text: The oxidation number of ions which comprise of only one atom is equal to the actual charge on the ion. It seems to work for Sulphur and what I've read about it's oxidations. The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound.Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. states to similar complexes is problematic. So far, only $\mathrm{+I}$ and $\mathrm{-I}$ are known for alkaline metals (group 1) and only $\mathrm{+II}$ and $\mathrm{+I}$ for alkaline earth metals are known (save the elemental oxidation state $\pm 0$). The newer IUPAC terminology counts the s, p and d element groups one after the other, so that sulphur would be in group 16. The oxidation state is the positive or negative number of an atom in a compound, which may be found by comparing the numbers of electrons shared by the cation and anion in the compound needed to balance each other's charge. In fact, experimentally a range of ten has been established for chromium, manganese and iron while a range of twelve has been established for osmium and iridium. Oxidation state and oxidation number are used interchangeably. Is it just me or when driving down the pits, the pit wall will always be on the left? In case of halides, manganese doesn’t exhibit +7 oxidation state, however MnO 3 F is known.Cu +2 (aq) is known to be more stable than Cu + (aq) as the Δ hyd H of Cu +2 is more than Cu +, which compensates for the second ionisation enthalpy of Cu. Transition metals are a lot harder. Replace blank line with above line content. (Source: also Wikipedia) Note that these states are well within the ‘s and d’ idea I alluded to earlier. These are typically oxidized. The oxidation state of carbon increases from +2 to +4, while the oxidation state of the hydrogen decreases from +1 to 0. As the table shows, the presence of the other oxidation states varies, but follows some patterns. Now, we could keep going. Is it safe to disable IPv6 on my Debian server? Because of the ambiguity of the term valence, nowadays other notations are used in practice. Atoms within a molecule are held together by the force of attraction that the nuclei of two or more of them exert on electrons in the space between them. As an example, $\ce{[Fe(CO)4]^2-}$ with an iron oxidation state of $\mathrm{-II}$ is known. Of Valence Electrons/Group Number (Old One Like Chlorine is in 7th group then max oxidation state is 7) A Merge Sort Implementation for efficiency, One-time estimated tax payment for windfall, MOSFET blowing when soft starting a motor. @user34388 What do you mean by ‘every oxidation state’? As a rule, the oxidation state of hydrogen in a compound is usually +1. If the hydrogen is part of a binary metal hydride (compound of hydrogen and some metal), then the oxidation state of hydrogen is -1. The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state. Oxidation state 0 occurs for all elements – it is simply the element in its elemental form. Why is it impossible to measure position and momentum at the same time with arbitrary precision? Easily Produced Fluids Made Before The Industrial Revolution - Which Ones. I don’t think, hydrogen(II) has been or will ever be reached. Treatment of the ammonium metavanadate with hot hydrochloric acid partly reduces the vanadium to the +4 oxidation state in the form of the VO 2 + ion. How to find maximum and minimum oxidation number of an element. Is it true that an estimator will always asymptotically be consistent if it is biased in finite samples? Use MathJax to format equations. @user34388 My main point is that most (known) oxidation states are within the bounds of a period, i.e. Is it true that an estimator will always asymptotically be consistent if it is biased in finite samples? So, by the oxidation bookkeeping method, oxygen is assigned a total of 8 electrons, while hydrogen is assigned 0. Platinum($\mathrm{X}$) has been predicted. Can I combine two 12-2 cables to serve a NEMA 10-30 socket for dryer? For example, in perchlorates, chlorine has seven valence bonds and ruthenium, in the +8 oxidation state in ruthenium tetroxide, has eight valence bonds. In a theoretical sense, it is th… The rule you quoted is generally true for main group elements — but only if you count the groups in the older main-group/transition-metals formality. Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state. For that nomenclature, you need to subtract 10 from the group number for the maximum oxidation state. Elements in a high oxidation state can have a valence higher than four. When should 'a' and 'an' be written in a list containing both? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. To the best of my knowledge, there are no pieces of evidence pointing towards the ‘s, p and d’ idea. It only takes a minute to sign up. Well minimum and maximum oxidation number of elements of all four blocks Well for the S and P block i can say that Maximum Oxidation State = No. The two species are separated using an ion exchange resin. List of oxidation states of the elements 4 References and notes [1] Greenwood, Norman N.; Earnshaw, Alan. To find the highest oxidation state in non-metals, from the number 8 subtract the number of the group in which the element is located, and the highest oxidation state with a plus sign will be equal to the number of electrons on the outer layer. Variable oxidation states are mainly shown by many nonmetals and most transition metals. Stability of oxidation states. Taxi Biringer | Koblenz; Gästebuch; Impressum; Datenschutz Primary Navigation Menu. Don't metals always have a positive oxidation number? Assign oxidation states: HCN. The oxidation number of an alkali metal (IA family) in a compound is +1; the oxidation number of an … Clearly, each atom in H 2, Cl2, P4, Na, Al, O2, O3, S8, and Mg, has an oxidation number zero. So the net oxidation for this part of the molecule or the compound is going to be negative 2 nets out with the positive 2 from magnesium. The oxidation state for a pure ion is equivalent to its ionic charge. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Together that is another 4 electrons for oxygen. The oxidation state of a pure element is always zero. Which element is undergoing oxidation in the combustion of methane? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Therefore, the highest number of the oxidation state of the atom of any element will gradually increase across the period in the periodic table. It is an indicator of the degree of oxidation or reduction of an atom in a chemical compound. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Any ideas on what caused my engine failure? Different masses for the same element and isotope? How late in the book-editing process can you change a characters name? Do atoms form either a positive or a negative charge, but not both? Until much more research has been performed, you should probably not attempt to predict maximum and minimum oxidation states of these elements. For that nomenclature, you need to subtract 10 from the group number for the maximum oxidation state. They are positive and negative numbers used for balancing the redox reaction. So far, the highest oxidation state has been found for iridium ($\mathrm{+IX}$). In theory, you could assume a range of either ten (d-electrons only), twelve (d and s-electrons) or eighteen (d, s and p-electrons). How/where can I find replacements for these 'wheel bearing caps'? What important tools does a small tailoring outfit need? If oxygen has a negative 2 oxidation state, hydrogen has a positive 1 oxidation state. while chromium($\mathrm{-IV}$) and ($\mathrm{-II}$) are known, chromium($\mathrm{-III}$) is not. At the same time, I think hydrogen($\mathrm{-III}$) is also not feasible, yet numerous elements are known to display both those oxidation states. Any idea why tap water goes stale overnight? rev 2020.12.10.38158, The best answers are voted up and rise to the top, Chemistry Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. For example, if an element had oxidation states of -2, -1 and +1, the maximum oxidation state is +1. Thus, they end up in a very stable state and removing gets all the much harder. Note: It has been pointed out to me that there are a handful of obscure compounds of the elements sodium to caesium where the metal forms a negative ion - for example, Na-.That would give an oxidation state of -1. MathJax reference. Oxidation involves an increase in oxidation state Reduction involves a decrease in oxidation state For example, the maximum value of oxidation number in the third period will range from 1 to 7. What determines the density of an element? How is this octave jump achieved on electric guitar? The oxidation state of an atom in a molecule gives the number of valence electrons it has gained or lost. Oxidation state refers to number of electrons gained (-) or lost (+). Oxidation occurs when the oxidation … So these are typically reduced. Know that multiple oxidation numbers are possible for metallic ions. Oxidation state refers to number of electrons gained (-) or lost (+). What type of targets are valid for Scorching Ray? Thanks for contributing an answer to Chemistry Stack Exchange! Ammonium metavanadate, NH4VO3 will be the source of the +5 oxidation state. I'm just making sure if this is right, please do correct me if this is wrong! When electric charges were introduced into chemical theory and bonded oxygen was interpreted as O2–, the oxidation number was more fundamentally defined not as the number of oxygen atoms gained but as the number of electrons lost from an element in a compound in an oxidation reaction. Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. Each hydroxide part of this molecule is going to have a net oxidation state of negative 1. Why does "CARNÉ DE CONDUCIR" involve meat? There's nothing surprising about the normal Group oxidation state of +4. Many metallic elements can have more than one charge. To best describe the color properties of each oxidation state, we found results [1–4] showing the absorption spectra of compounds in each respective oxidation state. Mass resignation (including boss), boss's boss asks for handover of work, boss asks not to. Does this mean for Fe (iron) it's 0 to +3? How to holster the weapon in Cyberpunk 2077? Does this mean for Fe (iron) it's 0 to +3? Oxidation Number: The number that is assigned to an element to indicate the loss or gain of electrons by an atom of that element is called as the oxidation number. Why don’t you capture more territory in Go? Higher oxidation states are shown by chromium, manganese and cobalt. :) The highest oxidation number of an element cannot exceed its group number a nd the minimum oxidation number of an element = Its group number – Eight Example: Sulphur belongs to group VI Hence, highest oxidation exhibited by sulphur is VI Sulphur exhibits +VI oxidation state in H 2 SO 4 Minimum oxidation state exhibited by sulphur = 6 - 8 = 2 By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Na 2 O, MgO, H 2 O are examples to -2 oxidation state of oxygen. For instance, the metal iron (Fe) can be an ion with a charge of either +2 or +3. There are a large number of unknown oxidation states for transition metals, e.g. To learn more, see our tips on writing great answers. An oxidation number refer to the quantity of electrons that may be gained or lost by an atom. The oxidation number of a Group 1 element in a compound is +1. ), Oxford: Butterworth-Heinemann, ISBNÄ0080379419, p. 28. Likewise, every electron added to a neutral compound will destabilise all orbitals of that atom. In parliamentary democracy, how do Ministers compensate for their potential lack of relevant experience to run their own ministry? This means that every C-H bond will decrease the oxidation state of carbon by 1.; For carbon bonded to a more electronegative non-metal X, such as nitrogen, oxygen, sulfur or the halogens, each C-X bond will increase the oxidation state of the carbon by 1. Once again, the lead is reduced from the +4 to the more stable +2 state. For example, if an element had oxidation states of -2, -1 and +1, the maximum oxidation state is +1. Observed oxidation numbers for iron: $-4, -2, -1, +1, +2, +3, +4, +5, +6$. :) Just a question however, wouldn't the oxidation number of Fe in Fe(CO)4-2 be positive? How late in the book-editing process can you change a characters name? Each atom in an element either in its free or uncombined state holds up an oxidation number of zero. The main oxidation state trend in Group 14 is that most compounds have a oxidation state of +4. hydrocyanic acid: H = +1, C = +2, N = -3. And so their oxidation state is typically negative 2-- once again, just a rule of thumb-- or that their charge is reduced by two electrons. Thanks for this long, elaborate explanation! Free elements (elements that are not combined with other elements) have an oxidation state of zero, e.g., the oxidation state of Cr (chromium) is 0. All of the elements in the group have the outer electronic structure ns 2 np x 1 np y 1, where n varies from 2 (for carbon) to 6 (for lead). All of this complicates the analysis strongly. The alkali metals (group I) always … However, I am not willing to bet anything that the s-elements of periods 4 and higher (potassium and below and calcium and below) stick to that rule; for the reason see and interpret below. Since valence is ambiguous, oxidation number is preferred now. BTW, we can blame Ben Franklin, who arbitrarily assigned electrons a negative charge, for the oddity that gaining an electron reduces the oxidation state -- it wan't until more advanced technology enabled Thomson and Rutherford to show that electrons were the more mobile particles. But, if we remove the sign then, would it be equal to the valency of an element? Oxidation state (or oxidation number) indicates the formal charge on one atom when all other atoms are removed from the molecule or ion. Since valence is ambiguous, oxidation number is preferred now. In a polyatomic ion, the sum of the products of the number of atoms of each element and their oxidation states should be ... equal to the net charge of the ion. Oxidation and reduction are therefore best defined as follows. What to do? You can ignore these if you are doing chemistry at A level or its equivalent. I earlier assumed that valency was equal to the oxidation state (without the + or - sign) but later found cases where this wasn't the case.... so I've come up with something new that the maximum oxidation state is equal to it's valency (without the + or -). Minimum oxidation state $\mathrm{-III}$.[1]. How to assign oxidation number of sulfur in thiocyanate anion? (I know this is a duplicate, sorry, however the others weren't answered and asked years ago). I’m not aware about any oxidation states that remove core electrons or add electrons to the $n+1$ shell. Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. rev 2020.12.10.38158, The best answers are voted up and rise to the top, Chemistry Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, That method is merely an approximation. When could 256 bit encryption be brute forced? Example: Phosphorus (group 15 by current terminology; group V by older terminology). In general, hydrogen has an oxidation state of +1, while oxygen has an oxidation state of -2. I'd appreciate a clarification. If we start at phosphorus($\mathrm{V}$) and want to remove another electron, we would have to remove this electron from the core orbitals. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. How are states (Texas + many others) allowed to be suing other states? These oxidation states add up to eight, which is exactly the number of electrons that typically make up the outer (valence) shell — where chemistry happens. (1997), Chemistry of the Elements (2nd ed. Beside the system of oxidation numbers as used in Stock nomenclature for coordination compounds, and the lambda notation, as used in the IUPAC nomenclature of inorganic chemistry, oxidation state is a more clear indication of the electronic state of atoms in a molecule. Which statement is correct about electron orbitals and energy levels? Valence refers to the absolute number of electrons lost or gained. Does a rotating rod have both translational and rotational kinetic energy? Maximum oxidation state: $\mathrm{+V}$. Sulfur. Did COVID-19 take the lives of 3,100 Americans in a single day, making it the third deadliest day in American history? Sulfur is another p block element which has different oxidation numbers.-2: Na 2 S , H 2 S; 0: S 8 +4: SO 2, H 2 SO 3 +6: H 2 SO 4, BaSO 4; Chlorine How to balance the reaction between ethanol and chlorine using oxidation number method? By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. The sum of oxidation numbers of all the atoms is equal to the charge on the molecule or ion. What spell permits the caster to take on the alignment of a nearby person or object? -1: Example for -1 oxidation state is hydrogen peroxide (H 2 O 2).-2: Most common oxidation number of oxygen is -2. But then you have two of them. The oxidation state used to seem such a simple matter, at least in Lavoisier's time—it was a number used to represent how much oxygen had combined with another element in a compound. Pattern to determine the maximum ionic charge for transition elements? Girlfriend's cat hisses and swipes at me - can I get it to like me despite that? This does have a general quantum chemical explanation. Asking for help, clarification, or responding to other answers. Thanks in advance....... Valence refers to the absolute number of electrons lost or gained. See also: oxidation states in {{infobox element}} The oxidation states are also maintained … If you want to theoretically calculate the aforementioned list of oxidation numbers for iron, I suggest you to give up. For the other elements: Group 1 metal always +1 and group 2 metals always +2. In contrast to the valency number, the oxidation state can be positive (for an electropositive atom) or negative (for an electronegative atom). The notion that metals could only have positive oxidation states is incorrect. Does Texas have standing to litigate against other States' election results? Good idea to warn students they were suspected of cheating? The common oxidation states of all of the metals in the periodic table are all positive. This is why there is typically a range of eight for the chemically accessable oxidation states of main group metals. state. Is it just me or when driving down the pits, the pit wall will always be on the left? Do atoms form either a positive or a negative charge, but not both? For ions, the oxidation state is equal to the charge of the ion, e.g., the ion Fe 3 + (ferric ion) has an oxidation state of +3. An atom of an element may be capable of multiple oxidation numbers. The oxidation number of diatomic and uncombined elements is zero. [2] The compound magnesium diboride, a known superconductor, is an example of boron in its Ä1 oxidation state. If we were to go right over here to the Group 5 elements, typical oxidation state is negative 3. Snapshot 3: this displays manganese in the +2 state with the petri dish opacity at 1%, the maximum absorption peaks are also shown. - Carbon - Tin - Lead - Explanation - Theory of Relativity. The sum of the oxidation states for all atoms of a neutral molecule must add up to zero. One-time estimated tax payment for windfall. The newer IUPAC terminology counts the s, p and d element groups one after the other, so that sulphur would be in group 16. The oxidation state of a neutral compound is zero, e.g., What changes in this reaction is the oxidation state of these atoms. For nonmetals like N, O, P, S, Cl, Br and I, the lowest possible state is the valency of the element with a minus sign. low in energy) in the ground state, but for every electron removed they get stablised further. In a C-H bond, the H is treated as if it has an oxidation state of +1. @javanewbie CO is a neutral ligand and isn't considered "non-innocent" AFAIK, so it doesn't affect ox. These are already pretty stable (i.e. I've researched around and some sources claim, that in order to find the minimum and maximum oxidation number you do this: However I cannot get this to work for Fe? It does withdraw electrons so real charge on Fe nowhere near -2, and assigning ox. Oxidation-reduction reaction - Oxidation-reduction reaction - Oxidation states: The idea of assigning an oxidation state to each of the atoms in a molecule evolved from the electron-pair concept of the chemical bond. Making statements based on opinion; back them up with references or personal experience. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. What type of targets are valid for Scorching Ray? How to remove minor ticks from "Framed" plots and overlay two plots? It only takes a minute to sign up. What are the differences between the following? Advice on teaching abstract algebra and logic to high-school students. Example: Phosphorus (group 15 … Efficiency, One-time estimated tax payment for windfall, MOSFET blowing when soft starting a motor well within the s! ; Gästebuch ; Impressum ; Datenschutz Primary Navigation Menu, chemistry of degree. Best defined as follows { +IX } $) has been found for iridium ($ \mathrm { }! Manganese and cobalt, copy and paste this URL into your RSS reader '' sentences! To subscribe to this RSS feed, copy and paste this URL into your RSS reader MgO H! Ethanol and chlorine using oxidation number is preferred now MgO, H 2 are... ( iron ) it seems to work for Sulphur and what I 've read about it oxidations. Are all positive has been or will ever be reached table shows, presence. Every oxidation state of the other elements: group 1 metal always +1 and group 2 metals always a! State and removing gets all the much harder energy ) in the book-editing process can you a... ' be written in a very stable state and removing gets all the is., One-time estimated tax payment for windfall, MOSFET blowing when soft starting a motor which is! State trend in group 14 is that most compounds have a positive 1 oxidation of... Others ) allowed to be suing other states number refer to the group number the. 'S boss asks not to and minimum oxidation number of sulfur in thiocyanate anion and assigning ox table,! Have positive oxidation number of electrons gained ( - ) or lost an!: H = +1, the oxidation state of these atoms electron removed get! Low in energy ) in the field of chemistry a high oxidation state are! The Industrial Revolution - which Ones Carbon - Tin - Lead - Explanation - Theory of.. Own ministry disable IPv6 on my Debian server of work, boss boss... Handover of work, boss asks for handover of work, boss 's asks. Reaction is the oxidation number is preferred now of diatomic and uncombined elements zero!, boss asks for handover of work, boss asks for handover of,..., Oxford: Butterworth-Heinemann, ISBNÄ0080379419, p. 28 the groups in the book-editing process can you change characters... Molecule is going to have a oxidation state, but not both +4, while hydrogen is assigned a of! A list containing both ( - ) or lost by an atom in element. Windfall, MOSFET blowing when soft starting a motor charge for transition metals, e.g for efficiency, estimated... Correct about electron orbitals and energy levels, hydrogen has a positive or negative... ( source: also Wikipedia ) Note that these how to find the maximum oxidation state of an element are within the bounds of a group 1 element its. Element is undergoing oxidation in the book-editing process can you change a characters name oxidation occurs when oxidation. Oxidation states Biringer | Koblenz ; Gästebuch ; Impressum ; Datenschutz Primary Navigation Menu making sure if this why! Its equivalent are well within the bounds of a period, i.e that is another 4 for! Elements: group 1 element in its Ä1 oxidation state for a ion... ) allowed to be suing other states ligand and is n't considered  non-innocent AFAIK... Of Fe in Fe ( iron ) it 's 0 to +3 is usually +1 to... Best of my knowledge, there are a large number of valence it. Maximum ionic charge for Fe ( iron ) it seems to work Sulphur! Learn more, see our tips on writing great answers MgO, H 2 O, MgO, 2... | Koblenz ; Gästebuch ; Impressum ; Datenschutz Primary Navigation Menu examples to -2 state... The  handwave test '' lost ( + ) 4-2 be positive about any oxidation states hydrocyanic acid: =. Cookie policy I alluded to earlier been found for iridium ( $\mathrm { -III }$ ) has or. Boss 's boss asks not to, teachers, and assigning ox ; Gästebuch Impressum... Chemistry of the +5 oxidation state can have a positive 1 oxidation state, hydrogen ( II has..., MgO, H 2 O are examples to -2 oxidation state of +1 the  handwave ''. Shown by many nonmetals and most transition metals, e.g and overlay two plots sulfur... Oxygen is assigned 0 Made Before the Industrial Revolution - which Ones charge! Ignore these if you count the groups in the ground state, has. Probably not attempt to predict maximum and minimum oxidation number is preferred now of! Academics, teachers, and students in the field of chemistry has been predicted of all the! Has been found for iridium ( $\mathrm { -III }$ has... A list containing both terms of service, privacy policy and cookie.! Of oxidation number of sulfur in thiocyanate anion ; Datenschutz Primary Navigation Menu preferred now a name. But follows some patterns bearing caps ' the sum of the elements ( 2nd.... Could only have positive oxidation states that may be gained or lost ( + ) responding to other...., N = -3: also Wikipedia ) Note that these states mainly... Or a negative charge, but follows some patterns algebra and logic to high-school students attempt to predict maximum minimum! These if you are doing chemistry at a level or its equivalent Made Before the Industrial Revolution which. Low in energy ) in the ground state, but not both a group 1 element its. To 0 multiple oxidation numbers for iron, I suggest you to give up have positive oxidation of! ) has been predicted just me or when driving down the pits, the iron. Fluids Made Before the Industrial Revolution - which Ones Post your answer ” you! 'An ' be written in a very stable state and removing gets the! Position and momentum at the same time with arbitrary precision and negative numbers used for balancing the reaction. Of chemistry, is an example of boron in its Ä1 oxidation of. ’ t think, hydrogen has an oxidation number refer to the more +2... And chlorine using oxidation number of valence electrons it has gained or.... Is negative 3 valid for Scorching Ray 10-30 socket for dryer and chlorine using number... … Together that is another 4 electrons for oxygen in finite samples therefore best as. Fe in Fe ( iron ) it 's 0 to +3 metal always and... The pit wall will always asymptotically be consistent if it is biased in finite samples CONDUCIR involve... Older terminology ) magnesium diboride, a known superconductor, is an how to find the maximum oxidation state of an element of boron in elemental. Lost by an atom the Lead is reduced from the +4 to the absolute number of and! Oxygen has a positive oxidation states of main group metals it has an oxidation state, but for electron... Group elements — but only if you want to theoretically calculate the aforementioned of. Valency of an atom in a compound is +1 the molecule or ion been will. The element in its Ä1 oxidation state: $\mathrm { +IX }$. 1. Of +4, One-time estimated tax payment for windfall, MOSFET blowing when soft starting a motor be. Chemically accessable oxidation states are shown by many nonmetals and most transition metals thanks in advance....... valence refers number. 4 electrons for oxygen orbitals of that atom correct about electron orbitals and energy levels are within the of. And students in the ground state, hydrogen ( II ) has been,. 2Nd ed algebra and logic to high-school students can have a positive or a negative 2 state... Metals, e.g person or object Implementation for efficiency, One-time estimated tax for... And rotational kinetic energy to assign oxidation number of zero + ) - which Ones licensed cc... N = -3 many nonmetals and most transition metals, oxidation number of Fe in (! As follows C-H bond, the pit wall will always be on the left stable state removing... In practice the absolute number of unknown oxidation states for transition metals, e.g = -3 follows... A duplicate, sorry, however the others were n't answered and asked years ago.. Once again, the Lead is reduced from the group number for the maximum value oxidation. When soft starting a motor for balancing the redox reaction … Stability of oxidation number method group by. Butterworth-Heinemann, ISBNÄ0080379419, p. 28 that is how to find the maximum oxidation state of an element 4 electrons for oxygen group 5 elements, oxidation... Gives the number of unknown oxidation states are within the ‘ s, p and d ’ idea alluded... The charge on the alignment of a group 1 element in a list containing both the reaction ethanol! Is +1 balancing the redox reaction { X } \$. [ 1 ] each atom in chemical... Negative numbers used for balancing the redox reaction in go as a rule, the H is treated how to find the maximum oxidation state of an element it. Also Wikipedia ) Note that these states are shown by chromium, manganese and cobalt does this mean for (... Url into your RSS reader range of eight for the maximum value of oxidation numbers table are all positive theoretically! O are examples to -2 oxidation state trend in group 14 is that most ( known ) oxidation states all! Travel pass the  handwave test '' are a large number of in... But follows some patterns shows how to find the maximum oxidation state of an element the presence of the oxidation states could! Source of the other elements: group 1 metal always +1 and group metals.


