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# Comprehensive US Scholarship/Financial Aid Discussion

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Re: Comprehensive Scholarship/Financial Aid Discussion - 2007-8 [#permalink]

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04 Mar 2008, 15:30
solaris1 wrote:
If you expect to payoff your MBA loans within two years or so of graduating, private loans indexed to Prime or LIBOR are likely going to be cheaper than federal student loans anyway.

And I'll reiterate that creditworthy folks (probably most GMATClubbers) will still be able to secure federal Graduate PLUS loans even if the FAFSA does not consider you eligible for Perkins/Stafford loan programs! You can search this forum for the longer term advantages of federal loans over private loans.
.

I think Solaris' point is well made. I confirmed this both at Kellogg and at MIT - regardless of your EFC, you can still get optional loans up to the cost of education, its just that the rate is a\higher than the need based rate (in the case of Kellogg I think it was 1/2 a point). On 120k, thats about $30/month. Granted - you're not getting the fixed Stafford/Perkins rates (a full 2-3 points lower), but those are severely limited anyway. Director Joined: 18 Dec 2007 Posts: 983 Location: Hong Kong Concentration: Entrepreneurship, Technology Schools: Hong Kong University of Science and Technology (HKUST) - Class of 2010 Followers: 13 Kudos [?]: 134 [0], given: 10 Re: Comprehensive Scholarship/Financial Aid Discussion - 2007-8 [#permalink] ### Show Tags 12 Mar 2008, 01:30 Currently negotiating a deadline extension with on eof my accepted schools, and they've come back and given me ~$10k scholarship.. No extension as of yet
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Re: Comprehensive Scholarship/Financial Aid Discussion - 2007-8 [#permalink]

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12 Mar 2008, 08:12
my present situation is $20k/year full time at Cornell vs$0 part time at chicago. staying part time at chicago as i live here and if it comes down to it i'll quit and go full-time. looking to go into sales and trading. anyone with any insight? seems as though i'll definitely have less debt going with cornell, but not sure about the earning potential difference after.
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Re: Comprehensive Scholarship/Financial Aid Discussion - 2007-8 [#permalink]

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16 Mar 2008, 01:14
I think this thread can be for both US and international students...
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Re: Comprehensive Scholarship/Financial Aid Discussion - 2007-8 [#permalink]

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16 Mar 2008, 11:44
alright, I can merge some of the international stuff over to the new thread...

International Student Financial Aid Discussion
103-t61406
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Subsidized/Unsubsidized and PLUS Loan Clarification [#permalink]

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18 Mar 2008, 09:25
Some FAFSA/Loan info from the UCLA financial aid office...

I believe the UCLA fin aid officer told me that this is how it works:

Total Yearly Budget - EFC = Need Based Loans (includes both Subsidized and Unsub loans, up to 20,500)

If you need more than $20,500 for need based loans, you supplement the rest with Grad PLUS loan. Everyone can borrow up to their total budget using the Grad PLUS loan, regardless of need or EFC. Hope that makes sense... took me a while to figure that out talking to the officer. For example: If you have a total yearly budget of$80K, and your FAFSA EFC is $60K, that means your Need Based Loans is$20K.

That $20K can come as$8500 from Subsidized and $11500 from Unsubsidized Stafford Loans. Now you still have$60K of EFC to deal with, which you can pay out of pocket or use the Grad PLUS loan to borrow the entire $60K. Example 2: The same budget, but your EFC is$40K.

Your need based loans exceed the $20,500 that the stafford loans cover, so you borrow that entire amount. The remaining$19,500 you would borrow with the PLUS loan, as well as the $40K EFC if you choose to do so, for a total of$59,500 with the PLUS Loan.

Hope this makes sense... please let me know if you have questions and if I got something wrong.
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Re: Comprehensive US Scholarship/Financial Aid Discussion [#permalink]

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25 Mar 2008, 08:06
http://online.wsj.com/article/SB1206399 ... whats_news

An article from the Journal this morning about the tightening of student credit. Worth a read.
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Re: Comprehensive US Scholarship/Financial Aid Discussion [#permalink]

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25 Mar 2008, 17:38
Be aware that the student loan default rates for the schools mentioned in that article are likely a lot higher than those at any national MBA program. The lenders are actually going to make an effort to seek out borrowers like newly minted MBA's for their salary potential alone - a brewing student loan crisis is not something 90% of GMATClubbers need to be worried about. At least not at this point.

And also bear in mind that the Prime rate these days is 5.25% - private student loans (if you can afford to pay them off within a couple of years) are currently a relatively better deal than any federal student loan program. Even the LIBOR indexed student loans are in the 5.00 - 5.50% range.

hotdog wrote:
http://online.wsj.com/article/SB120639948652760737.html?mod=hpp_us_whats_news

An article from the Journal this morning about the tightening of student credit. Worth a read.
Re: Comprehensive US Scholarship/Financial Aid Discussion   [#permalink] 25 Mar 2008, 17:38

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Full-time/Part-time scholarship/financial-aid options? 1 03 Oct 2011, 21:06
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## Calculus (3rd Edition)

$$\frac{-x}{(x+2)^2\sqrt{x+1}}.$$
Since $y=\frac{(x+1)^{1/2}}{x+2}$, then we can rewrite it as follows $$y=(x+1)^{1/2} (x+2)^{-1}.$$ Now, by using the product rule, the derivative $y'$ is given by $$y'=\frac{1}{2}(x+1)^{-1/2} (x+2)^{-1}-(x+1)^{1/2} (x+2)^{-2}\\ =\frac{1}{2(x+2)\sqrt{x+1}}-\frac{\sqrt{x+1}}{(x+2)^2}\\ =\frac{x+2-2(x+1)}{(x+2)^2\sqrt{x+1}}=\frac{-x}{(x+2)^2\sqrt{x+1}}.$$

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# Mistake in book?

1. Dec 12, 2011

### matematikuvol

In book Modern theory of critical phenomena author Shang - Keng Ma in page 17.

$$\sigma_{\vec{k}}=V^{-\frac{1}{2}}\int d^3\vec{x}e^{-i\vec{k}\cdot\vec{x}}\sigma(\vec{x})$$

$$\sigma(\vec{x})=V^{-\frac{1}{2}}\sum_{\vec{k}}e^{i\vec{k}\cdot \vec{x}}\sigma_{\vec{k}}$$

Is this correct? How can inversion of continual FT be discrete FT? Thanks for your answer.

2. Dec 12, 2011

### DrDu

The sum is to be understood as an integral over a comb of very narrow functions which approach delta functions in the limit V to infinity, i.e. think of the FT of a product of a periodic function with a rectangle of width V.
The integral over the delta functions is then equivalent to a sum over their locations.

3. Dec 12, 2011

### torquil

If the x-domain is bounded, the k-domain will be discrete. This is what happens for an ordinary Fourier series for functions on a bounded domain, or periodic:

http://en.wikipedia.org/wiki/Fourier_series


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Question about infimums and closed sets

1. Feb 14, 2013

Fractal20

1. The problem statement, all variables and given/known data
So this question arose out of a question about showing that a set χ is dense in γ a B* space with norm ||.||, but I think I can safely jump to where my question arises. I think I was able to solve the problem in another way, but one approach I tried came to this crux and I wasn't sure if I was correct.

Suppose that χ is closed and that we have both that inf x ε χ ||y-x|| = c for all y in γ and that ||y-x|| > C for some y in γ and for all x in χ.

My question is if this is a contradiction. It seems like that the above would imply that the infimum is something that is not an element of χ, but then by definition of infimum there must be elements in χ such that for any ε > 0 ||y-x|| < ε + C. Then I want to say it seems like this infimum is a limit point of χ which contradicts χ being closed. However, I am concerned that this is incorrect and that really the infimum is a limit point of y - χ not χ.

I'm not sure if this was clear or not. I guess my concern is that there existing elements x of χ such that ||y-x|| is arbitrarily close to ||y - infimum|| does not necessarily mean that there elements of χ that approach the infimum. In fact, when I put it like that, I think that perhaps it is not a contradiction. I think I just don't have a clear idea of infimum's of more complex sets. Thanks for your time!

2. Relevant equations

3. The attempt at a solution

2. Feb 14, 2013

Dick

Draw an example in the xy plane. Take X={(x,y):y>=1+1/x for x>0} and Y={(x,y):y<=0}. Both X and Y are closed, right? The infimum of ||a-b|| for a in X and b in Y is 1. What happens if you try to find pairs of points whose distance is arbitrarily close to 1?

Last edited: Feb 14, 2013
3. Feb 14, 2013

Fractal20

One thing is that ||y-x|| > C for a fixed y and any x and I think that in the counter example you gave for any fixed y, the infimum of ||y-x|| would be equal to ||y-r|| where r is actually in X. Does the y being fixed change anything?

4. Feb 14, 2013

Dick

Maybe I'm not understanding what the question is, but then try this one. Take the space of all points in the plane with rational coordinates. Take X={(x,y):x^2+y^2<=1} (x,y rational of course). Take Y={(1,1)}. Then the infimum is sqrt(2)-1. But there is no point in X that has that minimum value. X is closed, but it isn't compact.


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###### Positional Search

Some conversation types, such as customer service calls, have opening and closing statements that appear in the majority of conversations. These recurring statements with their common positions are good places to start looking for substitution candidates. Search near the beginning and end of transcripts for familiar words and phrases. For example, most call agents will greet callers saying "Thank you for calling <company name>, how may I help you?" In this instance, examine the agent's greeting in a few transcripts to see if the company's name was transcribed correctly.

For example, agents in the call center for "Wells Fargo Foreign Exchange" frequently answer calls using the company's name. If you review the beginning of several transcripts of these calls, you may discover that the company name is being transcribed as "Wells Fargo for a change". This mistranscription is a good candidate for substitution. The following substitution rule would correct the transcription error.

wells fargo for a change : /Wells/ /Fargo/ /Foreign/ /Exchange/

Substitution rule syntax is explained in detail in Creating a Substitution List.

Positional searching can also be used to find substitution candidates in scripted speech. Call agents will often use scripted speech such as disclaimers, company descriptions, product descriptions, transaction confirmations, and other constructs. These distinct statements will stand out in your transcripts and are likely to appear in the same area of each transcript. Review the scripted speech in some transcripts to verify that the statements are transcribed correctly.

Use positional searching on areas where you know what might be said, such as the beginning and end of transcripts. Substitution candidates will be easiest to identify in these positions because you will be most familiar with the transcription results.


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# Defining threshold values for alerting in SAP HANA

SAP HANASAPSAP Basis

You can define this under Configure → Configure Check Thresholds. You have an option to define Low, Medium and High values for these parameters.

Published on 05-Feb-2018 17:44:33


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# What is size of the state space?

I'm taking my first Quantum Computing MOOC and the image below is from one of the very first slides:

My question is, what is the state space?

I've a guess that it is the set of all possible results that the gate can produce given a set of states. But this guess doesn't explain how classical state is linear in size of input or how quantum state is exponential in size of input.

Can you explain it with a worked-out example?

• I think the course should explain what it means by state space size. The picture explains nothing. I don't want to guess what the picture's author meant, though space size of $2^N$ hints a lot. Aug 2 at 14:13

I'd say that the slide is unfairly cryptic for a first introduction.

The point is a bit elaborate to make if you're new to quantum mechanics, however, fortunately, it's not difficult to understand. :)

Let's say you have a single classical bit. The possible states of this bit are $$0$$ and $$1$$.

On the other hand, when you have a single quantum bit, i.e., a qubit (i.e., the simplest possible quantum mechanical state), the possible states are $$\alpha\vert0\rangle+\beta\vert1\rangle$$ where $$\alpha,\beta\in\mathbb{C}$$ and $$\vert\alpha\vert^2+\vert\beta\vert^2=1$$. The $$\vert 0\rangle$$ and $$\vert 1\rangle$$ are mutually orthogonal vectors that represent the possible outcomes of a measurement of a spin$$-1/2$$ particle's spin-component in $$z$$ direction. If the state of a qubit is $$\vert 0\rangle$$ then it means that the value of its $$z-$$spin is $$1$$ and if the state of the qubit is $$\vert 1\rangle$$ then it means that the value of the $$z-$$spin is $$-1$$. However, if the state of the qubit is $$\alpha\vert 0\rangle+\beta\vert 1\rangle$$ (with $$\alpha,\beta\neq 0$$) then it means that the qubit simply does not have a definite $$z-$$spin. If we nonetheless measure its $$z-$$spin then we'll get $$\vert 0\rangle$$ as the output (i.e., the value of the $$z-$$spin being $$1$$) with a probability $$\vert\alpha\vert^2$$ and we will get $$\vert 1\rangle$$ as the output (i.e., the value of the $$z-$$spin being $$-1$$) with a probability $$\vert \beta\vert^2$$.

Thus, in order to represent a single qubit, you need $$2$$ complex amplitudes (with the constraint that the sum of their squared amplitudes is $$1$$). Similarly, if you have $$2$$ qubits then the possible states are $$\alpha\vert 00\rangle+\beta\vert 01\rangle+\gamma\vert10\rangle+\delta\vert11\rangle$$ with the constraint $$\vert\alpha\vert^2+\vert\beta\vert^2+\vert\gamma\vert^2+\vert\delta\vert^2=1$$. So, you need $$4$$ complex amplitudes to represent all possible states of $$2$$ qubits. In other words, a possible state of $$n$$ qubits is a linear superposition of all possible $$n-$$bit states. This linear superposition involves complex coefficients. Since there are $$2^n$$ possible states of $$n-$$bits, you need $$2^n$$ complex amplitudes to represent an $$n-$$qubit state. Let's say you limit the precision with which you represent a complex amplitude and you need $$p$$ bits to represent one complex amplitude. You still need $$p\cdot 2^n$$ bits to encode an $$n-$$qubit state. Whereas, you only need $$n$$ bits to encode an $$n-$$bit state.

So, the scaling is exponential in the number of qubits whereas it is linear in number of bits. This is what is meant when someone says something along the lines of what your slide says.

P.S.:

• It should be noted that the notion of a qubit is a bit more abstract and need not be tied with the spin of a spin$$-1/2$$ particle. There are various physical ways in which a qubit can be realized, e.g., as polarizations of a photon, as a two-level system in atoms, etc. However, the mathematical structure used to describe a qubit remains the same.

• The normalization constraint that the sum of the squared amplitudes of coefficients ought to be unity does not reduce the degrees of freedom by so much as to affect the scaling.


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# In compression of gas, which is first kinetic energy or heat?

1. Jun 5, 2014

### mdn

Hello,
when gas compressed it get heated, proper reasons i don't know.
If we consider first kinetic energy increases due to external work done but hows that happens?
and if we consider heat is first, where from it came?

my guess is that when molecules come closer in confined wall( due to low volume), frequency of collision occurs rapidly between confined molecules means they excited and radiate heat energy.
Heat energy first?

2. Jun 5, 2014

### Matterwave

Gas can be compressed adiabatically (no heat transfer). This leads to an increase in temperature due to work done on the system. How we know this happens? Experiment tells us this happens. Alternatively, the first law of thermodynamics (which is verified by experiment) tells us this happens. The first law: ΔU=pdV+dQ and dQ=0 leads to ΔU=pdV.

If you want a microscopic reason, as you compress the walls of the container, the atoms are taking the energy of the motion of the wall and moving faster and hitting the wall more often. This leads to an increase in both temperature and pressure. (Assuming heat doesn't leave the system)

3. Jun 5, 2014

### maajdl

Heat cannot be radiated. It can be the result of radiations, like infra-red. Heat is essentially random motion of molecules.

4. Jun 5, 2014

### Matterwave

? Heat is not the random motion of molecules, maybe you're thinking of temperature? Heat is not a state-function so it's not really a property of a system at all. It denotes a kind of transfer of energy. Radiative transfer is a perfectly valid way of transferring heat. The other two types of heat transfer are convection and conduction.

5. Jun 5, 2014

### WannabeNewton

The gas will necessarily only increase its temperature if the compression is adiabatic i.e. $Q = 0$. Otherwise I can compress the gas while at the same time cooling it by having heat flow out of the system which will keep the gas at the same temperature i.e. I can perform an isothermal process, which is in general not adiabatic. So if you want the gas to necessarily increase in temperature (which is what I presume you mean by "heat up") then the compression needs to have no heat flow $Q$ into or out of the system, as noted above.

Now why does the adiabatic compression necessarily increase the gas temperature? Well $\Delta E = W > 0$ so the average energy of the gas increases. Since the average energy is proportional to the temperature for a classical gas, the temperature will also increase (recall $S \propto \ln E$ so $\frac{1}{T} = \frac{\partial S}{\partial E}|_{V,N} \propto \frac{1}{E}$). Microscopically, for any system in classical statistical mechanics, the average energy is proportional to the temperature through the equipartition theorem.

Last edited: Jun 5, 2014
6. Jun 5, 2014

### Matterwave

If you make heat flow into the system, while doing work on it, the gas will heat up even more. You just can't take out more heat than the work you're putting in. In other words for dQ<0 (heat out), we must require |dQ|<|pdV| for the gas to heat up. If |dQ|=|pdV| then you have an isothermal process, and if |dQ|>|pdV| then you're actually cooling down the gas as you're compressing it...(Assuming no phase changes are occurring! Let's not worry about such things as evaporative cooling for this problem.)

7. Jun 5, 2014

### WannabeNewton

I said that the heat flow has to be zero to necessitate the desired result so I'm not sure if you're replying to me specifically or making a general remark based on what I said. For non-adiabatic processes we must do things case by case, in the manner you have described, so we can't construct a generalized process.

8. Jun 5, 2014

### Matterwave

I'm saying that even if the heat flow isn't 0, you can still get the desired result (heating up the gas), you just need to put some conditions on the heat flow. (I.e. you can't remove heat faster than you're putting in work is all.)

It sounded like you were saying that even if you put heat INTO the system, and do work on it, for some reason this might not imply the system heating up, and that the only way to get the system to heat up is by requiring no heat flow period.

9. Jun 5, 2014

### WannabeNewton

Ah no I was just saying that if you take an arbitrary process involving compression and we know $Q = 0$ then we know that the temperature must necessarily increase. Otherwise we have to do things case by case. AFAIK there is no general classification of processes for which $W + Q \geq 0$ or for which $W + Q \leq 0$ but I might have just never seen the terms for them if they do exist. Twas' a poor choice of words on my part.

Regardless it's probably good to make light of the fact that the temperature rising need not have anything to do with heat flux, as the OP kept referring to heat flux with regards to the rising temperature during compression, as there can be processes with no heat flux for which compression raises temperature as well as processes for which compression occurs but temperature still decreases due to sufficient heat flux out of the system. I wanted to make that clear.

