Early Universe 3. The Doppler Effect and the Special Theory of Relativity

Published on May 12, 2018

Early Universe 3. The Doppler Effect and the Special Theory of Relativity

    On the website of free lectures MIT OpenCourseWare posted a course of lectures on cosmology by Alan Gus, one of the creators of the inflationary model of the universe.

    Your attention is invited to the translation of the third lecture: "The Doppler effect and the special theory of relativity."


    Non-relativistic Doppler shift

    At the end of the last lecture, we began to discuss the Doppler shift and introduced notation. It was a case when the observer is motionless, and the source moves with speed$ v $. We considered sound waves that had a fixed speed relative to a certain medium.


    The speed of the wave relative to the medium is denoted $ u $, $ v $ means the speed of removal of the source, as shown in the figure. $ Δt_s $ - the time interval between the wave crests emitted by the source, that is, the period of the wave at the source. $ Δt_o $denotes the period of the wave at the observer. We need to calculate the relationship between$ Δt_o $ and $ Δt_s $.

    The figure shows the different stages of this process. At the first stage, the source moves to the right and emits the first crest of the wave. So far, nothing particularly interesting.

    In the second stage, the source emits a second wave crest. But during this time the source has moved, this movement is highlighted in yellow. The time between the emission of the wave crests is$ Δt_s $. Therefore, the distance that the source will pass during this time is equal to$ vΔt_s $. Call this distance$ Δl $.
    This is a really important step, he explains the Doppler shift. It can be seen that the second crest of the wave must pass slightly more than the first crest, by the amount$ Δl $.
    The third stage - the wave has passed the distance between the observer and the source. At this stage, the first ridge has just come to the observer. The fourth stage - the second ridge came to the observer.

    To understand what the Doppler shift is equal to, it must be noted that if both objects were stationary, there would be no difference in the period of the wave between the observer and the source. Each wave crest would hit the observer with some delay equal to the time during which the sound wave travels from the source to the observer. But, in the absence of movement, this delay is the same for each ridge. So, if the source is not moving$ Δt_o $ = $ Δt_s $.
    But due to the movement of the source, the second ridge will have to go a greater distance by$ Δl $. The difference between the periods will be equal to the time it takes the wave to travel this distance.

    $ Δt_o = Δt_s + \ frac {Δl} u $


    We know what it is $ Δl $. $ Δl $ - it's simple $ vΔt_s $. Substituting in our equation we get:

    $ Δt_o = Δt_s + \ frac {vΔt_s} u $


    This equation shows the relationship between $ Δt_o $ and $ Δt_s $. Can find relationship$ Δt_o $ and $ Δt_s $.

    $ \ frac {Δt_o} {Δt_s} = 1+ \ frac vu $


    This ratio is also the ratio of the wavelength of the observer. $ λ_o $ and at the source $ λ_s $because the wavelength is simply equal to the wave speed multiplied by its period $ Δt $.
    There is a standard definition for describing the Doppler or redshift.

    display
    \ frac {λ_о} {λ_s} = 1 + z
    display


    $ z $called doppler or redshift. Astronomers subtract a unit from the ratio of wavelengths, so that when both objects are stationary,$ z $ it turned out to be equal to 0. Such a case corresponds to the absence of a redshift and means that the wavelength is the same for the source and for the observer.

    display
    \ frac {λ_o} {λ_s} = \ frac {Δt_o} {Δt_s} = 1+ \ frac vu = 1 + z
    display


    Thus, we get the redshift for non-relativistic motion, or sound wave, in the case when the source moves:

    $ z = \ frac vu $


    We now turn to another simple case, when the observer moves, and the source is motionless. The source is still on the right and the observer is on the left. But this time the observer moves with speed$ v $. In both cases$ v $ - this is the relative speed between the source and the observer.


    The first stage is quite simple again. The source radiates the first wave crest. Stage number two — the second crest of the wave is emitted by the source. Stage number three - the first crest of the wave reaches the observer. Stage number four - the second crest of the wave reaches the observer.

    Between the time when the first ridge comes to the observer, and the time when the second ridge comes to the observer, that is, the time between the third and fourth stages the observer has moved. He moved an equal distance$ v $multiplied by the time between these stages. The time between these stages is just the time that passes between the receiving of two crests by the observer. This is what we have identified.$ Δt_o $- wave period measured by the observer. The distance traveled is simple.$ vΔt_o $. Everything you need to get an answer happens inside the yellow rectangle at the last stage.

