Relativity tangential light clock

Question

I summarised the question in the final para, if you prefer to skip all my rationale and trig.

btw, I have looked at

• Using light clocks, can one derive the length contraction formula without the 'bouncing' of the photon?
• How do I derive the Lorentz contraction from the invariant interval?
• What if a light clock travels perpendicular to mirrors that make up the clock?

I have also done some off site research, (not to mention the luxury of being taught some of this - a long long time ago).

None of which inputs seem to cover my precise point. I hope you will also find this is not a duplicate. There is necessarily some scene setting and context.

Reasoning

Consider two light clocks, A and B, in the same moving frame of reference and an observer, O, at rest.

These are initially one way light clocks in which the light travels from an emitter to a detector. They are not two way clocks in which light travels back to its origin, although that is discussed as a check and to show the reasoning can derrives the Lorentz transform correctly.

Clock A: Tangential to Velocity

The treatment of light clock A is the familiar clock from every special relativity text book, I recap it here.

I used this diagram even though it is a two way clock.

Light clock A is oriented so its light moves tangentially to the observed direction of motion. Light clock B is oriented so its light moves parallel to the observed direction of motion.
In special relativity the length contraction effect can be derived by applying basic trigonometry to the light path. This is well established in text books, the distance $D$ that O observes the light in clock A to travel is $$D^2=\frac{L^2}{\bigg(1-\frac{v^2}{c^2}\bigg)}$$
$L$ is the distance from light clock detector to mirror (in its own frame of reference)
$D$ is the distance the light travels in the observer's rest frame $v$ is the frame's velocity
$c$ is the assumed speed of light

O's necessary assumption the speed of light is constant means that he observes clock A running slower in proportion to the greater path length; and so we get length contraction.

Clock B: Parallel to Velocity

I now try to apply this logic to clock B. The distance that O observes the light in clock B to travel.
In the forward direction the distance that the detector moves is $$v\delta t$$ The distance the light moves must be this plus L (but length contracted) $$D=L/\gamma+v\delta t$$ where for the observer again $$\delta t=D/c$$ Unfortunatly that returns $$D=\frac{L}{\gamma\bigg(1-\frac{v}{c}\bigg)}$$

D is bigger than L, in O's frame light takes longer to travel D, and so time in the clock frame appears dilated to O.

I go on and do the two way clock.

If Clock B was Two Way Clock Parallel to Velocity

OK let's determine the length of the return journey of the light and add it to length of the outbound journey of the light, $D$ in the previous equation. I can say $$c\delta t+v\delta t=L/\gamma$$ (two "objects" moving together from a separation of L, one at v, one at c) This gives us $\delta t$ and so the length of the light's return journey is $$D_{return}=\frac{cL}{(c+v)}=\frac{L}{\gamma(1+v/c)}$$ Adding this up, then $$D_{bothways}=\frac{L}{(1+v/c)}+\frac{L}{\gamma(1-v/c)}$$ so this works out to $$D_{bothways}=\frac{2L}{\gamma(1-v^2/c^2)}$$ Now I didn't derrive it here but we can do that eleswhere $$\gamma=1/\sqrt(1-v^2/c^2)$$ So we get, which is twice as big for Clock A and correct. $$D_{bothways}=\frac{2L}{\sqrt(1-v^2/c^2)}$$

Conclusion

The clock A and clock B equations don't agree... so that under certain (contrived?) circumstance, if

• the effect being viewed is mono-directional and
• runs parallel to the the motion of the frame, (depending on whether it is in the direction or counter the direction of motion)

You will get different magnitudes of time dilation effects. OK, I better ask about this...

The Question

As a community we've already accepted length contraction as directional. Given that time dilation is a manifestation of the same effect, it seems only logical to accept that (under some conditions) relativistic temporal effects can also be directional e.g. one way light clocks (if such a thing can exist).

What is wrong with my application of maths?

Consequences

Just to try and make this sem relevant, we now start to get weird behavior if we take two differently oriented directional light clocks on a hypothetical relativistic journey. They will start to tell different times. I understand that GPS satellite clocks are relativisitcally compensated. If the (I believe atomic) clocks have a "directional" component, is it possible this effect could be significant?

Answer 1

Well , I see some deficiencies in your calculations and explanations:

1- I think in your first equation you must replace $L$ with $D$ and vice versa or you have to use $(1-\frac{v^2}{c^2})^{-1}$ instead of $(1-\frac{v^2}{c^2})$, because from O's viewpoint $D$ should be greater than $L$ according to your denotations for $D$ and $L$.

2- A one way light clock is meaningless. A clock must have its own "Tick"s and "Tack"s that represent a complete period. Moreover, tick and tack must both occur in one point where the point observer is located, regardless of the path which light travels, to illustrate a point clock for a point observer. Remember that the path of light can be any arbitrarily chosen path not essentially a straight line even it can be a function of a closed curvilinear using a set of infinitesimal mirrors and I proved this long ago that such a clock complies with Einstein's time dilation factor (the traditional gamma factor)!