The Twins Paradox is a consequence of the Lorentz transformation equations in the context of Special Relativity, in which an observer considers a moving clock to run slow, and hence a moving observer to appear to age more slowly. Because of this, each of two twin observers in relative motion regards the other as being the one in motion, and each, therefore, expects the other's clock to be running more slowly, and the other to be younger. Clearly, of course, two clocks cannot each be slower than the other, and two persons cannot each be younger than the other.

One attempt to solve this contradiction involves identifying a difference between one twin and the other. In the case, for example, where one twin remains on earth, and the other travels to a nearby star and back, one of them engages in using his engines to create accelerations and decelerations during the round trip, while the other never fires his engines, and does not accelerate, and this is used to apparently prove that only one twin is to be regarded as younger than the other.

What is not properly dealt with here is that the travelling twin sees the earth, and the star, and the whole galaxy, to be the accelerated reference frame, rather than himself. He sees the Earth, the stationary twin, and the star, all together leaving his stationary reference frame and returning; which is a perfect and contradictory counterpart of what the stationary twin sees from his earthbound point of view. This, of course, can be called ridiculous, but calling it so does not in any way modify the equations, or translate an explanation into their context, and thus does not get rid of the paradox they suggest. (For those who think in terms of the 'extremal aging' concept, each twin will argue that his clock is the one that defines the 'extremal' time accumulation). The answer usually given to this is that the twin that fires his engines can detect his acceleration from within his reference frame, by means of an accelerometer, for example, whereas the other twin cannot detect any acceleration within his frame. A difference between the twins is presumed to have been established, in this way, that will be sufficient to solve what is regarded as only an apparent paradox. However, this argument does not clearly explain how a non-inertial frame must be the one to follow a different spacetime path to the inertial frame, especially from the perspective that can view acceletation as consisting of a series of incremental changes from one inertial frame to another at a slightly higher velocity. I don't intend to discuss these matters here, since what follows will make such considerations to be no longer relevant.

However, I shall proceed, for the moment, with an analysis of the above apparent explanation, which involves the view that the travelling twin follows a different 'spacetime path' than the stationary twin, between the event when he leaves earth and the event when he returns. Such a different spacetime path, involving a different accumulated total distance in the direction of a time coordinate, allows the travelling twin to accumulate a lesser total time on his clock than the stationary twin. This view is interpreted mathematically by integrating, or accumulating, clock times to show that the accelerated twin accumulates less time on his clock than the other. By thus pursuing this argument, I don't myself declare it to be clear and understandable and, furthermore, it will not be the result arrived at below.

The integration process used can be described as follows: Start with the Minkowski spacetime metric equation, or Lorentz transformation equation, which are equivalent c2d2 = c2dt2 - ds2 or d2 = dt2 - ds2 = (1/)2dt2 where = (1 - v2/c2)-1/2, or 1/ = (1 - v2/c2)1/2, and c = 1 which gives the Lorentz equation d = (1/)dt d, here, is one small time interval recorded on the moving twin's clock (while travelling to the star and back). It is also the value that will be associated with the moving clock by the twin that remains stationary on the earth. And it is equally the value that will be understood by any other observers moving with different velocities, and all will agree on the value of d. This also includes the case of an observer moving with zero velocity relative to the travelling twin; that is, it includes the travelling twin himself. d is thus also the time the travelling twin, as an observer at zero velocity, associates with his own clock, and can be made to be the time he actually sees on his own clock. d is thus an invariant, i.e. independent of any relative velocity, and is referred to as the travelling twin's own 'proper time'. dt, however, is the corresponding time measured on his own stationary clock by the twin on earth, and also by any of the other observers clocks, which are equally stationary in their reference frames, and each, using his own version of this equation, will measure a different value of dt according to his own clock, depending on his velocity relative to the travelling twin. That is, dt is not an invariant, but depends on velocity. If, for example, all clocks are physically identical, and if d represents one tick of the travelling clock, other observers might measure 1.x, or n.x ticks etc on their clocks, depending on