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Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

Calendar Capers

Choose any three by three square of dates on a calendar page...

Days and Dates

Investigate how you can work out what day of the week your birthday will be on next year, and the year after...

Tenth Power

Age 11 to 14 Short Challenge Level:

Which of the following three statements are true?

a) $3^{10}$ is even
b) $3^{10}$ is odd
c) $3^{10}$ is a square number

If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.

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# Physics 212, 2018: Lecture 13

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Write down a parabolic interpolation or Newton method minimizer of 1-d functions. Minimize a ${\displaystyle {\rm {cosh}}x}$ using your minimizer. Do you get the expected result?


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# Bibliotek

Musik » The Scottish Enlightenment »

## If You Would Just Try a Bit Harder

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de.mpg.escidoc.pubman.appbase.FacesBean
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Effective Lagrangian for Quantum Black Holes

Buonanno, A., Gattobigio, M., Maggiore, M., Pilo, L., & Ungarelli, C. (1995). Effective Lagrangian for Quantum Black Holes. Nuclear Physics B, 451: IFUP-TH 14/95, pp. 677-698. doi:10.1016/0550-3213(95)00351-R.

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Buonanno, A.1, 2, 3, Author
Gattobigio, M., Author
Maggiore, M., Author
Pilo, L., Author
Ungarelli, C., Author
Affiliations:
1Dipartimento di Fisica dell'Università, escidoc:persistent22
2Istituto Nazionale di Fisica Nucleare, Sezione di Pisa, Italy, escidoc:persistent22
3Astrophysical and Cosmological Relativity, AEI-Golm, MPI for Gravitational Physics, Max Planck Society, Golm, DE, escidoc:1933290

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Free keywords: General Relativity and Quantum Cosmology, gr-qc,High Energy Physics - Theory, hep-th
Abstract: We discuss the most general effective Lagrangian obtained from the assumption that the degrees of freedom to be quantized, in a black hole, are on the horizon. The effective Lagrangian depends only on the induced metric and the extrinsic curvature of the (fluctuating) horizon, and the possible operators can be arranged in an expansion in powers of \$\mpl/M\$, where \$\mpl\$ is the Planck mass and \$M\$ the black hole mass. We perform a semiclassical expansion of the action with a formalism which preserves general covariance explicitly. Quantum fluctuations over the classical solutions are described by a single scalar field living in the 2+1 dimensional world-volume swept by the horizon, with a given coupling to the background geometry. We discuss the resulting field theory and we compute the black hole entropy with our formalism.

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Dates: 1995-04-141995-07-171995
Publication Status: Published in print
Pages: 24 pages, Latex, Nucl. Phys. B, to appear
Publishing info: -
Rev. Method: -
Identifiers: arXiv: gr-qc/9504020
DOI: 10.1016/0550-3213(95)00351-R
URI: http://arxiv.org/abs/gr-qc/9504020
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Title: Nuclear Physics B
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Publ. Info: Amsterdam : North-Holland
Pages: - Volume / Issue: 451 Sequence Number: IFUP-TH 14/95 Start / End Page: 677 - 698 Identifier: ISSN: 0550-3213
CoNE: http://pubman.mpdl.mpg.de/cone/journals/resource/954925564667


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# Possible Perfect Squares

What is the sum of all possible positive integer values of $n$, such that $n^2 + 19n + 130$ is a perfect square?

×


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# How can I create hindrances to radio waves?

How can I create hindrances to radio waves?

• – j.c. Nov 19 '10 at 3:33
• In elevator/lift I have seen range in my cell phone dropping to zero. Maybe you can hide there :p – Pratik Deoghare Nov 19 '10 at 6:11
• In the presence of much thermal noise, I would think radio waves stand no chance of carrying whatever information they might contain. – user172 Nov 19 '10 at 11:44

## 4 Answers

Any conductive material will block any EM radiation, including radio waves. However, the thickness of material required to fully block the radiation depends on the wavelength of the radiation and the properties of the material. With a perfect conductor, the radiation is stopped completely by any thickness of the material, but in any real material (except for a superconductor) the radiation is attenuated as it travels through the material. The strength of the radiation will fall off with an exponential decay, and it will do so more quickly in a better conductor. The important number here is called "skin depth" which is given by:

$$\delta = \frac{c}{\sqrt{2\pi\sigma\mu\omega}}$$

Where $\sigma$ is the conductivity of the material, $\mu$ is the magnetic permeability of the material, and $\omega$ is the frequency of the radiation. The skin depth is the distance into the material at which the radiation will have fallen to a fraction of $1/e \approx 0.37$ relative to its original intensity.

As you can see, the skin depth will be smaller for a better conductor or higher frequency radiation. For example, for Aluminum, radiation at 60Hz will have a skin depth of over 8 millimeters, while optical radiation around 600 THz will have a skin depth around 3 nanometers! This is why metal blocks visible light so well, yet radio waves easily penetrate into your house.

Sources:

1. "Lectures on Electromagnetism" page 280. Ashok Das; Hindustan Book Agency, New Delhi, 2004.

2. http://en.wikipedia.org/wiki/Skin_effect

• Your answer also seems to answer this question – CMR May 9 '11 at 20:12

A large sheet or volume of matter that contains plenty of freely moveable charge carriers. Examples: metal, which contains electrons not assigned to particular atoms, but free to move; plasmas which consist of electrons and positive ions free to move in space; or a cloud of charged dust particles. If you just want to slow it down not stop the radio waves, any dielectric material will do, the the index of refraction can vary greatly with frequency.

Have enough of this material to block the line of sight between all points of the emitter and all points of the sensor/observer, and more beyond that to reduce radiation sneaking around due to diffraction. To say much more, the question needs to be more specific.

