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# The Lorentz Contraction, or, How Motion Affects Space (Part 1)

We are now well informed and knowledgeable about special relativity’s *time dilation* phenomenon. Furthermore, we have realized that the relativistic *distance contraction* phenomenon is a direct result of, and *always* ‘accompanies’, or complements, the time dilation phenomenon – these two physical effects are forever interwoven. Those gritty, high speed muons provided us with an excellent example of this reality. If we recall, from ** our** point of view, it was

*time dilation*that caused the muons to age much more slowly than they

*normally*would, thus enabling them to traverse the entire expanse of our atmosphere and slam into the earth well

*after*their disintegration should have occurred. But from the

**point of view, it was the**

*muons’**length contraction*of the atmosphere that enabled the earth to slam into them well

*before*their impending decay,

**, from the**

*because**muons’*perspective, time, for

*them*, was passing by at the

*normal*rate.

## Thought Experiment #2 – How Does Motion Affect Space?

We have already shown that the* principle of relativity*,* *that is, *the first postulate*, and the *constancy of the speed of light*, that is, *the second postulate*, combine to cause time dilation, but we have not done the same for the phenomenon of relativistic length contraction, at least, we have not done so with the same “thoroughness”. So we ask ourselves, “Can length contraction be deduced using *only* special relativity’s two fundamental postulates, just as time dilation was? Specifically, from the standpoint of the postulates, *why* does length contraction happen in the first place?”.

First, we repeat our earlier thought experiment, but this time, we ask Scotty to bring with him a *second* photon clock which is completely identical to his original one. He synchronizes both photon clocks so that they are ticking with the *same** cadence*, and promptly boards his train (again), taking them with him. He then positions *both* clocks so that they are standing *vertically* and ** side by side** on the passenger car’s floor, such that their photons are traveling

*up and down*between their respective mirrors as they tick, just like before. Scotty now leaves his

*first*photon clock standing in the

*vertical*position. We will call this

*“Clock V”*, since its photon, from

*Scotty’s*point of view, is traveling in the

*vertical*direction, which is a direction that is

*perpendicular*to the direction of motion of his train. He then

*re-positions*his

*second*clock so that, although

*still**directly beside**Clock V*, it is now lying

*horizontally*on the passenger car’s floor, such that its photon is now traveling

*forward and backward*between the mirrors as it ticks. We will call this

*“Clock H”*, since its photon is traveling in a

*horizontal*direction that is

*along,*or

*in line with,*the direction of motion of Scotty’s train. And so, as Mr. Scott observes both of his clocks blissfully ticking away, the T.O. Special shoots through the train crossing once more, traveling with uniform motion from

*left to right*at the relative speed

*v*, which is close to the speed of light

*c*. Now, from

*our*vantage point standing at the train crossing, we

*again*see

*Clock V’s*photon take its usual

*double diagonal*path as it completes each tick, as shown in

*Figure 1.1*by the dashed blue lines.

And *just* as before, we observe that *Clock V* is ticking at a *slower* rate than *our* stationary photon clock by a factor of

that is, the *relativistic time dilation factor*.

Also, we recall that ** we** had previously derived, from our

*first*thought experiment about time dilation, that the amount of time it takes, as observed from

*point of view, that is,*

**our***as measured by*,

**us**using**our**watches**for**

*Clock V’s*photon to complete 1 tick, is given by the equation

Here, *“V”* denotes *Clock V*, *h* is the distance between the reflective surfaces of the mirrors, *v* is the relative speed of Scotty’s train, and *c* is, of course, the speed of light (to refresh our memory, we can quickly review our derivation of the *time dilation factor*, in which we labeled the above equation for *t _{V}* as

*“Eqn (2)”*). Nothing new here, just reminiscing ‘old times’. However, for reasons that will become self-evident, we will change the form of our equation slightly, by dividing

*both*the numerator and the denominator by the speed of light,

*c*. Doing so obviously does

**change our equation’s**

*not**value*whatsoever, but yields the new form:

We will soon be using this equation and the equation for *Clock H*, which we will now derive, to surprising effect, so let’s keep the former well wedged in our brains as well.^{[1]}

## Thought Experiment #2 – Clock H’s Photon’s Left To Right Journey

“That’s nice, but what about *Clock H?*”, we ask. Well, our next logical step is to derive the equation for the *total* amount of time that it takes, again, as observed from ** our** point of view, that is,

