How Does the Plane Wave Equation Define Wavefronts?

iScience
Messages
466
Reaction score
5
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.
 
Physics news on Phys.org
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
 
  • Like
Likes 1 person
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.
 
  • Like
Likes prosteve037
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Back
Top