Light Energy Gain???

Discussion in 'Physics & Math' started by dav57, Nov 3, 2004.

  1. dav57 Extraordinary Thinker Thingy Registered Senior Member

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    As you (the observer) approach a beam of light, the faster you go, the greater the measured frequency of the beam of light goes up.

    Now, relativity states that the speed of light relative to the observer is invariant. But if this is the case then ponder the following:

    If you capture the energy of the beam as you travel towards it, you ought to be able to “extract” energy from it. Lets say that over a given amount of time you extract an amount of energy equal to 1 joule.

    If you then accelerate to a much higher speed (and not forgetting that the relativists will still claim that the measured velocity relative to the beam is unchanged) you will be able to collect more energy in the same time (relative to the observer).

    What is going on here? If we can extract more energy then where is it coming from? Surely it’s because the relative speed of the beam with respect to the observer has changed?
     
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  3. thed IT Gopher Registered Senior Member

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    Here's a thought, if you two observors at two different speeds then they measure two seperate blue shifts (frequencies). Can it be that the light wave actually has two seperate speeds? What of 100 observors or even a 1000?

    May be the speed of the observor has a lot to do with it?
     
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  5. geodesic "The truth shall make ye fret" Registered Senior Member

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    Any attempt to interact with the light beam and remove energy from it will slow down the observer removing energy. Thus the decrease in frequency is due not to the loss of energy from the photon, but the loss of energy of the observer.
    As an analogy, imagine holding a (small) wind turbine. If you are stationary, thenthe wind makes the turbine rotate, but if you run into the wind, the turbine rotates faster due to the energy you are putting in.
     
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  7. 2inquisitive The Devil is in the details Registered Senior Member

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    Nice analogy with the wind turbine, geodesic, but it does not correlate with an
    observer in an inertial frame in space. In the inertial frame, the observer considers
    himself at rest and requires no energy to remain so. Do you state the observer in
    the inertial frame is required to have a 'memory' of the non-inertial (accelerating)
    frame to extract the energy from?
     
  8. geistkiesel Valued Senior Member

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    You are correct. Look at it with yourself at rest and the different peaks of the energy cross the plane of your eye as you count the peak to peak of the wavelength. By then moving toward the beam the frequecny increases also, but the wave length does not increase. The "standard model' of Doppler mesurements, for instance, tells us that if the source of he emitted radiation is moving in the direction of the radiation (light) that the source "compresses" the beam by virtue of the motion of the source and the light.

    This is impossible. If the source compresses the wavelengths then a force must necessarily be applied to the light. In other words the wavelength will not shorten and still put up no resistance to the compression by the moving source. The matter comprising the wavelength is not a form of "superconductivity" that relinquishes its space without resistance.

    So if the source compresses the wavelength then the matter of the beam within the wavelength must feel acceleration, that is a force. It is as if the sinusoidal string of the beam got an inertial nudge from the source (which is necessarily continuous as long as the radiation continues.

    We may see the preceding as a manifestation of Newton's law of inertia, that an object at rest or moving uniformly in a straightline wil continue in that state of motion until acted on by an outside force.

    If the matter within the wavelength feels the force then the "light" feels the force (they are the same), which would impose a component of the source's motion onto the speed of the light, which by experiment has been found not to occur. Remember, the speed of light is cnstant at C.

    Likewise, the standard model of Doppler radiation tells us that a source moving away from the light will be stretched out byapplying a stretching force, which extends the wavelength of the beam, which should slow the speed of the light arriving at the plane of your eye.

    Try some simple calculations. A string of boxcars move past you at 320 km/hr = 360/3600 = 1/10 km/sec = 100 meters/sec. You time how long it takes one boxcar to pas which you calculate as .05 sec. = .1 x .05 = .005 km = 5 meter. The frequency of boxcars passing is 1/.05 = 20/sec.

    Now you begin to move in the opposite direction of the train at 1/10 the speed of the train or .1 x .t = .01 km/sec = 10 meters/ sec. Now you measure the time it takes for one boxcar to pass while you are in motion which would be 5 meters/ 110 meters/sec = 5/110 = .045 sec, for a measured frequency of 22/sec. Multiplying the frequency x the wavelength = 22 x 5 = 110, the correct total relative velocity between observer and the boxcars.



    If we assume the speed of the train has nothing to do with the motion of the observer and we use the measured frequency of 22, then the calculated wavelength is 100/22 = 4.55 meters, which is a wavelength shorter than the known wavelength that was measured and knows as 5 meters. Calculating he contribution of he moving observer we see that 10/22 = .45. This added to the calculated wavelength that ignores the observer motion is 4.55 + .45 = 5 the correct wavelength.

