text:
# Isotopes of rutherfordium

Rutherfordium (Rf) is an artificial element, and thus a standard atomic mass cannot be given. Like all artificial elements, it has no stable isotopes. The first isotope to be synthesized was either 259Rf in 1966 or 257Rf in 1969. There are 15 known radioisotopes from 253Rf to 268Rf (2 of which, 266Rf and 268Rf, are unconfirmed) and 4 isomers. The longest-lived isotope is 267Rf with an estimated half-life of 5 hours. The longest directly measured half-life is 263Rf at 11 minutes, and the longest-lived isomer is 261mRf with a half-life of 81 seconds.

## Table

nuclide
symbol
Z(p) N(n)
isotopic mass (u)

half-life decay
mode(s)[1][n 1]
daughter
isotope(s)
nuclear
spin
excitation energy
253Rf 104 149 253.10044(44)# 13(5) ms SF (51%) (various) (7/2)(+#)
α (49%) 249No
253mRf 200(150)# keV 52(14) µs
[48(+17-10) µs]
(1/2)(-#)
254Rf 104 150 254.10005(30)# 23(3) µs SF (99.7%) (various) 0+
α (.3%) 250No
255Rf 104 151 255.10127(12)# 1.64(11) s SF (52%) (various) (9/2-)#
α (48%) 251No
256Rf 104 152 256.101152(19) 6.45(14) ms SF (96%) (various) 0+
α (6%) 252No
257Rf 104 153 257.102918(12)# 4.7(3) s α (79%) 253No (1/2+)
β+ (18%) 257Lr
SF (2.4%) (various)
257mRf 114(17) keV 3.9(4) s (11/2-)
258Rf 104 154 258.10343(3) 12(2) ms SF (87%) (various) 0+
α (13%) 254No
259Rf 104 155 259.10560(8)# 2.8(4) s α (93%) 255No 7/2+#
SF (7%) (various)
β+ (.3%) 259Lr
260Rf 104 156 260.10644(22)# 21(1) ms SF (98%) (various) 0+
α (2%) 256No
261Rf 104 157 261.10877(5) 5.5(25) s α (76%) 257No 3/2+#
β+ (14%) 261Lr
SF (10%) (various)
261mRf 70(100)# keV 81(9) s β+ 261Lr 9/2+#
α (rare) 257No
262Rf 104 158 262.10993(24)# 2.3(4) s SF (99.2%) (various) 0+
α (.8%) 258No
262mRf 600(400)# keV 47(5) ms SF (various) high
263Rf 104 159 263.1125(2)# 11(3) min SF (70%) (various) 3/2+#
α (30%) 259No
265Rf[n 2] 104 161 265.11668(39)# 2.5 min SF (various)
266Rf[n 3][n 4] 104 162 266.11817(50)# 10# h 0+
267Rf[n 5] 104 163 267.12179(62)# 5# h
[2.3(+980-17) h]
SF (various)
268Rf[n 3][n 6] 104 164 268.12397(77)# 1# h 0+
1. ^ Abbreviations:
EC: Electron capture
SF: Spontaneous fission
2. ^ Not directly synthesized, occurs in decay chain of 285Fl
3. ^ a b Discovery of this isotope is unconfirmed
4. ^ Not directly synthesized, occurs in decay chain of 282Uut
5. ^ Not directly synthesized, occurs in decay chain of 287Fl
6. ^ Not directly synthesized, occurs in decay chain of 288Uup

### Notes

• Values marked # are not purely derived from experimental data, but at least partly from systematic trends. Spins with weak assignment arguments are enclosed in parentheses.
• Uncertainties are given in concise form in parentheses after the corresponding last digits. Uncertainty values denote one standard deviation, except isotopic composition and standard atomic mass from IUPAC which use expanded uncertainties.

## History of synthesis of isotopes by cold fusion

This section deals with the synthesis of nuclei of rutherfordium by so-called "cold" fusion reactions. These are processes which create compound nuclei at low excitation energy (~10-20 MeV, hence "cold"), leading to a higher probability of survival from fission. The excited nucleus then decays to the ground state via the emission of one or two neutrons only.

208Pb(50Ti,xn)258−xRf (x=1,2,3)

This reaction was first studied in 1974 by the team at Dubna. They measured a spontaneous fission activity assigned to 256Rf. [2] The reaction was further studied in 1985 by the GSI team who measured the decay properties of the isotopes 257Rf and 256Rf. The team were able to determine some initial spectroscopic properties of 257Rf and found that the alpha decay pattern was very complicated.[3]

After an upgrade of their facilities, they repeated the reaction in 1994 with much higher sensitivity and detected some 1100 atoms of 257Rf and 1900 atoms of 256Rf along with 255Rf in the measurement of the 1n,2n and 3n excitation functions. The large amount of decay data for 257Rf allowed the detection of an isomeric level and the construction of a partial decay level structure which confirmed the very complicated alpha decay pattern. They also found evidence for an isomeric level in 255Rf. [4] The GSI team continued in 2001 with the measurement of the 3n excitation function. In 2002, scientists at the Argonne University in Illinois began their first studies of translawrencium elements with the synthesis and alpha-gamma spectroscopy of 257Rf.[5] In 2004, the GSI began their spectroscopic studies of the 257Rf isotope. In 2007, the Lawrence Berkeley National Laboratory (LBNL) studied the 2n product, 256Rf, in a search for K-isomers and discovered three such isomers.[6]

Currently suggested decay level scheme for 255Rf from the study reported in 2007 by Hessberger et al. at GSI[7]
207Pb(50Ti,xn)257−xRf (x=2)

This reaction was first studied in 1974 by the team at Dubna. They measured a spontaneous fission activity assigned to 255Rf. The reaction was further studied in 1985 by the GSI team who measured the decay properties of the isotope 255Rf. A further spectroscopic study was reported in 2000 which led to a first decay level scheme for the isotope.[8] The isomeric level proposed in 1994 was not found. In 2006, the spectroscopy was continued and the decay scheme was confirmed and improved. [9]

206Pb(50Ti,xn)256−xRf (x=1,2)

The team at GSI first studied this reaction in 1994 in an effort to study neutron deficient isotopes of rutherfordium. They were able to detect 255Rf and 144 atoms of the new isotope 254Rf, which decayed by spontaneous fission.[4]

204Pb(50Ti,xn)254−xRf (x=1)

The team at GSI first studied this reaction in 1994 in an effort to study neutron deficient isotopes of rutherfordium. They were able to detect 14 atoms of the new isotope 253Rf, which decayed by spontaneous fission.[4]

208Pb(48Ti,xn)256−xRf (x=1)

In 2006, as part of a program looking at the effect of isospin on the mechanism of cold fusion, the team at LBNL studied this reaction. They measured the 1n excitation function and determined that the change of a Ti-50 projectile to a Ti-48 one significantly reduced the yield, in agreement with predictions.[10]

124Sn(136Xe,xn)260-xRf

In an important study, in May 2004, the team at GSI attempted the symmetric synthesis of rutherfordium by attempting to fuse two fission fragments. Theory suggests that there may be an enhancement of the yield. No product atoms were detected and an upper limit of 1000 pb was estimated for the yield of this reaction.[11]

## History of synthesis of isotopes by hot fusion

This section deals with the synthesis of nuclei of rutherfordium by so-called "hot" fusion reactions. These are processes which create compound nuclei at high excitation energy (~40-50 MeV, hence "hot"), leading to a reduced probability of survival from fission. The excited nucleus then decays to the ground state via the emission of 3-5 neutrons.

238U(26Mg,xn)264−xRf (x=3,4,5,6)

The hot fusion reaction using a uranium target was first reported in 2000 by Yuri Lazarev and the team at the Flerov Laboratory of Nuclear Reactions (FLNR). They were able to observe decays from 260Rf and 259Rf in the 4n and 5n channels.[12] They measured yields of 240 pb in the 4n channel and 1.1 nb in the 5n channel. In 2006, as part of their program on the study of uranium targets in hot fusion reactions, the team at LBNL measured the 4n,5n and 6n excitation functions for this reaction and observed 261Rf in the 3n exit channel.[13][14]

244Pu(22Ne,xn)266−xRf (x=4,5)

This reaction was reported in 1996 at LBNL in an attempt to study the fission characteristics of 262Rf. The team were able to detect the spontaneous fission (SF) of 262Rf and determine its half-life as 2.1 s, in contrast to earlier reports of a 47 ms activity. It was suggested that the two half-lives might be related to different isomeric states.[15] The reaction was further studied in 2000 by Yuri Lazarev and the team at Dubna. They were able to observe 69 alpha decays from 261Rf and spontaneous fission of 262Rf.[16] Later work on hassium has allowed a reassignment of the 5n product to 261mRf.

