18 replies to this topic

[mwr]

M type stars show typically intense and broad TiO molecular absorption bands. These stars are thus very rewarding targets for low resolution spectroscopy using the Star Analyser. The band spectrum of TiO falls into the category of molecular spectrum known as vibronic. The structural features of this spectrum result from transitions involving vibrational substates of electronic states. These structural features can be, indeed, resolved using a SA-200 grism. However, the assignments of the structural features to distinct electronic and vibrational states can be quite cumbersome because the classical books in spectroscopy for amateur astronomers do not contain the relevant information (e.g. Walker's Spectroscopic Altas or Gray's and Corbally's Stellar Spectral Classification). A search in the primary literature is necessary and this task was very welcome to me in times of Corona lock-downs and cloudy nights:

The star for this purpose was U Ceti, which is a Mira variable star. It's spectrum was captured 01/10/21. Differential photometry using the green channel of my DSLR yielded an apparent brightness of 8.4 and shows that U Ceti is close to its maximum when compared with AAVSO data:

 

 

H gamma and H delta lines are clearly in emission due to shock waves:

 

 

The transitions between the electronic energy states could be assigned to the corresponding bands on the basis of this paper: http://adsabs.harvar...ApJS...26..313P

 

 

 

The resolution of the SA-200 was high enough to resolve the vibrational structure of most of the bands and allowed an assignment on the basis of this paper: https://iopscience.i...owBJzibF1DHah3E

 

 

There are still some unassigned TiO bands in the near infrared part of the spectrum around 8500 A and hints for other publications on this topic would be very welcome.

Edited by mwr, 13 January 2021 - 10:10 AM.

[Tangerman]

In my quantum chemistry course I took last semester, we calculated the transitions of bimolecular species using Morse potentials for the ground and excited states. With good Morse parameters, I can calculate the transitions and their relative intensities to a very high precision. If I remember, I'll look for some Morse parameters for TiO later today and come back with a bit more information. If I don't remember, feel free to message me. 

[c2m2t]

Good Morning!

When it comes to spectrum analysis, I am way out of my league. I happened to see this post in the "Recent Topics" column, and it caught my eye. I am a double star imager and Mira happens to be one of those special systems that exhibits both variability and colour. Oddly enough, the variable nature of it was something I had not really become aware of until circumstances of late 2017 and early 2018 unfolded. I live in Eastern Ontario, Canada where skies for amateur astronomers, like myself, have been steadily degrading in the last 30 years. I became sensitized to this when I took up the hobby about 13 to 14 years ago. As a result, I began hatching plans to make trips into the south-central and southwestern States during November new moon periods to indulge myself in the absorption of photons. These trips became intensive imaging sessions to capture double stars that, at 45 degrees latitude, were hugging my local horizon...a murky area not well suited for imaging. 2017 was my first trip south and I ended up at Copper Breaks State Park, Texas.

 

The image I captured of Mira, aka H VI 1, a double star system first catalogued by William Herschel in 1782... you will recognize it by the Nov. 17, 2017 date...the cropped image in the upper right. It was not a particularly good  image, I thought, the transparency was not particularly good that night, but I did enjoy the very red colour that was captured. So, thinking that it was not a very good image, I decided to re-image it again Jan. 13, 2018 from my home observatory. The later image turned out much better, in spite of having Mira closer to my local horizon. Oddly enough, I did not give it much thought at that point and attributed the redness of the earlier image a result, more of the hazy skies that night in Texas than of it actually showing that much red at what was apparently being near its minimum. 

 

Mira is actually a 4 star system (visible stars). None of the pairings made up of A,B,C and D, are physical systems. The only true binary pair is the very compact pair that make up the primary star "A". They have the WDS (Washington Double Star Catalogue) designation JOY 1Aa,Ab. For you spectrum analysts, it would be interesting to know if and how much this companion star impacts the variability of the system. The WDS provide component magnitudes of 6.80 and 10.40 with a separation of 0.5 arc-seconds (2014). I have also attached a screen capture from Stelle Doppie showing some of the notes about Joy 1 that are contained in the WDS.

 

I have also attached an image of the Mira system.

 

Cheers, Chris.

Attached Thumbnails

• 
• 

[Tangerman]

Okay, looking more closely at your post, the TiO is going between various electronic states, which would mean finding way too many Morse potentials, which actually wouldn't be too useful. Doing what I had suggested is good for differentiating between different vibrational states, not good for differentiating between electronic states. So, here's what I have instead: https://webbook.nist...37201&Mask=1000

This is a link to the nist database, which is quite useful for spectroscopic information of simple molecules. The very left column tells you which state. The bottom row here is the ground state. The v₀₀ (very right column) tells you the energy between the ground electronic state and the excited electronic state. This column is in wavenumbers (cm^-1, a common unit for spectroscopists, but nobody else uses it. It makes nice numbers when you're doing IR spectroscopy). You can convert to nm by looking up wavenumber to nm, or you can do it yourself. For example, if the column says 10000 cm^-1, then the wavelength is 1/(10000 cm^-1) = 1 cm/10000. Then multiply by 10⁷ nm/cm, and you have it in nm (and 10⁸ Å/cm if you prefer). If this is unclear, please let me know.

So, 850 nm is about 11765 cm^-1, which is pretty close to the E ³∏ transition. That is an allowed transition from the ground state, and thus I would expect it to be intense enough to see. Hopefully this clears it up, and in the future if you have questions about simple molecules and spectral assignments, you have a new resource.

[mwr]

 

So, 850 nm is about 11765 cm^-1, which is pretty close to the E ³∏ transition. That is an allowed transition from the ground state, and thus I would expect it to be intense enough to see. Hopefully this clears it up, and in the future if you have questions about simple molecules and spectral assignments, you have a new resource.

Thanks Tangerman! With the help of the NIST database and the references given there I was able to find "THE INFRARED TELESCOPE FACILITY (IRTF) SPECTRAL LIBRARY" for cool stars. It is indeed the E ³∏ transition.

[mwr]

Thanks Tangerman! With the help of the NIST database and the references given there I was able to find "THE INFRARED TELESCOPE FACILITY (IRTF) SPECTRAL LIBRARY" for cool stars. It is indeed the E ³∏ transition.

I have revised the assignment of the electronic transitions and have included the missing E ³∏ transition (the so called epsilon transition) which partially overlaps with the gamma  transition as noted in the IRTF library:

 

"[...] two additional band heads of similar depths exist at 0.8508 and
0.8582 μm. Together, the six-band heads appear to form two
sets of triplet band heads.

The 1–1 and 2–2 bands of the epsilon

system exhibit band heads at these wavelengths and therefore it
is likely that these features arise from the epsilon system alone or are
a combination of the gamma and epsilon systems."

 

 

When the spectrum of U Ceti is compared with a M4 III standard star spectrum from the Pickles library the weakness of Ca I and Ca II lines is obvious:

 

 

This kind of line-weakening, not due to an abundance effect, is called "veilling" and a typical phenomenon in Mira type stars. However, I'm not sure if this is really the case here. Comments are welcome. 

[Organic Astrochemist]

Like many terms in astronomy "veiling" probably isn't very helpful and is applied in multiple contexts.

 

One thing that is going on is that the higher Balmer lines are in emission and H-epsilon should also be in emission, but it falls on top of the Ca II H line, so those photons pump the emission that you see in Ca II at 8662

here for example

 

You have rectified your spectra, but I find that the bluer continuum seems stronger near maximum light. I'm not sure if the continuum is greater or there are just many shock-induced emissions. In any case, I can see how these could fill in absorptions and cause "veiling".

Suggested here for T Tauri stars

 

The whole idea of what is a photosphere is kind of complicated for Miras both because it changes so much based on wavelength and based on pulsation cycle. If a shorter-wavelength absorption (near the photosphere for the shorter wavelength) is blocked by molecules or dust at much greater distance, has the absorption been "veiled"?

 

I found these older lists of fluorescent lines for atoms and molecules (Ca II at 8662 isn't listed).

atomic lines

molecular lines

[Organic Astrochemist]

Okay, looking more closely at your post, the TiO is going between various electronic states, which would mean finding way too many Morse potentials, which actually wouldn't be too useful. Doing what I had suggested is good for differentiating between different vibrational states, not good for differentiating between electronic states. So, here's what I have instead: https://webbook.nist...37201&Mask=1000

This is a link to the nist database, which is quite useful for spectroscopic information of simple molecules. The very left column tells you which state. The bottom row here is the ground state. The v₀₀ (very right column) tells you the energy between the ground electronic state and the excited electronic state. This column is in wavenumbers (cm^-1, a common unit for spectroscopists, but nobody else uses it. It makes nice numbers when you're doing IR spectroscopy). You can convert to nm by looking up wavenumber to nm, or you can do it yourself. For example, if the column says 10000 cm^-1, then the wavelength is 1/(10000 cm^-1) = 1 cm/10000. Then multiply by 10⁷ nm/cm, and you have it in nm (and 10⁸ Å/cm if you prefer). If this is unclear, please let me know.

So, 850 nm is about 11765 cm^-1, which is pretty close to the E ³∏ transition. That is an allowed transition from the ground state, and thus I would expect it to be intense enough to see. Hopefully this clears it up, and in the future if you have questions about simple molecules and spectral assignments, you have a new resource.

One aspect of these types of spectra that mwr and myself have discussed is the shape of these TiO bands. Specifically, how the bandhead is on the left or blue side (in the P-branch). 

 

The data you have given us here shows why. For these transitions the rotational constant Be is smaller in the lower state than in the higher state. To me this is a little surprising because some of the higher states actually have shorter internuclear distances, re (which is a bit less of a surprise to me).

 

This reference suggests that "the rotational constant  in the upper state can now either be larger or smaller than in the lower electronic state. This depends on the binding energies and the equilibrium distances Re in the two

states."

Demtroeder

 

I guess to find the bandhead to the right or on the red side (in the R-branch) the upper state would need to have both a shorter internuclear distance than the lower state and a also have a binding energy that was only slightly smaller than the lower state. I think this can only happen if the lower state is NOT, in fact, the ground state.

 

Any thoughts on this?

Edited by Organic Astrochemist, 16 January 2021 - 02:04 AM.

[mwr]

Like many terms in astronomy "veiling" probably isn't very helpful and is applied in multiple contexts.

 

One thing that is going on is that the higher Balmer lines are in emission and H-epsilon should also be in emission, but it falls on top of the Ca II H line, so those photons pump the emission that you see in Ca II at 8662

here for example

 

 

This reference is very helpful and the scattering  of H epsilon photons by the Ca II H line out to IR wavelengths via the Ca II 8662 A line is highly interesting. I had to consult the corresponding Grotrian diagram of Ca II to really understand what's actually going on:

 

 

So the "veilling" of the Ca II lines in this case is actually a "filling-in". Thanks Jim for pointing this out! 

Edited by mwr, 16 January 2021 - 07:55 AM.

[mwr]

 

 

I guess to find the bandhead to the right or on the red side (in the R-branch) the upper state would need to have both a shorter internuclear distance than the lower state and a also have a binding energy that was only slightly smaller than the lower state. I think this can only happen if the lower state is NOT, in fact, the ground state.

 

Any thoughts on this?

What about the CH methylidyne bands of CH carbon stars? Ground state is X²P with r_e=1.1199. The band head at 4320 Angström can be assigned to the R₂ branch of the A²D - X²P transition. For the A²D state r_e is 1.1019.

https://webbook.nist...Mask=1000#ref-7

Edited by mwr, 16 January 2021 - 12:43 PM.

[Tangerman]

One aspect of these types of spectra that mwr and myself have discussed is the shape of these TiO bands. Specifically, how the bandhead is on the left or blue side (in the P-branch). 

 

The data you have given us here shows why. For these transitions the rotational constant Be is smaller in the lower state than in the higher state. To me this is a little surprising because some of the higher states actually have shorter internuclear distances, re (which is a bit less of a surprise to me).

 

This reference suggests that "the rotational constant  in the upper state can now either be larger or smaller than in the lower electronic state. This depends on the binding energies and the equilibrium distances Re in the two

states."

Demtroeder

 

I guess to find the bandhead to the right or on the red side (in the R-branch) the upper state would need to have both a shorter internuclear distance than the lower state and a also have a binding energy that was only slightly smaller than the lower state. I think this can only happen if the lower state is NOT, in fact, the ground state.

 

Any thoughts on this?

I can think of a few things that may explain the band head being on the blue side. First, the Frank-Condon principle. The overlap of the states might be best for going to, say, v'=7 (where v' is the excited vibrational state). Then, you might see the band going to v'=6, which would be lower energy, and more bands, all the way to v'=0. Because of anharmonicity, the vibrational states get closer together as you go higher, so there's a larger (potentially much larger) difference between going to v'=0 and v'=7 than there is between v'=7 and v'=14. These higher bands would appear overlapped and not be readily visible, so the band head, where all the overlapping bands are contributing to some intensity of the transition, would be on the blue side, with the red side more spread out. 

 

Additionally, we certainly are not bound to the ground vibrational state. If you have enough energy to get to excited electronic states, you have enough to get to excited vibrational states, so you are correct, we may not be starting from the ground state, although I suspect we are starting from the ground electronic states. 

 

Third, that rotational constant. When the equilibrium bond length becomes larger, as is often the case in excited electronic states, the rotational constant then often decreases dramatically, which causes some odd things. The R branch can actually reverse on itself (so R(20) is to the red of R(1)), which can form a band head with the rest of the spectrum on the red side. If the rotational constant dramatically increases, then the P branch can reverse on itself and form a band head, with the rest of the spectrum to the blue side. 

 

So I think you are by and large correct, organic astrochemist, that to get a band head on the blue side would be most easily accomplished if you aren't going from the ground state, but are instead decreasing the bond length by going to the new state. However, I'm not sure if this would be enough to overcome Frank-Condon factors. Without the ability to see the fine structure of the electronic transitions here (not an issue with the spectra you've taken mwr, I'd only expect to be able to see this with gas-phase spectroscopy, and most easily with supersonic jet spectroscopy), I can't definitively say what causes that band head to be on the blue side. 

[Organic Astrochemist]

What about the CH methylidyne bands of CH carbon stars? Ground state is X²P with r_e=1.1199. The band head at 4320 Angström can be assigned to the R₂ branch of the A²D - X²P transition. For the A²D state r_e is 1.1019.

https://webbook.nist...Mask=1000#ref-7

OK so I'm really not sure I understand this but this is a great example to test that.

 

I looked up CH carbon stars in Gray and Corbally and figure 8.20 shows HD 26 which has CH bands in the R-branch, the P branch and the Q-branch! What is going on?

 

My limited understanding would suggest that the transition from the ground state X²Π to the A²Δ state would be in the P branch because Be for X²Π = 14.457 24 25 9 and for A²Δ = 14.934 20 9 so the higher state is greater than the lower state (curiously written as B''<B'). Gray and Corbally seem to agree with me labelling the band redward of the G-band as P. How can you predict the wavelength of the transition?

 

It does seem like the transition from X²Π to B²Σ+ could have a bandhead in the R-branch. Corbally and Gray show a bandhead in the R-branch, something has to cause that.

 

The Q branch is supposed to happen when B''=B'. The transition from X²Π to C²Σ+ has very similar Be values and seems like an allowed transition. Could this be the origin of the G-band? I think the Q-band is supposed to be common in free radicals.

 

Do I have any of this right?

 

Here is probably more detail than I can use.

CH

[Organic Astrochemist]

I can think of a few things that may explain the band head being on the blue side. First, the Frank-Condon principle. The overlap of the states might be best for going to, say, v'=7 (where v' is the excited vibrational state). Then, you might see the band going to v'=6, which would be lower energy, and more bands, all the way to v'=0. Because of anharmonicity, the vibrational states get closer together as you go higher, so there's a larger (potentially much larger) difference between going to v'=0 and v'=7 than there is between v'=7 and v'=14. These higher bands would appear overlapped and not be readily visible, so the band head, where all the overlapping bands are contributing to some intensity of the transition, would be on the blue side, with the red side more spread out.

Additionally, we certainly are not bound to the ground vibrational state. If you have enough energy to get to excited electronic states, you have enough to get to excited vibrational states, so you are correct, we may not be starting from the ground state, although I suspect we are starting from the ground electronic states.

Third, that rotational constant. When the equilibrium bond length becomes larger, as is often the case in excited electronic states, the rotational constant then often decreases dramatically, which causes some odd things. The R branch can actually reverse on itself (so R(20) is to the red of R(1)), which can form a band head with the rest of the spectrum on the red side. If the rotational constant dramatically increases, then the P branch can reverse on itself and form a band head, with the rest of the spectrum to the blue side.

So I think you are by and large correct, organic astrochemist, that to get a band head on the blue side would be most easily accomplished if you aren't going from the ground state, but are instead decreasing the bond length by going to the new state. However, I'm not sure if this would be enough to overcome Frank-Condon factors. Without the ability to see the fine structure of the electronic transitions here (not an issue with the spectra you've taken mwr, I'd only expect to be able to see this with gas-phase spectroscopy, and most easily with supersonic jet spectroscopy), I can't definitively say what causes that band head to be on the blue side.

Thanks Tangerman,
I’m (still) trying to get a physical understanding of these electronic transitions. If we were only talking about rotation and vibration I understand that higher excited states would have greater equilibrium distances and smaller rotational constants.

[mwr]

Gray and Corbally seem to agree with me labelling the band redward of the G-band as P. How can you predict the wavelength of the transition?

 

It does seem like the transition from X²Π to B²Σ+ could have a bandhead in the R-branch. Corbally and Gray show a bandhead in the R-branch, something has to cause that.

 

 

Hi Jim. Here is my understanding of the CH band assignment:

 

The bandhead in the R-branch in Gray and Corbally's example (Figures 8.20 and 8.21) can be rather assigned to the ground state X²Π to the A²Δ state transition:

 

 

taken from https://onlinelibrar..._phYQGZXSHTGVp9

 

v₀₀ is at 4310 Angström (see NIST) with the bandhead at approx. 4320 A. 

 

So this transition from the ground state has P, Q and R branches (see Table II here: http://bernath.uwate...les/1991/86.pdf).

 

The complete assignments of the CH, C2 and CN bands in carbon stars would be another nice exercise for cloudy nights....

[Tangerman]

Thanks Tangerman,
I’m (still) trying to get a physical understanding of these electronic transitions. If we were only talking about rotation and vibration I understand that higher excited states would have greater equilibrium distances and smaller rotational constants.

Electronic transitions are weird. There are some things that usually happen (like usually the equilibrium bond length increases, usually molecules are less tightly bound in excited electronic states), but there are so many exceptions that I don't think anyone can predict without computational modeling what will happen at an excited electronic state. As an example, take the excimer laser (which neither utilizes an excimer, but rather an exciplex, nor is generally regarded as a true laser by some). XeCl and other similar compounds used in these lasers have a bound excited state and a dissociative ground state, which just doesn't stand to reason. Yet, it happens. So don't feel bad for not understanding electronic transitions very well, nobody does. If it's the Frank-Condon principle you want to understand more about, I can provide some physical interpretation of what's happening there.

[Organic Astrochemist]

This has been an incredibly educational thread. Thanks to mwr and Tangerman.

 

I think I understand much better how to identify electronic and vibrational transitions. I am not sure what they mean.

 

What, if anything,  can we infer about the Franck Condon overlap (the effect of internuclear distance) or the effect of temperature on the spectrum?

 

 

I've drawn a red box on mwr's spectrum. This shape seems consistent with a transition between two electronic states with similar internuclear distances in the lowest vibrational level. Is that a fair inference?

 

What would it take to produce a band with the most intense absorption on the right rather than on the left? A shorter internuclear distance in the excited state?

 

The TiO absorptions are known to change in the pulsational cycle. Is the effect of temperature on vibronic spectra apparent in our spectra?

 

Thanks.

[mwr]

This has been an incredibly educational thread. Thanks to mwr and Tangerman.

 

 

Thanks to you for your sound input and precise literature recommendations. Always highly appreciated!

 

 

 

I've drawn a red box on mwr's spectrum. This shape seems consistent with a transition between two electronic states with similar internuclear distances in the lowest vibrational level. Is that a fair inference?

 

 

You have hit the sweet spot: The transitions of exactly this band had been labelled with a question mark in the original publication:

 

 

 

However, the 0,0-transition ( v' - v'' = 0) points indeed to rather small internuclear distances when compared to other transitions with v' - v'' < > 0. 

[Tangerman]

What, if anything,  can we infer about the Franck Condon overlap (the effect of internuclear distance) or the effect of temperature on the spectrum?

Those are very good questions. Honestly, I'd never thought too much about what we can infer from the Franck-Condon overlap, I'd really just used it as a justification for why certain bands have the strength that they have. If you have really good spectra, you could fill out something called a Deslandres table and find various physical parameters, such as vibrational constants, anharmonicity, and the energy gap between excited electronic states. To get the internuclear radius, I think that rotational constants are used, so we'd need good enough resolution to see individual rotational bands. The problem with trying to use the Franck-Condon overlap to get internuclear distance is that a ground state distance of 1.5 Å and excited state distance of 1.6 Å would give the same result as a ground state of 3.0 Å and excited state of 3.1 Å.

 

I've drawn a red box on mwr's spectrum. This shape seems consistent with a transition between two electronic states with similar internuclear distances in the lowest vibrational level. Is that a fair inference?

 

What would it take to produce a band with the most intense absorption on the right rather than on the left? A shorter internuclear distance in the excited state?

Properly analyzing the band that you've circled would take a lot of time. I think that there are too many variables for me to say for certain if that's a good inference or not.

I do think to produce a band with the most intense absorption on the right, you would need a shorter internuclear distance in the excited state, which is most easily done by going from one excited state to another, but then again, I'm sure some few transitions from the ground state to an excited state exist which shorten the internuclear distance. I just can't think of any.

 

The TiO absorptions are known to change in the pulsational cycle. Is the effect of temperature on vibronic spectra apparent in our spectra?

I think that the most obvious effect of temperature here is the strength of those -1 bands, indicating losing vibrational energy. To get those, you have to start in an excited vibrational state, and the higher the temperature, the more molecules will be in excited vibrational states. So at a higher temperature, those bands would be even more intense, and at a lower temperature, they would be less intense. The temperature could also shift what each band looks like, although that would be a smaller change and harder to predict the effect.

[Organic Astrochemist]

Those are very good questions. Honestly, I'd never thought too much about what we can infer from the Franck-Condon overlap, I'd really just used it as a justification for why certain bands have the strength that they have. If you have really good spectra, you could fill out something called a Deslandres table and find various physical parameters, such as vibrational constants, anharmonicity, and the energy gap between excited electronic states. To get the internuclear radius, I think that rotational constants are used, so we'd need good enough resolution to see individual rotational bands. The problem with trying to use the Franck-Condon overlap to get internuclear distance is that a ground state distance of 1.5 Å and excited state distance of 1.6 Å would give the same result as a ground state of 3.0 Å and excited state of 3.1 Å.

 

Properly analyzing the band that you've circled would take a lot of time. I think that there are too many variables for me to say for certain if that's a good inference or not.

I do think to produce a band with the most intense absorption on the right, you would need a shorter internuclear distance in the excited state, which is most easily done by going from one excited state to another, but then again, I'm sure some few transitions from the ground state to an excited state exist which shorten the internuclear distance. I just can't think of any.

 

I think that the most obvious effect of temperature here is the strength of those -1 bands, indicating losing vibrational energy. To get those, you have to start in an excited vibrational state, and the higher the temperature, the more molecules will be in excited vibrational states. So at a higher temperature, those bands would be even more intense, and at a lower temperature, they would be less intense. The temperature could also shift what each band looks like, although that would be a smaller change and harder to predict the effect.

I just happened to have recently acquired two spectra recently, one of Aldebaran (K5, a little hotter than U Ceti) and another of Mira (M5- M9, probably a little cooler than U Ceti). Here they are overlaid in the region of interest. This is at a little higher resolution, R ~1100.

 

The α₁ band seems more intense than α_-1 in Mira. This makes sense if there are relatively fewer molecules in higher vibrational states. 

 

In the case of Aldebaran, it is not so obvious that α₁ band is more intense than α_-1. The hotter temperature would favor higher vibrational states and favor the transition from higher vibrational states in the ground electronic state to lower vibrational states in the excited electronic state. This would favor α_-1.

 

I also think the shapes of these bands is very interesting. In Mira the TiO bands have their characteristic "checkmark" look. I think this is caused by relatively fewer transitions from higher  vibrational states (because higher vibrational states are less populated). But I think that the α₁ band in Aldebaran is quite different and has a much flatter look to it. To me, this makes sense if, due to the higher temperature, there aren't relatively more transitions from lower vibrational states (because the temperature causes higher vibrational states to become more populated). 

 

I'm curious about the effect of temperature and rotation on broadening the bands. It seems like the bands might be sharper at higher temperatures and are more broadened into a continuum at lower temperatures. Of course there are also a lot of atomic lines superimposed as well.