STUDY.Physics.SpecialRelativity
SpecialRelativity: Inertial Frames
InertialFrames version 1.0.0-alpha
STUDY.Physics.SpecialRelativity/InertialFrames
Physics case studies: Special Relativity: Inertial Frames
Copyright (C) 2020 Julio P. Di Egidio (http://julio.diegidio.name)
InertialFrames is part of the STUDY.Physics.SpecialRelativity collection
(see https://github.com/jp-diegidio/STUDY.Physics.SpecialRelativity).
InertialFrames is released under the terms of the GNU-GPLv3 license.
As usual, NO WARRANTY OF ANY KIND is implied.
The game
Bob and Alice pass by each other collinearly at a relative constant speed.
They reset their clocks to zero at the moment they are colocated.
(For an interactive version of these diagrams see the
interactive demo page.)
From the space-time diagrams we can readily see that, while each of Bob’s and Alice’s clock keeps marking the one and only universal (aka “proper”) time, in their own frames each has the other moving away faster in the future, as well as faster in space than the plain frame speed, exactly by a factor of gamma.
Incidentally, this shows how inertial relative motion already has the elements of time travel. But notice that there is no paradox here, the situation is perfectly symmetrical as expected: the so called Twin Paradox rather ensues because Alice inverts her motion at some point and comes back to base, which is not symmetrical.
Below we show derivations of the temporal and spatial coordinates and velocities of “particles” B and A (for Bob and Alice, respectively), and of the corresponding “projected particles”. We parametrise all formulas, besides on the relative (frame!) speed, on the (one!) proper time.
Formal setup
In SR in 1+1 dimensions, with c = 1, and
generic coordinates of the form P = (tP, xP), and
generic velocities of the form vP = (v_tP, v_xP).
Given PARAMETER b (for "beta") a speed in [0, 1] sub Reals:
Let g (for "gamma") be 1/sqrt[1 - b^2] in [1, g] sub Reals.
// (i.e. in [1, oo], but we will keep gamma throughout as a scale factor)
For PARAMETER t (for "tau") the proper time in [0, 1] sub Reals:
(Notice that we keep track of the domains of definition of the various quantities just in order to help eventually render the two frames as finite diagrams: this has no relevance on the physics. Also notice that we use a somewhat simplified notation, most notably t instead of tau, because only ASCII characters are allowed here and in code.)
FRAME B
This is the frame of reference in which Bob is at rest at the origin of space.
We derive coordinates (whence velocities) in this frame by imposing the relevant physical constraints, then by plain geometry.
Particle B
This is Bob’s actual space-time location (in frame B).
We derive the coordinates from the constraint that Bob is at rest at the origin of space, hence xB = 0, then imposing the constraint that Bob’s proper distance from the origin of space-time is t (i.e. tau), hence t = sqrt[tB^2 - xB^2] = tB, hence tB = t:
B = (t, 0)
vB = (1, 0)
with the coordinates as derived above being:
tB(t) = t in [0, 1]
xB(t) = 0 in [0, 0]
and, by taking derivatives of the coordinates w.r.t. t, the velocities being:
v_tB(t) = d(tB(t))/dt = 1
v_xB(t) = d(xB(t))/dt = 0
Particle A
This is Alice’s actual space-time location.
We derive the coordinates from the constraint that Alice’s (frame) velocity is b, hence xA = b*tA, then imposing the constraint that Alice’s proper distance from the origin of space-time too is t (i.e. tau), hence t = sqrt[tA^2 - xA^2] = sqrt[tA^2*(1 - b^2)] = tA/g, hence tA = gt and xA = gbt:
A = (gt, gbt)
vA = (g, gb)
with the coordinates being:
tA(t) = gt in [0, g]
xA(t) = gbt in [0, g]
and the velocities being:
v_tA(t) = d(tA(t))/dt = g
v_xA(t) = d(xA(t))/dt = gb
Particle A_sim
This is the (moving) point of intersection of Alice’s world line with Bob’s line of simultaneity. It represents what Bob can measure about Alice in his own plane of simultaneity, e.g. here he finds length contraction and time dilation.
