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This week, in the MAT8181 Time Series course, we’ve discussed unit root tests. According to Wold’s theorem, if $(Y_t)$ is  (weakly) stationnary then

$Y_{t}=\sum _{{j=0}}^{\infty }\psi_{j}\varepsilon _{{t-j}}+\xi _{t}$

where $(\varepsilon _{{t}})$ is the innovation process, and where $(\xi _{{t}})$ is some deterministic series (just to get a result as general as possible). Observe that

$\sum _{{j=0}}^{{\infty }}|\psi_{{j}}|^{2} < \infty$

as discussed in a previous post. To go one step further, there is also the Beveridge-Nelson decomposition : an integrated of order one process, defined as

$\Delta Y_{t}=(1-L) Y_t=\sum _{{j=0}}^{\infty }\psi_{j}\varepsilon _{{t-j}}+\xi=\Psi(L)\varepsilon _{{t}}+\xi$can be represented as

a linear trend $+$ a random walk $+$ a stationary remaining term

i.e.

$Y_{t}=\underbrace{Y_0 + \xi t }+\underbrace{\Psi(1)\sum_{j=1}^t\varepsilon _{{i}}}+\underbrace{\tilde\Psi(L)\varepilon_0-\tilde\Psi(L)\varepsilon_t}$

where $\tilde\Psi(\cdot)$ is the polynomial with terms $\tilde\psi_j$, where

$\tilde\psi_j =\sum_{i=j+1}^\infty\psi_i$

For unit-root tests, we will use various representation of the process. In order to illustrate the implementation of those tests, consider the following series

> E=rnorm(240)
> X=cumsum(E)
> plot(X,type="l")

• Dickey Fuller (standard)

Here, for the simple version of the Dickey-Fuller test, we assume that

$Y_t=\alpha+\beta t+\varphi Y_{t-1}+\varepsilon_t$

and we would like to test if $\varphi=1$ (or not). We can write the previous representation as

$\Delta Y_t=\alpha+\beta t+[\varphi-1] Y_{t-1}+\varepsilon_t$

so we simply have to test if the regression coefficient in the linear regression is – or not – null. Which can be done with Student’s test. If we consider the previous model without the linear drift, we have to consider the following regression

> lags=0
> z=diff(X)
> n=length(z)
> z.diff=embed(z, lags+1)[,1]
> z.lag.1=X[(lags+1):n]
> summary(lm(z.diff~0+z.lag.1 ))

Call:
lm(formula = z.diff ~ 0 + z.lag.1)

Residuals:
Min       1Q   Median       3Q      Max
-2.84466 -0.55723 -0.00494  0.63816  2.54352

Coefficients:
Estimate Std. Error t value Pr(>|t|)
z.lag.1 -0.005609   0.007319  -0.766    0.444

Residual standard error: 0.963 on 238 degrees of freedom
Multiple R-squared:  0.002461,	Adjusted R-squared:  -0.00173
F-statistic: 0.5873 on 1 and 238 DF,  p-value: 0.4442

Our testing procedure will be based on the Student’s t value,

> summary(lm(z.diff~0+z.lag.1 ))$coefficients[1,3] [1] -0.7663308 which is exactly the value computed using > library(urca) > df=ur.df(X,type="none",lags=0) > df ############################################################### # Augmented Dickey-Fuller Test Unit Root / Cointegration Test # ############################################################### The value of the test statistic is: -0.7663 The interpretation of this value can be done using critical values (99%, 95%, 90%) > qnorm(c(.01,.05,.1)/2) [1] -2.575829 -1.959964 -1.644854 If the statistics exceeds those values, then the series is not stationnary, since we cannot reject the assumption that $\varphi-1=0$. So we might conclude that there is a unit root. Actually, those critical values are obtained using > summary(df) ############################################### # Augmented Dickey-Fuller Test Unit Root Test # ############################################### Test regression none Call: lm(formula = z.diff ~ z.lag.1 - 1) Residuals: Min 1Q Median 3Q Max -2.84466 -0.55723 -0.00494 0.63816 2.54352 Coefficients: Estimate Std. Error t value Pr(>|t|) z.lag.1 -0.005609 0.007319 -0.766 0.444 Residual standard error: 0.963 on 238 degrees of freedom Multiple R-squared: 0.002461, Adjusted R-squared: -0.00173 F-statistic: 0.5873 on 1 and 238 DF, p-value: 0.4442 Value of test-statistic is: -0.7663 Critical values for test statistics: 1pct 5pct 10pct tau1 -2.58 -1.95 -1.62 The problem with R is that there are several packages that can be used for unit root tests. Just to mention another one, > library(tseries) > adf.test(X,k=0) Augmented Dickey-Fuller Test data: X Dickey-Fuller = -2.0433, Lag order = 0, p-value = 0.5576 alternative hypothesis: stationary We do have here also a test where the null hypothesis is that there is a unit root. But the p-value is quite different. What is odd is that we have > 1-adf.test(X,k=0)$p.value
[1] 0.4423705
> [email protected]$coefficients[4] [1] 0.4442389 (but I think it is a coincidence). • Augmented Dickey Fuller It is possible to had some lags in the regression. For instance, we can consider $\Delta Y_t=\alpha+\beta t+[\varphi-1] Y_{t-1}+\psi \Delta Y_{t-1}+\varepsilon_t$ Again, we have to check if one coefficient is null, or not. And this can be done using Student’s t test. > lags=1 > z=diff(X) > n=length(z) > z.diff=embed(z, lags+1)[,1] > z.lag.1=X[(lags+1):n] > k=lags+1 > z.diff.lag = embed(z, lags+1)[, 2:k] > summary(lm(z.diff~0+z.lag.1+z.diff.lag )) Call: lm(formula = z.diff ~ 0 + z.lag.1 + z.diff.lag) Residuals: Min 1Q Median 3Q Max -2.87492 -0.53977 -0.00688 0.64481 2.47556 Coefficients: Estimate Std. Error t value Pr(>|t|) z.lag.1 -0.005394 0.007361 -0.733 0.464 z.diff.lag -0.028972 0.065113 -0.445 0.657 Residual standard error: 0.9666 on 236 degrees of freedom Multiple R-squared: 0.003292, Adjusted R-squared: -0.005155 F-statistic: 0.3898 on 2 and 236 DF, p-value: 0.6777 > summary(lm(z.diff~0+z.lag.1+z.diff.lag ))$coefficients[1,3]
[1] -0.7328138

