- #1

- 6

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$$\frac {dt'} {dt}=\frac {d} {dt}\frac t {\sqrt {1 - \frac {v^2} {c^2}}}$$

Where ##v=\frac {dx} {dt}##. This gave me the following result:

$$\frac {dt'} {dt}=γ+\frac {tva} {γc^2}$$

Where ##γ=\frac 1 {\sqrt {1 - \frac {v^2} {c^2}}}## and ##a=\frac {d^2x} {dt^2}##.

For one, I found this pretty interesting already, especially the product ##tva## in the numerator of the second term. Rewriting this in terms of the position of the moving observer ##x(t)## should help with I'm asking for:

$$\frac {dt'} {dt}=\frac 1 {\sqrt {1 - \frac {x'^2} {c^2}}}+\frac {tx'x''} {c^2}\sqrt {1 - \frac {x'^2} {c^2}}$$

I'm interested in finding a solution for when the change in the time of the moving observer with respect to the time of the stationary observer is zero, or ##\frac {dt'} {dt}=0##. More specifically, I'm trying to find a position function ##x(t)## that satisfies the differential equation:

$$0=\frac 1 {\sqrt {1 - \frac {x'^2} {c^2}}}+\frac {tx'x''} {c^2}\sqrt {1 - \frac {x'^2} {c^2}}$$

I'm not too great with differential equations (especially of this caliber) so I'm having some trouble solving this. Can any of you help me out here to find a generalized solution for ##x(t)##?

Thank you!

(Also, I'd love to hear your interpretations of my result for ##\frac {dt'} {dt}##!)