We derive A_sim’s coordinates by intersecting A’s world line, i.e. X = bT, with B’s line of simultaneity, i.e. T = t (with X,T formal parameters), hence T = t and X = bt, whence tA_sim and xA_sim.
We also notice that, all relations being linear, the ratio of A_sim’s proper distance (from the origin of space-time) to A’s proper distance must be the same as the ratio of A_sim’s time coordinate to A’s time coordinate. A_sim’s proper distance is t’ = sqrt[tA_sim^2 - xA_sim^2] = sqrt[t^2*(1 - b^2)] = t/g (with t’ again just a formal parameter), while A’s proper distance is given as t, hence we get t’ = t/g. So below we rather compute A_sim’s coordinates by applying Alice’s actual coordinates to this “rescaled proper time” (and, of course, we get the same result as above). [That is poorly phrased: I will fix it when I find the time…]
A_sim = (t, bt)
vA_sim = (1, b)
with the coordinates being:
tA_sim(t) = tA(t/g) = t in [0, 1]
xA_sim(t) = xA(t/g) = bt in [0, 1]
and the velocities being:
v_tA_sim(t) = d(tA_sim(t))/dt = 1
v_xA_sim(t) = d(xA_sim(t))/dt = b
Particle B_sim
This is the (moving) point of intersection of Bob’s world line with Alice’s line of simultaneity. It represents what Alice can measure about Bob in her own plane of simultaneity, e.g. here she finds length contraction and time dilation.
We derive B_sim’s coordinates by intersecting B’s world line, i.e. X = 0, with A’s line of simultaneity, i.e. (T-tA) = b*(X-xA) <=> (T-gt) = b*(X-gbt), hence X = 0 and T = gt(1-b^2) = t/g, whence tB_sim and xB_sim.
Now, doing as above, B_sim’s proper distance is t’ = sqrt[tB_sim^2 - xB_sim^2] = sqrt[(t/g)^2] = t/g , while B’s proper distance is given as t, hence we get t’ = t/g. So below we rather compute B_sim’s coordinates by applying Bob’s actual coordinates to this “rescaled proper time”.
B_sim = (t/g, 0)
vB_sim = (1/g, 0)
with the coordinates being:
tB_sim(t) = tB(t/g) = t/g in [0, 1/g]
xB_sim(t) = xB(t/g) = 0 in [0, 0]
and the velocities being:
v_tB_sim(t) = d(tB_sim(t))/dt = 1/g
v_xB_sim(t) = d(xB_sim(t))/dt = 0
Particle A_app
This is the (moving) point of intersection of Alice’s world line with Bob’s past light cone. It represents what Bob can see by looking in Alice’s direction.
We derive A_app’s coordinates by intersecting A’s world line, i.e. X = bT, with B’s past light cone, i.e. T = t-X, hence T = t/(1+b) and X = bt/(1+b), whence tA_app and xA_app.
Now, doing as above, A_app’s proper distance is t’ = sqrt[tA_app^2 - xA_app^2] = sqrt[(t/(1+b))^2 - (bt/(1+b))^2] = sqrt[t^2(1 - b^2)]/(1+b) = t/(g(1+b)), while A’s proper distance is given as t, hence we get t’ = t/(g(1+b)). So below we rather compute A_app’s coordinates by applying Alice’s actual coordinates to this “rescaled proper time”.
A_app = (t/(1+b), bt/(1+b))
vA_app = (1/(1+b), b/(1+b))
with the coordinates being:
tA_app(t) = tA(t/(g(1+b))) = t/(1+b) in [0, 1]
xA_app(t) = xA(t/(g(1+b))) = bt/(1+b) in [0, 1/2]
and the velocities being:
v_tA_app(t) = d(tA_app(t))/dt = 1/(1+b)
v_xA_app(t) = d(xA_app(t))/dt = b/(1+b)
Particle B_app
This is the (moving) point of intersection of Bob’s world line with Alice’s past light cone. It represents what Alice can see by looking in Bob’s direction.
We derive B_app’s coordinates by intersecting B’s world line, i.e. X = 0, with A’s past light cone, i.e. (T-tA) = (X-xA) <=> (T-gt) = (X-gbt), hence T = g(1-b)t = t/(g(1+b)) and X = 0, whence tB_app and xB_app.
