# Differential equation from derivative of time dilation

• I
• bb1414

#### bb1414

Hi all! I was messing around with the equation for time dilation. What I wanted to do was see how the time of a moving observer ##t'## changed with respect to the time of a stationary observer ##t##. So I differentiated the equation for time dilation ##t'## with respect to ##t##:
$$\frac {dt'} {dt}=\frac {d} {dt}\frac t {\sqrt {1 - \frac {v^2} {c^2}}}$$
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}##!)

Your starting assumption is wrong. For a general observer for whom the velocity is not constant, the time-dilation is an integral relationship, i.e.,
$$\tau = \int^t \gamma^{-1} \, dt,$$
not ##\tau = \gamma t## (also note that it is ##\gamma^{-1}##, not ##\gamma##). Only for constant speed does this integrate to ##\tau = t/\gamma##.

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##.

This is never the case. That derivative will always be ##\gamma^{-1}##.

Dale
Your starting assumption is wrong. For a general observer for whom the velocity is not constant, the time-dilation is an integral relationship, i.e.,
$$\tau = \int^t \gamma^{-1} \, dt,$$
not ##\tau = \gamma t## (also note that it is ##\gamma^{-1}##, not ##\gamma##). Only for constant speed does this integrate to ##\tau = t/\gamma##.

This is never the case. That derivative will always be ##\gamma^{-1}##.

Thanks for clearing this up Orodruin. My answer did seem a little odd to me, and this makes much more sense.

Dale
Hi all! I was messing around with the equation for time dilation. What I wanted to do was see how the time of a moving observer ##t'## changed with respect to the time of a stationary observer ##t##. So I differentiated the equation for time dilation ##t'## with respect to ##t##:
$$\frac {dt'} {dt}=\frac {d} {dt}\frac t {\sqrt {1 - \frac {v^2} {c^2}}}$$
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}##.

I have a couple of concerns. Firstly, using your expression for t', I get
$$\frac{dt'}{dt} = \gamma +\frac{t\,v\,a}{\gamma^3 \, c^2}$$

I have some other concerns. The Lorentz transform gives

$$t' = \gamma \left( t - \frac{vx}{c^2} \right)$$

are you intending to evaluate ##\frac{\partial t'}{\partial t}## at x=0? This gives the answer that Orodruin mentioned if you work it out, namely

$$\frac{\partial t'}{\partial t} |_{x=0} = 1 / \gamma$$

In general ##\frac{\partial t'}{\partial t}## will depend along what worldline one evaluates it, and it's a partial differential equation, not a differential equation. One can avoid using PDE's if one manages to replace one of the time coordinates with proper time, ##\tau##.