Why are stellar absorption line spectra passed through a cold gas? The Next CEO of Stack OverflowWhy do stars have absorption spectra?Galaxy Spectra: Emission and Absorption Linesabsorption and emission lines in Cassiopeia A hydrogen spectrumWould it be possible to detect nuclear explosion on exoplanet?Why has the amount of star formation in the Universe decreased over time?Spectroscopy - Absorption linesPhotometric surveys vs. Spectroscopic surveysVariable speed of light impact on spectral absorption lines in distance luminous objects?Spectroscopy: The trustworthiness of reflected, refracted, and “mixed” light sourcesWhy are there dark lines in an absorption line spectrum from the Sun?

Is there a difference between "Fahrstuhl" and "Aufzug"?

What day is it again?

Is there a reasonable and studied concept of reduction between regular languages?

Which one is the true statement?

Why don't programming languages automatically manage the synchronous/asynchronous problem?

Getting Stale Gas Out of a Gas Tank w/out Dropping the Tank

How to find image of a complex function with given constraints?

How to avoid supervisors with prejudiced views?

Help/tips for a first time writer?

Players Circumventing the limitations of Wish

Reshaping json / reparing json inside shell script (remove trailing comma)

"Eavesdropping" vs "Listen in on"

Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?

Inexact numbers as keys in Association?

Airplane gently rocking its wings during whole flight

A question about free fall, velocity, and the height of an object.

Lucky Feat: How can "more than one creature spend a luck point to influence the outcome of a roll"?

Is it OK to decorate a log book cover?

What CSS properties can the br tag have?

Why did early computer designers eschew integers?

Reference request: Grassmannian and Plucker coordinates in type B, C, D

TikZ: How to fill area with a special pattern?

Spaces in which all closed sets are regular closed

Can someone explain this formula for calculating Manhattan distance?



Why are stellar absorption line spectra passed through a cold gas?



The Next CEO of Stack OverflowWhy do stars have absorption spectra?Galaxy Spectra: Emission and Absorption Linesabsorption and emission lines in Cassiopeia A hydrogen spectrumWould it be possible to detect nuclear explosion on exoplanet?Why has the amount of star formation in the Universe decreased over time?Spectroscopy - Absorption linesPhotometric surveys vs. Spectroscopic surveysVariable speed of light impact on spectral absorption lines in distance luminous objects?Spectroscopy: The trustworthiness of reflected, refracted, and “mixed” light sourcesWhy are there dark lines in an absorption line spectrum from the Sun?










2












$begingroup$


I am currently studying spectral lines and saw in my notes that the way in which the absorbtion line spectra for a star was measured was by taking its light and shining it through a cold gas. The remaining light would show a black body spectrum with dark lines at the wavelengths which correspond to an exact difference in energy levels between two orbitals for an electron.



My question is, isn't this sort of spectroscopy supposed to identify the elemental composition of the star? It seems to me this would identify what elements were in the cold gas instead.



Note: I am asking whether the measured absorption spectra represents the stars or the cold gas. If it measures the stars, why? Wouldn't changing the gas change the wavelengths at which electrons will make a transition?










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    I am currently studying spectral lines and saw in my notes that the way in which the absorbtion line spectra for a star was measured was by taking its light and shining it through a cold gas. The remaining light would show a black body spectrum with dark lines at the wavelengths which correspond to an exact difference in energy levels between two orbitals for an electron.



    My question is, isn't this sort of spectroscopy supposed to identify the elemental composition of the star? It seems to me this would identify what elements were in the cold gas instead.



    Note: I am asking whether the measured absorption spectra represents the stars or the cold gas. If it measures the stars, why? Wouldn't changing the gas change the wavelengths at which electrons will make a transition?










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      I am currently studying spectral lines and saw in my notes that the way in which the absorbtion line spectra for a star was measured was by taking its light and shining it through a cold gas. The remaining light would show a black body spectrum with dark lines at the wavelengths which correspond to an exact difference in energy levels between two orbitals for an electron.



      My question is, isn't this sort of spectroscopy supposed to identify the elemental composition of the star? It seems to me this would identify what elements were in the cold gas instead.



      Note: I am asking whether the measured absorption spectra represents the stars or the cold gas. If it measures the stars, why? Wouldn't changing the gas change the wavelengths at which electrons will make a transition?










      share|cite|improve this question











      $endgroup$




      I am currently studying spectral lines and saw in my notes that the way in which the absorbtion line spectra for a star was measured was by taking its light and shining it through a cold gas. The remaining light would show a black body spectrum with dark lines at the wavelengths which correspond to an exact difference in energy levels between two orbitals for an electron.



      My question is, isn't this sort of spectroscopy supposed to identify the elemental composition of the star? It seems to me this would identify what elements were in the cold gas instead.



      Note: I am asking whether the measured absorption spectra represents the stars or the cold gas. If it measures the stars, why? Wouldn't changing the gas change the wavelengths at which electrons will make a transition?







      astrophysics astronomy spectroscopy






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 26 at 8:21









      Johnny

      1031




      1031










      asked Mar 24 at 10:28









      Vishal JainVishal Jain

      1119




      1119




















          3 Answers
          3






          active

          oldest

          votes


















          11












          $begingroup$

          The idea of shining light from a (hot) source through a cold gas is a very basic model of what a stellar atmosphere is like.



          The radiation field emitted from hotter interior layers has to pass through cooler outer layers before it gets to us. This is how absorption lines become imprinted on what otherwise would be a featureless continuum. In other words, the star's radiation is self-absorbed by its outer, cooler layers.



          A good way to look at this is that the photons we receive come from the layer at which the optical depth (at that wavelength) is approximately unity. i.e. We see down into the atmosphere as far as a mean free path for a photon of that wavelength.



          Because the solar interior gets hotter as we go inwards and the radiation field approximates to a Planck blackbody function, then if the mean free path is long at a particular wavelength we see deeper, hotter and therefore brighter. On the other hand, if the mean free path is short (for example because there is a radiative transition of some type of atom at that wavelength and that element exists in the atmosphere), then our view will be to shallower depths and cooler temperatures, meaning less intense light.



          The mean free path will be proportional to the number density of that particular type of atom and thus by measuring the strength of absorption lines we get not just the indication of the presence of a particular species of atom, but also its number density (a.k.a. its abundance).






          share|cite|improve this answer









          $endgroup$




















            8












            $begingroup$

            One has not to make confusion between the way we can create an absorption spectrum in laboratory (making radiation with a continuum spectrum passing through a cold gas) and the way absorption spectra are formed in the photosphere of a star or by absorption by planet atmospheres or interstellar gases. In the astrophysical case, spectroscopists have only to collect the light coming from far sources.



            No need to use a further absorbing layer which could not say anything about composition of far objects. Absorption lines in solar spectrum were observed for the first time by Fraunhofer mid-19th century. Basic starting information about absorption spectroscopy can be found in this wikipedia page.






            share|cite|improve this answer









            $endgroup$




















              6












              $begingroup$


              I am currently studying spectral lines and saw in my notes that the way in which the absorbtion line spectra for a star was measured was by taking its light and shining it through a cold gas.




              The absorption spectrum of a star is generated when light coming from within the photosphere passes through the "cold" outer atmosphere of the star (where "cold" in this context merely means that the vast majority of the atoms are in the ground state, which can be quite hot indeed by human standards). To get the absorption spectrum of the star, we simply directly measure the light coming towards us.






              share|cite|improve this answer











              $endgroup$













                Your Answer





                StackExchange.ifUsing("editor", function ()
                return StackExchange.using("mathjaxEditing", function ()
                StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
                StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
                );
                );
                , "mathjax-editing");

                StackExchange.ready(function()
                var channelOptions =
                tags: "".split(" "),
                id: "151"
                ;
                initTagRenderer("".split(" "), "".split(" "), channelOptions);

                StackExchange.using("externalEditor", function()
                // Have to fire editor after snippets, if snippets enabled
                if (StackExchange.settings.snippets.snippetsEnabled)
                StackExchange.using("snippets", function()
                createEditor();
                );

                else
                createEditor();

                );

                function createEditor()
                StackExchange.prepareEditor(
                heartbeatType: 'answer',
                autoActivateHeartbeat: false,
                convertImagesToLinks: false,
                noModals: true,
                showLowRepImageUploadWarning: true,
                reputationToPostImages: null,
                bindNavPrevention: true,
                postfix: "",
                imageUploader:
                brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
                contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
                allowUrls: true
                ,
                noCode: true, onDemand: true,
                discardSelector: ".discard-answer"
                ,immediatelyShowMarkdownHelp:true
                );



                );













                draft saved

                draft discarded


















                StackExchange.ready(
                function ()
                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f468353%2fwhy-are-stellar-absorption-line-spectra-passed-through-a-cold-gas%23new-answer', 'question_page');

                );

                Post as a guest















                Required, but never shown

























                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                11












                $begingroup$

                The idea of shining light from a (hot) source through a cold gas is a very basic model of what a stellar atmosphere is like.



                The radiation field emitted from hotter interior layers has to pass through cooler outer layers before it gets to us. This is how absorption lines become imprinted on what otherwise would be a featureless continuum. In other words, the star's radiation is self-absorbed by its outer, cooler layers.



                A good way to look at this is that the photons we receive come from the layer at which the optical depth (at that wavelength) is approximately unity. i.e. We see down into the atmosphere as far as a mean free path for a photon of that wavelength.



                Because the solar interior gets hotter as we go inwards and the radiation field approximates to a Planck blackbody function, then if the mean free path is long at a particular wavelength we see deeper, hotter and therefore brighter. On the other hand, if the mean free path is short (for example because there is a radiative transition of some type of atom at that wavelength and that element exists in the atmosphere), then our view will be to shallower depths and cooler temperatures, meaning less intense light.



                The mean free path will be proportional to the number density of that particular type of atom and thus by measuring the strength of absorption lines we get not just the indication of the presence of a particular species of atom, but also its number density (a.k.a. its abundance).






                share|cite|improve this answer









                $endgroup$

















                  11












                  $begingroup$

                  The idea of shining light from a (hot) source through a cold gas is a very basic model of what a stellar atmosphere is like.



                  The radiation field emitted from hotter interior layers has to pass through cooler outer layers before it gets to us. This is how absorption lines become imprinted on what otherwise would be a featureless continuum. In other words, the star's radiation is self-absorbed by its outer, cooler layers.



                  A good way to look at this is that the photons we receive come from the layer at which the optical depth (at that wavelength) is approximately unity. i.e. We see down into the atmosphere as far as a mean free path for a photon of that wavelength.



                  Because the solar interior gets hotter as we go inwards and the radiation field approximates to a Planck blackbody function, then if the mean free path is long at a particular wavelength we see deeper, hotter and therefore brighter. On the other hand, if the mean free path is short (for example because there is a radiative transition of some type of atom at that wavelength and that element exists in the atmosphere), then our view will be to shallower depths and cooler temperatures, meaning less intense light.



                  The mean free path will be proportional to the number density of that particular type of atom and thus by measuring the strength of absorption lines we get not just the indication of the presence of a particular species of atom, but also its number density (a.k.a. its abundance).






                  share|cite|improve this answer









                  $endgroup$















                    11












                    11








                    11





                    $begingroup$

                    The idea of shining light from a (hot) source through a cold gas is a very basic model of what a stellar atmosphere is like.



                    The radiation field emitted from hotter interior layers has to pass through cooler outer layers before it gets to us. This is how absorption lines become imprinted on what otherwise would be a featureless continuum. In other words, the star's radiation is self-absorbed by its outer, cooler layers.



                    A good way to look at this is that the photons we receive come from the layer at which the optical depth (at that wavelength) is approximately unity. i.e. We see down into the atmosphere as far as a mean free path for a photon of that wavelength.



                    Because the solar interior gets hotter as we go inwards and the radiation field approximates to a Planck blackbody function, then if the mean free path is long at a particular wavelength we see deeper, hotter and therefore brighter. On the other hand, if the mean free path is short (for example because there is a radiative transition of some type of atom at that wavelength and that element exists in the atmosphere), then our view will be to shallower depths and cooler temperatures, meaning less intense light.



                    The mean free path will be proportional to the number density of that particular type of atom and thus by measuring the strength of absorption lines we get not just the indication of the presence of a particular species of atom, but also its number density (a.k.a. its abundance).






                    share|cite|improve this answer









                    $endgroup$



                    The idea of shining light from a (hot) source through a cold gas is a very basic model of what a stellar atmosphere is like.



                    The radiation field emitted from hotter interior layers has to pass through cooler outer layers before it gets to us. This is how absorption lines become imprinted on what otherwise would be a featureless continuum. In other words, the star's radiation is self-absorbed by its outer, cooler layers.



                    A good way to look at this is that the photons we receive come from the layer at which the optical depth (at that wavelength) is approximately unity. i.e. We see down into the atmosphere as far as a mean free path for a photon of that wavelength.



                    Because the solar interior gets hotter as we go inwards and the radiation field approximates to a Planck blackbody function, then if the mean free path is long at a particular wavelength we see deeper, hotter and therefore brighter. On the other hand, if the mean free path is short (for example because there is a radiative transition of some type of atom at that wavelength and that element exists in the atmosphere), then our view will be to shallower depths and cooler temperatures, meaning less intense light.



                    The mean free path will be proportional to the number density of that particular type of atom and thus by measuring the strength of absorption lines we get not just the indication of the presence of a particular species of atom, but also its number density (a.k.a. its abundance).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 24 at 12:27









                    Rob JeffriesRob Jeffries

                    70.2k7142243




                    70.2k7142243





















                        8












                        $begingroup$

                        One has not to make confusion between the way we can create an absorption spectrum in laboratory (making radiation with a continuum spectrum passing through a cold gas) and the way absorption spectra are formed in the photosphere of a star or by absorption by planet atmospheres or interstellar gases. In the astrophysical case, spectroscopists have only to collect the light coming from far sources.



                        No need to use a further absorbing layer which could not say anything about composition of far objects. Absorption lines in solar spectrum were observed for the first time by Fraunhofer mid-19th century. Basic starting information about absorption spectroscopy can be found in this wikipedia page.






                        share|cite|improve this answer









                        $endgroup$

















                          8












                          $begingroup$

                          One has not to make confusion between the way we can create an absorption spectrum in laboratory (making radiation with a continuum spectrum passing through a cold gas) and the way absorption spectra are formed in the photosphere of a star or by absorption by planet atmospheres or interstellar gases. In the astrophysical case, spectroscopists have only to collect the light coming from far sources.



                          No need to use a further absorbing layer which could not say anything about composition of far objects. Absorption lines in solar spectrum were observed for the first time by Fraunhofer mid-19th century. Basic starting information about absorption spectroscopy can be found in this wikipedia page.






                          share|cite|improve this answer









                          $endgroup$















                            8












                            8








                            8





                            $begingroup$

                            One has not to make confusion between the way we can create an absorption spectrum in laboratory (making radiation with a continuum spectrum passing through a cold gas) and the way absorption spectra are formed in the photosphere of a star or by absorption by planet atmospheres or interstellar gases. In the astrophysical case, spectroscopists have only to collect the light coming from far sources.



                            No need to use a further absorbing layer which could not say anything about composition of far objects. Absorption lines in solar spectrum were observed for the first time by Fraunhofer mid-19th century. Basic starting information about absorption spectroscopy can be found in this wikipedia page.






                            share|cite|improve this answer









                            $endgroup$



                            One has not to make confusion between the way we can create an absorption spectrum in laboratory (making radiation with a continuum spectrum passing through a cold gas) and the way absorption spectra are formed in the photosphere of a star or by absorption by planet atmospheres or interstellar gases. In the astrophysical case, spectroscopists have only to collect the light coming from far sources.



                            No need to use a further absorbing layer which could not say anything about composition of far objects. Absorption lines in solar spectrum were observed for the first time by Fraunhofer mid-19th century. Basic starting information about absorption spectroscopy can be found in this wikipedia page.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 24 at 12:43









                            GiorgioPGiorgioP

                            4,2501628




                            4,2501628





















                                6












                                $begingroup$


                                I am currently studying spectral lines and saw in my notes that the way in which the absorbtion line spectra for a star was measured was by taking its light and shining it through a cold gas.




