Speed C is Dependant Upon Source?

[omin]

1. The speed of light is constant.

2. Light may reflect off objects.

3. The speed of light is not determined by it's source.

If the speed of light is constant and this speed is unaffected by it's source, then the expression, the shape light is, is it's natural shape and it's speed simultaneously. Light's inertia is it's shape, it's physical expression and vice versa. Light speeds itself, objects don't speed the light.

But, light can be reflected. For something to change directions, acceleration must occur. For acceleration to occur, a change in velocity must occur. For velocity to occur, a change of speed and/or direction must occur. Can it be proven that only direction occurs when light is reflected? What rules out a change of speed assuredly during reflection? Is a mirror that reflects light a new source compared to the source where light last emerged?

If the source does not determine the speed of light, then why does the speed of light always acclerate away from a source at the speed of light consistently? How does the light know that the source is there to speed away from at the constant speed?

If two mirrors are traveling at different speeds compared to each other and light is reflected from one to another, shouldn't now light change speed dependent upon source or what happens?

Since light is constant, it must have an exact quantity between the mirrors at any given state, and this quantity of light definitely changes with the distance the mirrors are from each other. When the mirror's distance is compressed, where does this extra light go? Does it begin to reflect faster?

-Still trying to understand relativity.

[Tom Mattson]

1. The speed of light is constant.

2. Light may reflect off objects.

3. The speed of light is not determined by it's source.


So far, so good.


If the speed of light is constant and this speed is unaffected by it's source, then the expression, the shape light is, is it's natural shape and it's speed simultaneously. Light's inertia is it's shape, it's physical expression and vice versa. Light speeds itself, objects don't speed the light.


To be blunt, this makes no sense. What is the shape of light, and why do you think that this shape is identical to the inertia of light?


But, light can be reflected. For something to change directions, acceleration must occur. For acceleration to occur, a change in velocity must occur. For velocity to occur, a change of speed and/or direction must occur. Can it be proven that only direction occurs when light is reflected? What rules out a change of speed assuredly during reflection?



All you have to do is measure the speed of the incident beam and of the reflected beam. In every case, you'll find that they are the same. Actually, this has nothing whatsoever to do with relativity. If you bounce a tennis ball off a rigid wall, and the collision is elastic, then the speed of the tennis ball will be the same before and after the collision.


Is a mirror that reflects light a new source compared to the source where light last emerged?


Yes it is, because the photons are absorbed and re-emitted by the mirror.


If the source does not determine the speed of light, then why does the speed of light always acclerate away from a source at the speed of light consistently?


You've just answered your own question: The very fact that the light always moves (not accelerates) away from the source at the speed of light consistently is precisely why we say that the source does not determine the speed of light. That's what it means for the speed of light to be independent of the speed of the source.


How does the light know that the source is there to speed away from at the constant speed?


It doesn't know anything.


If two mirrors are traveling at different speeds compared to each other and light is reflected from one to another, shouldn't now light change speed dependent upon source or what happens?


No, it shouldn't. You said it yourself, the speed of light is independent of the speed of the source.


Since light is constant, it must have an exact quantity between the mirrors at any given state, and this quantity of light definitely changes with the distance the mirrors are from each other. When the mirror's distance is compressed, where does this extra light go? Does it begin to reflect faster?


What "exact quantity" are you talking about?

[chroot]

When a photon is reflected from a surface, it is not actually stopped and turned around. Instead, it is absorbed, and a new photon is emitted in the opposite direction.

- Warren

[Alkatran]

When a photon is reflected from a surface, it is not actually stopped and turned around. Instead, it is absorbed, and a new photon is emitted in the opposite direction.

- Warren

Ah, you beat me to it. :cry:

Sounds like someone needs to brush up on their relativity.

[russ_watters]

To put a finer point on what warren said: For something to change directions, acceleration must occur. For acceleration to occur, a change in velocity must occur. Light doesn't accelerate in the way you think it does. It doesn't change speed. It doesn't slow down to a stop, then reverse its course. It is absorbed and re-emitted, and never travels at any speed other than C.

[Chronos]

No acceleration. Photon's are forbidden to travel at any speed other than c.

[pmb_phy]

No acceleration. Photon's are forbidden to travel at any speed other than c.
Note: A photon can accelerate in a gravitational field.

Pete

[urtalkinstupid]

Note: A photon can accelerate in a gravitational field.

Note: That is why light changes frequency. Doppler affect.

[Integral]

Note: A photon can accelerate in a gravitational field.

Pete
The direction light is traveling can and will change in a gravitational field, thus an acceleration, but the magnitude of the velocity will remain a constant c.

[rayjohn01]

Perhaps someone can answer this related question. In the original 'train with mirror' relativity demonstration the traveller supposes his light to travel perpendicular to the train whilst the observer sees it travel a triangular path,from which the time relations are derived.
The hidden assumtion is that the DIRECTION of light has changed with the source ( not it's speed).
There then arises a question -- if light is absorbed by a mirror and then re-emitted it is assumed that it's momentum did not change ( hence the angle of reflection and reception are equal )
But in the above scenario the momentum did change -- so the question is does a moving mirror have equal angles or not ?

[grounded]

Yes it is, because the photons are absorbed and re-emitted by the mirror.


When a photon is reflected from a surface, it is not actually stopped and turned around. Instead, it is absorbed, and a new photon is emitted in the opposite direction.

- Warren


Ah, you beat me to it. :cry:

Sounds like someone needs to brush up on their relativity.


Is the above true?

When light strikes a mirror, almost all of it is reflected, a small amount is absorbed but it would only heat the mirror and not emit photons.

the energy of the oscillating electrons does not go back to the light. The energy instead goes toward increasing the motion of the atoms, which causes the material to heat up.
Microsoft® Encarta® Reference Library 2002. © 1993-2001 Microsoft Corporation. All rights reserved.

[urtalkinstupid]

What does your quote have to do with the reflection of light on mirrors? All your quote is doing is taking chroot's explanation a little further by saying a very small amount is not reemitted and heats up the mirror.

chroot, how exactly is a photon absorbed and a new photon emitted in the opposite direction?

[Hurkyl]

When light strikes a mirror, almost all of it is reflected, a small amount is absorbed but it would only heat the mirror and not emit photons.

That's not entirely accurate; all of it gets absorbed, and then almost all the energy is reemitted.

You have to be careful to distinguish between the macroscopic and microscopic terms.

[Tom Mattson]

Perhaps someone can answer this related question. In the original 'train with mirror' relativity demonstration the traveller supposes his light to travel perpendicular to the train whilst the observer sees it travel a triangular path,from which the time relations are derived.
The hidden assumtion is that the DIRECTION of light has changed with the source ( not it's speed).
There then arises a question -- if light is absorbed by a mirror and then re-emitted it is assumed that it's momentum did not change ( hence the angle of reflection and reception are equal )
But in the above scenario the momentum did change --


Right, some momentum is imparted to the mirror itself.


so the question is does a moving mirror have equal angles or not ?

Yes, the angles are equal.

[Entropy]

how exactly is a photon absorbed and a new photon emitted in the opposite direction?

When the photon is absorbed it exerts a force equivalant to all of it's energy. Then according to Newtons laws for every action (force on the mirror) there is an equal and opposite reaction, this being the emission of another photon in the opposite direction. Then when the mirror uses its own momentum to create a new photon in the opposite direction. I don't want to confuse the mircoscopic with the macroscopic but this is the same principle that causes a ball will bounce off a wall instead of just stopping once it hits the wall.

[urtalkinstupid]

So, photons are created by momentum? How is that possible?

[Entropy]

So, photons are created by momentum? How is that possible?

Well not exactly. You see since it absorbed the first photon it needs to exert an equal and opposite force back on to the photon. But since the photon is no longer there (it was absorbed) it simply emits a photon in the opposite direction. But photons do have momentum if thats what you mean.

[urtalkinstupid]

No, what I meant is what you previously said.

Then when the mirror uses its own momentum to create a new photon in the opposite direction.

How is a photon created in that process? If the photon is not created, where does it come from?

[beatrix kiddo]

stupid makes a good point. u say that the photon gets totally absorbed then a new photon is made from the mirror? that doesn't make sense... or are u saying that the mirror absorbs the photon then emits it?

[beatrix kiddo]

and if the mirror absorbs the photon and emits it then doesn't that mean the photon had to be slowed down, then accelerated in the opposite direction?

[Tom Mattson]

stupid makes a good point. u say that the photon gets totally absorbed then a new photon is made from the mirror? that doesn't make sense... or are u saying that the mirror absorbs the photon then emits it?

It makes plenty of sense. It's quantum mechanics at work (have you not taken chemistry?). A photon is incident on an atom. The atom absorbs the photon (IOW, the photon is destroyed) and an atomic electron is promoted to a higher energy level. The resulting state is unstable, and it quickly decays. In the decay, a new photon of the same wavelength is created and emitted.

[Alkatran]

stupid makes a good point. u say that the photon gets totally absorbed then a new photon is made from the mirror? that doesn't make sense... or are u saying that the mirror absorbs the photon then emits it?

and if the mirror absorbs the photon and emits it then doesn't that mean the photon had to be slowed down, then accelerated in the opposite direction?

