Infinity in Finite Proper Time

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
0 replies · 3K views
Messages
7,643
Reaction score
1,602
Working on pervect's "messy unsolved" problem has led me to an interesting result. Let [itex]\left( x , t \right)[/itex] be a global inertial coordinate system for Minkowski spacetime.

Consider the worldline given by

[tex]t \left( \tau \right) = \frac{\tau^3}{3} - \frac{1}{4 \tau}[/tex]

[tex]x \left( \tau \right) = -\frac{\tau^3}{3} - \frac{1}{4 \tau}.[/tex]

Then

[tex]\frac{dt}{d \tau} = \tau^{2} + \frac{1}{4 \tau^2}[/tex]

[tex]\frac{dx}{d \tau} =- \tau^{2} + \frac{1}{4 \tau^2}.[/tex]

Note that [itex]dt/d\tau > 0[/itex], and that

[tex] \begin{align}<br /> \left( \frac{dt}{d \tau} \right)^2 - \left( \frac{dx}{d \tau} \right)^2 &= \left( \tau^{2} + \frac{1}{4 \tau^2} \right)^2 - \left( - \tau^{2} + \frac{1}{4 \tau^2} \right)^2\\<br /> & = 1.<br /> \end{align}[/tex]

Therefore, [itex]\tau[/itex] is the proper time for a futute-directed timelike worldline.

Note also that when [itex]\tau = -1[/itex], both [itex]t[/itex] and [itex]x[/itex] are finite, but as [itex]\tau \rightarrow 0_-[/itex], both [itex]t[/itex] and [itex]x[/itex] wander off to positive infinity.

The situation is unphysical because the 4-acceleration is unbounded, although there are no hyperlight speeds.

Regards,
George

PS I think I have found an expression for the 4-acceleration of a specific example of pervect's problem, but I have to check to see if my solution really does satisfy the necessary criteria.
 
Last edited: