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Thread: Continuum spectrum objects

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    Continuum spectrum objects

    Are there any persistent heavenly objects which have blank continuum spectra and therefore no redshift?

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    The answer depends, to some extent, on what you mean by "blank continuum spectra." If one considers only some particular region of the electromagnetic spectrum -- say, radio waves -- then there are plenty of sources with featureless spectra. There are very likely some radio sources for which counterparts have not been identified at other wavelengths; these sources might then qualify.

    If you mean "featureless spectra across the entire electromagnetic spectrum," then no such sources exist, as all sources must stop producing photons at some high energy level.

    If you mean "featureless spectra across a very wide range of the spectrum", then there may be some sources, most likely some synchrotron radiation from a jet emitted by some AGN. But in most such cases, I would guess, astronomers would have at least some guess at the redshift by identifying a counterpart in the optical (the galaxy surrounding the AGN).

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    Quote Originally Posted by StupendousMan View Post
    If you mean "featureless spectra across the entire electromagnetic spectrum," then no such sources exist, as all sources must stop producing photons at some high energy level.
    I don't understand what this means.
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    Quote Originally Posted by BigDon View Post
    I don't understand what this means.
    I'm guessing that chornedsnorkack is looking for a plot device (for a story) where redshift/blueshift can't be used easily to detect an incoming object of of some particular type.
    StupendousMan pointed out that if you are looking for something across the entire spectrum, the answer is no, because we don't see continuous sources of photons above some cutoff energy of gamma ray, so it can't cover the "entire electromagnetic spectrum". In the old days, Type O stars were identified as having featureless spectra, but just in the optical range, and because our spectroscopes back then weren't as precise as they are now. Today, there is nothing known that fits the original request.
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    Quote Originally Posted by BigDon View Post
    I don't understand what this means.
    A spectrum with no absorption lines and no emission lines (or more accurately, all emission lines) would have a continuous spectrum, and thus would have no markers for determining Doppler shift.

    Of course, there can't be any such thing, there must be some top end-point beyond which no higher frequency is observed.

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    Dave, I think Anton is on to what I didn't know before. ( Well, one of the things.)

    Quote Originally Posted by antoniseb View Post
    StupendousMan pointed out that if you are looking for something across the entire spectrum, the answer is no, because we don't see continuous sources of photons above some cutoff energy of gamma ray, so it can't cover the "entire electromagnetic spectrum". In the old days, Type O stars were identified as having featureless spectra, but just in the optical range, and because our spectroscopes back then weren't as precise as they are now. Today, there is nothing known that fits the original request.
    The bolded part.

    Is that because the sources of higher end gamma rays are usually very brief? Because my mind is parsing this like you're saying above a certain extreme temperature matter no longer emits photons.

    And my mind is rebelling against that notion.
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    Quote Originally Posted by BigDon View Post
    ... Is that because the sources of higher end gamma rays are usually very brief? Because my mind is parsing this like you're saying above a certain extreme temperature matter no longer emits photons.
    Very high temperatures are short-term or are buried under layers of cooler matter (like the core of our Sun), and there are temperatures that can't be sustained by any physical process. Check out "Pulsar Wind Nebulae" as something that emits towards the high end of the sustained energy range, but these things fail the original request, because the objects do have many discrete energies that make them easy to find redshifts for.
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    Oh yes, there is necessarily some energy limit.
    Which does not necessarily give any redshift.

    After all, a black body spectrum has a characteristic peak energy - but that does not give any redshift.
    Because it is impossible to distinguish a cool black body from a hot but redshifted black body.

    Gamma ray flashes were a mystery with decades. Even though atoms have inner electrons and characteristic x-rays, and nuclei have gamma ray lines, it was impossible to find any redshift for a long time, leaving it unknown whether the flashes were nearby phenomena or distant powerful ones. Only finding optical afterglow finally gave discrete lines and redshifts.

    With gamma ray flashes resolved, are there any astronomical objects left which still do not have redshifts and whose distance therefore still is unknown by orders of magnitude?

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    Quote Originally Posted by chornedsnorkack View Post
    ... are there any astronomical objects left which still do not have redshifts and whose distance therefore still is unknown by orders of magnitude?
    FRBs, also UHECRs come from somewhere? but as individual events, we can't tell where (note that UHECRs are not photons).
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    Basically, if it doesn't have spectral lines or ionization edges, you can't get a Doppler shift. Even if it has a higher and lower limit to its spectrum, how would you know where they should be? So anything that is a good blackbody without a low-density outer layer to produce lines would suffice to mask a Doppler shift. The all-time champion at that is the CMB, we can only try to calculate its Doppler shift from theory.

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    I will guess that the real SEDs of stars, not BB profiles, would reveal much. Some claim if the peak temp. of the Sun determined it's color then the Sun would be green. But they are using a simple Wien formula and not bothering to look at any of the thousands of sp. irr. data clearly revealing a blue peak(s) in the blue.
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    Quote Originally Posted by George View Post
    I will guess that the real SEDs of stars, not BB profiles, would reveal much. Some claim if the peak temp. of the Sun determined it's color then the Sun would be green. But they are using a simple Wien formula and not bothering to look at any of the thousands of sp. irr. data clearly revealing a blue peak(s) in the blue.
    This snaggletoothed departure from a pure blackbody spectrum consists of spectral features of known wavelength. Another G2 star, strongly redshifted, would be recognizable as such at any range at which we could get a good spectrogram of it.

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    Quote Originally Posted by George View Post
    I will guess that the real SEDs of stars, not BB profiles, would reveal much. Some claim if the peak temp. of the Sun determined it's color then the Sun would be green. But they are using a simple Wien formula and not bothering to look at any of the thousands of sp. irr. data clearly revealing a blue peak(s) in the blue.
    Even a pure black body spectrum for the sun's surface temperature would not be green, unless one chooses a quite eccentric white point.

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    Quote Originally Posted by grant hutchison View Post
    Even a pure black body spectrum for the sun's surface temperature would not be green, unless one chooses a quite eccentric white point.
    Right, and that's their point since even the greatest intensity along a BB distribution, or any peak for that matter, will not serve to give you a star's color, though a red peak will often do so.

    My main point is that a star's variation from a pure Planck distribution might serve as signatures for those stars. So even if there are no spectral lines, these signatures will, I think, often serve to reveal what we need to know. The Sun has it's highest peaks in the blue portion, so wouldn't other G2V stars be similarly peaked? Or does metalicity screw these up too much fot their SEDs to be useful as fingerprints?

    I further assume, as a novice, that absorption lines would be a must for post MS stars as they actively puff gasses all over the neighborhood. PMS stars or YSOs will have their placental nebula gasses doing the same thing.
    Last edited by George; 2018-May-19 at 06:49 PM.
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    Quote Originally Posted by George View Post
    Right, and that's their point since even the greatest intensity along a BB distribution, or any peak for that matter, will not serve to give you a star's color, though a red peak will often do so.

    My main point is that a star's variation from a pure Planck distribution might serve as signatures for those stars. So even if there are no spectral lines, these signatures will, I think, often serve to reveal what we need to know. The Sun has it's highest peaks in the blue portion, so wouldn't other G2V stars be similarly peaked? Or does metalicity screw these up too much fot their SEDs to be useful as fingerprints?

    I further assume, as a novice, that absorption lines would be a must for post MS stars as they actively puff gasses all over the neighborhood. PMS stars or YSOs will have their placental nebula gasses doing the same thing.
    Those peaks and notches are from the absorption features created by metals, in particular the H and K lines of ionized calcium. If we had only hydrogen and helium, we would have much closer to a blackbody spectrum. The Balmer lines of hydrogen are weak in G stars, and the helium lines are primarily in the ultraviolet if I am not mistaken.

    I would imagine that Alpha Centauri A and all other main sequence G stars will show the same snaggletooth pattern as the Sun, and that any redshift or blueshift will be readily measurable. I have seen photos of spectra of the massed light of remote galaxies with large redshifts, and the calcium lines were clearly visible.

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    Quote Originally Posted by George View Post
    Right, and that's their point since even the greatest intensity along a BB distribution, or any peak for that matter, will not serve to give you a star's color, though a red peak will often do so.
    Did that point need to be made? "If lots of colours are present, the colour seen will not correspond to any of the individual colours"?

