Journey Through Time
Page 13
We therefore here apply the Lorentz transformation and Einstein’s theory of relativity.
In 1887, Albert Michelson and Edward W. Morley set up an experiment to test or detect ether to confirm that light moves with such speed due to ether. The philosophy behind the experiment was that light would have a higher speed if both the Earth and light moved in the same direction in relation to ether (other reference frame) or a lower speed if light moved in the opposite direction to Earth’s movement in relation to ether. The same principle applied to the train and ball-throwing in the train as explained in that example previously. The test confirmed that the speed of light remains the same and that ether does not exist. It confirmed that light travels at the same speed irrespective of whether the source of light is moving or the place from where it is measured is moving.
All the progress made in science and physics to that moment, and particularly tests confirming the constant speed of light, set the scene for Einstein to make a drastic change in how space and time is perceived. He developed his theory of special and general relativity.
The Special Theory of Relativity was published in 1905. This theory is basically related to changes taking place in different inertial reference frames in relation to each other. It does not involve acceleration. Therefore, it applies only to movement with constant velocity in a straight line as any change of direction (bending or circular movement) would imply acceleration and would not be an inertial reference frame. Obviously, standing or positioned at rest is also an inertial reference frame.
The General Theory of Relativity deals with non-inertial reference frames or with cases where acceleration is involved. It actually explains gravitation to be caused by curvature of a space-time made by matter, which has a large mass density. The larger mass density and pressure exerts the larger curvature of space-time.
Both theories are basically referring to time and space, which are not constant. Their quantities are different to two observers who do that observation from different frames of reference to each other. How different these quantities will be depends on how large the difference in speed is between two reference frames.
THE SPECIAL THEORY OF RELATIVITY
Einstein published The Special Theory of Relativity in 1905. His theory is based on two postulates. They are:
1.The laws of nature are the same in all inertial frames of reference
2.The speed of light in a vacuum is the same in all inertial frames of reference
Before my attempt to explain this theory in my way, I would like to start with one short story.
Alison was preparing to go to a business meeting at a business centre, which she needed to attend on that day at 2pm. She arrived at a luxury 5 star hotel near the business centre the day before. She was given instructions for the exact location of the business centre. She needed to go forward to the end of the high street (around 400 metres) and then turn right for another 200 metres where she would reach the business centre. She needed to go by lift up to the third floor where the meeting would be taking place at 2pm.
What Alison was given were coordinates of a precise location in a space, consisting of three dimensions. She needed to be there that day at 2pm, which is another dimension or time dimension. Four dimensions, three space dimensions and one time dimension therefore determine the location of the object. We would not be able to allocate Alison at the business centre at 11am, as she would not be there before 2pm.
Alison left her hotel at 1.30pm and walked at a steady speed, reaching the business centre at exactly 2pm. She was walking through three dimensions of space and dimensions of time with the constant velocity at a speed of 1200 metres per hour. She was able to pass 600 metres in half an hour, from her hotel to the business centre. The trajectory of her movement can be traced through time and space if we make a diagram. It is very difficult to draw a diagram of space outlining all three dimensions and particularly if we need to add another dimension: time dimension. The diagram can be simplified by using only one dimension of space, say only from left to right or right to left, and time dimension.
Picture 6.04
Left to right space dimension is outlined with a horizontal axis while time dimension is outlined with a vertical axis as shown in the illustration above.
We can make diagrams of movements in space and time of many different objects that move at different speeds. In other words, we can make and compare different reference frames in a time space diagram. However, in order to be able to do so, we have to have the same unit for time and space. We can do that by using light year, hour or second for distances which the light passes in a year, hour or second. We can do that because light has a constant speed and therefore there is an exact distance it will pass in a particular unit of time. In the diagram of space-time below, I used light seconds for measuring distance in space. Each light second actually represents a distance of 300 000 km which is the distance the light passes in a second. If we now plot the location the photon has after 1 second passing a distance of 1 light second and continue to do so, then we can draw a line through these locations the photon was at each second and the distances passed during this second. The drawn line is the trajectory of photon movement through space and time at the speed of light. This line is at 45 degrees in relation to space dimension and it is important as it represents the speed of light.
Picture 6.05
Using this kind of diagram, nothing can travel with a trajectory of less than 45 degrees as this would imply speed, which is faster than the speed of light (No. 1 diagram in illustration below). In this case, this object will pass larger distance for a shorter time than light does and this is not possible as there is nothing faster than the speed of light. This is therefore not allowed. Also, moving in the opposite direction (No. 3 in illustration) is not allowed as this would imply moving backwards in time or going to the past and that is not possible.
