Relativity: Bell Spaceship Paradox (robphy) - v3.0
Bell Spaceship Paradox (robphy) - v3.0 A famous puzzle in Special Relativity that is probably best resolved by drawing a Minkowski spacetime diagram.
What this visualization shows
This is the LAB frame's spacetime diagram, where the LAB is at rest. Using the usual conventions, time runs upwards and light signals are drawn at 45 degrees. First we discuss some features of this spacetime diagram, followed by a discussion of the Bell Spaceship Paradox.
- Alice is riding in a uniformly accelerated spaceship. Her worldline is a hyperbola. At time t=0 in the LAB frame, Alice's position is given by event A and, for the purposes of this example, Alice is momentarily at rest. At event A, the tangent line at A is parallel to the LAB worldline. When "VELAlice" is checked, the tangent line is drawn through the draggable event X along Alice's worldline. [You can also drag that event X by selecting it and using keyboard arrow keys.]
- Check "Bob" to reveal another uniformly accelerated observer Bob with the same [proper] acceleration as Alice. You can drag event B to event A to show that their worldlines are congruent. Separate events A and B to obtain distinct worldlines for Alice and Bob. At time t=0 in the LAB frame, Bob is also at rest. To see this, check "SIMLAB" to identify the events that are simultaneous in the LAB frame with the draggable event X on Alice's worldline. Then, check "VELLAB" to show the tangent lines through those events. Note that Alice and Bob always have the same velocity at any instant t in the LAB frame, and that velocity is zero at t=0.
- The meaning and representation of simultaneity is very important in this problem. Note that this line of simultaneity for the LAB frame is perpendicular [in the spacetime-sense] to the LAB's worldline. In this case for the LAB frame, it looks like ordinary Euclidean perpendicularity. A spacetime-diagrammatic way to define simultaneity for an observer is by reflecting the observer's tangent line across the light signal line to obtain the observer's line of simultaneity. Temporarily check "Lab LightCone" to see this.
- Which events are simultaneous for Alice? Let's unclutter by unchecking "Bob" and "SIMLAB". Check "VELAlice", "Alice LightCone", then check "SIMAlice". By dragging event X to event A, we see that at time t=0 in the LAB frame, Alice's line of simultaneity coincides with the the LAB's line of simultaneity. Afterwards, for later events along Alice's worldline, they no longer conicide.
- As you move event X on Alice's worldline, note that her line of simultaneity goes through the same event CAlice (on the LAB's t=0 line). This is a property of a uniformly accelerated observer. This event CAlice is the intersection of the lightlike asymptotes of her worldline and is called the "center" of the hyperbola (when you include the opposite branch on the other side of CAlice). One can say that, for a uniformly accelerated observer, the radius from the center to event X is perpendicular [in the spacetime-sense] to the tangent line at event X. Check "Alice's Horizon" to see the lightlike asymptotes. [If she maintains the value of her acceleration, Alice cannot receive signals from events to left of CAlice.]
- Uncheck everything except for "SIMAlice" and "VELAlice", then check "Bob". Alice and Bob are accelerated from rest in the LAB frame with the same uniform acceleration. 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?
- Note that, in Alice's frame, event SIMAlice, on Bob is simultaneous with event X on Alice's worldline. More importantly, in Alice's frame, Bob is not at rest, but is moving away from Alice. The tangent line to Bob's worldline at SIMAlice, on Bob is generally not parallel to the tangent line at event X on Alice's worldline. A string connecting Alice and Bob cannot maintain the length it had when event X coincided with event A. The string will break. Temporarily check "SIMLAB" and "VELLAB" to recall the situation viewed from the LAB frame (where Alice, Bob, and the string are in motion). By checking "Bob's LightCone" and "Bob's Horizon", we see a line of events simultaneous according to Bob. Although X and SIMAlice, on Bob are simultaneous according to Alice, they are not simultaneous according to Bob. According to Bob, with his set of line of simultaneous events, Alice is moving away from Bob. Again we conclude that the string will break.
- So, what we have is that: even though in the LAB frame, Alice and Bob have identical accelerations and identical velocities during the motion, they do not have the same velocities in Alice's frame or in Bob's frame. A moving string joining Alice and Bob cannot maintain its original length. It must break.
- Uncheck everything except for "SIMAlice", "VELAlice", and "Bob". Now check "Alice2". Alice2 is a uniformly accelerated observer that is, according to Alice, always at rest with respect to her. Alice2 has the same "center" as Alice but has a different value of acceleration. Think of the analogous case of concentric circles in Euclidean geometry. A string connecting Alice and Alice2 won't break... even though in the LAB frame, after t=0, Alice2 does not have the same velocity as Alice.
- Slide event A2 to event B to compare the worldlines of Alice2 and Bob. Slide event X on Alice's worldline from event A and onward into the future. Note the growing gap between Alice2 and Bob.