    You can write equations for this case. This time everything is a little more complicated. Let's start with the same idea.$ Δt_o $ would be equal $ Δt_s $if there was no movement. But$ Δt_o $getting a little longer due to the extra distance the second ridge goes. This additional distance is called again.$ Δl $. The delay time will be again$ Δl $ divided by $ u $wave speed.

    But this time we have another formula for$ Δl $. This time$ Δl $ equals $ vΔt_o $, but not $ vΔt_s $as in the previous case.

    display
    Δt_o = Δt_s + \ frac {Δl} u = Δt_s + \ frac {vΔt_o} u
    display


    The equation gets a little harder because $ Δt_o $appears on both sides of the equation. However, this equation with one unknown, is easily found$ Δt_o $. After simple algebraic transformations we get:

    $ \ frac {Δt_o} {Δt_s} = (1- \ frac vu) ^ {- 1} $


    Subtracting one we get the final equation for $ z $, again for the non-relativistic case, when the observer moves:

    $ z = \ frac {Δt_o} {Δt_s} -1 = (1- \ frac vu) ^ {- 1} - 1 = \ frac {v / u} {1-v / u} $


    It is worth noting that when speed $ v $ small compared to the wave speed, which is often the case if we consider a light wave, but also occurs in the case of sound propagation, then both formulas for $ z $almost the same. They are both proportional$ v / u $, if a $ v / u $few. The only difference is the denominator.

    In the second case, we have a denominator$ 1-v / u $. In the first case$ z $ just equal $ v / u $and there is no denominator. If a$ v / u $is small, the denominator in the second case is close to 1. Thus, the two formulas will be almost the same. You can describe it a little more accurately by calculating the difference between z in both cases. Having done simple calculations we get:

    $ z_ {source \ _moving} - z_ {observer \ _moving} = \ frac {(v / u) ^ 2} {1- \ frac vu} $


    The formula clearly shows that the difference between $ z $ proportional to $ (v / u) ^ 2 $and not just $ v / u $. If a$ v / u $equal to one thousandth, the difference will be one millionth. Therefore, for slow speeds, it does not matter whether the source moves or the observer moves. But the answers, of course, will be very different if the speed$ v $ comparable to $ u $.
    STUDENT: Does this not violate Galileo’s principle of relativity?

    TEACHER: Not really. For our calculations, the air in which the sound wave moves is critical. In both cases, the air is resting relative to the figure. If you make the transformation of Galileo from one picture to another, then after the transformation, the air will move, and you will not get the same picture.

    Therefore, everything is consistent with the Galilean theory of relativity. It must be remembered that the air here plays a crucial role. When we say that an observer or source is at rest, in reality it means that he is at rest in relation to the medium in which the wave moves.

    STUDENT: I ​​noticed that if$ v $ more $ u $, in the first case, the answer is always positive, everything is in order. But if$ v $ more $ u $in the second case, it turns out a negative answer. It seems strange to me.

    TEACHER: Yes, if$ v $ more $ u $then in case of movement of the observer the answer becomes negative. This means that the wave will never reach the observer. If the observer moves faster than the speed of the wave, the wave will never catch up with him. Therefore, it turns out such an unusual answer. If the source moves faster than the wave speed, the wave still reaches the observer. Therefore, in the first case, we get the correct answer.

    Relativistic time dilation
    Let's now move on to the relativistic case. We will need some facts from the theory of relativity. Since there are specialized courses in the theory of relativity, I do not want our lectures to become such a course. However, I want our course to be fully understood by people who have not passed the theory of relativity. Knowledge of the special theory of relativity is not a prerequisite for our course. Therefore, my goal will be to tell you enough about the special theory of relativity so that you can understand further. I will not display the results, their output can be found in other courses. If you do not want to visit them, then it's okay too. But I want my course to be coherent.

    So, we will consider the consequences of the special theory of relativity, not trying to connect them directly with the fundamental ideas of the special theory of relativity. However, I remind you where the special theory of relativity came from. It originated in Albert Einstein's head when he considered the Galilean theory of relativity, which was asked a minute ago. The Galilean theory of relativity says that if you look at any physical process in a frame of reference that moves at a uniform speed relative to another frame of reference, then in both report systems the laws of physics must be described in the same way.

