# 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

## 2 thoughts on “Does a photon experience time?”

1. Enjoyed this one, Rich!

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2. Ken Olcott

Well, the equations stretched my ability to comprehend, but I get that this is how theoretical physics does its explaining. I was more expecting a discussion of light-speed and time compression. Einstein’s Bubble is a new concept to me – worth more research, next chance I get (LOL). Thanks, Bro.

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