Let's discuss Mathematics

@Samson , about the +- presentation, another thought of how it can be presented using not just the real number line but distance (absolute value). So Xo is used as a pivot point to max/min x, which turns it (in this world) into an analogue of 0. For the special case, Xo is 0. I included also one use within secondary education math. Other uses include limits (by just having an inequality instead of an equality).

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A question. The following is my proof that if you have two equations of the form:
ax^2+bx+c
cx^2+bx+a
with the coefficients having the same value (c,a,b) and just c and a changing positions,
also with c(a) not zero

then it follows (a) that of their two roots -as c(a) is not zero, there are two roots and none is zero-, one root of the first will be the reciprocal of one root of the second
and also (b) that actually both of their roots will be reciprocals of the other roots.

I wish to ask:
1) is this true? (that always both roots of the first will have a reciprocal in roots of the second)
2) can you think of a different way to prove this? (eg with identities to do with b and the root of the discriminant)
3) I suspect that the reciprocal is always one with altered position in regards to -b/coefficient of the stable term being added or subtracted from the square root of the discriminant - eg if x1=(-b+sqrootD)/2a, it will always be the reciprocal of x2'=(-b-sqrootD)/2c. Is this true?

Here are my solutions:

1738769720238.png


Thanks for any help :) Eg @a pen-dragon
 
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This is indeed true and there is an easier (and more general) proof, providing a similar statement for any polynomial:

We want to solve a*x²+b*x+c=0. (Or the roots of an arbitrary polynomial)
Let r be the reciprocal of x. Then x=1/r and we obtain:
a*(1/r)²+b*(1/r)+c=0
If we multiply this by r² we get our final result:
c*r²+b*r+a=0

The condition that a and c must not be equal to zero emerges from demanding r to be finite and non-zero. a or c equal to zero will always produce a root at zero.

More generally this procedure will always reverse the order of the coefficients.


Concerning 3), this is indeed true, as can easily be checked by multiplying the standard solution of the roots. Note the order of the signs in the first terms is exchanged. Both cases result in the same.
(-b+-\sqrt(b²-4ac))/(2a)*(-b-+\sqrt(b²-4ac))/(2c) = (b²-(b²-4ac))/4ac = 1
 
This is indeed true and there is an easier (and more general) proof, providing a similar statement for any polynomial:

We want to solve a*x²+b*x+c=0. (Or the roots of an arbitrary polynomial)
Let r be the reciprocal of x. Then x=1/r and we obtain:
a*(1/r)²+b*(1/r)+c=0
If we multiply this by r² we get our final result:
c*r²+b*r+a=0

The condition that a and c must not be equal to zero emerges from demanding r to be finite and non-zero. a or c equal to zero will always produce a root at zero.

More generally this procedure will always reverse the order of the coefficients.


Concerning 3), this is indeed true, as can easily be checked by multiplying the standard solution of the roots. Note the order of the signs in the first terms is exchanged. Both cases result in the same.
(-b+-\sqrt(b²-4ac))/(2a)*(-b-+\sqrt(b²-4ac))/(2c) = (b²-(b²-4ac))/4ac = 1
Thanks. For some bizarre reason I didn't bother with the direct multiplication to find a proof for question 3 :D
And yes, the proof for question1 is certainly more condensed and alludes to more general use too - although I constructed my own meaning to use it for question3...
 
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I am trying to get a simple overview of the geometry of gravity waves. The best explaination I found is Slides 7 - 18 of the presentation below, but I got a bit lost at the last few stages with gauge conditions and polarization tensors and stuff. Am I right about the below?

- If I draw a right angle isosceles triangle on a flat plane I know the two sides adjacent to the right angle are the same length because of Euclid.
- If I draw a right angle isosceles triangle on a piece of paper and measure the two sides adjacent to the right angle with a wooden ruler they always seem the same length because any gravity wave affects the paper and the ruler similarly
- If I draw a right angle isosceles triangle with some hanging mirrors and measure the two sides adjacent to the right angle with a laser interferometer I have a gravity wave detector, and what I measure are real world deviations from the two sides of an isosceles triangle being the same length.

How do the arms vary in length as a wave goes through? There is these two diagrams in the presentation, but it does not make it clear how they relate to one another. Where is the source of the gravitational wave in that picture of a detector with arrows?

aPBPnsO.png

MyaXSgr.png


Slides 7 - 18 of this presentation
 
Keep in mind that gravitational waves' lowest angular momentum excitations are quadruple ones, corresponding to spin-2 (quasi-)particles. This means that the polarization is a different one from the one of light, which is composed of spin-1 photons and has a dipolar polarization.

Equivalently to photons the polarization is perpendicular to the propagation direction, due to the wave propagating with the speed of light.

For how to represent them, let me say here that a quadrupole can be thought to be two dipoles with opposite orientation and infinitesimal distance. For other visualizations check the Wikipedia articles for dipoles, quadrupoles and gravitational waves.

The blue arrows in the interferometer picture are such a visualization of a quadrupole oscillation. Remember that the oscillation is perpendicular to the propagation, and you see that the propagation is vertically to the interferometer.
 
A minor question on the above - or rather on a very specific part of them. How was the issue of no perfect right angle being constructed, overcome? In other words, why does indeed a (with very very many decimal parts met) approximation not interfere with this impressive tech?
(I tried to google for the answer, but due to being completely unaware of physics at such a level, it didn't succeed).
 
A minor question on the above - or rather on a very specific part of them. How was the issue of no perfect right angle being constructed, overcome? In other words, why does indeed a (with very very many decimal parts met) approximation not interfere with this impressive tech?
(I tried to google for the answer, but due to being completely unaware of physics at such a level, it didn't succeed).
The right angle is generated by the beam splitter. Could it be that that has more decimal places of accuracy than the interferometer?
 
A minor question on the above - or rather on a very specific part of them. How was the issue of no perfect right angle being constructed, overcome? In other words, why does indeed a (with very very many decimal parts met) approximation not interfere with this impressive tech?
(I tried to google for the answer, but due to being completely unaware of physics at such a level, it didn't succeed).
I think you would be interested in the whole thing, but they have they graph of sources of error:

wpTiFR1.png
 
The angle of the interferometer really does not have to be that precise. A small deviation of the angle from 90° would result in a small systematic uncertainty, but not hinder the total observation. Think of it as stretching the overall spectrum. You would see slightly different frequencies, but you would definitely still see oscillations, albeit with lower sensitivity.

Also the 90° angle is not a design necessity, it just provides the greatest sensitivity with two beam lines. The planned next generation detectors I know of will all be using three interferometers in an equilateral triangle setup. For reference look up the Einstein telescope and LISA.
 
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