    Galilean relativity theory played a very important role in the history of physics. The key question in the time of Galileo was whether the earth moved around the sun or the sun around the earth. Galileo took an active part in this dispute. One of the arguments proving that the Sun should move around the Earth, and not vice versa, was such that if the Earth moves around the Sun, then this means that we move with the Earth at a very high speed. The speed of the earth around the sun is high by ordinary standards. People at that time believed that obviously such a movement should be felt. This was proof that the earth is still and the sun is moving. Because, otherwise, the effect of the rapid movement of the Earth would be felt.

    For Galileo’s point of view that it is the Earth that is moving, it is critically important that we do not notice such a movement. If we move evenly, then the laws of physics remain exactly the same as they would be if we were left alone. This is the essence of Galileo’s theory of relativity. She was very clearly outlined by Galileo in his writings.

    All this was true for mechanical phenomena. However, in the 1860s, Maxwell derived his equations. Rather, he concluded their conclusion, most of these equations already existed. It followed from Maxwell’s equations that light must move at a fixed speed, which can be expressed in terms of the electric and magnetic constants$ ε_0 $ and $ µ_0 $. We denote this speed.$ c $. Now imagine that you hit a spacecraft that moves at, say, half the speed$ c $and chased a ray of light. According to physics, which was known at that time, it turned out that from the point of view of a spacecraft moving at a speed$ c / 2 $, the light pulse will move away from it all with speed $ c / 2 $. But this means that in the frame of reference of such a fast-moving spacecraft, the laws of physics must somehow be different. Maxwell's equations must differ from the standard form.

    There was some tension between Maxwell’s physics and Newton’s physics. Tension, but not contradiction. It is quite possible to imagine that there is a fixed frame of reference in which the Maxwell equations have a simple form. But Newton's equations have the same form in all inertial reference systems. To explain why this is happening, physicists invented the idea of ​​the ether, that is, the medium in which light waves propagate, like air, in which sound waves propagate. The frame of reference, in which the Maxwell equations have a simple form, is the frame of reference in which the ether rests. If we are moving relative to the ether, then the equations become different. That is what people thought in 1904. It was a consistent point of view, but it meant that there was a duality between electromagnetism and mechanics.

    Einstein thought that physics might not be so illogical. Maybe there is a more elegant way that can explain everything. He realized that if we modify the equations that are used to transform between different reference systems, then we can make Maxwell's equations invariant. It is possible to make Maxwell's equations valid in all reference systems. Let's go back to our example with a ship chasing a light beam. According to the new transformation equations proposed by Einstein, it turns out, although it strongly contradicts the intuition that the light pulse moves away from the ship at a speed$с$. Although the ship itself is moving at a speed$ c / 2 $trying to catch up with the light pulse.

    It is not obvious how this can be. But it turns out that this is exactly what happens. Basically it was Einstein's guess. He suggested that there is no ether, that the laws of physics, and electromagnetism, and mechanics are the same in all reference systems. In order to do this, the transformation equations between different reference systems must be different from those used by Galileo.

    These transformations are called Lorentz transformations. In this lecture, we will not write them out. In this lecture we will talk about the three physical effects that follow from the Lorentz transformations. One of these effects is time dilation. A little later we will discuss two other main effects that are necessary for understanding the special theory of relativity and explaining how it can be that the speed of light is the same for all observers, even for those who move.


    The slowdown in time is that if you watch a moving clock, the moving clock “looks” going slower. I note that I put the word "look" in quotes. We will come back to this and discuss in detail what is meant by the word "look." However, moving clocks will look, in my frame of reference, always slower in an absolutely predictable number of times. This number is a well-known expression in the special theory of relativity.$γ$:

    $ γ = \ frac 1 {\ sqrt {1-β ^ 2}} $


    Where $β$ Is just a designation for $ v / c $, the speed of movement of the clock, divided by the speed of light. If a$ v / c $ a little, then a slowdown is also a little, $γ$almost equal to 1. Slowing down time by 1 time means that time does not slow down at all. If a$γ$close to 1, the effect will be slight. But moving clocks will always go slower.

    Let's now go back to the word "look." There is a subtlety. Last year, Brian Greene's four-part film “Cosmos Fabric” was released on PBS . He tried to illustrate time dilation. He showed a man sitting in a chair, and a man walking toward him and carrying a watch over his head. The camera showed that the person sitting in the chair will see how the clock begins to move more slowly when moving. This is not true. This is not what he actually sees. And this is the key problem of the word "look."

    When we say that a moving clock is slower, we do not mean that the observer really sees it. The complexity of the situation lies in the fact that when you look at something, you register light pulses that come to your eyes at a given point in time. As the light travels the path in a finite time, it means that you see different things at different times. For example, if there is an object, say, a laser pointer flying towards me, I will see its back where it was at an earlier time than the front. Because the light that is emitted from the back takes more time to reach my eye than the light emitted from the front of the pointer.

    Therefore, when an object approaches me, I will see its different parts at different points in time. This complicates things. What I see, taking into account the special theory of relativity, is quite difficult. This can be calculated, but for this there is no simple expression. It is necessary to figure out step by step what I will see at any given time. It is absolutely not like a simple picture.

    Thus, the statement that clocks go slower in$γ$times based not on what the observer actually sees. It is based on what the reference system sees, not a particular person. This ultimately leads to a simpler picture. The reference system can be represented as a set of rulers connected to each other, so that they form a coordinate grid, and a set of clocks, located everywhere inside this grid.

    In this case, all observations are made locally. That is, if we want to measure time in some kind of reference system, we do not use the central clock, waiting until the light pulse reaches this central clock. Instead, the reference system is filled with clocks that have been synchronized with each other from the very beginning. If we want to know at what time some event happened, we look at the clock next to it. This clock shows when this event occurred.

    As a rule, this is how we work with different coordinate systems. If we want to understand what a particular observer sees, the picture becomes more complicated. We must take into account the speed of light. Only by eliminating the time for the propagation of light and calculating that the local clock will show, we will see a time dilation in a simple form, that a moving clock always goes slower.

    In particular, in the example of a man sitting in a chair, and a clock approaching him. The person will experience what we are discussing in this lecture - the Doppler shift. As the clock approaches him, he will experience a blue shift rather than a red one. He will see that the clock is moving faster, not slower, exactly the opposite of what was shown in the television program. It will seem to him that the clock is moving faster due to the fact that each subsequent light pulse travels a smaller distance, as the clock approaches the observer. This effect makes a greater contribution than the effect of slowing down a moving clock, when compared with fixed clocks that are directly next to each other.

    STUDENT: If the clock very quickly flies past us, could we see their slowdown when they are strictly perpendicular to us?

    TEACHER: Yes, you are absolutely right. When a clock flies past the observer and is strictly opposite it, the speed of the clock in its frame of reference is perpendicular to the speed of the photons that it sees. At the same time, he will see the net effect of time dilation.

    I want to add that I and several other people from MIT participated in the making of the Brian Green film. We discussed this issue for a long time with Brian Green via email. We all said it was wrong. However, Brian Green took the position that this was done intentionally, that he tried to illustrate the effect of time dilation, without discussing the Doppler shift. Since he did not want to talk about the Doppler shift, he simply ignored the fact of its existence. We all thought it was wrong from a pedagogical point of view. But we could not convince Brian of this.

    Relativistic Doppler shift

    Now we will again calculate the Doppler shift, this time considering that the moving clock is slower in$γ$time. We will deal with the relativistic case where the wave is a light wave. And the speeds can be comparable to the speed of light. This time the effect of slowing down time is large enough to take it into account.

    This time both answers should be the same. If the answers are different, then it turns out that our picture of the world is wrong, contradictory. It should not matter whether the source moves or the observer moves. Previously, it mattered, and we explained this by the fact that air was involved in the process. If you make a transformation in order to move from one case to another, from the case when the source moves, to the case when the observer moves, the air will have different speeds in different cases. In one case, he will be stationary, in the other case he will move. Therefore, we did not plan to receive the same answer.

    But now, when we move from the case where the source moves, to the case where the observer moves, the ether should move with a different speed. But the main axiom of the special theory of relativity is that there is no ether, at least there are no physical effects arising from the ether. So you can pretend that it does not exist. Therefore, in the special theory of relativity, we must receive the same answer, whether it be a moving source or a moving observer. This is actually the same situation, only viewed from different frames of reference. The special theory of relativity asserts that it does not matter in which reference frame we do the calculations. We will use the same pictures, but this time we will take into account the fact that moving clocks go slower in$γ$ time.


    To begin with, let's think at what stage it is important for us to slow down the time of a moving clock? On the second. It is at this stage that the source measures the period between the emission of two wave crests by moving clocks. You can simply imagine that the source emits a series of pulses, where each pulse represents a wave crest. For me, it looks a little simpler, because you don’t need to think about the sine wave that the source actually creates.

    The time between these pulses, measured by the source clock, is what we denoted as$ Δt_s $. The source moves in our picture. All calculations will be made in our reference system. This is very important, since the transformations between reference systems are a bit complex in the special theory of relativity. When you solve a problem, it is very important to choose a reference system that you will use to describe the problem, and stick to it. If something is originally described in a different frame of reference, you need to understand how it looks in your frame of reference. To then relate this to other events that are described in your frame of reference.

    For our task, our frame of reference will be the frame of reference of the picture, the frame of reference, which is at rest relative to the observer. You can also call it the observer's reference system. With respect to this reference system, the source is moving. The source emits a sequence of pulses. You can imagine that the source is just a clock. Any phenomenon that repeats at regular intervals is a clock. Thus, the source is a moving clock that moves more slowly in$γ$time.

    For the rest, nothing changes. The observer also has a clock that he uses to measure the time between ridges. But the watch of the observer is in our frame of reference. Thus, there is no time delay associated with the observer’s clock; there is only time delay associated with the source clock. And again, everything important is depicted inside the yellow rectangle. Now we need to look at the equations and see how they change.

    Last time, the time interval measured by an observer was the sum of two members. As the first member was$ Δt_s $, it would be the only member if the source rested. This is also true in our case. But time at the source is slower in$γ$time. That is, if we do not take into account changes in the length of the path - we will take into account these changes in the next term - then the period measured by the observer will differ from the period measured by the source in$γ$time. But you need to find out whether$γ$stand in the numerator or denominator. A mental example can help.

    So, the source clock is slower. Suppose we are talking about a time interval of one second. If the source's clock is slower, it means that we should have more time for the source to pass a second. Let's say that the clock is going twice slower. This means that the source will take only one second every two seconds. This means that the period we will see will be longer than$ Δt_s $ at $γ$time. Thus, before the first member we put the multiplier$γ$. The second term is still equal$ Δl / u $.

    display
    Δt_o = γΔt_s + \ frac {Δl} u
    display


    But the expression for $ Δl $ also changing. $ Δl $- This is the time interval that the light pulse needs to travel an additional distance. The extra distance is proportional to the time between pulses. This time varies due to the slower time of the source clock. So the second term also increases in$γ$ time.

    display
    Δt_o = γΔt_s + \ frac {Δl} u = γΔt_s + \ frac {vγΔt_s} u = γ (1+ \ frac vu) Δt_s
    display


    Thus, the entire response increases in $γ$time. Considering that

    $ γ = \ sqrt {\ frac 1 {1 - {(\ frac vu)} ^ 2}} $


    and

    $ 1 - {(\ frac vu)} ^ 2 = (1- \ frac vu) (1+ \ frac vu) $


    after algebraic transformations we get

    display
    Δt_o = \ sqrt \ frac {1 + β} {1-β} Δt_s
    display


    So, we have received the answer, taking into account the special theory of relativity, in the case of movement of the source. When taking into account the theory of relativity, our response increased in$γ$time. We expect that the answer will not depend on whether the source or the observer is moving, but, of course, this needs to be checked with the help of calculations.


    We will take as a basis the calculation that we have already done for the non-relativistic case, with a moving observer. We will try to calculate the relativistic case. Now watch watcher go slower. They go slower in relation to us, in relation to our frame of reference, where our frame of reference is, by definition, the frame of reference of our picture.

    The most important thing is happening again in the yellow rectangle. The source is motionless, therefore$ Δt_s $- it's just the wave period measured by our watch. But the period measured by the observer,$ Δt_o $will be different. Therefore, we write our equation differently, replacing the expression for$ Δl $. For$ Δl $ instead $ vΔt_o $ we will write $ vΔt '$.

