Here's a diagram for wavefront refraction in the case where V2 < V1 :

[ just reverse the direction to see the other case ]

and the least time idea :

the 'x' value is the one to vary to find the least time. If x is too far to the left the light spends too little time in the fast medium. If x is too far to the right the light spends too much time in the fast medium. ( Calculus gives the full exact answer, and there is also a neat argument based on geometric symmetries that I can't quite remember ).

Cheers, Mike.

( edit ) Anyone for calculus? Probably not ... :-)[pre]Let
Total time = T
= time in first medium + time in second medium
= sqrt(h^2 + x^2) / V1 + sqrt(b^2 + (a-x)^2) / V2
So dT/dx = 2x / [V1 * sqrt(h^2 + x^2)] + 2(a-x)(-1) / [V2 * sqrt(b^2 + (a-x)^2)][/pre]where dT/dx = 0 for an extremum. I know its a minimum but strictly speaking you'd take another derivative and find that positive for the value of x that makes dT/dx = 0.

What I don't understand is when light travels through a material, it travels slower due to the electrons. When it leaves the medium, it speeds up. Where does the energy for that acceleration come from???

What I don't understand is when light travels through a material, it travels slower due to the electrons. When it leaves the medium, it speeds up. Where does the energy for that acceleration come from???

From the last atom that absorbed it. It was captured by it, so that atom went to a higher energy state, then returns ( some of ) that energy on re-emitting. The path the light takes has all these exchanges of energy happening as it goes along - energy going to and fro between the photons and electrons.

Cheers, Mike.

( edit ) Oh, and as regards what you see with curvature, recall what was found with the total solar eclipse ( 1919 and since ) with light from distant stars grazing the edge of the Sun. They didn't see two images of the one star or some curved shiny path across the sky - they just saw the star but in a different position compared to when the Sun wasn't present in that direction ( by comparing to photos taken at other times ). The presence of the Sun simply wiggles the apparent direction from which we see the light arriving here on Earth.

So if you can imagine being able to ignore the glare of the Sun, and simply focus upon where you would have to look to see the nearby stars : you would have to move your line of sight slightly away from the Sun to locate the star when the Sun is nearby compared to either before or after when the Sun wasn't in the vicinity of that direction. That change discloses what we have called curvature or deviation from 'flat' space.

( edit ) If there had of been two stars in the same direction in the sky from our viewpoint here on Earth, so that one was behind the other, that would still appear that way to us whether the Sun was nearby or not. If there were twenty all lined up exactly one behind the other .... same deal. That's what I meant by looking precisely down one side of a triangle and seeing a point only ( a line 'end on' ) regardless of which type of curvature I'm in.

( edit ) Whoops. Didn't mention 'mean free path' which is basically how far on average does a particle travel before it hits something. So inside the Sun, for photons starting at the core this is from 1 to 10 millimeters ( I just looked that up ). It has to go about 700,000 km to get to the surface - so that's a long trip. On the other hand, a neutrino hardly interacts with anything, so it's mean free path in lead is some twenty or so light years!!! :-)

I have made this letter longer than usual because I lack the time to make it shorter.Blaise Pascal

From the last atom that absorbed it. It was captured by it, so that atom went to a higher energy state, then returns ( some of ) that energy on re-emitting. The path the light takes has all these exchanges of energy happening as it goes along - energy going to and fro between the photons and electrons.

There must be some loss of energy for every time energy is transferred. I cannot see how the light can return to c after leaving the medium considering the energy lost.

There must be some loss of energy for every time energy is transferred.

Why? The only way for a photon to lose energy is to lower its frequency (change colour).

Quote:

I cannot see how the light can return to c after leaving the medium considering the energy lost.

It never left c. There was no energy lost (fluorescence aside), only the time needed for absorption and subsequent re-emission (at the same frequency).

GruÃŸ,
Gundolf

Computer sind nicht alles im Leben. (Kleiner Scherz)

There must be some loss of energy for every time energy is transferred. I cannot see how the light can return to c after leaving the medium considering the energy lost.

Energy comes in a number of forms, and strictly speaking it's not something you measure directly - one measures some quantities and then you calculate a number. Each 'type' of energy has it's own formula.