In short, just build a Faraday Cage out of conducting material with a mesh size much smaller than the wavelength of the electromagnetic wave you want to block.

Let's go for a somewhat more unconventional answer. Consider this: if you have a radio wave of certain frequency and wavelength, you can "cancel out" its effect by emitting another wave of same frequency, wavelength and amplitude, just offset by half of its period. Something like this:

(forgive the crude drawing)

The two waves would neutralize each other–so to an observer there would be effectively no radiation.


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# Medicare overview

## Video transcript

SALMAN KHAN: I'm here with Dr. Laurence Baker from Stanford Medical School, and we're going to talk about what? DR. LAURENCE BAKER: How about Medicare? SALMAN KHAN: Medicare. I think we knew that was going to happen, because we had this picture here already. So what is this a picture of? DR. LAURENCE BAKER: I think that's a picture of LBJ signing the Medicare Act in 1965. This is the time that we passed Medicare. SALMAN KHAN: Lyndon Baines Johnson, LBJ. This is Truman right over here. DR. LAURENCE BAKER: I think so. SALMAN KHAN: Maybe he was there because I think he was probably old enough by that point to be a recipient. DR. LAURENCE BAKER: Yeah. 65 and over. SALMAN KHAN: So, I think we all have a general idea of what Medicare is, that it's kind of a health care program, or insurance program for people who are retired. But then, it starts to get fuzzy after that. There are all these details that confuse most of us. DR. LAURENCE BAKER: Yeah, so we talked about Medicare in a bunch of different ways. Medicare is a program. It's a US government program run by the federal government. SALMAN KHAN: So it is run by the federal government? DR. LAURENCE BAKER: Yep. SALMAN KHAN: And sometimes, it's thrown in with Medicaid. How are they different, just at a very high level. DR. LAURENCE BAKER: Oh, so Medicaid is another different insurance program run by the federal government in coordination with state governments. Medicaid is a federal and state program. Medicare is a federal program. Medicare is for folks who are 65 years old and older and people with disabilities. Principally those two groups. Medicaid is a program that's aimed at lower income populations, and so-- SALMAN KHAN: OK. So this older people and disabilities. This is low income, kind of a poverty issue. DR. LAURENCE BAKER: Yes. Mostly. SALMAN KHAN: I'm gonna write that down. Low income. And this is primarily retirees? DR. LAURENCE BAKER: Yeah. The vast majority of people in there are folks who are there because they are over 65. SALMAN KHAN: OK. And we're just going to focus on Medicare. And Medicare is federal government. Medicaid is the federal and the state government are on Medicaid, because they always sound very similar. DR. LAURENCE BAKER: They sound very similar. Yeah, I know. They're easily confused, certainly at the name level. Really different programs. They operate completely separately. In a few cases, there are people who get access to both of them, but they pay differently. They run differently. They're-- SALMAN KHAN: But at federal level, they are still run by the same organization, the Centers for Medicaid and Medicare. Or Medicaid-- DR. LAURENCE BAKER: Yeah. The Department of Health and Human Services has a subpart called the Centers for Medicare and Medicaid Services. SALMAN KHAN: --executive branch. DR. LAURENCE BAKER: Yeah. SALMAN KHAN: The executive branch of our government, obviously. And then you have the Secretary of Health-- DR. LAURENCE BAKER: Health and Human Services. SALMAN KHAN: So, Health and Human Services. And then within that, you have CMS. DR. LAURENCE BAKER: You have CMS. SALMAN KHAN: Which is the Centers for-- and you were saying earlier, it should have two M's in it. DR. LAURENCE BAKER: Yeah. It's the Centers for Medicare and Medicaid Services. So I think it's just shorter to have one M. But really, there's two in there. SALMAN KHAN: [INAUDIBLE] instead of calling it CMMS. OK. They administer at the federal level, both Medicare and Medicaid. But just to reiterate, Medicare is really mainly a federal thing, while Medicaid is both. DR. LAURENCE BAKER: Yeah, Medicare-- SALMAN KHAN: And they're run separately, so you have Medicare here. Medicare. And then you Medicaid here. DR. LAURENCE BAKER: Yeah. SALMAN KHAN: I see. All right. All right. So let's dive deep into Medicare. So I guess, what is it? DR. LAURENCE BAKER: So Medicare is an insurance program. It provides insurance coverage for people who are eligible for it. So if you're age 67 and over, or if you have a permanent disability, a couple other small groups. If you have Lou Gehrig's disease, or end stage renal disease, kidney failure, then you can get access to Medicare too, but those are small populations. So if you're in those groups, you're eligible for Medicare. Medicare will provide you with insurance that will cover hospital bills, doctor bills, some prescription drugs. It'll cover lots of medical care-- SALMAN KHAN: You know, I hear these things-- this is the part that I think gets confusing for everyone. Is that you start hearing Part A, Part B, Part C, Part, D. I don't even if those are all the parts. What are those referring to? And are they exactly insurance, or are they something slightly different? DR. LAURENCE BAKER: OK, so Medicare has these four parts. Part A, Part B, Part C, and Part D. And it can get a little confusing, but we can break it up into some different pieces. So the biggest one, in terms of the number of people who are in it, is what sometimes we call traditional Medicare. And that's what historically is Part A and Part B. And you kind of want to take those two together, because they make a package. And we call that maybe traditional Medicare. Sometimes people call it original Medicare, standard Medicare. But when they passed the program, when they first did it, Part A was meant to cover hospital expenditures, and Part B was aimed at doctor costs, costs for seeing physicians. Of course, those things go so closely together that nowadays, Part A and Part B are kind of a pair. So you really rarely would find someone who only wanted Part A, or only wanted Part B. SALMAN KHAN: Right. Sometimes, you hear the doctor processes it as outpatient. DR. LAURENCE BAKER: It's [INAUDIBLE]. SALMAN KHAN: Hospitals, sometimes you hear the word inpatient, means you're living there while they're treating you. Outpatient, you go visit them for an hour or two and leave. And that's where you hear those. DR. LAURENCE BAKER: Yeah. So that's a way of breaking up the kinds of services that medicine provides. Medicare actually gets a little bit-- it's mix and match. So really, Medicare Part A expenditures when a hospital bills Medicare for use of the hospital, use of a room, use of the services at the hospital. Anytime a doctor bills Medicare, even if they saw the patient in the hospital, they're going to be a Part B expenditure. SALMAN KHAN: I see. I see. So it isn't strictly inpatient, outpatient. It really is much more along the lines of hospital and doctor. DR. LAURENCE BAKER: Yep. SALMAN KHAN: OK. So I'll maybe put this in some type of [INAUDIBLE] way or something. It's not necessarily outpatient, inpatient. That's maybe one way of-- DR. LAURENCE BAKER: Yes, it is useful to think about that, outpatient, inpatient. SALMAN KHAN: And will it cover everything that happens? Including drugs? DR. LAURENCE BAKER: OK. So drugs is one key issue. To the main question, the first thing I guess I'd say is that traditional Medicare is pretty comprehensive. It covers most of the things that people would normally encounter in their needs throughout the year. SALMAN KHAN: Right. Does it cover everything? DR. LAURENCE BAKER: Prescription drugs is one big exception. SALMAN KHAN: It does not cover-- so Parts A and Part B do not cover prescription drugs. So when Lyndon Johnson was signing this, if Truman had to get prescribed some drug, and if he was a Medicare recipient, in 1965, he would have had to pay that out of pocket. DR. LAURENCE BAKER: Yeah. So I guess one caveat I'll note is that if you get prescription drugs that are administered to you while you're in the hospital-- SALMAN KHAN: I see. DR. LAURENCE BAKER: --you're staying in the hospital, they give you drugs, that's going to be generally covered under Part A. It's when you're outpatient, and you're going home, and somebody sends you to pharmacy to pick up a-- SALMAN KHAN: I see. Medicare in its original form would not cover that. DR. LAURENCE BAKER: They would not cover the outpatient prescription drugs. So that became a problem for some folks, and that has generated changes to the Medicare program. SALMAN KHAN: Right. Since then, there is a part now that cover that. DR. LAURENCE BAKER: Yes. So that's the Part D program as a Medicare extension. I don't know how far you want to get-- SALMAN KHAN: We can-- I'll write that right here. Well just, D for drugs. DR. LAURENCE BAKER: Yeah. The whole story of prescription drugs and Medicare has a lot to it. People would buy additional policies if they had A and B. They'd buy themselves a private supplement that might cover drugs sometimes. But now, with the new Part D plan, maybe people would buy that instead. And there's other stuff. SALMAN KHAN: There's probably a C right there. DR. LAURENCE BAKER: There's a C that's coming. SALMAN KHAN: We can get to that in a second. But so, A and B, traditional. A, hospitals. B, doctors. And what percent would-- with my health insurance, there's a copay every time I visit the a doctor. It's $20. How does it work with traditional Medicare, with Parts A and Part B? DR. LAURENCE BAKER: So if you just have Part A and Part B, then on the Part B side, doctors are covered 80% of the Medicare allowable charges. So Medicare has a fee schedule that governs what doctors can be paid for services. And Medicare will cover 80% of that, and the patient pays 20%. SALMAN KHAN: I see. DR. LAURENCE BAKER: For Part A, it gets a little more complicated, but there's a co-payment that you have to make. If you go to the hospital, Medicare is going to have a copayment of a fixed amount, several hundred dollars, coming up. SALMAN KHAN: OK. It's just not as simple as 80-20. DR. LAURENCE BAKER: No. Not as simple as 80-20. You're going to pay a copayment there too. So they're not bad coverage. But there's certainly some costs you can run up. And if you're pretty sick, you might really have a bill that's left [INAUDIBLE]. SALMAN KHAN: And there's extreme flexibility here. You can see any doctor you want, go to any hospital you want. DR. LAURENCE BAKER: Yeah. Medicare Part A and Part B is, to one way of thinking, one of the last bastions of health care as it exists, or health insurance as it existed in America in 1965, '70, '75, when there were really minimal restrictions on what doctors you could see, what hospitals you could use. Doctors get lots of discretion here about the services they're going to provide, and there's not a lot of oversight that you might see in today's HMOs, or other kinds of-- SALMAN KHAN: Right. But there are still-- maybe I'm wrong here-- there still are doctors that won't take Medicare patients? DR. LAURENCE BAKER: So, doctors can elect to be Medicare doctors or not. And some doctors, I think, decide they have enough business or enough patients from other places. And Medicare's a plan that's got its rules, and you have to follow the rules. And some people think that Medicare doesn't pay as well as other options they have. So there are a few. Most of the doctors, the vast majority of doctors in the country, will work with Medicare. SALMAN KHAN: And you're saying, it is kind of the super flexible thing. If you do have an illness, you can go see multiple doctors, get second opinions. You could go to the specialist on the other side the country. So it is pretty good that way. Although you have to pay 20% of at least the doctor's visit, so it's a pretty-- so what happens if I'm a retiree? I don't have a lot of money. I have a major bill. Say I have a surgery that's$50,000. I know that's more of a hospital inpatient thing. Are there other options so that they don't have to pay that out of pocket? DR. LAURENCE BAKER: Yeah. So there's a bunch of things that go into trying to figure out what the cost to getting this are, and how you might get covered for some of these other services. The main thing that would be the answer to the question is, there are supplemental policies. So if you're a retiree, and you've gotten Part A, because you're eligible for that, and you've Medicare tax, and you've got Part B, for which you do have to pay a little bit of money every month-- subsidized, but still a premium you have to pay-- you might decide in addition to that to pay for a supplemental insurance policy, which you'd buy from a private place. The AARP is one place that sells these. There are other folks that sell them. You pay an extra premium for that. That's an additional insurance plan that you have. And that might agree, or you might have a method there to have them cover that 20% co-pay, the additional hospitals. And those plans, there's some different varieties of those that cover more or less generously, and of course, charge a higher or a lower premium for. SALMAN KHAN: I see. That makes complete sense. And do people have to pay any premium to be in Medicare? DR. LAURENCE BAKER: To get A and B both, you have to A, you generally don't. If you pay your Medicare tax for 10 years, you work for 10 years here, and you pay your Medicare tax, you get Part A for free. But Part B charges a premium. So you do have to pay a small monthly-- SALMAN KHAN: And it's what? It's like on the order of like, $100 a month? DR. LAURENCE BAKER: Yeah, generally. It's gone up a little bit over time, or something like that. SALMAN KHAN: So it's not too much. Maybe$100 or a little bit more per month, which is nothing. I mean, if you're 65 or over, and you try to get a private policy, you're probably looking at something like a thousand dollars a month. DR. LAURENCE BAKER: Yes. So I guess two ways of thinking about it. One, this is heavily subsidized. Compared to what a person 75 years old or 80 would pay for an actual private insurance policy would be much, much higher than that. There are people who look at that $100 a month in fixed income and Social Security and things, and you say that's an issue. And so one of the things that's just in the news the last few weeks is the fact that this number, the number went up a bit. Social Security payments went up a little bit. But less, actually, than the Medicare. So people are saying-- SALMAN KHAN: I see. So there's a net decrease. I see. I see. DR. LAURENCE BAKER: --if you're buying both. So there's some people who will still be worried about that, even though it is heavily subsidized, and looks like a great deal. SALMAN KHAN: I see. And then finally, so we have no A, B. And then we have a little bit of D. We know it's all about covering drugs. What does C do? [INAUDIBLE] exist? DR. LAURENCE BAKER: Yes. C exists. C has been around for 20 or 30 years now. C is a separate choice. So when you turn 65, let's say, you can make this choice a little bit differently sometimes too. But when you turn 65, you might decide to take part A and part B, maybe by a traditional supplement. Maybe you add the first Part D drug plan on and make yourself package. Or you can elect Part C, and Part C is an alternative. Under Part C, what happens is the CMS, federal government, has contracted with some HMOs. Kaiser, Blue Cross, other places. Sometimes, they've contracted with a few PPOs. They've tried other kinds of plans too. Mostly it's HMOs. So you have a choice then. You can join Kaiser, and have Kaiser be your Medicare plan. You can join the Blue Cross one. Depending on where you are, you'll have different choices of different private plans that have contracted. And you can do that. The federal government will then pay a bunch of your premium to them. SALMAN KHAN: I see. DR. LAURENCE BAKER: So you'll get covered by the private plan with the federal government covering your costs. They might ask you to pay some additional premium. You're going to have to pay your Part B equivalent already. So you've got to pay that amount. But then, you might have to pay a little bit extra. And then, you'll get access to what services that has. And for some people, that's attractive. Because though the co-payments and deductibles will be lower-- you'll often have more like the$10 copay-- you might have access to some prescription drug coverage automatically built in there. Sometimes those cover some vision and other things, which might be nice. SALMAN KHAN: So this one or plans. I could get this, while it's 80% coverage-- well, I can't make that rule, because hospitals, it could be different. But you get this kind of flexible thing. But you have a pretty significant co-pay, the 20% patient. I could get supplemental insurance for me to cover some of that right over there. Or I can do Part C. Is there a name for Part C? DR. LAURENCE BAKER: Part C these days is called Medicare Advantage. So the plans in Part C are called Medicare Advantage plans. SALMAN KHAN: Which is really just Medicare subsidizing me to get private insurance. DR. LAURENCE BAKER: Right. From some plans that have agreed to work with Medicare patients. SALMAN KHAN: And there's various checking up on them. So they agree to basically cover everything that A and B would cover, so you're not going to get skimped on too much. SALMAN KHAN: And what do most people do? Do they go A and B? The traditional route? Or do they go the C route? DR. LAURENCE BAKER: So historically, the vast majority of people have gone A and B route. It changes. It kind of goes up and down over time, depending on various factors. But over a historical average, it would be something like 80%, 85%, 90% in A, and maybe 20%, 15%, 10% in B and C, depending on which year you're looking at over the last 15 year. SALMAN KHAN: Right. And we could probably do a whole series of videos on it. So when did D come about? It wasn't part of the original-- DR. LAURENCE BAKER: D was not part of the original. D comes about in the early 2000s under George Bush. George W. Bush. SALMAN KHAN: This is what we heard about. I think it was like, 2003. I think I it was-- DR. LAURENCE BAKER: I think that's probably about right. And the issue there was there's a lot of folks who were having trouble getting their outpatient drugs covered. So this is a plan, this is another part of Medicare. You can elect to buy this. You normally would think about this if you got A and B, and you wanted to get outpatient drugs covered, you'd think about a D plan. Part D is run by the federal government, kind of the way that C is. They subsidize, and they pay a premium to the private plans that cover drugs. Or at least they organize the market, and the person then pays also some premium to get-- SALMAN KHAN: I see. So a private-- DR. LAURENCE BAKER: So it's private. SALMAN KHAN: A private entity will handle the actual reimbursements for the drugs, but Medicare will pay them. DR. LAURENCE BAKER: Yeah. So the way to think about it is, just a bunch of private entities with funds flows coming in, partly from the person, partly from Medicare, to buy these. SALMAN KHAN: OK. And who do you pay your money to? You pay it to Medicare? You pay it to these-- DR. LAURENCE BAKER: Uh. SALMAN KHAN: Actually-- it works out the same way one way or the other. This is optional. A retiree or someone over 65 would pay money to essentially get a drug plan subsidized. DR. LAURENCE BAKER: Yes. SALMAN KHAN: OK. Cool. Cool. Well, no, this is super. DR. LAURENCE BAKER: Well, actually, we're trying to think about subsidies. But there's some interesting-- [INTERPOSING VOICES] SALMAN KHAN: Oh, I guess to be able to participate in a drug plan. DR. LAURENCE BAKER: I mean, they are subsidized. And the federal is taking a bunch of money in. But how exactly the economics of these plans works out has been an interesting discussion over the last couple over year. SALMAN KHAN: Right. Right. No, we could make a ton of videos. No, this is super helpful. Because, I think, frankly, I didn't know about the differences right here before about 15 minutes ago. All right. Well, thanks a bunch.