    $ Δt '$ not equal $ Δt_o $. $ Δt '$- This is the time elapsed between the third and fourth stages, that is, the time elapsed between the arrival of two adjacent wave crests to the observer, measured in our frame of reference. We describe everything from the point of view of our reference system.$ Δt '$ differs from $ Δt_o $ at $γ$ because the watches are slower in relation to us $γ$time.

    Again, you need to think a little where it should be$γ$in the numerator or denominator. We know that the observer’s clock is slower than ours. This means that the time it takes for one observer to pass one second should take more than a second. therefore$ Δt '$ = $ γΔt_o $. For example, during the time that the observer’s clock passes one second, two seconds pass by us.

    We will repeat the calculation we did for the non-relativistic case when the observer was moving. But in the calculation we add the time dilation, which will make this calculation true. First we write out the equations as they appear in our frame of reference, that is, they use the interval$ Δt '$:

    $Δt'=Δt_s+\frac {vΔt'}c$


    Now we can perform transformations similar to those we performed for the non-relativistic case and get an expression for $Δt'$:

    $Δt'=(1-\frac vc)^{-1}Δt_s$


    Substituting the expression for $Δt_o$ we get:

    display
    Δt_o = \ frac 1γΔt '= \ sqrt {(1 + β) (1-β)} \ frac 1 {1-β} Δt_s
    display


    or:

    display
    Δt_o = \ sqrt \ frac {1 + β} {1-β} Δt_s
    display


    This expression is true both in the case of the movement of the source and in the case of the movement of the observer.

    Redshift$z$ in the relativistic case it turns out:

    display
    z = \ frac {Δt_o} {Δt_s} -1 = \ sqrt \ frac {1 + β} {1-β} -1
    display


    So we got what we expected. That the result is consistent with the principles of the theory of relativity. Our answer does not depend on whether the source or observer is moving, since it does not matter in which reference system we perform the calculations.

    Other effects of the special theory of relativity

    Now I want to talk about two other kinematic effects of the special theory of relativity, namely the Lorentz contraction and the change in the concept of simultaneity. But before addressing these effects, there is another issue that we need to discuss. This is a clock that moves with acceleration.

    The special theory of relativity describes inertial reference systems and what transformations are performed during the transition from one inertial system to another. If we know how clocks go at rest in one frame of reference, the special theory of relativity fully describes how the clocks will go in a frame of reference moving at a uniform speed with respect to the original frame of reference. Or in other words, it describes how the clock will go if it moves at a constant speed.

    However, in the real world, we have very few hours that can be considered inertial. Any clock that we see around us - the clock on the wall that moves with the Earth, or my wristwatch, is constantly being accelerated. We want to be able to work with clocks that accelerate and move at a relativistic speed. This, for example, occurs in satellites. The GPS system, as you probably know, will not work if the calculations do not take into account the effects of the special theory of relativity and even the general theory of relativity. Thus, the study of the behavior of a moving clock is a critical technological task.

    What can we say about accelerating clocks? There is a common myth that a general theory of relativity is needed to describe acceleration. Therefore, we must postpone the conversation about accelerating hours until we pass the course of the general theory of relativity. In fact, it is not. The general theory of relativity is the theory of gravity, which states that gravity and acceleration are closely related. In this context, acceleration appears in the general theory of relativity.

    However, the special theory of relativity is sufficient to describe any system that is described by equations that are consistent with the special theory of relativity. Special relativity does not describe gravity. Therefore, in a situation where gravity is important, the special theory of relativity is not able to give the correct results. But while gravity is absent, while we are dealing only with electromagnetic forces, no one prevents us from using the equations of the special theory of relativity.

    We must use the equations of dynamics in the special theory of relativity, which show how bodies react to forces. Whenever a force is applied, acceleration appears. Such equations exist. We can combine, for example, electromagnetism with relativistic mechanics to describe a system of particles that interact with the help of electromagnetic forces, in full accordance with the special theory of relativity. And, despite the fact that these particles are accelerating, we can count everything we want for them.

    In particular, if there are clocks made of parts, the physics of which we understand, the special theory of relativity can tell us how these clocks will behave, even when they accelerate. However, this calculation can be very, very complex. Because the physics of any real watch, for example, my wristwatch, is very complex. But we do not need to write out the equations describing my wristwatches in order to understand how they will behave during acceleration.

