- #1

#### Rex_chaos

Suppose there is a plane wave u=exp(-i k x), where k is a wavenumber. How to determine it's moving direction?

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- #1

Suppose there is a plane wave u=exp(-i k x), where k is a wavenumber. How to determine it's moving direction?

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- #2

In general, you can find the direction of the power flow of an EM wave by calculating the Poynting vector:

**P** = **E x H**.

Anyway, for this case (the case of a TEM plane wave), the electric field may be expressed as:

**E(R)** = **E**0 exp(-i**k** dot **R**).

The corresponding magnetic field may be expressed as:

**H(R)** = (1/eta) **an x E(R)**.

Here, eta is the intrinsic impedance of the medium.

I will assume that your wave is linearly polarized which makes

**E**0= **ay** E0 cos(2[pi]f t)

and

**H**0 = **az** (E0 / eta) cos(2[pi]f t).

You should be able to prove to yourself that the power flow is in the +x direction.

eNtRopY

Anyway, for this case (the case of a TEM plane wave), the electric field may be expressed as:

The corresponding magnetic field may be expressed as:

Here, eta is the intrinsic impedance of the medium.

I will assume that your wave is linearly polarized which makes

and

You should be able to prove to yourself that the power flow is in the +x direction.

eNtRopY

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- #3

- 36

- 0

Perhaps i missed something during my physics classes but the k in this equation is the wavevector which gives you the direction. In general the solution to a 1D wave equation will be of the form:

Aexp(ikx)+Bexp(-ikx)

Where the plus is for moving to the right en the minus for moving to the left. In higher dimensions the k is a vector pointing in the direction of travel...

Aexp(ikx)+Bexp(-ikx)

Where the plus is for moving to the right en the minus for moving to the left. In higher dimensions the k is a vector pointing in the direction of travel...

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- #4

Aexp(ikx)+Bexp(-ikx)

Where the plus is for moving to the right en the minus for moving to the left...

Yes, I know the result. However, how to prove that +k corresponding to a wave moving to the right?

- #5

- 36

- 0

It has a definite value at each point in space which is fixed in time.

Only if you have :

Aexp(i(kx-wt)) + Bexp(-i(kx+wt))

you can have a moving wave. Note the difference of the sign in front of w.

To show in which direction it moves it suffices to study the argument being zero:

kx-wt=0 --> x=ct moving to +x

kx+wt=0--> x=-ct moving to -x

where c=w/k the velocity of the wave.

- #6

Pete

- #7

- #8

Originally posted by heumpje

Perhaps i missed something during my physics classes but the k...

It is true that in the trivial case which has the form E ~ exp(-kx) the wavevector gives the direction of the traveling wave. However, examine a case with superpositioned wave patterns having unequal direction, orientation and magnitude. You will see that calculating the Poynting vector gives the most straight-forward method for determining the direction of an EM wave.

eNtRopY

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- #9

Originally posted by pmb

Pete

It is common practice to absorb the time dependence in the magnitude coefficient as I shown above. I assume Rex_chaos meant:

u [pro] exp(-i k x)

rather than

u = exp(-i k x).

Otherwise, his problem is too trivial.

eNtRopY

- #10

Sometimes that's true. But to determine the direction you have to know what it was that was absorbed.

Pete

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