# Sagnac effect- the maths

1. Jul 4, 2004

### MathematicalPhysicist

in this webpage: http://www.mathpages.com/rr/s2-07/2-07.htm we have an explanation to the saganc effect and in page 2 and 3 i got stumbled:
i cant get to the quadratic equation in page 3 i.e:
[c^2-R^2*w^2*cos(theta)]*T^2+-[2R^2*w*sin(theta)]*T-2R^2[1-cos(thata)]=0
from this:
c^2*T^2=2R^2*[1-cos(w*T)*cos(theta)+-sin(wT)sin(theta)]
in page 2.

i tried to use limits because in the paper the said that w*T is extremely small, but what i got is
c^2*T^2=2R^2*[1-cos(theta)]
(i took limits to both the sin and cos functions with w*T as the angle), but as you see it doesnt match, then what have idone wrong?

2. Jul 5, 2004

### MathematicalPhysicist

no one knows?

3. Jul 5, 2004

### MathematicalPhysicist

for those too much lazy or dont understnad the equations i wrote here they are in the attachment.

Last edited: Jan 4, 2006
4. Jul 5, 2004

### MathematicalPhysicist

sorry but the other picture is too much big (anyway you can see it in the link i first gave i know :yuck: ).

5. Jul 5, 2004

### DW

Its just math. Perhaps no one cares, but why assume no one knows? That is "your" speculation. Try again, but this time as it said, keep powers of omega*T up to second order, NOT JUST FIRST ORDER.

6. Jul 5, 2004

### Chronos

w is the angular velocity of the device and T is the time required for the light pulse to travel from one mirror to the other in the foward and reverse direction. w*T therefore is a measure of distance, not angle. You need to plug in the radius of the ring to derive the angle of rotation, which is microscopic.

7. Jul 6, 2004

### MathematicalPhysicist

when you mean omega*T up to second order do you mean that the deravartive of the cos and sin of w*t will be of 2degree?

and to chronos when you multiply omega with T you get the unit of rad which can be translated to degrees with which you measure the length of the arc.
anyway chronos, when you mean the radius of the ring are you reffering to capital R in this paper?

8. Jul 6, 2004

### DW

No, I mean that you have to carry out the taylor series expansion of the cos function to second order in omega*t, not just to first. The sin function's second order term is zero so even though you are really to carry them both out to second order be sure to pick up the second order term from the cos expantion.
In other words use:
$$sin(\omega T) \approx \omega T$$
and
$$cos(\omega T) \approx 1 - \frac{\omega ^{2}T^{2}}{2}$$

Last edited: Jul 6, 2004
9. Jul 6, 2004

### MathematicalPhysicist

now that's more clear, thanks.

10. Feb 22, 2010