    I note that many of you already have a lot of experience with accelerating watches, because many of you are wearing watches that accelerate all the time. And they usually work. We usually assume that, although the watch is accelerating, it is made well enough to withstand the acceleration that gives it to your wrist and show the correct time.

    On the other hand, one can imagine the opposite situation. If you take a mechanical clock, and throw them into the wall, they will break against the wall and stop. When they crash against a wall, they experience a very large acceleration. If the acceleration is large enough, we can predict what will happen to the clock, even if it is a complex interaction. If the acceleration is large enough, it will simply break the clock and it will stop. This is one of the possible effects that acceleration can have on a watch.

    Other effects are similar to this one. If the movement of my hand affects the work of a wristwatch, this is a mechanical effect that can be calculated by understanding the mechanics of the watch, rather than using the principles of the general theory of relativity. The difference with the special theory of relativity here is that the special theory of relativity can make an accurate prediction of how the clock will behave if they move at a constant speed without even knowing anything about the device of this clock. The special theory of relativity can make such a prediction, because there is a symmetry, the Lorentz symmetry, which connects the moving and resting hours. This is the exact symmetry of nature. No matter what the clock is made of, if it moves at a constant speed, the special theory of relativity asserts$γ$ time.

    On the other hand, neither in the special theory of relativity nor in the general theory of relativity there is a similar principle concerning acceleration. How acceleration acts on a clock depends, of course, on how big the acceleration is, how the clock is arranged, and how the acceleration affects the various internal parts of the clock. The bottom line is that when we talk about accelerating clocks, we always assume that clocks are made well enough so that acceleration does not affect how fast they go. We assume that this is an ideal watch, that they are made perfectly well. When we say that acceleration does not affect the speed of the clock, we mean that at every moment of time the clock goes at exactly the same speed as other watches that move simultaneously with our clock at the same speed, but without acceleration .

    At any time my watch will have a certain speed. The rate of their progress will be very slightly affected.$γ$which in our case will be very close to 1. If we consider my watches to be ideal clocks, we assume that at any moment they go at the same speed as the clock, which does not accelerate, but which move with the same speed as a wristwatch. So the multiplier$γ$will remain, but there will be no acceleration effect. The speed of the clock will be determined only by their speed relative to our reference system.

    Now I want to talk a little about other consequences of the special theory of relativity. A little later we will talk about the dynamic consequences of the special theory of relativity, which include well-known equations, such as$e=mc^2$. But before we talk about dynamic quantities, such as energy and momentum, we will end with the consideration of the kinematic effects of the special theory of relativity. By kinematics, I mean the consequences of the special theory of relativity for measuring time and distance.

    If we confine ourselves to the consequences for measuring time and distances, by kinematic effects, then there are exactly three such consequences from the special theory of relativity. The whole special theory of relativity, in a sense, is embodied in these three effects. Time dilation is one such effect.


    The second consequence is another well-known effect of the special theory of relativity, the Lorentz contraction, or sometimes called the Lorentz-Fitzgerald contraction. In his description the word “looks” will appear again. I will always write this word in quotes to remind you that this is not exactly what the observer will see. Any rod that moves with speed$v$ along its length relative to a given frame of reference, it will “look” for an observer in this frame of reference shorter than its length in $γ$time. The length of a rod that moves perpendicular to its length does not change. This is all shown in the figure.

    This is a very famous consequence of the special theory of relativity. It means that the rocket gets shorter and shorter as it moves faster and faster. Again, you need to remember that this is not what you actually see. This is what happens if measurements are made by local observers, and then the rocket length is calculated from these measurements.


    The third and last effect is a little harder to describe. But this is a very important effect. The first two effects would not be consistent if there were no third effect. The third effect is a change in the concept of simultaneity, or relativity of simultaneity.

    Suppose we have a system consisting of two hours that are synchronized in their frame of reference, relative to which they rest. Let them also be connected by a rod that has a certain length in their frame of reference, which we call$l_0$. If the whole system moves relative to us at a speed$v$along the bar, for us this clock does not look synchronized, despite the fact that it is synchronized in its frame of reference.

    In particular, the back of the watch will look a bit running ahead at the time$βl_0/c$. I recall that$β=v/c$. $l_0$ - the distance between the clocks measured in the clock reference system. $c$- this is, of course, the speed of light. On the other hand, if the clock moves in the direction perpendicular to the line connecting them, then the clock looks synchronized.