242Pu(22Ne,xn)264−xRf (x=3,4?,5?)

The synthesis of element 104 was first attempted in 1964 by the team at Dubna using this reaction. The first study produced evidence for a 0.3 s SF activity tentatively assigned to 260104 or 259104 and an unidentified 8 s SF activity. The former activity was later retracted and the latter activity associated with the now-known 259104.[17] In 1966, in their discovery experiment, the team repeated the reaction using a chemical study of volatile chloride products. The group was able to identify a volatile chloride decaying by short spontaneous fission with eka-hafnium properties. This gave strong evidence for the formation of [104]Cl4 and the team suggested the name kurchatovium. Although a half-life was not accurately measured, later evidence suggested that the product was most likely 259104.[17] In 1968, the team searched for alpha decay from 260104 but were unable to detect such activity. In 1970, the team repeated the reaction once again and confirmed the ~0.2 s SF activity. They also repeated the chemistry experiment and obtained identical results to their 1966 experiment and calculated a likely half-life of ~0.5 seconds for the SF activity. In 1971, the reaction was repeated again and 0.1 s and 4.5 s SF activities were found. The 4.5 s activity was correctly assigned to 259104.[17] A chemistry experiment in the same year reaffirmed the formation of a 0.3 SF activity for an eka-hafnium product.[17] Later, the 0.1−0.3 s SF activity was retracted as belonging to a kurchatovium isotope but the observation of eka-hafnium reactivity remained and was the basis of their successful claim to discovery.[17] The reaction was further studied in 2000 by Yuri Lazarev at Dubna. They were able to observe 261Rf in the 3n channel, later reassigned to 261mRf.

242Pu(20Ne,xn)262−xRf

This reaction was first studied in 1964 to assist in the assignments using the analogous reaction with a Ne-22 beam. The Dubna team were unable to detect any 0.3 s spontaneous fission activities.[17] The reaction was later studied in 2003 at the Paul Scherrer Institute (PSI) in Bern, Switzerland. They detected some spontaneous fission activities but were unable to confirm the formation of 259Rf.[18]

248Cm(22Ne,αxn)266−xRf (x=3?)

This reaction was studied in 1999 at the University of Bern, Switzerland in order to search for the new isotope 263Rf. A rutherfordium fraction was separated and several SF events with long lifetimes and alpha decays with energy 7.8 MeV and 7.9 MeV were observed. A second experiment using a study of the fluoride of rutherfordium products also produced 7.9 MeV alpha decays.[19]

248Cm(18O,xn)266−xRf (x=3?,5)

This reaction was first studied in 1970 by Albert Ghiorso at the LBNL. The team identified 261Rf in the 5n channel using the method of correlation of genetic parent-daughter decays. A half-life of 65 s was determined. [20] A repeat later that year using cation exchange chromatography indicated that the product did not form a +2 or +3 cation and behaved as eka-hafnium. A study of the properties of rutherfordium isotopes was performed in 1981 at the LBNL. In a series of reactions, a 1.5 s SF activity was identified and assigned to a fermium descendant although later evidence indicates a possible assignment to 262Rf. In contrast, in a subsequent review of isotope properties by Somerville et al. at LBNL in 1985, a 47 ms SF activity was assigned to 262Rf. This assignment has not been verified. [21] The reaction was further studied in 1991 by Czerwinski et al. at the LBNL. In this experiment, spontaneous fission activities with long lifetimes were observed in rutherfordium fractions and tentatively assigned to 263Rf. In 1996, chemical studies on the chloride of rutherfordium were published by the LBNL. In this experiment, the half-life was improved to 78 s. A repeat of the experiment in 2000 assessing the volatility of the bromide further refined the half-life to 75 s.

248Cm(16O,xn)264−xRf (x=4)

This reaction was studied in 1969 by Albert Ghiorso at the University of California. The aim was to detect the 0.1–0.3 s SF activity reported at Dubna, assigned to 260104. They were unable to do so, only observing a 10–30 ms SF activity, correctly assigned to 260104. The failure to observe the 0.3 s SF activity identified by Dubna gave the Americans the incentive to name this element rutherfordium.[17]

246Cm(18O,xn)264−xRf

In an attempt to unravel the properties of spontaneous fission activities in the formation of rutherfordium isotopes, this reaction was performed in 1976 by the FLNR. They observed an 80 ms SF activity. Subsequent work led to the complete retraction of the 0.3s - 0.1s - 80 ms SF activities observed by the Dubna team and associated with background signals.[17]

249Bk(15N,xn)264−xRf (x=4)

This reaction was studied in 1977 by the team in Dubna. They were able to confirm the detection of a 76 ms SF activity. The assignment to rutherfordium isotopes was later retracted. The LBNL re-studied the reaction in 1980 and in 1981 they reported that they were unable to confirm the ~80 ms SF activity. The Dubna team were able to measure a 28 ms SF activity in 1985 and assigned the isotope correctly to 260104.[17]

249Cf(13C,xn)262−xRf (x=4)

This use of californium-249 as a target was first studied by Albert Ghiorso and the team at the University of California in 1969. They were able to observe an 11 ms SF activity which they correctly assigned to 258104.[22]

Currently suggested decay level scheme for 257Rfg,m from the study performed in 2004 by Hessberger et al. at GSI[citation needed]
249Cf(12C,xn)261−xRf (x=3,4)

In their 1969 discovery experiments, the team at University of California also used carbon-12 beam to irradiate a californium-249 target. They were able to confirm the 11 ms SF activity found with a carbon-13 beam and again correctly assigned to 258104. The actual discovery experiment was the observation of alpha decays genetically linked to 253102 and therefore positively identified as 257104.[22] In 1973, Bemis and his team at Oak Ridge confirmed the discovery by measuring coincident X-rays from the daughter 253102.[23]

## Synthesis of isotopes as decay products

Isotopes of rutherfordium have also been identified in the decay of heavier elements. Observations to date are summarized in the table below. EC refers to electron capture.

Evaporation residue Observed Rf isotope
288Uup 268Rf (possible EC of 268Db)
291Lv, 287Fl, 283Cn 267Rf
282Uut 266Rf (EC of 266Db)
269Sg 265Rf
271Hs 263gRf
263Db 263mRf (EC of 263Db)
266Sg (possibly 266mSg) 262Rf (possibly 262mRf)
277Cn, 273Ds, 269Hs, 265Sg 261mRf, 261Rf
271Ds, 267Hs, 263Sg 259Rf
269Ds, 265Hs, 261Sg 257Rf
264Hs, 260Sg 256Rf
259Sg 255Rf

## Chronology of isotope discovery

Isotope Discovered Reaction
253Rf 1994 204Pb(50Ti,n) [4]
254Rf 1994 206Pb(50Ti,2n) [4]
255Rf 1974? 1985 207Pb(50Ti,2n)
256Rfg 1974? 1985 208Pb(50Ti,2n)
256Rfm1 2007 208Pb(50Ti,2n)
256Rfm2 2007 208Pb(50Ti,2n)
256Rfm3 2007 208Pb(50Ti,2n)
257Rfg,m 1969 249Cf(12C,4n) [22]
258Rf 1969 249Cf(13C,4n) [22]
259Rf 1969 249Cf(13C,3n) [22]
260Rf 1969 248Cm(16O,4n)
261Rfa 1970 248Cm(18O,5n) [20]
261Rfb 1996 208Pb(70Zn,n) [24]
262Rf 1996 244Pu(22Ne,4n) [15]
263Rfa 1990? 248Cm(18O,3n)
263Rfb 2004 248Cm(26Mg,3n) [25]
264Rf unknown
265Rf 2010 242Pu(48Ca,5n)
266Rf? 2006 237Np(48Ca,3n) [26]
267Rf 2003/2004 238U(48Ca,3n) [24]
268Rf? 2003 243Am(48Ca,3n) [27]

## Isomerism in rutherfordium nuclides

### 263Rf

Initial work on the synthesis of rutherfordium isotopes by hot fusion pathways focused on the synthesis of 263Rf. Several studies have indicated that this nuclide decays primarily by spontaneous fission with a long half-life of 10–20 minutes. Alpha particles with energy 7.8-7.9 MeV have also been associated with this nucleus. More recently, a study of hassium isotopes allowed the synthesis of an atom of 263Rf decaying by spontaneous fission with a short half-life of 8 seconds. These two different decay modes must be associated with two isomeric states. Specific assignments are difficult due to the low number of observed events. It is reasonable to tentatively assign the long life to a meta-stable state, namely 263mRf, and the shorter life to the ground state, namely 263gRf. Further studies are required to allow a definite assignment.