This value is the one obtained using

> df=ur.df(X,type="none",lags=1)
> summary(df)

###############################################
# Augmented Dickey-Fuller Test Unit Root Test #
###############################################

Test regression none

Call:
lm(formula = z.diff ~ z.lag.1 - 1 + z.diff.lag)

Residuals:
Min       1Q   Median       3Q      Max
-2.87492 -0.53977 -0.00688  0.64481  2.47556

Coefficients:
Estimate Std. Error t value Pr(>|t|)
z.lag.1    -0.005394   0.007361  -0.733    0.464
z.diff.lag -0.028972   0.065113  -0.445    0.657

Residual standard error: 0.9666 on 236 degrees of freedom
Multiple R-squared:  0.003292,	Adjusted R-squared:  -0.005155
F-statistic: 0.3898 on 2 and 236 DF,  p-value: 0.6777

Value of test-statistic is: -0.7328

Critical values for test statistics:
1pct  5pct 10pct
tau1 -2.58 -1.95 -1.62

And again, other pckages can be used:

> adf.test(X,k=1)

Augmented Dickey-Fuller Test

data:  X
Dickey-Fuller = -1.9828, Lag order = 1, p-value = 0.5831
alternative hypothesis: stationary

Hopefully, the conclusion is the same (we should reject the assumption that the series is stationary, but I am not sure about the computation of the p-value).

• Augmented Dickey Fuller with trend and drift

So far, we have not included the drift in our model. But this is simple to do (this will be called the augmented version of the previous procedure): we just have to include a constant in the regression,

> summary(lm(z.diff~1+z.lag.1+z.diff.lag ))

Call:
lm(formula = z.diff ~ 1 + z.lag.1 + z.diff.lag)

Residuals:
Min       1Q   Median       3Q      Max
-2.91930 -0.56731 -0.00548  0.62932  2.45178

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)  0.29175    0.13153   2.218   0.0275 *
z.lag.1     -0.03559    0.01545  -2.304   0.0221 *
z.diff.lag  -0.01976    0.06471  -0.305   0.7603
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.9586 on 235 degrees of freedom
Multiple R-squared:  0.02313,	Adjusted R-squared:  0.01482
F-statistic: 2.782 on 2 and 235 DF,  p-value: 0.06393

The statistics of interest are obtained here considering some analysis of variance outputs, where this model is compared with the one without the integrated part, and the drift,

> summary(lm(z.diff~1+z.lag.1+z.diff.lag ))$coefficients[2,3] [1] -2.303948 > anova(lm(z.diff ~ z.lag.1 + 1 + z.diff.lag),lm(z.diff ~ 0 + z.diff.lag))$F[2]
[1] 2.732912

Those two values are the ones obtained also with

> df=ur.df(X,type="drift",lags=1)
> summary(df)

###############################################
# Augmented Dickey-Fuller Test Unit Root Test #
###############################################

Test regression drift

Call:
lm(formula = z.diff ~ z.lag.1 + 1 + z.diff.lag)

Residuals:
Min       1Q   Median       3Q      Max
-2.91930 -0.56731 -0.00548  0.62932  2.45178

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)  0.29175    0.13153   2.218   0.0275 *
z.lag.1     -0.03559    0.01545  -2.304   0.0221 *
z.diff.lag  -0.01976    0.06471  -0.305   0.7603
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.9586 on 235 degrees of freedom
Multiple R-squared:  0.02313,	Adjusted R-squared:  0.01482
F-statistic: 2.782 on 2 and 235 DF,  p-value: 0.06393

Value of test-statistic is: -2.3039 2.7329

Critical values for test statistics:
1pct  5pct 10pct
tau2 -3.46 -2.88 -2.57
phi1  6.52  4.63  3.81

And we can also include a linear trend,

> temps=(lags+1):n
> summary(lm(z.diff~1+temps+z.lag.1+z.diff.lag ))

Call:
lm(formula = z.diff ~ 1 + temps + z.lag.1 + z.diff.lag)