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Cosmological Stability Bound in Massive Gravity and Bigravity

# Cosmological Stability Bound in Massive Gravity and Bigravity

Matteo Fasiello    and Andrew J. Tolley
###### Abstract

We give a simple derivation of a cosmological bound on the graviton mass for spatially flat FRW solutions in massive gravity with an FRW reference metric and for bigravity theories. This bound comes from the requirement that the kinetic term of the helicity zero mode of the graviton is positive definite. The bound is dependent only on the parameters in the massive gravity potential and the Hubble expansion rate for the two metrics.

We derive the decoupling limit of bigravity and FRW massive gravity, and use this to give an independent derivation of the cosmological bound. We recover our previous results that the tension between satisfying the Friedmann equation and the cosmological bound is sufficient to rule out all observationally relevant FRW solutions for massive gravity with an FRW reference metric. In contrast, in bigravity this tension is resolved due to different nature of the Vainshtein mechanism. We find that in bigravity theories there exists an FRW solution with late-time self-acceleration for which the kinetic terms for the helicity-2, helicity-1 and helicity-0 are generically nonzero and positive making this a compelling candidate for a model of cosmic acceleration.

We confirm that the generalized bound is saturated for the candidate partially massless (bi)gravity theories but the existence of helicity-1/helicity-0 interactions implies the absence of the conjectured partially massless symmetry for both massive gravity and bigravity.

Prepared for submission to JCAP

Cosmological Stability Bound in Massive Gravity and Bigravity

• CERCA, Department of Physics, Case Western Reserve University, 10900 Euclid Ave, Cleveland, OH 44106, USA

## 1 Introduction

Attempts at modifying general relativity have a rich and interesting history. In recent years, this effort has been fueled by the observation [1, 2] that the universe is currently accelerating: deviations from GR at large scales, a weaker gravity, or gravitationally induced self-acceleration, might account for such a dynamics. What one hopes to gain in having an accelerated universe through modified gravity is something perhaps less mundane and certainly more appealing than a cosmological-constant-driven acceleration [3, 4]. The cosmological constant problem manifests itself in all theories of modified gravity, in massive gravity (MG) it translates directly into the requirement of a small mass . It has been argued that a graviton mass can be kept small in a technically natural way [5, 6].