Picture 6.06
If we measure the speed and movement of other objects in such a time-space diagram, we have to use the light year as a space measure. As light is the fastest that anything can travel through space and time, then the speed of another object is or can be measured as a fraction of light speed. For example, if we said that an object travels at 1/10 of the speed of light that means that it takes 10 years for this object to pass the distance the light passes in 1 year. If an object moves at 1/5 of the speed of light it will takes 5 years for this object to pass the distance the light passes in 1 year. If the object travels at 3/5 of the speed of light it will take the object 5 years to pass the distance the light passes in 3 years.
I will now move to special relativity, which applies only to objects that are moving in inertial reference frames (they are moving either with constant velocity or are at rest).
Let’s imagine that we have two men, each encapsulated in a closed container with a glass window so that they can see each other (like in two spaceships facing each other). If these two containers are in a space so they cannot see stars, Earth or are not able to see any other objects apart from each other and both containers are in inertial reference frames then they would not be able to detect which one of them is moving (this obviously applies if one of them is in an inertial reference frame applying to a constant velocity). In such a scenario, neither of them will be able to know which one is moving or whether their movement is the result of the movement of both of them. This is the focus of the special theory of relativity from where we have also what is called the twin paradox.
Basically, we have the change of the quantity of the time and length for an event when those two men, who belong to two different reference frames in relation to each other, measure this event.
In my following example, I will try to explain it in a way that it is similarly explained in almost any book you read about it. Perhaps a slight difference will be that I will overemphasise bits, which make me confused when I read about it initially. In other words, I pa
y attention to those things which help me to make sense of or understand this topic better. I have used mathematics in this section but obsessively will try to explain calculations in each step supporting this by using numbers on some occasions to make it more understandable.
I will use an example which demonstrates how time slows down for a man travelling on a train as perceived by a man who is at rest and who watches a train passing.
A clock will be used as a beam of light which leaves the floor of the train and hits the ceiling of the train. We can imagine this to be a photon. The speed of light is constant. If the distance from the floor to the ceiling of the train is constant and unchangeable then the interval of time that passes from the moment the photon leaves the floor to the moment the photon reaches the ceiling is always the same. We have therefore constant rhythms with the same pace of time separating two events in time (the moment the photon leaves the floor to the moment it hits the ceiling). With the constant unchangeable rhythm we can measure time passing. If something produces a bang noise every 1minute and does not change its rhythm then after we hear a bang 10 times we know 10 minutes have passed.
To go back to our light and photon, we will have the same rhythm or time between the photon leaving the floor and hitting the ceiling as the time needed for the photon to bounce back from the ceiling and hit the floor from where it starts its journey. This time interval will always be the same for the man standing in the train.
Let’s imagine the impossible for the sake of understanding this topic. Let’s imagine that it takes 1 photon 1 second to travel from the floor to the ceiling in the train. Let’s imagine that another photon does not leave the floor until the previous one reaches the ceiling and such a pattern goes on. We will have a constant pace of time where after 1 second every successive photon will hit the ceiling 1 second apart. After 10 such events we would know that 10 seconds in time have passed. That is time which has passed for the man standing in the train and for the man watching time passing in that train from the ground. What is very important to clarify here is that this statement of the same time passing for the man on the train and perception of this time passing in the train for the man on the ground applies only if both the man on the train and the man on the ground are at rest (in the same reference frame) as in the illustration below.
Picture 6.07
If the train moves at a high speed then we have a completely different scenario. For the man who stands in the train there will be no change in the time needed for the photon to reach the ceiling from the moment it leaves the floor. The time passing or clock ticking represented by photons is not changed for the man on the train as it is in the same reference frame with the clock (in this case, photon) which is on board the train. However, for the man on the ground the clock ticking on the now moving train has slowed down.
Why does this happen?
The reason is that for the man on the ground at the moment when a photon leaves the floor, the train moves a bit forward as it is not at rest any more. By the time the photon reaches the ceiling, the train will have moved so much that light or photons will travel diagonally in order to reach the ceiling. This distance still remains the same for the man on the train and goes up in a straight light, as both the photon and the man are moving at the same speed. Time has not slowed down for the man on the train as he has still the same rhythm or time passed for the photon to travel from the floor to the ceiling. However, for the man on the ground the distance from the floor to the ceiling in the train is not any more an upwards straight line. It is now diagonal which is a longer distance.
