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## Sunday, May 13, 2012

### Nope, Nordhaus is still (mostly) right

[NB: See the UPDATE at the bottom of this post.]

Following our previous tête-à-tête on the Heartland Affair, here's a response to Bob Murphy's latest post on climate change.

In the interests of keeping this as a brief as possible, let me just say that my post follows a back-and-forth involving different groups of people. My aim here isn't to give an overview to the entire discussion, but rather to reply directly to Bob's key points. However, the four act summary is as follows...
• ACT 1: A group of 16 self-proclaimed climate skeptics write an op-ed in the Wall Street Journal, "No need to panic about global warming"
• ACT 2: William Nordhaus, professor at Yale University and one of the founding fathers of climate change economics, pens a widely cited essay in response, “Why the Global Warming Skeptics Are Wrong.” Among other things, Nordhaus takes umbrage with the fact that these skeptics have bungled his own research in trying to argue that climate change is not cause for concern.
• ACT 3: Enter Bob Murphy, who comes to defend the honour of the original skeptics in his article, "What Nordhaus Gets Wrong".
• FINAL ACT: Stickman Grant McDermott sweeps in from the shadows to set right the wrongs of the blogging world. Order is restored, women swoon and the sound of childrens' laughter fills the air. Murphy curses and retreats to his secret lair, vowing revenge.

Now that you've seen the plot spoiler, let's go through some details:

1) Global Temperatures Have Been Flat for a Decade

Bob claims that Nordhaus is using a rhetorical slight of hand in addressing the skeptic claim that global temperatures have not increased in 10 years. Nordhaus does this by referring to the instrumental temperature record, which dates back to the 19th century and shows a clear warming trend. Bob claims that this isn't fair, since "no warming in the last decade" and "no warming since 1880" are different statements.

To be frank, this is simply dancing around the issue. If the original claim was meant to serve no purpose beyond itself, then perhaps that would be okay. (Vacuous, but okay.) However, the implication and inference of the skeptics' original statement was abundantly clear: Global warming has stopped, so all this talk about dangerous increases in future temperatures is probably bunk.[*] Nordhaus rightly points out that this is a meaningless observation and that only long-term trends can provide proper context. However, Bob tries to head off this approach too, by providing a long-term graph with rescaled Y-axis; ostensibly to show that temperatures have hardly changed in absolute terms over the last 150 years.

This is simply disingenuous. Let's conduct a simple thought experiment to show why. Imagine that observed warming since the Industrial Revolution isn't caused by mankind, but rather some combination of natural factors. Let's posit that the observed long-term trend due to this natural forcing trend continues unabated, so that temperatures eventually exceed pre-industrial levels by more than 2°C. Will there be economic costs associated with this, relative to a world without warming? Well, according to virtually every single study on the matter, the answer is a clear "YES". If it wasn't, then we wouldn't even be having this discussion... Why do you think adaptation through economic growth is taken so seriously -- not least of all, by prominent contrarians? (Also, see #3 below.)

The take home message is this: Seemingly small changes on the global scale will have tremendous impacts on the climate and natural systems that we depend on. For instance, a 3°C rise in average global temps will correspond to a world that probably hasn't existed for millions of years, entailing massive unknowns and downside risks. Further, the changes in regional temperatures will actually be far more acute, even if these "balance out" to some extent at a global level. (This point is often sadly overlooked by the focus on global averages.)

2) Actual Global Warming Has Been Smaller Than What the Models Predicted

Bob largely bases his comments here on a Master Resource blog post, which in turn cites a paper by Santer et al. (2011). Now, the main purpose of this study is quite technical in that it aims to show how different timescales can affect the ability to distinguish between climate "noise" and "signal". (Ironically, their bottom line is that we gain no understanding of climate change drivers simply by looking at the temperature record from a single decade.) However, we don't really need to concern ourselves with that here. The relevant issue -- and the one that Bob limits his discussion to -- relates to the below graph. It shows temperature measurements of the lower troposphere (TLT), as "observed" by two satellites (RSS and UAH) and predicted by climate models. Without going into specific details, Bob uses this as evidence to suggest that climate models exaggerate the warming effect in comparison to what reality will turn up.

Allow me to make two comments in response. The first thing to note is that satellite temperature measurements appear to suffer from an unresolved cooling bias.[**] Santer et al. (ibid, p. 10) actually highlight this issue in their paper:
“Given the considerable technical challenges involved in adjusting satellite-based estimates of TLT changes for inhomogeneities [Mears et al., 2006, 2011b], a residual cool bias in the observations cannot be ruled out, and may also contribute to the offset between the model and observed average TLT trends.”
In other words, the observed temperatures in the above graph are probably lower than they "should" be. Santer et al. go on to list a bunch of other factors that could also explain the (slight) difference between these satellite measurements and the model predictions. I wish to emphasise that this difference is, indeed, "slight". More fundamentally, this relates to my second point: The models are still remarkably accurate. Again, let's turn to the actual paper (p.10):
“There is no timescale on which observed trends are statistically unusual (at the 5% level or better) relative to the multimodel sampling distribution of forced TLT trends. We conclude from this result that there is no inconsistency between observed near-global TLT trends[...] and model estimates of the response to anthropogenic forcing.”
Translation: The models have performed exactly within the bounds of what we would hope for. I find it... interesting that Chip Knappenberger (author of the MR post) chooses to ignore these caveats and findings, instead offering his readership an alternative conclusion that the models are on the "verge of failing". But then what do Santer & co. know? I mean, they just authored the study.

As for the assertion that Nordhaus is arguing against a strawman by focusing on the role that human activity has played in driving climate change, versus the alternative (weaker?) claim that the models have simply overestimated warming full stop... Let's consider the original WSJ op-ed again: "The lack of warming for more than a decade[...] suggests that computer models have greatly exaggerated how much warming additional CO2 can cause." (Emphasis added.) You could perhaps argue that the statement is ambiguous, but I certainly don't think that Nordhaus is unjustified in highlighting the fact that, only by including CO2 alongside natural forcings, can we square model predictions with observed warming.

For more on the track record of computer models in predicting actual climate outcomes (including reconstructions of the past), see here.

3) CO2 Is Not a Pollutant / CO2 Poses No Harm to Humans

Now we get down to actual economics. Bob declares his unwillingness to wade into a semantic battle of how we define a pollutant... A smart move on his part, since defending the banal assertion that "CO2 can't be harmful because we, like, exhale it and stuff" -- a point that the WSJ skeptics effectively make -- could only lead to much wailing and gnashing of teeth. Instead, he thinks that he's caught Nordhaus at his own game by taking a deeper look at a study that he (i.e. Nordhaus) references in support of his arguments. The paper in question is a literature survey by Richard Tol (2009) entitled "The Economic Effect of Climate". The key figure is below, which Bob uses to claim that CO2 rise will bring net benefits; at least up until two degrees warming. [UPDATE: Bob Ward points out several typos in Tol's paper, which would bring the onset of (expected) negative effects forward. See here for an updated version of the below figure on the And Then There's Physics blog.]

