Meet the Beatles!

  Here’s the opening blog of a quartet.  The theme: giving you a better understanding of the behavior of the most familiar objects in our sky: The Sun, the Moon, the brightest “star”, Venus, and the constellations that make up the zodiac.  First up, though: The Sun.

As with most of my generation, I grew up listening to Beatles music.  “Help!” was the first album my older sister ever bought, and we proceeded to almost wear the vinyl out on her mid-60’s record player.  Surprisingly, though, I never paid much attention to Beatles lyrics; I suppose I thought they were largely nonsensical (though frequently clever) plays on words that John Lennon created for his own amusement.

My inattention to Beatles lyrics extended even to McCartney classics like The Fool on the Hill, even though I think the melody is absolutely gorgeous.   Sing it with me, won’t you?

Day after Day
Alone on a hill
The man with the foolish grin is keeping perfectly still
And nobody wants to know him
They can see that he's just a fool
And he never gives an answer

But the fool on the hill
Sees the sun going down
but the eyes in his head
see the world spinning round....

Ooh, ooh,
Round and round and round and…

Just typical Beatles nonsense, right? Certainly nothing too profound.  At least, that's what I used to think.

But that was before I started what has become almost a weekday ritual in my life.  I happen to live right beside Steven's Creek Trail, which parallels the section of a creek that flows out of the Santa Cruz Mountains and into the south end of San Francisco Bay. Most days after work, I combine walking and running along the trail for about two miles out and back.  The exercise is great, of course, but it also gives me a daily chance to think and ruminate about topics to post on the Whabblog.  

As great good luck would have it, Steven’s Creek (and the trail) is oriented almost exactly north-south, at least along my portion.  That chance alignment means that I’m facing virtually due north during the 20 minutes or so that I’m running away from my place, and virtually due south on the way back.

What does this have to do with "The Fool on the Hill"? Well, the chance alignment of the creek with the Earth's axis affords an extended opportunity for me to visualize things.  Astronomical things, quite often, like the relationship between the Earth and nearby objects like the Sun, Moon, and even Venus. 

The opportunity to visualize comes in handy.  Take the behavior of the Sun, for instance.  Every day, the Sun rises at some point along the eastern horizon (somewhere off to your left, if you stand outside and face due south, like I do while I’m running back to “McCann Manor”).  In a pattern that’s as familiar to us as the back of our hands, the Sun moves steadily across the sky until finally disappearing somewhere off to our right at sunset.  

Of course, intellectually, we all know that the movement of the Sun is just an illusion.  It doesn’t really “rise” in the East, or “move” across the sky, or “set” in the West.   Instead, the Sun’s apparent motion is brought about by the fact that the Earth spins on its axis, completing one full rotation every day.

But it's one thing to SAY that the movement of the sun is an illusion; it's quite another to fully “grok” the reality behind the illusion.  What I mean is, it’s actually quite difficult to square sunrise, sunset, and everything in between with the reality that the Sun doesn’t budge one inch.   To smash the illusion, imagine yourself outside, facing due south, right at sunrise. Imagine further that the Earth is spinning in an easterly direction, which forces you (and everything around you) to move continuously to your left.  Visualize that spin literally forcing the Sun (the stationary object) to move in the opposite direction and track across the sky to the west.  Eventually, the continual steady spin pushes the Sun all the way to the western horizon, and then past it.

Now comes the fun part.  Just because the Sun has set, don’t stop now.   Continue to imagine you rotating in that easterly (leftward) direction.  What effect does the rotation have on the position of the Sun after it sets?

We’ve seen that when the Sun was “up”, your local direction of movement was “pushing” it ever further to the west.  Once the Sun sets sets, though, that leftward spin starts to “pull” the Sun back toward you; that is, it now “pulls” the Sun in the same direction that you’re spinning.  Expressed another way, your rotation is yanking the Sun ever closer to the eastern horizon, with sunrise as the inevitable result.

So you think you have this concept nailed? Here's the ultimate test.  Stay up to about midnight, go outside, and face due south. Then, even thought you can’t see it, point exactly to where the sun is.  If you get it right, you’ll be pointing straight down into the ground below you, at a position roughly aligned with the North-South meridian. If you stayed up later, you'd still be pointing into the ground, but at a location that's starting to slide off to your left.   Are you with me, here?  Stay up even later, and eventually you'd be pointing to a position way to your left, just below the eastern horizon. And, then: presto! The constant eastward spin of the Earth forces the Sun back into view along the eastern horizon, and you’re pointing at the sunrise.

This ability, to imagine where the Sun is and how it behaves at night, is the key to abolishing the illusion of a stationary Earth and a moving Sun.  Of course, you don't actually have to go outside, or wait until midnight (or later) to visualize it. Try, instead, going out right at sunset, the next time the sky is clear, and face south. Then imagine you are suddenly spinning to the left (east) at much faster rate than the speed that the Earth actually rotates. What happens to the Sun?  It’s actually quite easy to imagine it racing around that circle through the ground beneath you, closing in quickly on the point that intersects the location on the eastern horizon where the Sun rises.

And then you'll have done it, patient reader: You'll have matched the observational capabilities of the Fool on the Hill!  Day after day, for the rest of your life, the eyes in YOUR head will see, not a moving ball of fire, but a world (our world) continuously spinning round and round and round. And YOU will have a gut-level understanding of a truth about the relationship between the Earth and the Sun that eluded humankind for almost the entire time that we’ve existed on the planet. 

The form of visualization exercise I’ve asked you to engage in will pay dividends in the rest of the blogs in this series, too.  In the meantime, let’s sit back and appreciate the genius of the Beatles’ lyrics, in addition to their melodies.

The End of the Beginning

Have we reached the tortured end? Yep, almost (at least for the moment…. hehehehe).  But before I leave the topic of time dilation and special relativity, I’d like to address time dilation in the real world, as opposed to the imaginary world of Einstein-style thought experiments featuring vaguely vagina-shaped spacecraft carrying excited male celebrities. 

Did you know that, every second of every day, subatomic particles, mostly protons, strike our atmosphere?  You may have heard them referred to as cosmic rays.  They are most definitely not rays, though; protons are bona fide particles with bona fide masses.  And despite the name, cosmic ray, they’re hardly exotic; there are one or more protons inside every atom in your body, and inside every atom in the universe. 

As cosmic rays, though, protons travel to us over phenomenal distances often after being spun up by black holes to about 0.99 of the speed of light.  When these naked protons (by naked, I mean they’re not part of an atom) strike the gas molecules in our upper atmosphere, about 15 kilometers above the ground, they immediately decay.  One of the byproducts of proton decay is a particle called a muon, which continues through the atmosphere in roughly the same path as the originating proton at roughly that same high rate of speed.  Now, one thing about muons is that they’re very short-lived particles.  They take an average of just two-millionths of a second to decay into other particles. That’s just an average, mind you.  Some muons survive for a little more time than two-millionths of a second, while others survive for a little less. 