    Quote Originally Posted by George View Post
    My main point is that a star's variation from a pure Planck distribution might serve as signatures for those stars. So even if there are no spectral lines, these signatures will, I think, often serve to reveal what we need to know. The Sun has it's highest peaks in the blue portion, so wouldn't other G2V stars be similarly peaked? Or does metalicity screw these up too much fot their SEDs to be useful as fingerprints?
    As Hornblower points out, all those variations from the Planck distribution are due to spectral lines.

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    Quote Originally Posted by Hornblower View Post
    Those peaks and notches are from the absorption features created by metals, in particular the H and K lines of ionized calcium. If we had only hydrogen and helium, we would have much closer to a blackbody spectrum. The Balmer lines of hydrogen are weak in G stars, and the helium lines are primarily in the ultraviolet if I am not mistaken.
    Ok, and that makes perfect sense given the CMBR is a near perfect BB distribution. It's essentially a H & He only 3000K star, but redshifted a tad.

    I would imagine that Alpha Centauri A and all other main sequence G stars will show the same snaggletooth pattern as the Sun, and that any redshift or blueshift will be readily measurable. I have seen photos of spectra of the massed light of remote galaxies with large redshifts, and the calcium lines were clearly visible.
    So in the imaginary case of the OP, w/o spectral lines, I'm guessing the snaggletooth will serve reasonably well. Is this likely?
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    Quote Originally Posted by grant hutchison View Post
    Did that point need to be made? "If lots of colours are present, the colour seen will not correspond to any of the individual colours"?
    Yes, it's a burden for every heliochromolist. But it's germane to the OP if indeed the SED is enough w/o absorption or emission lines to guide us. A BB approach won't work on its on because it lacks fingerprint character given the absence of these lines. The fact the the Sun peaks in blue when it's Planck distribution peak is green gives the Sun and all stars character perhaps sufficient for, for whatever reason, a line-less spectrum.
    Last edited by George; 2018-May-19 at 08:47 PM.
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    Quote Originally Posted by George View Post
    Yes, it's a burden for every heliochromolist. But it's germane to the OP if indeed the SED is enough w/o absorption or emission lines to guide us. A BB approach won't work on its on because it lacks fingerprint character given the absence of these lines. The fact the the Sun peaks in blue when it's Planck distribution peak is green gives the Sun and all stars character perhaps sufficient for, for whatever reason, a line-less spectrum.
    My bold. I cannot make any sense out of that sentence. Please elaborate on what you mean by "character sufficient for a lineless spectrum".

    Once again, if a hypothetical luminous body is devoid of any spectral feature whose rest wavelength is known, I don't see any way to disentangle the temperature and redshift dependence of the observed wavelength of an overall peak.

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    Quote Originally Posted by Hornblower View Post
    My bold. I cannot make any sense out of that sentence. Please elaborate on what you mean by "character sufficient for a lineless spectrum".
    Your snaggletooth description for stellar SEDs is what I assume might be enough of a distinguishing feature to allow the star's parameters to be determined regardless of Doppler shift. [The simple demonstration of this is the blue vs. green peak for the Sun.] Your earlier response suggests I might be right in this assumption.

    Once again, if a hypothetical luminous body is devoid of any spectral feature whose rest wavelength is known, I don't see any way to disentangle the temperature and redshift dependence of the observed wavelength of an overall peak.
    But, again, would the SED's snaggletooth pattern itself serve like a DNA pattern to expose its real identity?
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    Quote Originally Posted by George View Post
    Your snaggletooth description for stellar SEDs is what I assume might be enough of a distinguishing feature to allow the star's parameters to be determined regardless of Doppler shift. [The simple demonstration of this is the blue vs. green peak for the Sun.] Your earlier response suggests I might be right in this assumption.

    But, again, would the SED's snaggletooth pattern itself serve like a DNA pattern to expose its real identity?
    No absorption lines, no snaggletooth. The star is thermalized just below the photosphere, at least as I understand it. It is absorption features (and sometimes emission features) of known wavelength that cause it to depart from a blackbody at and just above the photosphere.

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    Quote Originally Posted by Hornblower View Post
    No absorption lines, no snaggletooth. The star is thermalized just below the photosphere, at least as I understand it. It is absorption features (and sometimes emission features) of known wavelength that cause it to depart from a blackbody at and just above the photosphere.
    I think I misunderstood your meaning of snaggletooth. I'm suggesting that a mountain range can be recognized by merely its silhouette, no colorful slopes, etc. Can't a SED's profile alone reveal what we need to know about the star?

