform, and rotate back.

The thinking that led to the Lorentz transform was carried out by Hendrik Lorentz, George FitzGerald, Henri Poincare, and Joseph Larmor, among others, beginning in 1887, to make sense of the negative result of the Michelson-Morley experiment. Poincare noticed that the transformation constituted a mathematical group, and gave it its name. Einstein reformulated it in 1905 with his Special Theory of Relativity, and is credited with showing that relativity was not about how things behaved but about how space and time themselves behave.

An interesting application of the Lorentz transform is in the addition of two velocities. When two Lorentz transforms, one for speed v and one for speed w, are applied one after the other, to what speed does the resulting transform correspond?

Or, more simply, if someone in a vehicle traveling at speed v throws an object with (relative) speed w, what will that object's speed be relative to a "stationary" observer?

Let the vehicle's reference frame use primed coordinates. If the thrown object's speed is w, its position might be given as ${\displaystyle x'=wt'\,}$. Plugging that into the Lorentz transform, we get

${\displaystyle x={\frac {(w+v)t'}{\sqrt {1-v^{2}/c^{2}}}}\qquad \qquad t={\frac {(1+vw/c^{2})t'}{\sqrt {1-v^{2}/c^{2}}}}}$

The speed, as seen by a stationary observer, is x/t, or

${\displaystyle {\frac {w+v}{1+vw/c^{2}}}}$

This is the velocity addition law. One can see that, for speeds much less than the speed of light, this is close to the classical value of v+w. Also, if either speed is c, the result is c. This is just the principle of the constancy of the speed of light. One can also check that, if v and w are both less than c, the result is less than c.

Hint: Multiply out ${\displaystyle (1-w/c)(1-v/c)\,}$

The next article in this series is Special relativity/momentum.