Light is special. A photon never slows down, even when traveling between interactions in a material. However it's energy is not kinetic ( dependent on speed ), there is a formula for it and c is not involved :

Energy = Planck's constant * frequency

[ or in symbols : E = hf ]

This really says that a photon's frequency is it's energy, with h present to translate the measurement units. Cycles per second to Joules, say. So, as Gundolf says, to change a photon's energy the frequency must alter.

I guess the next question becomes : what is frequency for a photon? By that I mean : what actually is wiggling in some cyclical fashion that we deem some repetition rate or 'frequency'? That's really quite a difficult one to answer fully and/or consistently, but ultimately it's yet another thing called 'phase'. However we can't measure any particular phase, only compare phases - which is what the LIGO interferometers ( read 'phase comparators between different paths' ) is about.

The history of science shows gradual increasing use of the idea of energy. So as new phenomena are studied then new energy types are added to help in description. Or put another way, because we like the idea of energy conservation ( it never gets lost or magically appears ) we tack on extra terms in the accounting as needed to make sure of that!! :-)

My earlier comment about 'dark energy' compared to the energy of cars and water jugs, was a poke at the general program of inventing energy types for new experiences. The problem I foresee with folding dark energy into the description is simply how can one convert it to/from any other form. A wallaby can hit my car and convert energy of movement to sound, heat, structural changes etc - all accountable energy wise with sufficient care. But I haven't yet heard of a mechanism to do likewise with dark energy, so that some observable effect ( other than cosmic expansion rates ) can be employed to test it's presence. So far all that is really on the table is : we reckon there's a process at cosmic scale, don't know the detail though, and assuming the principle of energy conservation holds then it must have an energy associated with it. And we've got a new name for it too .... err, what word/terms would fit .... that hasn't been used so far .... try 'dark' energy?

Cheers, Mike.

( edit ) Sigh, I retract .... it's been done already. This product demonstrates the use of 'vacuum force technology'. :-)

I have made this letter longer than usual because I lack the time to make it shorter.Blaise Pascal

Light is special. A photon never slows down, even when traveling between interactions in a material. However it's energy is not kinetic ( dependent on speed ), there is a formula for it and c is not involved :

Energy = Planck's constant * frequency

Ahh, this I did not know. Counter-intuitive, isn't it? One of the benefits of mass=0 I suppose. There must be more to it than this. It seems like we are only working with some of the pieces of the puzzle.

Quote:

( edit ) Sigh, I retract .... it's been done already. This product demonstrates the use of 'vacuum force technology'. :-)

Was that an add??? LOL. Looks like a crappy product to me. The things people come up with...

By cheating. Define the path that light takes ( when not interacting with stuff ) to be 'straight'! :-)

In everyday terms the idea is to think of three points, and how they can be viewed in certain ways. If I look directly from the first point to the second point ( so the second point lies exactly behind the first ) and see the more distant third point precisely overlapped also, then I say they all lie along one and the same line. ...

Einstein basically said the shortest distance between two points is the path light takes. Always. However Pythagorus' Theorem and similiar derived traditional geometric properties of triangles don't always apply.

What of the example of gravitational lensing around a star?

You have the gravitationally 'deflected' path. However, is not the shortest straight line path straight through the centre of gravity through the star itself (if the star was to be transparent)?

Quote:

Quote:

Would refraction occur at all if a particle could be "infinitely thin"?

(To be particular (particles) about it all, that is...)

Ah, this is a deep one. There are several ways of answering. Try these .....

The classical way, really done properly first by Huygen's, was to have the width idea - but of a wavefront. Consider some straight wavefront approaching some ( also straight ) boundary between two materials. ...

The effect is to turn the wavefront to become more parallel to the boundary than it was initially. ...

[ Snell's Law expresses the geometry of this, relating ( sines of ) those angles with the normal to the two speeds. ]

As shown in your following diagram.

There is also an assumption there that a wavefront always propagates along the normal. What mechanism dictates that? Why not... (See the following diagram to yours.)

