Relativistic time dilation with vectors (any velocity direction) and seeming contradiction

I've studied the existing derivation for relativistic time dilation, namely the light clock example that results in the following well-known equation:

$$\Delta t = \Delta t_0\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$$

This usually begins with a stationary light clock with a photon moving up and down. Time dilation comes in when the clock's reference frame is moved along in a "forward" direction (i.e. to the right in the typical diagram) at some velocity u.

My first question: Why is it assumed that the photon and velocity directions are ultimately perpendicular?

What would happen if the photon were moving in the same direction as the velocity vector? I wanted to know, so I did the following math (excuse the ASCII art).

Stationary frame

Assume we have a photon emitter pointing "forward":

      d
(( ------>
|


The time it takes the photon to reach a distance d is given by:

$$\Delta t_0=\frac{d}{c}$$

Moving frame

Imagine that this photon emitter is now placed onto a train, also moving "forward" at some velocity u.

             d
(( ------>
uΔt     |
----> -----
O   O


Assuming the speed of light, c, is identical in both frames, the time it takes for the photon to reach distance $$d+u\Delta t$$ (from the perspective of an external resting frame) is given by:

\begin{aligned} \Delta t&=\frac{d+u\Delta t}{c} \\ &=\frac{d}{c}+\frac{u\Delta t}{c} \\ &=\Delta t_0+\frac{u\Delta t}{c} \\ \Rightarrow \Delta t-\frac{u\Delta t}{c}&=\Delta t_0 \\ \Rightarrow \Delta t\left(1-\frac{u}{c}\right)&=\Delta t_0 \\ \Rightarrow \Delta t&=\Delta t_0\frac{1}{1-\frac{u}{c}} \\ \end{aligned}

The end result is obviously very different from the standard equation given at the top of this post. Note that if you point the emitter upwards, you'll end up with that more common equation (you don't actually need two mirrors in the light clock scenario for the math to work out the same).

A more general equation

I carried this further and did the vector-based math (for vectors of any dimension) for an emitter at an arbitrary angle $$\theta$$ with respect to the direction of movement. I'll leave out the ASCII art this time.

Let $$\vec{d}$$ represent the travel of the emitted photon in the stationary frame. The time it takes this photon to traverse the vector is given by:

$$\Delta t_0=\frac{\left|\vec{d}\right|}{c}$$

Now for the vector-based inclusion of a moving frame. Let $$\vec{u}$$ represent the frame's velocity vector. Again assuming the speed of light, c, is identical in both the stationary and moving frames, the time it takes for the photon to traverse its new path (from the perspective of an external resting frame) is given by:

$$\Delta t=\frac{\left|\vec{d}+\vec{u}\Delta t\right|}{c}$$

Doing some algebra, we can solve this for $$\Delta t$$:

\begin{align} \Delta t^2&=\frac{\left|\vec{d}+\vec{u}\Delta t\right|^2}{c^2} \\ &=\frac{\left<\vec{d}+\vec{u}\Delta t,\vec{d}+\vec{u}\Delta t\right>}{c^2} \\ &=\frac{\left<\vec{d},\vec{d}\right>+2\left<\vec{d},\vec{u}\Delta t\right>+\left<\vec{u}\Delta t,\vec{u}\Delta t\right>}{c^2} \\ &=\frac{\left|\vec{d}\right|^2+2\Delta t\left<\vec{d},\vec{u}\right>+\Delta t^2\left<\vec{u},\vec{u}\right>}{c^2} \\ &=\frac{\left|\vec{d}\right|^2}{c^2}+\Delta t\frac{2\left|\vec{d}\right|\left|\vec{u}\right|\cos{\theta}}{c^2}+\Delta t^2\frac{\left|\vec{u}\right|^2}{c^2} \\ &=\Delta t_0^2+\Delta t\frac{2\Delta t_0 \left|\vec{u}\right|\cos{\theta}}{c}+\Delta t^2\frac{\left|\vec{u}\right|^2}{c^2} \\ \end{align}

By combining like terms of $$\Delta t$$, applying the quadratic formula (keeping only the positive root), and quite a bit of simplifying, $$\Delta t$$ becomes:

$$\Delta t = \Delta t_0\frac{1}{\sqrt{1-\frac{\left|\vec{u}\right|^2}{c^2}\sin^2{\theta}}-\frac{\left|\vec{u}\right|}{c}\cos{\theta}}$$

It can be verified that for $$\theta=\frac{\pi}{2}$$ (emitter pointing perpendicular to movement) $$\Delta t = \Delta t_0\frac{1}{\sqrt{1-\frac{\left|\vec{u}\right|^2}{c^2}}}$$, and for $$\theta=0$$ (emitter pointing in the same direction as movement) $$\Delta t=\Delta t_0\frac{1}{1-\frac{\left|\vec{u}\right|}{c}}$$.

My next question (albeit essentially a rewording of the first): Why isn't a more general equation like this used? Without it, it seems a contradiction can arise.

Let's return to the forward-pointing emitter scenario and compare the traditional equation with the more general one derived above. We'll start with the general equation (without vectors, for simplicity, since this scenario is one-dimensional):

\begin{align} \Delta t&=\Delta t_0\frac{1}{1-\frac{u}{c}} \\ \Rightarrow \Delta t_0&=\Delta t\left(1-\frac{u}{c}\right) \\ \end{align}

The photon's total velocity (from the perspective of an external resting frame) is:

\begin{align} \frac{d}{\Delta t}&=\frac{c\Delta t_0+u\Delta t}{\Delta t} \\ &=\frac{c\Delta t\left(1-\frac{u}{c}\right)+u\Delta t}{\Delta t} \\ &=c\left(1-\frac{u}{c}\right)+u \\ &=c-u+u \\ &=c \\ \end{align}

This is expected. Now, let's run the same experiment using the traditional equation:

\begin{align} \Delta t&=\Delta t_0\frac{1}{\sqrt{1-\frac{u^2}{c^2}}} \\ \Rightarrow \Delta t_0&=\Delta t\sqrt{1-\frac{u^2}{c^2}} \\ \end{align}

The photon's total velocity (from the perspective of an external resting frame) becomes:

\begin{align} \frac{d}{\Delta t}&=\frac{c\Delta t_0+u\Delta t}{\Delta t} \\ &=\frac{c\Delta t\sqrt{1-\frac{u^2}{c^2}}+u\Delta t}{\Delta t} \\ &=c\sqrt{1-\frac{u^2}{c^2}}+u \end{align}

It's not immediately apparent, but this total velocity is not always $$c$$. For frame movement velocities $$u=0$$ and $$u=c$$, the result becomes $$c$$ (easily verified by mental substitution). However, for $$u$$ values in between $$0$$ and $$c$$, the total velocity exceeds $$c$$. For instance, $$u=\frac{1}{2}c$$ yields a total photon velocity of $$c\frac{1+\sqrt{3}}{2}\approx 1.366c$$.

Plotting the total velocity in proportion to $$c$$ reveals this "hump" more visually. One can see that the plot begins and ends with a value of $$1.0$$, yet remains strictly greater than $$1.0$$ in the middle.

This brings me to my final question: What am I missing? I'd be foolish to assume there's a flaw in the traditional equation before considering the more likely scenario that I've got a gap somewhere (especially given I've never taken a single physics course). I just can't find it...

Thanks for bearing with me through this long post! I hope it was interesting at the least, and I'm looking forward to learning from your responses.

• – PM 2Ring Nov 30 '20 at 10:57
• Also, Why is the photon clock equivalent to all clocks?. The key point is that all clocks (in good working order), no matter their orientation or mechanism, have to experience the same time dilation. Any clock discrepancy would contradict the axiom that velocity is relative. – PM 2Ring Nov 30 '20 at 11:07
• @PM 2Ring What if it’s not actually time dilation they experience? For instance, if you assume that time dilation is impossible, and space itself just contracts, you can have different contractions at various angles without problems. I’ve done some math for this and will probably post it as a separate but related question. – Andrew M. Nov 30 '20 at 16:10