the twin's velocity relative to each separately. For the travelling twin himself, of course, =1, and his dt = d. Some people might assert that understanding this is straightforward but, in fact, it is not so simple a matter to identify the true interpretation of the time-related intervals. The Minkowski equation is a 'metric' equation, and is therefore geometric in nature, while time is normally measured by a number of repetitions, i.e. the number of ticks on a clock. These are converted into a length by multiplying by velocity c, as in cdt. If, however, clock times can appear to be altered, a possibility that the 'lengths' of the ticks themselves, and not only the number of ticks, could be altered is immediately and inevitably suggested. Thus a clock tick, or period, itself acquires a geometric character, and a time interval becomes composed of a number of clock ticks, and a length of each tick. This automatically complicates the interpretation of the equations. I therefore discuss this in more detail at the following link, and will proceed to state only the conclusion here: The Interpretation of Time Intervals. The point of view adopted here, as argued in the link, is that the tick length of all clocks remains invariant across all inertial reference frames, which means that there is a direct relationship between a time interval, such as dt, and the number of ticks that define it. Thus, a lesser value of dt means a lesser number of ticks, with the tick length assumed constant. To get the total time accumulated by the travelling clock, the Lorentz equation is used to integrate d in terms of dt as d = (1/)dt and this is less than the accumulated time on the stationary clock, which is said to be simply dt because, for the stationary clock, since its velocity in its own frame is zero, = 1 , and (1/)dt = dt. The problem with this is that the first of these equations cannot be integrated as it stands. If I want to integrate a function f(x)dx, I cannot do so in the normal way if dx is not an arbitrary or constant interval, but is itself a function of x. Instead, dx must first be converted as, for example, dx = f'(x)dx', where dx' is an arbitrary or constant interval, independent of x. Thus, I must integrate f(x)f'(x)dx'. Such a situation can occur in an integration or summation, dy = f(x)dx, where always dy = f(x)dx, if dy is not allowed to function as a resultant value, or as a dependent variable, but has its value fixed in advance. In the equation d = (1/)dt, where always d = (1/)dt, d is specified in advance as a constant, independent of , and therefore cannot be regarded as a resultant value, or dependent variable. Since acceleration can be regarded as a stepwise series of inertial reference frames, each at a slightly different velocity, the values of dt change incrementally with velocity. Therefore the integral equation must really be written d = (1/(v))dt(v), or it can be written d = (1/(t))dt(t) A way to get around this is to convert dt to an interval that, like d, will remain an invariant, independent of , throughout the summation. Then the relationship between d and dt will effectively be the same as if d was a dependent variable. An objection to this may be raised, however, in that the above equations involve a comparison between dt and (1/)dt, if the same dt is considered to be used in both. Therefore, what does it matter if dt is a function of velocity, since it is the same measure used for both integrals, and therefore the comparison between them is valid without the need to convert dt to an interval independent of velocity. There is here, however, a fallacy involved in the use and interpretation of the time interval dt, and its relationship to d. I invite the reader, therefore, to proceed, as follows, to consider the nature of the integration (1/(v))dt(v) by converting dt to a constant interval, as this will cast a new light on the nature of the time interval dt, that will otherwise not be understood. To convert dt, start with a second Minkowski, or Lorentz transformation, equation, and this time specify dt to be a constant interval, dt', independent of velocity. We then have two transformation equations, as follows: d2 = dt2 - ds2 or d = (1/)dt, which is the same equation as before, with d independent of velocity d'2 = dt'2 - ds'2 or d' = (1/)dt', which is the new equation, with dt' now independent of velocity. The equations are to be considered written in terms of the same velocity, so that remains the same. In the second equation, since dt' is specified to be independent of velocity, d' must now be dependent on velocity, because is. One might therefore ask, "what does d' actually mean, or refer to?" Since this will not be important in the following calculation, let us dispose of this question by saying "let it mean nothing in particular, or nothing of importance". Now, divide the two equations into one another, to obtain d'/d = dt'/dt which leads to dt = (d/d')dt' which converts dt to an equation containing the constant interval dt' Putting this in the original equation (d = (1/)dt), we have d = [(1/)(d/d')]dt' It will not be necessary to go further into this equation, but only point out that, since d and dt' are both independent of velocity, the term in square brackets