Quote:

OK ... the second explanation probably comes out of the depths of quantum mechanics, but also has the classical Fermat's Principle of Least Time too. I like Richard Feynman's explanation ... It turns out that the most probable overall path for the photon is the one that takes the least time ( all nearby paths take longer ) and this path yields angles corresponding to the relationship that is Snell's Law.

This least time principle can be related to The Principle of Least Action, where 'action' is a quantity having units of momentum times distance ( or energy times time ).

Now that is a "beautiful" observation of something fundamental. Very interesting! Is that attributed to Feynman?

Planck's constant has these units and is thus an amount of action. What action 'is' no-one really knows, nor why minimising is a key issue. ...

Is that always true? Is that all bundled in with what is 'seen' for entropy?

Are we observing first hand effects or are we actually conjuring up abstractions that describe what we observe for what actually are second-level effects of underlying mechanisms?...

Quote:

Well, we can't have you in the dark! :-)

Except that the improved glow of the candle only illuminates a yet expanded sphere of mystery! :-p

Here's a diagram for wavefront refraction in the case where V2 green lines are straight lines.)

Aside: Should there be a small force of radiation pressure associate with the refractive deflection?...

Quote:

( edit ) Anyone for calculus? Probably not ... :-)[pre]

... which is Snell's Law!

Where multiple paths of insights all converge on the same representation, that has just got to be 'interesting'!

But then also... Do we see repeated structures for the many equations in physics because our world naturally follows those structures? Or are we merely seeing an artefact of the mathematical tools that we are using?... Is there some underlying assumption that then forces all our tools to be "hammers" that then can only see "nails" regardless of the job?

What of the example of gravitational lensing around a star?

You have the gravitationally 'deflected' path. However, is not the shortest straight line path straight through the centre of gravity through the star itself (if the star was to be transparent)?

Fair point, this lensing causes convergence of many otherwise divergent paths ( if the star wasn't there ). So you get an increase in intensity because photons that would have gone right on by the observer on some paths, have come together at the observers location instead. That is, there are many ways to go from A to B. The least principles don't say how many solutions you get, just that any actual path will be a minimum. It's not necessarily unique. Gravity is like that! If you take a black hole then there's lots of light paths that encircle the volume around it ( ie. outside the event horizon ) - you can even have images that simultaneously show opposite sides of an object behind the hole! :-)

Quote:

As shown in your following diagram.

There is also an assumption there that a wavefront always propagates along the normal. What mechanism dictates that? Why not... (See the following diagram to yours.)

We define it that way! We have a direction of propagation at some point in space and define the wavefront to be a surface perpendicular to that. Mind you that's not necessarily a flat plane, if you have diverging 'rays' of light from a point source, say, then the wavefront is an expanding sphere centered on that source.

A wavefront is a mathematical/geometrical abstraction ( there's alot of those in physics! ). Let's think in one dimension for the moment. Without getting too tense about factors of 2 and PI and whatnot, a wave is generally mathematically represented by something like this :

Amplitude = Max_amplitude * sin [distance_variable / wavelength - frequency * time_variable]

OR more compactly

W = Wmax * sin[k*x - w*t]

with Wmax being the 'height' of the wave, and where

k = 2 * PI / wavelength = 2 * PI / l

w = 2 * PI * frequency = 2 * PI * f

As sine[anything] is between -1 and +1 then W ranges between -Wmax and +Wmax. The entire quantity within the sine brackets is the phase

phase = p = k*x - w*t

here k and w are a constant for a given wave, whereas x and t are variables. So the value of the wave amplitude ( which is just a number ) depends on two things - where you are and what time it is. The minus sign indicates that this wave goes towards the right ( increasing values of x on the axis ), a plus means the other way.

You can sense the behaviour by asking the question : what can I do to x and/or t to increase the phase? If I hold t constant - take a snapshot at a given moment, looking to left and right along the wave - then phase increases with increasing values of x. If I hold x constant - stay at some point and watch the wave go by - then phase decreases with increasing time. Or put another way : if I want to catch an earlier part of the wave I can go back in time at some point OR I can go further positively along the x axis ( because it's already passed me by ). OR if I want to view a later part of the wave I go forward in time ( obviously ) or go back along the x-axis ie. it's coming from that direction. [ You'll probably have to think for a few minutes about this ].

