Proper Frame of a Tachyon

Consider a tachyons having a superluminal velocity of u > c in the +x-direction of a reference frame pictured to be at rest. Also consider a frame of reference moving with the subluminal velocity v<c also in the +x-direction relative to the rest frame. Then the equation of motion of the tachyons in the rest frame is x = ut. Inserting this equation into the first and last Lorentz transformation equations gives

x‘ = g(u – v)t

t’ = g (1 – uv/c²)t

If we let uv = c², then the equation of motion of the particle in the moving system has x’ taking on all possible values at time t’ = 0. In a plot of x’ versus t’ this is a horizontal line, showing that the particle is everywhere along the x-axis simultaneously. In other words, the particle travels with infinite velocity along the x-axis and is simultaneously everywhere for just an instant.

The frame of reference for which uv = c² is called the proper frame of the superluminal particle. In this proper frame, the velocity of the superluminal particle is infinite.

If uv ≠ c², then eliminating time from the above two equations gives the equation of motion of the tachyon in the moving frame:

Since u is always greater than v, the numerator of this expression is always positive. Therefore, if v < c²/u, the tachyon moves in the +x’-direction, the same direction it moves in the rest frame. But if v > c²/u, the tachyon moves in the -x’-direction, opposite the direction it moves in the rest frame.

As the velocity of the reference frame v approaches –c, the superluminal particle velocity u’ approaches +c, and as v approaches +c, u’ approaches –c. Therefore, by simply changing from one reference frame to another the superluminal particle can be observed to move in either direction along the x’-axis at any speed greater than the speed of light, including at an infinite speed.

In our example, u is positive to the right (in the +x-direction), so our tachyon must have been emitted (created) on the left and detected (annihilated) on the right as it moved forward in time relative to the rest frame. But there exists another reference frame where the same tachyon moves to the left. Since the creation process occurs on the left and the annihilation process on the right, then, in this new frame, the particle is annihilated before it is created. Since the “lifespan” of the particle logically must go from its birth (creation) to its death (annihilation), we conclude that the tachyon must actually be traveling to the right and moving backward in time, in spite of the fact that it appears to travel forward in time to the left.

Therefore, we conclude that superluminal particles can travel forward in time, backward in time, or even be simultaneously present everywhere at once along a line.

Questions:

Q1. In the proper frame of a tachyon, the speed v of the tachyon satisfies the relationship: (A) v = 0 (B) 0 < v < c (C) v = c (D) (E) .
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ModPhy1/Unit1/SpecialRelativity/RelativeView/Time/Simultaneity/Causality/Superluminal/