Residuals:
Min       1Q   Median       3Q      Max
-2.87727 -0.58802 -0.00175  0.60359  2.47789

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)  0.3227245  0.1502083   2.149   0.0327 *
temps       -0.0004194  0.0009767  -0.429   0.6680
z.lag.1     -0.0329780  0.0166319  -1.983   0.0486 *
z.diff.lag  -0.0230547  0.0652767  -0.353   0.7243
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.9603 on 234 degrees of freedom
Multiple R-squared:  0.0239,	Adjusted R-squared:  0.01139
F-statistic:  1.91 on 3 and 234 DF,  p-value: 0.1287

> summary(lm(z.diff~1+temps+z.lag.1+z.diff.lag ))$coefficients[3,3] [1] -1.98282 > anova(lm(z.diff ~ z.lag.1 + 1 + temps+ z.diff.lag),lm(z.diff ~ 1+ z.diff.lag))$F[2]
[1] 2.737086

while R function returns

> df=ur.df(X,type="trend",lags=1)
> summary(df)

###############################################
# Augmented Dickey-Fuller Test Unit Root Test #
###############################################

Test regression trend

Call:
lm(formula = z.diff ~ z.lag.1 + 1 + tt + z.diff.lag)

Residuals:
Min       1Q   Median       3Q      Max
-2.87727 -0.58802 -0.00175  0.60359  2.47789

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)  0.3227245  0.1502083   2.149   0.0327 *
z.lag.1     -0.0329780  0.0166319  -1.983   0.0486 *
tt          -0.0004194  0.0009767  -0.429   0.6680
z.diff.lag  -0.0230547  0.0652767  -0.353   0.7243
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.9603 on 234 degrees of freedom
Multiple R-squared:  0.0239,	Adjusted R-squared:  0.01139
F-statistic:  1.91 on 3 and 234 DF,  p-value: 0.1287

Value of test-statistic is: -1.9828 1.8771 2.7371

Critical values for test statistics:
1pct  5pct 10pct
tau3 -3.99 -3.43 -3.13
phi2  6.22  4.75  4.07
phi3  8.43  6.49  5.47
• KPSS test

Here, in the KPSS testing procedure, two models can be considerd : with a drift, or with a linear trend. Here, the null hypothesis is that the series is stationnary.

With a drift, the code is

> summary(ur.kpss(X,type="mu"))

#######################
# KPSS Unit Root Test #
#######################

Test is of type: mu with 4 lags.

Value of test-statistic is: 0.972

Critical value for a significance level of:
10pct  5pct 2.5pct  1pct
critical values 0.347 0.463  0.574 0.73

while it will be, in the case there is a trend

> summary(ur.kpss(X,type="tau"))

#######################
# KPSS Unit Root Test #
#######################

Test is of type: tau with 4 lags.

Value of test-statistic is: 0.5057

Critical value for a significance level of:
10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216

One more time, it is possible to use another package to get the same test (but again, a different output)

> kpss.test(X,"Level")

KPSS Test for Level Stationarity

data:  X
KPSS Level = 1.1997, Truncation lag parameter = 3, p-value = 0.01

> kpss.test(X,"Trend")

KPSS Test for Trend Stationarity

data:  X
KPSS Trend = 0.6234, Truncation lag parameter = 3, p-value = 0.01

At least, there is some kind of consistency, since we keep rejecting the stationnary assumption, for that series.

• Philipps-Perron test

The Philipps-Perron test is based on the ADF procedure. The code is here

> PP.test(X)

Phillips-Perron Unit Root Test

data:  X
Dickey-Fuller = -2.0116, Truncation lag parameter = 4, p-value = 0.571

with again, a possible alternative with the other package

> pp.test(X)

Phillips-Perron Unit Root Test

data:  X
Dickey-Fuller Z(alpha) = -7.7345, Truncation lag parameter = 4, p-value
= 0.6757
alternative hypothesis: stationary
•  Comparison

I will not spend more time comparing the different codes, in R, to run those tests. Let us spend some additional time on a quick comparison of those three procedure. Let us generate some autoregressive processes, with more or less autocorrelation, as well as some random walk, and let us see how those tests perform :

> n=100
> AR=seq(1,.7,by=-.01)
> P=matrix(NA,3,31)
> M1=matrix(NA,1000,length(AR))
> M2=matrix(NA,1000,length(AR))
> M3=matrix(NA,1000,length(AR))

> for(i in 1:(length(AR)+1)){
+ for(s in 1:1000){
+ if(i==1) X=cumsum(rnorm(n))
+ if(i!=1) X=arima.sim(n=n,list(ar=AR[i]))
+ library(urca)
+ M2[s,i]=as.numeric(pp.test(X)$p.value) + M1[s,i]=as.numeric(kpss.test(X)$p.value)
+ }}

Here, we would like to count how many times the p-value of our tests exceed 5%,

> prop05=function(x) mean(x>.05)
+ P[1,]=1-apply(M1,2,prop05)
+ P[2,]=apply(M2,2,prop05)
+ P[3,]=apply(M3,2,prop05)
+ }
> plot(AR,P[1,],type="l",col="red",ylim=c(0,1),ylab="proportion of non-stationnary
+ series",xlab="autocorrelation coefficient")
> lines(AR,P[2,],type="l",col="blue")
> lines(AR,P[3,],type="l",col="green")
> legend(.7,1,c("ADF","KPSS","PP"),col=c("green","red","blue"),lty=1,lwd=1)

We can see here how poorly Dickey-Fuller test behave, since a 50% (at least) of our autoregressive processes are considered as non-stationnary.


text:
# Compression

A single joint pose has ten f32s of data, or 40 bytes of data:

• Three f32s for translation
• Four f32s for rotation
• Three f32s for scale

Assuming you want to display a 1-second clip sampled at 30 samples per second, this equates to:

\begin{align} &4 \text{ bytes per f32} \\ &\times 10 \text{ f32s per joint} \\ &\times 30 \text{ samples per second} \\ &= 1200 \text{ bytes per joint per second} \end{align}

This is a lot! Considering a simple skeleton could have 100 joints in it and a modern game could have hundreds of seconds of animation clips.

## Compression Techniques

### Store less data per joint

• If you don't want to support scaling in your game engine (or if most joints don't need it), then omit it entirely, or reduce it to just a single f32 uniform scaling factor.
• If you don't think a joint should be able to stretch out and shrink in (imagine the neck joint translating away (not just rotating) from the pelvis joint), then you should also just drop it.
• If we can make certain assumptions about the data, we could omit "the last" f32 of a piece of data. For example, since the four f32s that make up the Quaternion for rotation are usually assumed to be normalized, we could omit the last f32 and derive it later when we decompress it by subtracting the other components from 1.
• If a pose doesn't change over the course of many samples (it is constant), then a further optimization could be to just store it once with a flag to indicate that it won't change over the course of the sample.

### Quantize f32s into u16s

• Each component of a quaternion falls in the range [-1, 1]. At a magnitude of 1, the 8-bits that comprise of the exponent bits in a floating point f32 are all zero. The remaining 23-bit mantissa provides accuracy to the 7th decimal place. Quaternions can be effectively encoded with 16 bits of precision, so we are wasting 16 bits per float for our quaternions.
• Quantization is the process of converting a 32-bit IEEE float into an n-bit integer representation. It is lossy.
• Encoding is f32 to u16.
• Decoding is u16 to f32. This is the part that is an approximated value of the original f32. This process is lossy, since we lose some precision to store the value.
• Quantization involves dividing a range of possible input values to N equally sized intervals. The f32 input maps to the integer index of the interval it falls inside. To decode, we cast the integer index into a f32 format it refers to, and then shift and scale it back to its original range.
• Rounding the float to the center of the enclosing interval and returning the left-hand side of the interval to which our original value was mapped is the best choice for how to do the quantization.

### Use less samples

• Some animations look fine at 15 samples per second. Doing this would reduce our animation data size in half.
• If the animation operates in a straight line, LERP could be a good stand-in for removing some samples.

### Use curve-based compression

• Storing animation data as regularly spaced sequence of nonuninform, nonrational B-splines to describe the paths of a joint over time allows channels with a lot of curvature to be encoded using only a few data points.

### Use wavelet compression

Signal processing theory has a technique called wavelet compression.

A wavelet is a function whose amplitude oscillates like a wave but with a very short duration. An animation curve is decomposed into a sum of orthonormal wavelets.

This is an advanced technique, TBD how it works later.

Further reading on wavelet compression: Animation Compression: Signal Processing

### Don't use an animation clip at all

• If an animation clip has no chance of being seen, i.e., it only plays on certain levels of your game or if the player has achieved a certain level/class of their character, then don't load the memory at all until it is needed.
• Most games have a core set of animation clips loaded when the game boots. All other animations are loaded on an as-needed basis.

format: plain


text:
# A central limit theorem for sojourn time of strongly dependent 2-dimensional gaussian process

• Jeon, Tae-Il
• Published : 1995.11.01
• 39 5

#### Abstract

Let $X_t = (X_t^(1), X_t^(2))', t \geqslant 0$, be a real stationary 2-dimensional Gaussian process with $EX_t^(1) = EX_t^(2) = 0$ and $$EX_0 X'_t = (_{\rho(t) r(t)}^{r(t) \rho(t)}),$$ where $r(t) \sim$\mid$t$\mid$^-\alpha, 0 < \alpha < 1/2, \rho(t) = o(r(t)) as t \to \infty, r(0) = 1, and \rho(0) = \rho (0 \leqslant \rho < 1)$. For $t > 0, u > 0, and \upsilon > 0, let L_t (u, \upsilon)$ be the time spent by $X_s, 0 \leqslant s \leqslant t$, above the level $(u, \upsilon)$.

#### Keywords

central limit theorem;sojourn;diagram formula


text:
Good thinking! We know that water gets converted to water vapour all the time. The water's temperature is definitely not as high as $$100^{\circ} \, \mathrm{C}$$!
It is true that if you heat water to $$100^{\circ} \mathrm{C}$$, all the water definitely gets converted to steam. But, even if the temperature is not that high, some random molecules in the water can still escape the force of attraction of the other molecules. This is called Evaporation.
Evaporation is very different from boiling (vaporisation) because evaporation occurs at all temperatures above $$0^{\circ} \mathrm{C}$$, not just at $$100^{\circ} \mathrm{C}$$


text:
Paper: Search for Lepton Flavor Violation in the emu Continuum

Comments due in CDS by: 20th Feb.

1.69: Motivate or reference "0.72" and "0.28"

2.13 and 2.14: It is just stylistic, but for consistency with places above where eta is mentioned, use the smaller sign instead of the equal sign.

2.37: It is just stylistic, but if you say 0.5 GeV instead of 500 MeV then all energy/momentum units in the paper will be GeV and makes things consistent. Also, the brain compares easier 0.5 GeV with 25 GeV than it would compare 500 MeV with 25 GeV.

2.51: It is just stylistic, but you can replace "transverse energy" with its symbol "E_T", to be consistent with the phrase above about the "p_T". It also saves space.

2.58: You say "missing transverse momentum" and denote it "ETmiss". Then it should be "missing transverse energy".

No, it's a vector quantity and in ATLAS we properly call it missing transverse momentum, if they call it missing transverse energy they'll be asked to change it back. (James F)

2.65: How does the single top process produce two prompt leptons, one being an electron and one being a muon and both leptons having opposite electric charge?

3.50: "two hypothesis" -> "two hypotheses"

3.54: "convoluting" might be in fact "convolving"?

3.59: "Fig 1." -> "Fig 1. a)"

3.66: "Fig 1." -> "Fig 1. b)"

4.16: "and in" -> "and"

Ref 15: For JHEP papers, the journal name must have 4 digits, where typically the first two are the digits of the year. For this paper might be 1012. A paper is found there on Spires, but must be double checked this is the paper indeed intended to reference.

Ref 18: Same as above. "07" -> "0207"

Ref 24: Same as above, but I could not guess what the year is from "001", also because the page is atypically only with two digits. Needs double check.

The paper is generally pretty good and presents an interesting search for RPV SUSY.

A) The paper neglects to mention of searches at HERA which are sensitive to the same physics:

stop production via R-parity violating SUSY:

This is a 90 degree rotation of the diagram for 1. The HERA experiments set limits on $\lamba_{131}$. See

H1 Collab., F.D. Aaron et al., Eur.Phys.J.C71 (2011) 1572

ZEUS Collaboration; S. Chekanov et al. Eur. Phys. J. C 50 (2007) 269

it would be nice to at least reference them.

B) Couldn't this process also occur via t-channel exchange of a LFV leptoquark?

like those in:

H1 Collab., F.D. Aaron et al., Phys. Lett. B701 (2011) 20 ,

would the present search also be able to improve on those exclusions?

C) What about signal processes like $d \bar{d} \rightarrow \tau \bar{\mu}$ with leptonic tau decay, were they considered?

D) What about processes such as $u \bar{b}\rightarrow e \mu$ are they perhaps constrained by B-physics measurements, if not why ignore them? (apart from the relatively small PDFs)

page 1 line 42-43: "The diagram ...has the same cross section" probably better as "The process ... has the same cross section"

page 1 line 65-66: sloppy phrasing here, the strange PDFs don't suppress the $ss \rightarrow e\mu$ process, without them it wouldn't happen at all, please reword.

page 1 line 69: More needs to be said about the 0.72 and 0.28, e.g. which PDF set is used to get this weighted sum, what is the associated uncertainty? How stable vs choice of alternative PDF? We need to know this to judge whether the limit vs coupling plot makes any sense.

references:

I agree with Adrian on the JHEP references, I don't know however if the paper is following some established ATLAS convention.

o p1 line 6:
I thought LPV is a general feature of GUT models, an even more convincing
extension. Not sure how to get a ref.
o p1 line 33:
Why is the top squark taken to be the lightest up type squark? Is it because
limits are set lowest? I assume b squark is more massvive. So I agree with James on this.
o line 35, page 3: Standard ATLAS tools needs a ref or description.
o Fig 2: dont understand the "total background". Signal hard to see
at all on dphi and nJet plot. The preselection tells one that
the top is understood but the final selection has no top visible. The issue
is to ensure tha tthe kinmatic distribution of diboson is correct.
Also, the instrumental suffers low statistics and the tails could
be an issue. It does not seem these plots are of any use in the final analysis.
o Fig 3: seems that in the end this is just a counting experiment
and it is hard to see where the kinematic cuts help.

Agree with others that a reference is needed for 0.72 and 0.28 pdfs, and that "using standard ATLAS tools" instead needs a description or a reference. Few minor comments:

p1, line 46 pluralised \lambda's could be confusing - would \lambda' indices be better?

p1, line 63 This sentence could be moved to line 51 - in its current position it reads as though the number of variables the cross section is dependent on has been simplified by ignoring s\dbar and d\sbar, but these are also only dependent on \lambda'_231 and \lambda'_132. Or change the sentence to have |\lambda'_131\lambda'_231| , |\lambda'_132\lambda'_232| and m_t~ as the three variables.

p2, line 30 "and with et>25Gev and to lie inside..." sentence too long, I think "and to lie" doesn't make sense. Perhaps "...reconstructed in the ID. They are required to have Et>25GeV and to lie inside..."

p3, line 34: "and were calculated" -> "were calculated"

4, line 33,: should reference 13 be a footnote instead?

