How to determine the direction of propgation of a plane wave?

  • Thread starter Rex_chaos
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  • #1
Rex_chaos
hi all,
Suppose there is a plane wave u=exp(-i k x), where k is a wavenumber. How to determine it's moving direction?
 

Answers and Replies

  • #2
eNtRopY
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) = E0 exp(-ik 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

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

and

H0 = 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
 
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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...
 
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  • #4
Rex_chaos
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
This wave does not move...
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
pmb
To determine in which direction the wave propagates you need to specity the time dependance as well as the space dependance. You've only given the spatial dependance of the phasor. There are two choices of a time dependance corresponding to two choices of the sign of "wt".

Pete
 
  • #7
damgo
If you're talking QM in the time-independent case, then it's just convention that the + waves goes right and the - goes left. It's actually erroneous to say they're going anywhere; "time-independent" scattering is a sort of vague approximation they teach you in beginning classes -- but of course they never explain why it works. :)
 
  • #8
eNtRopY
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
eNtRopY
Originally posted by pmb
To determine in which direction the wave propagates you need to specity the time dependance as well as the space dependance. You've only given the spatial dependance of the phasor. There are two choices of a time dependance corresponding to two choices of the sign of "wt".

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

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

Pete
 

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