Time Is Where You Find It

A familiar footstep in the hall outside my office, “C’mon in, Vinnie, the door’s open.”

“Got a few minutes, Sy?”

More than just “a minute.” This sounds serious so I push my keyboard aside. “Sure, what’s up?”

“I’ve been thinking about different things, putting ’em together different ways. I came up with something, sorta, that I wanted to run past you before I brought it to one of Cathleen’s ‘Crazy Theories‘ parties.”

“Why, Vinnie, you’re being downright diffident. Spill it.”

“Well, it’s all fuzzy. First part goes way back to years ago when you wrote that there’s zero time between when a photon gets created and when it gets used up. But that means that create and use-up are simultaneous and that goes against Einstein’s ‘No simultaneity‘ thing which I wonder if you couldn’t get around it using time tick signals to sync up two space clocks.”

“That’s quite a mix and I see why you say it’s fuzzy. Would you be surprised if I used the word ‘frame‘ while clarifying it?”

“I’ve known you long enough it wouldn’t surprise me. Go ahead.”

“Let’s start with the synchronization idea. You’re not the first to come up with that suggestion. It can work, but only if the two clocks are flying in formation, exactly parallel course and speed.”

“Hah, that goes back to our first talk with the frame thing. You’re saying the clocks have to share the same frame like me and that other pilot.”

“Exactly. If the ships are zooming along in different inertial frames, each will measure time dilation in the other. How much depends on their relative velocities.”

“Wait, that was another conversation. We were pretending we’re in two spaceships like we’re talking about here and your clock ran slower than mine and my clock ran slower than yours which is weird. You explained it with equations but I’ve never been good with equations. You got a diagram?”

“Better than that, I’ve got a video. It flips back and forth between inertial frames for Enterprise and Voyager. We’ll pretend that they sync their clocks at the point where their tracks cross. I drew the Enterprise timeline vertical because Enterprise doesn’t move in space relative to Enterprise. The white dots are the pings it sends out every second. Meanwhile, Voyager is on a different course with its own timeline so its inertial frame is rotated relative to Enterprise‘s. The gray dots on Voyager‘s track show when that ship receives the Enterprise pings. On the Voyager timeline the pings arrive farther apart than they are on the Enterprise timeline so Voyager perceives that Enterprise is falling farther and farther behind.”

“Gimme a sec … so Voyager says Enterprise‘s timer is going slow, huh?”

“That’s it exactly. Now look at the rotated frame. The pink dots show when Voyager sends out its pings. The gray dots on Enterprise‘s track show when the pings arrive.”

“And Enterprise thinks that Voyager‘s clock is slow, just backwards of the other crew. OK, I see you can’t use sync pulses to match up clocks, but it’s still weird.”

“Which is where Lorentz and Minkowski and Einstein come into the picture. Their basic position was that physical events are real and there should be a way to measure them that doesn’t depend on an observer’s frame of reference. Minkowski’s ‘interval‘ metric qualifies. After converting time and location measurements to intervals, both crews would measure identical spacetime separations. Unfortunately, that wouldn’t help with clock synchronization because spacetime mixes time with space.”

“How about the photons?”

“Ah, that’s a misquotation. I didn’t say the time is zero, I said ‘proper time‘ and that’s different. An object’s proper time is measured by its clock in its inertial frame while traveling time t and distance d between two events. Anyone could measure t and d in their inertial frame. Minkowski’s interval is defined as s=[(ct)²‑d²]. Proper time is s/c. Intuitively I think of s/c as light’s travel time after it’s done traversing distance d. In space, photons always travel at lightspeed so their interval and proper time are always zero.”

“Photon create and use-up aren’t simultaneous then.”

“Only to photons.”

~~ Rich Olcott

Thinking in Spacetime

The Open Mic session in Al’s coffee shop is still going string. The crowd’s still muttering after Jeremy stuck a pin in Big Mike’s “coincidence” balloon when Jim steps up. Jim’s an Astrophysics post‑doc now so we quiet down expectantly. “Nice try, Mike. Here’s another mind expander to play with. <stepping over to the whiteboard> Folks, I give you … a hypotenuse. ‘That’s just a line,’ you say. Ah, yes, but it’s part of some right triangles like … these. Say three different observers are surveying the line from different locations. Alice finds her distance to point A is 300 meters and her distance to point B is 400. Applying Pythagoras’ Theorem, she figures the A–B distance as 500 meters. We good so far?”

A couple of Jeremy’s groupies look doubtful. Maybe‑an‑Art‑Major shyly raises a hand. “The formula they taught us is a2+b2=c2. And aren’t the x and y supposed to go horizontal and vertical?”

“Whoa, nice questions and important points. In a minute I’m going to use c for the speed of light. It’s confusing to use the same letter for two different purposes. Also, we have to pay them extra for double duty. Anyhow, I’m using d for distance here instead of c, OK? To your next point — Alice, Bob and Carl each have their own horizontal and vertical orientations, but the A–B line doesn’t care who’s looking at it. One of our fundamental principles is that the laws of Physics don’t depend on the observer’s frame of reference. In this situation that means that all three observers should measure the same length. The Pythagorean formula works for all of them, so long as we’re working on a flat plane and no-one’s doing relativistic stuff, OK?”

Tentative nods from the audience.

“Right, so much for flat pictures. Let’s up our game by a dimension. Here’s that same A–B line but it’s in a 3D box. <Maybe‑an‑Art‑Major snorts at Jim’s amateur attempt at perspective.> Fortunately, the Pythagoras formula extends quite nicely to three dimensions. It was fun figuring out why.”

Jeremy yells out. “What about time? Time’s a dimension.”

“For sure, but time’s not a length. You can’t add measurements unless they all have the same units.”

“You could fix that by multiplying time by c. Kilometers per second, times seconds, is a length.” His groupies go “Oooo.”

“Thanks for the bridge to spacetime where we have four coordinates — x, y, z and ct. That makes a big difference because now A and B each have both a where and a when — traveling between them is traveling in space and time. Computationally there’s two paths to follow from here. One is to stick with Pythagoras. Think of a 4D hypercube with our A–B line running between opposite vertices. We’re used to calculating area as x×y and volume as x×y×z so no surprise, the hypercube’s hypervolume is x×y×z×(ct). The square of the A–B line’s length would be b2=(ct)2+d2. Pythagoras would be happy with all of that but Einstein wasn’t. That’s where Alice and Bob and Carl come in again.”

“What do they have to do with it?”

“Carl’s sitting steady here on good green Earth, red‑shifted Alice is flying away at high speed and blue‑shifted Bob is flashing toward us. Because of Lorentz contractions and dilations, they all measure different A–B lengths and durations. Each observer would report a different value for b2. That violates the invariance principle. We need a ruggedized metric able to stand up to that sort of punishment. Einstein’s math professor Hermann Minkowski came up with a good one. First, a little nomenclature. Minkowski was OK with using the word ‘point‘ for a location in xyz space but he used ‘event‘ when time was one of the coordinates.”

“Makes sense, I put events on my calendar.”

“Good strategy. Minkowski’s next step quantified the separation between two events by defining a new metric he called the ‘interval.’ Its formula is very similar to Pythagoras’ formula, with one small change: s2=(ct)2–d2. Alice, Bob and Carl see different distances but they all see the same interval.”

Minus? Where did that come from?”

~~ 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.Minkowski diagram
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).

space VennThe 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?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