-- DeepakKar - 2012-02-14


text:
# Atomic spectra and atomic structure by Gerhard Herzberg

1. Oct 24, 2005

### Galaxy33

Hi! to everyone on the forum. I am new and did not really know where i should of posted this thread its not homework its just a question i have.

A friend of mine asked me a question about a book he read about (atomic spectra and atomic structure by Gerhard Herzberg). Gerhard Herzberg said that, in reality the electron revolves, not about the nucleus itself, but about a commen center of gravity; also the nucleus revolves about that center.

The question is, in 2005 is this view by Gerhard Herzberg still true? I did not know the answer so i posted it here.....

Galaxy......

2. Oct 24, 2005

### Conehead

Great question. Some physicists, during the advent of quantum mechanics, spent some time double checking but Herzberg is still correct.

Though the atomic nucleus greatly outweighs the electrons, the electrons still do some pulling--enough anyway to yank the whole atom (electrons and all) out of whack, even if just the smallest bit.

3. Oct 24, 2005

### nbo10

If you view the system classically, then yes the electron does revolve around the CM.

If you take the Qunatum view, the correct representation, then the question is ill-formed and can not be answered.

4. Oct 24, 2005

### ZapperZ

Staff Emeritus
And in what kind of observation/experiment/phenomenon does this "smallest bit" effect manifests itself?

Zz.

5. Oct 24, 2005

### Conehead

I'm assuming you're talking about HUP. Fine, this is a theoretical question and I'm going to keep assuming (unless asked otherwise) that Galaxy doesn't require proof of speed and position of the particle at the same time.

And, I might add, just because we can't nail down a mathematical snapshot of said particle doesn't mean that we don't know what it's doing. Feynman spends some time on this topic.

6. Oct 24, 2005

### nbo10

It has nothing to do with the Uncertainty principle. The electron, described by it's wavefunction, has a probability distribution that doesn't "revolve" around anything.

7. Oct 24, 2005

### ZapperZ

Staff Emeritus
What "mathematical snapshot" did you have in mind? The mathematical description of an atom has no such trajectory. One only needs to look at the solution fo the hydrogen atom to know this. And these were not derived out of the HUP either.

Zz.

8. Oct 24, 2005

### Conehead

Where is the electron distributed? Does it not have a position?

Based on the responses I'm way off on this, somehow. If that's the case give the proper answer. I'd hate argue bad info any longer than I have to.

Last edited: Oct 24, 2005
9. Oct 24, 2005

### ZapperZ

Staff Emeritus
This goes to the foundation of QM and why the Schrodinger Cat thought experiment came into being!

It does NOT have a definite position till it is measured. The s-orbital is a spherical distribution of ONE electron. You get this by solving the orbital part of the Schrodinger equation. The electron IS distributed all over the place simultaneously. This is what makes QM highly non-intuitive for anyone who skip the mathematical formalism.

How do you know such a description is valid? Besides the fact that QM gave unbelievably accurate energy spectrum of many atoms and molecules (something classical mechanics could not), we also have evidence from how bonding forms, especially in the formation of bonding and antibonding states. Such phenomenon has no intuitive counterpart in classical mechanics.

Zz.

10. Oct 24, 2005

### Conehead

So, I'm assuming that a thrown photon's trajectory is also a matter of probability?

11. Oct 24, 2005

### Gokul43201

Staff Emeritus
To say something about the OP's question : Yes, you can decompose the Schrodinger Equation for the H-atom into two parts - one dealing with the center of mass motion and the other dealing with the relative motion - using the standard change of variables :

$$\mu = \frac {m_1m_2} {m_1+m_2}~;~~M = m_1 + m_2~;~~r = |r_1 - r_2|~;~~R = \frac{m_1 r_1 + m_2 r_2}{m_1+m_2}$$

This makes - in the case of the H-atom - a very tiny change to the Hamiltonia.

Last edited: Oct 24, 2005
12. Oct 24, 2005

### ZapperZ

Staff Emeritus
I wouldn't know, since I have no idea what a "thrown photon" is.

Zz.

13. Oct 24, 2005

### Conehead

When an electron drops to a lower energy level it releases a photon

Did I do something to irritate you?

14. Oct 24, 2005

### ZapperZ

Staff Emeritus
No. I'm responding to what you have said. Is it wrong for me to get clarification of what you are saying or claiming? I have never seen the phrase "thrown photon" in all my years in this field. I have also haven't seen any experimental evidence to back your claim earlier of ".... electrons still do some pulling--enough anyway to yank the whole atom (electrons and all) out of whack, even if just the smallest bit.. "

So a photon emitted by an atomic transition is what you called "thrown photon"? I'm not sure why this would be relevant for this thread. Once photons are emitted one can very much invoke classical optics.

Zz.

15. Oct 24, 2005

### nbo10

Don't take thing's personally.

Life is in the details. When you ask a vague question you're going to get a vague answer. When you ask a general, non-specific, or ill-worded question, you'll find the answer you get isn't going to be helpful.

Photons are very different from electrons.

16. Oct 24, 2005

### Conehead

Oh, I wasn't taking anything personally. Just making sure someone else wasn't. I have very colloquial conversations with working physicists everyday with far better results than just now. But I have a way of not making sense sometimes.