At a velocity of about 0.99 the speed of light, or 297,000 kilometers per second, how far do muons travel through the atmosphere before they decay?

Distance equals speed multiplied by time.  In the muon’s frame of reference, 297,000 kilometers multiplied by two-millionths of a second (their average time of existence) yields roughly 0.6 of a kilometer.  This means that the average muon doesn’t even come close to covering the distance between the top of the atmosphere, where it was created, and the ground, before it decays.  Even the muons that beat the average, and survive for a little longer than two-millionths of a second, don’t make it.

That’s a good thing, because muons are radioactive.  Exposure to them can cause cancer.  But, is it really the case that muons never travel far enough to cause us any harm?  So far, our discussion about how long a muon exists, and how far it travels through the atmosphere, has been from the muon’s frame of reference.  In our (the Earth’s) frame of reference, which is at rest compared to the muons, we’ve got to factor in the extra distance that the muon is going to cover due to time dilation.  For a muon traveling at 0.99 light speed, time is stretched by a factor of about four.  In our frame of reference, therefore, instead of existing (and traveling) for about two-millionths of a second, the average muon exists for about eight-millionths of a second.  That is enough time to traverse about two and a half kilometers before decaying.

But that’s just the average muon.  For muons that decay more slowly than average, time dilation makes just enough of a difference that they do manage to reach the surface, and maybe even hit you or I.  It’s not a lot of muons; statistically, only about four out of every hundred last long enough for the combination of their relative longevity, compared to other muons, and time dilation, to make it all the way down.

From our earlier discussions, you’ll recall that, for the object traveling quickly, time is locally the same as it is for you, and the phenomenon of time dilation is manifest as space compression.  In the muon’s frame of reference, it lasts exactly as long as it should (on average, only two-millionths of a second).  Thus, the (statistically) most durable muons manage to make it all the way through the atmosphere, not because time dilation extends their lifetime, but because the distance between the top of the atmosphere and the ground is compressed by a factor of four.  

 

You could verify the presence of muons reaching the surface with a simple Geiger counter.  Being radioactive little beasties, some of the clicks you would hear when you turned on the counter would be in response to their presence.  If you climbed a mountain, and turned the Geiger counter on up there, you would hear more frequent clicks, because more muons make it that far.

The bottom line: Time dilation is real, influencing real events in our real lives. And that’s a wrap on the topic, at least for now.  I just have to “close the loop” between time dilation and my late friend Karl, my original motivation to research and write these blogs.  Frankly, the connection is kind of a downer.  I’ve had four different surgeries in the last couple of years for skin cancer.  Would I have escaped the cancer scourge if not for time dilation?  I don’t know.  Karl died of complications due to surgery necessitated by the strong suspicion that he had colon cancer.  I wonder if he would be alive today, too, if the phenomenon didn’t exist.

Ah, well.  Time dilation is built into the very fabric of our reality, so there’s no point in wishing.  I just hope you’ve enjoyed finding out what it’s all about.  I know Karl would have!

 

A Baseball... and a Terrible Accident

We’ve arrived at the penultimate time dilation blog, faithful readers!  Having explained the nuts and bolts of the phenomenon already, this time I’m going to feast on some of the implications of the effect, such as the solution to the conundrum I posed yesterday.  Before we get to that, though, I’d like to take care of an important piece of unfinished business. I want to be sure you understand that time dilation is not confined to the duration of events involving light pulses; it occurs for any and all events on board Shirley. 

What is an “event”, when you really think about it?  My definition is pretty simple: anything that has a beginning and an end, and takes a measurable amount of time to occur, qualifies as an event.  We’ve discovered that when an event occurs in a frame of reference that’s moving relative to you, during the time that separates the beginning of the event from its end, the objects involved travel farther in your frame of reference than in theirs.  Since Shirley is moving at the same speed in both frames of reference, the only way that she can possibly cover more space for you than for George is for the event to take more time.

To appreciate the point further, let’s return to yesterday’s version of the thought experiment.   Shirley was revved up to very close to the speed of light, and the critical event (the light flash traveling up and down her shaft) took almost 13 minutes (as measured by you).  Now, suppose you and George decide to do something completely different.  Ignoring the searchlight, George stands up on Shirley’s floor with a baseball in his hand.  At just the point where Shirley passes over you (still traveling at the same 299,999 kilometers per second) George lofts the baseball straight up in the air, with just enough speed that it travels upwards for exactly one second, and back down for another second.

Suppose also that a closed-circuit television lets you watch the baseball go up and down from your location on the ground.  What would you see?  Well, the television signal is a form of light that also travels at 300,000 kilometers per second.  Since the television signal is beaming directly from the floor of the spacecraft to your television set, what you see is determined by how far the signal has to travel, which, in turn, is dependent on how far away Shirley is when the images are emitted (these days, about 120 images are emitted every second).   We saw last time that in two seconds, extra space is being created at a furious rate by Shirley’s high speed (the distance along Line B is stretching out quickly).  This means that each successive television image of the baseball has to travel a longer and longer distance, which means a successively longer delay between when the onboard television camera records the image of the baseball, and when you see that image on your screen.  The net result is that you would see the ball moving very slowly upwards, as if you were watching film of the moving baseball in slow motion.  As time continued to pass, the movement of the ball would slow down more and more, until it became imperceptible.

The baseball wouldn’t stop moving entirely, though, and if you took a break and went to the bathroom, when you came back you would see that the ball had shifted position.  13 minutes after George released the baseball, you would finally see it return to his hand, and the event would be over.

Actually, in 13 minutes you could take a lot of breaks from watching your television.  You could go to the bathroom, have a brief conversation with your neighbor, watch the Kentucky Derby, and read a Whabblog.  Meanwhile, George would have no time to do any of these things; he would be fully occupied by throwing the ball up and catching it almost immediately.  A text from him to you might say something like: “Really, really rushed!  Just barely had time to throw the ball up before I had to catch it again!”

Not so for you, and not so for everyone sharing your frame of reference (which is everybody on the Earth).  Imagine all the things that happen across the world in 13 minutes.  Thousands of people die; thousands more are born; thousands exchange wedding vows; thousands get notices of a hiring or a firing; and there’s a distinct possibility that a big natural disaster like an earthquake occurs (and the first reports about it came in on CNN).  In short, a busy little chunk of everyday life, full of scores of individual events, passes by.

What would happen if you didn’t stop Shirley after the ball-throwing task (or the light pulse measurement task)?  Suppose you just let her continue moving at the same speed for, say, two weeks of ship time? You would just go about your normal daily activities during that period, and so would George onboard.  What would your texts to each other look like at the end of the period?