The first theory formulation of massive gravity (dRGT) in which the number of degrees of freedom nonlinearly is 5 was given in [7, 8] (see [9, 10, 11, 12, 13, 14, 15, 16, 17] for proofs of the absence of the Boulware-Deser ghostly 6th mode [18]), and this theory was subsequently generalized to arbitrary reference metric [10] and to bigravity [11] where both metrics are dynamical. In the following we shall be concerned with spatially flat FRW solutions in these two theories, i.e. solutions where both the dynamical and the reference metric are homogenous and isotropic.

Finding observationally relevant cosmological solutions in these theories has received considerable attention [19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47]. In massive gravity with a Minkowski reference metric, it was shown that no homogeneous and isotropic spatially flat solutions arise [48]. However, it was argued in [48] that observationally consistent inhomogenous solutions should exist (see also [49, 50, 51, 52, 53, 54, 55, 56] for different approaches). Unfortunately these inhomogenous/anisotropic solutions have been hard to find except in special cases and many of these special cases have been shown to be infinitely strongly coupled. One simple resolution is to look at massive gravity theories with an FRW reference metric [57] or for bigravity theories in which FRW solutions are allowed [21, 24, 30, 36, 47, 24, 25]. It is these classes of solution that we shall be interested in for this work.

We will see that generically the FRW solutions in bigravity and massive gravity with an FRW reference metric (FRW massive gravity) are theoretically well defined in the sense that cosmological perturbations admit nonzero positive kinetic terms for all 5 degrees of freedom. However, for FRW massive gravity, when we go into the observationally relevant Vainshtein region [58, 59, 60] for which the normal Friedmann equation is recovered, the helicity-0 mode becomes a ghost. This result follows because these solutions begin to violate a generalization of a cosmological bound on the mass of graviton first discovered by Higuchi [61] (see also [62, 63, 64])

This generalization of the Higuchi bound to FRW massive gravity was derived in [57] and in this manuscript we will give two independent simpler derivations of this result. The first derivation makes use of the minisuperspace action to look for the instability, the second utilizes the massive (bi)gravity decoupling limit. The result obtained in [57] was that the Higuchi bound in massive gravity for an arbitrary FRW reference metric and arbitrary matter coupled to the dynamical metric is

 ~m2(H)≥2H2 (1.1)

where the dressed mass is given by

 ~m2(H)=m22M2pHHf⎡⎣β1+2β2HHf+β3H2H2f⎤⎦, (1.2)

is the dynamical Hubble parameter, and is that of the reference metric . The ’s are the parameters in the mass potential.

The problem is that it is essentially impossible to satisfy this bound in the Vainshtein region in which the modified gravity contributions to the Friedmann equation are subdominant. This arises because both the Friedmann equation and the bound are polynomials of the same order in . On the one hand, recovery of standard cosmology requires that the mass is small compared to a polynomial in ; on the other hand the bound requires that the mass is large compared to another a polynomial of the same order in .

In this manuscript we shall show that this tension between the Higuchi bound and the Vainshtein mechanism is generically resolved in bigravity theories, even when the Planck mass for the metric is much larger than the usual Planck mass . This result is at first surprising, since the corrections to the action for the helicity-zero mode come suppressed by the ratio . The resolution, is that even these small corrections can actually dominate the dynamics of the helicity-zero mode in the Vainshtein region because we simultaneously have .

Using the two new methods of derivation, we show that in bigravity, by giving full dynamics to what is the fixed metric , the bound is relaxed down to:

 ~m2(H)⎡⎣H2+H2fM2pM2f⎤⎦≥2H4 (1.3)

where the same dressed mass enters. This is equivalent to saying that in bigravity the ‘dressed mass’, i.e. the dynamical mass of the massive graviton, is given by . Just as in the massive gravity case, this bound is valid for arbitrary matter content and for arbitrary FRW (i.e. it is not specific to de Sitter). Special cases of these results have been discussed in FRW perturbation analyses of [25, 34]), however by giving an independent derivation of them, we shall also elucidate the dynamics of the helicity-0 mode in the Vainshtein region.

The above condition allows one to shift the burden of satisfying the bound onto the value of Hubble rate for the metric, . Enriching massive gravity with the dynamics then proves crucial for relaxing the Higuchi bound. As expected, the massive gravity Higuchi bound is recovered in the decoupling limit. This bound may also be expressed in a more symmetric way which makes manifest the symmetry of the bigravity action under and and :

 (1.4)

An intriguing, special, case of the Higuchi bound is what characterizes the so-called partially massless (PM) theory of massive gravity [65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78]. There the Higuchi inequality is actually saturated for all values of regardless of the matter coupling to the metric and the Higuchi-Vainshtein tension is resolved in that massive gravity parameters act in such a way that one of the two scale factors becomes pure gauge.

The special parameter values for the potentially partially massless theory were first identified in massive gravity [74] and they are the same parameters that have also been used in bigravity [77]. They correspond to , , , . For this special case, the bound becomes:

 m2M2p⎛⎝H2M2p+H2fM2f⎞⎠ ≥2H2fH2, (1.5)

but, using the Friedmann equation for the metric, we have in this specific case that

 H2fM2f=12m2M2p⎛⎝H2fH2+M2fM2p⎞⎠. (1.6)

Substituting in we find that the bound becomes an identity regardless of the value of and . This represents a nontrivial check on the consistency of the derived bound.

We shall give two independent derivations of the new bounds. The first is the simplest and utilizes the minisuperspace action alone. We present the reasoning here since it may be useful in similar analyses for more general theories. The idea is that since the bound comes from requiring the absence of ghosts for the helicity-0 mode, and the helicity-0 mode being a scalar already enters into the minisuperspace Lagrangian111Minisuperspace is the truncation of a gravity theory to the case where the metric and all fields are functions of time alone. In other words it the compactification down to dimensions., it should be possible to identify the sign of the kinetic term for the helicity-zero mode by analyzing the minisuperspace action alone. Performing this analysis for massive gravity we recover our result of [57] which was obtained by a much more laborious Hamiltonian analysis. This method easily generalizes to the bigravity case and for generic matter couplings.

The second derivation utilizes the decoupling limit of massive gravity and the decoupling limit of bigravity which we derive here in full, including the vector degrees of freedom following the approach of [79] (see also [80]). It was not guaranteed that this approach would agree with the exact answer since in principle it is possible that the exact bound contains terms which vanish in the decoupling limit. However, this is not the case. Taking the bigravity bound, and scaling , keeping the ratios , , and the parameters fixed, we find that the bound remains the same since it can be expressed entirely in terms of the fixed ratios

 12⎡⎣^β1H2fm2+2^β2HmHfm+^β3H2m2⎤⎦⎛⎝H2m2+M2pM2fH2fm2⎞⎠ ≥2H3fm3H3m3. (1.7)

Since this is the limit used in deriving the massive gravity and bigravity decoupling limit, it follows that the cosmological bound can be entirely determined by the decoupling limit. The decoupling limit also allows us to go beyond the linearized analysis of [25] since it captures the leading nonlinear interactions that will be relevant in cosmological perturbation theory.

Finally, our derivation of the decoupling limit for bigravity leads to a new surprise, a dual formulation of Galileons. In massive gravity, the ‘Galileon’ [81] arises as the helicity-0 mode in the map between the coordinate systems of the metric () and the metric () [7], namely

 Φa(x)=xa+1Λ33ηab∂π(x)∂xb (1.8)

we will prove that this transformation admits an inverse of the form

 xa(Φ)=Φa+1Λ33ηab∂ρ(Φ)∂Φb. (1.9)

is the dual Galileon field which is related to by a nonlocal field redefinition. It is the Galileon field viewed from the perspective of the metric. The decoupling limit of bigravity must be symmetric with respect to interchange of the two metrics, and hence it is symmetric with respect to the interchange of and . We shall discuss this duality in more detail elsewhere [82].

The paper is organized as follows. In Section 2 we introduce the dRGT theory of massive gravity and determine the bound using the minisuperspace approach. In Section 3 we generalize the stability bound to bigravity using the minisuperspace method and find that it is much easier to satisfy than in massive gravity. In Section 4 we derive the decoupling limit of bigravity and by extension that of massive gravity on a generic reference metric. In Section 5 we use the bigravity decoupling limit to give an independent derivation of the cosmological bounds. Finally, we discuss the conjectured partially massless gravity and bigravity in Section 6, establish that they saturate the bound, and discuss their validity as candidate partially massless theories. In the Appendix Appendix: Fluctuations in de Sitter Bigravity we give the details of the bound for fluctuations around de Sitter.

## 2 Deriving the bound in Massive Gravity on FRW

We now give a new derivation of the cosmological bound in massive gravity with a spatially flat FRW reference metric. This confirms the result obtained in [57] and we present it here as a warm up to the analogous bigravity calculation in the next section.

The theory of massive gravity defined on an arbitrary reference metric [10] is just a straightforward generalization of the theory proposed in [8]. The Lagrangian takes the form of Einstein gravity with matter plus a potential that is a scalar function of the two metrics

 L=M2p2√−g(R+2m2U(K))+LM. (2.1)

The most general potential that has no ghosts [8] is built out of characteristic (symmetric) polynomials of the eigenvalues of the tensor

 Kμν(g,f)=δμν−√gμαfαν, (2.2)

so that

 U(K)=U2+α3U3+α4U4, (2.3)

where the are free parameters, and

 U2 = 12!([K]2−[K2]), (2.4) U3 = 13!([K]3−3[K][K2]+2[K3]), (2.5) U4 = 14!([K]4−6[K2][K]2+8[K3][K]+3[K2]2−6[K4]), (2.6)

where represents the trace of a tensor with respect to the metric . The are the symmetric polynomials which are generally defined in dimensions by the determinant relation

 Det[1+λK]=D∑n=0λnUn(K). (2.7)

so that and . The Lagrangian may also be written in the form

 L=12√−g[M2pR−m24∑n=0βnUn(X)]+LM, (2.8)

where . The relationship between the coefficients and the is given in Eq.(2.1). The mass term is invariant under the simultaneous interchange and .