Thanks for the information. Ciao.

17. Oct 24, 2005

### Galaxy33

Thanks to everyone who answered my question, but can i assume that the answer was yes, as Conehead stated (some physicists during the advent of quantum mecanics, spent some time double checking but Herzberg is still correct) ?

Titana.............

Last edited: Oct 25, 2005
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# Prime number characterisation using congruences

I want to prove that $n$ is prime. From the Wilson's theorem it follows that $n$ is prime if and only if $$(n-1)! + 1 \equiv 0 \pmod{n}$$

However, in my proof, I reduce the congruences to the following form:

$$24((n-1)! + 1) \equiv 0 \pmod{n}$$

I believe I can divide the congruence by 24 only if 24 is coprime to $n$, so I can safely deduce that if $n$ is coprime to 24, it is also prime. Otherwise I do not know what can I find out from the following congrunce. Numerical calculations suggest that this successfully characterises primes grater than 24, but I fail to see why. Can anyone please explain to me which (probably obvious and simple) step I am missing?

Edit:

Perhaps the question was not clear enough. It can be expressed in shorter form, without the background:

Given the congruence

$$24((n-1)! + 1) \equiv 0 \pmod{n}$$

what can one tell about the primality of $n$?

• Indeed I did misunderstand your question, although some of what I say in my answer looks relevant to what you actually meant. I am still slightly confused though: where did the factor of $24$ come from?? – Pete L. Clark Apr 22 '13 at 19:50
• In my opinion a "less localized" question would be: prove or disprove: for any positive integer $a$, all but finitely many positive integers $n$ satisfying the congruence $a((n-1)!+1) \equiv 0 \pmod{n}$ are prime. – Pete L. Clark Apr 22 '13 at 20:06
• It appears to me in the last stage of the proof for the following characterisation of the prime triplets of form n, n+2, n+6: $4320(4((n-1)!+1)+n) + 361n(n+2) \equiv 0 \pmod{n(n+2)(n+6)}$ Or, if you prefer, it's given in the question - it's the only problem there, otherwise I could safely deduce that $n$ is prime if congruence holds and not, if otherwise. – jureslak Apr 22 '13 at 20:06
• Yes, that is indeed more general version of my question. – jureslak Apr 22 '13 at 20:07

Yours is special case $\rm\ A = 24\$ in the following generalization.

Theorem $\rm\,\ \ N\mid A\,(1\!+\!(N\!-\!1)!)\iff N\mid A\ \ or\ \ N\:$ is prime.

Proof $\,\ (\Rightarrow)\$ $\rm\:d := gcd(N,A).\,$ By Euclid, $\rm\:d=1\:\Rightarrow\:N\mid(1\!+\!(N\!-\!1)!)\:\Rightarrow\:N\,$ prime by Wilson.
Else $\rm\:d>1\:$ and $\rm\: N = n d,\ A = a d,\:$ for $\rm\:n,a\in \Bbb N.\:$ Cancelling $\rm\:d\:$ yields $\rm\: n\mid \color{#c00}a\,(1\! +\! (\color{#0a0}{nd\!-\!1)!\,}).\$
Note $\rm\:d>1\:\Rightarrow\: n\mid\color{#0a0}{(n d\!-\!1)!\,},\$ so $\rm\ n\mid \color{#c00}a,\$ so $\rm\: n d\mid a d,\:$ i.e. $\rm\:N\mid A.\:$ $\ (\Leftarrow)\$ is clear, by Wilson.

Remark $\ \$ The above is the special case $\rm\ f(N)\, =\, 1+(N\!-\!1)!\$ of the following

Theorem $\ \$ If $\rm\,\ f\, :\, \Bbb N\to \Bbb N\:$ satisfies $\rm\ d>1\:\Rightarrow\:\color{#c00}{gcd(n,f(nd))= 1}\$ for all $\rm\:n,d\in\Bbb N,\$ then

$$\rm N\mid A\,f(N)\ \iff\ N\mid A\ \ or\ \ N\mid f(N)\qquad\qquad$$

Proof $\ \$ Let $\rm\ d = (N,A),\,\ N = nd,\,\ A = ad.\$ Canceling the factor $\rm\:d\:$ from $\rm\ N,A\$ yields

if $\rm\:\ d> 1\$ then $\rm\ N\mid A\,f(N)\iff \color{#c00}n\mid a\,\color{#c00}{f(nd)\stackrel{Euclid}{\iff}} n\mid a\:\iff nd\mid ad\iff N\mid A.$

If $\rm\ d = 1\$ then $\rm\ N\mid A\,f(N) \stackrel{Euclid}{\iff} N\mid f(N),\$ by $\rm\ (N,A) = d = 1.$

• +1: this is a typically clean solution. I admit though that my answer was intentionally suboptimal in an attempt to leave something for the OP to do. – Pete L. Clark Apr 23 '13 at 2:57

The proposed congruence $24((n-1)! + 1) \equiv 0 \pmod{n}$ does not hold for all positive integers $n$: the first two counterexamples are $n = 9$ and $n = 10$.

In fact:

$\bullet$ If $2^4 \mid n$, then $24((n-1)!+1)$ is not divisible by $16$, so the congruence fails.

