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Electromagnetic Radiation Propagation and Efficiency

  1. May 4, 2008 #1

    uby

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    Howdy,

    I'm no physicist, but I am a scientist by trade and some basic physics questions always bug me because I don't have any bases of understanding of the concepts. I apologize if these questions seem a bit abstract:

    Suppose an event occurs that causes the emission of EM radiation (for example, a transition in an electron's energy state that causes a photon to be released to satisfy conservation of energy).

    1) What is the driving force for the propagation of this energy? Is it simply by conservation of momentum that it will continue to propagate even in the absence of a medium? If you, by some convention, assign a momentum to any EM radiation to explain its propagation -- then don't you also have to say that work must be done to move the radiation through any distance in space?

    2) Related to the above scenario, when EM radiation propagates through space, is there any efficiency loss as the energy travels? If you assume that it performs no work on its surroundings, then I would also guess you'd have to conclude that it is perfectly efficient in that no loss of energy occurs no matter what distance the radiation travels. This seems like quite an odd thing to me!
     
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  3. May 4, 2008 #2

    uby

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    PS - although bringing this up now is a bit premature ... the reason I ask this philosophical question is because of how I view the propagation of gravimetric radiation. that is, gravimetric radiation "propagates" not because of conservation of momentum but because of curvature of space. I would like to think, if only for reasons of symmetry and beauty, that EM radiation "propagation" could be formulated in a similar fashion -- rather than assigning momentum to the radiation, but by describing its effect on space. although i've never come across anything to suggest this to be the case.
     
  4. May 4, 2008 #3

    cepheid

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    The simple answer is that no driving force is required. EM radiation simply propagates at speed c. It's difficult if you don't understand vector calculus or partial differential equations, to get into the details of electromagnetic theory. However, it suffices to say that one of the results of electromagnetic theory is that the constituent components of EM waves, namely electric and magnetic fields are governed by a "wave equation", the same "wave equation" that describes the oscillation of a medium in which there is a mechanical wave. The difference is that there is no medium. It is the fields themselves that are oscillating. In fact, a time-varying electric field produces a time-varying magnetic field which in turn produces a time-varying electric field and so on...as a result the fields are self-sustaining and can continue to propagate independently of the source that produced them. Of course, something has to get the fields propagating in the first place. This requires accelerating charge. Unmoving charge produces a *static* (not varying with time) electric field, and steadily moving charge produces a static magnetic field. As a result, there will be no fields that are changing with time, and therefore no electromagnetic wave will be radiated. If the charge is accelerating, however, then EM radiation will be produced. One way to do this is to have an oscillating electric dipole.


    While it is true that electromagnetic waves do carry energy and momentum, these characteristics are not arbitrarily assigned to them. (You can't do that. Something either has momentum or it does not). On the contrary, it is a fundamental *property* of EM radiation that comes out of electromagnetic theory. Conservation of momentum is always true, but it has nothing do to with whether or not EM radiation is *allowed* to propagate (again, it propagates of its own accord, as a fundamental consequence of electromagnetic theory). Similarly, the presence or absence of a medium does not affect whether light can or cannot propagate, since EM waves are not "oscillations within a medium." However, a medium does affect the speed at which EM waves propagate, since light can interact with matter.


    First of all, EM radiation is not matter. It doesn't have mass. So talking about "doing work" to get it moving is not really correct. If we can segue awkwardly into the quantum description of light for a brief moment, the so called particles of light (known as photons) are massless. It is a consequence of the special theory of relativity that massless particles must move at speed c. *Always.* They can never be slowed or stopped (unless they are destroyed). So that's another way of looking at it. From the "photon" perspective, light must always be on the move, because its individual constituents are massless particles.

    This is going to be a brief digression from the topic of light, but I wanted to point out a possible physics misconception in your above quoted statement too. Even if we WERE dealing with matter, something that has mass (such as a spacecraft), I should point out that Newton's Second Law (the conservation of momentum you're so fond of referring to), says that an object in motion will continue in motion in a straight line at a constant velocity in the absence of any external forces. So, assuming the spacecraft was moving to begin with, then NO additional work would be required to keep it moving.

    If it performs no work on its surroundings, then yes, there is no energy lost. However, in real life we don't have a perfect vacuum. There is dust etc in the interstellar medium. Light is capable of interacting with matter. Light can be absorbed by dust etc. Exactly what happens to EM waves when they encounter matter is complicated and I won't get into it here. Needless to say, the EM fields can "do work" on the atoms in the medium.
     
    Last edited: May 4, 2008
  5. May 5, 2008 #4

    Andy Resnick

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    Interesting questions, to be sure. Yes, radiation does carry energy and momentum. But, when light travels through a *lossless* medium, there is no dissipative process to decrease the energy (intensity) or change the momentum (wavevector). When losses exist, some energy is lost from the radiation as transferred to the medium (i.e. absorption) or lost in other ways (incoherent scattering).

    Radiometry is the thermodynamic basis for electromagnetic waves. Radiometry makes some very odd predictions, one of which is that the spectrum of highly coherent light will change during propagation. I don't fully understand why this is so, but the results are well-documented.
     
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