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Posers and Puzzles

Posers and Puzzles

  1. Subscriber sonhouse
    Fast and Curious
    04 May '06 21:41
    Suppose like the other light problem here, you have a spacecraft going extremely close to C and ahead of the craft is a 100% reflective mirror with the shiny side facing the spacecraft. A laser beam of some wavelength, say 500 NM (somewhere around green I think) and it shoots to the mirror ahead and there is also a mirror on the spacecraft aiming to the other one. The 500 Nm would be blue shifted up to some shorter wavelength and make its way to the mirror. The mirror then shoots it back but this time the wavelenght would be downconverted in the red. So is it back to 500 NM? And if the wave is upconverted to a higher frequency, isn't that adding energy to the beam that wasn't there before? Where does that energy come from?
  2. 04 May '06 21:56
    THE ENERGY COMES FROM 42 MILK DUDES.
  3. 04 May '06 21:57
    Originally posted by sonhouse
    Suppose like the other light problem here, you have a spacecraft going extremely close to C and ahead of the craft is a 100% reflective mirror with the shiny side facing the spacecraft. A laser beam of some wavelength, say 500 NM (somewhere around green I think) and it shoots to the mirror ahead and there is also a mirror on the spacecraft aiming to the ot ...[text shortened]... n't that adding energy to the beam that wasn't there before? Where does that energy come from?
    First you have to tell us where the observer of this experiment is. If he is on the ship then no frequency shift will occur. That's the whole idea behind relativity: regardless of the speed you are travelling all laws of physics appear the same. It's just the same as bouncing the beam between two mirrors in a stationary (ish) laboratory on earth.
  4. Subscriber sonhouse
    Fast and Curious
    04 May '06 22:14
    Originally posted by howardbradley
    First you have to tell us where the observer of this experiment is. If he is on the ship then no frequency shift will occur. That's the whole idea behind relativity: regardless of the speed you are travelling all laws of physics appear the same. It's just the same as bouncing the beam between two mirrors in a stationary (ish) laboratory on earth.
    Yes, I see that, but what about from an observers viewpoint? The distant observer has a probe that can tell the frequencies of the beams coming and going and to an observer at rest the beams would be upconverted and then downconverted on the way back. Where would that energy come from in the upconversion process and where would that energy go in the downconversion?
  5. 04 May '06 22:19
    Originally posted by sonhouse
    Yes, I see that, but what about from an observers viewpoint? The distant observer has a probe that can tell the frequencies of the beams coming and going and to an observer at rest the beams would be upconverted and then downconverted on the way back. Where would that energy come from in the upconversion process and where would that energy go in the downconversion?
    So is what you're asking: if a source of light is approaching an observer it will be blue shifted - where does the energy come from?
  6. Subscriber sonhouse
    Fast and Curious
    05 May '06 03:28
    Originally posted by howardbradley
    So is what you're asking: if a source of light is approaching an observer it will be blue shifted - where does the energy come from?
    Yes, the energy starts with quantum processes that makes the laser light and that process produces green light, say 500 Nm. The spaceship is going so close to C that the light actually comes out as 50 Nm, deep UV. We know there is more energy in shorter wavelengths so where does this energy boost ultimately derive?
    When it hits the mirror which may be on a boom sticking in front of the spacecraft and the light bounces back, it now loses the energy it gained. Where did it go?
    I can also see that the distant observer, who notes the frequencies of the two beams can use the ratio to deduce the velocity of the spacecraft, the 1/v^2-C^2 relationship will allow you to work that out.
  7. 05 May '06 08:47
    Originally posted by sonhouse
    Yes, the energy starts with quantum processes that makes the laser light and that process produces green light, say 500 Nm. The spaceship is going so close to C that the light actually comes out as 50 Nm, deep UV. We know there is more energy in shorter wavelengths so where does this energy boost ultimately derive?
    When it hits the mirror which may be on a ...[text shortened]... duce the velocity of the spacecraft, the 1/v^2-C^2 relationship will allow you to work that out.
    It doesn't go anywhere.