E=mc^2

You can turn matter into energy and vice versa, so you can change a photon into momentum and momentum into a photon.

[urtalkinstupid]

Thank you for explaining, Tom. I thought matter or energy could not be created nor destroyed? Yes, I have taken chemistry. I owned at chemistry! I'm taking Chemistry II AP this coming school year. So the excitation of an electron causes decay. What kind of decay is this called? When an electron goes into decay it produces photons? Is that the only product of this decay?

[Alkatran]

Thank you for explaining, Tom. I thought matter or energy could not be created nor destroyed? Yes, I have taken chemistry. I owned at chemistry! I'm taking Chemistry II AP this coming school year. So the excitation of an electron causes decay. What kind of decay is this called? When an electron goes into decay it produces photons? Is that the only product of this decay?

The photon isn't destroyed, it's absorbed.

[urtalkinstupid]

Well, Tom told me it was destroyed. Wouldn't the conversion from momentum to energy or mass involve something other than E=mc^2? Like maybe E=pc, since light is theorized to have no rest mass? Or would you just use the whole total energy equation if you were to convert a photon to mass?

[beatrix kiddo]

tom, yes i have taken chemistry (i posted that in the thread u so graciously ended) and i've know how a photon is produced, but thanks for the reiteration. for the mirror to make a whole new photon, that means the other photon was destroyed and that doesn't make sense because everyone knows that energy cannot be created or destroyed. but u, mr. mattson said that the photon gets destroyed. and entropy says that the mirror creates a new photon, which also doesn't make sense. quote me if u must, but the description u're giving for this process contradicts what einstein said..

[Alkatran]

tom, yes i have taken chemistry (i posted that in the thread u so graciously ended) and i've know how a photon is produced, but thanks for the reiteration. for the mirror to make a whole new photon, that means the other photon was destroyed and that doesn't make sense because everyone knows that energy cannot be created or destroyed. but u, mr. mattson said that the photon gets destroyed. and entropy says that the mirror creates a new photon, which also doesn't make sense. quote me if u must, but the description u're giving for this process contradicts what einstein said..

ABOSRBED, not destroyed. You know that's what he meant.
and EMITTED, not created. Same Reason.

[beatrix kiddo]

how am i supposed to know what someone meant, alkatran? u told me that when we discussed that equation. mattson said destroyed and entropy said created. there's a difference...

[Alkatran]

how am i supposed to know what someone meant, alkatran? u told me that when we discussed that equation. mattson said destroyed and entropy said created. there's a difference...

"Creating" a photon might as well be the same as "emitting" one, except that emitting better represents what is actually happening. Now, if he said "a photon is confused by the mirror" we would have a similar situation to the disregards of PEDMAS in said equation.

[urtalkinstupid]

Where does this newly emitted photon come from, if it travels in the opposite direction of the absorbed photon?

[Alkatran]

Where does this newly emitted photon come from, if it travels in the opposite direction of the absorbed photon?

It doesn't necessarily travel in any direction. When you shine a light on a wall the photons get absorbed then emitted any which way (which is why the whole room lights up).

And you were already told it comes from the enery given to (whatever the incident photon hit).

[beatrix kiddo]

no, alkatran. creating isn't the same as emitting! according to tom, the mirror absorbs the photon and then that energy is given back off in the form of another photon. there is conservation, not creation. to create the photon, the mirror would have to create the energy itself, and it doesn't. it just gives it back off.

[Alkatran]

no, alkatran. creating isn't the same as emitting! according to tom, the mirror absorbs the photon and then that energy is given back off in the form of another photon. there is conservation, not creation. to create the photon, the mirror would have to create the energy itself, and it doesn't. it just gives it back off.

You just keep on twisting those words, alright?

My point was that you're arguing your point based on an obvious mis-wording.

[beatrix kiddo]

what words am i twisting, tran? u're the one who told me that we can all look at creation and emission as the same thing. i was arguing u on that, not tom's mis-wording...

[Alkatran]

what words am i twisting, tran? u're the one who told me that we can all look at creation and emission as the same thing. i was arguing u on that, not tom's mis-wording...

I said they could be seen as the same thing. But fine, you know what? Just to end this argument: I was wrong. There, now get on with the main topic.

[Chronos]

Hopeless. Quantum transactions are not the same as logical operators.

[stewarta]

the guy said that the photon was absorbed... he also said the it was destroyed in the absorbtion... he said that this exites the electrons, causing one to raise to a higher field, thus becoming unstable, and emmiting a new photon...

what a buch of confusing nonsense.... if the speed of light cannot be changed, then tell me what is happening in a prism.

the speed of light CAN be changed as long as it is returned to it's original speed. the photon passes through the glass, hits the silver backing and is stopped. the ENERGY not the PHOTON is absorbed. the energy is then emmited back to the photon and it travels back through the glass at speed c...

there is no need to nuke this.... no need to over explain. occhams razor will stand strong

[Tom Mattson]

Kiddies,

When a photon is absorbed, it is destroyed. It ceases to exist. It is no more. It is an ex-photon.

Conversely, when a photon is emitted, it is created at the moment of emission.

Thank you for explaining, Tom. I thought matter or energy could not be created nor destroyed?


All the energy is accounted for at all times. An unperturbed atom has energy E1 at time t1. A photon of energy Eγ comes in and promotes an atomic electron to a higher energy state
E2. Conservation of energy demands that E2=E1+Eγ.


Yes, I have taken chemistry. I owned at chemistry! I'm taking Chemistry II AP this coming school year.


OK, so check out the solar system model of an atom. It's a very crude approximation, but it will at least give some mental image of the promotion and de-excitation of an atomic electron (it goes from a lower orbit to a higher one, and back again).


So the excitation of an electron causes decay. What kind of decay is this called? When an electron goes into decay it produces photons? Is that the only product of this decay?

It is an electromagnetic transition of the atom. By the way, it is the atom that decays, not the electron. Electrons are stable. And yes, the photon is the only new particle that is created.

[Entropy]

I'm taking Chemistry II AP this coming school year.

I hope its not as hard as AP Chem II at my school! I didn't take it, but my friends (they use to like chem) who did told me it made their lives a living hell. They had complete labs with full lab write ups due every other day. Not that its a hard subject, I think its just the teacher. :wink:

[urtalkinstupid]

Haha, that Chem II AP at your school sounds like the BIO II AP I just took this year. Full labs with a write up on each one. Only grades were labs and tests. Haha, average of class was like 60 or 70, but I managed to keep an A. Chem II AP at our school is supposed to be hard as *%(#. :cry:

Tom:
Anyways, How does the new photon know exactly which direction to go in order to reflect the image? Does it take the vector of the destroyed photon and reverse it? I know about electrons and them absorbing energy. I was simply saying what you were saying. So, I'm assuming it is just photon and energy that is created. Are you also saying that anything that is completely absorbed, all of its energy is destroyed? Essentially the photon isn't destroyed, just it's energy is passed on. I don't know.

[beatrix kiddo]

tom, it hurts my feelings when u call me kiddie... atleast make it kiddo..

u really shouldn't use words like created or destroyed, though, because it gives a false representation of what is occuring on the mirror. a photon is energy release. since no energy gets created or destroyed, then, technically, the photon really isn't destroyed either. absorption and emission cannot be viewed as destruction and creation.
entropy, chem II ap at our school is pretty hard. the teacher is nice and all, but she gives really hard tests, and labs, etc. (i am taking it next yr... but this is what i've heard from her former students)

[russ_watters]

Anyways, How does the new photon know exactly which direction to go in order to reflect the image? A photon doesn't "know" anything. Conservation of momentum dictates in which direction the atom will emit the photon. So, I'm assuming it is just photon and energy that is created. Are you also saying that anything that is completely absorbed, all of its energy is destroyed? Essentially the photon isn't destroyed, just it's energy is passed on. I don't know. The photon is destroyed when it is absorbed, but the energy doesn't vanish, its incorporated into the atom that absorbed it. It manifests itself by the electron jumping to a higher energy level.

[Tom Mattson]

Anyways, How does the new photon know exactly which direction to go in order to reflect the image?


Ignoring the creation and destruction of photons for a minute, and just focusing on the photon states before and after the reflection, there is a total momentum and total energy in each state. In other words, this annihilation and creation of photons has to take place under the same constraints as a collision. The conservation of momentum in the direction parallel to the mirror imposes one constraint, and conservation of energy imposes another (the photon energy is unchanged in the reflection). That means that the momentum of the photon normal to the mirror can only be equal and opposite to the original normal component.

So now your question can be reduced to "Why are energy and momentum conserved in the process of reflection?"

And the answer is: "Because that's the way it is."


Are you also saying that anything that is completely absorbed, all of its energy is destroyed?


Not destroyed, but converted. It is transferred to the atom in the form of a promoted electron.


Essentially the photon isn't destroyed, just it's energy is passed on. I don't know.

The photon is destroyed. For as long as the excited atomic state exists, there is no photon. A new photon is emitted when the atom de-excites.

[Tom Mattson]

tom, it hurts my feelings when u call me kiddie... atleast make it kiddo..