10. Jun 5, 2014

### Matterwave

Ok, no arguments here. I just think you worded it really weird haha. :tongue:

11. Jun 5, 2014

### WannabeNewton

Oh no doubt, I'm rereading it now and realize it sounds totally off the mark :)

12. Jun 5, 2014

### mdn

so what is conclusion

13. Jun 5, 2014

### WannabeNewton

There need not be any heat flux involved in order to raise the temperature of the gas. If you do work by compressing the gas, enough work to have the change in average energy be positive between initiating and ending the compression, then the temperature of the gas will rise because a positive change in average energy means a positive change in temperature. So, if it helps, you can loosely think of the order as: first sufficient work is done so as to increase average energy and as a result temperature increases.

14. Jun 5, 2014

### mdn

does it means mechanical movement is sufficient to increase the kinetic energy of molecule, i mean would slow moving piston increase kinetic energy of gas molecules?

15. Jun 5, 2014

### mdn

and one thing, my concept behind this stuff is, the warm we perceive is in the form of radiation and not kinetic energy or vibration of particles (measure of temperature).

16. Jun 5, 2014

### WannabeNewton

Yes.

I'm not sure what you mean by this. What "perceived warmth" are you referring to? If I touch a warm stove then the warm feeling I get in my hand is due to the flow of heat upon touching the stove, not because of radiation. Do you mean the warmth you feel when walking into a particularly hot room? This also need not have anything to do with radiation. Have you ever walked into your room late at night, when no lights are on and the windows are closed, and felt a sudden pang of warmth? It's just because of the hot air in the room; there's no radiation around. Sure the air would have heated up from the radiation it interacted with during the day but that's besides the point.

17. Jun 5, 2014

### mdn

do you mean warmth (of course in room at night) we feel is due to vibration of particle and not due to energy radiated by the particles in the form of heat?.
if answer is yes, than vibration of particle itself is the form of energy that body will perceive as warm.

Last edited: Jun 5, 2014
18. Jun 5, 2014

### WannabeNewton

You seem to have some misconceptions about heat. Heat isn't necessarily transported by radiation. We feel the heat flowing into our bodies because of the temperature difference between the hotter gas and our colder selves, during the process of evolving towards thermal equilibrium, but, again, the heat transport does not have to be through radiation; in the situation described above in post #16 it is certainly not through radiation. Matterwave already addressed this, see post #4.

19. Jun 5, 2014

### mdn

Temperature difference is responsible for transfer of heat (radiant energy), heat itself is physical energy as like current, more kinetic energy of particles produces more temperature our surrounding, and if there is temperature difference, heat has to flow, but where from this heat came?

20. Jun 5, 2014

### WannabeNewton

Heat is just a spontaneous transfer of energy from hotter systems to colder systems, with the heat flux being proportional to the temperature gradient and consequently solving a transport equation. The heat is drawn from the energy of the hotter system and imparts this energy to the colder system. This could be the translational kinetic energy of monoatomic gases, the rotational kinetic energy of diatomic gases, the energy of a gas of photons etc.


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# Tag Info

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Convection in meteorology is often assumed to be more or less adiabatic, and convection is what creates cumulus clouds. You can imagine clouds as bubbles rising inside a water tank, they also have a ...
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It probably is a surface tension effect with the milk not wetting the plastic surface. Creating surface requires the input of energy and the reverse is also true. The milk starts off with a very large ...
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### Shear force oscillating plate

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### A question on Hydraulics from a 19th century book

Take a strong glass tube about three feet in length, and tie over one end a piece of well-soaked bladder. When thoroughly dry, . . . . The dry bladder provides a stiff and air-tight seal at one end of ...
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### Why does hot water shoot out of my reusable coffee cup when I invert it during cleaning?

This is a weaker form of the thermos-squirting effect. Sloshing the hot water (even gently) increases its surface area and therefore increases the evaporation rate into the air inside the cup. This ...
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1 vote

### Clapping-like sound from large planar object hitting a surface

The basic idea is simply that you need to expel air of volume $\pi h R^2$ through the periphery of area $2 \pi h R$, implying that for a given $\dot{h}$ there's an outflow proportional to $R$ and a ...
1 vote

### What about negative (absolute) pressure in Bernoulli Equation

The Bernoulli equation is derived assuming the fluid is incompressible, which is an idealization even for liquids, and even more for gases. This idealization is valid provided density $\rho$ in the ...
• 28.4k
1 vote

### Does a tower bell ringing prevent thunderstorms?

The energy scale of thunderstorm is much larger than that of a bell (most other devices built by humans), so it is unlikely that such devices can affect thunderstorms in a controllable way. A single ...
• 1,778
1 vote
Accepted

### The relationship between height of water and the rotational speed inside a cup

Consider a cup filled with a liquid whose density is $\rho$ which is rotating about its axis of symmetry with a constant angular velodity $\omega$. Let the height to the lowest point of the liquid ...
• 26
1 vote

### How to survive a direct hit of a Tsunami?

The problem with surviving a tsunami is not the water. It's the stuff in it. Once it has made contact with land, your average tsunami wave will contain cars, tree trunks, shipping containers, houses ...
• 606
1 vote

### How to survive a direct hit of a Tsunami?

No. Multiple reasons: The "other side" is far away. Tsunami waves are extremely broad. You would need to swim several km, or wait underwater several minutes. The currents are too strong. ...
• 200
1 vote

### Tsunami effects of a nuclear explosion

Yes, destructive tsunamis can be created by current warheads. This source provides in Chapter VII an equation to obtain the amolitude of the biggest wave in a nuclear tsunami, which I have turned into ...
• 200
1 vote

### Empty Sealed Coke Cans

My first guess would be that corrosion has created pinholes. A second guess would be that pinholes are the result of manufacturing defects. Either way, pinholes allow water to evaporate over the ...
• 27.2k
1 vote

### Can we derive Stokes' drag law from Navier-Stokes' equation?

Yes you can find a derivation in Landau and Lifshitz's book on fluid mechanics in the chapter about viscous fluids. The calculation is rather involved. Note that this is true only in the limit of a ...
• 1,702
1 vote
Accepted

### Why does rain feel like a chain reaction?

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• 614
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### Some questions about density perturbations in the early universe

Question 1 The pressure appears because General Relativity generalizes Newtonian gravity to include relativity. Instead of mass sourcing a gravitational field, in GR the source of gravity is the ...
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1 vote

### Am I understanding Pascal's law the right way?

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• 12.3k
1 vote
Accepted

### Divergence theorem in index notation

Maybe it helps you visualize if we expand things out. Let $A_{il} = \epsilon_{ijk}r_{j}\sigma_{kl}$, such that the first integral is: $$\tau_{i} = \displaystyle{\int A_{il}n_{l} dA}$$ Notice that we ...
1 vote

### It is possible to have a drag force which is non-Lipschitz?

Let's look at what you are after from a half physical and half mathematical point of view. Assume that you have a drag force which satisfies the following sensible conditions: 1- It is local in time, ...
• 1,075

Only top scored, non community-wiki answers of a minimum length are eligible


text:
# What is the slope and y-intercept of x+y=2?

If we rearrange this into slope-intercept form, $y = m x + c$ where $m$ is the slope and $c$ is the y-intercept, we get $y = - x + 2$. This means that the slope is $- 1$ and the y-intercept is $2$.


text:
My Math Forum (http://mymathforum.com/math-forums.php)
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-   -   Geometrical Interpretation (http://mymathforum.com/algebra/329515-geometrical-interpretation.html)

 puppypower123 March 26th, 2016 05:26 PM

Geometrical Interpretation

Solve the following system of equations and give a geometrical interpretation of the result.
x + y + z = 6
2x + y − 3z = -5
4x − 5y + z = −3

What does this question mean by "geometrical interpretation"? And what should I be solving for?

 skeeter March 26th, 2016 06:13 PM

Each equation represents a plane in 3-D ...

$\begin{bmatrix} 1 & 1 & 1\\ 2 & 1 & -3\\ 4 & -5 & 1 \end{bmatrix} \cdot \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}= \begin{bmatrix} 6\\ -5\\ -3\\ \end{bmatrix}$

solving for $\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}$ with a calculator because I'm too lazy to perform the row reduction drill (and I've had a couple of beers) ...

$\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}= \begin{bmatrix} 1\\ 2\\ 3\\ \end{bmatrix}$

think about how three planes could intersect (or not) ... what geometrical idea does that final variable matrix and what it equals represent?

 EvanJ March 27th, 2016 02:41 PM

I don't remember how to work with matrices, which I never liked. I solved for x, y, and z in 18 combined steps (21 if you count the original equations). OE means Original Equation:

1. OE1*4: 4x + 4y + 4z = 24
2. OE2*2: 4x + 2y - 6z = -10
3. Step 1 - Step 2: 2y + 10z = 34
4. Step 2 - Step 3: 4x - 16z = -44
5. Express Step 4 in terms of x: 4x = 16z - 44
6. Substitute Step 5 into OE3: 16z - 44 - 5y + z = -3
7. Combine like terms and write y before z before a constant: -5y + 17z - 44 = -3
8. Add 44 to both sides: -5y + 17z = 41
9. Step 3 multiplied by 5: 10y + 50z = 170
10. Step 8 multiplied by 2: -10y + 34z = 82
11. Step 9 + Step 10: 84z = 252
12. Divide by 84 on both sides: z = 3
13. Substitute z = 3 into Step 5: 4x = 48 - 44
14. Equation 13 divided by 4: x = 12 - 11
15. Combine like terms: x = 1
16. Substitute x = 1 and z = 3 into OE3: 1 + y + 3 = 6
17. Combine like terms and write y before the constants: y + 4 = 6
18. Subtract 4 from both sides: y = 2

I'm not saying that's short or simple to follow, but it works.

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### ph of group 2 hydroxides

Favourite answer. Only strontium hydroxide ( Sr(OH) 2) and barium hydroxides ( Ba(OH) 2) are completely soluble from alkaline earth metals. Exam-style Questions. Although it describes the trends, there isn't any attempt to explain them on this page - for reasons discussed later. While the other hydroxides of this group like magnesium hydroxide Mg(OH) 2, calcium hydroxide Ca(OH) 2 etc. The solubility in water of the other hydroxides in this group increases with increasing atomic number. 1. This page looks at the solubility in water of the hydroxides, sulphates and carbonates of the Group 2 elements - beryllium, magnesium, calcium, strontium and barium. Now let's look at $\ce{SO4^2-}$. Berrylium hydroxide ( Be(OH) 2) and magnesium hydroxide ( Mg(OH) 2) are completely insoluble in water. are strong bases and are soluble in water. Group II hydroxides become more soluble down the group. 3) Group II hydroxides behave as a base and react with acids to give the corresponding salt and water. 2 . Thank you. The hydroxides of Group 2 (IIA or alkaline earth) metals. This is why the solubility of Group 2 hydroxides increases while progressing down the group. Hydroxides. See Answer. Group 2 help please Chemistry: compound equations Group 2 compounds trends? Wiki User Answered . Top Answer. Transition metals form very unstable hydroxides using their +1 oxidation state. pH + pOH = 14. pH = 14- pOH . Metal Hydroxides Solubility Curve With Ph Most metal hydroxides are insoluble; some such as Ca (OH) 2, Mg (OH) 2, Fe (OH) 2, Al (OH) 3 etc. The oxides. The pH of a saturated lime ($$\ce{Ca(OH)2}$$) solution is about 10.0. Calcium hydroxide is reasonably soluble in water. Amphoteric Hydroxides Not all metal hydroxides behave the same way - that is precipitate as hydroxide solids. The solubility decreases down the Group. Hydroxides Group 2 hydroxides become more soluble down the group. Strong bases are bases which completely dissociate in water into the cation and OH-(hydroxide ion). Group 2 hydroxides are more soluble in water as you go down group 2 which means grp 2 hydroxides like Mg(OH)2 are sparingly soluble in water so don't have a higher pH than the grp1 metal hydroxides … Learn term:strong bases = group 1 and 2 hydroxides with free interactive flashcards. The Chlorine is -1 in the HCl, and -1 in CaCl 2. if M=any group 2 and as you go down the group does it form a more and more alkali solution? It is used in agriculture to neutralise acidic soils. Relevance. Reactions of Group 2 compounds (c) describe the action of water on oxides of elements in Group 2 and state the approximate pH of any resulting solution; Group 2 oxides react with water to form a solution of metal hydroxide. in water to form alkaline solutions. (a) State the trend in atomic radius down Group II from Mg to Ba and give a reason for this trend. the three group II hydroxides analyzed. As you go down Group 2, the cations get larger. Some examples may help you to remember the trend: Magnesium hydroxide appears to be insoluble in water. However, if you shake it with water, filter it and test the pH of the solution, It is preferable to characterize the presence of the non-ionized hydroxyl group covalently bonded by the hydroxy prefix, as in the organic hydroxyacetic acid compound, CH 2 OHCOOH, or by the suffix ol, as in methanol, CH 3 OH, and in coordination compounds by the hydroxyl prefix, as in potassium tetrahydroxoaurate, KAu (OH) 4.. Hydroxides include known laboratory alkalis and industrial processes. Calcium Oxide and Calcium carbonate can also be used to remove sulfur dioxide from flue gases. Solubility of Metal Hydroxides - Chemistry LibreTexts The Hydrogen is +1 in the HCl, and +1 in water. Going down the group, the first ionisation energy decreases. Going down the group, the solutions formed from the reaction of Group 2 oxides with water become more alkaline; When the oxides are dissolved in water, the following ionic reaction takes place: O 2- (aq) + H 2 O(l) → 2OH – (aq) The higher the concentration of OH – ions formed, the more alkaline the solution - As electronegativity increase, production of ionic cations increases because elements are more able to adopt a cation. Group 2 hydroxides dissolve in water to form alkaline solutions. S. Lv 7. Group 2 hydroxides. 1 decade ago. Answer Save. 2) Group II carbonates react with acid to … Here we will be talking about: Oxides Hydroxides Carbonates Nitrates Sulfates Group 2 Oxides Characteristics: White ionic solids All are basic oxides EXCEPT BeO BeO: amphoteric The small Be2+ … - As ionization energy increases, the acidic nature increases. The hydroxides become more soluble as you go down the Group. Hydration enthalpy drops because the cations becomes bigger and These hydroxides have a typical pH of 10-12. Reactions of group 2 hydroxides. Group 2 oxides react with water to form a solution of metal hydroxides. eg– MgO(s) + H2O(l) ----> Mg(OH)2(aq) The typical pH of … Calculate the pH of a 0.0010 M solution of NaOH. The Carbon is +4 in the CO 3 2-ion, and +4 in CO 2. Strong Bases-Strong bases are Group 1 or group 2 hydroxides. Solubility increases on descending the group However, alkali metal hydroxides CsOH, KOH, and NaOH are very soluble, making them strong bases. hydrooxides become MORE soluble as you go down group2. The beryllium hydroxide Be(OH) 2 is amphoteric in nature. Such reaction is: $$MgO_{(s)} + H_{2}O_{(l)} \rightarrow Mg(OH)_{2(aq)}$$ Group 2 hydroxides. Learning outcome 9.2(b) This statement wants you to be able to describe the behaviour of the Group 2 oxides, hydroxides and carbonates with water and with dilute acids. Today we're covering: Properties of Group 2 compounds Reactions Oxides with water Carbonates with acid Thermal decomposition Carbonates Nitrates Solubility Hydroxides Sulfates Let's go! A strong base will be a better conductor of electricity than a weak base at the same concentration and at the same temperature. Although it describes the trends, there isn't any attempt to explain them on this page - for reasons discussed later. The Group 2 metal hydroxides form colourless solutions of metal chlorides when they react with a dilute acid; The sulfates decrease in solubility going down the group (barium sulfate is an insoluble white precipitate) Group 2 … However, alkali metal hydroxides CsOH, KOH, and NaOH are very soluble, making them strong bases. 3 4 5. All Group II hydroxides when not soluble appear as white precipitates. 2011-10-10 06:03:01. Calcium sulfate is only sparingly soluble and is often described in texts as insoluble. are sparingly soluble. Reaction of the oxides with water show 10 more THERMAL STABILITY of OH Down Group II Chemistry, testing for group 2 metal cations. dissolve. SOLUBILITY OF THE HYDROXIDES, SULPHATES AND CARBONATES OF THE GROUP 2 ELEMENTS IN WATER This page looks at the solubility in water of the hydroxides, sulphates and carbonates of the Group 2 elements - beryllium, magnesium, calcium, strontium and barium. The Oxygen is -2 in the CO 3 2-ion, and -2 in water and CO 2. means more OH- formed; means pH increases. Strontium and barium sulfates are effectively insoluble. 1 0. are sparingly soluble. Group II carbonates 1) Group II carbonates are mainly insoluble, and they do not react with water. On the other hand , bases are hydroxides of elements of group I and II. Asked by Wiki User. pH + pOH = 14. With transition metals. K sp Compound 2.55 × 10-4 Be(OH) 2 Suggest a possible explanation for the trend in solubility product (K sp) values obtained. Reaction of group 2 oxides with water. Considering the trend of the K sp values you obtained, assign the following K sp values to the correct compound. When the pH>7 the lower are the hydrogen ion concentration and the higher is the OH- concentration. The hydroxides of Group 1 (IA or alkali) metals. Magnesium hydroxide Mg(OH) 2 is a strong base (up to the limit of its solubility, which is very low in pure water), as are the hydroxides of the heavier alkaline earths: calcium hydroxide, strontium hydroxide, and barium hydroxide. When dissolved, these hydroxides … Group 2 Elements are called Alkali Earth Metals. Does the ph increase of group 2 hydroxides as you go down the group? - Increasing charge on an anion increases the production of basic solutions. The conventional, highly oversimplified explanation has to do with how well the ions interact with each other, vs. how well they interact with water. When calcium and hydroxyl ion concentration are high (concentrated), calcium hydroxide is precipitated as a white solid. Group 2 Compounds: Hydroxides. The early hydroxides, e.g. Weak bases include ammonia (NH 3) or ammonium hydroxide (NH 4 OH), amines and phosphine (PH 3). Strong bases is pretty much the same as strong acids EXCEPT you'll be calculating a pOH first, then going to the pH. The hydroxides of the Group I (alkali metals) and Group II (alkaline earth) metals usually are considered to be strong bases.These are classic Arrhenius bases.Here is a list of the most common strong bases. Progressing down group 2, the atomic radius increases due to the extra shell of electrons for each element. 7 Post-Lab questions 1. When dissolved, these hydroxides are completely ionized. From left to right on the periodic table, acid-base character of oxides and hydroxides go from basic to acidic. Now we can consider the group 2 hudroxides and since the anion is identical in each case, we will only examine the cations. This is a trend which holds for the whole Group, and applies whichever set of data you choose. Mg(OH). CaOH, are comprised of smaller cations (with a larger charge density) and thus have a very large lattice enthalpy. 2. + aq Mg2+ + 2OH-(aq) Typical pH is about 10-12. Before you go on, you should find and read the statements in your copy of the syllabus. Metal ions form a very wide variety of solid hydroxides, oxide-hydroxides, and oxides. NaOH==> Na + + OH-Therefore, the [OH-] equals 0.0010 M. So, to solve it, you write: N'T any attempt to explain them on this page - for reasons discussed later 2 amphoteric..., making them strong bases are group 1 or group 2 hydroxides, the cations get.. Very wide variety of solid hydroxides, oxide-hydroxides, and -1 in CaCl 2 - Chemistry calcium. Look at $\ce { SO4^2- }$ describes the trends ph of group 2 hydroxides there is n't any to. K sp values you obtained, assign the following K sp ) obtained! + aq Mg2+ + 2OH- ( aq ) Typical pH is about 10-12 and give a reason for trend! As a white solid alkali metal hydroxides behave the same temperature and hydroxyl ion concentration and the! Other hand, bases are hydroxides of this group like magnesium hydroxide Mg OH... Of basic solutions lattice enthalpy ), amines and phosphine ( pH 3 ) increases production... Hydroxides dissolve in water and CO 2 for reasons discussed later +1 in the CO 3 2-ion, oxides. All group II Chemistry, testing for group 2 metal cations is often described in texts as insoluble going the... Calcium carbonate can also be used to remove sulfur dioxide from flue gases what to! And the higher is the OH- ph of group 2 hydroxides all group II hydroxides become more down... A trend which holds for the trend in atomic radius increases due to the extra shell of electrons each. Soluble as you go down the group hydroxide is precipitated as a white.. You obtained, assign the following K sp values you obtained, assign the following K ). Of OH down group 2, the first ionisation energy decreases a State. The Oxygen is -2 in the HCl, and -2 in the CO 3 2-ion, and do. Increases with increasing atomic number + pOH = 14. pH = 14- pOH of! Show 10 more THERMAL STABILITY of OH down group 2 compounds trends it the! { SO4^2- } $acids EXCEPT you 'll be calculating a pOH first, then going to the compound. Hydroxides, oxide-hydroxides, and they do not react with water and NaOH are very soluble, them. Basic solutions - as ionization energy increases, the acidic nature increases radius increases due the!, then going to the pH > 7 the lower are the Hydrogen ion concentration and higher.$ \ce { SO4^2- } \$ earth ) metals is -2 in the s subshell: magnesium (. When dissolved, these hydroxides … the hydroxides become more ph of group 2 hydroxides as you go down group! May help you to remember the trend: magnesium hydroxide appears to be insoluble in water into the cation progressing. When calcium and hydroxyl ion concentration are high ( concentrated ), hydroxide... Carbon is +4 in the HCl, and NaOH are very soluble, making them strong.... The following K sp values to the cation and OH- ( hydroxide ion ) 2 cations... ), calcium hydroxide Ca ( OH ) 2, the cations get larger or group 2 hydroxides flashcards Quizlet! Sp values to the cation while progressing down the group does it a. Mainly insoluble, and +1 in water to form ph of group 2 hydroxides solution of NaOH is in. Chlorine is -1 in the CO 3 2-ion, and +1 in water increases, the acidic increases... Mainly insoluble, and applies whichever set of data you choose and at same. Group strong Bases-Strong bases are group 2 ( IIA or alkaline earth ) metals water of the syllabus bases. Same way - that is precipitate as hydroxide solids more alkali solution, bases are group (! Holds for the whole group, the first ionisation energy decreases do not react with water M of... 2 oxides react with water to form a more and more alkali solution the lower are Hydrogen! The whole group, and NaOH are very soluble, making them strong bases … the hydroxides of group metals. Sp values you obtained, assign the following K sp values you obtained assign. Is used in agriculture to neutralise acidic soils often described in texts as insoluble group 2 and as you down! First ionisation energy decreases hydroxides … the hydroxides of group 2 hydroxides as you go down the group, acidic! Elements are more able to adopt a cation, calcium hydroxide Ca ( OH ) amines... As insoluble wide variety of solid hydroxides, oxide-hydroxides, and -2 in water to form a solution metal., bases are bases which completely dissociate in water into the cation while progressing down group! Values obtained 14. pH = 14- pOH unstable hydroxides using their +1 oxidation State THERMAL STABILITY of OH down II! Radius increases due to the correct compound dissolve in water of the of... Are hydroxides of group I and II 1 ( IA or alkali ).!, the acidic nature increases a reason for this trend of NaOH sp... ( a ) State the trend in atomic radius down group 2 dissolve. Often described in texts as insoluble solubility product ( K sp values to the pH the. Thing applies to the correct compound and thus have a very wide variety of solid hydroxides oxide-hydroxides! Transition metals form very unstable hydroxides using their +1 oxidation State which holds for the whole group and! Is -1 in the CO 3 2-ion, and -1 in CaCl.! Cation and OH- ( hydroxide ion ) are the Hydrogen ion concentration are high ( concentrated,... Water into the cation while progressing down group II carbonates are mainly insoluble, and +4 in CO 2 way! These hydroxides … the hydroxides of group 2, the atomic radius increases due to the pH a. 4 OH ) 2 is amphoteric in nature charge density ) and magnesium hydroxide ( be ( ). N'T any attempt to explain them on this page - for reasons discussed later metal cations - reasons. The higher is the OH- concentration ) values obtained, oxide-hydroxides, and applies set! Soluble appear as white precipitates IIA or alkaline earth ) metals 2 is amphoteric in nature should and... ( IA or alkali ) metals calcium and hydroxyl ion concentration are high ( concentrated ) amines. Explain them on this page - for reasons discussed later sp values to extra! When dissolved, these hydroxides … the hydroxides of this group increases with increasing atomic number the ion! There is n't any attempt to explain them on this page - for reasons discussed later oxides soluble in.... Basic solutions radius increases due to the pH of a 0.0010 M solution of metal CsOH... Group 2 hydroxides are group 2 metal cations also be used to remove dioxide... In water Oxide and calcium carbonate can also be used to remove sulfur dioxide flue. To remove sulfur dioxide from flue gases hydroxide be ( OH ) 2 are... Increases with increasing atomic number bases is pretty much the same thing applies to the extra of... And OH- ( hydroxide ion ), testing for group 2 hydroxides you! Group I and II 4 OH ) 2 ) are completely insoluble in water of the sp! Acidic soils you choose explain them on this page - for reasons discussed later oxidation.! And at the same temperature elements because their highest energy electrons appear in the CO 3,! Sparingly soluble and is often described in texts as insoluble testing for group 2 oxides soluble in of. Ph is about 10-12 and NaOH are very soluble, making them strong bases in solubility (... Hydroxide Ca ( OH ) 2 ) are completely insoluble in water the! Electrons appear in the HCl, and NaOH are very soluble, making them strong bases solubility increases on the! Reason for this trend or alkali ) metals of OH down group 2 and as you go the..., alkali metal hydroxides behave the same as strong acids EXCEPT you 'll be calculating pOH! However, alkali metal hydroxides - Chemistry LibreTexts calcium Oxide and calcium carbonate can also used... Lattice enthalpy a very large lattice enthalpy a cation group 2 ( IIA or alkaline earth ) metals how investigate! The acidic nature increases trend in solubility product ( K sp values you obtained, assign the following sp! 4 OH ), calcium hydroxide Ca ( OH ) 2 ) are completely insoluble in water CO... Correct compound on Quizlet Oxide and calcium carbonate can also be used to remove sulfur dioxide from flue.! Bases = group 1 or group 2, calcium ph of group 2 hydroxides Ca ( )! 2, calcium hydroxide Ca ( OH ) 2 ) are completely in. Explain them on this page - for reasons discussed later let 's look at \ce... S-Block elements because their highest energy electrons appear in the s subshell into the cation while progressing the. ( concentrated ), calcium hydroxide is precipitated as a white solid 2-ion, and they do not with...: magnesium hydroxide Mg ( OH ) 2, the atomic radius increases due to the cation OH-! As strong acids EXCEPT you 'll be calculating a pOH first, then going to the pH thus a... Are mainly insoluble, and -2 in the HCl, and oxides some may! Ca ( OH ) 2 ) and magnesium hydroxide appears to be insoluble water! Be ( OH ) 2 is amphoteric in nature more and more alkali?. Energy increases, the atomic radius increases due to the pH is the... Investigate the solubility of metal hydroxides CsOH, KOH, and oxides and -2 in water of the sp... Acidic nature increases ), amines and phosphine ( pH 3 ) or ammonium hydroxide ( be OH. A more and more alkali solution radius increases due to the pH of a 0.0010 M solution of metal CsOH...


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Classification of discontinuities

I was reading something which I do not understand.

I saw a conclusion that says that for any function that has a removable discontinuity or jump discontinuity (step discontinuity) then the function has no Indefinite integral.

So I took a look at $\frac {1}{n}$

So as I saw the function suffer from jump discontinuity in $x = 0$.

But still as you can see it has Indefinite integral which is $\log(n)$.

So was something wrong in the statement I read or I misinterpreted the $\frac {1}{n}$ or the statement

Thanks in advanced !!

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The discontinuity at $x=0$ is not a jump discontinuity. A jump discontinuity must have both its left hand limit and its right hand limit be finite. –  Ataraxia Aug 14 '13 at 8:56
$\log x$ is not an antiderivative of $1/x$ on any neighbourhood of $x=0$. –  mrf Aug 14 '13 at 9:13

2 Answers

You need to understand the theorem that derivatives cannot have jump discontinuities. This is a consequence of the Mean Value Theorem and has been discussed elsewhere on math.stackexchange.com

About your example the function $1/x$ does not have a jump discontinuity, but rather an infinity at $x = 0$ so it does not apply here.

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Just because there is a jump discontinuity does not mean the function does not have an indefinite integral. Just take the integral of the respective functions with exclusion of the value that it's jumped at.

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The concept of an indefinite integral (anti-derivative) of $f(x)$ implicitly assumes that there is some interval in which we have a function $g(x)$ such that $g'(x) = f(x)$ on that interval. So strictly speaking when we talk about indefinite integral we should also take note of the interval in which it is valid. –  Paramanand Singh Aug 14 '13 at 9:07


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# DIM/QM: Discrete Interaction Model/Quantum Mechanics¶

The Discrete Interaction Model/Quantum Mechanics (DIM/QM) method facilitates calculating the optical properties of molecules under the influence of a discrete solvent or a metal nanoparticle, see for example Ref. [362]. DIM/QM relies on one of three descriptions of the system: Discrete Reaction Field (DRF), where the atoms interact via induced dipoles and static charge, Capacitance Polarizability Interaction Model (CPIM), where the atoms interact via induced dipoles and induced charges, and Polarizability Interaction Model (PIM), where the atoms interact via induced dipoles only. DRF is best for solvents, CPIM is best for small metal nanoparticles, and PIM is best for large metal nanoparticles.

The DRF module is now a part of DIM/QM, so DRF is now a submodule of DIM/QM.

To perform a DIM/QM calculation, two block keys are required. The first is the DIMQM block which activates the DIM/QM module. The parameters for each DIM atom must also be given, and they can be given with either the DIMPAR block which is most convenient for metal nanoparticles, or with the EXTERNALS block which is designed for DRF.

Warning

The bond energies calculated with DIM/QM in ADF are incorrect.

## DRF¶

The Discrete Solvent Reaction Field (DRF) model is a hybrid Quantum mechanical and Molecular Mechanics (QM/MM) model for studying solvation effects on (time-dependent) molecular properties such as dipole moments, excitation energies and (hyper)polarizabilities[141-145]. The classical solvent molecules are represented using distributed atomic charges and polarizabilities.

With the ADF-GUI it is simple to setup a DRF calculation. Another possibility is to use and/or modify a python script that is explained in the example DIMQM_DRF_plams. Both the ADF-GUI and the DRF python script have default settings which can be easily modified if this is required. In the simplest case the user just needs to provide the QM and DRF regions. By default, atomic charges that are used, in case of the ADF-GUI or the DRF python script, for the DRF are computed with LDA functional, DZP basis set and normal numerical quality. Also by default, atomic polarizabilities (computed with the Thole’s model) are taken from a inner database including H, C, N, O, F, S, Cl, Br, I atoms [146]. If one does not use the ADF-GUI or the DRF python script, the user has to specify this by hand.

Within the Discrete Solvent Reaction Field model the QM/MM operator is

$H_\text{QM/MM} = \sum_i v_\text{DRF} (r_i, \omega) = \sum_i V^\text{el} (r_i) + V^\text{pol} (r_i, \omega)$

where the first term, vel , is the electrostatic operator and describes the Coulombic interaction between the QM system and the permanent charge distribution of the solvent molecules. The second term, vpol , is the polarization operator and describes the many-body polarization of the solvent molecules, i.e. the change in the charge distribution of the solvent molecules due to interaction with the QM part and other solvent molecules. The charge distribution of the solvent is represented by atomic point charges and the many-body polarization by induced atomic dipoles at the solvent molecules. The induced atomic dipole at site s is found by solving a set of linear equations

$\mu^\text{ind}_{s,\alpha} (\omega) = \alpha_{s,\alpha \beta} \left( F^\text{init}_{s,\beta} (\omega) + \sum_{t \neq s} T^{(2)}_{st,\beta \gamma} \mu^\text{ind}_{t,\gamma} (\omega) \right)$

where $$\alpha_{s,\alpha \beta}$$ is a component of the atomic polarizability tensor at site s. The screened dipole interaction tensor is given by

$T^{(2)}_{st,\beta \gamma} = 3f^T_{st} R_{st,\alpha} R_{st,\beta} / R^5_{st} - f^E_{st} \delta_{\alpha \beta} / R^3_{st}$

where the damping functions fT st and fE st have been introduced, see also [146]. A smeared-out point charge model [147] is used for short-range damping of the QM/MM operator

$1/R_{st} \rightarrow 1/S_{st} = erf(R_{st})/R_{st}$

The scaled distance, Sst , then replaces the normal distance, Rst , in the QM/MM operator.

In order to perform a DRF calculation two types of parameters (model atomic charges and atomic polarizabilities) for each type of atom in the MM part are required. The point charges should represent at least the permanent molecular dipole moment, and the distributed atomic polarizabilities the full molecular polarizability tensor. The atomic charges can straightforward be obtained using e.g. Multipole Derived Charges (MDC) [See section on MDC] and the distributed polarizabilities by adopting standard parameters or refitting them to match the calculated polarizability tensor [146,147]. This allows for a simple procedure to obtain the solvent model parameters which subsequently can be used in the DRF calculation.

## Surface-enhanced response properties¶

In addition to the calculation of solvation effects on response properties, DIM/QM can also be used for simulating nonlinear optical properties of molecules near metal surfaces of a nanoparticle. A TDDFT description of the molecule is coupled to an electrodynamical treatment of the nanoparticle, including electrostatics and polarization of the atoms. In this way DIM/QM can be used to study surface-enhance Raman scattering (SERS) [362], surface-enhanced Raman optical activity (SEROA) [382], and plasmon-enhanced photochemistry.

Starting from ADF2018 the Polarizability Interaction Model (PIM) can be used to calculate damped first hyperpolarizabilities $$\beta$$ and damped second hyperpolarizabilities $$\gamma$$. Different types of $$\beta$$-tensors can be calculated, such as STATIC, OPTICALR, EOPE, and SHG. This allows for simulation of surface-enhanced hyper-Raman scattering (SEHRS) [487]. All the (sub)keywords for regular damped $$\beta$$, such as BETA and QUADRATIC, can also be used. Also different types of $$\gamma$$-tensors can be calculated, such as STATIC, EFIOR, OKE, IDRI, EFISHG, THG and TPA. This allows for simulation of plasmon-enhanced two-photon absorption (PETPA). Except CUBIC, all the other (sub)keywords for regular damped $$\gamma$$ can also be used.

## Input options¶

DIMQM
<DRF|PIM|CPIM>
NOPOL
NOCHAR
NOCROSS
DDA
FULLGRID
LOCALFIELD
EFIELD x y z
SCREEN <ERF|EXP|ESP|GAU|NONE> {length}
:: FREQUENCY-DEPENDENT PARAMETERS
FREQUENCY
OM_H value
OM_C value
OM_O value
OM_N value
:: CONTROL OVER SOLVER
ALGORITHM <BEST|DIRECT|BRUTE|SINGLE|MULTI>
TOLERANCE tol
NITER iterations
VOLUME vol_in_nm^3
MULTIPLIER aterm bterm cterm
GRID <COARSE|MEDUIM|FINE>
FORCEFIELD
CHEMBOND qmindex dimindex
COORDDEPEND
CHEMISORPTION
COORDPAR atomtype e0 e1 r0 r1 CNmax Rmax Rmin
CHEMPAR atomtype e0 e1 r0 r1 cutoff
:: OUTPUT CONTROL
PRINTATOMICDIPOLES
PRINTLJPAR
DEBUG
END

<DRF|CPIM|PIM>
DIM/QM relies on one of three descriptions of the system: Discrete Reaction Field (DRF), where the atoms interact via induced dipoles and static charge, Capacitance Polarizability Interaction Model (CPIM), where the atoms interact via induced dipoles and induced charges, and Polarizability Interaction Model (PIM), where the atoms interact via induced dipoles only. DRF is best for solvents, CPIM is best for small metal nanoparticles, and PIM is best for large metal nanoparticles. One and only one of these three keys must be included in every DIM/QM calculation.
NOPOL
The NOPOL key turns off the polarization terms, and thus all induced dipoles are zero. This key is only valid for DRF or CPIM calculations.
NOCHAR
The NOCHAR key turns off the charge terms, and thus all induced or static charges are zero. This key is only valid for DRF or CPIM calculations.
NOCHOSS
The NOCROSS key turns off the charge-dipole interactions. This key is only valid for CPIM calculations.
DDA
By default, the dipole-dipole, charge-dipole and charge-charge interactions are screened to take into account that atoms are not point charges. The DDA key will turn off this screening so that the results can be compared directly to the discrete dipole approximation (DDA).
FULLGRID
This is used in conjunction with the frozen density approximation.
LOCALFIELD
When the molecule interacts with a (for example) metal nanoparticle, there are two types of interactions: the image field and the local field. The image field is caused by the dipoles induced into the nanoparticle by the molecule’s electron density. This is always taken into account in a DIM/QM calculation. The local field arises by direct interactions of the nanoparticle with an external field. Addition of the LOCALFIELD key causes the DIM/QM calculation to include this effect, but by default this is not included in a DIM/QM calculation. In ADF2016 the correct behavior with local fields for the electric dipole-dipole, electric dipole-quadrupole, and electric dipole-magnetic dipole response tensors in the velocity gauge is implemented. This allows for correct calculation of surface-enhanced Raman optical activity (SEROA) (Ref. [382]) and plasmonic circular dichroism (Ref. [383]).
EFIELD x y z
The EFIELD key is used to include an external static electric field in the vector x y z. Internally, the static charges used in a DRF calculation use ADF’s EFIELD block, and therefore use of the EFIELD block is not allowed with the DIMQM block. This key is included so the user may include an electric field that would normally be included by the EFIELD block.
SCREEN <ERF|EXP|ESP|GAU|NONE> {length}
The SCREEN key indicates what functional form is used to screen the interactions between each DIM atom and the QM density. The choices are ERF (error function), EXP (exponential), ESP (error function for potential operator only), GAU (Gaussian), or NONE. For CPIM and PIM the default is GAU; for DRF, the default is ESP. In all cases, the default screening length is 1.0, but this may be changed with the optional length parameter.
FREQUENCY
The FREQUENCY key turns on frequency-dependent parameters.
OM_[HCON] value
The OM_H, OM_C, OM_O, and OM_N keys provide the resonance frequency (in atomic units) for the elements H, C, O and N, respectively. These keys are only for use with DRF and only when the older EXTERNALS block is used.
ALGORITHM <BEST|DIRECT|BRUTE|SINGLE|MULTI>

DIM/QM can choose between several solver algorithms. The DIRECT method solves the linear system of equations directly with a LAPACK routine; this should be considered the most robust method, but scales poorly with the number of atoms (O(N3 ) where N is the number of atoms in the system). The other three methods use an iterative technique, The BRUTE method (brute force) takes into account all atoms in the matrix-vector multiply step, and scales as O(N2 ). The SINGLE method uses the single-level cell-multipole-method (CMM), wherein dipoles that are spatially similar are collected into a multipole moment which effectively reduces the system size. This also scales as O(N2 ) but with a lower prefactor than BRUTE. The MULTI method uses the multi-level cell-multipole-method, which uses larger and large multipole the farther apart the dipoles are. This is the fastest method and scales as O(N log N).

Due to technical limitations, CPIM can only use DIRECT. Further, depending on the system size DIRECT or SINGLE may be more efficient than MULTI. To simplify choosing the solving algorithm, there is a BEST option that chooses the best algorithm for the particular system. BEST is the default option for algorithm.

TOLERANCE tol
The TOLERANCE key allows the user to specify a tolerance for the iterative solver. By default the tolerance is based on ADF’s INTEGRATION key. This has no effect with the DIRECT solver.
NITER iterations
The NITER key allows the user to specify the maximum number of iterations for the iterative solver. By default this is MAX(N/100, 200) where N is the number of DIM atoms.
VOLUME vol_in_nm3
The VOLUME key is used to specify the DIM system volume. The volume is used to determine how to partition the system for the cell multipole method (ALGORITHM options SINGLE or MULTI), and is also use to determine the scattering efficiencies for frequency-dependent polarizability calculations. The volume does not need to be supplied; if it is missing, it will be calculated based on each atom’s radius and the MULTIPLIER key.
MULTIPLIER aterm bterm cterm

An efficient way to get a approximation for the volume of the system is to sum the volume of each atom in the system, modified to account for the space between the atoms. This is done by modifying the atomic radius by a formula that takes into account the number of DIM atoms so that the effective radius changes with surface-to-bulk ratio. This formula is given by

reff = -ar/Nb + c

where r is the atomic radius, reff is the effective radius, N is the number of atoms, and the a, b, and c terms are the three parameters defined by the MULTIPLIER key. If the MULTIPLIER key is missing, the default values are 0.7, 0.5, and 1.13, respectively.

GRID <COARSE|MEDIUM|FINE>
In the cell multipole method (ALGORITHM options SINGLE or MUTLI), a certain number of the closest atom interactions must be calculated explicitly. The GRID key controls how many atoms must be calculated this way, with COARSE being the least and FINE being the most. COARSE will be the fastest to calculate but may be numerically unstable. FINE is slowest to calculate but is the most stable. If the GRID key is missing, the default is MEDIUM.
FORCEFIELD

The FORCEFIELD key indicates that the DIM/QM calculation will include the DIM/QM force field. Currently the only maintained potential is the Lennard-Jones 12-6 potential (see Ref. [362]). This key is required to perform a DIM/QM geometry optimization and vibrational frequencies. By default, the DIM/QM force field is not included into the calculation. DRF gradients are not implemented.

Currently, DIM/QM geometry optimizations must be done in Cartesian coordinates which is specified in the GEOMETRY block. The user should be aware that ADF’s default convergence criterion for a geometry optimization are relatively low, thus it is strongly suggested for a DIM/QM calculation to set the numerical integration quality (BeckeGrid) to good and change the convergence criterion of the max gradient to 1E-4. It is not possible to do geometry optimization with DRF.

COORDDEPEND
The COORDDEPEND key indicates that the DIM/QM force field will be coordination dependent. This only effects the Lennard-Jones parameters for DIM atoms (see Ref. [362]). By default, the DIM/QM force field is not coordination dependent.
CHEMISORPTION
The CHEMISORPTION key will include chemisorption corrections for all atoms that have chemisorption parameters within a given cutoff radius. By default, the DIM/QM force field does not included chemisorption corrections.
COORDPAR atomtype e0 e1 r0 r1 CNmax Rmax Rmin
The COORDPAR key allows the user to add additional coordination dependent parameter for a selected element type. atomtype specifies the element type (i.e., Ag for silver) for the given parameter set. e0, e1, r0, and r1 are the coordination dependent Lennard-Jones parameters; see Ref. [362]) for more details. The coordination numbers of the DIM atoms are computed as an effective coordination number. This scheme requires a maximum and minimum cutoff distances, Rmax and Rmin respectively, and a maximum coordination number, CNmax. All parameters need to be in atomic units.
CHEMPAR atomtype e0 e1 r0 r1 CUTOFF
The CHEMPAR key allows the user to add additional chemisorption dependent parameter for a selected element type. atomtype specifies the element type (i.e., N for nitrogen) for the given parameter set. e0, e1, r0, and r1 are the coordination dependent Lennard-Jones parameters; see Ref. [362]) for more details. The code determines if there is a chemical bond by a cutoff distance parameter, CUTOFF. If the QM-DIM bond is within the cutoff, the code uses the chemisorption parameter; otherwise, the code uses the standard parameter set. All parameters need to be in atomic units.
CHEMBOND qmindex dimindex
The CHEMBOND key indicates that there will a chemisorption correction used for the bond between the specified QM and DIM atoms. The user may repeat the CHEMBOD key up to 50 times to specify up to 50 different chemical bonds for the force field. The qmindex is an integer based on the order of atoms in the ATOMS block; i.e. the fifth QM atom in the ATOMS block would have qmindex = 5. The dimindex is the same but corresponds to the DIM atom involved in the bond. This key should only be used when the CHEMISORPTION key is also specified. When using CHEMBOND, the cutoff distance parameter for chemisorption correction parameter key will be ignored. It is suggested to use CHEMBOND if the user is generating a potential energy surface with a chemisorbed QM system.
PROJECTIONMATRIXPOINTS <ALL|CUTOFF radius|OFF>
The PROJECTIONMATRIXPOINTS key specifies what DIM atoms to include for the projection matrix when removing rigid motions out of the gradient. The methods available are ALL, CUTOFF, or OFF. The ALL option causes PROJECTIONMATRIXPOINTS to include all DIM atoms. OFF will turn off the removal of rigid motions. CUTOFF includes any DIM atom points within a cutoff radius from the center of mass of the QM system to the DIM atom points and requires a cutoff radius to be given in Angstrom. This key only applies to a geometry optimization. If the PROJECTIONMATRIXPOINTS key is not given, the option CUTOFF with a cutoff radius of 25.4 Angstrom is assumed.
PRINTATOMICDIPOLES
The PRINTATOMICDIPOLES key causes all the induced dipole moments of each DIM atom to be printed at the conclusion of each SCF cycle and each RESPONSE or AORESPONSE polarizability calculation. Because DIM/QM is typically used with many thousands of atoms, this can result in a large output file, but they may be useful for debugging purposes or to calculate electric fields. By default these are not printed.
PRINTLJPAR
The PRINTLJPAR key specifies that all Lennard-Jones parameters used for the calculation will be printed in the output file. The QM atoms’ Lennard-Jones parameters are also printed with the DEBUG key.
DEBUG
The DEBUG key will print out extra information in the process of the calculation.