    Guess what happens when the calculations are performed using relativity theory? The wavelength of light is shortened (compressed) to conform to the error of ignoring the observer’s motion. The "compression" of the wavelength silliness is inserted to justify relativity theory.

    Another consideration: A source moving at 360 m/sec is supposed to compress a photon moving at 300000 km/sec or 299640 km/sec faster than the source speed, or proportionaltely nearly 100 times faster.

    Would the relativity folks ignore the motion of the observer when calculating the parameters of the boxcar scenario and still have a job afterwards?

    Read Grounded's post [post=607276]here[/post] for a very clear and cogent picture.

    Your observation about the energy in the beam was cogent. All one would have to do is put a flashlight on a linear osicillator and accelrate in the direction of the light, return to the starting point and accelerate again. Each time take a little of the energy gain and charge the battery and fuel the oscillator engine and use the rest to run your electric car, forever.
     
  9. dav57 Extraordinary Thinker Thingy Registered Senior Member

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    Thanks for the input geistkiesel. And I had read Grounded's post previously anyway.

    I'd still like to hear what the relativists say about the extra energy gain from the same beam of light when the observer is clearly in an inertial frame before and after the period of acceleration.

    How and where and why can we get more energy from the same beam of light if there is no difference in relative velocities?
     
  10. thed IT Gopher Registered Senior Member

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    I think you are missing the fundamental point of Relativity, it allows disparate (apparently paradoxical views) to be resolved because things are in relative motion.

    In this case the observor starts in some inertial frame , accelerates and ends in an intertial. The observor can tell the end and start frames are different. You would take your frame of reference from the starting position, whatever it was. From this you can deduce the energy gain in the light beam is due to your motion.

    The point I was trying to make people think about, above, was this. You can take 2 observors in two different inertial frames. They will measure two different blue shifts (or even a red and blue shift) and hence calculate two differing energies for that light beam. If they communicate they see a paradox. All Relativity is doing is allowing the observors to find a way of resolving the apparent disparate measurements. They can deduce there is relative motion between themselves. The obvious point being that inertial frames are not identical even if you can treat them that way.
     
  11. dav57 Extraordinary Thinker Thingy Registered Senior Member

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    Yes, I understand. But the point I'm trying to make is that a SINGLE observer has gained energy from the same beam of light due only to the fact that his relative velocity to the beam has altered (hence, if extra energy has been gained from the same light beam, then the speed of the observer relative to the beam is the culprit for energy gain) What else, apart from a change in velocity can account for extra energy gained? You need to forget about two observers - I'm only talking about ONE beam of light and ONE observer and the fact that if the relative velocity is always the same then why does the observer witness 1. A different frequency, 2. A different wavelength, and 3. The ability to collect more energy in the same given amount of time.

    Hmmm, seems strange to me

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  12. thed IT Gopher Registered Senior Member

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    As I read the OP you are talking about one observor at two velocities. If they stay at costant velocity then there is no change in the doppler shift of the observed light beam. As you introduced a higher velocity they can deduce this from the doppler shift.
     
  13. dav57 Extraordinary Thinker Thingy Registered Senior Member

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    And.....
     
  14. thed IT Gopher Registered Senior Member

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    1,105
    And what ...

    You are saying that there is an anomalous energy gain by increasing velocity (accelerating) then saying there is no acceleration.

    Actually, Relativity does not say the measured relative velocity is unchanged, if you accelerate. Then you totally contradict yourself by,

    Either you can't understand what invariant means or you can't understand yourself. Which is it?
     
  15. dav57 Extraordinary Thinker Thingy Registered Senior Member

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    Don't worry, Thed, you're right, I dodn't understand myself sometimes (well, most the time actually

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    )

    It's often quite difficult to be explicit and concise when you are trying to explain something quickly - so forgive me if I've confused you.

    My example is really very simple and involves a rocket extracting some energy from a star it is travelling towards at a constant velocity and over a given period. The next thing that happens is that the crew accelerate the rocket to a very fast speed and then cruise at a higher velocity relative to the star. The crew can then collect more energy from a beam of light (the same beam) than before. And I was wondering how this happened if the relative velocity between the beam of light and rocket remained the same according to relativity. Where does the energy come from?

    Not sure if that's any clearer, Thed.

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  16. thed IT Gopher Registered Senior Member

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    Not a problem. I was half expecting a roasting in response. Thank you.

    But that is the thing, the relative velocity at start and end, with respect to the light, is different and you can tell the difference. This is why I brought up invariance as well. What SR says is that the speed of light is always measured the same (in inertial frames etc.) no matter what speed you are travelling at. It does not say that you can not tell your difference in speed with respect to light. I.E. c-0.999c != c-0.001c.

    What pre-Relativistic theories says is that the measured speed of light would be your speed plus/minus the speed of light, depending on direction. This should have a serious effect on things like spectroscopy and other disciplines using light as a measure.

    Did you, or anyone else for that matter, know that our velocity through space causes a red and blue shift in the cosmic microwave background.
     
  17. Pete It's not rocket surgery Registered Senior Member

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    Hi dav57,
    Relativistically, kinetic energy isn't related to velocity in a simple way like it is at low velocities.
    In the relativistic equation for kinetic energy, a velocity of light-speed implies infinite kinetic energy, which poses a problem when you're working out how much energy you can collect from a beam of light.
    Fortunately, light also has zero mass. But this introduces a problem of its own:
    Zero mass implies zero kinetic energy.
    Light-speed implies infinite kinetic energy.
    So, what is the energy that can be collected from a light beam? Zero? Infinity? Zero times Infinity?

    The end result is that when dealing with light, the relative velocity of the energy collector and the light beam isn't useful - it does not allow us to predict the energy of the beam.

    Looking at it another way, it also means that if something without mass (like a light beam) imparts energy, then its velocity must be exactly light-speed. No matter what the energy content is, its velocity can't change. This means that if something happens to change the energy of the beam, we should not expect the relative velocity to change.


    So, obvious question - What does change the energy of the beam? How can we predict its energy if the velocity is useless?

    The answer is the freqency of the beam. Higher frequency = Higher energy.


    Next question - how does the frequency change if the relative velocity is the same?

    Consider a short radio pulse, fired from the source for 1 second at a frequency of 1MHz. There are 1 million waves in the pulse.
    Now, you're approaching the source at high speed. It takes you only 1 millisecond to traverse the pulse. You've passed 1 million waves in one millisecond, so in your frame the frequency is 1GHz, right?
    Now, the only way that the velocity of the beam can be unchanged is if the wavelength has been reduced... which happens due to length contraction (in an armwaving way. I think you need the math formalism to explain it accurately).

    If the beam has higher frequency, then it must have higher energy. And since the beam has zero mass, it must still have the same velocity.
     
  18. dav57 Extraordinary Thinker Thingy Registered Senior Member

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    Thanks, Pete - it seems that relativity has an answer for everything

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    I hope my questions aren't too stupid but I still find this whole concept rather puzzling.

    Your answer, by the way, satisfies me to the point whereby you say "Now, the only way that the velocity of the beam can be unchanged is if the wavelength has been reduced..." and this is where I debated this with James R but never felt satisfied with the answers.

    I was saying that the wavelength *APPEARS* to change due to the fact that we are measuring the wavelengths of light WITH light itself, which I believe could introduce strange effects during measurements and observations.

    Anyway, these debates are all tied in with one another and I'm still having some difficulties.

    I need to look at this length contraction thing in more detail to see how and why it was derived......
     
  19. geistkiesel Valued Senior Member

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    Thed,
    I understand the postulate that emitted in the direction of a moving frame of reference does not retain any component of the frame velocity. Is this postulate invariant under all possible physical conditions, as far as you know?

    What is the difference between this statement (below) and the postulate that the relative speed of light will always be measured as C? :

    It does not say that you can not tell your difference in speed with respect to light. I.E. c-0.999c != c-0.001c.
     
  20. thed IT Gopher Registered Senior Member

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    In essence yes. The case where it is not true is where Special Relativity does not hold, in the vicinity of large gravitational fields for example. Even then it is true very locally.

    Invariance basically means that light speed, in vacuo, is a yard stick you can use to measure against. That is, you can measure speed relative to light speed. It's why particle/cosmological speeds are often quoted as a fraction of c, rather than km/s relative to the lab frame.
     
  21. geistkiesel Valued Senior Member

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    So if I am on a space ship zooming along with some velocity v and I determine a light beam alongside my ship and I want to determine how much faster the light is going with respect to my ship (in the same context as a I want to determione how much faster a 60 mph car passing me moving 50mph would be. [here 10 mph], you are saying I can make a measurement from my ship frame of reference and come up with a number, say .5c if this was my measured velocity with respec to Ve, the embankment when leaving my home planet?
     

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