Rotating Frame said:
A body at rest moves through time at the speed of light. An body in motion has some of it's energy diverted away from the time dimension, into the spatial dimensions, causing warps.
Brian Greene actually uses language like that, but (as you can see in his books) it doesn't have anything to do with "warps" (i.e. gravity). He uses it to explain time dilation and other SR phenomena. Here's a quote from "The fabric of the cosmos".
And just as Bart’s speed in the northward direction slowed down when he diverted some of his northward motion into eastward motion, the speed of the car through time slows down when it diverts some of its motion through time into motion through space. This means that the car’s progress through time slows down and therefore time elapses more slowly for the moving car and its driver than it elapses for you and everything else that remains stationary.
Zmunkz said:
I've never been able to find a technical discussion of what this interpretation comes from,
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It was described more-or-less as above: every single object has a "total velocity" of c through space and time.
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is there any validity to that interpretation?
I think this way of looking at it was made popular by Brian Greene. He uses it in both "The elegant universe" and "The fabric of the cosmos". The technical explanation (in units such that c=1, and with a -++++ signature) is as follows.
In my own rest frame, my world line coincides with the time axis. So the tangent to my world line is in the 0 direction (axes numbered from 0 to 3, with "time" being 0). Every vector of the form
$$\begin{pmatrix}r\\ 0\\ 0\\ 0\end{pmatrix}$$ where r is a real number is a tangent vector to the world line. The tangent vector with Minkowski "norm" -1 (-c for those who don't set c=1) is called my
four-velocity. I will denote its coordinate matrix in my own rest frame by u. We have
$$u=\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix},\qquad u^2=u^T\eta u=\begin{pmatrix}1 & 0 & 0 & 0\end{pmatrix}\begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix}=-1.$$
Greene calls ##\sqrt{-u^2}## the
speed through spacetime. We have ##\sqrt{-u^2}=1##. If we restore factors of c, this is ##\sqrt{-u^2}=c##. (Greene uses the metric signature +--- instead of -++++, so when he does this, he gets ##u^2=c^2##, and is therefore able to write the speed through spacetime as ##\sqrt{u^2}##. His ##u^2## is equal to my ##-u^2##).
Let's boost my four velocity to the rest frame of an observer who has velocity -v in my coordinate system. I should have velocity v in his.
$$u'=\Lambda(-v)u=\gamma\begin{pmatrix}1 & v^1 & v^2 & v^3\\ v^1 & * & * & *\\ v^2 & * & * & *\\ v^3 & * & * & *\end{pmatrix}\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix}=\gamma\begin{pmatrix}1\\ v^1\\ v^2\\ v^3\end{pmatrix}.$$ The asterisks denote matrix elements that are irrelevant to what we're doing here. If anyone cares, they are the components of the 3×3 matrix
$$\frac{1}{\gamma}I+\left(1-\frac 1 \gamma\right)\frac{vv^T}{v^Tv}.$$ The velocity components can be calculated like this:
$$\frac{dx^i}{dt^i}=\frac{u'^i}{u'^0}=\frac{\gamma v^i}{\gamma}=v^i.$$ As expected, my velocity in the new coordinate system is minus the velocity of the boost. This result is the reason why the normalized tangent vector is called the four-velocity.
The world line is the range of a curve ##C:\mathbb R\to M## where M is Minkowski spacetime. Its representation in a global coordinate system ##x:M\to\mathbb R^4## is the curve ##x\circ C:\mathbb R\to\mathbb R^4##. The world line is said to be
parametrized by proper time if the curve C that we use to represent it has the property that for each point p on the world line, the number ##\tau## such that ##C(\tau)=p##, is the proper time along the curve from C(0) to p. Such a C has the advantage that the four-vector with components ##(x\circ C)^\mu{}'(t)## is automatically normalized. So if y is my rest frame, and x is the coordinate system we transformed to above, we have ##u^\mu=(y\circ C)^\mu{}'(\tau)## and ##u'^\mu=(x\circ C)^\mu{}'(\tau)##. It's conventional to denote ##(x\circ C)^\mu(\tau)## by ##dx^\mu/d\tau##, so we have
$$u'^\mu=\frac{dx^\mu}{d\tau}.$$ Now let's use the fact that ##u^2## is Lorentz invariant.
$$-1=u^2=u'^2 =-(u^0)^2+(u^1)^2+(u^2)^2+(u^3)^2 =-\left(\frac{dt}{d\tau}\right)^2+\sum_{i=1}^3 \left(\frac{dx^i}{d\tau}\right)^2.$$ Let's manipulate this result with some non-rigorous physicist mathematics. (These things can of course be made rigorous).
\begin{align}
&\frac{dt}{d\tau} =\sqrt{1+\sum_{i=1}^3\left(\frac{dx^i}{d\tau} \right)^2}\\
&\frac{d\tau}{dt} =\frac{1}{\sqrt{1 +\sum_{i=1}^3\left(\frac{dx^i}{d\tau} \right)^2}}\\
&1=\left(\frac{d\tau}{dt}\right)^2 \left(1+\sum_{i=1}^3\left(\frac{dx^i}{d\tau} \right)^2\right) =\left(\frac{d\tau}{dt}\right)^2+\sum_{i=1}^3 \left(\frac{dx^i}{dt}\right)^2.
\end{align} Greene calls the square root of the first term the
speed through time and the square root of the second term the
speed through space. (This is according to note 6 for chapter 2 (p. 392) of "The elegant universe"). This allows him to say that an increase of the speed through space must be accompanied by an decrease of the speed through space.
If we had been talking about the motion of a massless particle (i.e. light) instead of the motion of an observer, we would have had ##u^2=0## instead of ##u^2=-1##. It's easy to see that what this does to the calculation above is to eliminate the first term on the right-hand side above. Since ##\tau=0## along the world line of a massless particle, this means that the result
$$\left(\frac{d\tau}{dt}\right)^2+\sum_{i=1}^3 \left(\frac{dx^i}{dt}\right)^2=1$$ holds for massless particles too.