More precisely if I take the following ( partial ) derivatives of the phase :

dp/dx = d(k * x - w * t)/dx = k

as t held constant. And

dp/dt = d(k * x - w * t)/dx = - w

since x is held constant.

These say the same thing as the verbal description, but k and w now state quantitatively how the phase changes with either distance or time. Note that if the wavelength is longer, k is smaller and you get a gentler change along the x axis - as you'd expect for a longer wave. If the frequency is greater then so is w and the phase changes more rapidly with time - as you'd expect by the idea of frequency/repetition. Now if you just think of a fixed value of p ( where | something | = magnitude of something )

| dx/dt | = | dx/dp | * | dp/dt | = ( 1/ k ) * ( w ) = ( l / [2 * PI]) * ( 2 * PI * f ) = l * f = speed of the wave

This tells you how fast a given feature on the wave ( spot of constant phase ) shoots along the x axis. Now if you've absorbed all this, then the concept is that this point of constant phase is the one dimensional idea of a wavefront. It's going in the positive direction - with the same sign as k is represented in the phase term.

Now the generalisation to more directions is a bit fiddly, but we've covered the basics here. You have a 'slight' change to the notation ( I'm really cheating a great deal for the purists. But hey, go with it .... ) :

W = Wmax * sin[|k|*x - w*t]

Here k is a vector in the direction of propagation, with magnitude |k|. If you step along in the direction of k you get the maximum positive rate of change of phase. Or the reverse direction, -k, for the maximum negative rate of change of phase. If you go at right angles to k you get zero rate of change of phase. In 3D space what is at right angles to a given arrow? A plane - a flat ( infinite ) surface. So we have just described a plane wavefront. Just imagine the k vector shooting along in space ( like an arrow! ) with this plane attached to it's rear end and you've got a good image. All points on that plane have the same phase value.

You can extend the description to curved wavefronts. Here the wave varies or 'spreads' ( or 'scrunches' even ) in an angular sense as it travels, so the k vector now varies. That is, the direction of propagation now depends on which part of the wave you look at - which isn't the case with a plane wave, as all vectors perpendicular to some plane will point along the same direction. They are parallel if you stay on the 'same side' of the wavefront, or anti-parallel if you flip to the 'other side' of the wavefront. Thus a wavefront, by being is a surface of constant phase, separates all lesser phase values from all greater phase values. And yes, the maths is far more messy!! :-)

I'll come back later to your other questions! :-)

Cheers, Mike.

( edit ) Another re-statement is that a curved surface has planes that are 'tangent' to it at each point on the surface. And each k becomes a 'normal' vector to the curved surface ( but as above, beware whether it's pointing 'in' or 'out' ).

( edit ) I forgot to mention an ordinary converging lens - like in a camera say. What defines 'focus' or focal point is the ability of the lens to bring all light rays from some extended object to a single point ( or plane ) in phase. So that's why that type of lens is fatter in the middle and slimmer off axis. The speed is slower within the lens glass. You delay the central rays more than the peripheral ones, allowing the peripheral ones to 'catch up' in phase so to speak. They go via a longer route, being off axis, but suffer less distance through the glass. The exact shape of the lens determines for what point the focus occurs. Hence for certain choices of initial and final points there can be very many paths between the two of equal total time, which are also minimal.

I have made this letter longer than usual because I lack the time to make it shorter.Blaise Pascal

## Here's a diagram for

)

Here's a diagram for wavefront refraction in the case where V2 < V1 :

[ just reverse the direction to see the other case ]

and the least time idea :

the 'x' value is the one to vary to find the least time. If x is too far to the left the light spends too little time in the fast medium. If x is too far to the right the light spends too much time in the fast medium. ( Calculus gives the full exact answer, and there is also a neat argument based on geometric symmetries that I can't quite remember ).

Cheers, Mike.

( edit ) Anyone for calculus? Probably not ... :-)[pre]Let

Total time = T

= time in first medium + time in second medium

= sqrt(h^2 + x^2) / V1 + sqrt(b^2 + (a-x)^2) / V2

So dT/dx = 2x / [V1 * sqrt(h^2 + x^2)] + 2(a-x)(-1) / [V2 * sqrt(b^2 + (a-x)^2)][/pre]where dT/dx = 0 for an extremum. I know its a minimum but strictly speaking you'd take another derivative and find that positive for the value of x that makes dT/dx = 0.

Thus rearranging[pre] V2 * x / sqrt(h^2 + x^2) = V1 * (a-x) / sqrt(b^2 + (a-x)^2)

V2 * sin(theta_i) = V1 * sin(theta_t)[/pre]which is Snell's Law!

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

## What I don't understand is

)

What I don't understand is when light travels through a material, it travels slower due to the electrons. When it leaves the medium, it speeds up. Where does the energy for that acceleration come from???

## RE: What I don't understand

)

From the last atom that absorbed it. It was captured by it, so that atom went to a higher energy state, then returns ( some of ) that energy on re-emitting. The path the light takes has all these exchanges of energy happening as it goes along - energy going to and fro between the photons and electrons.

Cheers, Mike.

( edit ) Oh, and as regards what you see with curvature, recall what was found with the total solar eclipse ( 1919 and since ) with light from distant stars grazing the edge of the Sun. They didn't see two images of the one star or some curved shiny path across the sky - they just saw the star but in a different position compared to when the Sun wasn't present in that direction ( by comparing to photos taken at other times ). The presence of the Sun simply wiggles the apparent direction from which we see the light arriving here on Earth.

So if you can imagine being able to ignore the glare of the Sun, and simply focus upon where you would have to look to see the nearby stars : you would have to move your line of sight slightly away from the Sun to locate the star when the Sun is nearby compared to either before or after when the Sun wasn't in the vicinity of that direction. That change discloses what we have called curvature or deviation from 'flat' space.

( edit ) If there had of been two stars in the same direction in the sky from our viewpoint here on Earth, so that one was behind the other, that would still appear that way to us whether the Sun was nearby or not. If there were twenty all lined up exactly one behind the other .... same deal. That's what I meant by looking precisely down one side of a triangle and seeing a point only ( a line 'end on' ) regardless of which type of curvature I'm in.

( edit ) Whoops. Didn't mention 'mean free path' which is basically how far on average does a particle travel before it hits something. So inside the Sun, for photons starting at the core this is from 1 to 10 millimeters ( I just looked that up ). It has to go about 700,000 km to get to the surface - so that's a long trip. On the other hand, a neutrino hardly interacts with anything, so it's mean free path in lead is some twenty or so light years!!! :-)

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

## RE: From the last atom

)

There must be some loss of energy for every time energy is transferred. I cannot see how the light can return to c after leaving the medium considering the energy lost.

## RE: There must be some loss

)

Why? The only way for a photon to lose energy is to lower its frequency (change colour).

It never left c. There was no energy lost (fluorescence aside), only the time needed for absorption and subsequent re-emission (at the same frequency).

GruÃŸ,

Gundolf

Computer sind nicht alles im Leben. (Kleiner Scherz)

## RE: There must be some loss

)

Energy comes in a number of forms, and strictly speaking it's not something you measure directly - one measures some quantities and then you calculate a number. Each 'type' of energy has it's own formula.

Light is special. A photon never slows down, even when traveling between interactions in a material. However it's energy is not kinetic ( dependent on speed ), there is a formula for it and c is not involved :

Energy = Planck's constant * frequency

[ or in symbols : E = hf ]

This really says that a photon's frequency is it's energy, with h present to translate the measurement units. Cycles per second to Joules, say. So, as Gundolf says, to change a photon's energy the frequency must alter.

I guess the next question becomes : what is frequency for a photon? By that I mean : what actually is wiggling in some cyclical fashion that we deem some repetition rate or 'frequency'? That's really quite a difficult one to answer fully and/or consistently, but ultimately it's yet another thing called 'phase'. However we can't measure any particular phase, only compare phases - which is what the LIGO interferometers ( read 'phase comparators between different paths' ) is about.

