Can someone please explain to me why massless particles follow null geodesics?

[zeromodz]

Anything that is massless must follow null geodesics. Why is this?

 

[Penn.6-5000]

It so happens that "geodesic" means "locally straight line" and "null geodesic" means "locally straight line at the speed of light."

Your question is then: "Why do massless particles have to travel in locally straight lines at the speed of light?"
1. They travel in locally straight lines because of Newton's First Law. (If you apply an external force to the particle, it will stop moving along geodesics.)
2. They travel at the speed of light because of quantum mechanics.

A closer look at part one:
Newton's First Law says that a particle will keep a constant velocity unless someone applies a force to it. If something moves with a constant velocity it is clearly moving in a straight line (or not moving at all). When general relativity came along and we got curved space, we lost our ability to talk about "straight lines" as such. In curved space, you can imagine tracing out a path of a constant velocity from your starting position by taking a tiny step in the direction of your velocity, and then sliding your velocity vector along that tiny step (without rotating it), taking another tiny step in the direction of your velocity vector, sliding the vector up to the new position, etc. This vector is always tangent to your path, so we call it a "tangent vector," and the process of sliding a vector through curved space without rotating it is called "parallel transport." A geodesic curve is one that is formed in this manner, and in fact most books give its technical definition as "a curve that parallel-transports its own tangent vector." But if that explanation is too bizarre, you can think of a geodesic as a path that always feels like a straight line if you're walking along it, even if it eventually ends up changing directions because the rest of space is curved. All objects, massless or not, move in straight lines when they aren't being pushed around by other forces, and we knew that
Technically I haven't answered the question, "Why do particles move in straight lines when they don't experience external forces? I understand *that* they do it, but *why* do they do it?" The most popular answer to that is quantum mechanical but not especially useful here.

A closer look at part two:
A "null geodesic" is a geodesic of length zero. There's a quirk when measuring lengths in relativity because geodesics aren't paths from point A to point B; they're paths from <point A at time X> to <point B at time Y>. Instead of calling it a "spacetime distance between two point/time pairs," we usually call it the "interval between two events." If [tex]\Delta t[/tex] is the time elapsed between X and Y and [tex]\Delta x[/tex] is the distance between points A and B (and if c is the speed of light) then the interval between the two events is:

[tex]\hbox{Interval} = \sqrt{c^2\Delta t^2 - \Delta x^2}[/tex]

(Note: That formula works only in flat space. In curved space, you have to break the path up into a bunch of infinitesimal pieces, apply that formula to each piece, and then integrate to add them all up.)

When is the interval zero? When [tex]c^2\Delta t^2 = \Delta x^2[/tex], which is true when [tex]\Delta x/\Delta t = c[/tex]. What kind of trajectory covers that distance in that amount of time? That is, how does a particle move that distance in that amount of time? Obviously, by traveling at the speed of light. Therefore a null geodesic would be the path traced out by a particle moving at the speed of light.

So now we ask, "Why do massless particles travel at the speed of light?" As you might have guessed, this question also has a quantum-mechanical answer. But University of Illinois has a somewhat acceptable http://van.physics.illinois.edu/qa/listing.php?id=1354" that doesn't depend on quantum mechanics. I call it only "somewhat acceptable" because it doesn't really tell you why such particles travel at the speed of light; it just talks about why things would be awkward if they moved slower than light.

 

[bcrowell]

I don't think the answer to the question has anything to do with quantum mechanics.

Here's the way I like to think about it:
http://www.lightandmatter.com/html_books/genrel/ch02/ch02.html#Section2.4
http://www.lightandmatter.com/html_books/genrel/ch04/ch04.html#Section4.2 (see subsection 4.2.2)

 

[Penn.6-5000]

We have a fact that is true for all particles at all speeds:

[tex]E^2=m^2c^4+p^2c^2[/tex]

We also have a pair of facts that hold only for particles moving slower than light; when [tex]v\geq c[/tex] the gammas are infinite or undefined:

[tex]\halign{\hfil#&#\hfil\cr % I roll my own \TeX tables, biznatch
E &= mc^2\gamma\quad\hbox{(with $\gamma =1/\sqrt{1-v^2/c^2}}$)}\cr
p &= mv\gamma\cr}
[/tex]

The first approach we might try is to just plug m = 0 into the bottom two equations. We get that E = p = 0 and conclude that a massless particle has no kinetic energy and no momentum, regardless of how fast it's moving. (To check our work, we plug it into the top equation and see that it's a true statement, so we figure we did it correctly.) That makes intuitive sense, right? The less massive a particle is, the less kinetic energy and momentum it has, and in the limit it has none at all. The problem with this conclusion is that it doesn't match observation. Photons have no mass, but they have energy; otherwise, lasers wouldn't be cool at all.

Lightandmatter.com avoids getting zeros by hiding behind some more complicated derivations. First, it plugs m = 0 into the topmost equation and takes the square root of each side to get E=pc. Then it plugs in the other two equations to get mc²γ = mvγc. Here it cancels out the m's and one c to get cγ = vγ, and then argues that for these to equal each other in the limit, v must equal c. It's admirable that this site recognized the need for a limit because when v equals c it turns out that gamma is infinite. But it managed to miss (or ignore) the fact that it outright plugged zero in for one of the m's but allowed the other two to cancel as nonzero values. So the math here is meaningless even though the answer turns out to be right.

The University of Illinois link I provided correctly sidesteps the math that can be sidestepped but the end result is unsatisfying. They argue that if a massless particle travels slower then light, then plugging m = 0 into the formulas for E and p will give an energy and momentum of zero, and since a particle with no energy can't do anything, it's not worth thinking about further. But they note that the formulas only apply to particles moving slower than light. If the particle is moving at light speed, then the formulas for E and p cease to apply so anything's possible. Since slower than light is boring and faster than light is impossible for other reasons, they say that the only interesting solution is that massless particles move at exactly light speed. This is a nice argument because it's technically correct, but it doesn't actually say why massless particles should travel at light speed... It just applies the process of elimination to the other possible speeds.

The most satisfying argument for me on this subject was based on the Klein-Gordon equation, which is a quantum-mechanical description of relativistic particles. But I don't remember how exactly it went and I'm having a hard time finding it again. So I'm sort of stuck in the same position I was earlier this evening.

 

[zeromodz]

We have a fact that is true for all particles at all speeds:

Then it plugs in the other two equations to get mc²γ = mvγc. Here it cancels out the m's and one c to get cγ = vγ

This is wrong because you cannot cancel out the masses, since the masses equal zero (Division by zero). Its illegal and an invalid move.

 

[Penn.6-5000]

This is wrong because you cannot cancel out the masses, since the masses equal zero (Division by zero). Its illegal and an invalid move.

Yes, the rest of the paragraph acknowledges that fact: "But [lightspeed.com] managed to miss (or ignore) the fact that it outright plugged zero in for one of the m's but allowed the other two to cancel as nonzero values. So the math here is meaningless even though the answer turns out to be right."

The University of Illinois' derivation described in the next paragraph is the correct way to do it.