    Imagine you are travelling in a space ship toward a planet. In an attempt to signal you (or parhaps zap you) the occupants of the planet fire a pulse of laser light at you. The pulse lasts a 300,000,000th of a second and is therefore appears to be one metre long - as observed by the planets inhabitants. It will consist of 1m/500nm waves = 2,000,000

    Now what do you see in your ship? Well, because of your relative motion toward the laser, the light will be blue shifted, or in other words the wavelength will be shortened. There will still be 2,000,000 waves, and the speed of light is constant for all observers. Hence the beam will appear to be foreshortened to you, and will impinge on your ship for a shorter period of time.

    So, to you, the beam is at a higher frequency (energy) but it lasts for less time. The two effect will exactly cancel each other out - so the total energy of the beam will remain the same.
  8. Subscriber sonhouse
    Fast and Curious
    05 May '06 09:11 / 1 edit
    Originally posted by howardbradley
    It doesn't go anywhere.

    Imagine you are travelling in a space ship toward a planet. In an attempt to signal you (or parhaps zap you) the occupants of the planet fire a pulse of laser light at you. The pulse lasts a 300,000,000th of a second and is therefore appears to be one metre long - as observed by the planets inhabitants. It will consist of ...[text shortened]... fect will exactly cancel each other out - so the total energy of the beam will remain the same.
    Neat explanation! What about the speedometer? Doesn't that work?
    You have sensors in front and in back of the ship, wouldn't they see both original beam and the reflection at differant frequencies?
    Are you a physicist?
  9. 05 May '06 12:02
    Originally posted by sonhouse
    Neat explanation! What about the speedometer? Doesn't that work?
    You have sensors in front and in back of the ship, wouldn't they see both original beam and the reflection at differant frequencies?
    Are you a physicist?
    I'm afraid that for the observer on the ship's frame of reference the light will always be shifted. He/she will never see "the original beam". All he knows it that his craft was "struck" by some ultra violet light. With only this information he can make no inference about the relative speed of his ship. Of course, if the signal was pre-arranged and he knew it was a 500nm laser, then he could calculate his closing speed.

    "OK", I hear you asking, "how do we know certain galaxies are receding from us? How do we know they are red-shifted and not merely emitting red light in the first place?"

    This is because the spectra from such galaxies contain characterisic adsorbtion lines. By matching up the lines astrophysicists can work out by how much the spectrum has been shifted.

    I am not now, nor have I ever been a physicist. I did work, for a while, at an observatory where I came into contact with physicists. I'm sure my less-than-rigorous explanations of relativity would leave them aghast.
  10. Subscriber sonhouse
    Fast and Curious
    05 May '06 12:20
    Originally posted by howardbradley
    I'm afraid that for the observer on the ship's frame of reference the light will always be shifted. He/she will never see "the original beam". All he knows it that his craft was "struck" by some ultra violet light. With only this information he can make no inference about the relative speed of his ship. Of course, if the signal was pre-arranged an ...[text shortened]... I'm sure my less-than-rigorous explanations of relativity would leave them aghast.
    If the observer had sensors fore and aft of the craft, it seems to me it would see the UV signal and from behind it should see the green light too. I don't see how the signal reflected by the mirror would be seen as UV, it would be reflected backwards and therefore downshifted I would think, and any leakage of that signal going to the aft of the ship could be picked up by a sensor at rest, at least for a while.
  11. 05 May '06 12:57
    Originally posted by sonhouse
    If the observer had sensors fore and aft of the craft, it seems to me it would see the UV signal and from behind it should see the green light too. I don't see how the signal reflected by the mirror would be seen as UV, it would be reflected backwards and therefore downshifted I would think, and any leakage of that signal going to the aft of the ship could be picked up by a sensor at rest, at least for a while.
    Discussing these things via the medium of ASCII text is always tricky - it's at about this time I reach for the large bits of paper and coloured pens - so you'll have to forgive me, but I'm lost as to the relevant positions and motions of all the sensors and mirrors.

