# Superluminal Superman

Comic book and movie plotlines often make Superman accelerate up to lightspeed and travel backward in time.  Unfortunately, well-known fundamental Physics principles forbid that.  But suppose Green Lantern or Dr Strange could somehow magic him past the Lightspeed Barrier.  Would that let him do his downtimey thing?

A quick review of Light’s Hourglass.  According to Einstein we live in 4D spacetime.  At any moment you’re at a specific time t relative to some origin time t=0 and a specific 3D location (x,y,z) relative to a spatial origin (0,0,0).  Your spacetime address is (ct,x,y,z) where c is the speed of light.  This diagram shows time running vertically into the future, plus two spatial coordinates x and y.  Sorry, I can’t get z into the diagram so pretend it’s zero.

The two cones depict all the addresses which can communicate with the origin using a flash of light.  Any point on either cone is at just the right distance d=√(++) to match the distance that light can travel in time t.  The bottom cone is in the past, which is why we can see the light from old stars.  The top cone is in the future, which is why we can’t see light from stars that aren’t born yet.

If he obeys the Laws of Physics as we know them, Superman can travel anywhere he wants to inside the top cone.  He goes upward into the future at the rate of one second per second, just like anybody.  On the way, he can travel in space as far from (ct,0,0,0) as he likes so long as it’s not farther than the distance that light can travel the same route at his current t.

From our perspective, the Hourglass is a stack of circles (spheres in 3D space) centered on (ct,0,0,0).  From Supey’s perspective at time t he’s surrounded by a figure with radius ct that Physics won’t let him break through.  That’s his Lightspeed Barrier, like the Sonic Barrier but 900,000 times faster.

Suppose Green Lantern has magicked Supey up to twice lightspeed along the x-axis.  At moment t, he’s at (ct,2ct,0,0), twice as far as light can get.  In the diagram he’s outside the top cone but above the central disk.

Now GL pours on the power to accelerate Superman.  Each increment gets the Man of Steel closer to that disk.  He’s always “above” it, though, because he’s still moving into the future.  Only if he were to get to infinite speed could he reach the disk.

However, at infinite speed he’d go anywhere/everywhere instantaneously which would be confusing to even his Kryptonian intellect.  On the way he might run into things (stars, black holes,…) with literally zero time to react.

But the plotlines have Tall-Dark-and-Muscular flying into the past, breaching that disk and traveling downwards into the bottom cone.  Can GL make that happen?

Enter the Lorentz correction.  If you have rest mass m0 and you’re traveling at speed v, your effective mass is m=m0/√[1-(v/c)²]. That raises a couple of issues when you exceed lightspeed.

Suppose GL decelerates Superluminal Supey down towards lightspeed.  The closer he approaches c from higher speeds, the smaller that square root gets and the greater the effective mass.  It’s the same problem Superman faced when accelerating up to lightspeed.  That last mile per second down to c requires an infinite amount of braking energy — the Lightspeed Barrier is impermeable in both directions.

The other problem is that if v>c there’s a negative number inside that square root.    Above lightspeed, your effective mass becomes Bombelli-imaginary.  Remember Newton’s famous F=m·a?  Re-arrange it to a=F/m.  A real force applied to an object with imaginary mass produces an imaginary acceleration.  “Imaginary” in Physics generally means “perpendicular in some sense” and remember we’re in 4D here with time perpendicular to space.

GL might be able to shove Superman downtime, but he’d have to

1. squeeze inward at hiper-lightspeed with exactly the same force along all three spatial dimensions, to make sure that “perpendicular” is only along the time axis
2. start Operation Squish at some time in his own future to push towards the past.

Nice trick.  Would Superman buy in?

~~ Rich Olcott

# You can’t get there from here

In this series of posts I’ve tried to get across several ideas:

• By Einstein’s theories, in our Universe every possible combination of place and time is an event that can be identified with an “address” like (ct,x,y,z) where t is time, c is the speed of light, and x, y and z are spatial coordinates
• The Pythagorean distance between two events is d=√[(x1-x2)2+(y1-y2)2+(z1-z2)2]
• The Minkowski interval between two events is √[(ct1-ct2)2 d2]
• When i=√(-1) shows up somewhere, whatever it’s with is in some way perpendicular to the stuff that doesn’t involve i

That minus sign in the third bullet has some interesting implications.  If the time term is bigger then the spatial term, then the interval (that square root) is a real number.  On the other hand, if the time term is smaller, the interval is an imaginary number and therefore is in some sense perpendicular to the real intervals.

We’ll see what that means in a bit.  But first, suppose the two terms are exactly equal, which would make the interval zero.  Can that happen?

Sure, if you’re a light wave.  The interval can only be zero if d=(ct1-ct2).  In other words, if the distance between two events is exactly the distance light would travel in the elapsed time between the same two events.

In this Minkowski diagram, we’ve got two number lines.  The real numbers (time) run vertically (sorry, I know we had the imaginary line running upward in a previous post, but this is the way that Minkowski drew it).  Perpendicular to that, the line of imaginary numbers (distance) runs horizontally, which is why that starscape runs off to the right.
The smiley face is us at the origin (0,0,0,0).  If we look upwards toward positive time, that’s the future at our present locationLooking downward to negative time is looking into the past.  Unless we move (change x and/or y and/or z), all the events in our past and future have addresses like (ct,0,0,0).

What about all those points (events) that aren’t on the time axis?  Pick one and use its address to figure the interval between it and us.  In general, the interval will be a complex number, part real and part imaginary, because the event you picked isn’t on either axis.

The two orange lines are special.  Each of them is drawn through all the events for which the distance is equal to ct.  All the intervals between those events and us are zero.  Those are the events that could be connected to us by a light beam.

The orange connections go one way only — someone in our past (t less than zero) can shine a light at us that we’ll see when it reaches us.  However, if someone in our future tried to shine a light our way, well, the passage of time wouldn’t allow it (except maybe in the movies).  Conversely, we can shine a light at some spatial point (x,y,z) at distance d=√(x2+y2+z2) from us, but those photons won’t arrive until the event in our future at (d,x,y,z).

The rest of the Minkowski diagram could do for a Venn diagram.  We at (0,0,0,0) can do something that will cause something to happen at (ct,x,y,z) to the left of the top orange line.  However, we won’t be able to see that effect until we time-travel forward to its t.  That region is “reachable but not seeable.”

Similarly, events to the left of the bottom orange line can affect us (we can see stars, for instance) but they’re in our past and we can’t cause anything to happen then/there.  The region is “seeable but not reachable.”

Then there’s the overlap, the segment between the two orange lines.  Events there are so far away in spacetime that the intervals between them and us are imaginary (in the mathematical sense).  To put it another way, light can’t get here from there.  Neither can cause and effect.

Physicists call that third region space-like, as opposed to the two time-like regions.  Without a warp drive or some other way around Einstein’s universal speed limit, the edge of “space-like” will always be The Final Frontier.

~~ 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?

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.

One 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