When you and George were doing the light-pulse measurement, it took about 775 seconds for you to see the light pulse return to Shirley’s floor, versus a paltry two seconds for him.  Thus, for every second of time that was passing for George, approximately 387 seconds pass for you.  That same multiplier works for any time scale.  Thus, while only two weeks of time are passing for George, 775 weeks of your life unfold.

775 weeks is almost 15 years!  At the end of George’s two weeks, if you could exchange photos along with your text messages, he would look identical to before; nobody ages noticeably in only two weeks!  But you… you would look noticeably older; depending on how well he knew you, George might even have trouble recognizing you for a moment.  And imagine all the pages and pages of news you could put in your text message to him, with 15 years worth of your life to draw from!

If we extended Shirley’s trip a little longer, to say a month, almost 30 years of your life would pass.  While George would hardly have time to age at all, there would be a nontrivial possibility that you would have grown old enough to die.   And in fact, if you cranked Shirley up even more, to more than 299,999 kilometers per second, the time dilation would grow so extreme that in the month George spent on board Shirley, many thousands of years would pass here on Earth.  When he returned, you would be just a distant memory.   Shirley would have turned into a very effective time machine.

I told you time dilation was a really freaky phenomenon, didn’t I?  But there’s one last aspect to it – the answer to yesterday’s paradox – that may be freakiest of all.  Let’s return to our regular thought experiment with the light pulse.  From George’s point of view, in the two seconds it took for the light pulse to go up and down Shirley’s cylinder, she covered exactly 599,998 kilometers.  That’s not a trivial amount, granted: it’s about 1.5 times the distance from the Earth to the Moon.  But 600,000 kilometers or so is utterly insignificant from your point of view, because for you, Shirley has traveled 236 million kilometers, about a third of the way to the planet Jupiter!  By virtue of traveling so far, she and George could well have collided with an asteroid, ending both the time measurement experiment and poor George’s life.  Meanwhile, from George’s perspective, nothing of the sort would have happened.

One afternoon last spring, while pondering this conundrum during an afternoon run on Stevens’ Creek trail, I had an “aha” experience, and the last piece of the time dilation puzzle finally fell into place.  The light pulse moving up and down Shirley’s shaft is the same event for you and George; therefore, it has to have the same history in both frames of reference.  So how do we get around the “asteroid collision” paradox?  There’s only one way.  Shirley, George, and the light pulse have to be at the identical location in the solar system at every point along the pulse’s journey, both in your frame of reference, and in theirs’.  That way, if George and Shirley meet with an asteroid along the way, they do so in both frames of reference.

Let’s assume George and Shirley have the good fortune to avoid any asteroids, and the light pulse reaches the floor of Shirley’s shaft quite safely.  The speed of light is constant, so according to Shirley’s odometer, she has to have moved exactly 599,998 kilometers to your right when the light flash reaches the floor.  The only way to reconcile that fact with Shirley being all the way out in the asteroid belt is if, from George’s point of view, space itself is compressed – literally, scrunched - in the direction of Shirley’s movement.  That is the simple, but astonishing truth, folks: Space actually shrinks along Shirley’s direction of motion, by exactly the same factor that George’s time expands for you.  In other words, at a speed of 299,999 kilometers per second, everything in Shirley’s path, including her final destination out in the asteroid belt, becomes 387 times closer than it is for you here on Earth.  That’s why, from George’s perspective, it only takes two seconds to reach it!

We just saw that if you are part of a frame of reference that moves at a sufficiently high rate of speed relative to the Earth, you become a time traveler, able to take a very short trip and yet return many years in the future.  Spatial compression is the amazing flip side to this phenomenon; it means that, within your lifetime, you could travel to very remote destinations in the universe, including other stars and even other galaxies, which to us on Earth are so far away that they remain forever out of reach. 

You could argue that it’s totally absurd to have gone to all the trouble of exploring and explaining a phenomenon that never actually happens, because nothing ever goes that fast.  Not true, however.  There are actually things in our universe that move at close to the speed of light, and experience significant time dilation.  I’ll reveal what those things are in tomorrow’s blog.

 

To the Moon, Alice! And beyond!

In the third (and final) version of our little thought experiment, we’re going to add a few “dilithium crystals" to Shirley’s engines so that we can crank her up to 299,999 kilometers per second, just one kilometer below the speed of light.  By now, you’re familiar with the drill.  At the exact instant Shirley passes over you, George flicks the searchlight on and off.  Both you and George measure the time it takes for the light pulse to travel up and down Shirley’s shaft.  You record your times, and repeat the exercise over and over, until you both get a stable average.  Luckily, your stopwatches are smart, so they help by automatically subtracting your brain processing time, and in addition, your stopwatch automatically subtracts the time for the pulse to travel from Shirley’s transparent floor back to your eyes.

When Shirley was going “only” 150,000 kilometers a second, your stopwatch registered about three tenths of a second more for the light pulse to complete its journey than George’s did.  Now, with Shirley moving almost twice as fast, a perfectly reasonable conjecture would be that your stopwatch should register a delay that’s about twice as big, or something in the neighborhood of 0.6 seconds.  Consequently, when you and George start the first measurement, you go in half expecting to see the light pulse still arriving only a little later than he does.

But I’m afraid that doesn’t happen. You don’t see the flash after 2.6 seconds.  You don’t see it after twenty-six seconds.  You don’t even see it after two hundred and fifty seconds!  As the time keeps mounting, you begin to panic, worried that you’ve somehow missed the flash and spoiled the experiment.  Just as you’re ready to call the whole thing quits, lo and behold, the flash finally arrives!  Breathlessly, you look down at your stopwatch, only to see that seven hundred and seventy-five seconds have passed.  Almost 13 minutes! 

13 minutes?  For the first time since you started this whole crazy time measurement gig, you scarcely believe the outcome of your own experiment.  How can a two second event be taking 13 whole minutes?  Your sense of disbelief persists as you and George perform the measurement again and again, only to confirm: two seconds for him; hundreds of seconds for you.

To make sense of this result, we have to go back to – where else – Pythagoras. In the one second it takes the light flash to go from the searchlight to Shirley’s mirrored ceiling, in George’s world, Shirley has been racing away along the line stretching off to your right at 299,999 kilometers per second.  Geometrically, Shirley’s movement has constructed a Line B that is 299,999 kilometers long.  Eager to find out how much distance 299,999 kilometers adds to the length of the all-important Line A (the line that traces out the path the light pulse takes from your perspective), you frantically crunch through the familiar Pythagorean calculations.  Once again, I’ll spare you the numerical details: Line A turns out to be 424,263 kilometers long.  That’s fully 124,263 kilometers longer than Line C, which is equivalent to the path that the pulse takes for George.  The resulting triangle is shown in the figure right below.

After one second of George’s time, his light pulse has covered 300,000 kilometers, enough distance to reach the ceiling.  After one second of your time, the light pulse has also covered 300,000 kilometers (light travels at the same speed for everybody), but since it has to travel up the diagonal of the right-angle triangle, it has 124,263 extra kilometers still to cover. 