    That said, if all the mirrors/sensors are part of the ship ie in the same inertial frame, then all they will "see" is the UV.
  12. Subscriber sonhouse
    Fast and Curious
    05 May '06 13:20
    Originally posted by howardbradley
    Discussing these things via the medium of ASCII text is always tricky - it's at about this time I reach for the large bits of paper and coloured pens - so you'll have to forgive me, but I'm lost as to the relevant positions and motions of all the sensors and mirrors.

    That said, if all the mirrors/sensors are part of the ship ie in the same inertial frame, then all they will "see" is the UV.
    A blackboard would be very helpful here for sure. Here is what I picture:
    A spacecraft going very close to C has a laser shooting out in front, maybe it zaps incoming meteors, whatever. It has its 500 Nm output and shoots ahead of the ship. Now on a boom, maybe 100 meters long, there is a mirror where the beam from the ship will reflect back to the ship. Some of that light will go past the ship to the rear. Now there are already in space at more or less rest velocity, sensors that are well ahead of the ship and well behind it. Under those conditions it seems clear at least that the laser beam going forward from the ship will be in the UV and since the light would be going backwards from the mirror, it should be down converted just as if you had the same laser pointed to the rear of the ship, the sensors would see it as say 10 micron IR radiation. Why would there be any differance between the two situations? A mirror reflecting the outgoing beam or another laser shooting backwards, whats the differance?
  13. 05 May '06 16:50
    Originally posted by sonhouse
    A blackboard would be very helpful here for sure. Here is what I picture:
    A spacecraft going very close to C has a laser shooting out in front, maybe it zaps incoming meteors, whatever. It has its 500 Nm output and shoots ahead of the ship. Now on a boom, maybe 100 meters long, there is a mirror where the beam from the ship will reflect back to the ship. S ...[text shortened]... A mirror reflecting the outgoing beam or another laser shooting backwards, whats the differance?
    Right, gotcha.

    There would be no difference. A laser pointing backward and a laser pointing forward and reflecting off a mirror are equivelent - provided the mirror is stationary with respect to the laser.

    We now have three cases.
    Anybody aboard the ship and measuring the frequency/wavelength of the beam will get "green" as their answer. Regardless of where they take the measurement: as it leaves the laser (and is going forward) or after it has hit the mirror (and is going backward).

    An observer at the forward sensor will get the answer "UV" because the ship is approaching them - for the reasons we discussed before.

    An observer at the rear sensor will get the answer "IR" because, to them, the ship is receding and the light will appear to be red shifted.

    Is this the answer you expected?
  14. Subscriber sonhouse
    Fast and Curious
    05 May '06 17:25 / 1 edit
    Originally posted by howardbradley
    Right, gotcha.

    There would be no difference. A laser pointing backward and a laser pointing forward and reflecting off a mirror are equivelent - provided the mirror is stationary with respect to the laser.

    We now have three cases.
    Anybody aboard the ship and measuring the frequency/wavelength of the beam will get "green" as their answer. Regar receding and the light will appear to be red shifted.

    Is this the answer you expected?
    Yep, thats the one. Therefore, combining the info from the two sensors, the data goes to a number cruncher and out pops the velocity of the spacecraft, or so it would seem.
    So its the spacecraft that would be unable to tell its velocity by that method because it would see no differance between the incident and reflected waves, just like on earth. So the measurement is possible to the observer because he sees a relative motion revealed by the doppler shifts but the craft itself will see no such doppler shift.
  15. 13 May '06 18:42 / 1 edit
    Originally posted by sonhouse
    Suppose like the other light problem here, you have a spacecraft going extremely close to C and ahead of the craft is a 100% reflective mirror with the shiny side facing the spacecraft. A laser beam of some wavelength, say 500 NM (somewhere around green I think) and it shoots to the mirror ahead and there is also a mirror on the spacecraft aiming to the ot ...[text shortened]... n't that adding energy to the beam that wasn't there before? Where does that energy come from?
    According to Einstein's Theory of Relativity, Light passes you at C no matter what your speed. There is/would be no extra energy to accont for.