possible answers

Here we are looking for transparently-described solutions to the puzzlers on our "drops in from hyperspace" page, although a simple answer key might also be useful as a way to cross-check the result of those solutions. We can also assume that you have access to the tables in "the acceleration 3-vector" page above.

Question set 1

Solutions aside, for an answer key only one might be able to use for all legs of the detour a characteristic time τo = Sqrt[(γo+1)/2]c/α ≈ 1.09868 [years] (where γo = Sqrt[1+(wo/c)2] ≈ 1.41421) to show that:

1a) the enemy velocity matched the battleship velocity on docking of wB = ½(1+cosh[τ/τo])wo+τosinh[τ/τo]α and was therefore wE ≈ Sqrt[1.221812+1.14392] = 1.67372 [ly/ty] at roughly atan[1.22181/1.1439] ≈ 46.886o counterclockwise from right;

1b) the proper-acceleration given as α = 1 [ly/y2] ≈ 9.51287 [m/s2] amounts to (from units conversion) about 0.970043 or nearly one [gee];

1c & 1d) the docking segment was given to take ΔτB of about one [battleship year], which (using the equations in Table 2) translates to map-time elapsed Δt of about 1.58792 [map years] to finish up at displacement {Δx, Δy} = {0.535485, 1.07195} [lightyears] from ignition at the origin;

1e) per unit mass from the map-frame perspective during the speed up the total kinetic energy increase is ΔK/m = c2Δγ ≈ 0.535485c2 ≈ 4.8127×1016 [J/kg] while the momentum increase is Δp/m = Δw ≈ 0.673716c ≈ 2.01965×108 [m/s];

1f) per unit mass from the map-frame perspective on ignition the starting power output P/m ≡ (1/m)δE/δτB = atc2 = α2τosinh[τ/τo] is zero [W/kg] since at turnaround (τ=0) the force is perpendicular to velocity, while before docking P/m ≈ 1.1439c2 per year i.e. about 3.26451×109 [W/kg], and the initial frame-variant force ΣF/m = |a| = |½(sinh[τ/τo]/τo)wo+cosh[τ/τo]α| is 1[ly/y2] = α while just before docking ΣF/m ≈ 1.51938c per year i.e. ≈ 4.81792 [N/kg] or [m/s2] ≈ 0.506463 α, consistent with an earlier assertion of ours that |Σf| ≤ |mα| where the map-time version of the net frame-variant force is Σf ≡ δp/δt = (1/γ)δp/δτ = (1/γ)ΣF;

1g) the traveler time saved by the detour is 4.28781[ty]-4[ty] ≈ 3.45371 [traveler months], but there's a different story from the vantage point of couch potatoes awaiting your arrival, as I'm afraid they'll see the ship arrive 6.35169[y]-6.06388[y] ≈ 3.45371 [map months] late!

Question set 2

To determine the pre-chase relative proper velocity of the enemy ship with respect to the battleship, and conversely of the battleship with respect to enemy ship, we have to go beyond the 1-map approach to use of the "multi-map" vector proper-velocity addition equation set wAC = wAB* + wB*C. Both solutions are depicted as blue vectors in the figure at right.

Let R stand for our reference-map (or consensus-metric) frame, B for the pre-chase battleship frame, and E for the enemy ship frame. Then start with wBR = {0,1} [ly/ty] and wER ≈ {1.1439, 1.2218} [ly/ty]. Multi-map vector addition then gives for question 2a) that wEB ≈ {1.1439, -0.22185} [ly/ty], and for question 2b) that wBE ≈ {-1.1439, -0.22185} [ly/ty].

Note in particular that unlike the low-speed case, wEB ≠ -wBE (even though their magnitudes are the same) when drawn from the map frame perspective because the 2nd "reference frame" subscript in each case specifies a different definition of simultaneity. This result of shifting map-frames might be referred to as "frame-shift" or Wigner rotation. It is clearly an artifact of one's frame choice, since from the vantage point of both E and B one can say that wEB = -wBE!