### 261Rf

Early research on the synthesis of rutherfordium isotopes utilised the 244Pu(22Ne,5n)261Rf reaction. The product was found to undergo exclusive 8.28 MeV alpha decay with a half-life of 78 seconds. Later studies by the GSI team on the synthesis of copernicium and hassium isotopes produced conflicting data. In this case, 261Rf was found to undergo 8.52 MeV alpha decay with a short half-life of 4 seconds. Later results indicated a predominant fission branch. These contradictions led to some doubt on the discovery of copernicium. However, it is now understood that the first nucleus belongs to an isomeric meta-stable state, namely 261mRf and the latter to the ground state isomer, namely 261gRf.[28] The discovery and confirmation of 261gRf provided proof for the discovery of copernicium in 1996.

### 257Rf

A detailed spectroscopic study of the production of 257Rf nuclei using the reaction 208Pb(50Ti,n)257Rf allowed the identification of an isomeric level in 257Rf. The work confirmed that 257gRf has a very complicated spectrum with as many as 15 alpha lines. A level structure diagram was calculated for both isomers.

## Spectroscopic decay level schemes for rutherfordium isotopes

### 257Rf

This is the currently suggested decay level scheme for 257Rfg,m from the study performed in 2004 by Hessberger et al. at GSI

[4]

### 255Rf

This is the currently suggested decay level scheme for 255Rf from the study reported in 2007 by Hessberger et al. at GSI

## Unconfirmed isotopes

### 268Rf

In the synthesis of ununpentium, the isotope 288115 has been observed to decay to 268Db which undergoes spontaneous fission with a half-life of 29 hours. Given that the electron capture of 268Db cannot be detected, these SF events may in fact be due to the SF of 268Rf, in which case the half-life of this isotope cannot be extracted. [27]

### 266Rf

In the synthesis of ununtrium, the isotope 282113 has been observed to decay to 266Db which undergoes spontaneous fission with a half-life of 22 minutes. Given that the electron capture of 266Db cannot be detected, these SF events may in fact be due to the SF of 266Rf, in which case the half-life of this isotope cannot be extracted.[26]

## Retracted isotopes

### 265Rf

In 1999, American scientists at the University of California, Berkeley, announced that they has succeeded in synthesizing three atoms of 293118. These parent nuclei successively emitted seven alpha particles to form 265Rf nuclei. Their claim was retracted in 2001. This isotope was finally created in 2010.

### 255mRf

A detailed spectroscopic study of the production of 255Rf nuclei using the reaction 206Pb(50Ti,n)255Rf allowed the tentative identification of an isomeric level in 255Rf. A more detailed study later confirmed that this was not the case.

## Chemical yields of isotopes

### Cold fusion

The table below provides cross-sections and excitation energies for cold fusion reactions producing rutherfordium isotopes directly. Data in bold represents maxima derived from excitation function measurements. + represents an observed exit channel.

Projectile Target CN 1n 2n 3n
50Ti 208Pb 258Rf 38.0 nb, 17.0 MeV 12.3 nb, 21.5 MeV 660 pb, 29.0 MeV
50Ti 207Pb 257Rf 4.8 nb
50Ti 206Pb 256Rf 800 pb, 21.5 MeV 2.4 nb, 21.5 MeV
50Ti 204Pb 254Rf 190 pb, 15.6 MeV
48Ti 208Pb 256Rf 380 pb, 17.0 MeV

### Hot fusion

The table below provides cross-sections and excitation energies for hot fusion reactions producing rutherfordium isotopes directly. Data in bold represents maxima derived from excitation function measurements. + represents an observed exit channel.

Projectile Target CN 3n 4n 5n
26Mg 238U 264Rf 240 pb 1.1 nb
22Ne 244Pu 266Rf + 4.0 nb
18O 248Cm 266Rf + 13.0 nb

## Future experiments

The team at GSI are planning to perform first detailed spectroscopic studies on the isotope 259Rf. It will be produced in the new reaction:

$\,^{238}_{92}\mathrm{U}\ + \,^{24}_{12}\mathrm{Mg}\to \,^{259}_{104}\mathrm{Rf}\ + 3 \,^{1}_{0}\mathrm{n}$

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20. ^ a b Ghiorso, A.; Nurmia, M.; Eskola, K.; Eskola P. (1970). "261Rf; new isotope of element 104". Physics Letters B 32 (2): 95–98. Bibcode:1970PhLB...32...95G. doi:10.1016/0370-2693(70)90595-2.
21. ^ Somerville, L. P.; M. J. Nurmia, J. M. Nitschke, and A. Ghiorso E. K. Hulet and R. W. Lougheed (1985). "Spontaneous fission of rutherfordium isotopes". Physical Review C 31 (5): 1801–1815. Bibcode:1985PhRvC..31.1801S. doi:10.1103/PhysRevC.31.1801.
22. "Positive Identification of Two Alpha-Particle-Emitting Isotopes of Element 104", Ghiorso et al.., Phys. Rev. Lett. 22, 1317-1320 (1969). Retrieved on 2008-03-04
23. ^ "X-Ray Identification of Element 104", Bemis et al., Phys. Rev. Lett. 31, 647-650 (1973). Retrieved on 2008-03-04
24. ^ a b see copernicium
25. ^ see hassium
26. ^ a b see ununtrium
27. ^ a b see ununpentium
28. ^ "EVIDENCE FOR ISOMERIC STATES IN 261Rf", Dressler et al., PSI Annual Report 2001. Retrieved on 2008-01-29


text:
# Challenge

HP is doing research in mathematics . After doing lots of research, she struck in a problem . She found four numbers n,a, b and c .Now, She wants to know how many number exists which are less than or equal to n and are divisible by a ,b or c .

# Input :

4 integers denoting n,a,b and c separated by space

# Output :

intger which denotes the count of the numbers which are divisible a,b,c

## Scoring:

This is , shortest code in bytes wins.

## closed as unclear what you're asking by DennisJul 24 '17 at 4:55

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Can you add test cases? Does the count include 0 and negative numbers? – fireflame241 Jul 24 '17 at 4:25
• What are the ranges of the input parameters? – Peter Taylor Jul 24 '17 at 8:11

# Python 2, 72 65 bytes

Note: This solution works for positive integers:

lambda a,b,c,n:len([x for x in range(1,n+1)if x%a*(x%b)*(x%c)<1])

Try it online!

# Python 2, 52 bytes

lambda n,a,b,c:n/a+n/b+n/c-n/a/b-n/a/c-n/b/c+n/a/b/c

Try it online!

format: plain


text:
# Math Help - Function within a function problem

1. ## Function within a function problem

Hello again! I am again lost on another problem. I wasn't sure what to call this, so hopefully it's still somewhat understandable. Thank you very much in advance!

The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?

(A) g(8) = 1/3 and g'(8) = -1/9
(B) g(8) = 1/3 and g'(8) = 1/9
(C) g(8) = 3 and g'(8) = -9
(D) g(8) = 3 and g'(8) = -1/9
(E) g(8) = 3 and g'(8) = 1/9

2. f(g(x)) = x and g(f(x)) = x

This means that these functions are inverses.

Can you take it from there?

3. Originally Posted by mamori
Hello again! I am again lost on another problem. I wasn't sure what to call this, so hopefully it's still somewhat understandable. Thank you very much in advance!

The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?

(A) g(8) = 1/3 and g'(8) = -1/9
(B) g(8) = 1/3 and g'(8) = 1/9
(C) g(8) = 3 and g'(8) = -9
(D) g(8) = 3 and g'(8) = -1/9
(E) g(8) = 3 and g'(8) = 1/9
First from $g(f(x)) = x$ sub. x=3. Next differentiate $g(f(x)) = x$ wrt x so $g'(f(x))\cdot f'(x) = 1$, then subs x= 3 again.