must be independent of velocity also. Let us put it = K (1/)(d/d') = K and, consequently d = Kdt' Now, let us specify that dt' represents, by definition, one tick on my clock, if I am the stationary twin. I can specify dt' in this way, because the effect of changing dt' is only to give a different value to K, but leave the equations the same as before. So I can specify dt' to be whatever I wish, provided it is not dependent on velocity. d, as before, can also represent one tick on the moving clock, as the equations are created, or specified, independently. Let us also specify that both clocks are physically identical. What, then, is the value of K? The Minkowski metric equation, and the Lorentz transformation equation, are both such that they are valid for a velocity, v = 0. That is, for a velocity, v = 0, ds = 0, in the Minkowski equation, and = 1, in the Lorentz transformation equation. Thus, in the equation (1/)(d/d') = K, when the velocity, v = 0, (1/) = 1, and we have K = d/d'. When both clocks are at rest together, and the relative velocity is zero, if both clocks are identical, the lengths of their ticks are identical, and therefore d = dt', and K = d/d' = 1. Only if the clocks were differently calibrated would the factor K have a value other than 1. If K = 1 when the clocks are at rest, it must be true at any velocity, since K is independent of velocity. We can therefore say that d = dt' represents the proper time for both of us, no matter what our relative velocity. And if this applies to us, it applies likewise to all inertial frames in relative motion and, thus, the proper time for any is a universal proper time for all. What, then, is to be said of the time interval dt in the equation d = (1/)dt, for dt depends on the velocity and is > d, and hence also > dt', which is my proper time? It follows that, since dt refers to an interval that is larger than my proper time, it penetrates into my future and, although it can be calculated, it cannot be immediately read on my clock. It further follows that, where we refer to an accumulated time which is the integral, dt, specifying dt to be the same as in the equation d = (1/)dt, this time refers to an accumulated time that also I will only encounter, or observe on my clock, in my future. It is always greater than the actual accumulated time I can read on my clock, which is dt'. In fact, here dt is artificially being forced to be the same as the dt in the other equation, where it is dependent on velocity, which is what I meant by referring to a 'fallacy' in the interpretation of dt. My clock, or any clock, accumulates only its own proper time, and not a greater time. That is, all clocks of all observers accumulate the same proper time, independently of velocity, and changes in velocity, and 'spacetime paths' considerations have no effect. WHAT, THEN, OF THE TWINS PARADOX? The explanation referred to at the start, which depends on the integration d, as the accumulation of the time on my travelling twin's clock, being less than the accumulated time, dt, on my clock, cannot be correct, because the greater accumulated time, dt, supposedly appearing on my clock, will not be visible until some time in my future, after my travelling twin has already arrived home. When he does arrive home, he will see the accumulation of my proper time dt', which will be the same as the accumulation of his proper time, d. In general, the value dt is only a projection of a time interval into a 'stationary', or 'coordinate' reference frame, and not the real, present time in that frame. Thought it can be useful for transformation purposes, in the usual way, it has no immediate physical significance. The universal proper time concept thus immediately disposes of the twins paradox. The twins really always remain the same age, no matter how they may see one another when in relative motion, and will again see one another to be the same age when they come to rest in the same reference frame once more. Let us ask, "how do the twins see one another when in relative motion?" To attempt an answer to this, consider, again, the first equation d = (1/)dt It has been shown above that d = dt', the proper time on my own clock, if I am the stationary twin, and dt > dt' is a time interval partially penetrating into my future. Consequently, although d, on the travelling clock, can be inferred, and hence dt calculated, it makes no sense to say that I can actually see d, since I cannot actually see dt. Consider, therefore, the second equation d' = (1/)dt' Here dt', my proper time, is a time I always actually see on my clock, and therefore it makes sense to suppose that, if I look at the time on a moving clock, I will see d', and not d, although the moving observer himself sees d. This means that I cannot see the moving clock as it really is in the universal proper time, because I see d' < d, which is less than the proper time, and thus the moving clock appears to run slow. It follows that the invariant d, though all observers can agree on its value, as usual, is not the value that coordinate frame observers can actually see on a moving clock, and each of them sees the moving clock to read a different time, depending on the velocity of the clock relative to him. Since, however, there is only one moving clock, how do I see it differently to what it really is? In answer to this, I make the suggestion that I see it not directly as it is, but as it was. That is, I see it via a present manifestation of its past history. When, therefore, I see my travelling twin to be younger than myself, and he sees me to be younger than him, we are each seeing the other as he was, and not as he is. When, however, we decelerate in order to come to rest again in the same reference frame, we will each see the other aging rapidly during the deceleration until, when at rest again together, we see one another to be the same age again as, if fact, we really always were. This applies independently of whether only one of us engaged in acceleration, or both. This interpretation also serves to explain how an observer sees a moving clock to 'run slow', notwithstanding that the tick length of a clock remains an invariant, as argued before. Instead of seeing the tick length of a clock increased, and the number of ticks in a given time interval decreased, which describes a change in the rate of a clock, an observer sees a moving clock reading an earlier, historical time, which creates the same effect as a reduction in the clock rate. Another way to view this, which enables a clock to currently display an earlier, historical time, is to suppose that the observer is observing the clock via a scale change in the direction of the time dimension. I would add that this 'past history' interpretation should not be thought to mean that, by interfering with my travelling twin's reference frame, I can alter his past history. Even if I thus create an effect of some such kind, I still cannot disturb his actual past history. This can be seen immediately in that, if I changed a person's past, the fact that I changed it would itself be a historical fact, necessitating the preservation of the original past as well. Clock Times Illustration THE EXTENDED LIFETIMES OF MUON PARTICLES Considering the phenomenon of the extended lifetimes of muon particles adds a further dimension to the above understanding of a universal proper time. Muon particles are created, at high velocity, in the upper atmosphere, by cosmic rays from space. Their lifetimes are such that they should never last long enough to reach the surface of the Earth. The fact that they do is explained by the consideration that, at high velocity, they are seen to age more slowly, as specified by the Lorentz transformation equation, and their thus extended lifetimes enable them to last long enough to reach the surface of the Earth. In the above interpretation, involving a universal proper time, a muon particle really does decay in the upper atmosphere, in the time in which it would at zero velocity. The muon particle that reaches the surface of the Earth is explained as an extended view of its past history. This, however, clearly cannot be merely a window into its past because, in its past, it never did reach the surface of the earth. This extended view of its past must clearly exist as a real particle, independently of its original past. I have therefore referred to it as a 'present manifestation' of its past history. THE HAFELE-KEATING EXPERIMENT The result of the Hafele-Keating experiment would appear to contradict the above conclusions regarding a universal proper time, in that it is reputed to show that a clock, taken on an aircraft around the world, and compared with a similar clock left stationary on the earth, shows a change in its accumulated time predicted by the 'spacetime paths' explanation of the twins paradox. The universal proper time explanation, above, predicts that, excluding effects of gravitation etc., the travelling clock will always show the same time as the stationary clock, when it returns. An engineer who obtained the original data of the experiment has shown that the result is severely flawed. In fact, if it shows anything, the experiment tends to show that the travelling clock does not accumulate a time difference, and rather supports than contradicts the universal proper time interpretation. I have commented on this in more detail in the above link, and also linked from there to the analysis by the engineer (AG Kelly), so I need not discuss it further here. By this reference to the Hafele-Keating experiment, I do not intend to make the case that it should be regarded as a satisfactory experimental proof of the existence of a universal proper time. In my view, a more rigorous experiment would be required to establish a definite experimental proof, one way or the other. Subsequent to posting this page I have posted another, linked below, that contains a separate argument that the spacetime interpretation of Special Relativity, represented by the usual spacetime diagram, illustrating effects like nonsimultaneity and foreshortening, cannot be a correct representation of a really existing instantaneous relationship of two inertial frames in relative motion in actual spacetime. A Lightlike Interpretation of Special Relativity

Einstein's Derivation of the Transformation Equations in Special Relativity

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Update March 2004, March 2007
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