    This effect is very important for the integrity of the whole picture. We will not argue that a special theory is consistent. We could well do it, but we will not do it, because our course is not devoted to a detailed study of the special theory of relativity. However, it may seem that between the consequences of the special theory of relativity - that moving clocks go slower, and the postulate that the same laws of physics are true for all inertial observers, there is a rather obvious discrepancy. This means that if you are moving relative to me, then for me your watch will go slower. But at the same time, my watch is slower for you. Because, from your point of view, you are at rest, and I move towards you. From your point of view, my watch is moving. And my watch should go slower.

    It seems to me that your watch is going slower. It seems to you that my watch is going slower. It seems that this is a contradiction. What happens if we just put the clock next to each other and compare how it goes? Which hours will go faster? How can we negotiate with each other? Of course, we can not keep the clock next to each other, and at the same time move them relative to each other. This is one of the reasons for resolving the contradiction. Recall what I really mean when I say that your watch is slower. I do all my measurements without directly observing your watch, because then the effect of a signal propagation delay occurs, which complicates the picture. I do all my measurements with the help of many local observers who surround me and are at peace with me. They give me their results. Only after receiving and combining their results, do I get a single picture of what, where and when it happened.

    Therefore, when I say that your watches are slow, I mean that I have many watches that are at rest with me. When your watch flies past me, local observers compare your watch with your watch. Then they pass the results to me. If your clock is slower, say, twice, it means that when your watch flies past the clock of my observer and his watch shows one second, your watch will only show half a second. When they fly past the more distant clock of my reference system, and the clock of my reference system shows two seconds, your clock will show one second, and so on. In this sense, your watch is slower.

    This should be compatible with the fact that according to your point of view, my watch also goes slower. If you assume that the clock in my frame of reference is synchronized, then you come to the conclusion that my clock is going faster. Because when your watch shows half a second, my watch shows one second. When your watch shows a second, my watch shows two seconds. According to this direct comparison, it turns out that my watch is going faster.

    But at the same time, we know that this is not true. You should get the same result as me. If we move in relation to each other, you should think that my watch is going slower. The way out of this difficult situation lies in the relativity of simultaneity. From your point of view, the sequence of clocks from my frame of reference, when they fly past you, really shows a longer time compared to your clocks. However, from your point of view, my watch is not synchronized with each other. Therefore, you cannot determine how fast my watch is going by measuring time on different clocks.

    If you want to find out how fast my watch is going, you have to watch one of my watches and watch how the readings change over time. You should not compare the readings of different clocks, because my clocks are not synchronized with each other, from your point of view. But if you watch one of my watches, using the dials of your watches, which are motionless towards you, just as I used the dials of my watches, when I measured the speed of your watches, then everything will fall into place. You will see my watches go slower. I will see your watch go slower. Since we disagree about which events occur at the same time, there is no contradiction. Thus, the relativity of simultaneity is crucial, otherwise we would get a glaring contradiction in the whole picture.

    This is all that I planned to tell in today's lecture. We discussed the kinematic consequences of the special theory of relativity. As I said, we will not try to withdraw them. If you are interested in how they are obtained, then you can take a specialized course on the special theory of relativity.

    Later we will discuss the implications of the special theory of relativity for momentum and energy, which will be important to us. Energy and momentum are of interest to us only as long as they are defined in such a way that they are stored values. That is why energy and momentum are important in physics. For a closed system, the total energy and momentum do not change. Energy and momentum can be transferred from one part of the system to another. But energy and impulse can neither be created nor destroyed.

    If we take the definitions of energy and momentum from Newtonian mechanics and use them in relativistic kinematics, it turns out that, for example, when particles collide, energy and momentum would be stored in one frame of reference and would not be saved in another frame of reference. The conservation laws would depend on the reference system used.

    Therefore, Einstein somewhat changed the definitions of energy and momentum in such a way that if they are saved in one frame of reference, then they are saved in any other frame of reference associated with the first transformations of the special theory of relativity. As soon as we change the kinematics of the transition from one reference system to another, we also need to change the definitions of energy and momentum so that the conservation laws are valid in all reference systems. In the following, we introduce some slightly modified, somewhat non-Newtonian definitions of the energy and momentum of moving particles.