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Using emission lines to determine redshift of a quasar


Can redshift be measured using fourier?Do contracting objects show redshift?How Do we know about redshift?Evidence of CMB redshiftRedshift quantizationRedshift observation at different wave lengthDoes wavelength affect redshift caused by the metric expansion of space?Photometric redshiftCorrectly scaling Power Spectrum and Correlation Function of Galaxies from Mock CataloguesHow to differentiate elements that have same spectral lines in a star?













2












$begingroup$


I am attempting this past paper question and am not sure how to tackle this, this is not homework! :)




The spectrum of a distant quasar shows two broad emission lines with observed
wavelengths of 317.7nm and 404.7nm. The strongest lines are most likely to be either Lyman-α with a rest wavelength 121.6nm, CIV with a rest wavelength of 154.9nm or MgII with a rest wavelength of 280.0nm.



(a) What is the redshift of the quasar? Please show your working.



(b) What two lines have been detected?



(c) The two lines have widths of 10nm and 20nm respectively. If I have measured
the mass of the black hole in the quasar to be 10^9Msun, how far from the centre of the quasar are the regions of ionised gas responsible for the emission lines?
Please specify any assumptions that you make to obtain your result.




I am assuming I must find z to be the ratio of the change in wavelength to the rest wavelength. However I'm unsure of how to identify these in the question.










share|improve this question







New contributor



Minish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$
















    2












    $begingroup$


    I am attempting this past paper question and am not sure how to tackle this, this is not homework! :)




    The spectrum of a distant quasar shows two broad emission lines with observed
    wavelengths of 317.7nm and 404.7nm. The strongest lines are most likely to be either Lyman-α with a rest wavelength 121.6nm, CIV with a rest wavelength of 154.9nm or MgII with a rest wavelength of 280.0nm.



    (a) What is the redshift of the quasar? Please show your working.



    (b) What two lines have been detected?



    (c) The two lines have widths of 10nm and 20nm respectively. If I have measured
    the mass of the black hole in the quasar to be 10^9Msun, how far from the centre of the quasar are the regions of ionised gas responsible for the emission lines?
    Please specify any assumptions that you make to obtain your result.




    I am assuming I must find z to be the ratio of the change in wavelength to the rest wavelength. However I'm unsure of how to identify these in the question.










    share|improve this question







    New contributor



    Minish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$














      2












      2








      2





      $begingroup$


      I am attempting this past paper question and am not sure how to tackle this, this is not homework! :)




      The spectrum of a distant quasar shows two broad emission lines with observed
      wavelengths of 317.7nm and 404.7nm. The strongest lines are most likely to be either Lyman-α with a rest wavelength 121.6nm, CIV with a rest wavelength of 154.9nm or MgII with a rest wavelength of 280.0nm.



      (a) What is the redshift of the quasar? Please show your working.



      (b) What two lines have been detected?



      (c) The two lines have widths of 10nm and 20nm respectively. If I have measured
      the mass of the black hole in the quasar to be 10^9Msun, how far from the centre of the quasar are the regions of ionised gas responsible for the emission lines?
      Please specify any assumptions that you make to obtain your result.




      I am assuming I must find z to be the ratio of the change in wavelength to the rest wavelength. However I'm unsure of how to identify these in the question.










      share|improve this question







      New contributor



      Minish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$




      I am attempting this past paper question and am not sure how to tackle this, this is not homework! :)




      The spectrum of a distant quasar shows two broad emission lines with observed
      wavelengths of 317.7nm and 404.7nm. The strongest lines are most likely to be either Lyman-α with a rest wavelength 121.6nm, CIV with a rest wavelength of 154.9nm or MgII with a rest wavelength of 280.0nm.



      (a) What is the redshift of the quasar? Please show your working.



      (b) What two lines have been detected?



      (c) The two lines have widths of 10nm and 20nm respectively. If I have measured
      the mass of the black hole in the quasar to be 10^9Msun, how far from the centre of the quasar are the regions of ionised gas responsible for the emission lines?
      Please specify any assumptions that you make to obtain your result.




      I am assuming I must find z to be the ratio of the change in wavelength to the rest wavelength. However I'm unsure of how to identify these in the question.







      redshift wavelength spectrum






      share|improve this question







      New contributor



      Minish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.










      share|improve this question







      New contributor



      Minish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      share|improve this question




      share|improve this question






      New contributor



      Minish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      asked Jun 10 at 10:58









      MinishMinish

      132




      132




      New contributor



      Minish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




      New contributor




      Minish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Steve Linton answers correctly that the line ratios must be (nearly) identical and hence identifies the two lines as Lyman $alpha$ and C IV, but it's not true that the two obtained redshifts are suspiciously alike. In fact, when you do the calculation, you get (from $z equiv lambda_mathrmobs/lambda_mathrmrest - 1$),
          $$
          beginarrayrcl
          z_mathrmLyalpha & = & frac317.7,mathrmnm125.67,mathrmnm & simeq & 1.6134 \
          z_mathrmC,IV & = & frac404.7,mathrmnm154.9,mathrmnm & simeq & 1.6127.
          endarray
          $$

          That is, Lyman $alpha$ is slightly more redshifted than C IV. This is often seen, and is due to Lyman $alpha$ scattering resonantly on the neutral hydrogen enshrouding the quasar / host galaxy. This effect is more pronounced the higher the redshift, and by $zsimeq6$, the blue part of the spectrum is completely gone (this is the so-called Gunn-Peterson trough).



          Relating the widths $Deltalambda$ of the lines to the motion around the black hole, the velocity of the gas responsible for emitting the lines is given by $v/c = Deltalambda / lambda$, so the velocities are
          $$
          beginarrayrcl
          v_mathrmLyalpha & = & c frac10,mathrmnm317.7,mathrmnm & simeq & 9,400,mathrmkm,mathrms^-1 \
          v_mathrmC,IV & = & c frac20,mathrmnm404.7,mathrmnm & simeq & 14,800,mathrmkm,mathrms^-1.
          endarray
          $$

          From simple dynamics, this can be converted to a distance $r$ from the black hole, knowing its mass $M_bullet$:
          $$
          beginarrayrcl
          r & = & fracG M_bulletv^2 & Rightarrow \
          r_mathrmLyalpha & simeq & 0.05,mathrmpc & simeq & 58,mathrmlighttext-days \
          r_mathrmC,IV & simeq & 0.02,mathrmpc & simeq & 23,mathrmlighttext-days,
          endarray
          $$

          which is consistent with the fact that these lines are formed in the so-called broad line region which has dimension from tens of light-days to several parsec (in contrast, the narrow line region has dimensions of several 100s to ~1000 light-years).






          share|improve this answer









          $endgroup$




















            4












            $begingroup$

            So the two observed lines must be two of the three suggested lines, red-shifted by the same amount. That means the ratio of their wavelengths will be unchanged, so we need two of the suggested lines whose wavelength is close to the ratio $404.7/317.6$. It's easy to check that only the first two are close to that ratio, so they must be those two. So the redshift should be equal to $317.7/121.6 = 2.61$ and to $404.7/154.9$ which is also $2.61$ (such close agreement in real data would be a bit unlikely, and lead one to check the authenticity of the data.






            share|improve this answer









            $endgroup$








            • 1




              $begingroup$
              You forgot to subtract $1$ from the ratio, so the redshift is only 1.61. More precisely, $z_mathrmLyalpha = 1.6134$ and $z_mathrmC,IV = 1.6127$, so Lyα is a bit more redshifted than C IV which is not unusual, and may be expected due to to resonant scattering.
              $endgroup$
              – pela
              Jun 10 at 17:52











            Your Answer








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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






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            active

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            2












            $begingroup$

            Steve Linton answers correctly that the line ratios must be (nearly) identical and hence identifies the two lines as Lyman $alpha$ and C IV, but it's not true that the two obtained redshifts are suspiciously alike. In fact, when you do the calculation, you get (from $z equiv lambda_mathrmobs/lambda_mathrmrest - 1$),
            $$
            beginarrayrcl
            z_mathrmLyalpha & = & frac317.7,mathrmnm125.67,mathrmnm & simeq & 1.6134 \
            z_mathrmC,IV & = & frac404.7,mathrmnm154.9,mathrmnm & simeq & 1.6127.
            endarray
            $$

            That is, Lyman $alpha$ is slightly more redshifted than C IV. This is often seen, and is due to Lyman $alpha$ scattering resonantly on the neutral hydrogen enshrouding the quasar / host galaxy. This effect is more pronounced the higher the redshift, and by $zsimeq6$, the blue part of the spectrum is completely gone (this is the so-called Gunn-Peterson trough).



            Relating the widths $Deltalambda$ of the lines to the motion around the black hole, the velocity of the gas responsible for emitting the lines is given by $v/c = Deltalambda / lambda$, so the velocities are
            $$
            beginarrayrcl
            v_mathrmLyalpha & = & c frac10,mathrmnm317.7,mathrmnm & simeq & 9,400,mathrmkm,mathrms^-1 \
            v_mathrmC,IV & = & c frac20,mathrmnm404.7,mathrmnm & simeq & 14,800,mathrmkm,mathrms^-1.
            endarray
            $$

            From simple dynamics, this can be converted to a distance $r$ from the black hole, knowing its mass $M_bullet$:
            $$
            beginarrayrcl
            r & = & fracG M_bulletv^2 & Rightarrow \
            r_mathrmLyalpha & simeq & 0.05,mathrmpc & simeq & 58,mathrmlighttext-days \
            r_mathrmC,IV & simeq & 0.02,mathrmpc & simeq & 23,mathrmlighttext-days,
            endarray
            $$

            which is consistent with the fact that these lines are formed in the so-called broad line region which has dimension from tens of light-days to several parsec (in contrast, the narrow line region has dimensions of several 100s to ~1000 light-years).






            share|improve this answer









            $endgroup$

















              2












              $begingroup$

              Steve Linton answers correctly that the line ratios must be (nearly) identical and hence identifies the two lines as Lyman $alpha$ and C IV, but it's not true that the two obtained redshifts are suspiciously alike. In fact, when you do the calculation, you get (from $z equiv lambda_mathrmobs/lambda_mathrmrest - 1$),
              $$
              beginarrayrcl
              z_mathrmLyalpha & = & frac317.7,mathrmnm125.67,mathrmnm & simeq & 1.6134 \
              z_mathrmC,IV & = & frac404.7,mathrmnm154.9,mathrmnm & simeq & 1.6127.
              endarray
              $$

              That is, Lyman $alpha$ is slightly more redshifted than C IV. This is often seen, and is due to Lyman $alpha$ scattering resonantly on the neutral hydrogen enshrouding the quasar / host galaxy. This effect is more pronounced the higher the redshift, and by $zsimeq6$, the blue part of the spectrum is completely gone (this is the so-called Gunn-Peterson trough).



              Relating the widths $Deltalambda$ of the lines to the motion around the black hole, the velocity of the gas responsible for emitting the lines is given by $v/c = Deltalambda / lambda$, so the velocities are
              $$
              beginarrayrcl
              v_mathrmLyalpha & = & c frac10,mathrmnm317.7,mathrmnm & simeq & 9,400,mathrmkm,mathrms^-1 \
              v_mathrmC,IV & = & c frac20,mathrmnm404.7,mathrmnm & simeq & 14,800,mathrmkm,mathrms^-1.
              endarray
              $$

              From simple dynamics, this can be converted to a distance $r$ from the black hole, knowing its mass $M_bullet$:
              $$
              beginarrayrcl
              r & = & fracG M_bulletv^2 & Rightarrow \
              r_mathrmLyalpha & simeq & 0.05,mathrmpc & simeq & 58,mathrmlighttext-days \
              r_mathrmC,IV & simeq & 0.02,mathrmpc & simeq & 23,mathrmlighttext-days,
              endarray
              $$

              which is consistent with the fact that these lines are formed in the so-called broad line region which has dimension from tens of light-days to several parsec (in contrast, the narrow line region has dimensions of several 100s to ~1000 light-years).






              share|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                Steve Linton answers correctly that the line ratios must be (nearly) identical and hence identifies the two lines as Lyman $alpha$ and C IV, but it's not true that the two obtained redshifts are suspiciously alike. In fact, when you do the calculation, you get (from $z equiv lambda_mathrmobs/lambda_mathrmrest - 1$),
                $$
                beginarrayrcl
                z_mathrmLyalpha & = & frac317.7,mathrmnm125.67,mathrmnm & simeq & 1.6134 \
                z_mathrmC,IV & = & frac404.7,mathrmnm154.9,mathrmnm & simeq & 1.6127.
                endarray
                $$

                That is, Lyman $alpha$ is slightly more redshifted than C IV. This is often seen, and is due to Lyman $alpha$ scattering resonantly on the neutral hydrogen enshrouding the quasar / host galaxy. This effect is more pronounced the higher the redshift, and by $zsimeq6$, the blue part of the spectrum is completely gone (this is the so-called Gunn-Peterson trough).



                Relating the widths $Deltalambda$ of the lines to the motion around the black hole, the velocity of the gas responsible for emitting the lines is given by $v/c = Deltalambda / lambda$, so the velocities are
                $$
                beginarrayrcl
                v_mathrmLyalpha & = & c frac10,mathrmnm317.7,mathrmnm & simeq & 9,400,mathrmkm,mathrms^-1 \
                v_mathrmC,IV & = & c frac20,mathrmnm404.7,mathrmnm & simeq & 14,800,mathrmkm,mathrms^-1.
                endarray
                $$

                From simple dynamics, this can be converted to a distance $r$ from the black hole, knowing its mass $M_bullet$:
                $$
                beginarrayrcl
                r & = & fracG M_bulletv^2 & Rightarrow \
                r_mathrmLyalpha & simeq & 0.05,mathrmpc & simeq & 58,mathrmlighttext-days \
                r_mathrmC,IV & simeq & 0.02,mathrmpc & simeq & 23,mathrmlighttext-days,
                endarray
                $$

                which is consistent with the fact that these lines are formed in the so-called broad line region which has dimension from tens of light-days to several parsec (in contrast, the narrow line region has dimensions of several 100s to ~1000 light-years).






                share|improve this answer









                $endgroup$



                Steve Linton answers correctly that the line ratios must be (nearly) identical and hence identifies the two lines as Lyman $alpha$ and C IV, but it's not true that the two obtained redshifts are suspiciously alike. In fact, when you do the calculation, you get (from $z equiv lambda_mathrmobs/lambda_mathrmrest - 1$),
                $$
                beginarrayrcl
                z_mathrmLyalpha & = & frac317.7,mathrmnm125.67,mathrmnm & simeq & 1.6134 \
                z_mathrmC,IV & = & frac404.7,mathrmnm154.9,mathrmnm & simeq & 1.6127.
                endarray
                $$

                That is, Lyman $alpha$ is slightly more redshifted than C IV. This is often seen, and is due to Lyman $alpha$ scattering resonantly on the neutral hydrogen enshrouding the quasar / host galaxy. This effect is more pronounced the higher the redshift, and by $zsimeq6$, the blue part of the spectrum is completely gone (this is the so-called Gunn-Peterson trough).



                Relating the widths $Deltalambda$ of the lines to the motion around the black hole, the velocity of the gas responsible for emitting the lines is given by $v/c = Deltalambda / lambda$, so the velocities are
                $$
                beginarrayrcl
                v_mathrmLyalpha & = & c frac10,mathrmnm317.7,mathrmnm & simeq & 9,400,mathrmkm,mathrms^-1 \
                v_mathrmC,IV & = & c frac20,mathrmnm404.7,mathrmnm & simeq & 14,800,mathrmkm,mathrms^-1.
                endarray
                $$

                From simple dynamics, this can be converted to a distance $r$ from the black hole, knowing its mass $M_bullet$:
                $$
                beginarrayrcl
                r & = & fracG M_bulletv^2 & Rightarrow \
                r_mathrmLyalpha & simeq & 0.05,mathrmpc & simeq & 58,mathrmlighttext-days \
                r_mathrmC,IV & simeq & 0.02,mathrmpc & simeq & 23,mathrmlighttext-days,
                endarray
                $$

                which is consistent with the fact that these lines are formed in the so-called broad line region which has dimension from tens of light-days to several parsec (in contrast, the narrow line region has dimensions of several 100s to ~1000 light-years).







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Jun 10 at 22:55









                pelapela

                18.1k3965




                18.1k3965





















                    4












                    $begingroup$

                    So the two observed lines must be two of the three suggested lines, red-shifted by the same amount. That means the ratio of their wavelengths will be unchanged, so we need two of the suggested lines whose wavelength is close to the ratio $404.7/317.6$. It's easy to check that only the first two are close to that ratio, so they must be those two. So the redshift should be equal to $317.7/121.6 = 2.61$ and to $404.7/154.9$ which is also $2.61$ (such close agreement in real data would be a bit unlikely, and lead one to check the authenticity of the data.






                    share|improve this answer









                    $endgroup$








                    • 1




                      $begingroup$
                      You forgot to subtract $1$ from the ratio, so the redshift is only 1.61. More precisely, $z_mathrmLyalpha = 1.6134$ and $z_mathrmC,IV = 1.6127$, so Lyα is a bit more redshifted than C IV which is not unusual, and may be expected due to to resonant scattering.
                      $endgroup$
                      – pela
                      Jun 10 at 17:52















                    4












                    $begingroup$

                    So the two observed lines must be two of the three suggested lines, red-shifted by the same amount. That means the ratio of their wavelengths will be unchanged, so we need two of the suggested lines whose wavelength is close to the ratio $404.7/317.6$. It's easy to check that only the first two are close to that ratio, so they must be those two. So the redshift should be equal to $317.7/121.6 = 2.61$ and to $404.7/154.9$ which is also $2.61$ (such close agreement in real data would be a bit unlikely, and lead one to check the authenticity of the data.






                    share|improve this answer









                    $endgroup$








                    • 1




                      $begingroup$
                      You forgot to subtract $1$ from the ratio, so the redshift is only 1.61. More precisely, $z_mathrmLyalpha = 1.6134$ and $z_mathrmC,IV = 1.6127$, so Lyα is a bit more redshifted than C IV which is not unusual, and may be expected due to to resonant scattering.
                      $endgroup$
                      – pela
                      Jun 10 at 17:52













                    4












                    4








                    4





                    $begingroup$

                    So the two observed lines must be two of the three suggested lines, red-shifted by the same amount. That means the ratio of their wavelengths will be unchanged, so we need two of the suggested lines whose wavelength is close to the ratio $404.7/317.6$. It's easy to check that only the first two are close to that ratio, so they must be those two. So the redshift should be equal to $317.7/121.6 = 2.61$ and to $404.7/154.9$ which is also $2.61$ (such close agreement in real data would be a bit unlikely, and lead one to check the authenticity of the data.






                    share|improve this answer









                    $endgroup$



                    So the two observed lines must be two of the three suggested lines, red-shifted by the same amount. That means the ratio of their wavelengths will be unchanged, so we need two of the suggested lines whose wavelength is close to the ratio $404.7/317.6$. It's easy to check that only the first two are close to that ratio, so they must be those two. So the redshift should be equal to $317.7/121.6 = 2.61$ and to $404.7/154.9$ which is also $2.61$ (such close agreement in real data would be a bit unlikely, and lead one to check the authenticity of the data.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Jun 10 at 11:08









                    Steve LintonSteve Linton

                    3,7431425




                    3,7431425







                    • 1




                      $begingroup$
                      You forgot to subtract $1$ from the ratio, so the redshift is only 1.61. More precisely, $z_mathrmLyalpha = 1.6134$ and $z_mathrmC,IV = 1.6127$, so Lyα is a bit more redshifted than C IV which is not unusual, and may be expected due to to resonant scattering.
                      $endgroup$
                      – pela
                      Jun 10 at 17:52












                    • 1




                      $begingroup$
                      You forgot to subtract $1$ from the ratio, so the redshift is only 1.61. More precisely, $z_mathrmLyalpha = 1.6134$ and $z_mathrmC,IV = 1.6127$, so Lyα is a bit more redshifted than C IV which is not unusual, and may be expected due to to resonant scattering.
                      $endgroup$
                      – pela
                      Jun 10 at 17:52







                    1




                    1




                    $begingroup$
                    You forgot to subtract $1$ from the ratio, so the redshift is only 1.61. More precisely, $z_mathrmLyalpha = 1.6134$ and $z_mathrmC,IV = 1.6127$, so Lyα is a bit more redshifted than C IV which is not unusual, and may be expected due to to resonant scattering.
                    $endgroup$
                    – pela
                    Jun 10 at 17:52




                    $begingroup$
                    You forgot to subtract $1$ from the ratio, so the redshift is only 1.61. More precisely, $z_mathrmLyalpha = 1.6134$ and $z_mathrmC,IV = 1.6127$, so Lyα is a bit more redshifted than C IV which is not unusual, and may be expected due to to resonant scattering.
                    $endgroup$
                    – pela
                    Jun 10 at 17:52










                    Minish is a new contributor. Be nice, and check out our Code of Conduct.









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                    Middle Expansion Olielle Resaix Definition: Uttering songs of triumph shouting with joy triumphant exulting Sejunction Journal 붙다 달 고급 품목 외출 The stretch trades the screeching tin. Definition: The act of speaking with a drawl a drawl Cough Sand Definition: An uproar a quarrel a noisy outbreak Shake Iron Publicize Horse House Baby 사과 Resaix Flaggy Jelly Temporary Unequaled Puppet A drop in the bucket Shrew 성격 회원 성질 미팅 The burn frames the tacky quality. Materialistic The smoke reduces the way. Yammoe Nondescript Cheek 얼굴 배 약하다 날리다 타다 The illegal country shows the iron. Help Rule Drearien Smoke Teaching Meaty Wasp Abraham Lincoln Jaws 진심 수리하다 Size Cork Idea Convert Think Lark John Lennon 거울 청소 군 추천하다 아이스크림