# Time shown on the face of a moving clock

1. Aug 19, 2014

### m4r35n357

I would appreciate it if some senior member would give me feedback about this result that I have been using in my raytracing simulations of relativistic flight, https://www.youtube.com/playlist?list=PLvGnzGhIWTGR8QYYvMCweXPHtZPnsjrw8.

For simplicity I am assuming that the primed and unprimed coordinate origins concide, and that $c = 1$.

Starting from the general Lorentz transform in 2+1 spacetime for an observer in the primed frame moving along the $x$ axis at velocity $v$ through a "scene" at rest in the unprimed frame:
$$\left(\matrix{% t' \cr x' \cr y' }\right) = \left(\matrix{% \gamma & -v\gamma & 0 \cr -v\gamma & \gamma & 0 \cr 0 & 0 & 1 }\right) \left(\matrix{% t \cr x \cr y }\right)$$
Which can be used straightforwardly to derive the exresssions for aberration, doppler shift etc. The light travel delay from any point ($x, y$) in the unprimed frame to a stationary observer at the origin is given by the $t$ component, and to a moving observer is given by the $t'$ component. If we apply the light cone constraint (light delay is just the radial distance from the observer to a point,$R = \sqrt(x^2 + y^2)$, similarly for $R'$ in the primed frame) to this we have:
$$\left(\matrix{% R' \cr R' \cos \alpha' \cr R' \sin \alpha' }\right) = \left(\matrix{% \gamma & -v\gamma & 0 \cr -v\gamma & \gamma & 0 \cr 0 & 0 & 1 }\right) \left(\matrix{% R \cr R \cos \alpha \cr R \sin \alpha }\right) = \left(\matrix{% \gamma R (1 - v \cos \alpha) \cr \gamma R (\cos \alpha - v) \cr R \sin \alpha }\right)$$
where $\frac{y}{x} = \tan \alpha$ and $\frac{y'}{x'} = \tan \alpha'$, so that:
$$R' = \gamma (1 - v \cos \alpha) R$$
which gives the light travel delay in terms of quantities in the unprimed frame (this is simpler). Note that the ratio of $R'$ to $R$ is numerically identical to the doppler factor. The time $T$ seen on the clock face by the moving observer is then given by subtracting $R'$ from the coordinate time in the unprimed frame:
$$T = t - \gamma (1 - v \cos \alpha) R = t - \frac{\sqrt(x^2 + y^2) (1 - v \cos \alpha)}{\sqrt(1 - v^2)}$$
In words; the time on the clock face seen by the moving observer is the coordinate time in the rest frame, delayed in line with the transformed light cone in the observer's frame.

As I said, comments and corrections welcome!

Last edited: Aug 19, 2014
2. Aug 21, 2014

### m4r35n357

Not even wrong, huh?
OK I've spotted my mistake, which is to apply the light cone calculations to events. I am currently testing the real answer (which is much simpler) and will correct this post when I am happy with it.

Last edited: Aug 21, 2014
3. Aug 22, 2014

### m4r35n357

Apologies fot the intial post, for the record this is how I am doing it now, and it seems to be correct, ie. it gives the right answers for the twin paradox.

"Home" is the stationary frame, "ship" is moving frame. Home clock is at $(X, Y, Z)$. For each moving object, log all events $(t, x, y, z)$ against the corresponding proper time, $\tau$.

Define:
$R = \sqrt((x - X)^2 + (y - Y)^2 + (z - Z)^2)$
$T_1 = t + R =$ Home time when $\tau$ is seen on the ship's clock.
$T_2 = t - R =$ Home time seen on ship's clock at $\tau$

Log $T_1, T_2$ against the corresponding $\tau$ and event.

Plot $T_2$ aginst $\tau$.
Plot $\tau$ aginst $T_1$

Done! Plots attached for twin paradox @ $v = 0.8c$, separation = 4 units, total time in rest frame = 10 units.

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Last edited: Aug 22, 2014
4. Aug 22, 2014

### m4r35n357

OK, just to wrap this up, here are some real clock times for a twin "paradox" run over 20 light years at an acceleration of (+-)0.103 (representing 0.1g).
And here is the video
Just for the hell of it I have added a clock at the "far/away" end for the Ship time graph.

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