- #71

- 10,319

- 1,492

This is an example of geometric units. We measure all distances in light years, and all times in years. As a consequence, the numerical value of "c" is 1, as is the numerical value of "a".

We predict from the formula that ##dx/d\tau## must equal ##\sqrt{3}## for an object to have the same energy as an object moving with ##dx/d\tau=0## at a height of 1. At a height of 1, ##\gamma_g = 1+ax/c^2 = 1+1 = 2##, meaning an object at that height has twice its rest energy.

In order for ##dx/d\tau## to be ##\sqrt{3}##, we find ##v = dx/dt = \sqrt{3}/2## , giving ##dx/d\tau## = (dx/dt) \, (dt/d##\tau##) = ##\sqrt{3}##.

Here's the plot I did on a Rindler chart. The chart charts the course of the two rockets, our reference rocket starting at d=1, and a second rocket 1 light year "higher" in the accelerated frame. These plots are done in the inertial frame co-moving with the rockets at t=0.

These two rockets are plotted on the graph in black. The projectile launched at a velocity of ##\sqrt{3} / 2## follows a straight line inertial trajectory of ##d = \frac{\sqrt{3}}{2}\,t+1##, plotted in green.

The plot confirms the prediction. The green line representing the launched projectile just catches up to the black line of the higher rocket.

The equations are fairly simple if one wants to go into more depth. The reference rocket follows a hyperbola ##d^2-t^2=1##. The rocket one light year above the reference rocket follows a hyperbola of ##d^2-t^2 = 4##. There are various articles on "hyperbolic motion" that should explain this more - I'm not going to look for any further references unless there are questions. The green line does indeed meet the black line at d=4, t=##2 \,\sqrt{3}##