How Does the Plane Wave Equation Define Wavefronts?

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Homework Help Overview

The discussion revolves around the plane wave equation, specifically how the condition \(\vec{k}\cdot\vec{r}=Const.\) relates to the concept of wavefronts and the nature of plane waves in the context of both space and time dependencies.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the condition \(\vec{k}\cdot\vec{r}=Const.\) and question whether it limits the information provided by the wave equation. Some express confusion about the relationship between the wave function's dependence on time and space.

Discussion Status

Some participants have expressed understanding of the concept, while others are seeking further clarification on the book's intent and the derivation of the wavefront condition. There is an ongoing exploration of the definitions and implications of wavefronts in relation to the plane wave equation.

Contextual Notes

Participants are discussing the nature of wavefronts and the assumptions underlying the plane wave equation, including the interpretation of the condition \(\vec{k}\cdot\vec{r}=Const.\) and its implications for the phase of the wave.

iScience
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Plane wave equation:

$$\psi(t) = \psi_0e^{i(\vec{k}\cdot\vec{r}-\omega t)}$$

The part that makes the domain of \psi(t_i) a plane is the k dot r part.

I'm reading a book that takes this term and imposes the following condition:

$$\vec{k}\cdot\vec{r}=Const.$$

which, i understand its necessity, but if we just plug in the LHS of the equation, the information on the RHS is lost no? i mean, we didn't use it; we just got rid of it. Can someone clarify this part for me please.
 
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iScience said:
Plane wave equation:

$$\psi(t) = \psi_0e^{i(\vec{k}\cdot\vec{r}-\omega t)}$$

The part that makes the domain of \psi(t_i) a plane is the k dot r part.

I'm reading a book that takes this term and imposes the following condition:

$$\vec{k}\cdot\vec{r}=Const.$$

which, i understand its necessity, but if we just plug in the LHS of the equation, the information on the RHS is lost no? i mean, we didn't use it; we just got rid of it. Can someone clarify this part for me please.

##\vec k\cdot\vec r = \text{const.}## would mean that ##\psi## is a function of time alone.
Isn't a plane wave also a function of space?
http://en.wikipedia.org/wiki/Plane_wave#Arbitrary_direction

You don't "plug in" the LHS of that equation - the equation is a definition of what the LHS means. If you already know what ##\psi(t)## is, then what extra information could the RHS possibly supply?
 
aha! i get it! thanks
 
Hm, I don't get it. What's the book intending to derive/demonstrate? Could you quote more details?
 
iScience said:
Plane wave equation:

$$\psi(t) = \psi_0e^{i(\vec{k}\cdot\vec{r}-\omega t)}$$

The part that makes the domain of \psi(t_i) a plane is the k dot r part.

I'm reading a book that takes this term and imposes the following condition:

$$\vec{k}\cdot\vec{r}=Const.$$

A wave is a traveling disturbance, and the disturbance described by the function ψ depends both on place and time.
$$\vec{k}\cdot\vec{r}=Const.$$ is the equation of a wavefront, a plane, where the phase of the wave is the same at each point. Consider a wavefront where C=0 at t=0, that is, ##\vec{k}\cdot\vec{r}=0## . The equation represents a plane at the origin that is perpendicular to the wave vector ##\vec k##. At a later time t, the points where the phase is zero are on the plane
$$\vec{k}\cdot\vec{r}-ωt=0$$ In case ##\vec k ## is parallel with the x axis, ##\vec k =k\hat e_x##, the plane is perpendicular to the x-axis and its position is determined by ##k x -ωt=0##, that is, at ##x=ω/k t ##: the wavefront travels in the positive x dirction, with speed ω/k. ω/k is the propagation velocity or phase velocity of the wave.

ehild
 
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ehild said:
A wave is a traveling disturbance, and the disturbance described by the function ψ depends both on place and time.
$$\vec{k}\cdot\vec{r}=Const.$$ is the equation of a wavefront, a plane, where the phase of the wave is the same at each point. Consider a wavefront where C=0 at t=0, that is, ##\vec{k}\cdot\vec{r}=0## . The equation represents a plane at the origin that is perpendicular to the wave vector ##\vec k##. At a later time t, the points where the phase is zero are on the plane
$$\vec{k}\cdot\vec{r}-ωt=0$$ In case ##\vec k ## is parallel with the x axis, ##\vec k =k\hat e_x##, the plane is perpendicular to the x-axis and its position is determined by ##k x -ωt=0##, that is, at ##x=ω/k t ##: the wavefront travels in the positive x dirction, with speed ω/k. ω/k is the propagation velocity or phase velocity of the wave.

ehild

What they are saying is NOT that ## \vec{k} \cdot \vec{r} = constant ## everywhere. What they are saying is this: the vector ## \vec{k} ## is a constant, and the vector ## \vec{r} ## can be anything. Now, you pick constant C. Then all the points satisfying the condition ## \vec{k} \cdot \vec{r} = C ## lie on a plane, right? (and not that that plane will be perpendicular to ##\vec{k}##) What we know is that everywhere on that plane the wave function has the same phase at any given instant (fixed t). So all the points on that plane correspond a fixed phase. This is the definition of a plane wave.
 
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