Now if we can have a situation where light does not have a constant speed and can adjust to a different situation than in this diagonal distance, the light can speed up to reach the ceiling at the same time as it does when it passes a shorter distance in the straight upwards line. In this situation again, both men (on the train and the ground) will measure the same time. However, light has constant speed and cannot speed up or slow down. Because of that, it will now take the photon a longer time to reach the ceiling, as the distance from the floor to the ceiling is longer. It means that we will have slower rhythm of that photon clock for the man standing on the ground. Although time did not change for the man on the train, the man on the ground perceived an increased time needed for the photon to reach the ceiling, which means an increase of the interval between two ticks and therefore he perceived that time has slowed down for the man on the train. This is only perceived from the position or observation made by the man who is standing on the ground.
In the illustration below I have tried to make it clear. I did not draw the other man on the train in order to make the picture clear, but imagine that the other man is on the train. The man on the train will not have a change of time as rhythm will not slow down, as for him the distance from the floor to the ceiling will remain a straight upwards distance. For the man on the ground, the rhythm will slow down as light now has to travel a longer distance, which will require a longer time to reach the ceiling. This will slow down rhythm which means that time will slow down for the man on the train as perceived by the man on the ground.
Picture 6.08 The blue cross on the diagonal line roughly corresponds to the distance a photon travels in a straight upwards line. The straight upwards line (not diagonal) is perceived by the man on the train. (The man on the train is not drawn here for the sake of getting a clearer picture.) So when we look now at the diagonal line which the photon has to travel to reach the ceiling (as perceived by the man on the ground), the distance, which corresponds to the straight up line, is up to the blue cross. The rest of the diagonal line is the additional distance that the photon needs to travel to reach the ceiling as perceived by the man on the ground.
We now have a triangle with sides where we can use Pythagoras’ theorem to find out the size of the hypotenuse.
Picture 6.09
The hypotenuse is the distance from the floor to the ceiling or diagonal line the photon is travelling as perceived by the man on the ground. The initial distance from the floor to the ceiling perceived by the man on the train is a straight up line and can be marked as d. The diagonal line can be marked as d’. If we know the constant speed of a moving object then we can work out the distance if we multiply it with time:
d = v x t (velocity multiplied by time )
An example is that with a constant speed of 20 km/hour, in 2 hours we will pass 40 km.
Therefore, the distance the train will pass by the time a photon reaches the ceiling for the man standing on the ground is v times t, so we have defined all three sides of the triangle as d, d’ and v times t’.
The velocity of this train is not known but the velocity of the photon is, which is the speed of light. So d is equal to c times t, while c times t’ is equal to d’. So to find out d’ can be formulated as:
(ct’)2 = (ct)2 + (vt’)2
I forgot to mention that Pythagoras’ theorem states that squared on hypotenuse is equal to the sum of squares of the two other sides of the triangle, which is in this case:
d’2 = d2 + ( vt’)2 (look for this equation in the earlier illustration with the drawn triangle in it). Now we can go back to the last written equation, which can be written in such a way that we move ct squared from the right side to the left side of the equation:
(ct)2 = (ct’)2 – (vt’)2
The logic behind it can easily be seen if we use numbers as an example:
6 = 4 + 2 so if we move 4 from the right side to the left side we get the equation where:
4= 6 – 2
We can now write the above handwritten equation as:
c2t2 = c2t’2 – v2t’2
Again, it is mathematically correct and it can be checked by using numbers as an example:
(2 . 3)2 = (6)2 = 36
or
22 . 32 = 4 . 9 = 36
Now the written equation above can be expressed as:
c2t2 = t’2 (c2 – v2)
That is correct and I will not use numbers again to check it, but if you are willing to do so, please do. In the next operation we can divide the equation by c squared:
c2t2 = t’2 (c2 – v2) divided by c2 is
We can now remove square by applying squared root and in doing so, obtain the following equation:
The above equation shows the relation between time elapsed in one inertial reference frame t (the man standing on the ground) and the time t’ elapsed for the man standing in the train as perceived by the man standing on the ground. This is basically the Lorentz transformation applied to time difference as a result of a different speed between our reference frame and another reference frame when we measure the time in our reference frame and the time in the other reference frame. The equation shows that the higher the speed of the other reference frame, the slower the time is for that reference frame as perceived by us.
We can see that in the situation where v reaches the speed of light, then t’ is equal zero and this makes perfect sense.
We remember our girl on the train moving at the speed of light and looking in the mirror. Photons from her face move at the speed of light for her which is 300 000 km/sec. These photons move at the same speed for the man standing on the ground. As she herself moves at the speed of light (she is on a train moving at such a speed) then the only way the man on the ground can see her photons moving at that speed is if time has stopped for the photons leaving her face. In other words, if time for these photons is zero as well as the distance from the mirror and her face. In that case, the photons from her face move in zero time and pass zero distance for the man on the ground but are perceived as moving at the speed of light because her train is moving at that speed. Mathematically, if v is equal c we have: