Absorption of light energy

Discussion in 'Astronomy, Exobiology, & Cosmology' started by timojin, Mar 20, 2017.

  1. timojin Valued Senior Member

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    Light gets dissipated or refracted traveling in the space ?
     
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  3. DaveC426913 Valued Senior Member

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    Neither.

    Can you be a little more specific about the circumstances?
     
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  5. timojin Valued Senior Member

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    Let say a star is emitting light . The space between the emitter and the receiver say ( Hubble ) for the sake eliminating our atmosphere absorption . So the emitted ( white light ) scattered, attracted in its path ?
     
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  7. karenmansker HSIRI Registered Senior Member

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    Light loses intrinsic energy to the environment (even a vacuum) as it traverses distance. Velocity? NOT! . . . . Energy? . . . . YES! Resolves as attenuated-amplitude and lower-frequency (Red-shift look-alike?)
     
  8. DaveC426913 Valued Senior Member

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    No it doesn't.

    Cosmological red shift is not intrinsic loss.
     
  9. DaveC426913 Valued Senior Member

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    The intensity of light for a given area will fall off simply due to the geometry of radiating out from a point.

    A star radiates light in all directions. The intensity of light (imagine it as the number of photons) will fall off as the square of the distance.

    Let's take a tiny slice of the output of the star.

    Say, 1,000 photons were intercepted by the 1 metre-on-a-side square opening of your telescope, which is placed at 1 million km distance.

    If you moved your scope out to 2 million km (twice the distance), those same 1,000 photons would now fill an area 2 metres-on-a-side (four times the area).

    But since your scope has not changed size. it will only intercept 250 (one fourth) of those photons. (The photons have not lost any energy, there are simply fewer of them.)

    The intensity of light received at your telescope has been reduced for no other reason than the geometry of the area of an expanding cone.

    There is no intrinsic loss of intensity in a hard vacuum.
     
    Last edited: Mar 20, 2017
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  10. karenmansker HSIRI Registered Senior Member

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    Yes, it does. ANY 'shift' (either way) involves an energy change (increase or decrease).
     
  11. karenmansker HSIRI Registered Senior Member

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    Dave: Please define "hard" vacuum . . . . . free of mass? . . . . free of energy? . . . free of fields? . . . . . free of strings? . . . is there really such a thing . . . as a "hard" vacuum?
     
  12. DaveC426913 Valued Senior Member

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    That is not what the OP asked.
     
  13. DaveC426913 Valued Senior Member

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    Hard vacuum means no mass such as dust or particles.
    We can explore an area through which light may pass that is small enough that it encounters no atoms, if the volume is small enough. And in that volume, the light will not get lost.

    While light loss due to intervening gas and dust does happen, I want to separate that out from the OP's question about some sort of intrinsic loss.
     
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  14. timojin Valued Senior Member

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    There will be a shift in the red light because of the large wavelength ( lover frequency ) ?
     
  15. timojin Valued Senior Member

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    Forget, what was the opening ,I like to hear your answer karenmasker
     
  16. karenmansker HSIRI Registered Senior Member

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    timojin: YOU posted the opening post, reread . . . . Look at the info below (cut from internet). If, for example, you start with UV light travelling thru space and it is observed having 'red shifted' to IR (extreme example) , one will record a frequency (inverse wavelength) shift that equates to a net loss of energy. Though I won't do so here, one can estimate the quantitative energy differential using the E=hv relationship (note actual calculation is a little more involved, but this is an approximation). Others may argue.

    Useful Info:
    Ultraviolet
    (UV) light falls in the range of the EM spectrum between visible light and X-rays. It has frequencies of about 8 × 1014 to 3 × 1016 cycles per second, or hertz (Hz), and wavelengths of about 380 nanometers (1.5 × 10−5 inches) to about 10 nm (4 × 10−7 inches).
    Infrared radiation extends from the nominal red edge of the visible spectrum at 700 nanometers (nm) to 1 mm. This range of wavelengths corresponds to a frequency range of approximately 430 THz down to 300 GHz. Below infrared is the microwave portion of the electromagnetic spectrum.
     
  17. DaveC426913 Valued Senior Member

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    As you wish.
     
  18. karenmansker HSIRI Registered Senior Member

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    E=hv: This equation says that the energy of a particle of light (E), called a photon, is proportional to its frequency (v), by a constant factor (h). This means that photons with low frequencies, like radio waves, have lower energies than photons with high frequencies, like x-rays.

    Assume that E is conserved (constant) and h is constant. Then v must also be constant. If v decreases (i.e., wavelength increases via observed red shift, or other), then either E or h must vary. h is a universal constant (we think!), so (intrinsic) E must decrease. The degree of red shift observed basically reflects the distance light travelled and thus the E differential. If E did not change over distance, then no red shift would be observed, ergo, E, h, and v would remain the same.
     
    Last edited: Mar 21, 2017
  19. karenmansker HSIRI Registered Senior Member

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    . . . . .re: OP . . . . . refraction and dissipation of traveling light . . .

    Think also of gravitational lensing, gravitational red shift, etc. . . . . affecting photons over long distances . . .
     
  20. exchemist Valued Senior Member

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    I must admit this does not feel quite satisfactory to me, when we are considering explanations of red shift.

    If you observe light emitted from a body that is receding from you, the energy of the photons you measure is lower than what you would have measured if you had been moving with the emitter. My understanding has always been that this is linked to the frame-dependence of kinetic energy. Some of the energy of the photons, as measured by you, is effectively going into overcoming the velocity difference between you and the emitter. The velocity of the light is no different of course, but its energy will be a function of differences between reference frames.

    So it does not indicate any energy loss en route, as it were. The energy measured is simply different in the two cases, just as the kinetic energy in a cricket ball, thrown to you from a receding car, will be measured by you as less than by the thrower of the ball.

    In fact the idea that the observed astronomical red shift is due to energy loss during transit, the so-called "tired light" hypothesis, seems to be dismissed fairly authoritatively here, by a prof at UCLA: http://www.astro.ucla.edu/~wright/tiredlit.htm
     
    Last edited: Mar 21, 2017
  21. karenmansker HSIRI Registered Senior Member

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    'authoritatively'? . . . . IMO, Wright suffers from preaching to himself. His preconceived biases regarding red shift and CMB causality are apparent (to me!) LOL

    // Mod Note: KSM is 100% off base here. Other posts moved.
     
    Last edited by a moderator: Mar 22, 2017
  22. exchemist Valued Senior Member

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    Are you, then, a proponent of the "tired light" hypothesis?

    If so, how do you rebut the points he makes in that link?

    I have to confess his CV quite impressed me: http://www.astro.ucla.edu/~wright/intro.html
    Didn't look to me like someone who would be talking out of his arse. And he is a university teacher of course. So perhaps "preaching to himself" is a little unfair. I imagine his students, at least, read what he puts online.

    But I'm only a chemist, of course. You?
     
    Last edited: Mar 21, 2017
  23. timojin Valued Senior Member

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    All that is fine and dandy ,but there are asteroids at the receiver (earth ) Don;t they absorb some of the energy. and deflect some of the incoming light ?
    If so, how those effects astronomical measurements ?
     

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