That’s a lot of extra distance: About 10 times the diameter of the Earth, in fact.  Even traveling at 300,000 kilometers per second, 124,263 kilometers is going to take the pulse almost half a second - 0 .414 seconds to be exact - to traverse.  In 0.414 seconds, though, Shirley, zipping along at that blinding 299,999 kilometers per second, moves an additional 124,262 kilometers down the line (0.414 times 299,999), extending Line B by the same amount.  By the Pythagorean theorem, that, in turn, lengthens Line A by 95,000 kilometers or so. 

Just as it was when Shirley was going 150,000 km per second, successive cycles of space/time creation have begun.  The extra time needed for your light pulse to cover that additional 95,000 kilometers along Line A is about 0.317 seconds.  But in 0.317 seconds, Shirley moves 80,000 kilometers further along.  These numbers are pretty big. Moreover, exactly the same-sized cycles will be created by Shirley’s movement during the one second of George’s time that the light pulse is traveling back down Shirley’s shaft, stretching Line A for the mirror-image triangle by identical amounts (I’ve illustrated that in the figure above, too).   So, it’s already clear that the time you’re going to measure for the light flash to reach Shirley’s floor will be larger - quite a bit larger – than when Shirley was traveling at “just” half the speed of light.

Still, just as we saw at that slower speed, the numbers in each cycle are shrinking rapidly.   You would be forgiven for thinking that the cycles would, again, collapse fairly quickly to insignificance.  However, you’d be wrong!   As the cycles start to pile on, an important aspect of the geometry of the situation begins to exert an ever-increasing influence.  To understand that aspect, I’m afraid we need to revisit the details of the Pythagorean theorem (yeah, I know.  Back to math.  Sorry about that).

Recall that according to the theorem (shown again above), to get the length of Line A, you first square the lengths of Line B and Line C.  Then, you add the squared values together.  Finally, you take the square root of the sum.  Remember when Shirley was moving only one kilometer per second, and I talked about the relative contributions that Line B and Line C make to the overall length of Line A?  Remember how I took pains to emphasize that squaring the lengths of Lines B and C, and then adding them together, has the effect of inflating any initial difference between them when it comes to how much they donate to the length of Line A?  Simply stated, when the lengths of Lines B and C are very discrepant, with one relatively long and the other relatively short, squaring the numbers before adding them together ensures that the length of Line A is almost completely controlled by the length of the longer side. 

 The figure to the right once again illustrates the geometry of the situation back when Shirley was moving only one kilometer per second (slow relative to the speed of light) making Line B vastly shorter than Line C.   With Line C so much more in control of the length of Line A than Line B, the math “forced” Line A to be virtually the same length as Line C.  Remember, too, that I took some time out to illustrate the case of a triangle where Lines B and C are of equal lengths?  In that case, their squared values are equal too, and the fact that you simply sum their squared values means that they start donating equal amounts of their own lengths to the length of Line A.  

That’s the situation we’re in right now.  With Shirley moving at 299,999 kilometers per second, after just one second Line B and Line C are virtually the same length (Line B is just one kilometer shorter).  Consequently, Line C is supplying just a tiny trifle more than 71% of its length to Line A, and Line B just a tiny trifle less.

In the very next cycle of time/space creation, though, Line B increases by another 124,262 kilometers, making it considerably longer than Line C.  As the space/time creation cycles pile on, Line B grows ever longer, while Line C (of course) stays the same.  Consequently, it’s Line B that begins to control a larger and larger proportion of the overall length of Line A.   Critically, this means that Line B also starts to donate a larger and larger proportion of the portion of its length that was just added in the latest space/time creation cycle.  

The top triangle of the three triangles in the figure below illustrates how the situation has evolved by the 20^th cycle of space/time creation, close to the number of cycles that completely “closed things out” when Shirley traveled at only 150,000 kilometers per second.  Shirley has moved almost 1.5 million kilometers down the line, making Line B almost five times as long as Line C.   In the very next (21^st) cycle, illustrated in the middle triangle of the figure, Shirley moves an additional 30,941 kilometers down the line.  Crucially, with the big (and growing) imbalance between the lengths of Line B and Line C, the vast majority of that additional length is donated directly to Line A.  Specifically, Line A lengthens by about 30,303 kilometers, only a fraction less than the increase in Line B itself.

It takes just slightly more than a tenth of a second for the light flash to cover that extra 30,000 or so kilometers.  In that amount of time, Shirley moves another 29,999 or so kilometers away, and again, virtually all of that additional length is donated to Line A. As the cycles of space/time creation continue to mount, and Line B grows ever longer compared to Line C, the proportion of the additional length of Line B that’s donated to A grows every larger too, until they’re virtually identical.  The result is illustrated in the bottom triangle of the figure above by the relative spacing between the large number of light pulses along the diagonals (those spacings are obviously not completely accurate; they are meant to just illustrate the general point).  Eventually, things almost settle into perfect equilibrium, where on each cycle, almost as much new distance is added to Line A as was added on the immediately previous cycle.  As a result, on each cycle almost as much new space is created for the light pulse to have to cover in the future as it just covered in the past.   Even though it’s traveling at 300,000 kilometers per second, the light pulse makes very little headway in its journey toward the top of the triangle (Shirley’s mirrored ceiling), or (after one second of George’s time) in its journey back down to the floor. 

Luckily for your patience, though, the amount of Line B that gets donated to Line A never - quite - equalizes.  On every successive cycle, the temporal window of opportunity for Shirley to move further down the line grows just a little bit smaller, and the old girl covers just slightly less additional distance compared to the cycle before. 

Eventually, the cycle IS choked off, and the light pulse DOES reach the ceiling (floor).   As shown by the pile-up of light pulses toward the end, though, it takes many, many, many extra cycles for that to happen.  The final Line B distance says it all: by the time the pulse reaches the ceiling, Line B has stretched to over 116 million kilometers, and by the time the pulse reaches the floor, Shirley has covered twice that distance, for a total of 232 million kilometers (in your frame of reference), an impressively large fraction of the distance to Jupiter! 

Compared to that colossal distance, the length of line C, which remains fixed at 300,000 kilometers, pales into insignificance.  This is why the shape of the triangle is so squashed, and why Line A is now almost identical in length to Line B (of course, the true triangle is a great deal more squashed than this figure can do justice).

Let’s pause here to take stock.  We have exactly the opposite geometry of the situation we found when the spacecraft was traveling at only one kilometer per second, and it was Line B whose length was insignificant compared to Line C.  See, now, why time dilation is magnified by such a colossal extent when the spacecraft gets very close to the speed of light?  Line B has an opportunity to get so long compared to C that virtually all of the length of Line B is donated to A; thus, as Line B grows, so grows Line A, creating ever more distance for your version of the light flash to have to cover.