Lighten up, kiddo. :wink:


u really shouldn't use words like created or destroyed, though, because it gives a false representation of what is occuring on the mirror.


No, it doesn't. And I use those words because those are the words that are used in quantum field theory, which is the best theory we have for the interactions between photons and electrons.

Actually, it's the best theory we have, period. (It has been verified to 10+decimal places).


a photon is energy release. since no energy gets created or destroyed, then, technically, the photon really isn't destroyed either.


No, the photon "technically" is destroyed. If it were not, then the atom could not be promoted to a higher energy level.


absorption and emission cannot be viewed as destruction and creation.


Can too. :tongue:

[beatrix kiddo]

Lighten up, kiddo. :wink:

hehehe...

i still disagree. i mean a photon is energy right? well if the energy isn't destroyed then the photon isn't destroyed. maybe stewarta is correct. maybe the photon gets stopped by the mirror and then the photon is accelerated at c, back off the mirror.

[urtalkinstupid]

Oh, no, Janna, that is contradicting with the constancy of light's velocity. Let's not do that!! It's bad to go against theories, even though they aren't full proof. I'd have to agree with you! The photon can not be destroyed. Well, I believe matter and energy can be destroyed and created, but I thought there was some kind of law stating that they can not be destroyed or created? Maybe I'm just hearing things. *looks around really quick*

I believe the photon is stopped. The energy that is emitted to an electron excites that electron. Once that electron returns to it's original energy state, it emitts the energy back out accelerating the photon in the opposite direction. WOO, another one of my outrageous ideas!

[beatrix kiddo]

hahahaha.. there we go. now let's really get the ball rolling and just come and say that anyone who says the photon is destroyed contradicts einstein! i don't agree with sir albert (on many levels), but tom does. so make a choice. either the photon gets destroyed or the velocity of light changes.

[urtalkinstupid]

Now, we know the velocity of light can't change unless there is a shift in frequency!! :cry:

[russ_watters]

I believe the photon is stopped. See, this is what I was talking about in the other thread. That's just not how science works. You can't just pull things out of the air and choose to believe them (or not believe them). You're setting yourself up to take a huge fall here. An active imagination is a good thing, but don't let it get the better of you. Your imagination isn't reality.

[urtalkinstupid]

NOTICE HOW I CORRECTED MYSELF BY SAYING FREQUENCY SHIFT! That compensates for the decceleration.

[russ_watters]

hahahaha.. there we go. now let's really get the ball rolling and just come and say that anyone who says the photon is destroyed contradicts einstein! i don't agree with sir albert (on many levels), but tom does. so make a choice. either the photon gets destroyed or the velocity of light changes. That's nonsense.

[urtalkinstupid]

I thought there was a law that said nothing can be created or destroyed. Energy and matter wise. Is that true?

[beatrix kiddo]

how is it nonsense russ?? the velocity of light changes because the photon (the light) is stopped. or do u believe the photon gets destroyed? and if u believe the photon gets destroyed, but u don't think the energy gets destroyed, that is in itself a contradiction because a photon is energy.

[ArmoSkater87]

A photon cant be destroyed, it can be converted into an electron/positron pair. (At least gamma can, i think)

[urtalkinstupid]

It doesn't get converted. Doesn't it decay?

[Bobert]

sorry about this.

[russ_watters]

I thought there was a law that said nothing can be created or destroyed. Energy and matter wise. Is that true? Matter and energy can't be created or destroyed, only converted from one to the other. Photons aren't matter and they aren't energy (though they carry energy). Think of it this way: a sound wave is created at your speaker and destroyed when it is absorbed by your eardrum. The energy is conserved. the velocity of light changes because the photon (the light) is stopped. or do u believe the photon gets destroyed? and if u believe the photon gets destroyed, but u don't think the energy gets destroyed... I don't believe, I know. There is a big differnece. Light does not stop, photons are destroyed, and energy is conserved. This has been explained now about half a dozen times. ...a photon is energy... Maybe this is the problem. See my sound wave analogy above.

[zoobyshoe]

In the early measurements of the speed of light, I know, they interpreted the interference patterns created by a split beam. Do they still measure it this way, or has a higher tech system been developed that uses a different method?

[urtalkinstupid]

I just thought of something. Take an object and accelerate it to .8c and then let it go about a constant velocity. Right when that object reaches .8c and light beam next to it wille mitt. How fast will that light beam be going with respect to the object. If the speed of light does not change, would it be c? I dont think it would c. I dunno though. Maybe there is a way around this paradox.

[zoobyshoe]

the velocity of light changes because the photon (the light) is stopped.
The photon keeps going full throttle till all of a sudden there is no more photon. Instead there is a more energetic electron. There is no time lag between the transition in which you could say there is a photon with no velocity, a "stopped" photon.

All the energy the photon had is now possessed by the electron which will jump to a higher orbit, meaning it is traveling in a circle of larger diameter than before (to use the solar system model). It takes more energy to travel in a circle of larger diameter. The electron can do this now because it has all the energy the photon once had, on top of the energy it, itself, had when the photon hit it. The photon's energy is not destroyed.

If you swing a rock tied to a string around it takes a certain amount of energy to keep it swinging. If you let more string out you'll find you have to swing harder to keep the rock in motion: it takes more energy to maintain a larger diameter orbit. Same with electrons. If you zap the electron with more energy by hitting it with a photon it just naturally jumps to the higher orbit. And there's no more photon.

Since it takes less energy for a smaller diameter orbit, the electron releases its extra energy, when it falls back to the lower orbit, in the form of a photon. Is this the same photon that it absorbed in the first place? They haven't figured out a way to tell, but there is nothing to indicate that it is. The electron has a certain total amount of energy and there is no reason to suppose any part of that energy can be distinguised from any other part. The new photon is simply whatever part of it's total energy that is convenient for it to release when it drops down to the lower orbit.

On the level of a single photon it is by no means certain that it will fly off at an angle equal to the angle of incidence. Light only seems to do that on average. I have just been reading about this in QED by Feynman. The best they can do for a given individual photon is calculate a probability for its direction. Only on the level of masses and masses of photons do they all add up to the neat and tidy angle of incidence equaling the angle of reflection.

[terrabyte]

for lasers too?

that's fascinating...

[nautica]

For the "kiddies" or "kiddoes" of this board, I found this site for you which may help explain it a little more on your level (I only say this b/c your argument sounds like that of my 10 year old son.)

It sounds like you must enjoy science and be relatively intelligent since you are in AP classes. But it appears you are to busy arguing over semantics instead of learning from the people on this board.

Nautica

[Chronos]

Pitcher throws photon to mirror. Mirror hands photon to umpire. Umpire discards it and and hands new photon to mirror. Mirror tosses new photon back to pitcher.

[Alkatran]

I just thought of something. Take an object and accelerate it to .8c and then let it go about a constant velocity. Right when that object reaches .8c and light beam next to it wille mitt. How fast will that light beam be going with respect to the object. If the speed of light does not change, would it be c? I dont think it would c. I dunno though. Maybe there is a way around this paradox.

In this scenario you have two observers. Observer B is moving at .8c according to observer A. Observer A sees ("sees" being used as a loose term, as you can't see a photon at a distance) the photon moving away at c and B moving away at .8c. For A the speed between the photon and B is .2c.

FRAME SWITCH!

B sees the photon moving away at c and A moving the other direction at .8c. For B the speed between the photon and A is 1.8c.

How can this possibly be true? That's what the theory of special relativity covers. B is moving through (space-)time at a different rate than A and B's space is contracted according to A (A's is contracted according to B).

In fact, something that happens at the same time for A will NOT happen at the same time for B.


For some reason I get the feeling we're about to be back to arguing with RAM1024 and GeistKessel over this... only their names have changed and they're a few years younger...

[russ_watters]

Pitcher throws photon to mirror. Mirror hands photon to umpire. Umpire discards it and and hands new photon to mirror. Mirror tosses new photon back to pitcher.Outstanding.

[russ_watters]

I just thought of something. Take an object and accelerate it to .8c and then let it go about a constant velocity. Right when that object reaches .8c and light beam next to it wille mitt. How fast will that light beam be going with respect to the object. If the speed of light does not change, would it be c? I dont think it would c. I dunno though. Maybe there is a way around this paradox. [emphasis added] Yes. Its called Special Relativity. It states that the speed of light is constant for all observers regardless of their inertial frame of reference and that the laws of the universe are the same everywhere.

[BobG]

Solar Radiation Pressure. For high altitude satellites, the overall perturbations are so small that the solar radiation pressure begins to become a significant factor affecting satellite orbits.

Light hitting a highly reflective surface exerts a force on the satellite. Light hitting an absorbant surface exerts an even smaller force on the satellite.

I've taken this as a given without worrying to much about the details, but this discussion kind of piques my curiosity about why this happens.

It would seem to me that, in the interests of energy conservation, absorbant materials would change the momentum more than reflective materials.

Any comments on why the actual behavior is so counter-intuitive?

Oops! Edited to remove inaccurate statement - totally absorbant materials still receive some force from light energy, just less.

Actual equation is:

F=\frac{F_sA}{cm}(1+q)cosi

where F_s is solar pressure or 1367 \frac{W}{m^2}
c is speed of light
m is mass of satellite
q is reflectivity (1 for totally reflective, 0 for totally absorbant)
i is incidence angle

On an asymetrical satellite design, similar to the NOAA GOES weather satellite, the torque from solar pressure would be in the range of 5.54 x 10^-4 Newton meters. Small, but still the largest factor affecting satellite attitude.

[Doc Al]

Light hitting a highly reflective surface causes a small change in the satellite's momentum.
Right.
Light hitting an absorbant surface causes an even smaller momentum change in the satellite
Right.
(a totally absorbant surface would impart no momentum change to the satellite).
Says who?
It would seem to me that, in the interests of energy conservation, absorbant materials would change the momentum more than reflective materials.

Any comments on why the actual behavior is so counter-intuitive?
It's no different than bouncing a ball against a wall. The ball has incoming momentum +p. If it bounces back (like a superball), then it's final momentum will be -p, so \Delta p = -2p. If it's absorbed (like a putty ball), then the final momentum is 0 (assume a massive wall) and \Delta p = -p. The ball that bounces off transfers more momentum to the wall that the one that sticks.

[BobG]

Right.

Right.

Says who?

It's no different than bouncing a ball against a wall. The ball has incoming momentum +p. If it bounces back (like a superball), then it's final momentum will be -p, so \Delta p = -2p. If it's absorbed (like a putty ball), then the final momentum is 0 (assume a massive wall) and \Delta p = -p. The ball that bounces off transfers more momentum to the wall that the one that sticks.

Makes sense as long as I stay on the surface. The photons being absorbed by atoms and then released adds some confusion factor.

But, when I think about it, it could be thought of as the photon's kinetic energy has been converted to potential energy when the electron level is raised. When the electron falls back into its original orbit, the potential is reconverted back to kinetic energy.

[Alkatran]

Solar Radiation Pressure. For high altitude satellites, the overall perturbations are so small that the solar radiation pressure begins to become a significant factor affecting satellite orbits.

Light hitting a highly reflective surface exerts a force on the satellite. Light hitting an absorbant surface exerts an even smaller force on the satellite.

I've taken this as a given without worrying to much about the details, but this discussion kind of piques my curiosity about why this happens.

It would seem to me that, in the interests of energy conservation, absorbant materials would change the momentum more than reflective materials.

Any comments on why the actual behavior is so counter-intuitive?

Oops! Edited to remove inaccurate statement - totally absorbant materials still receive some force from light energy, just less.

Actual equation is:

F=\frac{F_sA}{cm}(1+q)cosi

where F_s is solar pressure or 1367 \frac{W}{m^2}
c is speed of light
m is mass of satellite
q is reflectivity (1 for totally reflective, 0 for totally absorbant)
i is incidence angle

On an asymetrical satellite design, similar to the NOAA GOES weather satellite, the torque from solar pressure would be in the range of 5.54 x 10^-4 Newton meters. Small, but still the largest factor affecting satellite attitude.

It's pushed more because the reflective surface not only takes the photon hit, it gets pushed by the photon taking off in the other direction.

[Tom Mattson]

i still disagree. i mean a photon is energy right?


No: A photon has energy. A photon is not to be identified with energy.


well if the energy isn't destroyed then the photon isn't destroyed.


The energy is not destroyed, but the photon is. It has to be in order for the atom to attain the higher energy state.


maybe stewarta is correct. maybe the photon gets stopped by the mirror and then the photon is accelerated at c, back off the mirror.


This cannot be, because EM fields (the only thing around that can possibly exert appreciable forces on things) do not interact with photons.

[BobG]

Makes sense as long as I stay on the surface. The photons being absorbed by atoms and then released adds some confusion factor.

But, when I think about it, it could be thought of as the photon's kinetic energy has been converted to potential energy when the electron level is raised. When the electron falls back into its original orbit, the potential is reconverted back to kinetic energy.

I take back the comment about thinking it made some sense.

If the satellite were massive and did not move, it would make sense. The light leaves with the same amount of energy it left with. The fact that the satellite does move due to solar pressure means some of the energy has been transferred to the satellite.

I'm having trouble figuring out how everything balances out.

Of course, the energy of light depends not upon its speed, which is constant, but upon its intensity and frequency. So, if the satellite moves away from the light source, the light reflected must be emitted at a lower frequency than it had when it first struck the surface.

Close?

[Alkatran]

I take back the comment about thinking it made some sense.

If the satellite were massive and did not move, it would make sense. The light leaves with the same amount of energy it left with. The fact that the satellite does move due to solar pressure means some of the energy has been transferred to the satellite.

I'm having trouble figuring out how everything balances out.

Of course, the energy of light depends not upon its speed, which is constant, but upon its intensity and frequency. So, if the satellite moves away from the light source, the light reflected must be emitted at a lower frequency than it had when it first struck the surface.

Close?

Sounds right, the photoelectric effect and all that.

[russ_watters]

If the satellite were massive and did not move, it would make sense. The light leaves with the same amount of energy it left with. The fact that the satellite does move due to solar pressure means some of the energy has been transferred to the satellite.

I'm having trouble figuring out how everything balances out. Take a speed and reverse it, you get -v. Since the energy equation includes v^2, the minus signs cancel out and you get the exact same energy as you had before, but in the opposite direction.

Momentum, on the other hand, has just v....

[Chronos]

It's no different than bouncing a ball against a wall. The ball has incoming momentum +p. If it bounces back (like a superball), then it's final momentum will be -p, so \Delta p = -2p. If it's absorbed (like a putty ball), then the final momentum is 0 (assume a massive wall) and \Delta p = -p. The ball that bounces off transfers more momentum to the wall that the one that sticks.

The used to sell light pinwheels in novelty shops [still do, i suppose]. The 'clockwise' face of each vane is painted white, the opposing face is painted black. It is suspended in an evacuated glass container to eliminate air resistance. When exposed to bright light, the wheel spins clockwise.

[BobG]

The used to sell light pinwheels in novelty shops [still do, i suppose]. The 'clockwise' face of each vane is painted white, the opposing face is painted black. It is suspended in an evacuated glass container to eliminate air resistance. When exposed to bright light, the wheel spins clockwise.

These pinwheels actually work on heat differential, not solar pressure. It's virtually impossible (at least with today's technology) to test the effects of solar pressure on Earth. To come close, the evacuated glass container would have to be a perfect vacuum, vs. the extremely low pressure it actually is.

I'm not alone in trying to figure out how to correlate solar pressure with the idea that light will leave a 'perfect' mirror with the same energy it entered.

Thomas Gold (Center for Radiophysics and Space Research, Cornell University) has some doubts, as well.
http://www.arxiv.org/html/physics/0306050

I have some problems with his paper, though. The biggest is that satellites reportedly do experience a force from solar pressure. Solar pressure can cause torques which disturb satellite attitude and changes the eccentricity of geosynchronous satellites, given enough time.

Louis Friedman, working on the Cosmos I, a satellite which will use solar sail propulsion defends the idea that solar pressure does exert a force on satellites:
http://www.planetary.org/solarsail/ss_and_physics.html

I have some doubts about how well the experiment will work so close to Earth and even they realize the observed effects are likely to be very small. (Other effects, such as gravity gradients, tend to dwarf the effect of solar pressure for low orbiting satellites.

If solar pressure does exert a net force on a satellite, then there has to be energy conservation. The light has to lose some energy in transition. This letter comes closest to explaining how energy balances out:
http://solarsails.info/news/newscientistletter.html

Since the energy of light depends on its intensity and frequency, the light reflected off the mirror has to see a reduction in one or the other.

[BobG]

Take a speed and reverse it, you get -v. Since the energy equation includes v^2, the minus signs cancel out and you get the exact same energy as you had before, but in the opposite direction.

Momentum, on the other hand, has just v....

If you reduce v, don't you also reduce v^2?

[zoobyshoe]

The used to sell light pinwheels in novelty shops [still do, i suppose]. The 'clockwise' face of each vane is painted white, the opposing face is painted black. It is suspended in an evacuated glass container to eliminate air resistance. When exposed to bright light, the wheel spins clockwise.
That's a Crookes' Radiometer. It doesn't rotate because of the light hitting the vanes, though. It is chronically sold with that misinformation in accompanying literature. This site explains how it actually works:

Crookes radiometer
Address:http://www.fact-index.com/c/cr/crookes_radiometer.html

[Doc Al]

That's a Crookes' Radiometer. It doesn't rotate because of the light hitting the vanes, though. It is chronically sold with that misinformation in accompanying literature.
Right! If light pressure were the cause of the vanes spinning, the vanes would spin in the opposite direction than they do.

[russ_watters]

If you reduce v, don't you also reduce v^2? Well sure - but who said anythign about reducing v? All you are changing is the sign. The magnitude stays the same.

[BobG]

Well sure - but who said anythign about reducing v? All you are changing is the sign. The magnitude stays the same.

I guess I just missed what you were trying to say.

I agree that reflected light has to leave at the same speed it entered. But something has to change in order to account for motion caused by solar pressure. If a more reflective surface results in a greater force exerted on the surface, then it would seem that the only thing that can change is the light's frequency. If solar pressure were causing the object to move away from the light source, then the frequency would change by default (Doppler effect).

This almost has to be the answer to what I was asking (how do you account for the motion of the object), but was asking for confirmation that this was what 'balanced' the equation.

[omin]

Two gods have bats made made of pure mirrorium. The pitcher throws photons at constant c, but each gods swings their bats at different speeds.

The photon enters the mirror with a speed based upon the perspective of it's last source, but this same photon does not leave the mirror it entered. You're all saying another photon leaves. This photon is manifested in the new reflection source and leaves at a rate based upon where it manifested. It follows Newtons laws launching forth from it's new source, forgeting it's previous launch conditions.

It sounds correct to say light's speed is constant, but it's constant from a source which means lights speed is determined by a combination of the source where it's manifested and the properties of light itself.

If a light is reflected, the source where reflection occurs can increase the lights velocity to a orginating source faster than light left the source originally, just by the newly reflecting source simply accelerating during the reflection process in the direction toward the previous source.

Correct or incorrect?

[BobG]

Two gods have bats made made of pure mirrorium. The pitcher throws photons at constant c, but each gods swings their bats at different speeds.

The photon enters the mirror with a speed based upon the perspective of it's last source, but this same photon does not leave the mirror it entered. You're all saying another photon leaves. This photon is manifested in the new reflection source and leaves at a rate based upon where it manifested. It follows Newtons laws launching forth from it's new source, forgeting it's previous launch conditions.

It sounds correct to say light's speed is constant, but it's constant from a source which means lights speed is determined by a combination of the source where it's manifested and the properties of light itself.

If a light is reflected, the source where reflection occurs can increase the lights velocity to a orginating source faster than light left the source originally, just by the newly reflecting source simply accelerating during the reflection process in the direction toward the previous source.

Correct or incorrect?

Incorrect. Each photon leaves the bat at the speed of light. Since the bat is moving the same direction as the photon, each photon is closer together. In other words, the frequency increases, not the speed.

Edit: Wow! This was the original question. The discussion became so sidetracked I almost forgot there was an original question.

The bat receives photons at a certain rate. It emits the photons at the same rate it received them. But, since the bat is moving, the photons are emitted closer together. That means another observer receiving the photons from the bat will receive them at a higher rate than the bat actually emitted them.

[omin]

The bat receives photons at a certain rate. It emits the photons at the same rate it received them. But, since the bat is moving, the photons are emitted closer together. That means another observer receiving the photons from the bat will receive them at a higher rate than the bat actually emitted them.

The pitcher now has to simultaneously pitch to two gods. Both gods step up to bat. The gods are exactly 60 feet from the pitcher. The pitcher winds up, then throws two photons. Both photons travel c toward both gods. One god is slightly distracted for a moment, but begins his swing just after the other god. The late swinging god compensates by accelerating his bat and in doing so hits the photon harder with his bat. One more thing, both bats hit their photons at the same time.

Do the photons absorb into each bat at the same rate? Does the new photon manifestation process occur at the same rate in each bat? Does each photon emerge from the bat at the same speed, relative to the playing field?

If photon absorbtion and manifestation is simultaneous, and emergece of photons from the bat have the same speed, where does the extra energy go that the late swing god exerted in his swing?

Speed c is determined by the source or what?

If the photon is unaffected by the instataneous intertia of the bat during it's emergence from the bat, then its speed c of emergence is relative to what?

If the photon is affected by the instantaneous inertia of the bat (it's physical origin now), then it should move speed c from the instantaneous inertia of the bat, right?

If the later is true, then late swinging god's photon is now moving faster than the other photon.

[BobG]

Since the late swinging god's bat is moving faster, they could only hit one photon each at the exact same time. Both would receive the photon at the speed of light - both would send the photon skyward at the same rate.

However, light consists of many photons. It has a wavelength (in fact, that's what distinguishes color). The late swinging, fast swinging god will receive photons at a faster rate than the early swining, slow swinging god. Both will emit photons at the same rate they were received (each photon emitted at the speed of light). Each bat moves between emission of a photon, so even though the photons are emitted at the same rate (frequency) and speed they were received, the photons are emitted closer together.

Players in the field will receive the fast swinging gods photons at a faster rate (frequency) than the slow swinging god, since there was less distance between the photons emitted by the fast swinging gods. Each photon is traveling at the speed of light.

Energy depends upon how many photons each batter is emitting at a time (intensity) and the rate they are emitting the photon packets (frequency). So the difference in energy between the fast swinging god and the slow swinging god is reflected by the rate the players are receiving the photons.

In other words, the energy of the swing is reflected by how many photons are received over a given amount of time, not the speed of each photon.

[terrabyte]

if it's just one photon, energy of the swing makes no difference.

photons have no mass to be accelerated in this fashion

[omin]

Since the late swinging god's bat is moving faster, they could only hit one photon each at the exact same time. Both would receive the photon at the speed of light - both would send the photon skyward at the same rate.

Hey Bob, you answered, thank you, but could you explain, because it's not common wisdom to those trying to understand, which is often the only effect of pat answers. This is the question you answered:

Does each photon emerge from the bat at the same speed, relative to the playing field?

An explanation I'm sure would answers these questions:

Do the photons absorb into each bat at the same rate? Does the new photon manifestation process occur at the same rate in each bat?

If photon absorbtion and manifestation is simultaneous, and emergece of photons from the bat have the same speed, where does the extra energy go that the late swing god exerted in his swing?

Speed c is determined by the source or what?

If the photon is unaffected by the instataneous intertia of the bat during it's emergence from the bat, then its speed c of emergence is relative to what?

If the photon is affected by the instantaneous inertia of the bat (it's physical origin now), then it should move speed c from the instantaneous inertia of the bat, right?

[omin]

if it's just one photon, energy of the swing makes no difference.

photons have no mass to be accelerated in this fashion

All things with physical properties can have action and reaction. (Newton Law III)

Photons do have action and reaction because the full process involved in human vision consciousness directly reacts to photons, which means humans sense photons.(Empiricism)

Only things with physical properties can be sensed, which is only directly by humans or human aided sense, instruments. All things with physical properties have the property of mass. (Empiricism, Technology, The Physical Property of Mass and NL III)

Therefore, photons have the property mass.

If photons don't have mass, then should I throw out Newtons Laws? Newtons laws seem to me to imply mass to anything sensed. Or what?

[russ_watters]

If photons don't have mass, then should I throw out Newtons Laws? Newtons laws seem to me to imply mass to anything sensed. Or what? Newton's laws have well known limitations and that's one of them.Only things with physical properties can be sensed, which is only directly by humans or human aided sense, instruments. All things with physical properties have the property of mass. (Empiricism, Technology, The Physical Property of Mass and NL III) Since when? That's an assumption based on a narrow-minded view.

All waves, by definition, have no mass.

omin, you asked a lot of questions about C in previous posts. Have you read what Special Relativity has to say about it? It's the second postulate you should be concerned with.

[omin]

Newton's laws have well known limitations and that's one of them. Since when? That's an assumption based on a narrow-minded view.

All waves, by definition, have no mass.

omin, you asked a lot of questions about C in previous posts. Have you read what Special Relativity has to say about it? It's the second postulate you should be concerned with.

Waves are psychologial theories, which are physically subjective as we speak. I insist they only become thoughts (theory in the mind), because they have a physically properties of a objective origin, unless wave theory has nothing to do with the physical world and is pure fantasy of the imaginative human subjective realm. Any valid theory derives from physics. Is wave theory valid, then? If so, are you saying wave theory is a figment of the mind only and represents only subjective fantastic order? If not, I momentarily put words in your mouth. I apologize on that point. The book Wu Li Masters claims physically objective originating cause of the subjective wave theory through descriptions of past experiments.

I've read the second postulate, but the explanations in my text and in the video I watched represent things in a way that is outside my understanding. That is why I am questioning for explanation. I could give you the pat answers the texts and video asserted, but that doesn't mean I understand. Understanding it is what I am after, not memorization.

Furthermore, narrow minded? I'm open for explanations, how is that narrow minded. I never asserted anything that asserts I am right and everything beyond that is wrong.

Come on man.

[Tom Mattson]

Waves are psychologial theories, which are physically subjective as we speak.


No, waves are propagating disturbances in a physical field.


The book Wu Li Masters claims physically objective originating cause of the subjective wave theory through descriptions of past experiments.


The book Quantum Mechanics by Sakurai claims that all physical phenomena are really wave phenomena. Does that mean that the objective world doesn't exist?

You're talking philosophy here, when everyone else in the joint is talking physics.

And you're wrong about photons not having "reactions". When an atom emits a photon, the atom recoils. This happens despite the fact that the photon has no mass.

[Tom Mattson]

All waves, by definition, have no mass.


Matter waves have mass.

[omin]

Interesting. We sense the atom recoil. Do atoms recoil between photons emiting or just when photons emit? If they only emit when the photon leaves, the photon is involved in the recoil.

The state of the atoms interia (in respect to another atom in non-uniform motion) does not determine the photons speed, but the photon emission process determines the recoil? This appears to represent a violation of N III Law.

The velocity of the body which the atom resides and the velocity of the atom in the body does not determine the photons speed from the atom? If this represents the physics, then the photon moves constant at c accross something like the field aether rather and the photon's speed being determined by the atom from which it was emitted.

If the atom does determine the photons c from it, the photon moves specificaly from the atom at c. If another body of atoms were moving slower than this body (in relation to a third point) that simultaneously emitted a photon, the body of atoms traveling faster will emit a photon that travels faster than the slower moving body of atom's emitted photon.

This implies light travels at same speed from atoms through space, but approaches objects in non-linear motion at different speeds.

Or what? I'm open to understand.

[russ_watters]

Matter waves have mass. You mean like sound (or maybe I don't know what a "matter wave" is...)? Sound waves ride on air, which has mass - sound waves themselves do not.

[Chronos]

You lost me there, Tom. Waves, by definition, have no mass, just mass potential.

[pmb_phy]

Matter waves have mass.

A matter wave is more of a verb than a noun. It is not a thing but rather something which describes a thing. The term matter wave usually refers to the wavelike characteristics of the probability distribution associated with a particle which has a non-zero proper mass. I spose that it can apply to a photon to since all the wavelike properties of a photon are identical to the wavelike properties of any other particle and as such they don't depend ont he particle's proper mass.

Pete

[pmb_phy]

Interesting. We sense the atom recoil. Do atoms recoil between photons emiting or just when photons emit? If they only emit when the photon leaves, the photon is involved in the recoil.

For simplicity consider a single isolated atom in an inertial frame of reference. In this frame the atom emits a photon. The atom recoils because the photons are emited. This can be described by saying that in order for the total momentum of the system to be conserved the atom must have momentum and that means it must move. The amout of energy the photon has is directly related to the final speed of the atom. The higher the photon energy the higher the final speed of the atom. If the atom is in a body then, for the purposes of analyzing the dynamics, you can simply consider the body to be that which emits the photon.

No matter what the final speed of the atom, the final speed of the photon is always the same, c. All that is different for photons with different momentum is the energy of the photon. The speed of the atom, however, is a function of the atoms final proper mass (the proper mass of the atom must have decreased in this process) and the atom's momentum.

Pete

[Tom Mattson]

Interesting. We sense the atom recoil.


I'm pretty sure that no one has ever "senses" an atom recoil.


Do atoms recoil between photons emiting or just when photons emit? If they only emit when the photon leaves, the photon is involved in the recoil.


They recoil as a direct result of emission (or absorption, for that matter) of photons. This is because the photon carries momentum and energy, and both are conserved quantities.


The state of the atoms interia (in respect to another atom in non-uniform motion) does not determine the photons speed, but the photon emission process determines the recoil? This appears to represent a violation of N III Law.


It doesn't violate Newton's third at all. In fact, it's entirely consistent with it. The change in momentum of the atom is precisely equal to the momentum of the photon.


The velocity of the body which the atom resides and the velocity of the atom in the body does not determine the photons speed from the atom?


No. The speed of the photon is c, regardless of the motion of the atom.

[Tom Mattson]

You mean like sound (or maybe I don't know what a "matter wave" is...)? Sound waves ride on air, which has mass - sound waves themselves do not.


You lost me there, Tom. Waves, by definition, have no mass, just mass potential.


What does QM teach us? It teaches us that particles act as waves. These waves are referred to as matter waves.


A matter wave is more of a verb than a noun. It is not a thing but rather something which describes a thing. The term matter wave usually refers to the wavelike characteristics of the probability distribution associated with a particle which has a non-zero proper mass.


I don't see how "matter wave" can be considered a verb. For instance, an electron is a matter wave. It doesn't do a matter wave.


I spose that it can apply to a photon to since all the wavelike properties of a photon are identical to the wavelike properties of any other particle and


Not really. As I'm sure you know, the wavelike properties of light are described by the EM wave equation. Not so with matter waves. If we want to get relativistic, then spin-0 bosons are described by the Klein-Gordon equation, and spin-1/2 fermions are described by the Dirac equation. There are more complicated equations for higher spins, but none of them is identical to the EM wave equation.


as such they don't depend ont he particle's proper mass.


Would you consider frequency and wavelength "wavelike properties"? If so, then the wavelike properties of matter waves depend on their proper mass.

[BobG]

They recoil as a direct result of emission (or absorption, for that matter) of photons. This is because the photon carries momentum and energy, and both are conserved quantities.

So this explains the (1+q) part of this equation?

F=\frac{F_sA}{cm}(1+q)cosi

In other words, all the photons received are received and some force is exerted to the surface receiving them.

Depending upon the reflectivity of the surface, only a certain percentage of the photons are emitted and only the departing photons affect the momentum of the surface.

The difference, the absorbed photons, have their kinetic energy converted to potential energy by raising the level of the electron's orbit.

Fairly close?

[Tom Mattson]

So this explains the (1+q) part of this equation?

F=\frac{F_sA}{cm}(1+q)cosi



I'm not familiar with the equation. What do all the variables stand for?

[BobG]

F=\frac{F_sA}{cm}(1+q)cosi


where F_s is solar pressure or 1367 W/m^2.
c is speed of light
m is mass of satellite
q is reflectivity (1 for totally reflective, 0 for totally absorbant)
i is incidence angle

The equation refers to the affect of solar pressure on high altitude satellites. Over time, it causes torques to build up on the spacecraft that have to be compensated for to keep the satellite's sensors/antennas/etc pointed accurately. It also affects the shape of the orbit, since solar pressure slows the satellite slightly when the satellite is headed towards the sun and speeds it up slightly when it is headed away from the sun.

[omin]

...The atom recoils because the photons are emited...for the total momentum of the system to be conserved the atom must have momentum... The amout of energy the photon has is directly related to the final speed of the atom...No matter what the final speed of the atom, the final speed of the photon is always the same, c. All that is different for photons with different momentum is the energy of the photon.

The change in velocity of atom (in recoil) is directly proportional to the amount of energy of the photon, but the photon's velocity isn't? How is that consistent with conservation of energy? In every physical collision, the velocity change of one entity is directly proportional to the change of velocity of the other entity, unless the quantity of change of velocity is internalized in the photon, making it unapparent to the viewer. Is this what the photon does?

I'm pretty sure that no one has ever "senses" an atom recoil.

Couldn't this be done through instrument aided sense?

[Tom Mattson]

The change in velocity of atom (in recoil) is directly proportional to the amount of energy of the photon,

Where are you getting this from?

First, the change in velocity is a vector, and the energy of a photon is not. There is no way that one could be proportional to the other. Second, even if we were talking about the change in the speed of the atom, this wouldn't be right. The change in speed of the atom is determined by:

(1/2)matom(vf2-vi2)=ΔEphoton.

That's not a direct proportion!


but the photon's velocity isn't?


Right. The photon's energy is directly proportional to its frequency.


Couldn't this be done through instrument aided sense?


Either the human senses it, or the instrument does. I stand by my original statement: humans don't sense atomic recoil. It's below the threshold of our sense of touch.

[BobG]

Well, I'm probably mixing apples and oranges, but it kind of feels like all the atoms in my skin are moving faster when I step into the sunlight.

This is a complicated subject when you try to figure out its relationship to things you observe.

You're receiving light. Some of the light is reflected. Some of the light exerts a force that can move objects (the same light that is reflected). Some of the light creates heat (the portion that isn't reflected). It almost does seem like we're getting too much energy out of the transaction.

[russ_watters]

What does QM teach us? It teaches us that particles act as waves. These waves are referred to as matter waves. Gotcha - fair enough. I thought you meant sound waves or the like.

[omin]

Where are you getting this from?

First, the change in velocity is a vector, and the energy of a photon is not. There is no way that one could be proportional to the other.

I get it from this simple equation.

KE = 1/2 m * v ^2

where m = mass of photon

v = speed of photon (velocity has direction, which is included in the energy equation.)

Therefore, if photon mass is consistent, a change in speed changes photons energy. If the photon doesn't change speed, then does it's mass change during recoil?

If the photon doesn't change speed or mass during recoil, then where does the energy go? Conservation of energy represents that a change of energy on one side of the equation is proportionate to the other side of the equation, where one side is a atom and it's property recoil and the other is a photon emission.

Please explain that equation. I'm not sure if it infers the photon has change in it's mass or velocity when the atom in the equation appear to.

Either the human senses it, or the instrument does. I stand by my original statement: humans don't sense atomic recoil. It's below the threshold of our sense of touch.

If an instrument senses recoil, for us to know the instrument senses recoil, we have to sense the instrument sensing recoil. If we know it sensed it, we indirectly sense recoil. Instrument aided sense. All sense is indirect really, we don't sense the objects we sense a peice of the chain of a chain reaction. I'm basing this on the fact that thought or sense of a physical object is an expression of the object rather than the object. Which assures me, we've sensed the photon, because it has mass. Without mass, it sends no expression properties of its existence.

[russ_watters]

If the photon doesn't change speed or mass during recoil, then where does the energy go? What you may be missing here is that the photon doesn't acceerate up to C like a car accelerating up to highway speed. At the instant of release, it is traveling at C. The "recoil" is therefore also an instantaneous change in energy. A photon doesn't undergo "changes" when it is emitted.

[Tom Mattson]

I get it from this simple equation.

KE = 1/2 m * v ^2

where m = mass of photon

v = speed of photon (velocity has direction, which is included in the energy equation.)


OK, so where are you getting that from?

That equation does not apply to photons. Firstly, it only applies to particles with mass (which photons don't). Secondly, it doesn't even apply to massive bodies if the speed v is comparable to c (and in the case of photons, the speed is c).


Therefore, if photon mass is consistent, a change in speed changes photons energy. If the photon doesn't change speed, then does it's mass change during recoil?


Since this is predicated on the above misconception, I have no answer for you.


If the photon doesn't change speed or mass during recoil, then where does the energy go? Conservation of energy represents that a change of energy on one side of the equation is proportionate to the other side of the equation, where one side is a atom and it's property recoil and the other is a photon emission.


I already told you: The photon energy is proportional to its frequency. For photons: E=hf (h=Planck's constant, f=frequency).


If an instrument senses recoil, for us to know the instrument senses recoil, we have to sense the instrument sensing recoil. If we know it sensed it, we indirectly sense recoil. Instrument aided sense. All sense is indirect really, we don't sense the objects we sense a peice of the chain of a chain reaction. I'm basing this on the fact that thought or sense of a physical object is an expression of the object rather than the object. Which assures me, we've sensed the photon, because it has mass. Without mass, it sends no expression properties of its existence.


The instrument doesn't aid our senses, it is replaces them. There is no such thing as "indirect sensation". Sensations are direct, first person experiences. But now we're talking philosophy again.

[Tom Mattson]

where F_s is solar pressure or 1367 W/m^2.
c is speed of light
m is mass of satellite
q is reflectivity (1 for totally reflective, 0 for totally absorbant)
i is incidence angle


OK, now I can tell you about (1+q). If a photon is absorbed by a body and never heard from again, then it transfers all its momentum to the body. This would correspond to q=0. This is a highly idealized state of affairs that is not realized in practice, but q=0 can be approached and it serves as a lower bound. Now for the other extreme: the photon is absorbed, and then promptly re-emitted in the exact opposite direction. The momentum change of the photon is not p, but 2p. This would correspond to q=1, which gives us a multiplier of 2 for the force on the body.

[BobG]

I get it from this simple equation.

KE = 1/2 m * v ^2

where m = mass of photon

v = speed of photon (velocity has direction, which is included in the energy equation.)

Therefore, if photon mass is consistent, a change in speed changes photons energy. If the photon doesn't change speed, then does it's mass change during recoil?

If the photon doesn't change speed or mass during recoil, then where does the energy go? Conservation of energy represents that a change of energy on one side of the equation is proportionate to the other side of the equation, where one side is a atom and it's property recoil and the other is a photon emission.

Please explain that equation. I'm not sure if it infers the photon has change in it's mass or velocity when the atom in the equation appear to.



If an instrument senses recoil, for us to know the instrument senses recoil, we have to sense the instrument sensing recoil. If we know it sensed it, we indirectly sense recoil. Instrument aided sense. All sense is indirect really, we don't sense the objects we sense a peice of the chain of a chain reaction. I'm basing this on the fact that thought or sense of a physical object is an expression of the object rather than the object. Which assures me, we've sensed the photon, because it has mass. Without mass, it sends no expression properties of its existence.

A photon has no mass. So the equation for finding the kinetic energy of an object of mass doesn't apply.

You need to use this equation:

E=\frac{hc}{\lambda}
where E = energy

c = speed of light (2.9979 x 10^8)
h = Planck's constant (6.626 x 10^-34)
and \lambda= wavelength

An alternative version is to use the frequency. Frequency is speed of light (corrected) divided by the wavelength. Energy is then Planck's constant times frequency.

The higher the frequency, the more energy. (or, the shorter the wavelength, the more energy)

Edit: Oops. Thinking faster than I could type.

[BobG]

OK, now I can tell you about (1+q). If a photon is absorbed by a body and never heard from again, then it transfers all its momentum to the body. This would correspond to q=0. This is a highly idealized state of affairs that is not realized in practice, but q=0 can be approached and it serves as a lower bound. Now for the other extreme: the photon is absorbed, and then promptly re-emitted in the exact opposite direction. The momentum change of the photon is not p, but 2p. This would correspond to q=1, which gives us a multiplier of 2 for the force on the body.

Right. I'm with you so far.

But, now, something should happen to the light when it is re-emitted in the opposite direction. If the light moved the mirror, then it seems like the light should lose some energy. In other words, the frequency of the reflected light would seem to have to have a lower frequency than the original received light.

[Tom Mattson]

Right. I'm with you so far.

But, now, something should happen to the light when it is re-emitted in the opposite direction. If the light moved the mirror, then it seems like the light should lose some energy. In other words, the frequency of the reflected light would seem to have to have a lower frequency than the original received light.

You're right. I failed to mention that q=1 is also an idealized case that is not realized in practice, but that it can be approached and that it is an upper bound.

[Doc Al]

But, now, something should happen to the light when it is re-emitted in the opposite direction. If the light moved the mirror, then it seems like the light should lose some energy. In other words, the frequency of the reflected light would seem to have to have a lower frequency than the original received light.
You're right. I failed to mention that q=1 is also an idealized case that is not realized in practice, but that it can be approached and that it is an upper bound.
Right, q=1 represents the case of an ideal reflector. But I don't think that this formula that BobG produced addresses his real concern, which is conceptual.

I think BobG is asking: Isn't it true that the reflected photons must lose energy? After all, they produce a radiation pressure on the reflecting surface, thus do work on the surface, thus supply energy to it. That energy must come from somewhere. If the photons reflect back with the same frequency as before, then something's fishy. (Is that an accurate statement of your reasoning, BobG?)

That reasoning looks good to me. I would say that the reflected photons must be redshifted (however slightly) compared to their original frequency. Comments? Am I missing something?

[BobG]

Yes, exactly.

Also, I take it that all the photons are absorbed by atoms in the surface of the object being struck. The difference between being converted to heat (increasing the motion of the atoms in a random pattern) vs. being reflected is just because of the direction the photons are re-emitted.

(I've got to read something on this, this has kind of got my curiosity up - especially if you go another step further and start substituting transparent or translucent surfaces for the 1 in 1+q).

[russ_watters]

That reasoning looks good to me. I would say that the reflected photons must be redshifted (however slightly) compared to their original frequency. Comments? Am I missing something?Its awful, but I don't know this one either. It seems like a situation where momentum is conserved but energy isn't. Maybe its just a matter of not taking into account mass differences, ie we tend to not think about how much velocity an individual photon will give to something with mass it hits.

Thinking about billiards balls, in a collision between a moving one and a stationary one, the moving one stops. What if the stationary one were larger than the moving one? The moving one bounces back, right? Taking the limit, if the stationary ball is almost infinitely large, it doesn't move much and the moving ball bounces back with almost the same speed/energy as it started with. Since light can't travel at almost C, it must be very, very, very slightly redshifted. Seems like maybe we just oversimplify and get rid of the "almost."

Am I close?

[omin]

The instrument doesn't aid our senses, it is replaces them. There is no such thing as "indirect sensation". Sensations are direct, first person experiences. But now we're talking philosophy again.


They call it physics because only physics can be sensed, the rest is religion and philosophical psychological order (pure theory).

If you say we don't sense the photon, then it's not physics. It's pure psychological mass in the brain.

Be aware, you are the one who is claiming I'm the philosopher, but you claim theories come from nonsense, which is no physical origin.

It's clear I"m after the physics here, which is only the senseable stuff!

[Tom Mattson]

They call it physics because only physics can be sensed, the rest is religion and philosophical psychological order (pure theory).


They call it physics because it's the study of things that exist. These things need not be sensed to be known. A detector whose workings are known is a valid substitute.


If you say we don't sense the photon, then it's not physics.


Wrong. Every day physicists do experiments with photons that are never sensed by humans. The detector registers their presence, and that is sufficient.


It's pure psychological mass in the brain.


What is "psychological mass"?


Be aware, you are the one who is claiming I'm the philosopher,


I said that you were drifting off into philosophy, but not that you are a philosopher. You've got a long way to go before you could be called a philosopher.


but you claim theories come from nonsense, which is no physical origin.


I never claimed that theories come from nonsense, nor did I claim that they come from "no physical origin". That's just your strawman argument.


It's clear I"m after the physics here, which is only the senseable stuff!

It's clear that you don't understand what you're talking about.

Anyway, do you have anything to say about the other corrections I made to your post? For instance, about the kinetic energy of the photon?

[omin]

A photon has no mass. So the equation for finding the kinetic energy of an object of mass doesn't apply.

You need to use this equation:

E=\frac{hc}{\lambda}
where E = energy

c = speed of light (2.9979 x 10^8)
h = Planck's constant (6.626 x 10^-34)
and \lambda= wavelength

The word order above are representations of thoughts. Thoughts only represent senses. Senses only represent the physically objective world. The physically objective world consists of either thought of the objective world or thought of thought existing in the mind (the subjective physical world). Therefore, these symbols are representation of physics of the objective world or they or only a theory of the mind which does not represent the objective world accurately.

