*Sandokhan* please read JackBlack's derivation again and

**explain in detail what is wrong with it!**Here is the diagram that accompanies it:

And here is a re-formatted version of JackBlack's derivation:

Now let's try 2 arcs, one at **R**_{1} and one at **R**_{2}, where the sections connecting the 2 arcs are along the radii.

What this means is that going between them is the same for both directions.

In each one you have it going to the inner arc along a radius, and going to the outer arc along a radius, so the path length and time taken will be equal for those sections.

The only thing giving rise to a difference will be the sections along the arc.

So now let's let α be the angle subtended by the arcs, and ω be the angular velocity.

So, one beam of light will propagate in the same direction as the big arc and the opposite direction as the little arc.

What this means is its trip along the big arc will be longer than if it was stationary and shorter along the little arc.

The distance it has to travel is **α R**_{2}+ω R_{2} t_{1a}+α R_{1}-ω R_{1} t_{1b}, where **t**_{1a} is the time taken to go along the big arc and **t**_{1b} is the time taken to go along the little arc.

This is because in **t**_{1a}, the arc will have moved along a bit, and the light needs to travel the length of the arc and that bit it has moved along, while for **t**_{1b} (from the perspective of the light) the arc has travelled back a bit, shrinking the distance.

Meanwhile, the light going the other way has to travel a distance of **α R**_{1}+ω R_{1} t_{2b}+α R_{2}-ω R_{2} t_{2a}, for similar reasons.

(Note: the **a** is the big arc, the **b** is the little arc, this is to make it simpler later on).

The total time for each will be the sum of the **a** and **b** part, so now we need to figure out what they are.

Well, in **t**_{1a}, the light (travelling at speed **c**) travels a distance α **R**_{2}+ω t_{1a}.

Thus **t**_{1a} c = α R_{2}+ω R_{2} t_{1a}

Thus **t**_{1a} = α R_{2}/(c-ω R_{2}).

Similarly, **t**_{2b} = α R_{1}/(c-ω R_{1}).

And **t**_{1b} c = α R_{1}-ω t_{1b}

Thus **t**_{1b} = α R_{1}/(c+ω R_{1})

and **t**_{2a} = α R_{2}/(c+ω R_{2}).

Now as I said, the total time for each one is given by:

**t**_{1} = t_{1a}+t_{1b}.

**t = t**_{2a}+t_{2b}.

And then we find the difference as (note: may give minus sign, I haven't checked, but the important part is the magnitude)

**dt = t**_{1}-t_{2}

** = (t**_{1a}+t_{1b})-(t_{2a}+t_{2b})

** = t**_{1a}+t_{1b}-t_{2a}-t_{2b}

** = t**_{1a}-t_{2a}+t_{1b}-t_{2b}

** = (t**_{1a}-t_{2a})+(t_{1b}-t_{2b})

Rather than try to solve it all at once, for simplicity we break it into 2 parts:

**d**_{ta} = t_{1a}-t_{2a}

And **d**_{tb} = t_{1b}-t_{2b}.

And thus **dt = d**_{ta}+d_{tb}

Now then:

**d**_{ta} = t_{1a}-t_{2a}

** = α R**_{2}/(c-ω R_{2})-α R_{2}/(c+ω R_{2})

** = α R**_{2} (1/(c-ω R_{2})-1/(c+ω R_{2}))

** = α R**_{2} ((c+ω R_{2})/(c-ω R_{2}) (c+ω R_{2})-(c-ω R_{2})/(c+ω R_{2}) (c-ω R_{2}))

** = α R**_{2} ((c+ω R_{2})-(c-ω R_{2}))/(c+ω R_{2}) (c-ω R_{2}))

** = α R**_{2} (c+ω R_{2}-c+ω R_{2})/(c^{2}-(ω R_{2})^{2})

** = α R**_{2} 2 ω R_{2}/(c^{2}-(ω R_{2})^{2})

** = 2 ω α R**_{2}^{2}/(c^{2}-(ω R_{2})^{2})

And then if we assume **ω R**_{2} = v << c (i.e. our system is moving much slower than the speed of light, and 30 km/s is still much slower than the speed of light at roughly 300 000 km/s, then we can simplify **(c**^{2}-(ω R_{2})^{2}) to c^{2}

And thus we end up with **d**_{ta} = 2 ω α R_{2}^2/c^{2}

Now we do the same for **d**_{tb}.

**d**_{tb} = t_{1b}-t_{2b}

** = α R**_{1}/(c+ω R_{1})-α R_{1}/(c-ω R_{1})

** = α R**_{1} (1/(c+ω R_{1})-1/(c-ω R_{1}))

(combining some steps from before for brevity, and noting that **ω R**_{1} will be tiny compared to **c** just like **ω R**_{2})

** = α R**_{1} (c-ω R_{1}-c-ω R_{1})/c^{2}

** = -2 ω α R**_{1}^{2}/c^{2}

Thus **dt = 2 ω α R**_{2}^{2}/c^{2}-2 ω α R_{1}^2/c^{2}

** = 2 ω α (R**_{2}^{2}-R_{1}^{2})/c^{2}

Now, can we simplify this any more?

I know, lets work out the area.

Note that α has always been in radians.

A circle has an area **π r**^{2}.

This circle is a circular sector which subtends an angle of **2 π**.

If it only subtends an angle of **α**, then it will only have an area of **α/(2 π)** of the circle.

Thus **A** (for a circular sector) is **α/(2 π r**^{2})

** = α r**^{2}/2

This means from the center to the outer arc you have an area of:

**A**_{2} = α R_{2}^{2}/2.

And for the inner one you have an area:

**A**_{1} = α R_{1}^{2}/2.

Thus the area between them is:

**A = α R**_{2}^{2}/2-α R_{1}^{2}/2

** = α (R**_{2}^{2}-R_{1}^{2})/2

Thus **2 A = α (R**_{2}^{2}-R_{1}^{2}).

Notice that was in the formula above?

That means we can sub it in.

As a reminder we had:

**dt = 2 ω α (R**_{2}^{2}-R_{1}^{2})/c^{2}

By subbing in the above we get:

**dt = 2 ω 2 A/c**^{2}

And thus:

**dt = 4 A ω/c**^{2}

Now that certainly looks like the formula you had (with one simplification).

But notice a key fact?

**It uses the area of the loop, not the area of the orbit.**

So no, you have repeatedly been using the wrong radius/area to try and pretend that the orbital Sagnac effect should be much greater.

You can do the same without the simplification, but it gets more complicated, and it still isn't based upon the area of the orbit.

But please note!

- This result,
**dt = 4 A ω/c**^{2}, agrees with the references that you gave. - The delay does not depend on the location of the centre of rotation, as is clearly stated in the references that you presented.
- As with all the derivations that I have seen in the literature the Sagnac delay depends on the
*area of the light path* and *not the total area swept*.

If you disagree, please present your own derivation.