And, of course, light always takes more time to cover more distance.

With this description, we’ve virtually finished the quest to understand the “time” in time dilation!  The key to the entire phenomenon lies in how quickly the spacecraft is traveling, and how quickly the light pulse can reach the floor of the spacecraft.  If the pulse can do so in a relatively small number of extra cycles of space/time creation, Line C will always win the competition with Line B for who donates the bigger proportion of their length to Line A, and things won’t get out of hand.  If, however, Shirley is moving fast enough, Line B wins the “who donates the most to Line A” sweepstakes, and the time/space creation cycles acquire a life of their own.

Believe it or not, there are only a few loose ends to tie up, now.  One of those ends is really nifty, however, so I’m going to leave you today with a teaser to it.  We’ve seen that, in your frame of reference, when Shirley travels at very close to light speed, she covers over 230 million kilometers before the light pulse returns to her floor.  That distance puts her way out in outer space: in the context of our solar system, 230 million kilometers away marks a location well inside the asteroid belt that lies between Mars and Jupiter.

Meanwhile, what distance would the odometer onboard Shirley herself read when the pulse reaches the floor?  599,998 kilometers, of course: Her speed, 299,999 kilometers per second, multiplied by the two seconds of elapsed time that it takes for George to see the flash.  But that is only about one and a half times the distance to the Moon, and is many millions of kilometers short of Mars, let alone the asteroid belt.  So. At the exact point in time when the light pulse returns to the spacecraft floor, time dilation appears to have created an enormous discrepancy between where Shirley is located in George’s world, compared to where she is in your world.

This discrepancy raises a major conundrum.  Since, from your perspective, Shirley travels all the way into the asteroid belt, it is entirely possible (though unlikely) that an asteroid lies somewhere along her path, and Shirley actually hits the asteroid in a high-speed collision that immediately pulverizes her, and poor George, into dust.  If that happened, you’d obviously never see the light pulse, because there’d be no glass floor around to reflect it back to you.  You wouldn’t care much about that, because you’d be mourning the loss of poor dedicated George.  Meanwhile, from George’s perspective, Shirley doesn’t travel nearly far enough to even enter the asteroid belt.  The high-speed collision never takes place, the light pulse reaches the floor with no problem, and George lives to record the time of that event.

Can exactly the same event have two such different histories? How can George both live and die?  How can Shirley get pulverized and not get pulverized? I invite you to speculate on how to resolve this paradox in today’s comment section!  Or, if you like, just sit tight and wait for me to provide the (in my humble opinion) quite mind-boggling answer! 

The Mystery of the Missing Time

All right, faithful readers!  Time’s a wasting, and we have some very important business to take care of today. You and George have been independently measuring the time for the light flash to travel up and down Shirley’s shaft while Shirley is traveling 150,000 kilometers per second (half the speed of light).  I captured the geometry in the illustration in the last blog, and I show it again in the left side of the figure below.  From your stationary point of view on the ground, in one second Shirley’s movement creates a Line B that’s 150,000 kilometers long.  Her motion thus forces your version of the light pulse to have to climb up and down the hypotenuse of the two adjoining right-angle triangles.  From Pythagoras, when Line B is 150,000 kilometers long, it adds almost 36,000 extra kilometers to Line A.  Adding the same amount to Line A on the way back down (the hypotenuse of the mirror-image triangle), a total of just less than 72,000 kilometers of extra distance is created. 

At the speed that light travels, that extra distance should have added about 0.236 extra seconds to your base measurement of two seconds.  Instead, though, your average measurement was around 2.301 seconds, or about seven tenths of a second longer than it should be. 

Not a big deal, you say?  Don’t count on it!  To understand the significance of this seemingly trivial discrepancy, I’ll repeat the straightforward question I posed at the end of the previous blog.  During the extra 0.236 seconds that the light pulse takes to cover the almost 72,000 extra kilometers along the two diagonals, what is Shirley doing?  Answer: She’s continuing to move away from you at the same 150,000 kilometers per second! 

But wouldn’t this constant movement make Line B even longer? 

Indeed it would!  In the 0.118 seconds it takes the light pulse to cover the extra 36,000 or so kilometers up the first hypotenuse, Shirley moves about 17,000 kilometers further away, stretching Line B from 150,000 kilometers, the length it is in the triangles on the left side of the figure above, to 167,000 kilometers, the length in the triangles on the right side of the figure (of course, Shirley is moving at the same speed while the light flash is going up to the ceiling as she is when the flash is going down, so her motion stretches Line B by the same amount while the light pulse is going in both directions).

From Pythagoras’ theorem, any time you add length to Line B, you’re adding length to Line A.  Doing the math (I’ll spare you the details), lengthening Line B by 17,000 kilometers adds roughly 8000 kilometers to Line A; this is the source of the number 8 at top of Line A on the “uphill” triangle.  But hold on a minute!  Isn’t your version of the light pulse going to have to take even more time to cover that additional 8000 kilometers?  Sure.  Since light travels so fast, its not much time, about 0.028 seconds to be precise (the time above the “8” in the figure).  Still, even that little sliver of additional time is enough for Shirley to slide an extra 4,000 kilometers to your right, which is also added directly to the length of Line B.  I haven’t shown this in the figure, but 4000 extra kilometers along Line B lengthens that all-important Line A by a little under 2000 kilometers, which takes the light pulse an additional 0.0066 seconds to cover, which allows Shirley to move even a little further away…  and so on.  

See the pattern?  Each additional increment in distance along Line A increases the travel time for your light pulse.  That increase in travel time creates an additional temporal “window of opportunity” for Shirley to move even further away, and further lengthen Line B.  The bottom line: your spacecraft is now traveling fast enough to set up repeated cycles of space and time creation!

At the same time, though, these cycles are shrinking very rapidly. Within only about 18 cycles in total (each one smaller than the last), the amount of additional distance being added to Line A is driven to virtually zero, choking off any opportunity for Shirley to move further away.  The whole space/time creation thing comes to a screeching halt and, even from your perspective, the light pulse reaches the ceiling.  Then, the whole routine of space/time creation is repeated while the light pulse comes back down the shaft.  Eventually, even in your frame of reference, the pulse reaches the floor, gets reflected back to your eyes, and you hit your stopwatch.

Adding up all the additions to the length of Line B donated by Shirley’s movement from the point of departure (which corresponds to the time when she passed directly over you, and you and George both started your stopwatches) to the return of the pulse to her floor yields a final length of about 173,000 kilometers for Line B, which adds a total of about 46,000 kilometers to Line A.  Multiplying 46,000 kilometers by two gives the total increase in the distance your version of the light pulse has to travel, compared to George’s version, of 92,000 kilometers.

How much extra time does it take light to cover that much extra distance?  You guessed it… approximately 0.301 extra seconds (actually, it’s .306666 seconds; I’m rounding to keep things straightforward).  In other words, exactly the amount of additional time you measured on your stopwatch for the pulse to complete its journey, over and above the flat two seconds measured by George!

We’re really starting to get somewhere, now, readers! The notion that, from your frame of reference, time and space mutually construct each other over successive cycles is kind of amazing.  And make no mistake: these cycles are perfectly real.  Your stopwatch doesn’t lie about the extra time that the event has taken.  As for space, suppose there was an odometer onboard Shirley that recorded the distance she covered from the point she passed over top of you to the point when the light pulse returned to the floor, two seconds later (as measured by George onboard).   Since Shirley is traveling 150,000 kilometers per second, the onboard odometer would read exactly 300,000 kilometers.  In your frame of reference, though, by the time the light pulse hits the floor, Shirley has traveled almost 50,000 kilometers further.  That’s a lot of extra distance out into space!

When you really think about this, it raises all sorts of puzzles, which I’ll explore in future blogs. To avoid straining your brain right now, you might be forgiven for saying to me: “What’s a few tens of thousands of kilometers, and less than a tenth of a second, among friends”?  True, there’s still not a dramatic discrepancy between your time measurements and those of George. Does that mean time dilation’s not that interesting, after all?  At this point, all I can say is: Hold on to your hats!  In the next blog, we’re going to pull out all the stops and crank up Shirley’s speed to 299,999 kilometers per second, just one kilometer less than the speed of light itself. 

299,999 kilometers per second is virtually twice as fast as Shirley was going in today’s blog.  That extra speed is going to do nothing to George’s measurements, of course.  It’ll be the same old boring two seconds for him.  However, anybody care to speculate what time you’re going to record on your stopwatch?

At Last! Some Time Dilation!

In our last blog, we discovered that when Shirley is moving and you are stationary, in your frame of reference the light pulse has to trace out a path up and down the hypotenuse of two right-angle triangles, and so the total distance traveled is more than in George’s frame of reference, where the pulse simply goes up and down in a straight line.  As Shirley was traveling only 1 kilometer per second, the additional distance was too short for you to reliably measure the increase in time it took for the pulse to cover it.  However, as we left our story, you and George were about to perform the time measurement experiment again, this time with Shirley moving smartly along at fully half the speed of light (150,000 km per second).  After your customary hundred or so repetitions of the experiment, George is climbing the walls with boredom, as he keeps measuring the same two seconds for the light pulse to go up and down Shirley’s shaft.  Your watch, though, is measuring a shade over 2.3 seconds.

Say what? How can the same event be taking different amounts of time?  The difference can’t be blamed on George.  He’s been well trained to toggle the searchlight on and off at the exact point he and Shirley pass over top of you.  Thus, you have supreme confidence that you and he are starting your stopwatches at the same time.  So where is the extra 0.3 seconds and change coming from?

The figure to the right shows where.  In the second it takes the light flash to reach her ceiling, Shirley moves exactly 150,000 kilometers along the line to your right, stretching Line B to 150,000 kilometers in length.  In George’s frame of reference, the pulse only has to go up and down a path that is the equivalent of Line C.  For you, it has to climb up and down the two Line A’s, the hypotenuse of the two right-angle triangles. Looking at the shape of these triangles, it is clear that your  “version” of the light pulse has to cover considerably more distance than George’s version.   

How much more?  Again, let’s work through the standard Pythagorean equation to find out.  Remember, to get the length of line A, you multiply the length of Line B by itself (square it), multiply the length of Line C by itself, add the resultant values together, and take the square root of that summed value. I’ll spare you having to go to your calculator:  150,000 (Line B) squared equals exactly 22 billion, five hundred million kilometers (again, see what happens when you square an already huge number?). We know already that squaring Line C gives 90 billion kilometers.  Adding these two big values together yields 112 billion, five hundred million kilometers.   The square root of that number is just slightly under 336,000 kilometers.  Therefore, both Line A going up and Line A going down are close to 336,000 kilometers long, for a total distance of almost 672,000 kilometers.

Whoa!  That is roughly 72,000 kilometers longer than the 600,000 kilometers the light flash travels from George’s perspective!  Clearly, your version is going to have to take more time to cover that extra distance.  How much more?  Easy.  Light travels at 300,000 kilometers per second, so covering an additional 72,000 or so kilometers takes your pulse about 0.236 seconds of additional time.

The figure illustrates another important way to imagine this situation.  For both you and George, the pulse travels at exactly the same rate: 300,000 kilometers per second.  For George, that’s enough to get all the way to the ceiling in just one second.  For your version of the pulse, traveling up Line A, one second of travel time puts the pulse 300,000 kilometers up the line.  However, as you can see from the position of the pulse along the line, that’s only 88% of the way to the top; the pulse is still 36,000 kilometers short of the ceiling.  Covering that extra distance is simply going to take more time.  There’s no getting around it.

Let’s pause for a minute to take stock.  Two different observers, you on the ground and George onboard Shirley, are measuring the duration of an identical event and finding that it takes different amounts of time to complete!   The general conclusion from your measurements is irrefutable.  When you determine the length of an event that occurs in the frame of reference of a spaceship (or anything else) that’s moving very fast with respect to you, the event covers a longer duration than it does when the event is measured from within the moving frame of reference itself (in this case, measured from “the moving frame of reference itself” means performing the measurement from onboard Shirley).  Mathematically, it’s the increasing length of side “B”, brought about by Shirley’s rapid motion, that is “causing” the whole thing.  The longer Line B becomes, the longer Line A is, and the more distance the light pulse has to cover.

But guess what?  There’s actually a slight problem with this analysis.  Recall that the average of the values you measured from your stopwatch was a little over 2.3 seconds, almost a tenth of a second longer than the 2.236 seconds you should have measured if the light pulse was traveling “only” 72,000 extra kilometers (36,000 km on the way up and 36,000 km on the way down).   You’re understandably quite eager to sweep this discrepancy under the rug, attributing it to just measurement error perhaps, since it’s pretty small.  But I can tell you now, blog readers, this difference is not to be trifled with.  As we’ll discover in the next blog, it holds within it the key to the most mind-blowing aspect of the whole time dilation phenomenon!

Would you care to speculate in today’s comment section about where this discrepancy is coming from, and why it is so important? Or just wait?  If you want to take a shot, here’s a hint in the form of a question: During the 0.236 seconds that it takes the light pulse to cover that extra 72,000 kilometers, what is Shirley doing?

Ladies and Gentlemen: Pythagoras!

  Picking up exactly where we left off, we’ve now solved the problem of how to ensure that you see the light pulse even though, when the pulse makes it back to Shirley’s floor, the old girl is two kilometers down the line to your right.  At risk of belaboring the point, the solution is illustrated (again) in the figure above.   The pulse travels up the cylinder to the top while Shirley travels one kilometer away to the right; it travels back down while she moves another kilometer to the right.  Meanwhile, the glass in Shirley’s floor has reconfigured itself to reflect the light right back to your eyes.   Your stopwatch automatically subtracts the extra time it takes the pulse to cover the distance from Shirley back to you. 

