Nisse said:
This is the sort of thing I was referring to, but did not express it as eloquently. I would be grateful for Frederik's comments on this.
Most of what Greene says are things I don't mind, e.g. "through space-time at one fixed speed", "all of the objects motion is used to travel through time", "it is the object’s speed in this generalized sense that is equal to that of light". This is his attempt to say the same things that I said about four-velocity, while consistently avoiding any reference to mathmatics. Unfortunately that last thing just makes it difficult, if not impossible, to understand what he's saying (if you don't know the mathematics already).
This is a part of the Greene quote that I don't like at all, because it's so easy to misinterpret: "Something traveling at light speed through space will have no speed left for motion through time".
Think of a spacetime diagram representing the inertial coordinates in which you are stationary. Time in the "up" direction, space in the "right" direction. (No need to consider more than one spatial dimension for this). What we mean by saying that an object that's stationary (in this frame) is moving only through time (in this frame) is that its four-velocity is a vector pointing straight up. Its components are (c,0). The four-velocity of an object moving with speed v to the right is \gamma(c,v). This vector isn't pointing straight up. It makes an angle arctan(v/c) with the time axis. This angle is in the interval [0,45°). That's what we mean when we say that the object is moving through both space and time. We're talking about the direction of its four-velocity vector
in our own rest frame.
Now look at the words "Something traveling at light speed
through space will have no speed left for motion through time". He's not talking about a particle with speed c. He's saying that if something has a four-velocity vector that makes a
90 degree angle with the time axis, it's not moving through time. This is an "object" (in some generalized sense) which is moving at
infinite speed, not light speed. The words "traveling at light speed" is a reference to the fact that a four-velocity vector always satisfies -(u^0)^2+\vec u^2=-c^2, and "through space" specifies that the projection of the vector he's talking about onto the time axis is zero.
Now let's talk about the
worst part of the Greene quote: "Thus light does not get old; a photon that emerged from the big bang is the same age today as it was then. There is no passage of time at the speed of light". This looks like complete crazy talk in this context, because he didn't even say that he has now abandoned the description of motion in terms of which way the four-velocity vector is pointing in
his rest frame, and is now talking about which way the four-velocity vector (c,0) of a stationary object will point if we apply a Lorentz transformation with speed v and take the limit v→∞. A Lorentz transformation with velocity -v takes (c,0) to \gamma(c,v), i.e. it tilts the time axis by an angle arctan v. It also tilts the x-axis in the opposite direction by the same amount. In the limit v→∞, both axes get tilted 45° in opposite directions, which makes them coincide. (They will point in the same direction or opposite directions, depending on the direction of the velocity).
There are several reasons why this inappropriate, and very misleading. The first is that he didn't
say that he's no longer considering the four-velocity vector in his own rest frame, and is instead trying to imagine what it looks like in "the photon's rest frame". The other reasons are the ones I have already stated (use the link in #2) for why it doesn't make sense to think of that limit as the photon's rest frame.
A claim has to make sense to be incorrect, and this one doesn't. So I don't think that's what those articles said. I'm assuming that they talked about the fact that the magnitude of the four-velocity (see e.g. 
