intuitive explanation why frequency is fixed


by pivoxa15
Tags: explanation, fixed, frequency, intuitive
pivoxa15
pivoxa15 is offline
#1
Dec19-06, 05:31 AM
P: 2,268
From treating light as a wave it is possible using Huygen’s theory to deduce that the frequency of the light will not change whether in vacuum or some other material. I have seen a mathematical proof of it and understand it but is there an intuitive explanation for it? Does it match Maxwell theory of light?

What about this explanation: Light is emitted by an accelerating charge that is also changing direction so after the light is emitted the frequency is fixed but wavelength change depending on the material. This explanation is just like treating light as a mechanical wave. Water waves are generated by a vibrator and if the medium it travels through changes the frequency is the same but wavelength changes. Correct? If so than its as if Huygen is treating light as a mechanical wave. Which is wrong as proved by Maxwell? So frequency is really not fixed according to Maxwell?
Phys.Org News Partner Physics news on Phys.org
Physicists consider implications of recent revelations about the universe's first light
Vacuum ultraviolet lamp of the future created in Japan
Grasp of SQUIDs dynamics facilitates eavesdropping
Meir Achuz
Meir Achuz is offline
#2
Dec19-06, 01:27 PM
Sci Advisor
HW Helper
P: 1,937
The boundary conditions of a wave at an interface can only be satisfied at all times if the incident wave and the reflected and transmitted waves all have the same frequency.
pivoxa15
pivoxa15 is offline
#3
Dec19-06, 02:42 PM
P: 2,268
Quote Quote by Meir Achuz View Post
The boundary conditions of a wave at an interface can only be satisfied at all times if the incident wave and the reflected and transmitted waves all have the same frequency.
Could you give an example?

So not only mechanical (i.e. sound, water) but electromagnetic waves also obey the law that frequency is fixed in any material?

Meir Achuz
Meir Achuz is offline
#4
Dec20-06, 09:04 AM
Sci Advisor
HW Helper
P: 1,937

intuitive explanation why frequency is fixed


If you tie two strings of differeing mass densilties together, the end of each stilng will oscillate at the frequency of the knot. The same is true for the E and B fields at a change of dielectric constant.
jtbell
jtbell is offline
#5
Dec20-06, 12:00 PM
Mentor
jtbell's Avatar
P: 11,254
Quote Quote by pivoxa15 View Post
Could you give an example?
Using Gauss's Law (one of Maxwell's equations), one can show that the component of the electric field E parallel to a boundary between two media must be continuous across the boundary. That is, it can't "jump" discontinuously as you cross the boundary. One can also show that at a boundary that carries no net surface charge, the perpendicular component of E must also be continuous across the boundary. See for example

http://farside.ph.utexas.edu/teachin...es/node59.html

Using Ampere's Law, one can come to similar (but sort of "opposite") conclusions about the magnetic field B: the perpendicular component must always be continuous, and the parallel component must be continuous across a boundary that carries no net surface current.

If an electromagnetic wave had different frequencies on the two sides of a refracting boundary, the E and B fields would have to be usually discontinuous at the boundary.
pivoxa15
pivoxa15 is offline
#6
Dec20-06, 03:24 PM
P: 2,268
Quote Quote by Meir Achuz View Post
If you tie two strings of differeing mass densilties together, the end of each stilng will oscillate at the frequency of the knot. The same is true for the E and B fields at a change of dielectric constant.
For a smooth oscillation of the two strings tied together, the place of the knot must oscillate smoothly and each string must also oscillate smoothly. Smooth oscillation imply constant frequency - correct? Hence the knot frequency must match the frequency of both strings. The knot is the end of one string and the start of the other. Therefore the frequency of the two strings are equal.

When applied to E and B fields upon entering two different media, consider the electric fields E1 and E2 in media 1 and 2 respectively. Even though they are different, they must be tied together and oscillate smoothly so from the above analogy must oscillate at a single constant frequency. Same applies for B.
pivoxa15
pivoxa15 is offline
#7
Dec20-06, 03:33 PM
P: 2,268
Quote Quote by jtbell View Post
Using Gauss's Law (one of Maxwell's equations), one can show that the component of the electric field E parallel to a boundary between two media must be continuous across the boundary. That is, it can't "jump" discontinuously as you cross the boundary. One can also show that at a boundary that carries no net surface charge, the perpendicular component of E must also be continuous across the boundary. See for example

http://farside.ph.utexas.edu/teachin...es/node59.html

Using Ampere's Law, one can come to similar (but sort of "opposite") conclusions about the magnetic field B: the perpendicular component must always be continuous, and the parallel component must be continuous across a boundary that carries no net surface current.

If an electromagnetic wave had different frequencies on the two sides of a refracting boundary, the E and B fields would have to be usually discontinuous at the boundary.

Could you explain a bit more about how the frequency ties in with this example? Should I be thinking about the E and B wave equations?
jtbell
jtbell is offline
#8
Dec20-06, 04:22 PM
Mentor
jtbell's Avatar
P: 11,254
Quote Quote by pivoxa15 View Post
Should I be thinking about the E and B wave equations?
Yes. For example, start with

[tex]E = E_{max} \sin (kx - \omega t + \phi_0) = E_{max} \sin \left( \frac {2 \pi x}{\lambda} - 2 \pi f t + \phi_0 \right)[/tex]

You have two waves like this, one on each side of the boundary.

[tex]E_1 = E_{1,max} \sin \left( \frac {2 \pi x}{\lambda_1} - 2 \pi f_1 t + \phi_{01} \right)[/tex]

[tex]E_2 = E_{2,max} \sin \left( \frac {2 \pi x}{\lambda_2} - 2 \pi f_2 t + \phi_{02} \right)[/tex]

For simplicity, let x = 0 at the boundary so the terms with x disappear. Now suppose [itex]f_1 \ne f_2[/itex]. Can you make [itex]E_1 = E_2[/itex] at all times t, while keeping [itex]E_{1,max}[/itex], [itex]E_{2,max}[/itex], [itex]\phi_{01}[/itex] and [itex]\phi_{02}[/itex] constant?
pivoxa15
pivoxa15 is offline
#9
Dec20-06, 06:48 PM
P: 2,268
So you first proved the electric fields in both media must equal at the boundary for all time.

Then one can see that the wave equation for E at a given location has one variable t. The constant scaling this variable is f. In order to keep both E equal at the boundary for all t, the freqeuncy must be the same for both E (otherwise as t changes the two E will differ). The other constants such as Emax, lambda, thi can be different for each E (note the term [tex]2\pi[/tex]ft will occur in both equations) provided they 'combine' in the end so that both E are equal

This would be Maxwell's way of showing that frequency is fixed in any media? So it agrees with Huygen's method by treating light as a water wave. So no matter what sort of wave the frequency will be the same in any media while the velocity and wavelength change.


Register to reply

Related Discussions
intuitive explanation of the gravity of a sphere General Astronomy 8
Please recommend me a book (intuitive) Atomic, Solid State, Comp. Physics 2
Intuitive way to explain |Q| = aleph zero? Set Theory, Logic, Probability, Statistics 5
intuitive explanation of gravity outside a sphere? Introductory Physics Homework 4
Intuitive physics General Discussion 29