                                The absorption spectrum of a star is generated when light coming from within the photosphere passes through the "cold" outer atmosphere of the star (where "cold" in this context merely means that the vast majority of the atoms are in the ground state, which can be quite hot indeed by human standards). To get the absorption spectrum of the star, we simply directly measure the light coming towards us.






                                share|cite|improve this answer











                                $endgroup$

















                                  6












                                  $begingroup$


                                  I am currently studying spectral lines and saw in my notes that the way in which the absorbtion line spectra for a star was measured was by taking its light and shining it through a cold gas.




                                  The absorption spectrum of a star is generated when light coming from within the photosphere passes through the "cold" outer atmosphere of the star (where "cold" in this context merely means that the vast majority of the atoms are in the ground state, which can be quite hot indeed by human standards). To get the absorption spectrum of the star, we simply directly measure the light coming towards us.






                                  share|cite|improve this answer











                                  $endgroup$















                                    6












                                    6








                                    6





                                    $begingroup$


                                    I am currently studying spectral lines and saw in my notes that the way in which the absorbtion line spectra for a star was measured was by taking its light and shining it through a cold gas.




                                    The absorption spectrum of a star is generated when light coming from within the photosphere passes through the "cold" outer atmosphere of the star (where "cold" in this context merely means that the vast majority of the atoms are in the ground state, which can be quite hot indeed by human standards). To get the absorption spectrum of the star, we simply directly measure the light coming towards us.






                                    share|cite|improve this answer











                                    $endgroup$




                                    I am currently studying spectral lines and saw in my notes that the way in which the absorbtion line spectra for a star was measured was by taking its light and shining it through a cold gas.




                                    The absorption spectrum of a star is generated when light coming from within the photosphere passes through the "cold" outer atmosphere of the star (where "cold" in this context merely means that the vast majority of the atoms are in the ground state, which can be quite hot indeed by human standards). To get the absorption spectrum of the star, we simply directly measure the light coming towards us.







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Mar 24 at 16:26

























                                    answered Mar 24 at 15:00









                                    probably_someoneprobably_someone

                                    18.6k12960




                                    18.6k12960



























                                        draft saved

                                        draft discarded
















































                                        Thanks for contributing an answer to Physics Stack Exchange!


                                        • Please be sure to answer the question. Provide details and share your research!

                                        But avoid


                                        • Asking for help, clarification, or responding to other answers.

                                        • Making statements based on opinion; back them up with references or personal experience.

                                        Use MathJax to format equations. MathJax reference.


                                        To learn more, see our tips on writing great answers.




                                        draft saved


                                        draft discarded














                                        StackExchange.ready(
                                        function ()
                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f468353%2fwhy-are-stellar-absorption-line-spectra-passed-through-a-cold-gas%23new-answer', 'question_page');

                                        );

                                        Post as a guest















                                        Required, but never shown





















































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown

































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown







                                        Popular posts from this blog

                                        Adding axes to figuresAdding axes labels to LaTeX figuresLaTeX equivalent of ConTeXt buffersRotate a node but not its content: the case of the ellipse decorationHow to define the default vertical distance between nodes?TikZ scaling graphic and adjust node position and keep font sizeNumerical conditional within tikz keys?adding axes to shapesAlign axes across subfiguresAdding figures with a certain orderLine up nested tikz enviroments or how to get rid of themAdding axes labels to LaTeX figures

                                        Tähtien Talli Jäsenet | Lähteet | NavigointivalikkoSuomen Hippos – Tähtien Talli

                                        Do these cracks on my tires look bad? The Next CEO of Stack OverflowDry rot tire should I replace?Having to replace tiresFishtailed so easily? Bad tires? ABS?Filling the tires with something other than air, to avoid puncture hassles?Used Michelin tires safe to install?Do these tyre cracks necessitate replacement?Rumbling noise: tires or mechanicalIs it possible to fix noisy feathered tires?Are bad winter tires still better than summer tires in winter?Torque converter failure - Related to replacing only 2 tires?Why use snow tires on all 4 wheels on 2-wheel-drive cars?