    I'm also assuming that we are talking about a real object who's spectrum just happens to look continuous. Perhaps if no snaggletooth pattern of lines can be determined then neither will its profile be useable. Is this the case?

    iPad use else I would have used a drawing to eschew obfuscation.
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    Quote Originally Posted by George View Post
    I think I misunderstood your meaning of snaggletooth. I'm suggesting that a mountain range can be recognized by merely its silhouette, no colorful slopes, etc. Can't a SED's profile alone reveal what we need to know about the star?
    You seem to be trying to separate things that are inseparable. The real Spectral Energy Distribution = black body - absorption lines (+ emission lines). You can't have a real SED that doesn't include absorption lines - if you take away the absorption (+ emission) you have a black body spectrum. The fact that the sun's spectrum characteristically peaks in the blue and not the green is because of absorption lines.

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    Quote Originally Posted by grant hutchison View Post
    You seem to be trying to separate things that are inseparable. The real Spectral Energy Distribution = black body - absorption lines (+ emission lines). You can't have a real SED that doesn't include absorption lines - if you take away the absorption (+ emission) you have a black body spectrum. The fact that the sun's spectrum characteristically peaks in the blue and not the green is because of absorption lines.
    Yes I understand this, but in the context of a story-line where only the profile is the useable data, for reasons unknown, is it not enough to reveal its temp., redshift, etc.? Will the mountain's silhouette alone not reveal the mountain?
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    Quote Originally Posted by George View Post
    Yes I understand this, but in the context of a story-line where only the profile is the useable data, for reasons unknown, is it not enough to reveal its temp., redshift, etc.? Will the mountain's silhouette alone not reveal the mountain?
    Rather than mountains and snaggletooths can you phrase this as a physics question without analogies? Is your question: "Can we tell the temperature of a black body spectrum independently of the redshift at which we observe it?" ?

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    Quote Originally Posted by George View Post
    Yes I understand this, but in the context of a story-line where only the profile is the useable data, for reasons unknown, is it not enough to reveal its temp., redshift, etc.? Will the mountain's silhouette alone not reveal the mountain?
    Yes, absorption lines in the spectral energy distribution will give us information about temperature which is lacking from a black body spectrum.

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    Quote Originally Posted by Shaula View Post
    Rather than mountains and snaggletooths can you phrase this as a physics question without analogies? Is your question: "Can we tell the temperature of a black body spectrum independently of the redshift at which we observe it?" ?
    My hiccup, I think, is that I'm limiting "absorption" to the dark portion of a spectrum, represented by the dark lines themselves. It is these lines, after all, that reveal redshift. In a fiction story where these lines are indeterminate, but, perhaps only in fiction, we are given the spectral energy values, I'm asking if this is enough? What I should have realized is that these values reveal where these dark or bright lines would be even if the imaging presented graphically failed to resolve them.

    So, if I give you several dozen, or more?, SED energy values and their respected measured wavelengths, then is this not all you need to determine redshift, temp., etc.? But, again, I see now that profile and lines are two expressions of the same thing...absorption. Using "profile" should suffice in a story line if no snaggletooth pattern is presented in the story, but, somehow, the spectral vales are resolved.
    Last edited by George; 2018-May-20 at 06:00 PM.
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    Here should show both the colorful absorption lines and the SED profile (blue line). Imagine in a movie that the colorful representation appeared continuous, but we still had the profile. [sorry, must run, but you see what I'm asking]
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    Quote Originally Posted by George View Post
    Here should show both the colorful absorption lines and the SED profile (blue line). Imagine in a movie that the colorful representation appeared continuous, but we still had the profile. [sorry, must run, but you see what I'm asking]
    No, I've no idea what you're asking.

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    Synchrotron radiation, like blackbody radiation, turns out to have a maximum at some characteristic energy, with slow decrease at lower energy and steeper dropoff at higher energies.

    In that case, the energy of maximum radiation is a function of the energy of the charge carriers and the magnetic field they are in.

    But featureless like a perfect black body. Hard to resolve whether the object is inherently a lower-energy-radiating one, or higher-energy-radiating one which is redshifted.
    One type with lack of spectra is said to be BL Lacertae objects.

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