Now, doing as above, B_app’s proper distance is t’ = sqrt[tB_app^2 - xB_app^2] = sqrt[(t/(g(1+b)))^2] = t/(g(1+b)), while B’s proper distance is given as t, hence we get t’ = t/(g(1+b)). So below we rather compute B_app’s coordinates by applying Bob’s actual coordinates to this “rescaled proper time”.
B_app = (t/(g(1+b)), 0)
vB_app = (1/(g(1+b)), 0)
with the coordinates being:
tB_app(t) = tB(t/(g(1+b))) = t/(g(1+b)) in [0, 1]
xB_app(t) = xB(t/(g(1+b))) = 0 in [0, 0]
and the velocities being:
v_tB_app(t) = d(tB_app(t))/dt = 1/(g(1+b))
v_xB_app(t) = d(xB_app(t))/dt = 0
Diagram boundaries
As can be extrapolated from above, coordinate boundaries for frame B are:
t in [0, g] // t here denoting the time coordinate
x in [0, g]
We can then draw a finite diagram by scaling down in both directions by a factor of gamma.
FRAME A
This is the frame of reference in which Alice is at rest at the origin of space.
We obtain coordinates in this frame by Lorentz-transforming by velocity b the corresponding coordinates in frame B. In fact, this is only to double-check the mathematical correctness of our previous derivations, the expected resulting diagram being perfectly symmetrical to the one found for frame B, so we could more efficiently obtain the coordinates in this frame (indeed we do, in code!) by just swapping the roles of (particles) B and A and inverting the sign of the space coordinates (and velocities) in all places.
Particle A’
Given that A = (gt, gbt):
tA'(t) = g(tA - b*xA) = g(gt - g(b^2)t) = t in [0, 1]
xA'(t) = g(xA - b*tA) = g(gbt - gbt) = 0 in [0, 0]
(Symmetrical to particle B in frame B as expected.)
Particle B’
Given that B = (t, 0):
tB'(t) = g(tB - b*xB) = g(t - 0) = gt in [0, g]
xB'(t) = g(xB - b*tB) = g(0 - bt) = -gbt in [0, -g]
(Symmetrical to particle A in frame B as expected.)
Particle B_sim’
Given that B_sim = (t/g, 0):
tB_sim'(t) = g(tB_sim - b*xB_sim) = g(t/g - b*0) = t in [0, 1]
xB_sim'(t) = g(xB_sim - b*tB_sim) = g(0 - b*t/g) = -bt in [0, -1]
(Symmetrical to particle A_sim in frame B as expected.)
Particle A_sim’
Given that A_sim = (t, bt):
tA_sim'(t) = g(tA_sim - b*xA_sim) = g(t - b*bt) = t/g in [0, 1/g]
xA_sim'(t) = g(xA_sim - b*tA_sim) = g(bt - b*t) = 0 in [0, 0]
(Symmetrical to particle B_sim in frame B as expected.)
Particle B_app’
Given that B_app = (t/(g(1+b)), 0):
tB_app'(t) = g(tB_app - b*xB_app) = g(t/(g(1+b)) - b*0) = t/(1+b) in [0, 1]
xB_app'(t) = g(xB_app - b*tB_app) = g(0 - b*t/(g(1+b))) = -bt/(1+b) in [0, -1/2]
(Symmetrical to particle A_app in frame B as expected.)
Particle A_app’
Given that A_app = (t/(1+b), bt/(1+b)):
tA_app'(t) = g(tA_app - b*xA_app) = g(t/(1+b) - b*bt/(1+b)) = t/(g(1+b)) in [0, 1]
xA_app'(t) = g(xA_app - b*tA_app) = g(bt/(1+b) - b*t/(1+b)) = 0 in [0, 0]
(Symmetrical to particle B_app in frame B as expected.)
Diagram boundaries
Again, as can be extrapolated from above, and as expected by plain symmetry with frame B, coordinate boundaries for frame A are:
t' in [0, g] // t' here denoting the time coordinate
x' in [0, -g]
This completes our calculations.