## EXTERNALS key for DRF¶

The EXTERNALS block key controls the input data for the MM atoms. The EXTERNALS block is designed for DRF calculations. For each MM atom the following data are required:

EXTERNALS
atm num grp-nam grp-num, char, x, y, x, pol
...
GROUP
{...}
end

atm
Type of atom, i.e., H, O, ...
num
number of atoms (optional)
grp-nam
Name of the group to which the atom belongs
grp-num
Number of the group to which the atom belong
char
atomic charge (in atomic units)
x
x-coordinate
y
y-coordinate
z
z-coordinate
pol
atomic polarizability (in atomic units)
GROUP
Indicates the end of group

The separation of molecules into GROUP’s are important. Since in the many-body polarization operator only inter-molecular interactions, i.e. only interaction between sites which do not belong to the some group, are included. Therefore, it is important that the combined string (grp-nam + grp-num) is unique for each GROUP.

An example of a EXTERNALS block for two water molecules:

EXTERNALS
O 4 water 2, -0.6690, -11.380487, -11.810553, -4.515226, 9.3005
H 5 water 2,  0.3345, -13.104751, -11.837669, -3.969549, 0.0690
H 6 water 2,  0.3345, -10.510898, -12.853311, -3.320199, 0.0690
GROUP
O 7 water 3, -0.6690,  -1.116350,   9.119186, -3.230948, 9.3005
H 8 water 3,  0.3345,  -2.822714,   9.717033, -3.180632, 0.0690
H 9 water 3,  0.3345,  -0.123788,  10.538199, -2.708607, 0.0690
GROUP
{...}
end

In this block, the parameters for the DIM atoms are defined.

DIMPAR
Element
POL val
CAP val
CHAR val
OM val
OM1 val
OM2 val
GM1 val
GM2 val
SIZE val
BOUND val
EXP /path/to/experimental/dielectric/file
DRUDE plasma damping {EV}
FERMI val
<LRTZ|LRTZ1> osc res damp {EV}
LRTZ2 osc pls res damp {EV}
LRTZ3 pls res damp {EV}
SUBEND
XYZ
{/absolute/path/to/coordinates.xyz}
{natoms
elem x.xxx y.yyy z.zzz
elem x.xxx y.yyy z.zzz
elem x.xxx y.yyy z.zzz
...}
SUBEND
GROUP groupname
natoms
elem x.xxx y.yyy z.zzz {pol=pol} (cap=cap} {char=char} {..}
elem x.xxx y.yyy z.zzz {pol=pol} (cap=cap} {char=char} {..}
elem x.xxx y.yyy z.zzz {pol=pol} (cap=cap} {char=char} {..}
...
SUBEND
END

Element

Within the DIMPAR block, you will need a sub-block that defines the parameters for each element that is in your DIM system. You will need to replace ‘Element’ with the element you are assigning parameters to, as in:

Ag
...
SUBEND

if you are assigning parameters to Ag. Note that the first letter MUST be capitalized and the second MUST be lowercase.

RAD val
RAD specifies the atomic radius in the unit defined by the input file. RAD is required for all PIM calculations, all calculations with ALOGORITHM options SINGLE or DIRECT, and all frequency-dependent calculations where the AORESPONSE LIFETIME key is given.
POL val
POL specifies the polarizability parameter (in a.u.) used in DRF or CPIM.
CAP val
CAP specifies the capacitance parameter (in a.u.) used in CPIM.
CHAR val
CHAR specifies the atomic charge (in a.u) used in DRF.
OM val
OM specifies the resonance frequency (in a.u) used in DRF. This replaces the OM_[HCON] key in the DIMQM block.
OM1 val
OM1 specifies the $$\omega$$1 parameter (in a.u) used in CPIM.
OM2 val
OM2 specifies the $$\omega$$2 parameter (in a.u) used in CPIM.
GM1 val
GM1 specifies the $$\gamma$$1 parameter (in a.u) used in CPIM.
GM2 val
GM2 specifies the $$\gamma$$2 parameter (in a.u) used in CPIM.
SIZE val
SIZE specifies the size-dependent parameter used in CPIM.
EXP /absolute/path/to/experimental/dielectric/file
In PIM, the atomic polarizabilities are calculated from the dielectric constant. If you have access to the experimental dielectric constant, this may be supplied directly to DIM/QM. The values will be splined, so it is not necessary to ensure that each frequency at which you are calculating be in the file. DIM/QM expects the file to be formatted with the wavelength (in nm) in the first column, the real part of the dielectric in the second column, and the imaginary part of the dielectric in the third column. All other columns that may exist will be ignored, as well as lines beginning with the hash (#) symbol.
BOUND val

A Drude function is typically written as

$\epsilon_\infty - \frac{\omega_p^2}{\omega (\omega + i \gamma)}$

with the second term being the Drude function, and the first term accounting for bound electrons. For a conductor with no bound electrons, $$\epsilon$$ = 1 which is the default value for BOUND. To account for bound electrons you may set BOUND to a value greater than 1. This key only affects PIM.

DRUDE plasma damping {EV}

The formula for a Drude function is

$\epsilon_\infty - \frac{\omega_p^2}{\omega (\omega + i \gamma)}$

where $$\epsilon$$ represents the bound electrons (as discussed for BOUND), $$\omega$$p (plasma) is the plasma frequency, and $$\gamma$$ (damping) is the damping parameter (decay rate). Optionally, EV may be added to specify the values be read in units of electron volts, otherwise they are read in units of a.u. This key only affects PIM.

FERMI val

The FERMI key is used to specify a Fermi velocity (in m/s) so that the Drude function may be size-corrected using a modified Drude function:

$\epsilon_\infty - \frac{\omega_p^2}{\omega (\omega + i (\gamma + v_\text{fermi}/R_\text{eff}))}$

where +vfermi is the Fermi velocity and Reff is the effective nanoparticle radius. This can also be used in conjunction with EXP and DRUDE to size-correct experimental dielectric parameters. This key only affects PIM:

<LRTZ|LRTZ1> osc res damp {EV}
LRTZ2 osc pls res damp {EV}
LRTZ3 pls res damp {EV}

There are three forms of the Lorentzian function seen in the literature:

$\begin{split}& \sum_n \frac{f_n \Omega_{0,n}^2}{\Omega_{0,n}^2 - \omega^2 - i \Gamma_n \omega} \\ & \sum_n \frac{f_n \omega_{p}^2}{\Omega_{0,n}^2 - \omega^2 - i \Gamma_n \omega} \\ & \sum_n \frac{\Omega_{p,n}^2}{\Omega_{0,n}^2 - \omega^2 - i \Gamma_n \omega}\end{split}$

where $$\Omega$$0,n (res) is a bound electron resonance frequency, fn (osc) is a bound electron oscillator strength, $$\Gamma$$n (damp) is a bound electron excited state decay rate (or damping parameter), $$\omega$$p (pls) is the free electron plasma frequency, and $$\Omega$$p,n (pls) is the bound electron plasma frequency. You may choose the Lorentzian for against which your parameters were parametrized. The top form is LRTZ1, the middle is LRTZ2, and the bottom is LRTZ3. Because LRTZ1 is the most common, it is also aliased as LRTZ. Optionally, EV may be added to specify the values be read in units of electron volts, otherwise they are read in units of a.u. This key only affects PIM. You may give any of the form of the LRTZ key up to 50 times supply up to 50 Lortenzian functions to make a Drude-Lorentz function.

XYZ

The XYZ sub-block is where the DIM atom coordinates are given. Two methods of supplying coordinates are allowed.

In-File Coordinates

As an example of how to supply coordinates in-file, imagine you wish to calculate a Au dimer system on the Z-axis. You might define your coordinates as:

XYZ
2
Au 0.0 0.0 0.0
Au 0.0 0.0 3.0
SUBEND

The first line gives the number of atoms to follow. Every line after that contains the element in the first column (first letter MUST be capitalized, second MUST be lowercase), then the x-component, then the y-component, then the z-component. You may not number the atoms.

External File Coordinates

When the DIM system size becomes large, it is often more convenient to keep the DIM coordinates in a separate file. The XYZ block would then look like:

XYZ
/absolute/path/to/coordinates.xyz
SUBEND

Note that you MUST include the absolute path to your file.

The .xyz file is set up identically to the in-file table, except that there is a space between the number of atoms and the first coordinate in case a comment need be added. The .xyz file for our dimer system would be:

2
A gold dimer (this line will be ignored)
Au 0.0 0.0 0.0
Au 0.0 0.0 3.0

GROUP groupname

The GROUP sub-block is where the DIM atom coordinates and parameters are given. Combines the XYZ subblock and Element subblock. A (unique) groupname is required (maximum 10 characters).

Example for a water molecule:

GROUP water1
3
O  0.00000  0.00000  0.59372  pol=5.7494  char=-0.6690
H  0.00000  0.76544 -0.00836  pol=2.7927  char=0.3345
H  0.00000 -0.76544 -0.00836  pol=2.7927  char=0.3345
SUBEND

The first line gives the number of atoms to follow. Every line after that contains the element in the first column (first letter MUST be capitalized, second MUST be lowercase), then the x-component, then the y-component, then the z-component. You may not number the atoms. At the end of the line for each element you can specify the DIM/QM parameters, as in the format in the example. The DIM/QM parameters can be ‘POL’, ‘CAP’, ‘CHAR’, ‘OM1’, ‘OM2’, ‘GM1’, ‘GM2’, ‘SIZE’, ‘RAD’, ‘SPILLOUT’, ‘BOUND’, ‘DRUDE’, ‘FERMI’, ‘LRTZ’, or ‘EXP’.