Relativity question

[jgreimer]

Good start but let's see if we can do this a little more scientifically. Suppose there is a spectral line that normally is in the green region with a wavelength of 5166 Angstroms, but due to the recession of the star we see that spectral line in the orange region at 5904 Angstroms. There is a formula that can tell us how much energy those photons have here:<br /><br />http://tpm.amc.anl.gov/NJZTools/EnergytoLambda.html<br /><br />That formula is λ = hc/E where λ is the wavelength in Angstroms, h is Planck's constant, c is the speed of light and E is energy in electron volts.<br /><br />We see that in the green region the spectral line has an energy of 2.4 electron volts (a voltage that is related to the voltage across a green LED by the way). When the spectral line is in the orange region its energy is only 2.1 electron volts.<br /><br />However we know that the photon hasn't really lost energy but that energy difference is due to the relative motions of the earth and the star. h is still the same and E is still the same so what must have changed is the velocity. The relative velocity between the earth and the star must be (2.4 - 2.1)/2.4 or about 12.5% of the speed of light. Yes, if that star were a pulsar its pulses would be stretched out by 12.5%, so as you mentioned the total energy we observe is the same because its stretched out over longer time.

 

[origin]

The last 2 post seem to be saying, if I understand correctly, that the energy seen on earth from a red shifted object is not diminished it is simply spread over a longer period of time. <br />I do not think this is correct. <br />If you look at it on a photon by photon basis it is clear that a red shifted photons have less energy than when they left the redshifted object relative to the observer on earth. <br />In the case of the pulsar, yes the time between pulses would be stretched out but the number of photons that left the pulsar would be the same it is just the flux would decrease. And each one of the photons would have less energy than if the pulsar were not red shifted. <div class="Discussion_UserSignature"> </div>

 

[alokmohan]

High physics not for lay man.

 

[origin]

<font color="yellow">High physics not for lay man.</font><br /><br />I am not a physicist, I am an engineer by profession so if this seems like high physics it is only due to my inability to effectively express myself. If you would like clarification on any of my points please ask a question and I will try to answer it better.<br /> <div class="Discussion_UserSignature"> </div>

 

[alokmohan]

How do y ou see a photon?

 

[siarad]

Therein lies the problem I think.<br />Our physics is based around the Milky Way or even more closely.<br />What happens in entirely different planes of reference I'm not sure.<br />I'm a retired electronics designer so don't take my words as gospel science <img src="/images/icons/smile.gif" />

 

[siarad]

Ah 'the measurement problem'

 

[origin]

<font color="yellow">How do y ou see a photon?</font><br /><br />The human retina at the back of the eye has two types of receptors known as cones and rods. <br /><br />The active substance in the rods is rhodopsin. Photons are absorbed by these molecules which change shape and chemically trigger a signal which is transmitted to the optic nerve. The signal is then sent to the brain, which we interpret as light.<br /><br />If you want to <i>detect</i> a single photon it is best to use a photomultipier tube.<br /><br /><br /> <div class="Discussion_UserSignature"> </div>

 

[SpeedFreek]

<font color="yellow">If you look at it on a photon by photon basis it is clear that a red shifted photons have less energy than when they left the redshifted object relative to the observer on earth.<br />In the case of the pulsar, yes the time between pulses would be stretched out but the number of photons that left the pulsar would be the same it is just the flux would decrease. And each one of the photons would have less energy than if the pulsar were not red shifted.</font><br /><br />But we know from tests in the laboratory, that looking at things on a photon by photon basis can change our results remarkably. <br /><br />If a certain number of photons leave a pulsar over a given time, say 1 second, and then due to cosmological expansion we receive those photons over a longer time, say 2 seconds, then we would receive the same number of photons over 2 seconds that were originally emitted over 1 second, would we not?<br /><br />The number of photons emitted by the pulsar stays the same as the number of photons detected, but they are detected over a longer time period. You say the flux decreases - flux is always measured either in terms of an amount per unit of <i>time</i> or an amount per unit of <i>space</i>. The time for detection of a given number of photons is, in this example, double the time for emission of that given number. If you go the other way, the amount of photons per unit of space is now half the amount it was to begin with. So the flux does indeed decrease.<br /><br />But why do the photons now have less energy too?<br /><br />If we treat the light as a wave that contained a given amount of energy per second of emission, we find that due to the cosmological expansion we will receive that wave of light over double the length of time, and so we receive that given amount of energy over double the length of time also. So we would infer that the amount of energy per second was halved. Everything makes sense here.<br /><br />But as soon as we think in sin <div class="Discussion_UserSignature"> <p><font color="#ff0000">_______________________________________________<br /></font><font size="2"><em>SpeedFreek</em></font> </p> </div>

 

[weeman]

Does the superpositioning of the photon suggest that it can exist in every conceivable point in space simultaneously? <br /><br />In other words, if you're not observing it, after leaving the light source, the photon makes a quick trip to the Andromeda galaxy before traveling through the slit. <div class="Discussion_UserSignature"> <p> </p><p><strong><font color="#ff0000">Techies: We do it in the dark. </font></strong></p><p><font color="#0000ff"><strong>"Put your hand on a stove for a minute and it seems like an hour. Sit with that special girl for an hour and it seems like a minute. That's relativity.</strong><strong>" -Albert Einstein </strong></font></p> </div>

 

[kyle_baron]

<font color="yellow"><br />Does the superpositioning of the photon suggest that it can exist in every conceivable point in space simultaneously? </font><br /><br />Yes, as long as you're not looking at it, and as long as the photon is in a wave form. <img src="/images/icons/smile.gif" /><br /><br /> <div class="Discussion_UserSignature"> <p><font size="4"><strong></strong></font></p> </div>

 

[siarad]

Good examples of 'the measurement problem'. Schrödinger did it with a cat <img src="/images/icons/wink.gif" />

 

[kyle_baron]

<font color="yellow"><br />As long as the distance (traveltime) from the emitter is the same, the photon can superposition "everywhere". It cannot superposition itself before or after itself..so to speak, but at the the same point in time seen from the emitter.</font><br /><br />Yes, I would agree. However, I was thinking about this the other day, and I came to the conclusion, that the photon's collapse from it's superposition (everywhere) is instantaneous, or FTL. This would also be an explanation for two entangled photon (waves). <div class="Discussion_UserSignature"> <p><font size="4"><strong></strong></font></p> </div>

 

[kyle_baron]

Yep, then we're in total agreement. <img src="/images/icons/smile.gif" /> I'm not sure where we go from here, to find an arguement. <img src="/images/icons/wink.gif" /> <div class="Discussion_UserSignature"> <p><font size="4"><strong></strong></font></p> </div>

 

[SpeedFreek]

Have a read of this article - The Quantum Eraser Experiment and see what you make of the final paragraph. <img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> <p><font color="#ff0000">_______________________________________________<br /></font><font size="2"><em>SpeedFreek</em></font> </p> </div>

 

[kyle_baron]

<font color="yellow"><br />Does matter particles surround itself with theese kind of timeslots produced from time of creation to present moment? </font><br /><br />Yes, matter particles (atoms) surround themselves with time slots (of planck time) from the time the atom is created, to the present.<br /><font color="yellow"><br /> Are they produced only when they emmit a photon, or are they produced continously</font><br /><br />The time slots are produced continuously, and naturally, at the speed of light, for each matter particle (atom). That's how Plank Time, and Planck Length work. Also, that's how time flows to the future (Times Arrow).<br /><font color="yellow"><br /> (photons) are carried away along with it? </font><br /><br />If your asking is the photon carried away with the time slot? I would answer that a photon striking a matter particle (atom) exists with in it's time slot (planck time). For the simple reason that we can now see the photon and the matter particle (atom). OTOH, before the photon reaches the matter particle (atom) or doesn't hit a matter particle, it exists as a wave in it's superposition (everywhere). And this would be outside of it's time slot. <br /><font color="yellow"><br />Are we describeing a fundamental field (space) or a field superimposed on an even more fundamental field? <br /></font><br /><br />I believe our 4-D space is a field (gravitational), superimposed on an even more fundamental field of nothingness (no time, no space) which would be beneath the Planck Length (in La-La Land) <img src="/images/icons/wink.gif" />.<br /><br />These were tough questions. They make my brain hurt. <img src="/images/icons/wink.gif" /><br /> <div class="Discussion_UserSignature"> <p><font size="4"><strong></strong></font></p> </div>

 

[freezian]

Could it be neutralised somehow? Considering that the Universe would be made up of the same material regardless of where you went? It could even be added to an element for an unknown reason,why not?

