All Photons Move at 300,000km/s.... But Don't?

Discussion in 'Pseudoscience' started by TruthSeeker, Jun 12, 2015.

  1. RajeshTrivedi Valued Senior Member

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    You just missed that, in the previous statement, i have referred to this momentum aspect formula (h/lambda) while talking about radiation pressure. I have not seen any reference wherein momentum of photon calculations play any role in ejection of electron..Not to my knowledge, may be you can give some. The color highlighted statement of yours afaik plays no role in electron ejection. Remember one thing Radiation Pressure can be explained classically (by Poynting Vector, Area and Absorption of Energy), and when we talk in QM the cumulative effect of momentum of all the photons do the job, but PEE cannot be explained classically and neither by cumulative effect of photon's momentum. The basic requirement is photon energy > certain minimum value [we don't say, momentum > some value]. So I do not think your reference to momentum in colored line is proper, still I am open you can provide any reference.

    Moreover by definition photon is quantum of energy, we don't say photon is quantum of momentum.
     
    Last edited: Jun 26, 2015
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  3. exchemist Valued Senior Member

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    Well you can do actually. Anyone familiar with QM will be fully aware that frequency and momentum go hand in hand, related by a simple proportionality factor. Hence the explanation of the uncertainty principle in terms of superposition of waveforms, for example.

    As for references here is de Broglie's relation: https://en.wikipedia.org/wiki/Matter_wave
     
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  5. RajeshTrivedi Valued Senior Member

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    ...

    I expected a better and thought out response from you. You appear to drag a point, which is not relevant.

    .....Pl give a link which associates momentum with Photo Electric Effect.

    PS: You are going ahead with a presumption that since we are talking about energy, wave length, wave-particle duality wherein momentum also pops in with some factor, so momentum must be playing some role in Photo Electric Effect. This is fallacy. Yes, momentum, light speed, energy, wave length and frequency of a photon are mathematically related, beyond that momentum cannot be linked with the effect per say.....
     
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  7. origin Heading towards oblivion Valued Senior Member

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    Right you are exchemist.
    The photoelectric effect indicates that the kinetic energy of the photoelectric electron is:
    \(K = hf - \phi\)

    So both the photoelectric effect and the momentum of the incident photon are directly proportional to the photons frequency.
     
  8. paddoboy Valued Senior Member

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    I actually had to look twice at your post Rajesh...I mean isn't it you that has been telling the forum one should not rely on references and links? Well telling me anyway.

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    Yet here you ask exchemist for references? While I have been asking you for references of your claims for ages to no avail.
    And of course you also got some WIKI info at your post 446 in regards to the Viking missions that may have been out by 15,000kms if allowances had not been made for the momentum of light/radiative pressure.
    Strange form of hypocrisy?

    And of course as others are telling you, momentum is most certainly an outcome and related to the photoelectric effect.
     
    Last edited: Jun 26, 2015
  9. paddoboy Valued Senior Member

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    27,543
    On the photon momentum/photoelectric effect, again exchemist has put it quite admirably...far better than I could.
    I would add though that photon momentum is obviously the momentum of the photon and just as obviously is related to the energy of the photon.
    EMR carries momentum and that's why we get radiation pressure as Rajesh raised in relation to solar panels/and sails and the ridiculous claim by jcc that photon momentum was non existent...or words to that effect.
    The photoelectric effect although primarily dependant on the frequency, would not exist if photons did not have momentum.
    And of course the photoelectric effect observations was the cause for the development of the particle model of light.
     
  10. RajeshTrivedi Valued Senior Member

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    Photo Electric Effect is a phenomenon, not any Physical parameter or quantity, so the question of it being directly proportional to photon frequency does not arise. The one of the associated parameters is the current (Amp), which is not linked to the momentum of incoming photon, the way you are suggesting.
     
  11. RajeshTrivedi Valued Senior Member

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    1. The link was given for jcc to start with, as such I do not trust the wiki beyond certain basics.

    The problem here is not that of any alternative theory, the point is that we are differing in the interpretation of a well established theory, so one side has to be at fault. It could be me or it could be all three of you..

    1. First and foremost the quantization of light was proposed earlier and used successfully by Max Planck in explaining the Black body radiation. Yes, Einstein explanation of Photo Electric Effect based on this pushed it further.

    2. The details about Photo Electric effects are available on the net and almost in all the high school books, formula for energy and momentum is given by Rpenner also in the starting part of this thread, so I will not repeat and I am not comfortable with typing formula as well.

    3. Photo Electric Effect also has a very peculiar issue associated with it, photon has to give up its entire energy, part of this energy is used to overcome the binding energy of the surface electron (which is dependent on the target material, low for Alkali Metals) [which is also the threshold or minimum energy required] and the left out is the Kinetic Energy of the Electron. If the Energy of the photon is less than the binding energy of the electron, no Photo Electric Effect.

    4. Now if we say that Photo Electric Effect is directly proportional to the momentum of incoming photon, then we must be able to explain the same effect, with the help of momentum. Try this.......You get no where.


    PS: i repeat just because momentum is associated with other parameters of photon (as per formula given by Rpenner), it does not mean, that it causes Photo Electric Effect.
     
  12. Dr_Toad It's green! Valued Senior Member

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    If y'all hate each other this much, start a fucking fan page on farcebook. Leave this place open to discussion instead of personal shit slinging. Damn, it gets tiresome reading past you two.
     
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  13. paddoboy Valued Senior Member

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    The photoelectric effect is an effect observed that gives rise to the model of light as a particle[photon, and asscoiated with frequency, energy and momentum.
    Again if light/photons had no momentum, we would not have the photoelectric effect or even light/photons
    That of course being only anything that appears to support your view?

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    You need to get used to the fact, that although the Internet is a source of plenty of rubbish, it is also the source of plenty of proper reliable acceptable knowledge of all kinds.
    Just as you yourself extracted knowledge from WIKI re allowances for radiative pressures on Viking of 15,000 miles.
    To reject links and references because in general they do not align with your interpretation is less than scientific and down right illogical.
    I see it as an issue of unecessary pedant. Photons by the very definition need to have momentum...no momentum, no photoelectric effect, no light'photons.
    Planck hypothesised that possibility from memory, Einstein via the scientific method developed it from the photoelectric effect observations.
    And frequency and all are related in some way.
    If the photon/light had no momentum, we would not have the photoelectric effect



    Momentum is a necessary property of light, as is frequency and energy.
    Without any one of them the photoelectric effect does not occur.
     
  14. paddoboy Valued Senior Member

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    http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.113.263005
    Photon Momentum Sharing between an Electron and an Ion in Photoionization: From One-Photon (Photoelectric Effect) to Multiphoton Absorption:

    ABSTRACT
    We investigate photon-momentum sharing between an electron and an ion following different photoionization regimes. We find very different partitioning of the photon momentum in one-photon ionization (the photoelectric effect) as compared to multiphoton processes. In the photoelectric effect, the electron acquires a momentum that is much greater than the single photon momentum ℏω/c [up to (8/5) ℏω/c] whereas in the strong-field ionization regime, the photoelectron only acquires the momentum corresponding to the photons absorbed above the field-free ionization threshold plus a momentum corresponding to a fraction (3/10) of the ionization potential Ip. In both cases, due to the smallness of the electron-ion mass ratio, the ion takes nearly the entire momentum of all absorbed N photons (via the electron-ion center of mass). Additionally, the ion takes, as a recoil, the photoelectron momentum resulting from mutual electron-ion interaction in the electromagnetic field. Consequently, the momentum partitioning of the photofragments is very different in both regimes. This suggests that there is a rich, unexplored physics to be studied between these two limits which can be generated with current ultrafast laser technology.

    • Received 28 July 2014
     
  15. paddoboy Valued Senior Member

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    27,543
    I think this is also relevant:
    http://van.physics.illinois.edu/qa/listing.php?id=29757
    Q & A: How does light have momentum without mass?
    Most recent answer: 05/06/2015
    Q:
    I read your statement about how light has momentum despite the fact that it has no mass. My question to you is regarding gravity in black holes. It is said that light can�t escape the enormous gravitational force in black holes; however, is it not true that gravity is directly proportional to the object�s MASS and inversely proportional to the distance between the two objects (Newtonian, I think). If so, light has no mass. So how would light be effected by this phenomenon??? Thanks for your enthusiasm in physics. Dan Sweeney
    - Dan Sweeney (age 16)
    Thayer Academy, Braintree MA, USA
    A:
    The use of words can make a lot of confusion. Unfortunately, the word "mass" has been used in two different ways in physics. One was the way Einstein used it in E=mc2, where mass is really just the same thing as energy (E) but measured in different units. This is the same "m" that you multiply velocity by to find momentum (p), and thus is sometimes called the inertial mass. It's also the mass that provides the source of gravitational effects. Light has this "m" because it has energy. So it is indeed affected by gravity- not just in black holes but in all sorts of less extreme situations too. In fact, the first important confirmation of General Relativity came in 1919, when it was found that light from stars bends as it goes by the Sun.

    The other way "mass" is often used, especially in recent years, is to mean "rest mass" or "invariant mass", which is sqrt(E2-p2*c2)/c2. This is invariant because it doesn't change when you describe an object at rest or from the point of view of someone who says it's moving. Obviously that's a good type of "mass" to give when you want to make a list of masses of particles. For a light beam traveling in a single direction, E=pc, so this "m" is zero. There is no point of view from which the light is standing still!

    However, once you consider light traveling in a variety of directions, the E's from the different parts just add up to give the total E but the vector p's don't. In fact the total p can be zero if there are beams traveling opposite ways. So for many purposes the older definition of m (the inertial mass) is more convenient than the invariant particle mass, since it's the inertial mass that's just the sum of the inertial masses of the parts. For light moving equally in all directions, like the light bouncing around inside a star, total p is zero, so both definitions just give m=E/c2.

    Mike W.
     
  16. exchemist Valued Senior Member

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    12,451
    However, I have now learnt what I wanted to know about Rajesh Trivedi………….

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    But don't worry, I'm now out of this, like the rest of us.
     
  17. arfa brane call me arf Valued Senior Member

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    Good grief (to put it mildly). Einstein "noticed" that electrons are emitted from the surface of a metal when light impinges on it. He found that the electron current was not related to the intensity, but to the frequency.

    Any physics student who learns about the photoelectric effect knows the cutoff parameter is frequency. Who is this Rajesh dude?

    As for banging on about momentum; a photon's energy is proportional to its frequency (this is 1st year physics, or even high school), so its momentum is too.
     
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  18. RajeshTrivedi Valued Senior Member

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    Ex Chemist....

    False and inaccurate statement. I am sorry you have miserably let the forum down after claiming yourself as Quantum Chemistry guy, you are out of your depths..
    On the context, you supported 'Origin' statement that Photo Electric effect is proportional to momentum of photon, another inaccurate statement..

    No, Ex Chemist is wrong ! And sorry your statement that Photo Electric Effect is directly proportional to photon frequency makes sense, but non scientific loose statement, the other statement that momentum is proportional to frequency is inaccurate statement.

    Great !! Generally I oppose your copy paste and links. but this one is an abstract of a scientific paper and a great one. This shows the kind of efforts you have put in, in classical literature you do not get terms like Photo Electric effect and momentum of photon together in the same para... Interesting paper and will be useful to all those readers who likes to get into intricacies. I really appreciate your enthusiasm and spirit.

    Without taking away any credit from you, I wish to state that this paper talks of (one of the aspect) about Conservation of Momentum during the Photo Electric Effect process. The Conservation of momentum in absence of external force is a basic Physics concept and this papers attempts to resolve that by linking remaining momentum with ion which is still bonded with the target metal. The paper does not prove that Photo Electric effect is related to momentum of photon, leave aside directly proportional to momentum. Nonetheless great effort from you. It s like this, Paddoboy, angular momentum needs to be conserved during BH formation by core collapse, but we do not say that Black hole Formation process is directly proportional to angular momentum or such things.



    Arf, you were doing great with those matrices, until you succumb to this. Yes, right, photon's energy is directly proportional to frequency under all circumstances, but can we say that momentum is also directly proportional to frequency under all circumstances...I am afraid, no. You appear to be a maths guy, Ex. Chemist supporting this inaccuracy probably motivated you.

    You are a nice jovial fellow. Yes personal fights are farcical, but inaccurate science by a self claimed experts in a science forum is blasphemy. I do not tolerate such farce.


    PS : I had undertaken that if I am proved wrong in any of these 3 responses (Q1,Q2,Q3 with Paddoboy), I will apologise to entire forum in particular to Paddoboy and walk out. I stand by that. Q3 is settled, Q2 is almost at the verge and will be settled by my next post where it will be proved that Photo Electric Effect has no dependence on Momentum of Photon as being stated by Ex.chemist, origin etc. (Hint : Try looking at Energy and momentum equations in a medium, Photon Momentum is always h/lambda, but it is hf/c only in vacuum. It is always correct to say that photon momentum is inversely proportional to wavelength, but it is not always correct to say that it is proportional to frequency.)
     
    Last edited: Jun 29, 2015
  19. paddoboy Valued Senior Member

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    If your ego blinds you to the fact that [1] I do not really care what you claim or think, and [2] that you are totally delusional, there's not much anyone here can do or say.
    That's sad.

    Ignoring the rest of this post and the usual farcical bullshit post, and chest puffing, the above really had me in stitches! I mean talk about pot calling kettle black!

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    Amazing how any one person can be so delusional and still keep a straight face.

    Considering that Q3 was not settled in the way that you imagined, but was just a usual pedantic farce with yourself, butting in on a total lame brain like jcc and what he said, and considering that you are also wrong with your photoelectric mission, and considering that you have been proven wrong in at least 8 or 9 other threads on virtually every issue you have raised, including your demolished paper, and considering you have yet to retract or admit defeat in your continuious game on this forum, and considering your handling of the truth in as many threads, I suggest the forum can take your dramatic emotional promises along with your equally dramatic claims with a grain of salt.
    tip: I don't believe you are impressing anyone Rajesh.
     
  20. paddoboy Valued Senior Member

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    Maybe you need to try and emulate that effort, because so far you are still behind the eight ball. Oh, and really, why drag BHs into this again? The delusions re BHs that you have expressed on this forum so far, makes it obvious you know very little on that area of cosmology.

    PS:
    I wouldn't really bank on too many replies Rajesh as at least two or three others, now have you on ignore for obvious reasons.
     
  21. paddoboy Valued Senior Member

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    That's all you need to remember Rajesh, plus the fact that if light had no momentum [which is a fallacy in itself, as any massless particle, must always travel at "c" by definition] than there would not be any photoelectric effect period.
     
  22. exchemist Valued Senior Member

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    Thanks Rajesh. I've got your number now, which was the object of the exercise as far as I'm concerned. Did you read the link, I wonder. Anyway, never mind. Have a nice day.
     
  23. origin Heading towards oblivion Valued Senior Member

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    11,888
    Really?
    The equation for the photoelectric effect: \(K=hf - \phi\)
    You are really trying to say that it is a nonscientific loose statement to say K is proportional to f? That is actually a completely accurate mathematical statement

    You get so high and mighty that you start embarrassing yourself.
     

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