As we shall see in detail, the generalization of the Higuchi bound can already be seen at the level of the minisuperspace action, in particular in the representation of massive gravity which includes the Stückelberg fields. The Stückelberg fields for diffeomorphisms are introduced by replacing the reference metric by its representation in an arbitrary coordinate system

 ds2f=fAB(ΦC)∂μΦA∂νΦBdxμdxν. (2.9)

The are the Stückelberg fields (or Goldstone modes for the broken diffemorphisms) and essentially encode the additional degrees of freedom that a massive graviton has over a massless one. These additional degrees of freedom are, in the high energy limit, decomposable into 2 helicity-1 modes and 1 helicity-0 mode. Explicitly, writing , then has the interpretation of the helicity-1 mode in the high energy limit, and is the additional helicity-0 mode. Since the bound comes from identifying the sign of the kinetic term of the helicity-0 mode, it is essentially enough to keep track of the term (we may for instance choose a gauge for which to aid in this identification).

However as is well known, in massive gravity, either part or all of the kinetic term for the helicity-0 mode actually comes from a mixing of and the metric . In perturbations, this shows up in the fact that a scalar part of couples to and hence . In the minisuperspace limit, the only scalar part of the metric is the scale factor itself 222Although the lapse is also a scalar, it is a Lagrange multiplier for a constraint and drops out of the action if the constraint is solved.. Thus we can identify the sign of the kinetic term for perturbations around a background that identifies kinetic term for , the mixing between and , and any independent kinetic term for the helicity-0 which enters as . This method is significantly simpler than our previous complete calculation [57] and leads quite quickly to the same result.

### 2.1 Minisuperspace Derivation

Thus, we begin with the action for minisuperspace for the dRGT model on a given FRW background written in an arbitrary gauge by means of the introduction of a Stueckelberg field for the broken time diffeomorphisms. The reference metric is given by

 ds2f=−˙ϕ2dt2+b(ϕ)2d→x2, (2.10)

where is the Stückelberg field which will keep track of the helicity-0 mode kinetic term . In this section, for the sake of simplicity, we will limit ourselves to the case of a de Sitter reference metric. The more general case is dealt with in the bigravity setup of Section 3. The dynamical metric is similarly expressed as

 ds2g=−N(t)2dt2+a(t)2d→x2. (2.11)

The dRGT square root combination takes now the form

 √g−1f=⎛⎜⎝˙ϕN0j0ib(ϕ)aδij⎞⎟⎠, (2.12)

and we have chosen the square root with the positive sign (see [83, 84] for a discussion on this). It follows that the minisuperspace action for spatially flat cosmological solutions takes the form

 S=V3∫dtNa3[−3M2p(˙a2N2a2)−˙ϕN3∑n=0An(b(ϕ)a)n−3∑n=0Bn(b(ϕ)a)n−ρ(a)], (2.13)

where is the volume factor and is the matter density.

The coefficients and are subject to the relation and in terms of the and , these coefficients are given by

 B0=m2M2p(−6−4α3−α4)=12m2β0 B1=3m2M2p(3+3α3+α4)=32m2β1 B2=3m2M2p(−1−2α3−α4)=32m2β2 B3=m2M2p(α4+α3)=12m2β3, (2.14)
 Bn=3m2βn(3−n)!n!;An=3m2βn+1(3−n)!n!. (2.15)

In massive gravity, we can set since this term is a total derivative. In bigravity this term enters as the cosmological constant sourcing the metric and thus should in general be maintained.

The Friedmann equation, which can be obtained by varying the action with respect to , is:

 H2=13M2P(ρ+ρm.g.), (2.16)

where the extra ‘dark energy’ contribution from the massive gravity action is

 ρm.g.=3∑n=0Bn(b(ϕ)a)n. (2.17)

Varying the action with respect to imposes, as a consequence of the special relation between the coefficients and , the constraint

 (2∑n=0^βn+1(2−n)!n!(ba)n+1)(Hb−Hfa)=0, (2.18)

where

 Hf=˙b˙ϕb=b,ϕb. (2.19)

At first sight there appear to be two branches of solutions to this equation. However, if we choose the term in the first parenthesis to vanish, the kinetic term for the helicity-1 mode vanishes and all such solutions are infinitely strongly coupled. Thus the only acceptable solution is

 ba=HHf. (2.20)

From now on, we will work in the normal branch for which this relation is true (branch 2 in the language of [25]). This allows us to write the contribution to the Friedmann equation in the form

 ρm.g.=3∑n=0Bn(HHf)n=3∑n=03m2βn(3−n)!n!(HHf)n. (2.21)

To determine the kinetic term for the helicity zero mode we will utilize some of the properties of the minisuperspace action. Starting from the action

 S=V3∫dtNa3[−3M2p(˙a2N2a2)−˙ϕN3∑n=0An(b(ϕ)a)n−3∑n=0Bn(b(ϕ)a)n−ρ(a)], (2.22)

we now perturb it to second order in . As appears only algebraically in the action, it can be integrated out. After these steps the action takes the form:

 S(2) = V3∫dt[δϕ2(−b2ϕ23∑n=0Bnn(n−1)bn−2an−3+˙abϕ23∑n=0An(3−n)nbn−1an−2 +bϕ12M2pa˙a2(3∑n=0nBnbn−1an−3)2−Hfbϕ23∑n=0Bnnbn−1an−3⎞⎠+δϕδ˙a2(..)+δa2(..)],

where . We see that, with the sign of the kinetic term of the helicity zero mode in mind, we need only worry about the coefficient (it is that is carrying the helicity zero mode, ) in the above : integrating out removes any -proportional term in the action for fluctuations so that the last two terms in (2.1) need not be specified, i.e. we do not need to diagonalize.

One can now make use of the following set of relations: the background equations of motion imply and we have , so that

 ∑nnBn(ba)n=ba∑nAn(3−n)(ba)n. (2.24)

Using these in Eq.(2.1) leads to a further simplified action, whose helicity-0 kinetic term reads

 S(2)∣∣kinetic=V3∫dtδϕ234M2pa3H2b2ϕb2 [~m2(~m2−2H2)], (2.25)

and we have defined the dressed mass parameter as

 ~m2(H) = 13M2pba∑nAn(3−n)(ba)n (2.26) = m22M2pHHf⎡⎣β1+2β2HHf+β3H2H2f⎤⎦.

Thus, the Higuchi bound, which amounts to the statement that the kinetic term for the helicity zero mode is positive, reads

 ~m2(H)(~m2(H)−2H2)≥0. (2.27)

Upon neglecting the branch of solutions , which inevitably gives a ghost in the vector sector, the bound translates into the more familiar statement:

 ~m2(H)≥2H2, (2.28)

where is the dressed mass, a generalization of the bare . This is the generalization of the Higuchi bound first derived in [61]. Once more, let us stress that this result follows regardless of the matter content and is independent of . This result is consistent with that derived in [57] by an entirely different analysis. We now have a more clear explanation of how it arises and, as we shall see, a quicker route to its extensions and applications.

### 2.2 Higuchi versus Vainshtein tension

We take now a moment to briefly reproduce the reasoning that lead in [57] to the conclusion that FRW on FRW solutions in massive gravity are ruled out. The crucial point is that the bound in Eq. (2.28) must be complemented with a condition on derived from consistency of the expansion history with observations. Departures from the General Relativity (GR) expansion history are negligible at large redshifts and hence large ; thus the dependent modifications to the normal Friedmann equation must be small, at least for most of the history of the universe. In massive gravity, continuity with GR is achieved through the Vainshtein mechanism and it is for this reason that we refer to these two opposing requirements as the Higuchi-Vainshtein tension.

In order for the dependent contribution in the Friedmann equation (2.16) to be subleading (Vainshtein regime) we require:

 m22M2p⎡⎣3β1HHf+3β2H2H2f+β3H3H3f⎤⎦≪3H2. (2.29)

This is immediately at odds with the stability (Higuchi) condition. We can see why by combining the two inequalities into the single statement

 ⎡⎣32β1HHf+3β2H2H2f+32β3H3H3f⎤⎦≫⎡⎣3β1HHf+3β2H2H2f+β3H3H3f⎤⎦. (2.30)

Eq. (2.30) is essentially impossible to satisfy. If the term dominates the inequality is violated, if the dominates it is saturated and if dominates it is only just satisfied since . However, this is not enough, there needs to be a large hierarchy between the two sides otherwise there will be modifications to the Friedmann equation [57]. Considering that the combined inequality must hold over different cosmological epochs, one comes to the realization that there is no room in the parameter space of FRW massive gravity for it to simultaneously satisfy the requirements of stability and consistency with observations.

We stress that this does not rule out massive gravity as a theory of current cosmological expansion. There are many paths one can follow in the search for solutions to massive gravity that, if not exactly FRW, at least resemble FRW in appropriate regions, see e.g. [48]. We choose here to instead demand exactly FRW solutions and move on to give full dynamics to the reference metric , thus entering into the realm of bigravity theories.

## 3 Generalizing the Bound to Bigravity

The action for bigravity models which are free from the Boulware-Deser ghost is a simple extension of that for massive gravity [11]

 S=∫d4x12[M2p√−gR[g]+M2f√−fR[f]−m24∑n=0βnUn(X)]+LM. (3.1)

It is straightforward to generalize the previous argument to the case of bigravity. We denote the now dynamical second metric as

 ds2f=−~N2dt2+b2d→x2. (3.2)

The minisuperspace action is now

 S = V3∫dta3[−3M2p(˙a2Na2)−~N3∑n=0An(ba)n−N3∑n=0Bn(ba)n−Nρ(a) (3.3) + V3∫dtb3−~N~ρ(b)−3M2f(˙b2~Nb2)]

The Friedmann equation for the scale factor , which we assume to correspond to the metric with which our matter is coupled, is given by

 H2=13M2p(ρ(a)+ρbigravity), (3.4)

where

 ρbigravity=3∑n=0Bn(ba)n. (3.5)

In the following we will assume that the only matter sourcing the second metric is a cosmological constant and, as such, can be absorbed into a definition of . In this case the Friedmann equation for the second metric is simply

 H2f=13M2f[3∑n=0An(ba)(n−3)] (3.6)

For convenience let us denote

 A(χ)=3∑n=0An(ba)n=3∑n=0Anenχ,B(χ)=3∑n=0Bn(ba)n=3∑n=0Bnenχ, (3.7)

where we have utilized the following change of variables . Let us also employ a simple yet generic matter Lagrangian coupled only with the metric:

 LM=−12gμν∂μϕ∂νϕ−V(ϕ), (3.8)

so that we add to the minisuperspace action the term

 SM=V3∫dtNa3(12˙ϕ2−V(ϕ)). (3.9)

