1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

If photons have no charge then how

  1. Apr 14, 2015 #1
    do they deflect / accelerate electrons in an antenna?

    For an electron to accelerate it has to experience an electric field or changing magnetic field.

    What I can tell from the standard model a photon carries no charge = no electric field.

    Further a classic EM wave is defined by an E vector giving the magnitude of the E field, EM waves are photons which have no charge therefore no E field - what gives?

    Obviously EM waves can accelerate electrons in antennas, I used a WiFi link to post this.

    How have I messed these ideas up.

    Apologies if this has been answered before or if it is a dumb question.
     
  2. jcsd
  3. Apr 14, 2015 #2

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Zero charge merely means that photons are not affected by external electric fields.
     
  4. Apr 14, 2015 #3
    1. then how is an E vector defined in an EM wave.

    2. what other forces act on an electron beside an E, M and EM force

    3. what force type is the photon imparting to the electron's mass to make it accelerate.
     
  5. Apr 14, 2015 #4
    Hello Houlahound,

    Introduction:
    A photon is a "packet" or quanta of electromagnetic radiation. It is the particle-like description of the electromagnetic wave.

    Analogy or Example:
    Imagine a calm body of water with a first beach-ball resting on it.
    Now imagine dropping (accelerating) a second beach-ball on (into) the water far away from the first beach-ball. The second beach-ball decelerates. Water-waves will ripple (or radiate) away from the second beach-ball and travel across the water's surface. The water is no longer calm, but is now disturbed!
    Now remember the first beach-ball (which was resting at some distance away from where the second beach-ball was dropped)?
    It can be seen that once the waves made by the second beach-ball ripple (or radiate) towards and reach the first beach-ball, the first beach-ball will start to move up and down (accelerate). Note that the water between the two beach-balls does not have the same property as the beach-balls.

    Experimental proof:
    Try this experiment in your kitchen sink with two smaller buoyant objects rather than using the beach-balls :p

    Summary:
    Charged bodies are like the beach-balls or buoyant objects. Space (or space-time, or the vacuum medium) is like the body of water (also a medium). When a (second) charged body is accelerated (it changes its velocity - it changes its speed or direction) it disturbs the medium of space with electromagnetic wave (or photons). Electromagnetic waves travel through space and this means that electric and magnetic fields "ripple" through space.
    Eventually, these ripples reach a (first, undisturbed and resting) charged particle and the wave's electric field applies a force on the (first) charged particle which was initially resting in space. The force accelerates the (first) charged particle.

    Conclusion:
    When one electron (charged particle) is accelerated, it sends a messenger (UNCHARGED electromagnetic wave) that travels through space to accelerate another electron (charged particle). Just like a client sending messages to his lawyer by means of a postal-worker; the postal worker carries the information between the client and lawyer but has NO authority (is NOT CHARGED) to act AS the client and have the lawyer work on his case.

    I hope this helps,

    Dr QUAW
     
  6. Apr 14, 2015 #5
    Your analogy works for me, thanks, so it's not right to think of the photon as a thing in itself, better to think of it as a message carrier but not the message ... sort of.

    Then defining an electric filed vector E for an EM wave which is just photons with E = 0 is very misleading and pedagogically unsound don't you think?
     
  7. Apr 14, 2015 #6
    Thanks Houlahound,

    Introduction:
    Yes, you will find that physicists often use different "models" to think about the Universe.
    Quantum Mechanics talks about "Wave-Particle Duality". Matter, (electron included) may behave like a particle or a wave.
    The photon IS a "thing". It has the following quantities: Momentum, Energy, Spin, Polarisation... but it is a "thing" born from the acceleration of charged particles and it must travel at the speed of light... so it appears to behave like a messenger or a consequence (It rarely stands around with nothing to do!).
    It MUST carry the message (Momentum, Energy, Spin, Polarisation) from the charged sender particle to the charged receiver particle but it does not have the "pen" (charge) to write a message of its own.

    Description:
    If you are talking about accelerating electrons, at a right-angle to the "flight path of the radiation beam, photon or wave" then the electric field vector and light's wavelike nature is what may be more important to you. This is what happens in antennas. The electrons "ride the electrical component of the electromagnetic wave" and stay in the antenna. Typically; radio-waves to infra-red wavelength radiation are implied here.
    Imagine a space with three axes within it: horizontal (x), vertical (y), and depth (z). Let's assume that a first electron floats in space and we move a second electron in the z-direction. We have accelerated the second electron and it will radiate electromagnetic waves in an ALMOST spherical volume. In fact, the volume will resemble an apple or doughnut with radiation weakest in the z-axis but strongest in the x-y plane or equator....

    {Experiment:
    Imagine pinching the z-axis or north and south poles of a sponge-ball; see the shape you'll make?}​

    ....Description Continued:
    The first electron (previously at rest) will also accelerate in the z-axis direction! This is because the electric field component of the radiation from the second electron is z-axis "polarised".
    Note that this is the due to the wave produced by the acceleration of a SINGLE electron (moving in a single direction). Most light sources are made up of numerous charged particles travelling in several random directions.
    Each photon (particle or "wavelet" or ripple) has an electric field component pointing in one particular direction (and moving at right angles to that direction!). However, an electromagnetic wave consists of SEVERAL photons, wavelets or ripples! So on average, the photons will have different electric field directions from one another and the wave (as a whole: A collection of photons and wavelets or ripples) will look like it has no NET or overall electric field direction. Its like throwing a hand-full of dice on the table and expecting them all to have "six" facing upwards. This is unlikely as the die are "unbiased" - just like the light is "unpolarised"
    When one looks through polarising film (sunshades) one notices that the light has been dimmed. This is because only oncoming light "vibrating in one direction" (electric field vibrating or oscillating in one direction) can pass through the sunshades. The other part of the light is blocked and that's how the shades keep the glare off ones eyes.

    ***Abstract & Further reading (Not immediately relevant, but just in case you wonder)
    If you are talking about knocking electrons OUT OF the antenna (ionising radiation), then the electrons may be scattered away from the antenna by visible light to gamma-ray wavelength radiation. The photoelectric effect and Compton scattering highlight photon energy and momentum transfer processes respectively (See first paragraph)****

    Don't hesitate to ask again if this doesn't suit!

    Best Regards

    Dr QUAW
     
  8. Apr 15, 2015 #7
    I think I got it, seems an unfortunate notation to label the amplitude of an EM wave with E the electric field with when it is not actually carrying a charge and therefore an electric field per se.

    thanks for help, I understand better now.
     
  9. Apr 15, 2015 #8

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    I don't think you do. There is a very real electric field associated with electromagnetic radiation. You can measure it with a meter.
     
  10. Apr 15, 2015 #9

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Hmm. How is it unfortunate? The model fits the facts very well. You can hardly do better than that. You can get an E field without actually having a nearby charge moving about and it would be really tiresome to use a model that demanded to relate the effect on an electron to whet happened to a charge that could, in fact, be millions of light years away (the original source of the em wave). I would say that the notion of a Field is a very useful one. The fact that the E must have been generated somewhere / somewhen is hardly relevant - and is independ of how the wave originated.
     
  11. Apr 15, 2015 #10

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    As sophiecentaur has asked, how is this unfortunate?

    I can have a waveguide with EM field in it, and there's no charge at all inside the waveguide!

    What is unfortunate is your mistaken a priori understanding that an E-field MUST contain charge. This is false. A charge may generate EM field, but it doesn't travel or accompany with that field! I can shake an electron up and down to generate EM field, and then I can stop shaking it. Yet, the pulse of EM field that I generated is still there traveling away to some distance (this is done in synchrotron light sources and RF antenna all over the world).

    You need to figure out where your misconception came from and where it actually fits into what we observe. Always do a "If this is true, then...." test.

    Zz.
     
  12. Apr 15, 2015 #11
    Hello Houlahound,

    All will become clear with Maxwell's equations. They are...

    (1) divergence of the electric field = div E = charge density divided by permittivity
    (2) divergence of the magnetic field = div H = zero
    (3) curl of the electric field = curl E = -dB/dt = negative rate of change of magnetic flux density with respect to time
    (4) curl of the magnetic field = curl H = J + dD/dt = conduction current density plus the displacement current (Displacement field D = E multiplied by permittivity)

    It seems that you are assuming that (1): The presence of the electric field in ANY region of space is due to a NEARBY charge. This is not true. Electromagnetic waves are generated by:

    A charge accelerating in position (a). According to equation (1) in reverse the stationary or moving charge always produces an electric field in the space around it which can be felt in position (b) also (remember this Houlahound. Static electricity can be felt at a distance; peoples hair stands on end before a thunderstorm!)

    The acceleration of the charge at position (a) will produce a magnetic field which can be felt at position (b) also (remember the Earth's magnetic core influences a magnetic compass?). This is equation (4) in reverse

    If the magnetic field just mentioned is time-varying (which it is, because when the charge was at rest there was no magnetic field, but when it moves, there is a magnetic field!). The time-varying magnetic flux density (B) gives rise to an electric field as shown in equation (3) in reverse.

    Equation (4) in reverse shows the time varying electric field just mentioned giving rise to a magnetic field (H) an so on and so forth...

    Summary:
    These changing charge speeds (acceleration) cause changing magnetic fields (H) which extend beyond position (a) and the changing H causes changing E. These changes radiate or "ripple" away from position (a) at the speed of light.

    Conclusion:
    E and H fields can be felt in position (b) which can be very distant from the accelerating charge at position (a).
    Although E field lines point towards (or away from) a charge (like pins in a pin-cushion), it doesn't mean that these fields cannot be felt at a distance from the charge (long pins can hurt you even if you're not touching the cushion!)
    Also, H field changing with time will produce an electric field or "electromotive force". A generator or dynamo works after all by moving a magnet (rotation) to induce an electric field and hence a current. The generator may not carry a NET or overall charge, but it can still produce an electric field as seen for equation (3). There are other ways to produce an electric field than using charges.

    For further help with future problems, mail my inbox and I can send you a presentation.

    Cheers

    Dr QUAW
     
  13. Apr 15, 2015 #12
    combined response to above responses, my misconception comes from Maxwell's equation: 1 above, and, the standard model that simply states that photons have zero charge.

    I think that is a pretty fair misconception to make given the info ie eqtn:1 + standard model requirement of photon.

    the distance thing is what I failed to get, which is not explicit anyway and I think from everyday experience most would conclude something light years away from a source is independent of it.

    at first blush to my literal mind these two things together combine to negate an E field on this independently existing real object called a photon that can be light years from it's source, in fact the source may have subsequently combined with an equal and opposite charge and electrically speaking field wise no longer exists, or it may have changed or ceased it's acceleration and is no longer associated with the EM wave it created.

    really what you guys are saying, (or how I interpret it) is that any/all EM wave/s altho free from a source antenna is/are actually still coupled to the source electron forever, that coupling may subsequently terminate on a receiving antenna and the photon now dies and loses it's history.

    is this respectable interpretation of what has been discussed?
     
  14. Apr 15, 2015 #13

    Nugatory

    User Avatar

    Staff: Mentor

    It is not :(

    There's really only one way to come to grips with this stuff, and that is to try to forget that you ever heard the word "photon" while you acquire a solid understanding of classical electrodynamics and can use Maxwell's equations to calculate the behavior of light waves.
     
  15. Apr 16, 2015 #14

    Drakkith

    User Avatar

    Staff: Mentor

    I agree with Nugatory. The concept of a photon does not replace the concept of an EM wave from classical physics, it enhances our understanding of it. This is similar to how learning that electrons behave in a wave-like fashion did not suddenly get rid of our understanding of an electron as a particle and make us start calling electrons 'electron waves'. I suggest learning how classical EM theory works first and then worrying about photons.
     
  16. Apr 16, 2015 #15

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    This is the same old problem. People feel the need to reconcile the QM and Classical approaches in all cases. There is just no point because they tread on each other's toes too much. (The approaches, not the people!!)
    A little learning is a dangerous thing, as they say, and School Science has a lot to answer for - as I have mentioned often.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook