Superman flying at lightspeed? Umm… no

Back when I was in high school I did a term paper for some class (can’t remember which) ripping the heck out of Superman physics.  Yeah, I was that kind of kid.  If I recall correctly, I spent much of it slamming his supposed vision capabilities — they were fairly ludicrous even to a HS student and that was many refreshes of the DC universe ago.FTL SupermanBut for this post let’s consider a trope that’s been taken off the shelf again and again since those days, even in the movies.  This rendition should get the idea across — Our Hero, in a desperate effort to fix a narrative hole the writers had dug themselves into, is forced to fly around the Earth at faster-than-light speeds, thereby reversing time so he can patch things up.

So many problems…  Just for starters, the Earth is 8000 miles wide, Supey’s what, 6’6″?, so on this scale he shouldn’t fill even a thousandth of a pixel.  OK, artistic license.  Fine.

Second problem, only one image of the guy.  If he’s really passing us headed into the past we should see two images, one coming in feet-first from the future and the other headed forward in both space and time.  Oh, and because of the Doppler effect the feet-first image should be blue-shifted and the other one red-shifted.

Of course both of those images would be the wrong shape.  The FitzGerald-Lorentz Contraction makes moving objects appear shorter in the direction of motion.  In other words, if the Man of Steel were flying just shy of the speed of light then 6’6″ tall would look to us more like a disk with a short cape.

Tall-Dark-And-Muscular has other problems to solve on his way to the past.  How does he get up there in the first place?  Back in the day, DC explained that he “leaped tall buildings in a single bound.”   That pretty much says ballistic high-jump, where all the energy comes from the initial impulse.  OK, but consider the rebound effects on the neighborhood if he were to jump with as much energy as it would take to orbit a 250-lb man.  People would complain.

Remember Einstein’s famous E=mc²?  That mass m isn’t quite what most people think it is.  Rather, it’s an object’s rest mass m0 modified by a Lorentz correction to account for the object’s kinetic energy.   In our hum-drum daily life that correction factor is basically 1.00000…   When you get into the lightspeed ballpark it gets bigger.

Here’s the formula with the Lorentz correction in red: m=m0/√[1-(v/c)2].  The square root nears zero as Superman’s velocity v approaches lightspeed c.  When the divisor gets very small the corrected mass gets very large.  If he got to the Lightspeed Barrier (where v=c) he’d be infinitely massive.

So you’ve got an infinite mass circling the Earth about 7 times a second — ocean tides probably couldn’t keep up, but the planet would be shaking enough to fracture the rock layers that keep volcanoes quiet.  People would complain.

Of course, if he had that much mass, Earth and the entire Solar System would be orbiting him.

On the E side of Einstein’s equation, Superman must attain that massive mass by getting energy from somewhere.  Gaining that last mile/second on the way to infinity is gonna take a lot of energy.

But it’s worse.  Even at less than lightspeed, the Kryptonian isn’t flying in a straight line.  He’s circling the Earth in an orbit.  The usual visuals show him about as far out as an Earth-orbiting spacecraft.  A GPS satellite’s stable 24-hour orbit has a 26,000 mile radius so it’s going about 1.9 miles/sec.  Superman ‘s traveling about 98,000 times faster than that.  Physics demands that he use a powered orbit, continuously expending serious energy on centripetal acceleration just to avoid flying off to Vega again.

The comic books have never been real clear on the energy source for Superman’s feats.  Does he suck it from the Sun?  I sure hope not — that’d destabilize the Sun and generate massive solar flares and all sorts of trouble.

Not even the DC writers would want Superman to wipe out his adopted planet just to fix up a plot point.

~~ Rich Olcott

Does a photon experience time?

My brother Ken asked me, “Is it true that a photon doesn’t experience time?”  Good question.  As I was thinking about it I wondered if the answer could have implications for Einstein’s bubble.

When Einstein was a grad student in Göttingen, he skipped out on most of the classes given by his math professor Hermann Minkowski.  Then in 1905 Einstein’s Special Relativity paper scooped some work that Minkowski was doing.  In response, Minkowski wrote his own paper that supported and expanded on Einstein’s.  In fact, Minkowski’s contribution changed Einstein’s whole approach to the subject, from algebraic to geometrical.

But not just any geometry, four-dimensional geometry — 3D space AND time.  But not just any space-AND-time geometry — space-MINUS-time geometry.  Wait, what?Pythagoras1

Early geometer Pythagoras showed us how to calculate the hypotenuse of a right triangle from the lengths of the other two sides. His a2+b2 = c2 formula works for the diagonal of the enclosing rectangle, too.

Extending the idea, the body diagonal of an x×y×z cube is √(x2+y2+z2) and the hyperdiagonal  of a an ct×x×y×z tesseract is √(c2t2+x2+y2+z2) where t is time.  Why the “c“?  All terms in a sum have to be in the same units.  x, y, and z are lengths so we need to turn t into a length.  With c as the speed of light, ct is the distance (length) that light travels in time t.

But Minkowski and the other physicists weren’t happy with Pythagorean hyperdiagonals.  Here’s the problem they wanted to solve.  Suppose you’re watching your spacecraft’s first flight.  You built it, you know its tip-to-tail length, but your telescope says it’s shorter than that.  George FitzGerald and Hendrik Lorentz explained that in 1892 with their length contraction analysis.

What if there are two observers, Fred and Ethel, each of whom is also moving?  They’d better be able to come up with the same at-rest (intrinsic) size for the object.

Minkowski’s solution was to treat the ct term differently from the others.  Think of each 4D address (ct,x,y,z) as a distinct event.  Whether or not something happens then/there, this event’s distinct from all other spatial locations at moment t, and all other moments at location (x,y,z).

To simplify things, let’s compare events to the origin (0,0,0,0).  Pythagoras would say that the “distance” between the origin event and an event I’ll call Lucy at (ct,x,y,z) is √(c2t2+x2+y2+z2).

Minkowski proposed a different kind of “distance,” which he called the interval.  It’s the difference between the time term and the space terms: √[c2t2 + (-1)*(x2+y2+z2)].

If Lucy’s time is t=0 [her event address (0,x,y,z)], then the origin-to-Lucy interval is  √[02+(-1)*(x2+y2+z2)]=i(x2+y2+z2).  Except for the i=√(-1) factor, that matches the familiar origin-to-Lucy spatial distance.

Now for the moment let’s convert the sum from lengths to times by dividing by c2.  The expression becomes √[t2-(x/c)2-(y/c)2-(z/c)2].  If Lucy is at (ct,0,0,0) then the origin-to-Lucy interval is simply √(t2)=t, exactly the time difference we’d expect.

Finally, suppose that Lucy departed the origin at time zero and traveled along x at the speed of light.   At any time t, her address is (ct,ct,0,0) and the interval for her trip is √[(ct)2-(ct)2-02-02] = √0 = 0.  Both Fred’s and Ethel’s clocks show time passing as Lucy speeds along, but the interval is always zero no matter where they stand and when they make their measurements.

Feynman diagramOne more step and we can answer Ken’s question.  A moving object’s proper time is defined to be the time measured by a clock affixed to that object.  The proper time interval between two events encountered by an object is exactly Minkowski’s spacetime interval.  Lucy’s clock never moves from zero.

So yeah, Ken, a photon moving at the speed of light experiences no change in proper time although externally we see it traveling.

Now on to Einstein’s bubble, a lightwave’s spherical shell that vanishes instantly when its photon is absorbed by an electron somewhere.  We see that the photon experiences zero proper time while traversing the yellow line in this Feynman diagram.  But viewed from any other frame of reference the journey takes longer.  Einstein’s objection to instantaneous wave collapse still stands.

~~ Rich Olcott

LIGO, a new kind of astronomy

Like thousands of physics geeks around the world, I was glued to the tube Thursday morning for the big LIGO (Laser Interferometer Gravitational-Wave Observatory) announcement.  As I watched the for-the-public videos (this is a good one), I was puzzled by one aspect of the LIGO setup.  The de-puzzling explanation spotlit just how different gravitational astronomy will be from what we’re used to.

There are two LIGO installations, 2500 miles apart, one near New Orleans and the other near Seattle.  Each one looks like a big L with steel-pipe arms 4 kilometers long.  By the way, both arms are evacuated to eliminate some sources of interference and a modest theoretical consideration.

LIGO3The experiment consists of shooting laser beams out along both arms, then comparing the returned beams.

Some background: Einstein conquered an apparent relativity paradox.  If Ethel on vehicle A is speeding (like, just shy of light-speed speeding) past Fred on vehicle B, Fred sees that Ethel’s yardstick appears to be shorter than his own yardstick.  Meanwhile, Ethel is quite sure that Fred’s yardstick is the shorter one.

Einstein explained that both observations are valid.  Fred and Ethel can agree with each other but only after each takes proper account of their relative motion.  “Proper account” is a calculation called the Lorenz transformation.   What Fred (for instance) should do is divide what he thinks is the length of Ethel’s yardstick by √[1-(v/c)²] to get her “proper” length.  (Her relative velocity is v, and c is the speed of light.)

Suppose Fred’s standing in the lab and Ethel’s riding a laser beam.  Here’s the puzzle: wouldn’t the same Fred/Ethel logic apply to LIGO?  Wouldn’t the same yardstick distortion affect both the interferometer apparatus and the laser beams?

Well, no, for two reasons.  First, the Lorenz effect doesn’t even apply, because the back-and-forth reflected laser beams are standing waves.  That means nothing is actually traveling.  Put another way, if Ethel rode that light wave she’d be standing as still as Fred.

The other reason is that the experiment is less about distance traveled and more about time of flight.

Suppose you’re one of a pair of photons (no, entanglement doesn’t enter into the game) that simultaneously traverse the interferometer’s beam-splitter mirror.  Your buddy goes down one arm, strikes the far-end mirror and comes back to the detector.  You take the same trip, but use the other arm.

The beam lengths are carefully adjusted so that under normal circumstances, when the two of you reach the detector you’re out of step.   You peak when your buddy troughs and vice-versa.  The waves cancel and the detector sees no light.

Now a gravitational wave passes by (red arcs in the diagram).  In general, the wave will affect the two arms differently.  In the optimal case, the wave front hits one arm broadside but cuts across the perpendicular one.  Suppose the wave is in a space-compression phase when it hits.  The broadside arm, beam AND apparatus, is shortened relative to the other one which barely sees the wave at all.

The local speed of light (miles per second) in a vacuum is constant.  Where space is compressed, the miles per second don’t change but the miles get smaller.  The light wave slows down relative to the uncompressed laboratory reference frame.  As a result, your buddy in the compressed arm takes just a leetle longer than you do to complete his trip to the detector.  Now the two of you are in-step.  The detector sees light, there is great rejoicing and Kip Thorne gets his Nobel Prize.

But the other wonderful thing is, LIGO and neutrino astronomy are humanity’s first fundamentally new ways to investigate our off-planet Universe.  Ever since Galileo trained his crude telescope on Jupiter the astronomers have been using electromagnetic radiation for that purpose – first visible light, then infra-red and radio waves.  In 1964 we added microwave astronomy to the list.  Later on we put up satellites that gave us the UV and gamma-ray skies.

The astronomers have been incredibly ingenious in wringing information out of every photon, but when you look back it’s all photons.  Gravitational astronomy offers a whole new path to new phenomena.  Who knows what we’ll see.

~~ Rich Olcott