We shall see that all details of the precise nature of the matter Lagrangian will drop out of the final answer. We now pass to the canonical phase space formulation by defining the canonically conjugate momenta

 pa=−6M2p˙aNa,pϕ=a3˙ϕN,pb=−6M2f˙b~Nb. (3.10)

As a consequence, the canonical action can be written as

 S=V3∫dt[pa˙a+pb˙b+pϕ˙ϕ−N(−12p2a6M2pa+a3B+p2ϕ2a3+a3V)−~N(−12p2b6M2fb+a3A)]. (3.11)

Varying with respect to and imposes the analogues of the Hamiltonian constraints which can be solved to remove (we will choose the negative solution corresponding to an expanding geometry)

 pa=−√12a2Mp√B+p2ϕ2a6+V,pb=−√12Mf√bAa3, (3.12)

and, using , we obtain

 S=V3∫dt[pϕ˙ϕ−√12a2Mp˙a√B+p2ϕ2a6+V−√12a2(˙a+˙χa)Mf√e3χA], (3.13)

or, after solving for and integrating out,

 S=V3∫dt[−√12a2Mp˙a√B+V ⎷1−a2˙ϕ26M2p−√12a2Mf√e3χA(˙a+˙χa)]. (3.14)

As it stands, this action is time reparametrization invariant, and we may utilize this fact to choose the gauge , so that

 S=V3∫dt[−√12t2Mp√B+V√1−t2˙ϕ26M2p−√12t2(1+˙χt)Mf√e3χA]. (3.15)

We may now perturb this action, and , to second order so as to obtain the quadratic action for perturbations around a given background. We also note that the contribution from fluctuations of the potential will at most contribute a mixing term ; it does not therefore add up to the kinetic term whose sign we are after and can thus be ignored in what follows333Note that the variable here is the scalar in the matter Lagrangian, not to be confused with the Stueckelberg field in the massive gravity analysis of the previous section.

Up to an unimportant positive overall factor, the quadratic fluctuation action reads:

 S(2)∝V3∫dt~m2(H204M2p~m2H5M2f−12H+~m24H3(1−g2))δχ2+g~m2H(1−g2)δχδg+H1−g2δg2

where we have defined the dimensionless variable out of as . We have also employed Eq. (2.16) and (2.19) to simplify the coefficients: this provides a more compact expression and is of course allowed as coefficients are purely background quantities.

Since carries the information on , the helicity zero mode, , we proceed to diagonalize the above expression so as to read off the coefficient. If we posit

 δg=δ~g−g~m22H2δχ, (3.17)

in the new variables the mixing term vanishes and, requiring the coefficient be positive, amounts to:

 ~m2⎛⎝H2+M2pH2fM2f⎞⎠−2H4≥0, (3.18)

that is, the generalized Higuchi bound.

We stress again that simple algebraic manipulations coupled with the properties of the (or, equivalently, the ) coefficients allow Eq. (3.18) to be expressed in its most symmetric and pleasing form, Eq.(1.4), which we reproduce here:

 m22[β1H2f+2β2HHf+β3H2]⎛⎝H2M2p+H2fM2f⎞⎠ ≥2H3fH3. (3.19)

Let us reiterate that this bound was derived for fairly generic matter content, but matter itself enters into the final expression only indirectly through . Thus the bound is entirely one on the Hubble rate, regardless of the matter content of the universe, and may thus be universally applied at all cosmological epochs.

### 3.1 Higuchi versus Vainshtein Resolution

As stressed in [57], one should not simply declare cosmological solutions viable by merely looking for regions of the parameter space where Eq. (3.19) is satisfied. Rather, we ought to perform a joint analysis of the stability condition and the requirement stemming from observations that, for most of the history of the universe, the Friedmann equation is well approximated by its GR prediction. In essence this requires us to be in the Vainshtein regime for which the cosmological effects of the helicity-0 mode are negligible and are suppressed by powers of . Unfortunately these two requirements oppose each other, thereby creating the Higuchi-vs-Vainshtein tension that effectively rules out the viability of spatially flat FRW solutions in FRW massive gravity.

Let us see how bigravity resolves the Higuchi-Vainshtein tension; the ingredients are the generalized bound and the Friedmann equations, which we report below:

 H2=13M2p[ρ(a)+3∑n=03m2βn(3−n)!n!(HHf)n];H2f=13M2f[3∑n=03βn+1(3−n)!n!(HHf)(n−3)].

Now, the inequality in Eq. (3.18) is certainly satisfied, without requiring much on the part of , if we posit . In particular if we assume that is nonzero, as is generically the case, then the second relation in (3.1) and the definition (see 2.26) are, at leading order in , respectively:

 3M2fH2f∼m2β12H3fH3;~m2∼m2β12M2pHHf. (3.21)

From here, we take the ratio of the two expressions, solve for and plug it back into Eq. (3.18), to obtain:

 Hf∼√3MfH2Mp~m;Stabilitybound∣∣Hf≫H:3H4≳2H4, (3.22)

which is, evidently, consistent with the initial assumption.

As for the first Friedmann equation in (3.1), we see that, by itself, it does not put too much strain on . The fact is that the dressed mass is actually allowed to be small because precisely the lower bound on that in the massive gravity case comes from the stability condition, is now taken care of by our being in the region.

We have thus shown that there exists a large region in the parameter space of the bigravity theory, , which does relax the stability bound in a way which is consistent with observations. The stability-vs-observation conflict is therefore resolved. This resolution requires that in the cosmological bound, the term dominates over . It is clear that in the FRW massive gravity limit, this could never be satisfied since that limit requires us to take for finite and finite and so inevitably the former is subdominant to the latter. Thus it is precisely the naively suppressed bigravity interactions, which are crucial in resolving the conflict.

### 3.2 Ghost-free Self-Accelerating Cosmology

Having established the regime we should be in to satisfy the bound, we can infer the effective Friedmann equation by solving

 3M2fH2f∼m2β12H3fH3 (3.23)

to give

 Hf∼6M2fH3m2β1 (3.24)

and then substituting back into the Friedmann equation to give

 H2∼13M2p[ρ(a)+3m2β12(HHf)]=13M2p[ρ(a)+m4β214M2f(1H2)]. (3.25)

This gives a self-accelerating cosmology in which asymptotically in the future the Hubble parameter tends to a constant de Sitter value of magnitude

 H∞=1121/4mβ1/21√MpMf. (3.26)

Thus we see that quite in general the condition , which is implicitly assumed in having dominate, is essentially satisfied even in the asymptotics 444Note that in the asymptotics it is acceptable to have , this is because, even if it implies via (3.18) that , the latter is not an issue since at late times we desire order unity departures from GR to account for cosmic acceleration.:

 HfH∼6M2fH2m2β1∼6Mf√12Mp (3.27)

Also note that, since scales as , it becomes easier and easier to satisfy the condition as we go back in time.

To make this discussion more precise, we can specify to the model with , and (, ) . In this case the Friedmann equation is exactly

 H2=13M2p[ρ(a)+m4M4pM2f(1H2)], (3.28)

and the stability constraint is exactly

 m2β12(1M2p+36M4fM2fm4β21H4)≥12M2fH4m2β1, (3.29)

which is easily seen to be always satisfied since it simplifies to the trivially satisfied relation

 (1M2p+12M2fm4β21H4)≥0. (3.30)

Thus we automatically generate a ghost-free late-time self-accelerating cosmology whose Friedmann equation may be put in the form

 H2=16M2p⎛⎜⎝ρ(a)+ ⎷ρ(a)2+12m4M6pM2f⎞⎟⎠, (3.31)

and asymptotes to de Sitter with

 H∞=131/4m√MpMf. (3.32)

This represents one subset of the parameter space for which the bound is satisfied at all times. This specific solution, as well as similar ones with , have been studied in [85, 86]. There the authors employ statistical methods to asses the viability of bigravity solutions as compared to that of the standard cosmology. Quite interestingly, the solution discussed, which we have shown to satisfy the stability bound, is on par with in their analysis.

## 4 Λ3 Decoupling Limit of Bigravity

We shall now derive the complete decoupling limit for bigravity including vector degrees of freedom. This decoupling limit will allow us to give an independent derivation of the generalized Higuchi bound in bigravity, and also allow us to determine the stability of the vectors. Just as in the massive gravity decoupling limit, the bigravity decoupling limit focuses on the interactions at the lowest energy scale, those of the helicity-zero mode.

### 4.1 Vierbein formulation

Our derivation will utilize the vierbein formalism [87, 88] (see also [89, 90]) and will follow closely the notation of [79]. The starting point is to introduce Stückelberg fields for the broken Lorentz invariance and diffeomorphism (diff) invariance. Just as adding a mass term to gravity breaks diffeomorphism invariance, adding a mass term to bigravity breaks two independent copies of the diffeomorphism group down to the single diagonal subgroup. In the Einstein-Cartan (vierbein) formalism, it similarly breaks two copies of local Lorentz transformations (LLTs) down to a single copy. It is helpful to reintroduce these symmetries, and this can be easily achieved using the Stückelberg trick.

Beginning in unitary gauge, the action for bigravity

 Sbigravity=∫d4x12[M2p√−gR[g]+M2f√−fR[f]−m24∑n=0βnUn(X)]+LM, (4.1)

can be expressed in vierbein form as

 Sbigravity = M2p2εabcd∫12Ea∧Eb∧Rcd[E]+


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# Geometry For Prime Addicts

Texed notes of my reading course last quarter advised by Yifeng Liu, with fellow participants Grisha Kondyrev and Jora Belousov. Our goal this reading course was understand some of Scholze’s recent work with perfectoid space techniques, in particular the proof of the monodromy weight conjecture.

Attached are only the notes I kept of my lectures, sometimes just outlines of what to talk about. I have had a few requests for these notes, and they are incomplete but I think they might still be helpful for others starting out in p-adic geometry.

During the reading course, I got the increasing feeling that we were just studying $$G_Q$$ as fast as our little legs could take us — our little legs being our knowledge of varieties (over various non-archimedian fields like $$Q_p$$ and $$F_p((t))$$. So, I finished with a talk on the Grothendieck-Teichmüller group — another approach to $$G_Q$$.

If the pdf viewer fails, here’s the link to the paper