The history of science shows gradual increasing use of the idea of energy. So as new phenomena are studied then new energy types are added to help in description. Or put another way, because we like the idea of energy conservation ( it never gets lost or magically appears ) we tack on extra terms in the accounting as needed to make sure of that!! :-)

My earlier comment about 'dark energy' compared to the energy of cars and water jugs, was a poke at the general program of inventing energy types for new experiences. The problem I foresee with folding dark energy into the description is simply how can one convert it to/from any other form. A wallaby can hit my car and convert energy of movement to sound, heat, structural changes etc - all accountable energy wise with sufficient care. But I haven't yet heard of a mechanism to do likewise with dark energy, so that some observable effect ( other than cosmic expansion rates ) can be employed to test it's presence. So far all that is really on the table is : we reckon there's a process at cosmic scale, don't know the detail though, and assuming the principle of energy conservation holds then it must have an energy associated with it. And we've got a new name for it too .... err, what word/terms would fit .... that hasn't been used so far .... try 'dark' energy?

Cheers, Mike.

( edit ) Sigh, I retract .... it's been done already. This product demonstrates the use of 'vacuum force technology'. :-)

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

## RE: Light is special. A

)

Ahh, this I did not know. Counter-intuitive, isn't it? One of the benefits of mass=0 I suppose. There must be more to it than this. It seems like we are only working with some of the pieces of the puzzle.

Was that an add??? LOL. Looks like a crappy product to me. The things people come up with...

## RE: RE: So what keeps the

)

What of the example of gravitational lensing around a star?

You have the gravitationally 'deflected' path. However, is not the shortest straight line path straight through the centre of gravity through the star itself (if the star was to be transparent)?

As shown in your following diagram.

There is also an assumption there that a wavefront always propagates along the normal. What mechanism dictates that? Why not... (See the following diagram to yours.)

Now that is a "beautiful" observation of something fundamental. Very interesting! Is that attributed to Feynman?

[edit] OK, so not Feynman but not so clear cut either: The history of the principle of least action. See also: [url=http://en.wikipedia.org/wiki/Action_(physics)#History_of_term_action]History of term action.[/url] [/edit]

Is that always true? Is that all bundled in with what is 'seen' for entropy?

Are we observing first hand effects or are we actually conjuring up abstractions that describe what we observe for what actually are second-level effects of underlying mechanisms?...

Except that the improved glow of the candle only illuminates a yet expanded sphere of mystery! :-p

Regards,

Martin

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See & try out your OS Freedom! Linux Voice

The Future is what We all make IT [url=http://www.gnu.org/copyleft/gpl.html](GPLv3

## RE: Here's a diagram for

)

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## RE: What of the example of

)

Fair point, this lensing causes convergence of many otherwise divergent paths ( if the star wasn't there ). So you get an increase in intensity because photons that would have gone right on by the observer on some paths, have come together at the observers location instead. That is, there are many ways to go from A to B. The least principles don't say how many solutions you get, just that any actual path will be a minimum. It's not necessarily unique. Gravity is like that! If you take a black hole then there's lots of light paths that encircle the volume around it ( ie. outside the event horizon ) - you can even have images that simultaneously show opposite sides of an object behind the hole! :-)

We define it that way! We have a direction of propagation at some point in space and define the wavefront to be a surface perpendicular to that. Mind you that's not necessarily a flat plane, if you have diverging 'rays' of light from a point source, say, then the wavefront is an expanding sphere centered on that source.

A wavefront is a mathematical/geometrical abstraction ( there's alot of those in physics! ). Let's think in one dimension for the moment. Without getting too tense about factors of 2 and PI and whatnot, a wave is generally mathematically represented by something like this :

Amplitude = Max_amplitude * sin [distance_variable / wavelength - frequency * time_variable]

OR more compactly

W = Wmax * sin[k*x - w*t]

with Wmax being the 'height' of the wave, and where

k = 2 * PI / wavelength = 2 * PI / l

w = 2 * PI * frequency = 2 * PI * f

As sine[anything] is between -1 and +1 then W ranges between -Wmax and +Wmax. The entire quantity within the sine brackets is the phase

phase = p = k*x - w*t

here k and w are a constant for a given wave, whereas x and t are variables. So the value of the wave amplitude ( which is just a number ) depends on two things - where you are and what time it is. The minus sign indicates that this wave goes towards the right ( increasing values of x on the axis ), a plus means the other way.