4. You're thinking way too much. =p

5. 1.

since g(f(x)) = x

g(f(3)) = 3

But f(3) = 8

so g(8) = 3

2. f(g(x)) = x

f and g are inverses so you have a theorem which tells you

g ' (8) = 1/ f'(3)

In general g ' (c) = 1/ f ' (a) where f(a) = c

Or From first principles:

[f(g(x)] ' = 1 for all x

But [f(g(x)] ' also equals f ' (g(x)) g ' (x)

at x = 3 f ' (g(8)) g ' (8) = 1

f ' (3) *g ' (8) =1

9* g ' (8) =1

g' (8) =1/9

f and g are inverses so you have a theorem which tells you

g ' (8) = 1/ f'(3)

6. Originally Posted by derfleurer
You're thinking way too much. =p
Originally Posted by Calculus26
1.

since g(f(x)) = x

g(f(3)) = 3

But f(3) = 8

so g(8) = 3

2. f(g(x)) = x

f and g are inverses so you have a theorem which tells you

g ' (8) = 1/ f'(3)

In general g ' (c) = 1/ f ' (a) where f(a) = c

Or From first principles:

[f(g(x)] ' = 1 for all x

But [f(g(x)] ' also equals f ' (g(x)) g ' (x)

at x = 3 f ' (g(8)) g ' (8) = 1

f ' (3) *g ' (8) =1

9* g ' (8) =1

g' (8) =1/9

f and g are inverses so you have a theorem which tells you

g ' (8) = 1/ f'(3)
Apparently me and Calc26

7. Hi everyone! Thank you so much for all your help! I did understand that the functions are inverses, but I forgot the theorem for the derivatives. So thanks for your explanation, Calculus26! I do have one question. Where does

But [f(g(x)] ' also equals f ' (g(x)) g ' (x)

come from? Another theorem? I never learned this theorem/rule, so it's a bit tough on understanding this.

8. That's just chain rule.

sin(x^2)' = cos(x^2)(2x)

And if you understood that they were inverses, you didn't really need to remember any calculus rules. Just know that the slope of the inverse is 1/m and that your domain/range swap. So f(3) = 8 yields f^-1(8) = 3.

9. Danny-- I thought we were paid by the number of words

10. Originally Posted by derfleurer
That's just chain rule.

sin(x^2)' = cos(x^2)(2x)

And if you understood that they were inverses, you didn't really need to remember any calculus rules. Just know that the slope of the inverse is 1/m and that your domain/range swap. So f(3) = 8 yields f^-1(8) = 3.
Thanks. =] I just understood everything on this page. Somehow that inverse slope rule slipped my mind. Thanks a million everyone!

11. Originally Posted by Calculus26
Danny-- I thought we were paid by the number of words
Don't we wish


text:
# Possible analytic images of the unit disc?

From Picard's theorem, the image of $\mathbb{C}$ under an analytic function has to the whole plane or $\mathbb{C}$ minus a single point. What about other open sets? This may be too broad, so how about the unit disc?

Let $B$ be the open unit disc. If $f:B\to \mathbb{C}$ is analytic, what can we say about $f(B)$?

We know that $f(B)$ has to be open. Can something further be said?

For instance, $f(B)$ need not be simply connected (consider $exp(\alpha z)$ for a large enough $\alpha$).

In fact $f(B)$ can be any open subset of $\Bbb C$. (Edit: Or rather any connected open set; thanks to Robert Israel for noticing that this is what I meant...)

Bonus: If $V$ is any open connected subset of the plane there exists $f$ with non-vanishing derivative such that $f(B)=V$, and in fact if $V\ne\Bbb C$ we can take $f$ to be a covering map.

Say $V\subset\Bbb C$ is open and connected. Assume first that $\Bbb C\setminus V$ contains more than one point. Now the Uniformiization Theorem shows that there is a simply connected Riemann surface $M$ and a holomorphic covering map from $M$ onto $V$. And $M$ is either the Riemann sphere, the disk or the plane. Except $M$ cannot be the Riemann sphere since $V$ is not compact, and the little Picard theorem shows that $V$ cannot be $\Bbb C$, so $M$ is the disk.

That does it except for $V=\Bbb C\setminus \{0\}$ and $V=\Bbb C$. There is a strip $S$ so that $\exp(S)=\Bbb C\setminus\{0\}$ and a map from the disk onto $S$. And there is a map from the disk onto the region $\Re z>-1$; the square of this map maps $B$ onto $\Bbb C$.

We're done except for the bonus part, and we're almost done with that. If $V\ne\Bbb C$ we've given a holomorphic covering map from $B$ onto $V$. If $V=\Bbb C$ there is no such covering map. But there does exist a holomorphic surjection with non-vanishing derivative:

Say $f$ is the entire function with $f(0)=0$ and $f'(z)=e^{\cos(z)}$. Then $f'(z+2\pi)-f'(z)=0$, so there exists $c$ with $f(z+2\pi)=f(z)+c$. Note that $c=f(0)>0$; in particular $c\ne0$.

Since $f(z+2\pi)=f(z)+c$ with $c\ne0$ the Big Picard theorem shows that $f$ takes every complex value infinitely many times. Hence $f(\Bbb C\setminus\{0\})=\Bbb C$. As above there is a covering map $g$ from $B$ onto $\Bbb C\setminus\{0\}$; now $f\circ g$ is a map from $B$ onto $\Bbb C$ with non-vanishing derivative.

• I think you mean any connected open set. – Robert Israel Dec 17 '15 at 19:01
• @RobertIsrael Eewps. Yeah, I'm pretty sure that's what I meant, thanks – David C. Ullrich Dec 17 '15 at 19:09

Bloch-Landau theorem states:
There exists an absolute constant $\ell$ such that, for any holomorphic function $f$ defined in the unit disc, with the property $|f'(0)| =1$, Then:
The image of the unit disc contains a ball with radius greater or equal to $\ell$.

Notice that the condition on the derivative is "just" a normalization constraint.

• An since $f'(0)$ can be smaller than $1$, what actually follows from this is that if $f$ is analytic in $B$ then $f(B)$ contains a disk... – David C. Ullrich Dec 17 '15 at 19:12
• @DavidC.Ullrich In the simple case where $f'(0)\ne 0$ then we can define $g(z)=f(z)/f'(0)$, so $g(z)$ containts balls with radius greater than landau-constant, and then , since $f(z)=g(z) \cdot f'(0)$ - $f$ must contains balls with radius factored by $f'(0)$. The other case is not so trivial to me , and i need to invest some more thinking into it. – Ranc Dec 17 '15 at 19:36
• Right. And a ball factored by $f'(0)$ can be any ball at all, so all this shows is that the image contains a ball. Which is not surprising; have you noticed the other answer showing that $f(B)$ can be any connected open set? – David C. Ullrich Dec 17 '15 at 19:38
• @DavidC.Ullrich Yes, but, I was merely suggesting a different approach to this question. The Bloch-Landau THM gives some information regarding any $f$ holomorphic. If you feel there is some duality ,or my answer is not so informative - I won't mind deleting it. – Ranc Dec 17 '15 at 19:55


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# Non singular black hole solutions

#### evanallmighty

Has anyone read this paper? If so, what is the interior solution it describes?

A non singular solution for spherical configuration with infinite central density
Fuloria, Pratibha; Durgapal, M. C.
Astrophysics and Space Science, Volume 314, Issue 4, pp.249-250
Abstract: A non-singular exact solution with an infinite central density is obtained for the interior of spherically symmetric and static structures. Both the energy density and the pressure are infinite at the center but we have e λ(0)=1 and e ν(0)≠0. The solution admits the possibility of receiving signals from the region of infinite pressure.
Keywords: General relativity, Exact solution, Astrophysics
http://www.springerlink.com/content/up024115727162w5/

Related Special and General Relativity News on Phys.org

#### bcrowell

Staff Emeritus
Science Advisor
Gold Member
Too bad that only the first page is viewable. Have you checked to see if it's on arxiv? It's hard to tell much from the first page, because they don't seem to be saying much about the type of matter involved. Presumably it's matter that violates an energy condition, because otherwise there'd have to be a singularity because of the Penrose singularity theorem...?

#### George Jones

Staff Emeritus
Science Advisor
Gold Member
This isn't a black hole solution, and the authors don't claim that it is. This paper matches a non-vacuum interior to a vacuum exterior such that both pressure and density become unbounded as r approaches zero, and that there is no event horizon.

#### evanallmighty

Yeah, I assumed that much. No, sadly its not on ArXiv... I was just posting to see if anyone has read it and knows the solution. It really stinks too that it only shows one page because there is only two.

Thanks

#### evanallmighty

This isn't a black hole solution, and the authors don't claim that it is. This paper matches a non-vacuum interior to a vacuum exterior such that both pressure and density become unbounded as r approaches zero, and that there is no event horizon.
Ah! Thank you. I suspected this, but since I thought little of it and probably went to fast as to make myself think it was a black hole I missed it. Do you know of any non singular black hole interior solutions?

EDIT---
How is it not a black hole solution, if the are getting rid of the singularity, how could it not be? Only black hole solutions have the gravitational singularity. It says that it admits the possibility if receiving signals from the region of infinite pressure, which if only a problem in black hole solutions, right? I think I am missing part of what your saying.

#### George Jones

Staff Emeritus
Science Advisor
Gold Member
Ah! Thank you. I suspected this, but since I thought little of it and probably went to fast as to make myself think it was a black hole I missed it. Do you know of any non singular black hole interior solutions?
A non-singular black hole solution would have to violate one of the assumptions of the Penrose-Hawking singularity theorems.
How is it not a black hole solution, if the are getting rid of the singularity, how could it not be? Only black hole solutions have the gravitational singularity. It says that it admits the possibility if receiving signals from the region of infinite pressure, which if only a problem in black hole solutions, right?
I am not sure what you're trying to say.

Inside the event horizon, a black hole spacetime is non-stationary (and thus non-static), but the first paragraph of the Introduction indicates that the paper treats static interior solutions. Their metric (after restoring the accidentally(?) omitted term) is the standard
$$ds^2 = e^{\nu\left(r\right)} dt^2 - e^{\lambda\left(r\right)} dr^2 - r^2 d\theta^2 - r^2 \sin^2 \theta d \phi^2.$$

They match vacuum exterior ton non-vacuum interior at $r = a$, and in equation (13) write
$$e^{\nu\left(a\right)} = e^{-\lambda\left(a\right)} = 1 -2u.$$
In the last paragraph before the discussion, they write "The solution thus remains valid for $u < 0.4$"
Clearly, nothing strange happens to the metric coefficients at the match value $r=a$, i.e., $t$ is timelike there and $r$ is spacelike.

PS Did you try clicking on the "Download PDF" above "Abstract" at the link that you gave?

#### evanallmighty

.I am not sure what you're trying to say.

Inside the event horizon, a black hole spacetime is non-stationary (and thus non-static), but the first paragraph of the Introduction indicates that the paper treats static interior solutions. Their metric (after restoring the accidentally(?) omitted term) is the standard
$$ds^2 = e^{\nu\left(r\right)} dt^2 - e^{\lambda\left(r\right)} dr^2 - r^2 d\theta^2 - r^2 \sin^2 \theta d \phi^2.$$

They match vacuum exterior ton non-vacuum interior at $r = a$, and in equation (13) write
$$e^{\nu\left(a\right)} = e^{-\lambda\left(a\right)} = 1 -2u.$$
In the last paragraph before the discussion, they write "The solution thus remains valid for $u < 0.4$"
Clearly, nothing strange happens to the metric coefficients at the match value $r=a$, i.e., $t$ is timelike there and $r$ is spacelike.

PS Did you try clicking on the "Download PDF" above "Abstract" at the link that you gave?
The Schwarzschild metric is a spherical static structure, but it describes a black hole. It is not possible in reality, but it is the most well known and simplest (black hole) metrics out there. If the paper is talking about resolving the singularity and the information paradox at r=0, than it must be talking about a black hole, because no star I know of has those problems.

I did click on the download PDF button but it wants 34 dollars for one page of a PDF.

#### George Jones

Staff Emeritus
Science Advisor
Gold Member
The Schwarzschild metric is a spherical static structure, but it describes a black hole.
No, the spacetime for a Schwarzschild black hole is static only outside the event horizon. Inside the event horizon, the spacetime for a Schwarzschild black hole is not static (it is not even stationary)
If the paper is talking about resolving the singularity and the information paradox at r=0, than it must be talking about a black hole, because no star I know of has those problems.
Again,
This isn't a black hole solution, and the authors don't claim that it is. This paper matches a non-vacuum interior to a vacuum exterior such that both pressure and density become unbounded as r approaches zero, and that there is no event horizon.

#### evanallmighty

No, the spacetime for a Schwarzschild black hole is static only outside the event horizon. Inside the event horizon, the spacetime for a Schwarzschild black hole is not static (it is not even stationary)
I've never heard that before. Papers I have read before on Schwarzschild interior/exterior solutions clearly say that the interior Schwarzschild solution represents a static, spherically symmetric mass of imcompressible fluid at rest:

http://www.google.com/url?sa=t&source=web&cd=3&ved=0CC4QxQEwAg&url=http://docs.google.com/viewer?a=v&q=cache:lB1mjOR9z-oJ:members.iinet.net.au/~housewrk/Papers/On%2520a%2520New%2520Interior%2520Schwarzschild%2520Solution.pdf+on+a+new+interior+schwarzschild+solution&hl=en&gl=us&pid=bl&srcid=ADGEEShRrYfZHNrWBpSXiqJEcc_Mx6WVmWcx60g8djEgB4B6bnWneFtiY8aSums2xi8PWgrqBofNm_VzF-B90LavFG2udjQmrxWgptZxWQGSjsKJR60iD5HsMzVnKupODEZqRTurEyV8&sig=AHIEtbRrEWpxBoF-Fa-tdaPDqQFt6raPOA&rct=j&q=on a new interior schwarzschild solution&ei=h9YITuiWIcrg0QGM58XHCw&usg=AFQjCNH-38bGeMJOUr0cWmJy8HdefqBFZQ&sig2=0byzsQTHI0wrPHKVkw8ILQ

and:

http://www.jstor.org/pss/78530
(I forgot where you can read the whole PDF)

Yes, I know these sources are only describing interior solutions where a>9/8 2m, which means there is no grav. collapse (meaning that it COULD be different from a black hole because it is describing a star...)BUT:

EVERY single source describes that SCINCE the Schwarzschild black hole HAS NO MOMENTUM/IS STATIC, that it's space time singularity is a point, not A RING as in other black hole solutions that have charge/spin, etc. (such as the Kerr or Kerr Newman metric) So How could the exterior and the interior even ever be matched if they are completely different?

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#### atyy

Science Advisor
I've never heard that before. Papers I have read before on Schwarzschild interior/exterior solutions clearly say that the interior Schwarzschild solution represents a static, spherically symmetric mass of imcompressible fluid at rest:

http://www.google.com/url?sa=t&source=web&cd=3&ved=0CC4QxQEwAg&url=http://docs.google.com/viewer?a=v&q=cache:lB1mjOR9z-oJ:members.iinet.net.au/~housewrk/Papers/On%2520a%2520New%2520Interior%2520Schwarzschild%2520Solution.pdf+on+a+new+interior+schwarzschild+solution&hl=en&gl=us&pid=bl&srcid=ADGEEShRrYfZHNrWBpSXiqJEcc_Mx6WVmWcx60g8djEgB4B6bnWneFtiY8aSums2xi8PWgrqBofNm_VzF-B90LavFG2udjQmrxWgptZxWQGSjsKJR60iD5HsMzVnKupODEZqRTurEyV8&sig=AHIEtbRrEWpxBoF-Fa-tdaPDqQFt6raPOA&rct=j&q=on a new interior schwarzschild solution&ei=h9YITuiWIcrg0QGM58XHCw&usg=AFQjCNH-38bGeMJOUr0cWmJy8HdefqBFZQ&sig2=0byzsQTHI0wrPHKVkw8ILQ

and:

http://www.jstor.org/pss/78530
(I forgot where you can read the whole PDF)

Yes, I know these sources are only describing interior solutions where a>9/8 2m, which means there is no grav. collapse (meaning that it COULD be different from a black hole because it is describing a star...)BUT:

EVERY single source describes that SCINCE the Schwarzschild black hole HAS NO MOMENTUM/IS STATIC, that it's space time singularity is a point, not A RING as in other black hole solutions that have charge/spin, etc. (such as the Kerr or Kerr Newman metric) So How could the exterior and the interior even ever be matched if they are completely different?
George Jones is correct and so are your sources. The terms they use, though similar, refer to different things. Your sources are not referring the part of a Schwarzschild black hole within its event horizon.

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#### bcrowell

Staff Emeritus
Science Advisor
Gold Member
Evanallmighty, please take a deep breath. The ALL CAPS come across as shouting. George Jones has a great deal of expertise in GR and has been very generous with his time in trying to help you. The angry, hostile tone is uncalled for.

#### evanallmighty

George Jones is correct and so are your sources. The terms they use, though similar, refer to different things. Your sources are not referring the part of a Schwarzschild black hole within its event horizon.
So they are describing the exterior solution? Kind of like when the schwarzschild exterior solution describes the singularity at r=0 even though it is the EXTERIOR and not the INTERIOR?

Is this the interior/internal and exterior/external argument?

#### evanallmighty

Evanallmighty, please take a deep breath. The ALL CAPS come across as shouting. George Jones has a great deal of expertise in GR and has been very generous with his time in trying to help you. The angry, hostile tone is uncalled for.
Oh, I'm sorry all, I didn't know it was coming off like that. Thanks for letting me know! (it was just supposed to kinda point out certain things that I don't get. I wasn't questioning his authority, just wondering for myself, because I am ignorant in this subject. By the way, if that ever happens again or if any questions/comments seem elementary/rude, it's because I am only 13.-Although that is not an exception)
Thanks everyone.

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#### evanallmighty

George Jones is correct and so are your sources. The terms they use, though similar, refer to different things. Your sources are not referring the part of a Schwarzschild black hole within its event horizon.
Ok, then... How? I really don't get that. In the paper: A New Interior Schwarzschild Solution by P.S Florides, it states: "The most celebrated exact solution of Einstein's vacuum field equations is the Schwarzschild (exterior) solution...It represents the grav. field in the exterior of a spherically symmetric distribution of matter of grav. mass m. The most familiar of Einstein's equations-(1.2) which is static, spherically symmetric, and forms an extension of the solution (Schwarzschild exterior) into the interior of the mass distribution is the Schwarzschild interior solution."
Isn't is saying that the interior is spherically symmetric/static? Or am I completely missing something big?

#### atyy

Science Advisor
Ok, then... How? I really don't get that. In the paper: A New Interior Schwarzschild Solution by P.S Florides, it states: "The most celebrated exact solution of Einstein's vacuum field equations is the Schwarzschild (exterior) solution...It represents the grav. field in the exterior of a spherically symmetric distribution of matter of grav. mass m. The most familiar of Einstein's equations-(1.2) which is static, spherically symmetric, and forms an extension of the solution (Schwarzschild exterior) into the interior of the mass distribution is the Schwarzschild interior solution."
Isn't is saying that the interior is spherically symmetric/static? Or am I completely missing something big?
Parts of the exterior can be joined to two sorts of solutions.

One sort is a matter solution, including the "interior Schwarzschild solution". There is no event horizon in this sort of solution. The "interior Schwarzschild solution" is static.

Another sort is a vacuum solution, such as the Kruskal-Szekeres extension of the exterior Schwarzschild. This is a solution with no matter, and it has an event horizon. The "interior Schwarzschild solution" is not the "interior Schwazrschild black hole solution". The latter is the same as the interior of the Kruskal-Szekeres black hole, also known as the maximal extension of the vacuum Schwarzschild solution. The interior of the Schwarschild black hole is not static.

The jargon is confusing, just watch for it.

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#### evanallmighty

Parts of the exterior can be joined to two sorts of solutions.

One sort is a matter solution, including the "interior Schwarzschild solution". There is no event horizon in this sort of solution. The "interior Schwarzschild solution" is static.

Another sort is a vacuum solution, such as the Kruskal-Szekeres extension of the exterior Schwarzschild. This is a solution with no matter, and it has an event horizon. The "interior Schwarzschild solution" is not the "interior Schwazrschild black hole solution". The latter is the same as the interior of the Kruskal-Szekeres black hole, also known as the maximal extension of the vacuum Schwarzschild solution. The interior of the Schwarschild black hole is not static.

The jargon is confusing, just watch for it.
Great! Thank you, and thank you all too! I understand it now. This is what I assumed but I was just clarifying/unsure. Now, not to ask more questions or take up anybody's time, but along with these two solutions the Schwarzschild exterior can always be matched with the spherically symmetric perfect fluid solutions too, right?

EDIT--
Oops! Never mind! Answered my own question. The perfect fluid solutions are part of the matter solutions.

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#### pervect

Staff Emeritus
Science Advisor
Question: if you have infinite pressure at the center, and it's not a black hole solution, can it satisfy the hydrostatic equilibrium equations? (Oppenheier-Volkoff eq's). Do the authors claim it satisfies the equations?

#### George Jones

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Gold Member
I am only 13
Sorry, I didn't realize this. Since you were asking questions about a technical paper. I thought that you wanted advanced, mathematical answers. The exterior/interior vacuum/non-vacuuum stuff is very confusing. I was more than twice your age before I learned about these things. Keep asking questions.

#### George Jones

Staff Emeritus
Science Advisor
Gold Member
Question: if you have infinite pressure at the center, and it's not a black hole solution, can it satisfy the hydrostatic equilibrium equations? (Oppenheier-Volkoff eq's).
I think that it is possible. From post #6, the solution is consistent with Buchdahl's theorem, $M/a < 4/9$.
Do the authors claim it satisfies the equations?
They don't seem to claim this explicitly.

#### evanallmighty

Sorry, I didn't realize this. Since you were asking questions about a technical paper. I thought that you wanted advanced, mathematical answers. The exterior/interior vacuum/non-vacuuum stuff is very confusing. I was more than twice your age before I learned about these things. Keep asking questions.
Well, not trying to sound smarter than I am or anything, but honestly, I have understood the answers given to me thus far, and the papers I have posted, etc. (With some hard thinking) It might take me awhile but I do feel confident after I get it/have been convinced.

I understand what you and Atyy are saying now, well, I figured that all along, but didn't realize it was the answer to my question. It is confusing, but I feel I have grasped the general concept. Correct me if that is impossible or wrong...

I'll keep taking the answers at the level they are now, haha!

Don't have any more questions for now... gimme a while

Thank you and I hope I am not taking up your time!

#### atyy

Science Advisor
I understand what you and Atyy are saying now, well, I figured that all along, but didn't realize it was the answer to my question. It is confusing, but I feel I have grasped the general concept.
So are there non-singular black hole solutions?

#### pervect

Staff Emeritus
Science Advisor
I think that it is possible. From post #6, the solution is consistent with Buchdahl's theorem, $M/a < 4/9$.

They don't seem to claim this explicitly.
My intuition is telling me - if you infinite pressure, in the absence of a coordinate singularity, you must have expansion. But I don't have any proof that this is correct, just a strong hunch at this point.

However, I wouldn't be terribly surprised if the authors have found a "new" solution by not bothering to include the hydrostatic equilibrium equations in their solution process.

#### TrickyDicky

No, the spacetime for a Schwarzschild black hole is static only outside the event horizon. Inside the event horizon, the spacetime for a Schwarzschild black hole is not static (it is not even stationary)
The interior of the Schwarschild black hole is not static.
This is the point where I always get stuck too, I just don't understand how a spacetime can be static only in part, I would have thought that the static property, being an invariant property based on the fact that the Lie derivative of the metric with respect to time vanishes, holds for any coordinate system so it shouldn't be abolished by a coordinate transformation.
So how exactly is the staticity of the Schwarzschild spacetime not a global property of the whole Schwarzschild spacetime manifold ?
Does anyone understand this and can explain it?

#### George Jones

Staff Emeritus
Science Advisor
Gold Member
A vector field doesn't have to have the same causal character everywhere, i.e., it can be timelike in one part of spacetime, lightlike in another part, and spacelike in another. For example consider Minkowski spacetime and the vector field given $x \mathbf{e}_x + t \mathbf{e}_t$.

Schwarzschild spacetime has a Killing vector field that is timelike outside the event horizon, lightlike on the event horizon, and spacelike inside the event horizon.

#### TrickyDicky

A vector field doesn't have to have the same causal character everywhere
Ok, but a static manifold has a unique timelike KV field, do you mean then that a manifold can have regions that fulfill this requirement and other regions that don't? What is that manifold called then, static and not static at the same time?


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# 4.2: Cooling and Mixing

## Newton’s Law of Cooling

Newton’s law of cooling states that if an object with temperature $$T(t)$$ at time $$t$$ is in a medium with temperature $$T_m(t)$$, the rate of change of $$T$$ at time $$t$$ is proportional to $$T(t)-T_m(t)$$; thus, $$T$$ satisfies a differential equation of the form

$\label{eq:4.2.1} T'=-k(T-T_m).$

Here $$k > 0$$, since the temperature of the object must decrease if $$T > T_m$$, or increase if $$T < T_m$$. We’ll call $$k$$ the temperature decay constant of the medium.

For simplicity, in this section we’ll assume that the medium is maintained at a constant temperature $$T_m$$. This is another example of building a simple mathematical model for a physical phenomenon. Like most mathematical models it has its limitations. For example, it is reasonable to assume that the temperature of a room remains approximately constant if the cooling object is a cup of coffee, but perhaps not if it is a huge cauldron of molten metal. (For more on this see Exercise 4.2.17.)

To solve Equation \ref{eq:4.2.1}, we rewrite it as

$T'+kT=kT_m. \nonumber$

Since $$e^{-kt}$$ is a solution of the complementary equation, the solutions of this equation are of the form $$T=ue^{-kt}$$, where $$u'e^{-kt}=kT_m$$, so $$u'=kT_me^{kt}$$. Hence,

$u=T_me^{kt}+c, \nonumber$

so

$T=ue^{-kt}=T_m+ce^{-kt}. \nonumber$

If $$T(0)=T_0$$, setting $$t=0$$ here yields $$c=T_0-T_m$$, so

$\label{eq:4.2.2} T=T_m+(T_0-T_m)e^{-kt}.$

Note that $$T-T_m$$ decays exponentially, with decay constant $$k$$.

##### Example 4.2.1

A ceramic insulator is baked at $$400^\circ$$C and cooled in a room in which the temperature is $$25^\circ$$C. After 4 minutes the temperature of the insulator is $$200^\circ$$C. What is its temperature after 8 minutes?

Solution

Here $$T_0=400$$ and $$T_m=25$$, so Equation \ref{eq:4.2.2} becomes

$\label{eq:4.2.3} T=25+375e^{-kt}.$

We determine $$k$$ from the stated condition that $$T(4)=200$$; that is,

$200=25+375e^{-4k}; \nonumber$

hence,

$e^{-4k} = {175\over 375} = {7\over 15}. \nonumber$

Taking logarithms and solving for $$k$$ yields

$k=-{1\over 4} \ln {7\over 15}={1\over 4}\ln {15\over 7}. \nonumber$

Substituting this into Equation \ref{eq:4.2.3} yields

$T=25+375 e^{-{t\over 4} \ln {15\over 7}} \nonumber$

(Figure 4.2.1 ). Therefore the temperature of the insulator after 8 minutes is

$\begin{array}{rl} T(8) & = 25+375 e^{-2 \ln {15\over 7}} \\[9pt] & = 25+375 \left({7\over 15}\right)^2 \approx 107^\circ \mbox{C}. \end{array}\nonumber$

##### Example 4.2.2

An object with temperature $$72^\circ$$F is placed outside, where the temperature is $$-20^\circ$$F. At 11:05 the temperature of the object is $$60^\circ$$F and at 11:07 its temperature is $$50^\circ$$F. At what time was the object placed outside?

Solution

Let $$T(t)$$ be the temperature of the object at time $$t$$. For convenience, we choose the origin $$t_0=0$$ of the time scale to be 11:05 so that $$T_0=60$$. We must determine the time $$\tau$$ when $$T(\tau)=72$$. Substituting $$T_0=60$$ and $$T_m=-20$$ into Equation \ref{eq:4.2.2} yields

$T = -20+\bigl(60-(-20)\bigr)e^{-kt}\nonumber$

or

$\label{eq:4.2.4} T = -20+80e^{-kt}.$

We obtain $$k$$ from the stated condition that the temperature of the object is 50$$^\circ$$F at 11:07. Since 11:07 is $$t=2$$ on our time scale, we can determine $$k$$ by substituting $$T=50$$ and $$t=2$$ into Equation \ref{eq:4.2.4} to obtain

$50 = -20+80e^{-2k}\nonumber$

This is shown in Figure 4.2.2 .

Hence,

$e^{-2k}={70\over 80}={7\over 8}.\nonumber$

Taking logarithms and solving for $$k$$ yields

$k =-{1\over 2} \ln {7\over 8} = {1\over 2} \ln {8\over 7}.\nonumber$

Substituting this into Equation \ref{eq:4.2.4} yields

$T = -20+80 e^{-{t\over 2}\ln {8\over 7}},\nonumber$

and the condition $$T(\tau)=72$$ implies that

$72 =-20+80 e^{-{\tau\over 2} \ln {8\over 7}};\nonumber$

hence,

$e^{-{\tau\over 2} \ln {8\over 7}} ={92\over 80} = {23\over 20}.\nonumber$

Taking logarithms and solving for $$\tau$$ yields

$\tau = -{2 \ln {23\over 20}\over \ln {8\over 7}} \approx-2.09\ \mbox{min}.\nonumber$

Therefore the object was placed outside about 2 minutes and 5 seconds before 11:05; that is, at 11:02:55.

## Mixing Problems

In the next two examples a saltwater solution with a given concentration (weight of salt per unit volume of solution) is added at a specified rate to a tank that initially contains saltwater with a different concentration. The problem is to determine the quantity of salt in the tank as a function of time. This is an example of a mixing problem. To construct a tractable mathematical model for mixing problems we assume in our examples (and most exercises) that the mixture is stirred instantly so that the salt is always uniformly distributed throughout the mixture. Exercises 4.2.22 and 4.2.23 deal with situations where this isn’t so, but the distribution of salt becomes approximately uniform as $$t\to\infty$$.

##### Example 4.2.3

A tank initially contains 40 pounds of salt dissolved in 600 gallons of water. Starting at $$t_0 = 0$$, water that contains 1/2 pound of salt per gallon is poured into the tank at the rate of 4 gal/min and the mixture is drained from the tank at the same rate (Figure 4.2.3 ).

1. Find a differential equation for the quantity $$Q(t)$$ of salt in the tank at time $$t > 0$$, and solve the equation to determine $$Q(t)$$.
2. Find $$\lim_{t\to\infty}Q(t)$$.

Solution a

To find a differential equation for $$Q$$, we must use the given information to derive an expression for $$Q'$$. But $$Q'$$ is the rate of change of the quantity of salt in the tank changes with respect to time; thus, if rate in denotes the rate at which salt enters the tank and rate out denotes the rate by which it leaves, then

$\label{eq:4.2.5} Q' = \mbox{rate in}-\mbox{rate out}.$

The rate in is

$\left({1\over 2}\ \mbox{lb/gal}\right) \times (4\ \mbox{gal/min}) = 2\ \mbox{lb/min}.\nonumber$

Determining the rate out requires a little more thought. We’re removing 4 gallons of the mixture per minute, and there are always 600 gallons in the tank; that is, we are removing $$1/150$$ of the mixture per minute. Since the salt is evenly distributed in the mixture, we are also removing $$1/150$$ of the salt per minute. Therefore, if there are $$Q(t)$$ pounds of salt in the tank at time $$t$$, the rate out at any time $$t$$ is $$Q(t)/150$$. Alternatively, we can arrive at this conclusion by arguing that

$\begin{array}{lcl} \mbox{rate out} & = & (\mbox{concentration})\times(\mbox{rate of flow out})\\[6pt] \mbox{}&=&(\mbox{lb/gal})\times(\mbox{gal/min})\\[10pt] &=&{Q(t)\over600}\times 4={Q(t)\over150}. \end{array}\nonumber$

We can now write Equation \ref{eq:4.2.5} as

$Q' = 2-{Q\over 150}. \nonumber$

This first order equation can be rewritten as

$Q'+{Q\over 150} = 2. \nonumber$

Since $$e^{-t/150}$$ is a solution of the complementary equation, the solutions of this equation are of the form $$Q=ue^{-t/150}$$, where $$u'e^{-t/150}=2$$, so $$u'=2e^{t/150}$$. Hence,

$u = 300e^{t/150}+c, \nonumber$

so

$\label{eq:4.2.6} Q=ue^{-t/150}=300+ce^{-t/150}$

(Figure 4.2.4 ). Since $$Q(0)=40$$, $$c=-260$$; therefore,

$Q=300-260e^{-t/150}.\nonumber$

Solution b

From Equation \ref{eq:4.2.6}, we see that that $$\lim_{t \to \infty}Q(t)=300$$ for any value of $$Q(0)$$. This is intuitively reasonable, since the incoming solution contains 1/2 pound of salt per gallon and there are always 600 gallons of water in the tank.

##### Example 4.2.4

A 500-liter tank initially contains 10 g of salt dissolved in 200 liters of water. Starting at $$t_0=0$$, water that contains 1/4 g of salt per liter is poured into the tank at the rate of 4 liters/min and the mixture is drained from the tank at the rate of 2 liters/min (Figure [figure:4.2.5}). Find a differential equation for the quantity $$Q(t)$$ of salt in the tank at time $$t$$ prior to the time when the tank overflows and find the concentration $$K(t)$$ (g/liter) of salt in the tank at any such time.

Solution

We first determine the amount $$W(t)$$ of solution in the tank at any time $$t$$ prior to overflow. Since $$W(0) = 200$$ and we are adding 4 liters/min while removing only 2 liters/min, there’s a net gain of 2 liters/min in the tank; therefore,

$W(t) = 2t+200.\nonumber$

Since $$W(150)=500$$ liters (capacity of the tank), this formula is valid for $$0 \le t \le 150$$.

Now let $$Q(t)$$ be the number of grams of salt in the tank at time $$t$$, where $$0 \le t \le 150$$. As in Example 4.2.3

$\label{eq:4.2.7} Q' = \mbox{rate in}-\mbox{rate out}.$

The rate in is

$\label{eq:4.2.8} \left({1\over 4}\ \mbox{g/liter}\,\right) \times (4\ \mbox{liters/min}\,) = 1\ \mbox{g/min}.$

To determine the rate out, we observe that since the mixture is being removed from the tank at the constant rate of 2 liters/min and there are $$2t+200$$ liters in the tank at time $$t$$, the fraction of the mixture being removed per minute at time $$t$$ is

${2\over 2t+200} = {1\over t+100}.\nonumber$

We’re removing this same fraction of the salt per minute. Therefore, since there are $$Q(t)$$ grams of salt in the tank at time $$t$$,

$\label{eq:4.2.9} \mbox{rate out} = {Q(t)\over t+100}.$

Alternatively, we can arrive at this conclusion by arguing that

$\begin{array}{lcl} \mbox{rate out} & = & (\mbox{concentration})\times(\mbox{rate of flow out}) =(\mbox{g/liter})\times(\mbox{liters/min})\\[10pt] &=&{Q(t)\over2t+200}\times 2={Q(t)\over t+100}. \end{array} \nonumber$

Substituting Equation \ref{eq:4.2.8} and Equation \ref{eq:4.2.9} into Equation \ref{eq:4.2.7} yields

$\label{eq:4.2.10} Q'=1-{Q\over t+100},\quad \text{so} \quad Q'+{1\over t+100} Q=1.$

By separation of variables, $$1/(t+100)$$ is a solution of the complementary equation, so the solutions of Equation \ref{eq:4.2.10} are of the form

$Q={u\over t+100}, \quad \text{where} \quad {u'\over t+100 =1}, \quad \text{so} \quad u'=t+100.\nonumber$

Hence,

$\label{eq:4.2.11} u = {(t+100)^2\over 2}+c.$

Since $$Q(0)=10$$ and $$u=(t+100)Q$$, Equation \ref{eq:4.2.11} implies that

$(100)(10) = {(100)^2\over 2}+c,\nonumber$

so

$c=100(10)-{(100)^2\over 2} =-4000\nonumber$

and therefore

$u = {(t+100)^2\over 2} -4000.\nonumber$

Hence,

$Q = {u\over t+200}= {t+100\over 2}-{4000\over t+100}.\nonumber$

Now let $$K(t)$$ be the concentration of salt at time $$t$$. Then

$K(t) = {1\over 4}-{2000\over(t+100)^2}\nonumber$

This is shown in Figure 4.2.6 .


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# 3.4: Evolution theory’s core concepts

So what are the facts and inferences upon which the Theory of Evolution is based? Two of its foundational observations are deeply interrelated and based on empirical observations associated with plant and animal breeding and the charateristics of natural populations. The first is the fact that whatever type of organism we examine, if we look carefully enough, making accurate measurements of visible and behavioral traits (this description of the organism is known as its phenotype), we find that individuals vary with respect to one another. More to the point, plant and animal breeders recognized that the offspring of controlled matings between individuals often displayed phenotypes similar to those of their parents, indicating that phenotypic traits can be inherited. Over many generations, domestic animal and plant breeders used what is now known as artificial selection to generate the range of domesticated plants and animals with highly exaggerated phenotypes. For example, beginning ~10,000 years ago plant breeders in Mesoamerica developed modern corn (maize) by the selective breeding of variants of the grass teosinte.68 All of the various breeds of dogs, from the tiny to the rather gigantic, appear to be derived from a common ancestor that lived between ~19,000 to 32,000 years ago (although as always, be skeptical; it could be that exactly where and when this common ancestor lived could be revised).69 In all cases, the crafting of specific domesticated organisms followed the same pattern.

Organisms with desirable (or desired) traits were selected for breeding with one another. Organisms that did not have these traits were discarded and not permitted to breed. This process, carried out over hundreds to thousands of generations, led to organisms that display distinct or exaggerated forms of the selected trait. What is crucial to understand is that this strategy could work only if different versions of the trait were present in the original selected population and at least a part of this phenotypic variation was due to genetic, that is heritable, factors. Originally, what these heritable factors were was completely unclear, but we can refer to them as the organism’s genotype, even though early plant and animal breeders would never have used that term.

This implies that different organisms have different genotypes and that different genotypes produce different phenotypes, but where genotypic differences came from was completely unclear to early plant and animal breeders. Were they imprinted on the organism in some way based on its experiences or were they the result of environmental factors? Was the genotype stable or could it be modified by experience? How were genotypic factors passed from generation to generation? And how, exactly, did a particular genotype produce or influence a specific phenotypic trait. As we will see, at least superficially, this last question still remains poorly resolved for many phenotypes.

## Contributors and Attributions

• Michael W. Klymkowsky (University of Colorado Boulder) and Melanie M. Cooper (Michigan State University) with significant contributions by Emina Begovic & some editorial assistance of Rebecca Klymkowsky.

3.4: Evolution theory’s core concepts is shared under a not declared license and was authored, remixed, and/or curated by Michael W. Klymkowsky and Melanie M. Cooper.


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2019
Том 71
№ 11

# Li–Yorke sensitivity for semigroup actions

Rybak O. V.

Abstract

We introduce and study the concept of Li–Yorke sensitivity for semigroup actions (dynamical systems of the form (X, G), where X is a metric space and G is a semigroup of continuous mappings of this space onto itself). A system (X, G) is called Li–Yorke sensitive if there exists positive ε such that, for any point xX and any open neighborhood U of this point, one can find a point yU for which the following conditions are satisfied:
(i) d(g(x), g(y)) > ε for infinitely many gG,
(ii) for any δ > 0; there exists hG satisfying the condition d(h(x), h(y)) < δ.
In particular, it is shown that a nontrivial topologically weakly mixing system (X, G) with a compact set X and an Abelian semigroup G is Li–Yorke sensitive.

English version (Springer): Ukrainian Mathematical Journal 65 (2013), no. 5, pp 752-759.

Citation Example: Rybak O. V. Li–Yorke sensitivity for semigroup actions // Ukr. Mat. Zh. - 2013. - 65, № 5. - pp. 681–688.

Full text


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시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 128 MB 7 5 4 66.667%

문제

When you read about environmental disasters such as the gulf oil spill, you frequently see mention of the damage to the “ecosystem” in the region (in the case of the Gulf of Mexico often with the adjective “fragile” in front of it). One can associate a number of different things with this term, but one dimension is that many different animals and plants depend on each other for survival (or population control), and introducing oil into the food chain could have all kinds of side effects. For instance, even if it didn’t affect Seymour directly, it might kill a lot of zooplankton, so the herrings wouldn’t have enough to eat, and in turn, there wouldn’t be enough food for Seymour. In turn, the squids might like this, because fewer seals means fewer predators. Here, you are to calculate the overall effect on the ecosystem.

We model this as follows: there are n species. Each species only eats species lower on the food chain, which means species with lower numbers. For each species, you will be given the initial number of members remaining after some of them have died directly because of the oil. Furthermore, you will be given their diet: how many members of each of the other species one member needs to eat to survive. That is supposed to describe a “balanced diet”: for instance, if we determine that a seal needs to eat a herring and a squid per day, then substituting two herrings and half a squid, or 10 squids, is not sufficient. If the supply of one or more of the food sources is already depleted, then the remaining members of the higher species starve and die (much to the delight of their other prey). For the purpose of this problem, to avoid complication, you will process the species in increasing order. For instance, if both seals (e.g., species 5) and dolphins (e.g., species 12) need herring in their diet, and the seals have eaten all the remaining herrings, then all the dolphins will die.

You are to output the number of surviving members of the species

입력

The first line contains the number K of data sets. This is followed by K data sets, each of the following form:

The first line contains an integer 1 ≤ n ≤ 100, the number of species.

This is followed by n lines, with line i describing species i, by giving i integer numbers. The first number is the current population of species i. This is followed by i − 1 integers, giving the number of members of species 1, 2,... ,i − 1 which a member of species i needs to eat to survive.

출력

For each data set, output “Data Set x:” on a line by itself, where x is its number. Then output the number of members of the species (one per line) which survive after one unit of time compared to the starting configuration. Each data set should be followed by a blank line.

예제 입력 1

1
6
200
30 2
1000 0 5
300 1 0 1
20 2 0 0 0
10 1 0 0 0 1

예제 출력 1

Data Set 1:
84
0
0
6
10
10