 

[kyle_baron]

<font color="yellow"><br />Could it be neutralised somehow? </font><br /><br />No, redshift just means that the energy is streched out over a wider area of space. The wavelength is streched out, and the frequency of the wavelength is lower. <div class="Discussion_UserSignature"> <p><font size="4"><strong></strong></font></p> </div>

 

[vandivx]

"Does the superpositioning of the photon suggest that it can exist in every conceivable point in space simultaneously?<br /><br />In other words, if you're not observing it, after leaving the light source, the photon makes a quick trip to the Andromeda galaxy before traveling through the slit."<br />---<br /><br />NO<br /><br />reason is the photon can be anywhere BUT only within its wave function domain which is spherical area spreading at speed of light as time passes<br /><br />it cannot show up or be detected at Andromeda because the wave function would spread that far only after some four light years passed and I assume you are not talking about such case of long experiment, in other words wave function in this regard doesn't breach light speed barrier, ie photon can't travel FTL<br /><br />vanDivX <div class="Discussion_UserSignature"> </div>

 

[kyle_baron]

<font color="yellow"><br /> if you're not observing it, after leaving the light source, the photon makes a quick trip to the Andromeda galaxy before traveling through the slit." <br />--- <br /><br />NO </font><br /><br />YES. Brian Greene when describing Feynman's "sum over paths"approach, p.110 "The Elegant Universe" uses the example for an electron: "It goes on a long journey to the Andromeda galaxy before turning back and passing through the left slit on its way to the screen."<br /><br />There is nothing in the statement about the speed of light, or a footnote showing further information.<br /><font color="yellow"><br />the photon can be anywhere BUT only within its wave function domain which is spherical area spreading at speed of light as time passes </font><br /><br />If that's your belief, fine. Otherwise, I'll need a web link for the photon's "wave function domain". Time does not pass, for the photon's wave function.<br /><br />To further clarify my point of view, concerning the super position of the electron wave:<br /><br />http://en.wikipedia.org/wiki/Quantum_superposition<br /><br /><i>The superposition principle is the addition of the amplitudes of waves from interference. In quantum mechanics it is the sum of wavefunction amplitudes</i><br /><br />So what does this mean? IMO, it means our individual electron wave becomes part of another wave, that streches all the way to the Andromeda galaxy (or end of the universe) and back.<br /> <div class="Discussion_UserSignature"> <p><font size="4"><strong></strong></font></p> </div>

 

[vandivx]

I don't agree with BG that's all I can say<br /><br />what I said is classic QM interpretation which I believe still stands<br /><br />there is potential that you could detect photon anywhere its wave function has spread, according to you that means one could possibly detect photon at Andromeda location while this photon was emited as part of lab experiment lasting very short time (on order of fraction of a second say as in double slit experiment) and that would mean FTL light travel which is flat contradiction in my books<br /><br />I don't know what BG meant and I don't read his books, I have once read one or couple of them before I decided those books of his aren't worth reading and that comes from one who's got sizable collection of 'science for laymen' books at home and finds them very valuable to read and re-read again and again<br /><br />not to confuse things, there is such a thing as non-locality in regard to wave function of particles which can be regarded as FTL or instantaneous travel but that means only 'travel' within the wavefunction and this (Schrodinger) particle wavefunction needs time to spread every time a particle is emited or otherwise measured and localized by this measurment (because wave function colapses to a point when particle is detected or measured)<br /><br />I don't care that time doesn't pass for photon from its point of view, all I know is that for example photons of light coming here from our Sun take some eight and a half minutes time to get here and that's a fact and those photons certainly can't make any 'detour via Andromeda' in any sense such as their wave function reaching there before it colapses when they land in my eye<br /><br />vanDivX <div class="Discussion_UserSignature"> </div>

 

[kyle_baron]

<font color="yellow"><br />there is potential that you could detect photon anywhere its wave function has spread, according to you that means one could possibly detect photon at Andromeda location while this photon was emited as part of lab experiment lasting very short time (on order of fraction of a second say as in double slit experiment) and that would mean FTL light travel which is flat contradiction in my books</font><br /><br />Because the photon (wave) doesn't travel in time, means it can travel anywhere through space, in "no time"(undetected and FTL). If it's undetected, there's no information exchanged, and therefore does not violate GR. A photon (particle) is detected at the speed of light.<br /><font color="yellow"><br />all I know is that for example photons of light coming here from our Sun take some eight and a half minutes time to get here and that's a fact </font><br /><br />No, that's an assumption, based on the photon particle traveling at c. All those photons traveling through space are waves (not particles), until they hit the atmosphere, or your eye. Otherwise, the empty space would be awfully bright! right? <img src="/images/icons/wink.gif" /><br /> <div class="Discussion_UserSignature"> <p><font size="4"><strong></strong></font></p> </div>