You can sense the behaviour by asking the question : what can I do to x and/or t to increase the phase? If I hold t constant - take a snapshot at a given moment, looking to left and right along the wave - then phase increases with increasing values of x. If I hold x constant - stay at some point and watch the wave go by - then phase decreases with increasing time. Or put another way : if I want to catch an earlier part of the wave I can go back in time at some point OR I can go further positively along the x axis ( because it's already passed me by ). OR if I want to view a later part of the wave I go forward in time ( obviously ) or go back along the x-axis ie. it's coming from that direction. [ You'll probably have to think for a few minutes about this ].

More precisely if I take the following ( partial ) derivatives of the phase :

dp/dx = d(k * x - w * t)/dx = k

as t held constant. And

dp/dt = d(k * x - w * t)/dx = - w

since x is held constant.

These say the same thing as the verbal description, but k and w now state quantitatively how the phase changes with either distance or time. Note that if the wavelength is longer, k is smaller and you get a gentler change along the x axis - as you'd expect for a longer wave. If the frequency is greater then so is w and the phase changes more rapidly with time - as you'd expect by the idea of frequency/repetition. Now if you just think of a fixed value of p ( where | something | = magnitude of something )

| dx/dt | = | dx/dp | * | dp/dt | = ( 1/ k ) * ( w ) = ( l / [2 * PI]) * ( 2 * PI * f ) = l * f = speed of the wave

This tells you how fast a given feature on the wave ( spot of constant phase ) shoots along the x axis. Now if you've absorbed all this, then the concept is that this point of constant phase is the one dimensional idea of a wavefront. It's going in the positive direction - with the same sign as k is represented in the phase term.

Now the generalisation to more directions is a bit fiddly, but we've covered the basics here. You have a 'slight' change to the notation ( I'm really cheating a great deal for the purists. But hey, go with it .... ) :

W = Wmax * sin[|k|*x - w*t]

Here k is a vector in the direction of propagation, with magnitude |k|. If you step along in the direction of k you get the maximum positive rate of change of phase. Or the reverse direction, -k, for the maximum negative rate of change of phase. If you go at right angles to k you get zero rate of change of phase. In 3D space what is at right angles to a given arrow? A plane - a flat ( infinite ) surface. So we have just described a plane wavefront. Just imagine the k vector shooting along in space ( like an arrow! ) with this plane attached to it's rear end and you've got a good image. All points on that plane have the same phase value.

You can extend the description to curved wavefronts. Here the wave varies or 'spreads' ( or 'scrunches' even ) in an angular sense as it travels, so the k vector now varies. That is, the direction of propagation now depends on which part of the wave you look at - which isn't the case with a plane wave, as all vectors perpendicular to some plane will point along the same direction. They are parallel if you stay on the 'same side' of the wavefront, or anti-parallel if you flip to the 'other side' of the wavefront. Thus a wavefront, by being is a surface of constant phase, separates all lesser phase values from all greater phase values. And yes, the maths is far more messy!! :-)

I'll come back later to your other questions! :-)

Cheers, Mike.

( edit ) Another re-statement is that a curved surface has planes that are 'tangent' to it at each point on the surface. And each k becomes a 'normal' vector to the curved surface ( but as above, beware whether it's pointing 'in' or 'out' ).

( edit ) I forgot to mention an ordinary converging lens - like in a camera say. What defines 'focus' or focal point is the ability of the lens to bring all light rays from some extended object to a single point ( or plane ) in phase. So that's why that type of lens is fatter in the middle and slimmer off axis. The speed is slower within the lens glass. You delay the central rays more than the peripheral ones, allowing the peripheral ones to 'catch up' in phase so to speak. They go via a longer route, being off axis, but suffer less distance through the glass. The exact shape of the lens determines for what point the focus occurs. Hence for certain choices of initial and final points there can be very many paths between the two of equal total time, which are also minimal.

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal