# Space as a 4-D MRI Scan?

by MattRob
Tags: scan, space
 Sci Advisor P: 2,470 Space as a 4-D MRI Scan? It's a little more complicated. The real reason is that everything always moves AT the speed of light. The angles change. But because of the way the time dimension works, the projection of the proper 4-velocity onto 3-space looks like normal 3-velocity with the norm capped at c. Ok, maybe I should start with this one from the beginning. First of all, let me work in natural units, where c=1. Also, lets drop 2 of the spacial coordinates, so that we have just two coordinates, one of which is time. The two coordinates are x(τ) and t(τ). τ is proper time, which you can think of as just a parameter in a parametric equation. More interestingly, if you have a clock that follows x(τ), t(τ) path, τ is the time that the clock will read. Hence the name. Proper velocity will also have two components. vx=dx/dτ and vt=dt/dτ. The later is of particular interest at this stage, because it tells us how far an object will move in time, relative to our coordinate system, vs how much time passes according to the clock moving with the object. In other words, this is time dilation. We define dt/dτ=γ. Proper velocity is not what we observe, however. What we are really watching is displacement with respect to our time. So v=dx/dt = vx/γ is what we are really looking at. It so happens that proper velocity of every physical particle has magnitude of the speed of light. Keeping Minkowski metric in mind, we have $$v_t^2 - v_x^2 = c^2 = 1$$ Since we have vt=γ, and vx² cannot be negative, the above sets limits on γ, which is greater or equal to 1. It also gives us vx in terms of γ. $$v_x = \pm \sqrt{\gamma^2-1}$$ Since we already have v=vx/γ, we can easily do the substitutions. $$v = \pm \sqrt{1-\frac{1}{\gamma^2}}$$ Two things you should be able to see here. First of all, you can solve for gamma. $$\gamma = \sqrt{\frac{1}{1-v^2}}$$ This looks exactly like your standard Lorentz boost formula with c=1. Second part is that we can establish limits on v. When γ=0, v = 0. When γ goes to infinity, v goes to ±1. There is no real value for γ that can make v>1. In the mean time, the proper velocity, vx goes to ±∞ as γ goes to infinity. That tells you that it's possible to travel any distance in finite amount of proper time. Also, notice that vt can take positive or negative value for any given γ. Standard model suggests that vt for particles and anti-particles has opposite sign. The choice of which is which is arbitrary.