The white arrows inside Shirley, showing the path of the light pulse from George’s perspective, illustrate the other important fact that I briefly touched on last time.  When the pulse of light reaches the floor, in George’s reference frame it has traveled along a perfectly straight path up to the ceiling and back down, just as it did when Shirley wasn’t moving.  This is because George is moving right along with Shirley.  It would be just like if George shone a flashlight on the ceiling of a moving train from within the dining car. Both the flashlight and the dining car are moving together, at the same speed, so the beam from the flashlight would go directly up and down.  Viewed another way, from George’s perspective, Shirley is perfectly stationary. 

From your perspective, though, Shirley is displaced two kilometers when the flash returns to the floor.  That simple geometric fact dictates one of the linchpins behind General Relativity and the whole phenomenon of time dilation: From where you’re sitting (actually, from where you’re lying, since you’re flat on the ground), the light pulse cannot have traveled straight up and down!  To see the path it’s had to take, look at the really narrow triangle on the left side of the figure to the right.  When the light pulse reaches her ceiling, Shirley is physically one kilometer down to your right.  To get to the ceiling, therefore, the pulse HAS to have traveled up the slanted side (the hypotenuse) of a very, very, narrow (very acute) right-angle triangle.  Then, after the pulse is reflected, it has to travel down the hypotenuse of an identical (but mirror-image) triangle to get back to the floor.  As the figure illustrates, the hypotenuse of the two right-angle triangles, which I’m going to label Line A, is ever so slightly longer than the straight line that defines the path that the pulse follows for George.  I’ve labeled the line in the triangle that is the equivalent length to George’s path as Line C.  That leaves the side of the triangle created by Shirley’s motion down the line to your right as Line B.

Now here’s the crux of the issue.  The speed of light is the same everywhere, in all frames of reference.  Since the light pulse has to cover more distance for you than it does for George, the pulse has to take more time to complete its journey.  How much more time is determined by how much longer Line A is than Line C.  Employing some simple grade-school geometry, let’s now compute that distance.

To work it out, we need to use the handy old Pythagorean theorem that you hopefully remember from boring high school math courses.  In plain English, the Pythagorean theorem provides a way to calculate the length of Line A based on the lengths of Lines B and C.   As shown in the formula below, you multiply the length of Line B by itself, and then multiply the length of Line C by itself, and then add (sum) those two values together.   Line A’s length is just the square root of that sum:

 For the triangle we’re concerned with, the length of Line B is just the distance Shirley travels in one second, or one kilometer.  Line C is the same distance that the light pulse covers from George’s perspective, or 300,000 kilometers.  300,000 multiplied by itself is a ridiculously large number, 90 billion in fact. On the other hand, as far as Line B is concerned, multiplying the number one by itself equals… LOL… one!  The sum of these two numbers is 90 billion and one.  Obviously, being so extremely close to 90 billion, the square root of 90 billion and one is virtually identical to the square root of 90 billion itself, or 300,000.  Thus, the extra distance along Line A compared to Line C is miniscule. 

Even thought these calculations may seem quite obvious, working through them enables me to draw your attention to a couple of things that will become important later on.  First, note how the math jives completely with the physical shape of the triangle in the figure.  My illustration doesn’t begin to do justice to the small size of the slope of line A; in reality, the slope is so small that if they were superimposed, A would overlay C almost perfectly.  Clearly, when line C is super long and B is super short, virtually all of the length of the hypotenuse (Line A) is determined (though I actually prefer the connotations of the word “donated”) by the much longer line C.

Turning this around, and comparing the figure to the math, the visual impression that Line A is almost equal in length to Line C jives perfectly with the Pythagoran theorem.  When you square a number that’s huge to begin with, you end up with a humungous number; when you square a very small number, the result stays pretty small.  When you add them together (and then compute the square root), the big number (the long line) contributes virtually everything to the result, while the small number (the short line) contributes virtually nothing.

Forgive me, patient readers because, at this point, I’d like to take a short detour from our main thread to more thoroughly illustrate how the relative lengths of B and C determine what proportion of the length of Line A is controlled by the one line compared to the other.  In the triangle we’ve been discussing up to now, Line B is very short compared to Line C.  But now, please direct your attention to the triangles depicted in the figure below, where Lines B and C are equal in length (both 300,000 kilometers long).  Since now, Line B equals the length of Line C, B squared and C squared are both 90 billion, and when you sum the two, for a total of 180 billion, and compute the square root, you find that they contribute exactly equal amounts to the length of side A (which turns out to be 424,264 kilometers).  In other words, of Line A’s total length, B and C both “donate” exactly half, 212,132 kilometers, or just over 71 percent of their own lengths. The crucial take-home message:  Once their lengths are equalized, Lines B and C contribute equal amounts to the length of Line A. 

Don’t believe me? Pick some examples of your own, say a triangle where B and C are just one kilometer long, run the numbers through the theorem, and confirm it for yourself!   And suppose you went back to our target triangle, where Line B is a tiny fraction of Line C, and used the Pythagorean theorem to calculate the length of Line A while gradually increasing the length of Line B.  What you’d find, of course, is that the longer Line B got relative to Line C, the more Line B would “donate” to the length of Line A.  

The importance of what happens when Line B approaches (and then exceeds) the length of Line C won’t become obvious until later blogs.  For now, let’s return to our main thread, where Shirley is traveling only one kilometer a second, and Line B is miniscule compared to Line C.  As I mentioned earlier, the tiny extra distance “donated” to Line A by the presence of Line B takes the light pulse such a tiny amount of extra time to traverse that it is way below the sensitivity of you and your stopwatch to measure.  For all extents and purposes, both you and Astronaut George are still in the same “time zone”, and Einstein has still not really shown up to the party. 

But that’s about to change.  In the next blog, you and George are going to crank up Shirley’s speed to something far more substantial: fully half the speed of light (150,000 kilometers per second).  That’s just absurdly fast, almost 2,000 times faster than as the fastest spacecraft ever launched.  It is so fast, in fact, that in just the single second it takes for the light pulse to travel up to Shirley’s ceiling, the old gal travels 150,000 kilometers along the line to your right (and creates a Line B that’s 150,000 kilometers long).  Add the additional second it takes the pulse to return to the spacecraft floor, and Shirley will travel a total of 300,000 kilometers away to your right! 

What will such a blinding speed do to the shape of our all-important triangles?  What will it do to the relative lengths of Lines B and C?   If you get really ambitious, you could easily run the numbers through the Pythagorean theorem yourself, and then compute the distance the light pulse will now have to travel.  What value will you, stationary on the ground, register on your stopwatch for how long it takes the pulse to complete its journey?  And what value will George register?