Now, Bob is qualitatively correct here; several studies indicate that there are net gains to be had from some moderate level of warming. (Incidentally, this is not something that Nordhaus would appear to dispute.) There are several reasons for such benefits, including increased agricultural production in certain parts of the world. However, some crucial points must be emphasised:
• Look at the graph and tell me where your "ideal" temperature increase lies. The answer should be obvious: Around 1°C, since this is where we maximise welfare. Anything beyond that and you are doing worse than you would at the 1°C optimum.
• The build-up of CO2 is characterised by tremendous inertia. Similarly, it takes thousands of years for CO2 released into the atmosphere to return to natural sinks. Of course, you would also need to consider the structural barriers and delays involved in reorganising your economy away from fossil fuels...
• Simply put, you almost certainly need to establish a carbon price long before you reach the 1°C turning point... let alone the sharp negative effects that we expect beyond 2°C. Tol actually highlights this in his paper (p. 34): "Policy steps to reduce emissions of greenhouse gases in the near future would begin to have a noticeable affect on climate sometime around mid-century — which is to say, at just about the time that any medium-run economic benefits of climate change begin to decline." To his credit, Bob acknowledges these issues... albeit in a footnote. To my mind, this is such a fundamental matter that I can't help but think that it deserves more than footnote status. (I'd also like to know how he squares these issues with his conclusion that immediate mitigation efforts -- presumably the beginning of some carbon price -- should be cast in "serious doubt".)
Another reason to be very cautious about the benefits of unimpeded CO2 emissions include disproportionately large downside risks (i.e. "fat tails") and, indeed, even regular loss aversion. If someone offers you the chance to flip a coin so that you either win $10,000 with 50% probability, or lose$9,500 with 50% probability, would you take it? I suspect that a clear majority of people would decline to take part in such a game of chance, even though the expected value is positive (i.e. $250). The idea that people value losses more than gains is very well established in the theoretical and empirical literature. Lastly, we would also need to consider the regional disparities in any presumed gains. Africa, for instance stands to be the big loser from climate change, while Eastern Europe wins due to increased agricultural output. Needless to say, this throws up some very thorny ethical issues around the potential benefits that some countries might enjoy at the expense of others. 4) Nordhaus’ Own Work Shows Harms of Government Intervention Okay, Bob is quick to say that the WSJ skeptics "screwed up" here. Anyone that has read Nordhaus's work knows that he as long called for Governments to put a price on carbon. Still, Bob now offers up one of this own papers to show that... well, it would seem that changing various factors in a climate model can lead to substantially different results. To be honest, I'm not entirely sure what to do with this. Admittedly, I haven't read Bob's paper -- I'll hopefully do so when I get time -- but in his blog post at least he is making some pretty unremarkable points. For instance, Al Gore apparently has some very bad policy recommendations. Okay, but Nordhaus never mentions Gore in his essay and neither do the 16 WSJ skeptics. (Personally, I'm not particularly interested in what AG has to say about the economics of climate change and I certainly can't think of any major nation pushing for 90 percent emission reductions by 2050.) As for the fact that Nordhaus "assumes" that his calibrated carbon tax will be perfectly implemented over time... Um, yes, that's how benchmarks work. We describe the first-best scenario and then look at how deviations from that optimum will impact results. Nevertheless, let me try to reply in kind by saying that I, too, have nagging problems with Nordhaus' methodology. In particular, using observed market rates for making normative judgments and implicitly assuming perfect substitutability between man-made and natural goods. I have previously written about these issues here and here. THOUGHT FOR THE DAY: The Wall Street Journal continues to push climate change opinion that is highly misleading and, at times, patently wrong. Bill Nordhaus did us all a service in picking apart some of the more egregious mistakes, as exemplified by this particular op-ed. Bob Murphy makes some interesting points in response, but I don't see that he offers any substantial rebuttals of the fundamental criticisms levied at the "WSJ 16". Nordhaus continues to have the right of this, as far as I am concerned. Obfuscation, begone! UPDATE: Bob and I are have a little back and forth about this at his blog (also at Daniel's). However what I really wanted to show you is this comment by Richard Tol about his own paper: It's easy to misinterpret Figure 1 from Tol (2009). Initial warming is indeed likely to be beneficial: CO2 fertilization of crops, reduced spending on heating homes, and fewer cold-related deaths are the main factors. However, totals do not matter. The incremental impact turns negative around 1.2K. If we were able to control climate, we would warm the planet by 1.2K and stop there. However, the momentum of the climate system and the energy system is such that, if you accept the mainstream view of the workings of the climate, we cannot avoid 1.2K warming, or 2.0K warming for that matter. The initial benefit is thus a sunk benefit: We will enjoy it regardless of what we do. Indeed. ___ [*] The original WSJ op-ed also makes use of the much misrepresented quote by Kevin Trenberth from the "Climategate" emails. That is: "The fact is that we can't account for the lack of warming at the moment and it is a travesty that we can't." For more on that issue, see here. [**] Part of this probably has to do with the fact that they employ a convoluted modeling procedure to produce temperature readings. For starters, satellites do not measure temperature directly, but rather various signal wavelengths that must then be mathematically inverted to obtain indirect inferences of temperature. I should also say that the UAH temperature set referenced by the Santer et al. study, which is produced by John Christy and Roy Spencer, is particularly contentious. You can read more about these issues here and here. #### 15 comments: 1. "For instance, a 3°C rise in average global temps will correspond to a world that probably hasn't existed for millions of years, entailing massive unknowns and downside risks." What approximate year do you expect this 3 degree Celsius rise to be achieved? 1. Mark, of course it depends on many factors. The IPCC has a nice summary of expected mean global temperature rise under different scenarios: http://www.ipcc.ch/publications_and_data/ar4/wg1/en/spmsspm-projections-of.html The short answer is that a "BAU" progression will take us to the three degree threshold (relative to 1980-1999 levels) towards the end of the century. 2. "The short answer is that a "BAU" progression will take us to the three degree threshold (relative to 1980-1999 levels) towards the end of the century." It's interesting that you should use the phrase "BAU," because it doesn't appear on the IPCC page you referenced. What does appear on the page are 6 scenarios. In four of the six scenarios, the warming for 2090-2099 (ralative to 1980-1999) is less than 3 degrees Celsius. But let's say you think that the warming for 2090-2099 relative to 1980-1999 is going to be 3 degrees Celsius. Does it bother you that, in all of recorded history, there has never been a single decade with warming of 0.3 degrees Celsius, but you are expecting warming to *average* 0.3 degrees Celsius or more for the next 10 decades? Don't you think your expectation is slightly unrealistic? For example, wouldn't you expect temperatures to rise in response to the logarithm of greenhouse gas concentrations? That is, wouldn’t you expect that a logarithmic increase of greenhouse gases would be necessary to sustain a linear increase in temperatures? And therefore, if greenhouse gas concentrations are increasing less than logarithmically, the temperature increase would be less than linear, rather than greater than linear? Also, with regard to Richard Tol’s analysis, it appears that the “best estimate” for damage to the economy caused by a 3 degree Celsius rise would be minus 2.5 percent of GDP. Why do you think we should care whether the people of 2090-2099 happen to experience a 2.5 percent GDP decline relative to if we could miraculously keep temperatures constant? For example, can’t the world GDP be expected to increase by more than 2.5 percent per year…so that trimming off 2.5 percent means essentially that the same GDP that would be reached in 2090 is now reached in 2091, and so on? And why the comparatively poor people of 2012 sacrifice anything for the comparatively rich people of 2090-2099. In 2012, the world per capita GDP is about$10,000 (PPP, 2012 dollars). What do you think the per capita GDP will be in 2095, PPP, 2012 dollars? The world-wide average life expectancy at birth in 2010 was 67 years. What do you think it will be in 2095?

Finally, you write, “Similarly, it takes thousands of years for CO2 released into the atmosphere to return to natural sinks.” How do you think that statement is relevant? What do you think the world per capita GDP will be in, say, 2162 (150 years from now)…again, in PPP, 2012 dollars?

3. It's interesting that you should use the phrase "BAU,"...

Why interesting? Have you read the scenario descriptions? If so, I'm not sure how you could find it objectionable to describe those upper prediction paths as "BAU", i.e. steady economic growth coupled with unregulated emissions. (Incidentally, we are currently on track with the A2 scenario, which is precisely what I have described as BAU...)

Does it bother you that, in all of recorded history, there has never been a single decade with warming of 0.3 degrees Celsius?... [W]ouldn’t you expect that a logarithmic increase of greenhouse gases would be necessary to sustain a linear increase in temperatures?

With due respect, you appear to be making some well-worn errors here. First, future emissions paths (and hence temperature rise) cannot, of course, be explained as simple linear extrapolations from the trend of previous decades. The relevant issue is the underlying forcing mechanisms; which we expect to rise at an increasing rate. As for logarithmic-vs-linear radiative forcing, here and here are good places to start.

Your second-to-last paragraph gets to a core issue of climate change economics. The short answer to your question about why we should/do care about these issues is: The discount rate. Future net benefits (costs) are discounted over the full period so that they can be reflected in present value terms. Of course, this is the standard way of evaluating any multi-period flow of gains and losses; discounting provides a dispassionate tool for evaluating a highly emotive subject. (That said, debates over the discount rate itself are far from dispassionate. If you are interested in a further discussion, this post might interest you.) The broader point, however, is that these damages lower future growth relative to its baseline indefinitely into the future. This point is crucial to understanding why those future GDP losses matter, even if they appear small to you. On that note...

Regarding your choice of phrasing: "if we could miraculously keep temperatures constant". To me, this indicates that you have fundamentally misunderstood the message in Tol's paper. Those are not the aims suggested by the likes of Nordhaus & co at all. For instance, here's Nordhaus (p. 14) laying out his "optimal" policy scenario:

"Our estimate of the optimal emissions-reduction rate for CO2 relative to the baseline is 15 percent in the ﬁrst policy period, increasing to 25 percent by 2050 and 45 percent by 2100. This path reduces CO2 concentrations, and the increase in global mean temperature relative to 1900 is reduced to 2.6°C for 2100 and 3.4°C for 2200"

Does that sound like a policy that would cripple the present world economy? (For the record, I personally think that Nordhaus understates the negative impact of warming on future wealth. The previous link on the discount rate provides one reason related to his implicit assumption of perfect substitutability between man-made and environmental goods.)

To finish, I don't understand your final comment. My statement was in relation to inertia of the carbon cycle and how our choices today will have no affect on the medium-term benefits that moderate warming is expected to bring (C.f. the "sunk" benefits described by Richard Tol in the post update.)

4. *effect, not affect in the last paragraph.

PS - Apologies if I come across as ornery in the above comment. I wrote it late last night after a long day. You do ask some important questions that anyone interested in this debate needs to grapple with. I hope that my answers provide some insight into why I hold the views that I do... as well as offer perspective on how economists evaluate the issues in climate change.

5. Hi Stickman,

"Why interesting?"

Your use of the phrase "BAU" is interesting because the IPCC goes to great lengths to point out that they do *not* have a "BAU" scenario. In fact, they are emphatic that they don't estimate the probability of occurrence of any of their scenarios:

"No judgment is offered in this Report as to the preference for any of the scenarios and they are not assigned probabilities of occurrence,..."

IPCC brags about their unscientific approach

This failure to assess the likelihood of particular scenarios allows people like you to make statements like, "Incidentally, we are currently on track with the A2 scenario, which is precisely what I have described as BAU..."

As Michael Jackson would say, "No, no...that's just ignorant." (No offense intended, and with due respect. ;-))

Seriously, there is no reason a person of your background (as far as I know of it) should have any idea why the A2 scenario is hilariously unrealistic. But ask anyone who has seriously studied energy and fuel trends, and they should be able to tell you why A2 is hilariously unrealistic.

Here is a nice webpage of the IPCC emission scenarios. Go to the A2 flavors. For example, A2 ASF, The World. Take a close look, and see if you see anything ridiculous. If you don't see it, I'll tell you. (Or you could ask Jesse Ausubel. ;-))

IPCC scenarios

Best wishes,
Mark

6. Hi Mark,

Apologies for ignoring you; I'm simply snowed with work at the moment. However, I do mean to have a go at properly responding to all your questions when I get a chance.

In the meantime, I've just clicked through to the IPCC scenario page that you provide above... specifically the A2-ASF-World link... And, yes, I would have to agree with your criticism that this cannot be described at "BAU" (even in the loose sense which I initially intended). For one thing, a population level of 15bn by 2100 is very much towards the upper probability tail of forecasts that I've seen. (I'd also have to take a closer look at that projected growth in coal use...) Mea culpa.

a) I initially put "BAU" in quotation marks, because it was meant to serve as a crude summary of scenarios where carbon emissions are left unregulated and economic growth remains fairly healthy.
b) I had assumed population growth rates of the B2 and A2 families were the same after scanning a document before posting my comment. My mistake and you are right to call me out on this.

7. Hi Stickman,

"For one thing, a population level of 15bn by 2100 is very much towards the upper probability tail of forecasts that I've seen. (I'd also have to take a closer look at that projected growth in coal use...)"

My reference was much more towards coal use. Converting the energy data in those Excel tables to coal use in billions of tonnes per year (using 29 GJ/Mg), scenario A2 averages 17 billion tonnes per year for 2010 to 2110, with a value in the year 2100 = 37 billion tonnes per year (the highest of any decade). Similarly, the A1F1 scenario averages 16 billion tonnes per year, with a peak value occurring in the year 2100, at 25 billion tonnes per year.

But a simple check of Wikipedia indicates that total world provable reserves life is only 118 years at the 2010 production rate of 7.273 billion tonnes per year. In other words, over the next 100 years, the A2 scenario envisions the world using about *2.3 times* the world's proven recoverable reserves. And the production value in 2100 is *still increasing,* at over 5 times the current production.

Wikipedia coal...see table on major producers

It simply doesn't make any sense. It's like nobody at the IPCC has even heard of the concept of "Hubbert's Peak."

Here's a paper by a group who has heard of the concept of Hubbert's Peak. They project global production peaking circa 2025...at "30 percent above current production." See Figure 3, for example. They predict production of hard coal will be under 3,000 million tons in 2100. Compare that with the IPCC A2 scenario, in which coal production in the year 2100 will be 41,000 million metric tonnes (or 45,000 million tons).

