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Phase, Phase Difference and Phase Shift

 
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May5-08, 10:08 AM   #1
 

Phase, Phase Difference and Phase Shift


What's the difference between Phase, Phase Difference and Phase Shift in terms of waves?

I've taken a look to Wikipedia and a few other sites already, so please do not forward me to them...
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May5-08, 10:37 AM   #2
 
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"phase" usually refers to the angular argument of a function. E.g.,
[tex]
f(\phi)=\sin(\phi)\;,
[/tex]
where [tex]\phi[/tex] is the phase. Or, e.g.,
[tex]
f(x,t)=\sin(kx-\omega t)\;.
[/tex]
Here, [itex]kx-\omega t[/itex] is the phase. If we have two different waves and the first wave has phase [itex]\phi_1[/itex] and the second wave have phase [itex]\phi_2[/itex] the the "phase difference" is
[tex]
\Delta \phi=\phi_2-\phi_1\;.
[/tex]
For example, if I have two wave of the same frequency (2\pi\omega) and wave length (2\pi/k) which travel over different distances (L_1 and L_2, respectively) then there will be a phase difference
[tex]
\Delta \phi=k(L_2-L_1)
[/tex]
between the waves.

"Phase Shift" is just what it sounds like--a shift in the phase of a wave. Often this comes up in scattering where the effect of a scatterer on a wave is just to shift the phase of the wave by a "phase shift" (usually denoted by [itex]\delta[/itex]). E.g., For scattering a wave of wavelength 2\pi/k off a (very small) hard-sphere of radius R (very small means kR<<1), the scattering phase shift is [itex]\delta=-kR[/itex]. This means that if I have a wave which is initially of the form
[tex]
\cos(kz)
[/tex]
and I scatter it off a very small sphere, the resultant wave is of the form
[tex]
\cos(kz)-\frac{R}{r}\cos(kr-kR)\;,
[/tex]
where R is the (small) size of the sphere and r is the (large) distance to the point of observation (viewing screen or whatever).
May6-08, 06:54 AM   #3
 
Imagining the function sin(2[tex]\pi[/tex]ft + [tex]\phi[/tex]) ...

...you're saying that the whole argument (2[tex]\pi[/tex]ft + [tex]\phi[/tex]) is what so called "phase"?

So why do people (my lecturers included) keep saying that the phase in the above function is the [tex]\phi[/tex]?
As you might imagine, this leads me to some confusion...

PS: Thanks for your reply
May6-08, 07:26 AM   #4
 
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Phase, Phase Difference and Phase Shift


Sometimes, when people write down a periodic function like [tex]sin (2\pi f t + \phi) [/tex], they implicity separate out the time-varying part, which could be the same for different waves, and the constant [tex]\phi[/tex] which is allowed to vary. For example, when a light wave passes through a lens, the color of the light stays the same (the [itex] 2\pi f t [/itex] part), but the wavefront changes shape (the [itex] \phi [/itex] part).

So, people sometime refer to the absolute phase and relative phase, but all these are just modifiers to 'the phase' of a wave.
May6-08, 07:36 AM   #5
 
Thats initial phase angel , which means now wave has some y (displacement) initally at point x=0 due to the phase angel.
actually phase is important concept when you are dealing with interference of two or more waves . Suppose two waves are having same frequency and same velocity but have to cover diff. distances then at last point they will surely have some diff displacement due to path diff. And if they will have some path diff then there will also be some phase diffrence in btw them which is

§ (phase diff.) = k (wave no.) * (L2-L1)(path diff.)

i hope this help bit!
May6-08, 11:32 AM   #6
 
Quote by unseensoul View Post
Imagining the function sin(2[tex]\pi[/tex]ft + [tex]\phi[/tex]) ...

...you're saying that the whole argument (2[tex]\pi[/tex]ft + [tex]\phi[/tex]) is what so called "phase"?

So why do people (my lecturers included) keep saying that the phase in the above function is the [tex]\phi[/tex]?
As you might imagine, this leads me to some confusion...

PS: Thanks for your reply
technically the phase is the whole argument of the sine function and the phase shift is phi. People seldom get this terminology right though. However, usually the context in which the terminology is used makes it easy to identify the meaning.
May9-08, 09:15 PM   #7
 
Thank you very much for your help. Finally, I got it ;)
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