Stay tuned!

Shirley's Magic Bottom

Let’s recap. You are trying to measure the duration of an event, the journey of a pulse of light up and down the shaft of your spacecraft, Shirley.  In the current version of the experiment, you’re stationary, looking up from the ground, while Shirley is moving at one kilometer per second along the line formed by your outstretched right arm.  You and George start your stopwatches at the same time, when George toggles the searchlight on and off while Shirley is directly above you.  The question is, do you and George measure the same amount of time for the light flash to travel up to the ceiling and back?

Before tackling that question, we first have to deal with an obvious problem that rears its ugly head now that Shirley is moving.  We know that from George’s perspective it takes the flash two seconds to travel up and down Shirley’s shaft.  By the time the flash gets back to the floor, therefore, Shirley is no longer directly above you; she’s two kilometers down the line to your right.  That’s no problem for George; being inside Shirley, he still sees the flash as he always has.  But obviously, you can no longer see it because, when it passes through the glass floor, you’re not there!  As things, stand, you don’t have any event on which to stop your stopwatch. 

To deal with that little complication, I’d like to take some poetic license and add a rather advanced capability to Shirley’s bottom.  Specifically, the molecules of the glass that make up her ass automatically realign such that, when the light pulse hits the floor, it is reflected at just the right angle to travel directly back into your eyes.  That way, both you and George see the return flash of light, and you both have an event to stop your stopwatches with.

 

Physically, this situation is illustrated in the figure above.  Since George is inside the spacecraft, he’s moving right along with the pulse.  From his perspective, the pulse goes straight up to the ceiling and straight back down, just like it did when Shirley (and George) weren’t moving. The white arrows going straight up and down inside Shirley illustrate the path that the pulse takes for George.  Meanwhile, when the flash reaches the floor it is steered immediately along the dotted line right back to you.  In this way, you continue to see the flash, too.

OK, how does the value on your stopwatch compare to the value on George’s?  Light takes only a tiny amount of extra time, 1/150,000 of a second to be exact, to cross the two kilometers separating you from Shirley.  Although the flash is reaching you just a teensy, tiny bit later than it reaches George, the difference is far too small to show up with a measuring device as crude as a human operating a stopwatch. Still, you’re a stickler for accuracy.  You know that, compared to George, the light has had to travel a longer distance, which means that you are not… quite… measuring the duration of the same physical event that George is.  But your smart stopwatch comes to the rescue! You simply program the watch to automatically subtract the additional time required for the flash to travel from Shirley to you (recall that your watch is already automatically subtracting your brain processing time).

Does the reconfigurable glass floor solve all the measurement problems that accompany the fact that Shirley is now moving?  Possibly not, because something else about this whole situation is starting to gnaw at you.  The figure provides a strong hint to the problem.  Now that Shirley is moving, what is the exact path through space that the light has to take from your perspective in order to get to the ceiling and back? Is it possible that this path is different from the path that it takes from George’s perspective (which is just straight up and down)?

If so, how is it different?  And why? The more you take a crack at answering that question, and “stay ahead” of the blogs, the less pain you’re going to have later on when Einstein finally crashes our little party!  Anybody care to tackle this issue in the comments section?

Introducing Astronaut George!

So, have you been doing your homework, faithful Whabblog readers? Have you been performing the thought experiment repeatedly, and always getting the same two second result (once your smart stopwatch automatically subtracts your “brain processing” time)?  If you have, then chances are you’ve gotten thoroughly sick and tired of the whole, dare I use the word, tedious, activity, because the outcome is always the same!  Your spreadsheet never shows anything but two-second entries.

To spice things up a little, you decide to hire an astronaut accomplice.  Since “accomplice” kind of sounds like you’ve decided to rob a bank or something, perhaps “collaborator” or even “confederate” is a better label.  Your “confederate”, who for entirely pernicious reasons is actor George Clooney, agrees to lie on the floor of the spacecraft in your stead.  Therefore, it is now Astronaut George who toggles the searchlight on and off, and measures how long it takes to see the pulse of light on its return from the ceiling.  When George does this, he gets the same two-second result you did (of course, his stopwatch is smart enough to automatically subtract his brain-processing time, too).

Meanwhile, where have you gone?  Not far, because you want to continue to measure the travel time of the light flash right along with Astronaut George.  Therefore, you arrange for Shirley to be tethered just a couple of feet off the ground – just high enough that you can lie flat on the ground underneath the transparent floor, and stare straight up through the floor in the direction of the ceiling (just like George is doing inside Shirley).  And just like George, you’re going to start your stopwatch at the exact point when he flicks the searchlight on (remember, the searchlight is bidirectional, so you immediately see the light coming through the floor), and you’re going to push your stopwatch again, halting it, when you see the return flash.  Of course, when the reflected flash reaches Shirley’s transparent floor, it only has to pass through the glass and travel another couple of feet to reach you – a negligible distance.  Thus, not only do you both start your stopwatches at the same time, you stop them at the same time, too.  Both stopwatches register exactly the same two-second duration for the light pulse to complete its journey.

Still, now it’s both of you, not just you yourself, who keep getting the same result, and two people are now getting restless and bored. The thought experiment, with its predictable two-second result on both stopwatches, is still rather tedious. This is all fine and dandy, you’re probably beginning to think, but what about Einstein?

Relax.  It’s always darkest just before the dawn!  Shirley is a spaceship, remember, so why not repeat the identical experiment with you remaining motionless in the same position on the ground, but with Shirley now in motion?  To accommodate this slightly more complex situation, you and Astronaut George work out the following arrangement.   Lying on the ground, you stretch your arms straight out to your left and right, so your body forms the shape of a cross.  George takes control of Shirley and, still hovering at the same short distance above the ground from you, maneuvers her to a point over on your left, directly along the line formed by your outstretched left arm.  Next, George accelerates Shirley to a constant speed, so that she (and he) move steadily towards you along the straight line.  Still holding the same constant speed, Shirley passes right over top of you and then continues down the line formed by your outstretched right arm.

For your first time measurement experiment with Shirley in motion, you decide to have her move at a steady one-kilometer per second along the line.  Although that’s actually pretty darn fast, about the speed of a supersonic military jet, it pales in comparison to the speed of the light itself:  One kilometer per second is only 1/300,000 of THAT ridiculous speed!  Still, it’s fast enough that Astronaut George has to take great pains to toggle the searchlight on and off at the exact instant Shirley passes over you.  That way, you still see the flash of light through the floor of the spacecraft when George toggles the searchlight on, and you both START your stopwatches in unison.

But do you both still see the light flash return to the spacecraft floor at the same time? Do you both stop your stopwatches together, and measure the same two seconds for the event to happen?  That is the question I’ll tackle in the next blog.  In the meantime, feel free to speculate on the answer!