Energy Watch Group paper on future coal production

It's just insane. The A2 and A1F1 scenarios simply do not seem credible. Carl Sagan popularized the great phrase: "Extraordinary claims require extraordinary evidence." It is truly an extraordinary claim that the A2 and A1F1 scenarios could possibly come to pass...that coal production in 2100 could be as much as 5 times current production and still rising. The IPCC provides no evidence (of which I'm aware) that their coal usage projections for these two scenarios are credible.

8. [1/2]

Hi Mark,

On a study break, so let me reply quickly to some points.

I hesitate to say this, but your calculations look a bit off to me... And I think you are missing an important distinction regarding proven reserves versus total resources.

Just to be sure, can I first check that we're both using the same data as per above? If so, then let's consider the A2-ASF-World case again and, specifically, coal consumption...

I've just run some conversions in the Excel file, using your rate of 29 GJ/metric tonne to ensure consistency. (For instance to find 2100 coal use in billion metric tonnes, I first take 904 EJ and multiply it by 10^9 to specify in term of GJ. I then divide by 29*10^9 to get in terms of billion tonnes, arriving at a figure 31.2.)

Doing so for the whole series, I get the following projected consumption rates for every tenth year:
2010: 3.6 bt (billion tonnes)
2020: 4.5 bt
2030: 6.3 bt
2040: 8.2 bt
2050: 10.1 bt
2060: 14.3 bt
2070: 18.5 bt
2080: 22.7 bt
2090: 26.9 bt
2100: 31.2 bt
"AVERAGE" (2010-2100): 12.8 BT

It could be due to some rounding errors, but the fact that we get such different figures to begin with (e.g. 31 bt vs 37 bt for the year 2100) is perhaps not entirely trivial.

9. [2/2]
...cont.

I have also put "average" in quotation marks because these are of course point estimates for a single year at the beginning of each decade; though we would not expect much difference if each year over the entire period was estimated separately. Indeed, if I calculate the implied growth rate for coal consumption over the 90 period, I get:

3.6*(1+g)^90 = 31.2
--> g = 2.41%

Using this growth rate, we can integrate over the full 90-year period (see here) to arrive at a total consumption level of:

INT 3.6*(1+0.0241)^t dt = 1,153 billion tonnes
(Implying an average of 12.8 bt per year)

Now, this total figure still exceeds -- by about 35% -- the estimated global coal reserves provided by the World Energy Council (WEC), which, as of their most recent report, stood at 860 billion tonnes.

However, that does not mean that these estimates are irreconcilable or even unrealistic. Most importantly, as always with these things, definitions matter a lot. The WEC figure is "limited" to proven recoverable reserves which is very distinct from "resources". The definition of proven reserves is quite strict, which obviously means that estimated reserves will evolve over time. For instance, the coal reserves had increased by 13 billion tonnes from the WEC's 2007 survey to it's 2010 survey (see p. 3). This also underscores why R-P ratios -- though useful -- can be very misleading. They certainly change over time and not simply because of how consumption rates evolve. (This is one reason why I pay far less attention to "peak" theorists than I did a few years ago... the goalposts keep shifting.)

Actually, here's a passage from the International Energy Agency (IAE's) World Energy Outlook 2011 (p. 402) that illustrates very well:

"At the end of 2009, world coal reserves – the part of resources estimated to be economically exploitable with current technology (Box 11.3) – amounted to 1 trillion tonnes, equivalent to 150 years of global coal output in 2009 (BGR, 2010). [However,] total coal resources are many times greater. Globally, coal resources beneath land are estimated at around 21 trillion tonnes but, as market conditions change and technology advances, more coal will be “proven” over time."

(If you can't access the WEO document, I can email it to you. There's also a nice discussion of the difference between reserves and resources on page 405.)

So... under the A2 scenario that we are discussing here, the IPCC appears to be assuming that additional reserves are going to be found and will be economically recoverable... As per the above, many analysts would hold this to be a perfectly justifiable assumption.[*]

Okay, enough from me. Back to studies!

[*] Less important than the above... However, it's also possible that the IPCC are (implicitly) using a different conversion rate from joules to tonnes of coal (i.e. other than 29 GJ / mt). If you have time, it might be worth having a look through their documentation to check.

10. Hi Stickman,

Thanks for checking my numbers. I admit I get sloppy trying to get a point across. (Especially when I don't think the sloppiness will affect the point I'm trying to make.)

I used 29 GJ/Mg from this site, I think:

Not-very-authoritative-site

But I think I actually used 24-25 GJ/Mg.

But let's try to get much more definitive. According the the A2-ASF spreadsheet, coal energy in 1990 was 92 EJ. According to this site, quoting EIA, there were 4,931,628 English tons of coal used in 1990. That would be 4,475,162 Mg.

Indexmundi coal consumption

That produces a value of 20.6 GJ/Gg.

Using this IPCC site for 2005, I get 120 EJ coal energy:

IPCC site

Using the Indexmundi site again, I get coal usage for 2005 of 6,498,573 short tons, or 5,897,072 Mg.

That gives me a value of 20.3 GJ/Mg.

Average them both, I get 20.5 Gj/Mg.

So that would change your numbers (which were based on my bad value of 29 Gj/Mg to):

2010: 5.1 bt (billion tonnes)
2020: 6.3 bt
2030: 8.9 bt
2040: 11.6 bt
2050: 14.3 bt
2060: 20.1 bt
2070: 26.0 bt
2080: 31.9 bt
2090: 37.9 bt
2100: 43.9 bt

These numbers are of course even larger, so they're even more insane. I realize you don't think they are insane, but they are. You provided the WEO document that refers to what reserves theoretically exist. But that basically has nothing to do with what will actually be produced in the 21st century.

11. Part 2...

David Rutledge of Cal Tech has done interesting and good work on this subject. A Powerpoint presentation is here:

David Rutledge Powerpoint

Here's a paper:

Rutledge coal paper

From the abstract:

"We show that where the estimates based on reserves can be
tested in mature coal regions, they have been too high, and that more accurate estimates can be made by curve fits to the production history. These curve fits indicate that total world production, including past and future production, will be 680 Gt. The historical range for these fits made on an annual basis from 1995 to 2009 is 653 Gt to 749 Gt, 14% in percentage terms. The curve fits also indicate that 90% of the total production will have taken place by 2070. This gives the time scale for considering alternatives. This estimate for total production is somewhat less than the current reserves plus cumulative production, 1163 Gt, and very much less than the amount of coal that the UN Intergovernmental Panel on Climate Change, or IPCC, assumes is available for its scenarios."

Here's an intesting piece from the Oil Drum website:

Chinese coal monster running out of puff

Note how, from 1965 to 2009, the rest of the world *except China* has only increased from about 1300 MToe (million tons of oil equivalent) to 1500 MToe. Do you *really* think the rest of the world is going to increase its coal production by like a factor of 6 in the next 88 years, when in the last 44 years it has only increased production by 15 *percent*? (Note: Production in the U.S. is down sharply since 2009...so there may be essentially ZERO increase for the rest of the world in the 47 years from 1965 to 2012).

Again, nobody who knows anything about the history of coal production thinks that the A2 and A1F1 scenarios for coal consumption are realistic. I've provided you the Energy Watch Group paper, the David Rutledge material, and the Oil Drum webpage. Here's another good one, by the way, that predicts "peak coal" in China:

Peak coal in China

You won't find anyone who says these people don't know what they're talking about, the the IPCC does with its A2 and A1F1 scenarios. If you think you've found someone who can dispute these people, send them here. I'll set them straight. ;-)

12. What a mess!

Couldn't even do that right. I multiplied your numbers by 29/20.6. I should have multiplied by 29/20.5. So the numbers are slightly larger...but in the third decimal place.

Again, it doesn't change the fact that A2 and A1F1 are ridiculous. (And now I see A1C_AIM is ridiculous, too, based on David Rutledge pointing that out.)

2. Hi "Stickman",

I apologize if I came across as ornery in my comments. Any anger I have (which is considerable...you could go to my blog to see some) is directed more at the IPCC than at anyone.

IPCC's unfalsifiable "projections"

I'll respond in more detail when time allows, but in the meantime, here's some potentially useful material regarding what I think about economics and climate change:

21st century economic growth

Implications of 21st century economic growth

Mark

3. Hi,

Some more comments, as I have time.

1) "The relevant issue is the underlying forcing mechanisms; which we expect to rise at an increasing rate. As for logarithmic-vs-linear radiative forcing, here and here are good places to start."

You may expect the "underlying forcing mechanisms to rise at an increasing rate." I don't...at least not throughout this century. First, you are almost certainly only considering CO2. Not methane, black carbon, HCFCs or CFCs. (And certainly not SO2...a cooling force that will be reduced in the near term, so lack of significantly increased warming will show that the SO2 cooling effect is overestimated...and therefore the CO2 warming effect is also overestimated.)

Regarding the webpages you referenced: I essentially count as not very informed any page that says that CO2 will increase from 390 ppm to 780 ppm in the next 60-70 years. ("Wanna bet? I didn't think so.")

But enough of engineering. Let's get to economics...and morality.

2) "The broader point, however, is that these damages lower future growth relative to its baseline indefinitely into the future. This point is crucial to understanding why those future GDP losses matter, even if they appear small to you."

I'm pretty sure by "indefinitely" you mean "forever." If so, I strongly disagree. That's why I asked you about GDP per capita in 2162. But even if the GDP losses *did* continue forever, I still don't see the morality of the relatively poor people of 2012 sacrificing for the much more wealthy and well-off people of 2100 and onward. That's why I asked you the questions about GDP and life expectancy in 2100. I realize I ask a lot of questions, but I would like to know your answers:

"In 2012, the world per capita GDP is about \$10,000 (PPP, 2012 dollars). What do you think the per capita GDP will be in 2095, PPP, 2012 dollars? The world-wide average life expectancy at birth in 2010 was 67 years. What do you think it will be in 2095?"

No anonymous comments please. (Pseudonyms are fine.)


text:
# Pressure at same height in same fluid?

## Homework Statement

A sealed has a partition stuck down in water, creating two separate air pockets. The pressure at A is measured to 95 kPa (absolute) and the temperature is 20 C. Find the pressure at B.

## The Attempt at a Solution

Why is the pressure at A not equal to B? Doing U-tube exercises one uses that the pressure at the same height in the same fluid is equal, which is not the case here. Is it because the tank is sealed?

#### Attachments

• VCNjRIb.png
49.5 KB · Views: 1,170

If I'm not mistaken:
Yes, it's because the tank is sealed. The water is at different heights so the pressure differential is given by:
$$\rho_A g h_1 - \rho_w g h_2 - \rho_B g h_3 \\ = g(2\rho_A - 2\rho_w) \\ = 2g(\rho_A-\rho_w)$$
assuming ##\rho_A = \rho_B##

its been a while since i've done fluids so i may indeed be mistaken, take it with a grain of salt :)

CWatters
Homework Helper
Gold Member
Yes.

When we think about the pressure under water we sometimes ignore the effect of the air above because it's roughly the same everywhere but that's not always valid. Changes in the air pressure above the water changes the pressure below the water. This can cause the sea to rise and fall in different places (an example is a storm surge).

Consider the pressure at the bottom of the tank. It must be the same pressure on the left as the right or fluid would flow horizontally. However the depth of water on the right is greater than that on the left by 2m. So something must be equalising the pressure at the bottom OK?

Imagine if the tank was open at the top. The water on the right would fall until it was equal both sides. So something else must be preventing this and equalising the pressure at the bottom. One way to think of it is that the water has already fallen a bit on the right and risen on the left. This increased air pressure on the left and reduced pressure on the right is what causes the pressure at the bottom to be equal.

Thanks! is it possible to prove that the pressure at the bottom of the tank HAVE to equal each other?

The fluid is static. ie. in equilibrium. Remember that you can think of pressure at a point as being exerted equally onto that point from all directions. So if the pressures were different at the bottom on the left and right, then there would be a net flow from high pressure to low pressure ie not static

CWatters

format: plain


text:
• Microstructure and mechanical properties of fine-grained thin-walled AZ91 tubes processed by a novel combined SPD process

• # Fulltext

https://www.ias.ac.in/article/fulltext/boms/040/07/1471-1479

• # Keywords

Severe plastic deformation; thin-walled tube; ultrafine-grained; AZ91; hydro-bulge test.

• # Abstract

A combination of parallel tubular channel angular pressing (PTCAP) and tube backward extrusion (TBE) asa novel combined severe plastic deformation (SPD) was applied on AZ91 alloy to produce ultrafine-grained (UFG) thinwalledtubes. The effects of combined SPD process were investigated on the microstructure refinement and mechanicalproperties. Also, hydro-bulge test was carried out to reveal the mechanical properties. The results showed a notable increasein ultimate strength, yield strength and microhardness of the thin-walled UFG tube were achieved compared to that fromPTCAP process. A remarkable grain refinement achieved. Applying three-passes-combined process refine the grain sizeto 8.8 $\mu$m from an initial value of $\sim$150 $\mu$m. For one pass- and two passes-processed thin-walled tubes, it was about 12.4 and 9.8 $\mu$m, respectively. Yield and ultimate strengths were increased notably to 150 and 354 MPa for 2P $+$ TBE tube, from the initial values of 86 and 166 MPa, respectively. The maximum microhardness was increased to about 105 Hv for the2P S+\$ TBE tube from the initial value of 56 Hv. Hydro-bulge test showed that the bursting pressure increased to 246 bar for2 passes-thin-walled tube from the initial value of 160 bar. It was 220 and 195 bar for 1 pass- and 3 passes-thin-walled tubes,respectively.

• # Author Affiliations

1. School of Mechanical Engineering, College of Engineering, University of Tehran, Tehran 11155-4563, Iran
2. Faculty of Mechanical Engineering, Shahid Rajaee Teacher Training University, Tehran 16788-15811, Iran

• # Bulletin of Materials Science

Current Issue
Volume 42 | Issue 6
December 2019

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019


text:
# Why are turbines built differently when a change of entropy or temperature is involved?

We are talking about turbines for extraction of energy from a fluid.

Consider a typical wind turbine, or a francis turbin for water - we have one hub with on set of vanes that are driven by the medium.

Now, consider a typical turbine in a steam engine - here we have alternating moving vanes and fixed counter-vanes. The fixed vanes redirect the flow after each stage, something that's apparently not neccessary in a water turbine for a high efficiency.

1 - My intuition tells me that this has something to to with the change of temperature or entropy along the turbine. Is this intuition correst?

2 - Why are the turbines built differently? Assuming it's not possible to extract a reasonable amount of energy with a one-stage turbine from steam, why is that so?

3 - Assuming an appreciable pressure drop without a big change of temperature along the turbine (maybe a pneumatically driven turbine), will I still need a multi vane design to extract most of the energy?

Bonus question - can the same reasoning be applied to pumps or compressors - a multi stage design is neccessary for decent efficiency if a significant temperature change is involved?

-
Ultimately, energy addition/extraction is a function of $\dot{m} \Delta V$; gas/steam compressors/turbines have a high $\Delta V$ to work with, but are restricted in $\dot{m}$, whereas wind/water pumps/turbines have a low potential $\Delta V$, so have to increase $\dot{m}$.


text:
# Global and blow-up radial solutions for quasilinear elliptic systems arising in the study of viscous, heat conducting fluids

@article{Ghergu2019GlobalAB,
title={Global and blow-up radial solutions for quasilinear elliptic systems arising in the study of viscous, heat conducting fluids},
author={Marius Ghergu and Jacques Giacomoni and Gurpreet Singh},
journal={Nonlinearity},
year={2019}
}
• Published 1 August 2018
• Mathematics
• Nonlinearity
We study positive radial solutions of quasilinear elliptic systems with a gradient term in the form \left\{ \begin{aligned} \Delta_{p} u&=v^{m}|\nabla u|^{\alpha}&&\quad\mbox{ in }\Omega,\\ \Delta_{p} v&=v^{\beta}|\nabla u|^{q} &&\quad\mbox{ in }\Omega, \end{aligned} \right. where $\Omega\subset\R^N$ $(N\geq 2)$ is either a ball or the whole space, $1 0$, $\alpha\geq 0$, $0\leq \beta\leq m$ and $(p-1-\alpha)(p-1-\beta)-qm\neq 0$. We first classify all the positive radial solutions in case…
7 Citations

## Figures from this paper

Classification of radial solutions for elliptic systems driven by the $k$-Hessian operator.
We are concerned with non-constant positive radial solutions of the system \left\{ \begin{aligned} S_k(D^2 u)&=|\nabla u|^{m} v^{p}&&\quad\mbox{ in }\Omega,\\ S_k(D^2 v)&=|\nabla u|^{q} v^{s} Liouville-Type Theorems for Sign-Changing Solutions to Nonlocal Elliptic Inequalities and Systems with Variable-Exponent Nonlinearities • Mathematics • 2020 We consider the fractional elliptic inequality with variable-exponent nonlinearity (-\Delta)^{\frac{\alpha}{2}} u+\lambda\, \Delta u \geq |u|^{p(x)}, \quad x\in\mathbb{R}^N, $$where N\geq 1, A Liouville-Type Theorem for an Elliptic Equation with Superquadratic Growth in the Gradient • Mathematics • 2019 Abstract We consider the elliptic equation - Δ  u = u q  | ∇  u | p {-\Delta u=u^{q}|\nabla u|^{p}} in ℝ n {\mathbb{R}^{n}} for any p > 2 {p>2} and q > 0 {q>0} . We prove a Liouville-type theorem, Asymptotic behavior of blowing-up radial solutions for quasilinear elliptic systems arising in the study of viscous, heat conducting fluids • Mathematics • 2021 In this paper, we deal with the following quasilinear elliptic system involving gradient terms in the form: { ∆pu = v |∇u| in Ω ∆pv = v β |∇u| in Ω, where Ω ⊂ R (N ≥ 2) is either equal to R or equal Solvability of second-order uniformly elliptic inequalities involving demicontinuous$$\psi $$-dissipative operators and applications to generalized population models • Mathematics • 2021 Based on a new developed theory for variational inequalities, the purpose of this article is to investigate existence and uniqueness of nonzero positive weak solutions for a class of general A priori estimates and Liouville type results for quasilinear elliptic equations involving gradient terms • Mathematics • 2022 . In this article we study local and global properties of positive solutions of − ∆ m u = | u | p − 1 u + M |∇ u | q in a domain Ω of R N , with m > 1, p, q > 0 and M ∈ R . Following some ideas used ## References SHOWING 1-10 OF 25 REFERENCES Nonexistence results and estimates for some nonlinear elliptic problems • Mathematics • 2001 AbstractHere we study the local or global behaviour of the solutions of elliptic inequalities involving quasilinear operators of the type$$L_{\mathcal{A}^u } = - div\left[ {\mathcal{A}\left(
A new dynamical approach of Emden-Fowler equations and systems
• Mathematics
• 2010
We give a new approach on general systems of the form \[ (G)\left\{ \begin{array} [c]{c}% -\Delta_{p}u=\operatorname{div}(\left\vert \nabla u\right\vert ^{p-2}\nabla u)=\varepsilon_{1}\left\vert
Eternal solutions to a singular diffusion equation with critical gradient absorption
• Mathematics
• 2013
The existence of nonnegative radially symmetric eternal solutions of exponential self-similar type $u(t,x)=e^{-p\beta t/(2-p)} f_\beta(|x|e^{-\beta t};\beta)$ is investigated for the singular
Coercive elliptic systems with gradient terms
• Mathematics
• 2016
Abstract In this paper we give a classification of positive radial solutions of the following system: Δ  u = v m , Δ  v = h  ( | x | )  g  ( u )  f  ( | ∇  u | ) , \$\Delta u=v^{m},\quad\Delta
Large Solutions for a System of Elliptic Equations Arising from Fluid Dynamics
• Mathematics
SIAM J. Math. Anal.
• 2005
This paper proves that the elliptic system (0.1) has a unique radially symmetric and nonnegative large solution with v(0) = 0 (obviously, v is determined only up to an additive constant) and study the asymptotic behavior of these solutions near the boundary of Omega and determine the exact blow-up rates.
Local behaviour of the solutions of a class of nonlinear elliptic systems
Here we study the behaviour near a punctual singularity of the positive solutions of semilinear elliptic systems in R (N 3) given by u+ jxj uv = 0; v + jxj uv = 0; (where a; b; p; q; s; t 2 R , p; q
Existence and a Priori Estimates for Positive Solutions of p-Laplace Systems
• Mathematics
• 2002
We use continuation and moving hyperplane methods to prove some existence and a priori estimates for p-Laplace systems of the form−Δp1u=f(∣v∣) in Ω,u=0 on ∂Ω,−Δp2v=g(∣u∣) in Ω,v=0 on ∂Ω, where


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# Angkor Watby *SantiBilly

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text:
FREELessons: 67Length: 8.9 hours

• Overview
• Transcript

# 6.4 Navigating Routes Using the Router

In this lesson, you'll see how we can use Angular's Router to programmatically navigate between different states in our application, rather than waiting for users to use a routerLink anchor.


text:
# [Moved] Reasons to Choose a Specific FFT Formula

Status
Not open for further replies.

#### sona_

##### Junior Member level 1
hello
m a newbie on delsig..can anyone tell
(1) why formula chosen is y = 4*sin(2*pi*tone_bin/Nfft*t);for sine wave
(2) why is t chosen as t = [0:Nfft-1] why not t = [0:Nfft];
thanks all

#### barry

##### Advanced Member level 5
Because 0:nfft will be 1 greater than desired. zero is, in fact, a number. If you count from zero to one, that's two numbers, not 1.

sona_

### sona_

Points: 2

#### sona_

##### Junior Member level 1
yes...got the answer...and what about the formulae for sine wave

#### barry

##### Advanced Member level 5
yes...got the answer...and what about the formulae for sine wave

#### sona_

##### Junior Member level 1
the input as y = 4*sin(2*pi*tone_bin/Nfft*t); why is the frequency chosen as tone bin/nfft

#### barry

##### Advanced Member level 5
the input as y = 4*sin(2*pi*tone_bin/Nfft*t); why is the frequency chosen as tone bin/nfft

I have no idea. You start out asking about FFTs, and then you post a single equation which has nothing to do, as far as I can tell, with computing an fft.

You have obviously posted one line from a more complex calculation. I'm afraid I've left my mind-reading hat at home.

#### sona_

##### Junior Member level 1
ya i would have posted the complete program..actually i was talking about delsig toolbox in which all sine wave assume the same formulae...one of the example is (which quantizes the sine wave using delsig toolbox)

Code dot - [expand]1
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nLev = 8;          % Number of quantizer levels
M = nLev-1      % Number of steps
Nfft = 2^13;    % Record length
tone_bin = 31;  % prime wrt Nfft
t = [0:Nfft-1];
y = 4*sin(2*pi*tone_bin/Nfft*t);
v = ds_quantize(y, nLev);
numOfSamples = 400;
n=1:numOfSamples;
plot(t(n), y(n), 'b');

Last edited by a moderator:

#### sona_

##### Junior Member level 1
i guess since we want the frequency to be around a certain bin say 31 in above case hence we select tone bin as the frequency to be selected over the entire nfft length...pl correct if i am wrong

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text:
# Anyone tried lindows or lindash

1. Apr 1, 2004

### wolram

has anyone tried lindows or lindash, as it now in Europe thanks to MS,
i have read some glowing reports about it, i know it isn't cheap but i
am used to windows and thought the transition wont be to horrendous,
i can also buy my new PC with it pre installed.

2. Apr 1, 2004

### dduardo

Staff Emeritus
The thing I don't like about lindows is that you have to pay for the click-n-run feature. But if you don't feel confortable installing software either from rpm/tgz/bz2/ whatever and updating yourself, by all means pay for the service. You could also try out SUSE. Their YaST system is supposed to be very good. And their is also Mandrake Linux.

The good thing about lindows is that they provide alot of scripts that make installing hardware/software and networking easy. If it preinstalled on your computer the better.

The one thing I should warn you about is to create a non-root user when you first get the computer. If you use the computer as root, bad things can happen!

3. Apr 1, 2004

### wolram

The one thing I should warn you about is to create a non-root user when you first get the computer. If you use the computer as root, bad things can happen!
problem no 1, i dint know what this means, p 2 is it complicated as
i am only just literate in computers.
thanks for help DDURADO it is very much appreciated.

4. Apr 1, 2004

### dduardo

Staff Emeritus
Last edited by a moderator: Apr 20, 2017
5. Apr 1, 2004

### Dagenais

Lindows? You're leaving Microsoft Windows for Lindows?

To tell you honestly, you'll probably be happier with Windows.

For a commercial product, you're better off with SuSE.

For an easy Linux distribution, probably Mandrake.

6. Apr 3, 2004

### Dr Transport

The operating system you want depend on your needs. If you want to do some serious programming, try Linux, if your the occasional programmer, stay with Windows. I work in both, I do my really serous stuff on a Unix type box and surf at the house with Windows. My family has no idea about anything but XP, they prefer to stay there, although they quickly became addicted to Tux Racer last week when they played it.

7. Apr 18, 2004

### kioria

A typical thread seen on many of the forums. Like Dagenais said, you wouldn't switch from Windows XP to Linspire (formerly known as Lindows), and pay \$ (although its cheap). There are other Commercial Linux distributions you would like to consider, ie., SuSE, RedHat WorkStation, and etc., Even if its not a commercial dist. there are plenty you would find useful for what ever purpose you use Linspire for: Fedora, Mandrake. I guess you want to "try" Linspire because you think its the most easiest transition from Windows -> Linux. However, that is wrong. Try, Mandrake, it is as easy as using Windows XP. Nothing much to learn, except a few linux (unix) cmds (which is also required if you're using Linspire). Fedora ain't that bad as well.

Cheers :)

8. Apr 18, 2004

### faust9

I say if you want to give Linux a go try Gentoo. Gentoo is somewhat of a pain to install, if you go the stage 1(?) route you have to compile and bootstrap the entire system. there are a stage 2 and 3 available which allow you to skip some or all of the complining and bootstrapping (Its been a year since I installed Gentoo so I may have the stages mixed up). I believe Gentoo has a live CD implementation also...

The reason I recommend going the less travel route is you learn quite a bit about what goes into making a Linux system work. If you go the long hard route with Gentoo (or LFS for that matter) you come out the other end with a fairly well optimized kernel (even the un-initiated can compile a kernel that will be more responsive than some commercial kernels--Mandrake 9.0 comes to mind). Also, you have this sense of satisfaction when your finished.

Mind you installing Gentoo takes some time. It took 24 hours to download, compile and all the rest on my PII 333 (including X, KDE, and the best desktop environment--Enlightenment), so the install as only a few minutes longer than a fresh XP install (XP zealots calm down it was a joke).

As far as the easy distros go, Lycoris is another good one IMO. Lycoris still uses an older KDE implementation(2.2.2) but the eye candy is pleasing none the less. Lycoris is also very easy to use and detected and configured my obscure sound card which the only other linux distro to do that on my box was College Linux (I two HD's in my PII 333, one Gentoo, and the other I use to play around with new distros).

Finally, if you do go with Linspire remember apt-get. Linspire is a Debian derivative and still has the apt-get command available. If you use apt-get you can download apps from the Debian repository instead of from click-n-run. I think Linspire does install a default account and a root account now but don't quote me on this.

A couple of reviews if your bored:
http://www.osnews.com/story.php?news_id=5999
http://www.desktopos.com/reviews.php?op=showcontent&id=3
http://www.osnews.com/story.php?news_id=5922

http://www.distrowatch.com/

Or Newsforge.
http://www.newsforge.com/ [Broken]

Gentoo docs.
http://www.gentoo.org/doc/en/index.xml#doc_chap2

How to use apt:
http://www.debian.org/doc/manuals/apt-howto/index.en.html
http://bazar.conectiva.com.br/~godoy/apt-howto/how-to-use-apt-get.html [Broken]

That's about it. Give Gentoo a try. It's more fun than a nail in the head.

Last edited by a moderator: May 1, 2017


text:
# Partial Derivatives of Index Straddle w.r.t. Index Constituents

#### Will_Everett

Hi everyone,

I'm an undergraduate Math/Econ student planning on going onto a PhD in Finance. I am trying to replicate the results of the Driessen et al Journal of Finance paper on pricing correlation risk.

In order to delta/vega hedge a short index straddle I need to find the partial derivative of the straddle with respect to the variance of each constituent security, and the partial derivative of the straddle with respect to the constituent stock.

$\frac{\partial O_I}{\partial S_i} \text{ and } \frac{\partial O_I}{\partial \phi_i^2}$

Where $$O_I$$ denotes the price of the straddle, $$i \in [1, \ldots, n]$$ are constituents of the index, and $$\phi_i^2$$ is the variance of the constituent.

As far as I understand Delta and Vega are linear functions, so can I just use the weighted greek of the straddle? Can I just take $$\omega_i \cdot \Delta_I$$ and similar to get Vega?

Thanks for any help you can offer. My knowledge of option pricing is fairly limited, but I am anxious to learn more, so if you can point me in the direction of a book or some papers on the topic it would be great.

As a side note: If anyone is familiar with the OptionMetrics data set and knows what secid should be used for SPY I would love to hear. The one I am currently using lacks the Greeks, which isn't really an issue since I can just calculate them myself, but since I am using the pre-defined values for all other options it would be nice to have uniformity.

Edit: I have been thinking a bit about this and here is what I have so far:

If we suppose that movement in each stock indexed by $$i \in [1,\ldots, n]$$ is governed by the following process:
$\frac{dS_i}{S_i} = \mu_i dt + \phi_i dW_i$
(I might be wrong here, but should we have a third term on the right: $$\sum_j \rho_{ij} \phi_j S_j dW_j$$ for each $$i \neq j$$?)

Then we have the index as a function:
$S_I(t, W_1,\ldots, W_n) = \sum_i S_i$
I am not quite sure about the application of Ito's Lemma here to get the process for $$dS_I$$... I took a finance class aimed at students looking to grad school in which we briefly covered the Ito's Lemma in the context of BS, but I don't feel confident in applying it to a multi-variate function as in this example. I would assume that we would just get a unit weighted linear combination of applications of ito's lemma w.r.t. each of the constituent processes.

Any tips on how I would proceed? Once I can get the process for $$dS_I$$ it seems straightforward to use Ito's Lemma to get the process for call and put options on index, then assuming this still works neatly into the heat equation I think I could get at the delta and vega.

Possibly I am overcomplicating this, and I should just plug $$S_I(t, W_1,\ldots, W_n) = \sum_i S_i$$ into the standard Black-Scholes formula. In this case, then it seems my intuition of weighting Delta and Vega by the index weights would be correct.

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# If we put the empty set into a function why do we get the empty set back?

For example if we have this function:

$$f(x) = 6x + 28$$

If I put the empty set in:

$$f(∅) = 6(∅) + 28$$

Why do I get the $$∅$$ back?


text:
# Faucet mystery (help )

1. Sep 26, 2006

### blueman

faucet mystery (help!!!!!)

