[b]Bell Spaceship Paradox (robphy) - v3.0[/b][br]A famous puzzle in Special Relativity that is probably best resolved by drawing a Minkowski spacetime diagram.

This is the LAB frame's spacetime diagram, where the LAB is at rest. [br]Using the usual conventions, time runs upwards and light signals are drawn at 45 degrees.[br]First we discuss some features of this spacetime diagram, [br]followed by a discussion of the [color=#9900ff]Bell Spaceship Paradox[/color].[br][br][br][list][*][i][b]Alice is riding in a uniformly accelerated spaceship. [br][/b][/i]Her worldline is a hyperbola. [br]At time t=0 in the LAB frame, Alice's position is given by event A[br]and, [u][i]for the purposes of this example,[/i][/u] Alice is momentarily at rest.[br]At event A, the tangent line at A is parallel to the LAB worldline.[br]When "[b]VEL[sub]Alice[/sub][/b]" is checked, the tangent line is drawn through [br]the draggable event X along Alice's worldline. [br][You can also drag that event X by selecting it and using keyboard arrow keys.][br][br][/*][*]Check "[b]Bob[/b]" to reveal [i][b]another uniformly accelerated observer Bob with the same [proper] acceleration as Alice.[/b][/i] [br]You can drag event B to event A to show that their worldlines are congruent.[br]Separate events A and B to obtain distinct worldlines for Alice and Bob.[br]At time t=0 in the LAB frame, Bob is also at rest.[br]To see this, check "[b]SIM[sub]LAB[/sub][/b]" to identify the events that are [br]simultaneous in the LAB frame with the draggable event X on Alice's worldline.[br]Then, check "[b]VEL[sub]LAB[/sub][/b]" to show the tangent lines through those events.[i][b][br]Note that Alice and Bob always have the same velocity [br]at any instant t in the LAB frame, and that velocity is zero at t=0.[br][br][/b][/i][/*][*][i][b]The meaning and representation of simultaneity is very important in this problem.[/b][/i][br]Note that this line of simultaneity for the LAB frame [br]is [b]perpendicular [in the spacetime-sense][/b] to the LAB's worldline.[br]In this case for the LAB frame, it looks like ordinary Euclidean perpendicularity.[br]A spacetime-diagrammatic way to define simultaneity for an observer[br]is by reflecting the observer's tangent line across [br]the light signal line to obtain the observer's line of simultaneity.[br]Temporarily check "[b]Lab LightCone[/b]" to see this.[br][br][/*][*][i][b]Which events are simultaneous for Alice?[/b][/i][br]Let's unclutter by unchecking "[b]Bob[/b]" and "[b]SIM[sub]LAB[/sub][/b]".[br]Check "[b]VEL[sub]Alice[/sub][/b]", "[b]Alice LightCone[/b]", then check "[b]SIM[sub]Alice[/sub][/b]".[br]By dragging event X to event A, we see that at time t=0 in the LAB frame,[br]Alice's line of simultaneity coincides with the the LAB's line of simultaneity.[br]Afterwards, for later events along Alice's worldline, they no longer conicide.[br][br][/*][*]As you move event X on Alice's worldline, note that[br]her line of simultaneity goes through the same event C[sub]Alice[/sub] (on the LAB's t=0 line).[br]This is a [i][b]property of a uniformly accelerated observer[/b][/i].[br]This event C[sub]Alice[/sub] is the intersection of the lightlike asymptotes of her worldline[br]and is called the [i][b]"center"[/b][/i] of the hyperbola[br](when you include the opposite branch on the other side of C[sub]Alice[/sub]).[br]One can say that, for a uniformly accelerated observer, [br]the radius from the center to event X is [b][br]perpendicular [in the spacetime-sense][/b] to the tangent line at event X.[br]Check "[b]Alice's Horizon[/b]" to see the lightlike asymptotes.[br][If she maintains the value of her acceleration, [br]Alice cannot receive signals from events to left of C[sub]Alice[/sub].][/*][/list][br][color=#9900ff][size=200]The Bell Spaceship Paradox.[br][/size][/color][list][*]Uncheck everything except for "[b]SIM[sub]Alice[/sub][/b]" and "[b]VEL[sub]Alice[/sub][/b]", then check "[b]Bob[/b]".[br][color=#9900ff][b]Alice and Bob are accelerated from rest in the LAB frame with the same uniform acceleration.[br]In the LAB frame, at each instant of time, Alice and Bob have the equal accelerations and equal velocities. Will a string connecting Alice and Bob break?[br][br][/b][/color][/*][*]Note that, in Alice's frame, event SIM[sub]Alice, on Bob[/sub] is simultaneous with event X on Alice's worldline.[br]More importantly, [b]in Alice's frame, Bob is not at rest, but is moving away from Alice.[/b][br]The tangent line to Bob's worldline at SIM[sub]Alice, on Bob[/sub] is generally[br]not parallel to the tangent line at event X on Alice's worldline.[br]A string connecting Alice and Bob cannot maintain the length it had [br]when event X coincided with event A. [b]The string will break.[/b][br][br]Temporarily check "[b]SIM[sub]LAB[/sub][/b]" and "[b]VEL[sub]LAB[/sub][/b]" to recall the situation [br]viewed from the LAB frame (where Alice, Bob, and the string are in motion).[br]By checking "[b]Bob's LightCone[/b]" and "[b]Bob's Horizon[/b]", [br]we see a line of events simultaneous according to Bob.[br]Although X and SIM[sub]Alice, on Bob[/sub] are simultaneous according to Alice,[br]they are not simultaneous according to Bob.[br][b]According to Bob,[/b] with his set of line of simultaneous events, [br][b]Alice is moving away from Bob.[/b] Again we conclude that [b]the string will break[/b].[br][br][/*][*]So, what we have is that:[br][b][i][color=#9900ff]even though in the LAB frame,[br]Alice and Bob have identical accelerations and identical velocities during the motion,[br]they do not have the same velocities in Alice's frame or in Bob's frame.[br]A moving string joining Alice and Bob cannot maintain its original length.[/color][/i][i][color=#9900ff][br]It must break.[/color][/i][br][/b][br][/*][*]Uncheck everything except for "[b]SIM[sub]Alice[/sub][/b]", "[b]VEL[sub]Alice[/sub][/b]", and "[b]Bob[/b]".[br]Now check "[b]Alice[sub]2[/sub][/b]".[br]Alice[sub]2[/sub] is a uniformly accelerated observer that is, [br]according to Alice, always at rest with respect to her.[br]Alice[sub]2[/sub] has the same "center" as Alice but has a different value of acceleration.[br]Think of the analogous case of concentric circles in Euclidean geometry.[br][b]A string connecting Alice and Alice[sub]2[/sub] won't break...[/b][br]even though in the LAB frame, after t=0, Alice[sub]2[/sub] does not have the same velocity as Alice.[br][br][/*][*]Slide event A[sub]2[/sub] to event B to compare the worldlines of Alice[sub]2[/sub] and Bob.[br]Slide event X on Alice's worldline from event A and onward into the future.[br]Note the [b]growing gap between Alice[sub]2[/sub] and Bob[/b].[/*][/list]