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I am tasked with this seemingly innocuous looking question:

Consider the electric field of an arbitrary plane electromagnetic wave: $$\vec E=\vec E_0 \exp{\left[i\left(\vec k \cdot \vec r-\omega t\right)\right]}$$ Write out the three components of $\vec E$.

So I give my answer as

$$E_x= E_{0x} \exp{\left[i\left(k\cdot x-\omega t\right)\right]}$$ $$E_y= E_{0y} \exp{\left[i\left(k\cdot y-\omega t\right)\right]}$$ $$E_z= E_{0z} \exp{\left[i\left(k\cdot z-\omega t\right)\right]}$$

such that $\vec E=E_x \hat x+E_y \hat y+E_z \hat z$.

Where $\hat x,\hat y, \hat z$ are the set of mutually orthogonal unit vectors.


But I was suprised to learn that the actual answer is

$$E_x= E_{0x} \exp{\left[i\left(k\cdot x+k\cdot y + k\cdot z-\omega t\right)\right]}$$ $$E_y= E_{0y} \exp{\left[i\left(k\cdot x+k\cdot y + k\cdot z-\omega t\right)\right]}$$ $$E_z= E_{0z} \exp{\left[i\left(k\cdot x+k\cdot y + k\cdot z-\omega t\right)\right]}$$


Firstly, this question seems a little subjective as the question specifically asks about the electric field (reason for boldface) but then uses $\vec E$ to denote the electromagnetic wave. So basically I would need to know if $\vec E$ is representing the electric field or electromagnetic wave?

Unfortunately, there is no way I can check this but supposing either is true:

If we are considering just the $x$ component of the electromagnetic/electric wave/field; Why do we need all three components ($x,y,z$) in the argument of the exponential?

Mathematically I understand why the answer is correct since the equation $\vec E=E_x \hat x+E_y \hat y+E_z \hat z$ would not hold if the arguments of the exponentials were different.

But physically, I cannot understand why the answer that I gave is wrong.

Why can't each component of the wave/field have a single component in the exponential?

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A quick partial answer: with overwhelming likelihood, you're misreading the solutions, which probably read \begin{align} E_x&= E_{0x} \exp{\left[i\left(k_x x+k_y y + k_z z-\omega t\right)\right]}\\ E_y&= E_{0y} \exp{\left[i\left(k_x x+k_y y + k_z z-\omega t\right)\right]}\\ E_z&= E_{0z} \exp{\left[i\left(k_x x+k_y y + k_z z-\omega t\right)\right]}, \end{align} where all that's happened inside the exponent is that they've broken out the dot product into component notation: $$ \vec k\cdot\vec r=k_x x+k_y y + k_z z. $$ Both forms are equivalent, so you could equally well read \begin{align} E_x&= E_{0x} \exp{\left[i\left(\vec k\cdot\vec r-\omega t\right)\right]}\\ E_y&= E_{0y} \exp{\left[i\left(\vec k\cdot\vec r-\omega t\right)\right]}\\ E_z&= E_{0z} \exp{\left[i\left(\vec k\cdot\vec r-\omega t\right)\right]}. \end{align}


However, this does differ from your version, \begin{align} E_x&= E_{0x} \exp{\left[i\left(k_x x-\omega t\right)\right]}\\ E_y&= E_{0y} \exp{\left[i\left(k_y y-\omega t\right)\right]}\\ E_z&= E_{0z} \exp{\left[i\left(k_z z-\omega t\right)\right]}, \tag{incorrect!} \end{align} which artificially changes the dependence of the electric field: you had an initial dependence of $\exp{\left[i\left(\vec k\cdot\vec r-\omega t\right)\right]}$ that was independent of the polarization, so you cannot now have the different components have different spatial dependence.

For more details on why, see the existing answers.

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  • $\begingroup$ Well done, the fault is mine. I have used the dots in a sloppy way to mean multiply instead of the 'dot product', sorry for this. However, at the end of your answer you write the $k$'s with subscripts; this is something that I did not do. Unless by "your version" you were referring to what the author of the question should have wrote? $\endgroup$ – BLAZE Jul 10 '17 at 18:17
  • $\begingroup$ @BLAZE It seems then that your notation is ill-defined to the point where it is unusable. You write $k\dot x$ ─ is that an inner product between vectors? (Hint: if it isn't, that dot has absolutely no business being there.) If $x$ is a vector, what is it? If $k$ just denotes a scalar, what happened to the vector dependence of $\vec k$? Or is $k$ just a stand-in for all the components of $\vec k$, and it takes different values within the same expression? (Or, in other words: can you see how a few notational mis-steps can completely sink a text to completely unreadable levels?) $\endgroup$ – Emilio Pisanty Jul 10 '17 at 18:28
  • $\begingroup$ That's an awful lot of questions; but basically my answer to them is 'I agree with you' and I should have used subscripts on the components of $\vec k$ to make them distinct from each other. $\endgroup$ – BLAZE Jul 10 '17 at 18:34
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The exponential is already a number, so it has no components, by definition. The only vector quantity in the expression of $\mathbf{E}$ is $\mathbf{E}_0$, therefore it is the only part that you have to project along the axes.

Remember that component does not mean "take the $x$", it means projecting a vector upon a basis.

Why can't each component of the wave/field have a single component in the exponential?

because that is not what the components of the original vector are: you can by all means define a new set of projections along the axes such that they will correspond to what you want, by the will not give rise to your initial vector, rather to another one (defined in fact by those new projections).

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Why do we need all three components (x,y,z) in the argument of the exponential?

There is nothing to prevent the $x$-component of the vector (become the exponential is a part of the $x$ component) to be a scalar function depending on $x,y$ and $z$. For instance the vector field (in 2d) $$ \vec E= \hat x E_x+\hat yE_y\, ,\qquad E_x=\sin(x)\cos(y)\, , \qquad E_y=\cos(x)\cos(y) $$ is a perfectly legitimate vector field.

And $\vec E$ refers to the electric field, but it describes a wave as it is a function of $\vec k\cdot \vec r-\omega t$: this is why the exponential is common to all three components. In this way $\vec E \sim e^{i(\vec k\cdot \vec r-\omega t)}\vec E_0$ with $\vec E_0$ constant describes a plane electromagnetic wave.

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    $\begingroup$ Yes that's much better thanks; in your answer you mention that "$\vec E$ refers to the electric field". How do you know this? $\endgroup$ – BLAZE Jul 10 '17 at 17:50
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    $\begingroup$ @BLAZE I've taught a course on this topic for 9 of the last 15 years... ;) plus the default guess for $\vec E$ is that it's an electric field. $\endgroup$ – ZeroTheHero Jul 10 '17 at 17:52
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    $\begingroup$ Woah, okay okay :-) I was just asking as it wasn't clear to me at all. But I'm thankful that it's clear to you. $\endgroup$ – BLAZE Jul 10 '17 at 17:56
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    $\begingroup$ @BLAZE no worries my friend... Wrapping one's head about this stuff is not so easy the first time around. $\endgroup$ – ZeroTheHero Jul 10 '17 at 17:59

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