What I have said is not philosophy, it's the basis of what we call theory that all physicists must understand to represent the physical world accurately.

Mass represents anything physical or a property of something physical. Or is mass also a theory existing in the mind that is outside the valid representation of the physical world?

If you guys undermine valid physical theory by asserting psychological order that represent no physics of the world is valid physics, then it's not physics and we can't talke senseably about this.

[Tom Mattson]

The word order above are representations of thoughts. Thoughts only represent senses. Senses only represent the physically objective world. The physically objective world consists of either thought of the objective world or thought of thought existing in the mind (the subjective physical world). Therefore, these symbols are representation of physics of the objective world or they or only a theory of the mind which does not represent the objective world accurately.


If we can at least agree that things that are detected by measurement apparatus (as opposed to direct sensation) constitute knowledge of the physical world, then we can move on from this issue of "sensing" and get back to your misconceptions about the kinetic energy of photons.


What I have said is not philosophy,


It very obviously is philosophy. You are defining what it means to be physical, to be objective, to be a representation of the physical world, and you are demarcating the line between the mind and the physical. That's metaphysics through and through. And your earlier comments on how sensation is prerequisite to knowledge of the physical universe is very clearly an epistemological theory. If this isn't philosophy, then nothing is.


it's the basis of what we call theory that all physicists must understand to represent the physical world accurately.


The theories that physicists develop to represent the world are already known to be accurate because they match measurements so well. The proof of the pudding is in the eating.


Mass represents anything physical or a property of something physical. Or is mass also a theory existing in the mind that is outside the valid representation of the physical world?


What difference does it make? If mass is a property of physical objects, then we can formulate physical theories in terms of it. What you are doing here is pondering the metaphysical meaning of "mass", which is a philosophical tangent issue.


If you guys undermine valid physical theory by asserting psychological order that represent no physics of the world is valid physics, then it's not physics and we can't talke senseably about this.

Will you please stop spouting off your opinions about who is undermining what theory until you demonstrate that you have mastered the theories yourself? You are still hung up on some very basic issues and you would benefit the most from getting those cleared up, rather than doing all this philosophizing (there's that word again) on things you clearly do not understand.

[omin]

Will you please stop spouting off your opinions about who is undermining what theory until you demonstrate that you have mastered the theories yourself? You are still hung up on some very basic issues and you would benefit the most from getting those cleared up, rather than doing all this philosophizing (there's that word again) on things you clearly do not understand.

Physical theory has no contradictions;there may only be categories of existence, which may be compared to arrive at categorization and quantification. Physics is very, very, very little the Boolean exist or don't exist misrepresentation. It's here or there and how much in comparison to this much of what's over here. We may only speak what exists, that's physics and when we reach the limit of our senses and our physically derived probabilities diminish in positive value, we say I don't know.

If a thing has no mass(boolean!), it may not have the property distance. If it doesn't have the property distance, it may not have the property speed. If it doesn't have the property speed, it may not have the property acceleration. So, if not mass, none of these may derived. If none of these may be derived, nor can energy be represented. (To arrive at a rate of speed, you must sense what is speeding.) Without acceleration (what happens when we mean force) an object is outside our sense because there is lack of contact; therefore you do not know that it exists or will never know it exists unless contact occurs, which demands the absolute prerequisite mass. Knowledge depends upon mass and velocity, and finally contact (force). Have you ever been hit, by only, and I emphasize only, the velocity of the sun? Mass is absolutely necessary for a change in velocity or (action and reaction) to occur.

To assert the thought of phenomenon c, you must use the basic physical theories that build up to the point that allow us to begin to think about phenomenon c. Fundamental concepts necessary to arrive at the thought of phenomenon c are atleast mass, speed, velocity and acceleration. These concepts are absolutley necessary to get to the concept c, so c depends upon them. Once we arrive at the discussion c (or energy), saying authoritively, "you may now not use the theory we used to arrive here", is self-negating. You may build the pyramid, but if you destroy the base, the peak comes down with it.

In calling this a philosophical point is accuarate, but it's more accurate that it's physics in this discussion, because it's expressive of the principle in N III Law as it relates to the physical phenomena in which we are discussing.

There is but one road to c and it is through Newton.

[omin]

The displacement effect the light has on the satelite occurs during:
1. the light impact
2. during photon absorbtion
3. during photon manifestation
4. during photon emision
5. or some mix?

Thinking about billiards balls, in a collision between a moving one and a stationary one, the moving one stops. What if the stationary one were larger than the moving one? The moving one bounces back, right? Taking the limit, if the stationary ball is almost infinitely large, it doesn't move much and the moving ball bounces back with almost the same speed/energy as it started with. Since light can't travel at almost C, it must be very, very, very slightly redshifted. Seems like maybe we just oversimplify and get rid of the "almost."

Can a change in frequency of light and the mass of the object from which the light reflects be used as variables to attempt to give us the mass of light? If you believe light has no mass, I've heard the lecture, nevermind, unless of course you have a more convincing lecture.

No matter what the final speed of the atom, the final speed of the photon is always the same, c. All that is different for photons with different momentum is the energy of the photon. The speed of the atom, however, is a function of the atoms final proper mass (the proper mass of the atom must have decreased in this process) and the atom's momentum.

The matter of an atom decreases proportionate to the energy of a photon it emits. Matter is neither created nor destroyed. Therefore, the photon represents this difference in matter of the atom?

An Instataneous C?
Clearly, things are touching other things in this circumstance(N III Law), where photons emit from atoms and the atoms recoil. The photon manifests in an atom that has a velocity less than c (unless inner structural elements are already at c). There appears to be elements in a state of pre-photonic form in the atom. In the process of manifesting a photon, things must be accelerated into form, since they are in pre-photonic form, not yet a photon. Formation requires N III Law, unless the photon's inertia in this example is entirely unaffected and the whole process is determined and occurs without acceleration. Once the photon is a photon, it is said to have instantaneous c. To claim instataneous state c is justified only as a state. All states before the photon instantaneous c state then must be acceleration states? Meaning photon forming states.

How can all the elements that collapse form and reform during this process always be separate, in an isolated inertia?

If photon from atom separation is instataneous and the photon is instataneously at c, it either moves away from the instataneious velocity of the atom at separation at c, moves along the aether at c, moves along some other resistive field at c, or (instant c means instant intertia) moves based upon it's own physical dimensions at c (which would be determined by the atom!)

[BobG]

Omin,

Are you an antique book collector? I'm thinking the last book printing your theory came out sometime before 1910 at the latest.

[Tom Mattson]

Omin,

If you had a point to your last post to me, you have successfully hidden it by couching it in gibberish. The only thing intelligible enough to comment on is this:

There is but one road to c and it is through Newton.

Do you mean "there is but one means to derive the invariance of the speed of light"? If so, then you are wrong. Einstein derived it from the covariance of Maxwell's equations (not any of Newton's laws).

[omin]

That's okay guys. Whatever.

Explanation requires competence. You are either competent enough to explain things from my so called primitive 1910 understanding or not. If it's so primitive, why is it so hard to explain? Having trouble with simple things are ya?

So far, I've gained very little from the discussion, and that is the positive amount. The ability to convey things to someone interested in this forum has been demonstrated. Some things have been conveyed. But, it's been unnecessarily cluttered with loads of insults and incompetent explanation. I'm less likely to respond to you now. I resent being taunted into idiotic insult games through your pompous tones in your responses. Not that I've never been involved in it my life, but it's cleary understood by me to be the defense mechanism of intellect weakness and have never been a efficient use of time, except the time it took to end it. I'm here to have attention to those who have respect for the discipline and have respect for those who are seeking an understanding of the discipline.

You can write what you want but I'd rather you not respond to my posts unless your ready to be serious and express some repsect to those who may stand on your shoulders, since you claim you are further ahead. That means stop being a "Mr. I Know Everything, You Don't Know What Your're Talking About and your 'You are Gibbering Claims'" and begin debate that gets somewhere. People don't walk into a classrom because they know everything thing. Did you know that? You proved you haven't been aware of that by your attitude toward me! I'll debate and try to understand, but I'm finished with this riduculous tone. If it behooves you to keep it up, I'll just ignore you.

And, furthermore, I'm intelligent enough to know this is valuable to you.

[omin]

What do you mean by invariance?

[Doc Al]

What do you mean by invariance?
A key property of light is that its speed is invariant, meaning that it has the same value as measured in any inertial frame.

[Tom Mattson]

Explanation requires competence.


So does learning.


You are either competent enough to explain things from my so called primitive 1910 understanding or not. If it's so primitive, why is it so hard to explain? Having trouble with simple things are ya?


I've got a news flash for you: A great deal of the responsibility for your education falls to you. You would do well to lose this attitude of "If you can't spoon feed this to me in terms that I can understand, then you aren't competent at explaining physics". Not only will it exasperate anyone who tries to help you, but it means that you are shirking your responsibility to yourself as one who claims to be serious about learning.


So far, I've gained very little from the discussion, and that is the positive amount.


Well then look in the mirror. You get out what you put in. That has always been true, and it always will be true.