The general equation of the superposition of orthogonal waves?

[patric44]

Homework Statement

derive the general formula of the superposition of two waves which are orthogonal ?

Relevant Equations

x^2-2xy*cos(δ)+y^2 = A^2 sin(δ)^2

hi guys
i was trying to derive the general formula of two orthogonal waves
$$x^{2}-2xycos(δ)+y^{2} = A^{2} sin(δ)^{2}$$
where the two waves are given by :
$$x = Acos(ωt)$$
$$y = Acos(ωt+δ)$$
where $δ$ is the different in phase , i know it seems trivial but i am stuck on where should i begin ?
my trial :
$$x^2 = A^2 cos(ωt)^2$$
$$y = A[cos(ωt)cos(δ)-sin(ωt)sin(δ)]$$
squaring $y$ :
$$y^2 = A^2[cos(ωt)cos(δ)-sin(ωt)sin(δ)]^2⇒ y^2 = A^2 [cos(ωt)^2cos(δ)^2+sin(ωt)^2sin(δ)^2-2cos(ωt)cos(δ)sin(ωt)sin(δ)] $$
it seems that it got more complicated where to continue from here , or is there another method to get :
$$x^{2}-2xycos(δ)+y^{2} = A^{2} sin(δ)^{2}$$

 

[etotheipi]

Don't stop now, you're basically there! You have$$\begin{align*}x^2 + y^2 &= A^2[\cos^2{\omega t} + \cos^2{\omega t}\cos^2{\delta} + \sin^2{\omega t} \sin^2{\delta}] - 2A^2 \sin{\omega t}\cos{\omega t}\sin{\delta}\cos{\delta} \\ \\ -2xy\cos{\delta} &= -2A^2 \cos^2{\omega t}\cos^2{\delta} + 2A^2 \sin{\omega t}\cos{\omega t}\sin{\delta} \cos{\delta}\end{align*}$$What do you get if you add them together? Now it's just a case of applying $\sin^2{\theta} + \cos^2{\theta} = 1$ a couple of times

 

[haruspex]

Umm.. what has this to do with orthogonality?

 

[patric44]

Umm.. what has this to do with orthogonality?

this is the true statement of the problem , i guess because one wave is on the x and other on the y and has nothing to do with the more sophisticated meaning of the "orthogonality"

 

[etotheipi]

Umm.. what has this to do with orthogonality?

I think we have two orthogonal waves, one defined by $x = A\cos{(\omega t - kz)}$ and another defined by $y=A\cos{(\omega t - kz + \delta)}$, both propagating along the $z$ axis. Then we consider the wave formed by superposing those two. For the purposes of this problem, it would suffice to just consider the interference $z=0$.

[I think a real life application of this type of thing is resolving the electric field of an EM wave passing through a birefringent material w.r.t. the fast and slow axes, for instance!]

 

[haruspex]

I think we have two orthogonal waves, one defined by $x = A\cos{(\omega t - kz)}$ and another defined by $y=A\cos{(\omega t - kz + \delta)}$, both propagating along the $z$ axis. Then we consider the wave formed by superposing those two. For the purposes of this problem, it would suffice to just consider the interference $z=0$.

[I think a real life application of this type of thing is resolving the electric field of an EM wave passing through a birefringent material w.r.t. the fast and slow axes, for instance!]

In that model, wouldn't the superposition be $A\cos{(\omega t - kz)}\hat x+A\cos{(\omega t - kz + \delta)}\hat y$?
The algebra in your solution does not seem to have used any property of the wave pair other than having the same amplitude and frequency.

 

[etotheipi]

In that model, wouldn't the superposition be $A\cos{(\omega t - kz)}\hat x+A\cos{(\omega t - kz + \delta)}\hat y$?
The algebra in your solution does not seem to have used any property of the wave pair other than having the same amplitude and frequency.

That's indeed the superposition we're after! But we're just interested in deriving that identity $x^{2}-2xy\cos(\delta)+y^{2} = A^{2} \sin(\delta)^{2}$ for the $x$ and $y$ values on the resultant wave corresponding to any chosen $z$, so I don't think it's more complicated than that.

 

[TSny]

It took me a while to realize what the equation $x^{2}-2xy\cos(\delta)+y^{2} = A^{2} \sin(\delta)^{2}$ represents. I'll make a comment in case anyone else was having the same problem.

If you define a vector $\vec V = A\cos{(\omega t )}\hat x+A\cos{(\omega t + \delta)}\hat y$ and draw the vector from the origin of the x-y system, then the tip of $\vec V$ traces out a curve as a function of time.

The parametric represent of the curve is $x(t) = A\cos{(\omega t )}$, $\,\,\,y(t) = A\cos{(\omega t + \delta)}$.

Eliminating the parameter $t$ yields the direct relation between $x$ and $y$. The result is the equation of the ellipse $x^{2}-2xy\cos(\delta)+y^{2} = A^{2} \sin(\delta)^{2}$.

 

[etotheipi]

If you plot the parametric curve $(x,y,z) = (\cos{(\omega t - kz)}, \cos{(\omega t - kz + \delta)}, z)$, representing the superposition of two orthogonal waves of equations $x = \cos{(\omega t - kz)}$ and $y=\cos{(\omega t - kz + \delta)}$, you end up with something like

If you vary $t$, the wave translates in the $z$ direction. If you vary $\delta$, you change the eccentricity of the projection of the wave onto the $x-y$ plane.

 

[TSny]

You can play with this visualization

 

[etotheipi]

You can play with this visualization

That's awesome! I found a nice summary of the relevant Physics here: https://scholar.harvard.edu/files/schwartz/files/lecture14-polarization.pdf

There's also the interesting case where you have birefringence, $\Delta n \neq 0$, which results in the two 'component waves' having different wavenumbers, and if we add the components in the normal way we can work how the plane of polarisation has changed after exiting the crystal!

I think I need to read it again to fully understand it, but I hope @patric44, you also find it useful!