I'm in a physics class and we have 3 sinks in the class room on the same wall.All the faucet are off. When you turn one faucet on all 3 faucet are on and the water is running from them.
So;
when you turn faucet 1 on , 2 and 3 are also on
when you turn faucet 2 on , 1 and 3 are also on
when you turn faucet 3 on , 1 and 2 are also on
:grumpy: :uhh: :shy:
It's impossible, but it's true.So we are tryinh to find an explanation for this mystery. Any help?? How does it work?? Why does it work??Please help,

Thank you,
blueman

2. Sep 27, 2006

### kleinjahr

Sounds like someone messed up the piping. Look under the sinks that'll, probably solve it.

3. Sep 27, 2006

### actionintegral

This has nothing to do with physics and should be posted in the puzzles section.

4. Sep 27, 2006

### blueman

if it isn't physics ,why should I post it in the puzzle section ?!?!
anyways, the pips are connected properly ...some maintenance guys came and checked the connections.....

5. Sep 27, 2006

### Danger

The only way that I can make any sense out of it is if you have electric controls on the taps (such as the IR sensors used in some public washrooms). If so, a wiring error could cause them to trigger each other.

6. Sep 27, 2006

### Dense

Everything has to do with physics. Sounds to me like an issue of fluid dynamics, hydraulics and vacuums.

Look for worn or poorly-fitted valves in the faucets, or venting problems causing too much pressure when the water flows. And call a decent plumber, not just some maintenance guys.

7. Sep 27, 2006

### Farsight

I think I saw this on TV.

8. Sep 27, 2006

### Danger

That doesn't make sense either. If a faucet is physically closed, then the actions of the others should have no effect. The only interaction between them should be a possible lessening of flow rate with more than one open, and then only if the line pressure is low. Physically opening one tap cannot physically open another one.

9. Sep 27, 2006

### B. Elliott

Unless there's some type of rod linkage connecting the three valves. Even then you'd expect the other faucets handles to move when you move one.

10. Sep 28, 2006

### cesiumfrog

Reminds me of those lateral thinking problems. So little background information is given that there are countless feasible explanations, which can't be distinguished unless further experiments are carried out. For one example, if water through one tap causes enough vibration, the other taps may leak while it is open (you'd want to listen how much noise the pipes make, to support or disprove this one). For another example, "blueman" could be making the whole thing up. You could investigate the plumbing, the fixture design, the teacher.. This is like trying to pick the curve of best fit (and is it linear? exponential? sinusoidal?) when you only have one or two data points.

11. Sep 28, 2006

### Andrew Mason

Why impossible? The taps are obviously in parallel and the faucets are in series with the water pipe.

AM

12. Sep 28, 2006

### Danger

13. Sep 28, 2006

### kmarinas86

This sounds like a joke.... :zzz:

14. Oct 4, 2006

### LHarriger

I've got the solution you all!
As we all know, divergence can be thought of as measuring an outflow from a source. So we should perform the following calculation:
$\int_{V} \bigtriangledown\cdot faucet\ d\tau$
being carful to make our volume big enough to pick up all faucets.
This will measure the outflow into our chosen volume from all the faucets
By Green's Theorem this should be equal to:
$\oint_{S}faucet\cdot da$
Which, of course, just represents the pipe flow to the sources as it passes through our surface.
Hence I propose we should vary our volume to find the point where the integrals no longer agree which will pinpoint the precise location at which the cause of this mysterious phenomenon exists.

Last edited: Oct 4, 2006
15. Oct 4, 2006

### actionintegral

What happens when you flush the toilet? :surprised

16. Oct 4, 2006

### LHarriger

Ahhhh Spin off research.......where all the great discoveries are made.

17. Nov 27, 2008

### ozgurakkas

Re: faucet mystery (help!!!!!)

As long as all the faucets have handles and they work properly and are closed while one of the faucets is working, there is no way all three will work simultanously. However faucets might have cencors so there might be a case with wiring.


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# Why did Kronecker develop the “adjoining a root” construction?

Kronecker is generally credited with the formalization of "adjoining a root to $f(x)=0$". Nowadays it is interpreted as the quotient $K[x]/(f)$, where $K$ is some appropriate algebraic structure in a wide range of structures. But if $K=\mathbb Q$ or $\mathbb Z$, some of the simplest examples possible, then the construction appears unnecessary, since our imaginary root is simply an element of $\mathbb C$, and there is no need to construct anything (assuming one is already satisfied that $\mathbb C$ exists).

Thus, for which structure(s) $K$ did Kronecker (or whoever actually first thought of this construction) develop his original construction, and why did he feel it was necessary?

According to Edwards, in his 1887 Fundamental Theorem of General Arithmetic Kronecker was working with polynomials with integer coefficients. He proclaimed, however, that "the so-called fundamental theorem of algebra is replaced" with the title theorem, and that he "put foundations of algebra in an entirely new form". Indeed the proof applies much more generally, providing that any polynomial can be split into linear factors in some extension of its field of coefficients. This is now known as the Kronecker's theorem, or the splitting field theorem, and that the proof does not depend on any analytic properties of $\mathbb C$ held great appeal in Kronecker's eyes. He names as an inspiration his "sharp critique of the method of defining quantities that began with Heine", a stand in for transcendental arguments of Cantor, Dedekind and Weierstrass.
Kronecker did not however come up with the modern quotient ring construction, being an early constructivist he had philosophical objections to Dedekind's ideals themselves, and even offered divisors as an alternative to them earlier. Instead, he formalized a more "explicit" construction of Galois, who called it "adjoining a root", arguably completing the long process of proving the FTA "algebraically". Gauss already pointed out the circularity of reasoning that roots are elements of $\mathbb C$ in Euler's and Lagrange's attempted proofs of FTA. They needed to perform computations with roots, which required treating them as elements of $\mathbb C$, in the course of proving that they are elements of $\mathbb C$. One either has to presuppose that the roots exist, which is circular, or to use continuity arguments to prove existence, which is "non-constructive".
Galois still tacitly assumed that computations with roots were valid, but he usually expressed the adjoined roots as polynomials with rational coefficients in a new indeterminate $t$, specifying $t$ to be a root of a known irreducible polynomial. Kronecker called this construction "the precious Galois principle", and proved that such presentation is always possible for polynomials with integer coefficients. Namely, given a polynomial $f(x)$ it is possible to find a polynomial $G(t)$ such that adjoining its root allows to express all roots of $f(x)$ constructively as polynomials in $t$. Thus, adjoining a single element splits $f(x)$ into linear factors, and computations with roots become justified.
• So are you saying that Kronecker introduced the construction because he wasn't satisfied with contemporary constructions of $\mathbb C$ ? – Jack M Apr 29 '15 at 2:18
• @Jack M From Kronecker's 1886 letter to Mittag-Leffler:"I have put the foundations of algebra in an entirely new form... I owe this beautiful and sure foundation of algebra to my sharp critique of the method of defining quantities that began with Heine, and to the precious Galois principle". Kronecker's point seems to be that whichever way $\mathbb{C}$ is originally constructed the splitting field has to be constructed independently. Then the roots can be legitimately manipulated, and proved to belong to $\mathbb{C}$. Otherwise, we have to presuppose that the roots exist, which is circular. – Conifold Apr 29 '15 at 3:08


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# I need help

Problem 64 Chapter 14

Natalie owns a condominium near Cocoa Beach in Florida. This year, she incurs the following expenses in connection with her condo:

Insurance $1,000 Advertising expense 500 Mortgage interest 3,500 Property taxes 900 Repairs & maintenance 650 Utilities 950 Depreciation 8,500 During the year, Natalie rented out the condo for 75 days, receiving$10,000 of gross income. She personally used the condo for 35 days during her vacation.

Assume Natalie uses the IRS method of allocating expenses to rental use of the property.

a) What is the total amount of for AGI (rental) deductions Natalie may deduct in the current year related to the condo?

b) What is the total amount of itemized deductions Natalie may deduct in the current year related to the condo?

c) If Natalie’s basis in the condo at the beginning of the year was $150,000, what is her basis in the condo at the end of the year? d) Assume that gross rental revenue was$2,000 (rather than $10,000). What amount of for AGI deductions may Natalie deduct in the current year related to the condo? e) Using the original facts, complete Natalie’s Form 1040, Schedule E, for this property. Also, partially complete Natalie’s 1040, Schedule A, to include her from AGI deductions related to the condo. Document Preview: Problem 64 Chapter 14 Natalie owns a condominium near Cocoa Beach in Florida. This year, she incurs the following expenses in connection with her condo: Insurance$1,000 Advertising expense 500 Mortgage interest 3,500 Property taxes 900 Repairs & maintenance 650 Utilities 950 Depreciation 8,500 During the year, Natalie rented out the condo for 75 days, receiving $10,000 of gross income. She personally used the condo for 35 days during her vacation. 64. Assume Natalie uses the IRS method of allocating expenses to rental use of the property. a) What is the total amount of for AGI (rental) deductions Natalie may deduct in the current year related to the condo? b) What is the total amount of itemized deductions Natalie may deduct in the current year related to the condo? c) If Natalie’s basis in the condo at the beginning of the year was$150,000, what is her basis in the condo at the end of the year? d) Assume that gross rental revenue was $2,000 (rather than$10,000). What amount of for AGI deductions may Natalie deduct in the current year related to the condo? e) Using the original facts, complete Natalie’s Form 1040, Schedule E, for this property. Also, partially complete Natalie’s 1040, Schedule A, to include her from AGI deductions related to the condo.

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# What is the binomial sum $\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n}$ in terms of zeta functions?

We have the following evaluations:

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\ &\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = \frac{1}{3}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} = -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right) \\&\sum_{n=1}^\infty \frac{1}{n^4\,\binom {2n}n} = \frac{17}{36}\,\zeta(4)\\ &\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n} = \,?\\ \end{aligned}

The paper gives the 3rd and 5th in terms of the Dirichlet L-functions, but does anyone know how to evaluate the 5th one in terms of the Hurwitz zeta function $\zeta(s,a)$?

Postscript: (A few hours later)

After Anon gave his answer, I did a little more sleuthing and found the case p = 7 in the Mathworld article on (which also had p = 5). The paper I cited was a bit old (1999) and the authors weren’t aware it was already found a year earlier by Plouffe. Hence,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n^5 \, \binom{2n}n} = -\frac{19}{3}\zeta(5)+\frac{2}{3}\zeta(2)\zeta(3)+\frac{\pi\sqrt{3}}{2^3\cdot3^2}\left(\zeta(4,\tfrac{1}{3})-\zeta(4, \tfrac{2}{3}) \right)\\ &\sum_{n=1}^\infty \frac{1}{n^7 \, \binom{2n}n} = -\frac{493}{24}\zeta(7)+2\zeta(2)\zeta(5)+\frac{17}{18}\zeta(3)\zeta(4)+\frac{11\pi\sqrt{3}}{2^5\cdot3^4}\left(\zeta(6,\tfrac{1}{3})-\zeta(6, \tfrac{2}{3}) \right)\\ \end{aligned}

With this “pattern”, I used an integer relations algorithm to try to find p = 9, 11, 13. No luck so far.