*as measured by*, for Scotty’s horizontal

**us**using**our**watches*Clock H*to

*complete*1

*tick*, just as we did for his

*Clock V*in our thought experiment on time dilation, with one of the results being our equation for

*t*. Since Scotty’s train is moving through the crossing from

_{V}*left to right*, our strategy will be to first calculate the travel time of

*Clock H’s*photon as it journeys from the left hand mirror to (strike) the right hand mirror, and then calculate its travel time as it thereafter journeys from the right hand mirror back to (strike) the left hand mirror, thereby registering 1

*tick*. Once we have calculated these two times, we can simply

*add them together*to obtain the

*total*time that it takes

*Clock H*to complete 1

*tick*, as observed from

**point of view standing at the train crossing.**

*our*And so, as the train speeds through the crossing and *we* carefully note the *path* of *Clock H’s* photon, ** what do we predict that we should observe?** Let’s have a look at

*Figure 1.2*to help us out.

We should observe that the photon first travels from the left hand mirror to the right hand mirror. But since Scotty’s train is *already* moving to the *right* at the very high relative speed *v* which is *close to* the speed of light *c*, and since the photon must *always* travel at the speed of light *c*, then it will take the photon *some time* to catch up to and strike the right hand mirror, because in essence, the mirror is *“trying to pull away”* from the photon at the very high relative speed *v*. Hence, the photon must travel a comparatively long distance to catch up to the mirror, as shown by the long, dashed blue line. “Okay, so how much time?”, we inquire. Well, let’s look at *Figure 1.2* in more detail and work it out…

We let *t*_{1} equal the amount of time, as measured by ** us** using

**watches, that it takes**

*our**Clock H’s*photon to travel from the left hand mirror to (strike) the right hand mirror. Now, recalling that

*DISTANCE TRAVELED = SPEED x TRAVEL TIME*, then the total distance traveled by the photon when it finally strikes the right hand mirror is thus equal to

*ct*

_{1}, as shown by the long, solid blue line in

*Figure 1.2*. But in this

**amount of time**

*same**t*

_{1}, the right hand mirror (and indeed, the entire photon clock itself) has also traveled to the right, but by a

*lesser*distance equal to

*vt*

_{1}, as shown by the long, solid green line, where

*v*is, of course, the speed of Scotty’s train

*relative to us*. Therefore, by studying the simple

*geometry*of

*Figure 1.2*, we can deduce that:

## Thought Experiment #2 – Clock H’s Photon’s Right To Left Journey

The photon then bounces off of *Clock H’s* right hand mirror and now moves *leftward* at the speed *c*, where it will *soon* hit the *oncoming* left hand mirror and consequently *complete its tick*, as shown in *Figure 1.2*. We say “soon”, because *Clock H* is *already* moving to the *right* at the very high relative speed *v*. Therefore, unlike the right hand mirror that was “trying to pull away” from the photon, the left hand mirror is, in essence, *“trying to ram into”* the photon at the very high relative speed *v*. And hence, the amount of time that it takes the photon to make its return journey from the right hand mirror to the left hand mirror will be considerably *less* than the amount of time *t*_{1}, traveling a comparatively shorter distance to do so, as shown by the short, dashed blue line. “Okay, fool, so how much time, for the second time?”, we inquire again. Cool it, you jive turkeys, because once more, *Figure 1.2* will provide the answer…

Likewise, we let *t*_{2} equal the amount of time, as measured by ** us** using

**watches, that it takes**

*our**Clock H’s*photon to travel from the right hand mirror to the left hand mirror. And again, recalling that

*DISTANCE TRAVELED = SPEED x TRAVEL TIME*, then the total distance traveled by the photon when it strikes the left hand mirror (and registers a tick) is equal to

*ct*

_{2}, as shown by the short, solid blue line in

*Figure 1.2*. But in this

**amount of time**

*same**t*

_{2}, the left hand mirror has traveled rightward by a distance equal to

*vt*

_{2}, as shown by the short, solid green line. Therefore, by again examining the geometry of the situation, as shown by

*Figure 1.2*, we can deduce that:

## Thought Experiment #2 – Clock H’s Photon’s Complete Left To Right To Left Journey

Consequently, having derived equations for the amounts of time *t*_{1} (for the photon’s left hand mirror to right hand mirror journey) and *t*_{2} (for the photon’s right hand mirror back to left hand mirror journey), we can now come up with an equation for the *total* amount of time that it takes, as observed from ** our** point of view, that is, as measured by

**using**

*us***watches, for**

*our**Clock H’s*photon to complete 1 tick. We will call this total amount of time

*t*, where

_{H}*“H”*denotes Scotty’s horizontal

*Clock H*. Hence, the total amount time,

*t*, that it takes

_{H}*Clock H*to complete 1 tick, as measured from

**point of view, is clearly just the**

*our**sum*of the amounts of time

*t*

_{1}and

*t*

_{2}. Expressing this mathematically gives us:

## Thought Experiment #2 – A Fundamental (and Hopefully Understandable) Mathematical Comparison

Okay. Let us now *mathematically* compare these two amounts of time *t _{V}* and

*t*that it takes Scotty’s

_{H}*Clock V*and

*Clock H*to complete their respective ticks, with

**amounts having been measured from**

*both***point of view. In doing so, we can understand how**

*our**t*and

_{V}*t*are related, and use this relationship to

_{H}**the**

*predict***of Scotty’s photon clocks, as observed from**

*behaviour***point of view. As we have previously derived, these amounts of time are given by the following equations:**

*our*1) The amount of time that it takes, as measured by ** our** watches, for

*Clock V*to complete 1 tick is given by:

2) The amount of time that it takes, as measured by ** our** watches, for

*Clock H*to complete 1 tick is given by:

The equations look similar, but they are clearly *not* the same. And just to remind ourselves, *h* is the distance between the reflective surfaces of the mirrors, *v* is the relative speed of Scotty’s train, and *c* is the *universally constant* speed of light.

Taking a closer look at these two equations, we notice that the *numerators* of the times *t _{V}* and

*t*are the

_{H}**, that is, they are both equal to the quantity**

*same*Now, recalling that

it is ** important** for us to

*remember*at this point that this quantity,

*2h*/

*c*, is therefore equal to

*the amount of time that it takes, as observed from*. And why is this so? Well, because of special relativity’s

**Scotty’s**point of view, that is, as measured by**him**using**his**watches, for**both**of his Clocks V and H to**complete 1 tick***first postulate*,

**has every justification to claim that**

*Scotty*

*he**and his train*are the ones

**, while we and the rest of the outside world are the ones**

*at rest***instead. Therefore, as observed from**

*moving***point of view on board his train,**

*Scotty’s**Clock V’s*photon travels

*only**up and down in the vertical direction, thus traversing a*

**of**

*total distance*

*only**2h*to complete 1 tick. And because of special relativity’s

*second postulate*,

**will measure**

*Scotty**Clock V’s*photon to traverse this total distance at the

**speed of light**

*universally constant**c*. Hence,

**will conclude that the amount of time that it takes**

*he**Clock V*to complete 1 tick is equal to

*2h*/

*c*. Likewise, as observed from

**point of view,**

*Scotty’s**Clock H’s*photon travels

*equal distances**to the right and then to the left in the horizontal direction, thus*

**traversing a total distance of only**

*also**2h*at the constant speed of light

*c*, to complete 1 tick. Hence,

**will again conclude that the amount of time that it takes**

*he**Clock H*to complete 1 tick is

**equal to**

*also**2h*/

*c*, just as that for

*Clock V.*

Therefore, and ** this is very important:** From

**point of view on board his train,**

*Scotty’s**Clock V*and

*Clock H*

*must and***. Although this fact may sound inconsequential at this juncture, as we’ll discover later on, it leads our current comparison of the times**

*will always remain in synchronization with each other**t*and

_{V}*t*, which are amounts of time that

_{H}**have calculated from**

*we***point of view, to a complete and utter destruction of the**

*our**first postulate*, whose resolution is the

*relativistic length contraction phenomenon*, that is,

*the Lorentz*

*Contraction*.Alright. Onward with our comparison… We now notice that the *denominators* of the times *t _{V}* and

*t*are, although similar,

_{H}**the same.**

*not**t*'s denominator is the quantity

_{V}*sqrt(1 - v*, while

^{2}/ c^{2})*t*'s is the quantity

_{H}*1 - v*. Similar, but

^{2}/ c^{2}*not*the same. Now we have already studied the “behavior” of the expression

*sqrt(1 - v*as the relative speed

^{2}/ c^{2})*v*

*increases*, in our discussion about time dilation. But for the purpose of unraveling the ins and outs of length contraction, it is nevertheless well worth investigating this expression, along with the expression

*1 - v*, in a little more detail this time around. Just before we do so, we note that we are, of course, excluding the “trivial” case where the relative speed

^{2}/ c^{2}*v*equals

*zero*from our comparison. And this is due to the fact that if

*v*= 0, then the object, that is, Scotty’s train, is

*at rest**with respect to the observer*, the observers being

**, and hence there can be neither length contraction nor time dilation at work,**

*us**as observed from*

*our**point of view*. Continuing on, we see that

*both*denominators contain the quantity

*1 - v*

^{2}/ c

^{2}, and since the relative speed

*v*of

*any*material object must

*always be*(the reason for which will be discussed in full later on), then the value of

**less**than the speed of light c*v / c*must

*always be*1. And if

**less**than*v / c*is always less than 1, then it follows that the value of

*v*must also

^{2}/ c^{2}*always be*1. Not to be repetitive, but if

**less**than*v*is always less than 1, then the final result in this logical chain is the fact that the quantity

^{2}/ c^{2}*1 - v*must also

^{2}/ c^{2}*always be*1. Putting it in plain English, “the result of subtracting a quantity whose value is always less than one from a value of one, must always be less than one itself” – makes sense, but it’s an ‘utter’ tongue twister.

**less**thanSo far, so good, we trust. Proceeding further, if a quantity is ** less** than 1, then

*its*

**square root***must be*1. For example, the quantity 0.81, which is

**greater**than the**original**quantity itself, but must still be**less**than**than 1, has a square root that is equal to 0.90, since 0.90 x 0.90 = 0.81. Thus, as we can see, the square root of 0.81 is**

*less***than 0.81 itself, but is still**

*greater***than 1.**

*less*^{[4]}

** And therefore**, having spewed that entire mouthful, we can conclude that: Since the quantity

*1 - v*is

^{2}/ c^{2}

*always**1, as we have just demonstrated, then*

__less__than*its*, which is the quantity

**square root***sqrt(1 - v*,

^{2}/ c^{2})

*must always be*__greater__than*1 - v*1, as we have also just demonstrated. Hence, for an object, such as Scotty’s train, that is

^{2}/ c^{2}itself, but must**also still be**__less__than

*moving at a given*__relative__speed*v*, the

*first*relationship that

** must always be true**. Additionally, we mention, for the sake of completeness, that the

*second*relationship that

*must also always be true**.*

Now, knowing this *first* relationship and taking a look back at our equations for the amounts of time *t _{V}* and

*t*, we are led to

_{H}**conclude that: The amount of time**

*finally**t*

_{H}**the amount of time**

*must ALWAYS be*__greater__than*t*. And why is this so? Because while their respective

_{V}*numerators*,

**of which equal**

*both**2h*/

*c*, are the

**,**

*same**t*'s

_{H}*denominator*, which equals

*1 - v*,

^{2}/ c^{2}

*must always be*__less__than*t*'s

_{V}*denominator*, which equals

*sqrt(1 - v*, as we have just proven.

^{2}/ c^{2})**from a**

*Then**purely mathematical*standpoint, a

__lesser__*means a*

**denominator****, hence our final conclusion that**

__greater__quotient*t*>

_{H}*t*.

_{V}And so at long last, **we** can make the following

**Because**

__prediction__:*the amount of time that it takes*1

**Clock H**to complete*tick*, which is

*t*, is

_{H}

*always*__greater__than*the amount of time that it takes*1

**Clock V**to complete*tick*, which is

*t*– with

_{V}**amounts of time being measured from**

*both***point of view, that is, measured by**

*our***using**

*us*

*our**watches – then*

*from*,__our__vantage point standing at the train crossing*,*__we__will observe that Scotty’s Clock H is ticking at a__slower__rate than his Clock V! And therefore*as a direct result of Clock H ticking more slowly than Clock V*,__we__will observe that Clock H__falls out of its original synchronization__with Clock V!## Notes:

[1] We bring to mind that we derived the equation

as part of our more general derivation of the *relativistic time dilation factor*, in our discussion about time dilation. As we know, the above equation for *t _{V}* is the equation for the amount of time, as observed from

*point of view, that is, as measured by*

**our****using**

*us***watches, that it takes Scotty’s Clock V to complete 1 tick. It is now worth mentioning that this equation can**

*our**also*be derived by directly using our newfound knowledge of the

*relativistic time dilation phenomenon*. To do so, we once again put ourselves into

**fragrantly smelling shoes. Now, remembering that**

*Scotty’s*then from ** Scotty’s** point of view, that is, as measured by

**using**

*him***watches, the amount of time that it takes his Clock V to**

*his***is therefore equal to**

*complete*1*tick**2h / c*instead. And why is this the case? Well, because Scotty has every right to claim that it is

**that are the ones**

*he and his train***, while**

*at rest***are the ones**

*we and the rest of the outside world***past him at the relative speed**

*moving**v*. Therefore, as observed from

**point of view, Clock V’s photon travels**

*Scotty’s*

*only**up and down*in the

*vertical direction,*thus traversing a

*total distance*of only

*2h*at the

*to complete 1 tick.*

**universally constant**speed of light c,Okay, let’s switch back to ** our** point of view. To us,

**are the ones**

*we***, while**

*at rest***are the ones**

*Scotty and his train***past us at the relative speed**

*moving**v*. Hence, due to the

*time dilation phenomenon*,

**will observe that**

*we***are all ticking at a**

*Scotty’s moving clocks***rate than**

*slower***(which are, to**

*our stationary clocks**us*, of course ticking at the

*normal*rate of time passage), by a factor of

which is the *relativistic time dilation factor*. Thus, from ** our** point of view, if

*we*observe that

*Scotty’s moving*photon

*Clock V*ticks Δ

*t*times,

*we*will observe that

*our stationary*photon clock ticks a corresponding

times, which must ** necessarily be greater** than Δ

*t*, because his clocks are ticking

*more slowly*than ours are (and also, because we already know that from a strictly algebraic standpoint, the value of the quantity

*sqrt(1 - v*is

^{2}/ c^{2})*always less than 1*). And therefore, if

**similarly observe an amount of time of**

*we**2h / c*elapse on

**(more**

*Scotty’s**slowly*ticking)

*moving*clocks – which is the amount of time, as measured from

**point of view, that it takes Clock V to complete 1 tick – then**

*his***must observe a**

*we***amount of time of**

*corresponding*__greater__elapse on *our** *(*normally* ticking)* stationary* clocks. And so, we clearly see that this last expression is *exactly equal* to our original expression for the amount of time *t _{V}*, which is the amount of time – as observed from

*point of view, that is, as measured by*

**our****using**

*us***watches – that it takes Scotty’s Clock V to complete 1 tick!**

*our*[2] Since the right hand mirror is moving to the right at the speed *v* relative to ** us**, and the photon is also moving to the right at the speed of light

*c*relative to

**(and to**

*us***), then the rate, or**

__everything else__!*speed*at which the

*distance between them**is*

*decreasing*must equal

*c – v*, as measured from

**point of view. And since the**

*our***distance between the mirror and the photon is equal to**

*initial**h*, then the

**that it takes the photon to catch up with and strike the right hand mirror, as measured from**

*amount of time***point of view, must be equal to**

*our**h / (c - v)*, since

*TIME = DISTANCE / SPEED*. This is an equivalent way of obtaining the equation for

*t*._{1}[3] Since the left hand mirror is moving to the right at the speed *v* relative to ** us**, but the photon is moving to the left at the speed of light

*c*relative to

**(and to**

*us***), then the rate, or**

__everything else__!*speed*at which the

*distance between them**is*

*decreasing*must equal

*c + v*, as measured from

**point of view. And since the**

*our***distance between the mirror and the photon is equal to**

*initial**h*, then the

**that it takes the photon to meet and strike the left hand mirror, as measured from**

*amount of time***point of view, must be equal to**

*our**h / (c + v)*, since

*TIME = DISTANCE / SPEED*. This is an equivalent way of obtaining the equation for

*t*.

_{2}[4] Actually, *any* real number greater than zero *always* has, in fact, *two* square roots – a *positive* square root and a *negative* square root. Thus, from our example, the number 0.81 has two square roots, specifically, +0.90 *and* -0.90, because 0.90 x 0.90 = 0.81, and (-0.90) x (-0.90) = 0.81 as well. However, for the level of our discussions, we will simply ignore, or discard the *negative* square root, since it has no “physical” connotation.

## Continue To Part 2

- The Lorentz Contraction, or, How Motion Affects Space (Part 2)

The 2nd in a series of articles on Albert Einstein's Special Theory of Relativity

## Go Back To "Does Motion Affect The Speed At Which Time Passes? (Part 1)"

- Does Motion Affect The Speed At Which Time Passes? (Part 1)

The 1st in a series of articles on Albert Einstein’s Special Theory of Relativity

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