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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?
$begingroup$
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.
redshift wavelength spectrum
New contributor
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add a comment |
$begingroup$
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.
redshift wavelength spectrum
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.
$endgroup$
add a comment |
$begingroup$
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.
redshift wavelength spectrum
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
redshift wavelength spectrum
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.
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
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Check out our Code of Conduct.
add a comment |
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2 Answers
2
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$begingroup$
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).
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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).
$endgroup$
add a comment |
$begingroup$
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).
$endgroup$
add a comment |
$begingroup$
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).
$endgroup$
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).
answered Jun 10 at 22:55
pelapela
18.1k3965
18.1k3965
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
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
add a comment |
Minish is a new contributor. Be nice, and check out our Code of Conduct.
Minish is a new contributor. Be nice, and check out our Code of Conduct.
Minish is a new contributor. Be nice, and check out our Code of Conduct.
Minish is a new contributor. Be nice, and check out our Code of Conduct.
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