Direction of travel of a plane wave given direction of electric field

[zenterix]

Homework Statement

If you have a plane EM wave whose electric field can be described by

$\vec{E}=-E_0\sin{(k(-y+vt))}\hat{i}$

Relevant Equations

What direction is the wave traveling?

Apparently, the direction of wave propagation is the direction of $\vec{E}\times\vec{B}$.

From what I have seen so far, given Maxwell's equations, the set of solutions giving plane waves has the characteristics that

1) electric field has only a component in the $y$ direction

2) magnetic field has only a component in the $z$ direction

3) the direction of wave propagation is the direction of $\vec{E}\times\vec{B}$

For $\vec{E}=-E_0\sin{(k(-y+vt))}\hat{i}$, it seems that the direction of propagation is the $+\hat{i}$ direction.

Thus, it seems we could have the magnetic field be in one of the directions $\pm\hat{j}$ or $\pm\hat{k}$.

 

[hutchphd]

Homework Statement: If you have a plane EM wave whose electric field can be described by

$\vec{E}=-E_0\sin{(k(-y+vt))}\hat{i}$
Relevant Equations: What direction is the wave traveling?

For E→=−E0sin⁡(k(−y+vt))i^, it seems that the direction of propagation is the +i^ direction.

Why do you think that. Details please.

 

[zenterix]

Why do you think that. Details please.

I was very confused when I wrote the OP but now less confused.

Consider the expression $\vec{E}(y,t)=-E_0\sin{(-ky+kvt)}\hat{i}=E_0\sin{(ky-kvt)}$.

Consider a specific time $t=0$. Then $\vec{E}(y,0)=E_0\sin{(ky)}\hat{i}$.

This is a function of the coordinate in the $\hat{j}$ direction.

At each $y$ we have an electric field vector pointing somewhere along the $x$ axis, ie, a multiple of $\hat{i}$.

At $y=0$, for example, we have $\vec{E}(0,0)=0$.

At $y=\pi/2$ we have $\vec{E}(\pi/2,0)=E_0\hat{i}$.

Here is a plot

So, what we have for $t=0$ is an entire sine wave along the $y$-axis.

Now let's consider some other time $t$.

Assume $k>0$, $v>0$.

Note that in $\sin{(ky-kvt)}$ the $kvt$ is just a constant and represents a negative phase angle, which means that we shift the plot seen above to the right (green plot below).

The wave, then, seems to be moving in the $+\hat{j}$ direction.

 

[zenterix]

I do have another doubt though.

Fix $y=0$ and consider $\vec{E}(t)=E_0\sin{(-kvt)}$.

The period of this oscillation seems to be $2\pi/kv$.

The period of the oscillation when we consider a fixed point in time and consider the oscillation as a function of $y$ is $2\pi/k$.

I am wondering if this difference comes from a mistake on my part or not. It seems like I must be making a mistake.

 

[hutchphd]

The period of the oscillation when we consider a fixed point in time

What does that even mean????......at a fixed point in time nothing moves oscillitory or not. You need to just relax and think a bit..

 

[zenterix]

What does that even mean????......at a fixed point in time nothing moves oscillitory or not. You need to just relax and think a bit..

We have an equation $\vec{E}(y,t)=E_0\sin{(ky-kvt)}\hat{i}$.

This is a vector-valued function of a vector.

That is, for each point in the $yt$-plane we have an associated vector $\vec{E}(y,t)$.

When we fix a time $t=t_1$ we get a function $\vec{f}(y)=\vec{E}(y,t_1)=E_0\sin{(ky-kvt_1)}$.

This is a sine which, as far as I know, is an oscillation.

It is not an oscillation in time, but rather in $y$.

Do you disagree?

I wrote of a $yt$-plane, but this is just the domain of the function. The image is 3d space.

Physically, at the instant $t_1$, $\vec{f}(y_1)$ tells us the electric field vector at every point in the plane $y=y_1$ located in 3d space. Thus, $\vec{f}(y)$ tells us the electric field vector at every point in space.

When we plot these vectors as a function of $y$ we get a sine function. Again, this is an oscillation.

My doubt is simply that I expected this oscillation to be the same as the oscillation we get when we consider the electric field vectors at one single plane as a function of time.

Is this really that stupid? I feel like there is probably just a constant correction for the fact that the domain in one case is $y$ and in the other case is $t$. From the formulas, this constant seems to be just $v$.

I mean, it is actually obvious mathematically from just the original equation for $\vec{E}(y,t)$.

I want to just make sense of it physically.

Suppose that instead of fixing a value for $t$, we fix a value for $y$. For simplicity, suppose that we investigate the function $\vec{g}(t)=\vec{E}(0,t)$.

$g$ tells us the electric field vector at every point on the plane $y=0$, at time $t$

$$g(t)=\vec{E}(0,t)=E_0\sin{(-kvt)}$$

which is a sine function.

 

[zenterix]

I realized that

- When I consider the period of an oscillation that is a function of $y$ and get $2\pi/k$, the unit is a length.

- When I consider the period of an oscillation that is a function of $t$ and get $2\pi/kv$, the unit is a unit of time.

So, already here there is no good reason to expect them to be the same in magnitude.

However, I think the intuitive explanation is the following.

At a fixed point in time, we have a snapshot of the wave, which has a certain wavelength measured in length units of $y$.

The wave, mathematically, is an oscillation that is a function of $y$.

The period of the oscillation is $2\pi/k$.

On the other hand, when we consider just one point and plot the electric field at that point as a function of time, then we need to consider that the wave (the one we took a snapshot of when we fixed $t$) is moving with some velocity.

Though the wave itself has a period that is constant, the oscillation of an individual point depends on how fast the wave is moving through that point.

$kv$ is simply the angular velocity (when $t$ is the independent variable) and so we divide by this to find the period of the oscillation at a fixed point as a function of time.

In the previous case, $k$ is the angular velocity when we consider $y$ as the independent variable.

 

[Orodruin]

Congratulations! You have discovered the difference between frequency and wavelength. As you have found they are related by the wave speed.

 

[hutchphd]

Yes the cycle of an harmonic wave in spacce is called the wavelength. The cyclic in time is called the period. For the OP: these are semantics, do you understand that any f(x,t) of the form g(x-vt) is a traveling wave?
Harmonic waves are particularly useful because they are affiliated with harmonic motion, which is how objects in equilibrium usually respond to perturbation from equilibrium