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This question cropped up while I was playing with the equation for time dilation. If I set the speed to be $i$ (imaginary unit) the answer from the equation still makes sense, but does that matter if the input data doesn't make sense? Or rather does the input data make sense?

Let $v=i$ and $c=2$ then $T' = T \sqrt{1 -\frac{i^2} {2^2}}$, or $T' = T \sqrt{1 -\frac{-1} {4}}$, or $T' = T \sqrt{\frac{5}{4}}$.

Am I completely out of bounds here or is it something that can be explained?

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  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic ♦
    Commented Jul 25, 2014 at 18:31

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In what sense does the answer make sense? Does it describe some physical phenomenon?

You've done the algebra right, and gotten a result, but I don't think your inputs (imaginary velocity) mean anything physically. So the answer can't be expected to mean anything physically.

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  • $\begingroup$ The physical phenomenon it describes is the time dillation $T'$ between two objects moving relative to eachother. The relative differense in speed between them is $i$. I.e. object $a$ travels away from object $b$ at the speed of $i$ units for 1 unit of time. The speed unit used defines the speed of light in a vacuum as 2. How much does the time differ from $a$ and $b$? In the case above it differs by $\sqrt{\frac{5}{4}}$ units of time. $\endgroup$
    – fimas
    Commented Jul 25, 2014 at 17:25
  • $\begingroup$ The experimental results for time dilation show that an observer moving relative to a clock sees the hands of a clock moving more slowly than an observer stationary with respect to the clock. Your analysis shows the opposite. Note that in your notation, $T'$ is the time interval between two clock ticks in the stationary frame. Your analysis does not describe the time dilation that is actually observed. It does not describe a physical phenomenon. $\endgroup$
    – garyp
    Commented Jul 25, 2014 at 17:48
  • $\begingroup$ I guess that answers it then, an imaginary/complex speed value does not make sense since it breaks physics. $\endgroup$
    – fimas
    Commented Jul 25, 2014 at 22:02
  • $\begingroup$ Another way to see that it can't be of interest physically is to consider the case where $v=c$, which is intermediate between $v<c$ and $v>c$. You get nonsense for $v=c$. $\endgroup$
    – user4552
    Commented Oct 2, 2014 at 21:49
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First off Gamma is $\frac{1}{\sqrt{1-v^2/c^2}}$ but that's beside the point.

While by itself it looks okay, if you plug i into a Lorentz transformation everywhere there's a velocity you will also need to either have your time component complex or one of your spatial components as a complex number. This results in that component reversing sign meaning that the proper time or proper separation between events will no longer be equal which was the whole point of a Lorentz transformation in the first place.

Of note is that occasionally time is defined as complex for calculations in general relativity but this is done for both T and T' (I assumed that you were only defining it for one side of the equation); while this does result in what appears to be an imaginary velocity it's just a mathematical trick so the physicist doesn't have to bother keeping the signs straight in the calculations, since at the end it does not change the physics.

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