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Time Dilation


Light-clock recap


We have used the light-clock as a thoughtmodel to illustrate the dilation of time in a moving frame, that is - anobserver at rest will see time passing slower in a moving frame. There are afew points that need to be clearly stated:


1.      There are two essential stepsin the argument about the light-clock:


a. when the observer looks at the movingclock, they see light travelling over a longer distance to reach the othermirror


b. light always travels at the same speed(Einstein's second principle of special relativity)


Because time is distance over velocity, andthe distance has increased while the velocity has stayed the same, it meansthat the observer must conclude that light takes longer to reach the othermirror when the clock is moving (as compared to the clock at rest).


2.      Often when one encounters thelight-clock example for the first time, there is some confusion about the"mechanism" of the clock. Why would light move on a diagonal if youare moving the mirrors?


To better visualize the setup think aboutthe light-clock as always sending light from dot A to dot B. When the clock ismoving (with constant, uniform velocity) the light ray still needs to travelfrom dot A to dot B, because nothing in the way the light-clock functions canchange. If the light ray wouldn't reach the dot B, then the principle of relativitywould be contradicted: because for an observer at rest the light on hislight-clock always moves from A to B, while for a moving observer the light onhis light-clock would go somewhere else, so he would know he is moving.


Alternatively, you could keep thelight-clock at rest, and move the observer. Nothing should change in ourreasoning. Now the moving observer will see light travelling on a diagonal pathbecause he sees everything moving in the opposite direction of his motion.


3.      It is important to notice thatwhat matters is not the motion of the clock per se (remember there is noabsolute motion in relativity). It is only the relative motion between theobserver and the clock.


4. Even though we have used the light-clockto derive our result, the dilation of time applies to all inertial movingframes, and it is not restricted only to light-clocks or clocks moving withvelocities close to the speed of light. It wouldn't make much sense if twoclocks (a light-clock and a mechanical clock) would show the same time whenthey are at rest, but different times when moving with the same velocity. Ifthis were to happen, it would directly contradict Einstein's first principle ofrelativity (because if you were to notice a difference between your mechanicalclock and your light-clock, you could infer that you are in a moving frame).Have a look at the following Brain Teaser for a more detailed discussion ofthis.


The Amount of Time Dilation


The light-clock is a great example not onlybecause it allows us to understand the dilation of time, but it also gives us away of easily deriving the amount with which time dilates. If you want to seethe mathematical derivation, check the supplement in the next unit. Thederivation is very basic and it only uses algebra and Pythagoras' theorem.


What we find from the light-clockderivation is this:


if an observer at rest compares the timekept by his rest clock with the time that he sees on a moving clock, he willconclude:


1.      Qualitative conclusion: onesecond on the moving clock takes more than a second on the rest clock, as seenby the rest observer


2. Quantitative conclusion: one second onthe moving clock takesseconds on the rest clock, as seen by the rest observer

( γ>1 for v>0 - see the derivation inthe exercise below).

 2。定量的结论:上一秒钟移动的时钟以γ= 11−V2C21秒的休息时钟,通过其他的观察者              (γ>1为了v0)参见下面练习中的推导)。

To visualize what we mean by this, imaginethat the rest observer uses light-clocks and that it takes 1 second for lightto travel from the lower mirror to the upper mirror. Then, if the two light-clocksare synchronized such that the light rays leave the bottom mirror at the sametime, then when light reaches the upper mirror on the rest clock, it onlytraveled part of the distance on the moving clock. And when it reaches theupper mirror on the moving clock, the light on the rest clock is already movingtowards the bottom mirror.


The relation between the amounts of time isthis:


a.      when 1 second has passed on therest clock, 1/γ seconds have passed on the moving clock, as seen by the restobserver


b.      when 1 second has passed on themoving clock, γ seconds have passed on the rest clock, as seen by the restobserver


If you find γ confusing, you can replace itby some arbitrary value, let's say 2 (you can easily compute what velocity thisactually corresponds to):


a.      1 second on rest clockcorresponds to 0.5 seconds on moving clock, as seen by the rest observer


b.      1 second on the moving clockcorresponds to 2 seconds on the rest clock, as seen by the rest observer


We always say for the observer at rest,because if there was a second observer moving with the moving clock, he wouldsee everything the other way around: for the second observer the moving clockwould be at rest (because he is moving with it), and the rest clock would beslowed down (because it's moving with respect to him).


If you replace the clocks by twins, youwill get the famous Twin Paradox (which we will cover in greater detail in afuture lesson): each twin sees their moving sibling as being younger.


The Gamma Factor


In a way, you can think about γ factor as aconversion coefficient between two units. E.g. 1 meter=3.28 feet. Thedifference is that we are actually talking about seconds in both cases, butmeasured in different frames of reference. When the clocks stop moving we recover1 second= 1 second.

在某种程度上,你可以把γ因子看成是两个单位之间的转换系数。1米=3.28英尺。不同的是,我们实际上是在两种情况下谈论秒,但在不同的参照系中测量。当时钟停止移动时,我们恢复1= 1秒。

Let's condense our discussion in a formula.An observer at rest wants to know how much is an interval of time Δt on hisrest clock versus a moving clock. Then he will write:


(See next unit for the derivation of γ).


Note: By interval of time we mean theamount of time between two events. Both the rest frame of reference and themoving frame of reference agree that the two events take place, but they won'tagree about the difference in time between them. However, they will agree thatthe ratio between the two intervals of time is given by γ.


Note: These are the only two relations thatyou need to know for this lesson and the quiz. However, to do well on the quizit is essential to understand the meaning behind the formula. When you approachthe quiz don't think about how to plug the numbers into the formula, but how tomake physical sense of the problem. Visualize each problem and try to expressit in your own words.


The aim of the quiz is not to make you proficientat solving this type of problems, but it should give you the ability ofchecking your final answers. You should be able to know if your answer makessense or not.




Imagine that an observer at rest is lookingat his rest clock and at a clock moving with a velocity of v=0.5 c. We willoften express velocities in terms of c, the speed of light, to ease yourcomputations. If the observer sees 1 second passing on his rest clock, how manyseconds would he see that have passed on the moving clock?

想象一个静止的观察者正在观察他的休息时钟和一个以v0.5 C的速度移动的时钟。我们通常用C,即光的速度来表示速度,以简化计算。如果观察者看到他的休息时钟经过了1秒,他看到移动时钟经过了多少秒?

We start by finding γ for the givenvelocity:


You can plug in this result in the aboveformula, or you can simply reason about it directly. If 1 second on the movingclock, is gamma seconds in the rest clock, how many seconds on the moving clockis 1 second on the rest clock (as seen by rest observer)?


Our result is less than one second as weexpected. Always check your intuition!


·The γ factor, , might seem very off-putting, so let'sbuild some intuition about it. The goal is to understand how γ depends on v.There are a few special cases to consider:


In the last case, v>c, γ does not exist(it does not have a real solution) because v2/c2>1 and1−v2/c2<0 and we cannot take the square root of anegative number without obtaining an imaginary part. The physicalinterpretation is that nothing can move faster than the speed of light. And ifa reference frame could move with the speed of light, everything would seemfrozen in time (nothing in such a frame could move, because then it would haveto move faster than light).