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2022-07-25 20:43:11
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16 Jan

lattice energy formula

Lattice energy calculation for ZrO2 molecule What is lattice energy ? The energy released in this process is known as lattice energy or lattice enthalpy. These have been made according to the syllabus 9701 and cover all the relevant topics for examination in March/June. r 0 = Closest ion distance. n = Born Exponent. We then plunge the ions into water: K(g)+ + nH2O→ K(aq)n+ +ΔHsolv(K+) Cl-(g) + nH2O→ Cl(aq)n- +ΔHsolv(Cl)- (They have combined these two steps into one.) Which one has higher lattice energy, MgF2 or CaF2 ? Evaluation Of Force Field Calculations Of Lattice Energies On A, Lattice Energy Definition Trends Equation Video Lesson, Born Haber Cycles And Lattice Energy Ppt Video Online Download, Lattice Energies Of Various Oxides Used For Calculating The, Solved Calculating Lattice Energy Example Problem Calcul, Solved Problems 1 Calculate Lattice Energy Of Kcl Ioniz, Lattice Energy Chemistry Video Clutch Prep, Given The Following Thermodynamic Data Calculate The Lattice, Born Haber Cycle Calculation Of Ionic Crystal Lattice Energy, Problem Set 6 1 Use The Following Data To Do The Chegg Com, The Calculation Of The Lattice Energy The Born Haber Cycle, Chemistry 101 Calculating Lattice Energy Using The Born Haber, How Do You Construct A Born Haber Cycle To Calculate The Lattice, Lattice Energy The Born Haber Cycle Chemistry Libretexts. The attraction of the two ions releases energy and the process is exothermic. Lattice energies cannot be measured directly but are obtained from a thermochemical cycle called the Born–Haber cycle, in which Hess’s law is used to calculate the lattice energy from the measured enthalpy of formation of the ionic compound, along with other thermochemical data. You CAN even get the proper results. Why is the lattice energy of ZrO2 so high ? Hydration energy lattice energy and solubility pdf An energy lattice can also help predict solubilities compounds. [42] M. Maldova n, C.K. When a salt, such as NaCl dissolves in water, the crystals disappear on the macroscopic scale. The electron affinity of is X -335 kJ/mol. Finally, calculate the lattice energy. This paper presents a new technique for estimation of lattice energies of inorganic ionic compounds using a simple formula. Cheetham, A.K. Lattice Energy Calculator. This video describes how to calculate proportional abundance to determine the simpson index value. The solid crystal is a three-dimensional entity. formula weight of ionic crystal is formed from the requisite number of gaseous cations and anions,is called lattice energy. Lattice energy,What is Lattice energy Definition of Lattice energy. After the formation of ions, they combine together to form ionic compound. The first point explains why MgO has a higher lattice energy than NaF. This article includes everything about lattice energy with its definition, trend, formula, etc. 6 answers. Determination of Lattice enthalpy . Beta version # BETA TEST VERSION OF THIS ITEM This online calculator is currently under heavy development. The following formula is used to calculate a lattice energy between ions. Solid State Chemistry. Direct observation techniques are based on imaging the lattice distortion with X-rays or electron beams. One way is the energy released when gaseous ions combine to form an ionic solid. By Hess’s Law . The question is "The lattice energies of FeCl3, FeCl2, and Fe2O3 are -2631,-5359, and -14,774 kJ/mol (in no particular order). Best Answer 100% (1 rating) Previous question Next question Get more help from Chegg. Still have questions? How To Calculate Body Fat Percentage Loss Like Big... How To Buy Fortnite Skins That Arent In The Item Shop, How To Calculate Isoelectric Point Of Histidine, How To Butterfly A Chicken Breast With Bone, How To Burn 1000 Calories At The Gym In An Hour. Lattice energy of MX = (505 + 117 + 425 + 119.5 - 305) kJ/mol = +861.5 kJ/mol. Determination of Lattice enthalpy . Born-Lande equition,Born-Haber cycle The Born-Haber cycle of sodium chloride crystal ( NaCl ) . With our tool, you need to enter the respective value for Atomic Radius and hit the calculate button. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Which one has higher lattice energy, MgF2, CaF2 or ZrO2 ? Join Yahoo Answers and get 100 points today. Question: What Is The Formula For Lattice Energy? After the development of particles, they join together to shape an ionic compound. Notes for the CIE O Level Chemistry – 20: Lattice Energy. ionic mixes are more steady due to their electrostatic power between the two inverse particles. In the three dimensional lattice (cubic lattice), the exponent characterizing the vanishing of the magnetization in the paramagnetic state is $$\beta \approx 0.326$$. Magnetization, energy and heat capacity as a function of the temperature for the 2D ising model, from cluster monte carlo simulations. Ask Question + 100. The amount of energy released, when one gm. Some energy quantities, such as the lattice energy of a mineral or the electron affinity of an atom, can be difficult to measure in the lab. Lattice energies cannot be measured directly but are obtained from a thermochemical cycle called the Born–Haber cycle, in which Hess’s law is used to calculate the lattice energy from the measured enthalpy of formation of the ionic compound, along with other thermochemical data. Kapustinskii Equation. The Lattice energy, $$U$$, is the amount of energy required to separate a mole of the solid (s) into a gas (g) of its ions. K = 2.31 X 10 -10 J.nm. Chemical Calculator Molar Mass Empirical Formula Degree of Unsaturation Unit Converter Lattice Energy d-d Spectrum d-d Spectrum (Jahn-Teller effect) Quadratic Equation Cubic Equation Quartic Equation Quintic Equation Radiation Dose Food Energy Requirements Crystal Structures Colour and Light Molecular Orbital Diagrams. Reaction 1: Ba ( s) → Ba ( g) ΔH1 = ΔHsub = 180.0kJ / mol More than twice as much energy is required to sublime barium metal (180.0 kJ/mol) as is required to sublime cesium (76.5 kJ/mol). How lattice energy influence melting point ? Why can’t I sleep it’s half 3 and I’m filled with energy? Furthermore, one is able to determine the ionic radii (or more properly, the thermochemical radius) using the Kapustinskii equation when the lattice energy is known. It may or it may NOT work correctly. LATTICE VIBRATIONS AND PHONONS In semiconductor crystals, the atoms are tightly coupled to one another, and the binding energy is called cohesive energy, which is deﬁned as the energy needed to separate the crystal into independent ions at a large distance from each other. The bond energy of X2 is 187 kJ/mol. Vol. what is the formula for lattice energy? Lattice Energy Formula Lattice energy refers to the energy that is released when two oppositely charged gaseous ions attract to each other and form an ionic solid. K is the constant (2.31 * 10 ^ -19 (J*nm) Q1 is the numerical ion charge of ion 1. e = Electron charge (1.6022 × 10-19 C) Z + and Z – = Cation and anion charge. The formation of a crystal lattice is exothermic, i.e., the value of ΔHlattice is negative because it corresponds to the coalescing of infinitely separated gaseous ions in vacuum to form the ionic lattice. It is not possible to calculate the lattice enthalpy directly from the forces of attraction and repulsion between ions but factors associated with crystal geometry must also be included. We will sum the energies of each step and that will be our total lattice energy. ϵ o = Permittivity of free space. The calculated lattice energy gives a good estimation for the Born-Landé equation; the real value differs in most cases by less than 5%. See the answer. Journal of Chemical Education. Lattice energy,What is Lattice energy Definition of Lattice energy. ionic mixes are more steady due to their electrostatic power between the two inverse particles. Performance & security by Cloudflare, Please complete the security check to access. Tap a snap you want to move. On the atomic scale, the Na + and Cl-ions in the crystal are released into solution. The energy delivered in this cycle is known as lattice energy or lattice enthalpy. Lattice Energy is directly proportional to the Charge on ion and inversely proportional to radius of atom . If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. Using the formula and the values from steps 1-3 we find the lattice energy to be, 1.54e^-19. Given the following thermodynamic data, calculate the lattice energy of CaBr2(s).Express your answer to four significant figures, and include the appr... Q. It can be evaluated considering different contributions to the potential energy. The lattice enthalpy is indirectly determined by the use of Born - Haber Cycle. The amount of energy released, when one gm. That means, Table shows lattice crystal energy in kJ/mol for selected ion compounds. How To Calculate Lattice Energy Formula DOWNLOAD IMAGE. Jang,E.L. Now automatically, a question comes to our mind: what is a crystal lattice? Determining the Lattice Energy of MX2. Lattice energy is also known as lattice enthalpy and can be stated in two ways. This new method demonstrates the relationship between chemical hardness and lattice energies of ionic compounds. One way is the energy released when gaseous ions combine to form an ionic solid. With our tool, you need to enter the respective value for Atomic Radius and hit the calculate button. Alternately, it can be defined as the energy that must be supplied to one mole of an ionic crystal in order to separate it into gaseous ions in a vacuum via an endothermic process . Join. What is unit of lattice energy ? Please help explain to me how I should get the answer! Consider the lattice energy of any ionic compound. In 1918, Born and Lande presented the following model for lattice energy: $E = - \frac {N_AMz^+z^-e^2}{4 \pi \epsilon_o r_o} (1-\frac {1}{n})$ A crystal is a highly transparent substance with a regular geometric shape. The electron affinity of is X -335 kJ/mol. and P. Day. The enthalpy of sublimation of M is 143 kJ/mol. We can present this as the sum of two processes: (1) the evaporation of salt to produce gaseous ions characterized by enthalpy lattice, and (2) the hydration of these ions for the production of the solution. High charges on the ions mean high lattice energy; Small separation means high lattice energy; Thus point 2 addresses point 2 in your question, K is bigger than Li, hence the separation is bigger in KF than LiF, hence KF has a lower lattice energy than LiF. Here chemical hardness values of ionic compounds are calculated via our molecular hardness equation. Lattice energy can be defined as the energy required to convert one mole of an ionic solid into gaseous ionic constituents. However, please VERIFY all results on your own, as the level of completion of this item is NOT CONFIRMED. Lattice Energy Formula Questions: 1. LE = (K * Q1 * Q2) / R. Where LE is the lattice energy. To calculate Lattice Parameter of FCC, you need Atomic Radius (r). Dependence of activation energy to lattice strain on TiO2 nanoparticles ... transformation from anatase to rutile by arrheniuse formula. Lattice energy can be a very complicated process but is often simplified by using Coulomb’s law. In science, when we say lattice, we usually refer to a crystal lattice. The Born-Haber cycle of calsium chloride crystal , CaCl2 . Match the appropriate formula to each lattice energy. r = the distance between the ion centers. Relationships among Ionic Lattice Energies, Molecular (Formula Unit) Volumes, and Thermochemical Radii H. Donald B. Jenkins* and Helen K. Roobottom Department of Chemistry, University of Warwick, Coventry CV4 7AL, West Midlands, U.K. Jack Passmore* Department of Chemistry, University of New Brunswick, Fredericton, New Brunswick, Canada, E3B 6E2 Leslie Glasser Centre for … To calculate the lattice energy we must think of all the steps necessary to make a crystal from it's elements. This is useful for rather complex ions like sulfate (SO 2− Lattice parameter of FCC is the edge length of FCC unit cell is calculated using Lattice Parameter of FCC=2*Atomic Radius*sqrt(2). Match the appropriate formula to the energy and explain. The enthalpy of formation of MX2 is -745 kJ/mol. Some energy quantities, such as the lattice energy of a mineral or the electron affinity of an atom, can be difficult to measure in the lab. It is not possible to calculate the lattice enthalpy directly from the forces of attraction and repulsion between ions but factors associated with crystal geometry must also be included. The 3-dimensional arrangement of atoms or ions (charged atoms) in a systematic pattern within a crystal gives rise to a crystal lattic… The first and second ionization energies of M are IE 705 kJ/mol and IE 1403 kJ/mol. The Born–Haber cycle can be used to predict which ionic compounds are likely to form. This is useful for rather complex ions like sulfate (SO 2− An ionic compound has the overall potential energy, which we refer frequently as the lattice energy. Explain. 3 0. Let's also assume that the ions are point charges - in other words that the charge is concentrated at the centre of the ion. Consider the lattice energy of any ionic compound. The bond energy of X2 is 187 kJ/mol. Your IP: 136.243.145.137 electronic reprint. You should talk about "lattice formation enthalpy" if you want to talk about the amount of energy released when a lattice is formed from its scattered gaseous ions. The enthalpy of formation of MX2 is -745 kJ/mol. The lattice energy of BaO, with a dipositive cation and a dinegative anion, dominates the Born–Haber cycle. Jenkins, H. Donald B. Lattice energy can be a very complicated process but is … This is the amount of energy given off when 1 mole of the solid is produced. This is the amount of energy given off when 1 mole of the solid is produced. The solid crystal is a three-dimensional entity. 20: Lattice Energy. Theoretical values for lattice energy. Explain. To calculate the lattice energy we must think of all the steps necessary to make a crystal from it's elements. It can also be defined as the amount of energy that is spent to separate an ionic crystal into its constituent ions. • Related Terms. The lattice enthalpy is indirectly determined by the use of Born - Haber Cycle. How To Calculate Depreciation Expense Using Units ... How To Calculate Real Interest Rate With Cpi, How To Bypass Activation Lock On Iphone Xr. Please help explain to me how I should get the answer! It can also be explained using . Born Haber Cycle Wikipedia. These have been made according to the syllabus 9701 and cover all the relevant topics for examination in March/June. Based on one more definition of the Lattice Energy (LE), it is required to break down the ionic solids and converting its components to the atoms of gaseous ions. U L = equilibrium value of the lattice energy. The lattice energy for ionic crystals decreases as the charge of the ions _____ and the size of the ions _____.a) Increases, increases b) In... Q. You may need to download version 2.0 now from the Chrome Web Store. Lattice Energy Formula per mole is symbolised as. The Born–Haber cycle can be used to predict which ionic compounds are likely to form. The first and second ionization energies of M are IE 705 kJ/mol and IE 1403 kJ/mol. Lattice parameter of FCC is the edge length of FCC unit cell is calculated using Lattice Parameter of FCC=2*Atomic Radius*sqrt(2).

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Next: Heisenberg's Uncertainty Principle Up: Wave Mechanics Previous: Probability Interpretation of Wavefunction

# Wave Packets

As we have seen, the wavefunction of a massive particle of momentum and energy , moving in free space along the -axis, can be written

 (1116)

where , , and is a complex constant. Here, and are linked via the matter wave dispersion relation (1103). Expression (1116) represents a plane wave that propagates in the -direction with the phase velocity . However, it follows from Equation (1104) that this phase velocity is only half of the classical velocity of a massive particle.

According to the discussion in the previous section, the most reasonable physical interpretation of the wavefunction is that is proportional to (assuming that the wavefunction is not properly normalized) the probability of finding the particle between and at time . However, the modulus squared of the wavefunction (1116) is , which is a constant that depends on neither nor . In other words, the previous wavefunction represents a particle that is equally likely to be found anywhere on the -axis at all times. Hence, the fact that this wavefunction propagates at a phase velocity that does not correspond to the classical particle velocity has no observable consequences.

How can we write the wavefunction of a particle that is localized in ? In other words, a particle that is more likely to be found at some positions on the -axis than at others. It turns out that we can achieve this goal by forming a linear combination of plane waves of different wavenumbers: that is,

 (1117)

Here, represents the complex amplitude of plane waves of wavenumber within this combination. In writing the previous expression, we are relying on the assumption that matter waves are superposable. In other words, it is possible to add two valid wave solutions to form a third valid wave solution. The ultimate justification for this assumption is that matter waves satisfy the linear wave equation (1102).

There is a fundamental mathematical theorem, known as Fourier's theorem (see Section 9.1 and Exercise 11), which states that if

 (1118)

then

 (1119)

Here, is known as the Fourier transform of the function . We can use Fourier's theorem to find the -space function that generates any given -space wavefunction at a given time.

For instance, suppose that at the wavefunction of our particle takes the form

 (1120)

Thus, the initial probability distribution for the particle's -coordinate is

 (1121)

This particular distribution is called a Gaussian distribution (see Section 9.1), and is plotted in Figure 79. It can be seen that a measurement of the particle's position is most likely to yield the value , and very unlikely to yield a value which differs from by more than . Thus, Equation (1120) is the wavefunction of a particle that is initially localized in some region of -space, centered on , whose width is of order . This type of wavefunction is known as a wave packet. However, a wave packet is just another name for a wave pulse. (See Chapter 9.)

According to Equation (1117),

 (1122)

Hence, we can employ Fourier's theorem to invert this expression to give

 (1123)

Making use of Equation (1120), we obtain

 (1124)

Changing the variable of integration to , the previous expression reduces to

 (1125)

where and . The integral in the previous equation is now just a number, as can easily be seen by making the second change of variable . Hence, we deduce that

 (1126)

where

 (1127)

If is proportional to the probability of a measurement of the particle's position yielding a value in the range to at time then it stands to reason that is proportional to the probability of a measurement of the particle's wavenumber yielding a value in the range to . (Recall that , so a measurement of the particle's wavenumber, , is equivalent to a measurement of the particle's momentum, ). According to Equation (1126),

 (1128)

This probability distribution is a Gaussian in -space. [See Equation (1121) and Figure 79.] Hence, a measurement of is most likely to yield the value , and very unlikely to yield a value that differs from by more than .

We have just seen that a wave packet with a Gaussian probability distribution of characteristic width in -space [see Equation (1121)] is equivalent to a wave packet with a Gaussian probability distribution of characteristic width in -space [see Equation (1128)], where

 (1129)

This illustrates an important property of wave packets. Namely, in order to construct a packet that is highly localized in -space (i.e., with small ) we need to combine plane waves with a very wide range of different -values (i.e., with large ). Conversely, if we only combine plane waves whose wavenumbers differ by a small amount (i.e., if is small) then the resulting wave packet is highly extended in -space (i.e., is large).

According to Section 10.1, a wave packet made up of a superposition of plane waves that is strongly peaked around some central wavenumber propagates at the group velocity,

 (1130)

rather than the phase velocity, , assuming that all of the constituent plane waves satisfy a dispersion relation of the form . For the case of matter waves, the dispersion relation is (1103). Thus, the associated group velocity is

 (1131)

where . This velocity is identical to the classical velocity of a (non-relativistic) massive particle. We conclude that the matter wave dispersion relation (1103) is perfectly consistent with classical physics, as long as we recognize that particles must be identified with wave packets (which propagate at the group velocity) rather than plane waves (which propagate at the phase velocity).

In Section 10.1, it was demonstrated that the spatial extent of a wave packet of initial extent grows, as the packet evolves in time, like

 (1132)

where is the packet's central wavenumber. Thus, it follows from the matter wave dispersion relation, (1103), that the width of a particle wave packet grows in time as

 (1133)

For example, if an electron wave packet is initially localized in a region of atomic dimensions (i.e., ) then the width of the packet doubles in about .

Next: Heisenberg's Uncertainty Principle Up: Wave Mechanics Previous: Probability Interpretation of Wavefunction
Richard Fitzpatrick 2013-04-08


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# Embedding finite groups in symmetric groups whilst preserving conjugacy

Let $G$ be a finite group and suppose that $f,g \in G$ are not conjugate in $G$. It is classical result that every finite group embeds into a finite symmetric group $S = S_{|G|}$. My question is: can this embedding be modified in a way so that $f \not\sim_G g$?

It can be easily seen that in general, the answer is negative - consider $C_p$, a cyclic group of prime order. However, is there some nice description for instances, when it will be possible?

• Do you mean $f \not\sim_S g$? Also "can this embedding be modified" is not clear. What kinds of modifications do you allow? – Derek Holt Feb 15 '16 at 17:49
• Basically, anything is fine as long as I remain within the class of finite symmetric groups. Of course I still require the map to be a monomorphism. – Michal Ferov Feb 15 '16 at 21:54

The class of groups (which can be embedded into an $S_n$ such that in the embedded image conjugacy is completely determined by cycle structure --lets call them CCS-groups for conjugacy cycle structure'') will be closed under sub direct products (duplicate, triplicate etc. the cycles in additional copies), so it is more than just $S_n$. CCS-groups also include (explicit test) the dihedral groups of order 8 and 12. (From trying out groups of small order, and using the following condition, I conjecture that this is all.)
The same argument as for $C_p$ (coprime powers must be conjugate) shows that a CCS-group must have an integral character table. (See https://mathoverflow.net/questions/134581/groups-in-which-all-characters-are-rational for this separate question) This property is necessary but not sufficient ($Q_8$ has an integral character table but is not CCS).


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## Statics- using sine and cosine

When you have to break vectors up into components...how do youknow whether to use the sine or cosine function?