$\bullet$ For any odd prime $p$, if $p^2 \mid n$ then $24((n-1)!+1) \equiv p \pmod{p^2}$, so the congruence fails.

$\bullet$ If $n$ is of the form $ap$ for a prime $p \geq 5$ and $a \geq 2$, then $24((n-1)!+1)$ is not divisible by $p$, so the congruence fails.

Can you finish it off from here?

Here's how my sister solved it (Not a valid proof though)

$N$ be any composite number. Such that $N$'s factorization of primes is $p_1p_2 \dots p_k$

$\{p_1, \dots p_k \} \subset \{1,2, \dots n-1 \}$, therefore $N | (N-1)!$ and therefore $(N-1)!$ can't be $k (\mod N)$, where $k \neq 0$. The case of $N$ being composite is thus dismissed.


text:
# What is time? (Again)

## Recommended Posts

Posted (edited)
44 minutes ago, Markus Hanke said:

But what if we think of f(x) as an (uncountably) infinite set of real numbers (which it is, mathematically speaking)? Or better still - a 1-parameter family of real numbers? That set would be a static construct, as would be the relationship between the elements in the set. Of course you could externally impose a notion of “going from one element to another”, but I don’t think that is inherent in the set itself.

I fully agree! A function y = f(x) is static. I thought that is what I am saying all the time. But as soon as you say 'y changes as function of the change of x', I think you are implying time. What you are in fact doing is looking at y(t) = f(x(t)).

This triggered the whole discussion:

On 7/22/2020 at 9:58 AM, Markus Hanke said:

you can always have quantities that change with respect to one another, without reference to any notion of time.

So, no, I would say, as soon as you use 'change' you are implying time.

Edited by Eise

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53 minutes ago, Markus Hanke said:

Interesting, I haven’t heard this term before!

What is also interesting is that there is more than one way to viw these relationships, which is why we have several (slightly) different terms.

The condition I meant may be illustrated in the standard equation of an ellipse

This is the locus of a point which moves under the one condition

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

Alternatively we can introduce what is known as a parameter often denoted t, though sometimes a Greek letter is used.

$x = a\cos t$

$y = b\sin t$

Note the first form has two independent variables and one condition or equation, between them.

The second has one independent variable (the parameter) and two conditions or equations.

Both refer to the same ellipse.

The parameter is often (but not always) time and corresponds to what I called the running variable.

You can see how deeply time is embedded in our psyche.

But in fact this parameter t is an angle!

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41 minutes ago, Eise said:

So, no, I would say, as soon as you use 'change' you are implying time.

While I can easily imagine functions ( or teacups ) changing according to parameters other than time, I suspect the problem is with our common definition of the concept of change; it necessarily involves time. Even though it is not a variable of the system under observation, it is necessary for the observer to detect 'change'.

Sure, the teacup 'changes' with respect to location ( from spout to bowl ), but for us to observe both aspects of the teacup, time must necessarily be involved ( first you observe the spout, then, the bowl ).

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Posted (edited)
20 minutes ago, MigL said:

While I can easily imagine functions ( or teacups ) changing according to parameters other than time, I suspect the problem is with our common definition of the concept of change; it necessarily involves time. Even though it is not a variable of the system under observation, it is necessary for the observer to detect 'change'.

Sure, the teacup 'changes' with respect to location ( from spout to bowl ), but for us to observe both aspects of the teacup, time must necessarily be involved ( first you observe the spout, then, the bowl ).

As a Physicist, would you say it makes no difference if we 'observe'  first the bulb then the tail or first the tail then the bulb or of a p orbital or would Physics be different if we did not experience the entire curve all together in any interaction?

If you don't like a quantum example how about being struck by a large jack (as in the game of jacks) shaped object ?

Do we experience the entire momentum at the moment of impact or do we have to wait for the legs to spin round and hit us?

Edited by studiot

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6 hours ago, michel123456 said:

But after more thinking when you are stating "when all the parts are present together", I suspect that you mean "all the parts are present together at the same time". Or maybe (I don't know to say otherwise) that your cup somehow is existing there in space "out of time". As if the concept of a cup "existing" could be understood without the concept o time. But the concept of "together" has already the idea of time inside it.

I think it might be more a matter of some “idea of time” being inside your thinking.

Which is perhaps why you add, “at the same time” to “all the parts are present together” which to me adds nothing.

Just for the hell of it, as an experiment, you might try throwing the word “time” out the window (and all its accompanying baggage), and focus on relationships that persist (such as distances between positions on a cup) and the inherent structure as a whole that arises out of those persisting relationships. Instead of thinking “the inherent structure exists at a time” just think: “the inherent structure exists”. Or “it persists”, not “it persists through time”.

Now if you cry fowl here and say, “The only way the structure can persist is through time”, I submit to you that you’re adding something vague and extraneous in the sense that it doesn’t contradict the notion of change as presented by Marcus. We have to be clear that we are talking about a parameter, not some other notion of time which you or I might espouse.

PS. Although I have to admit, I’m very curious as to how you would define “time” outside the time of physics. Maybe a new thread in the philosophy sub forum.

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No,no, Studiot, I have one for you...

If you inhabited a 'space' with just the three orthogonal dimensions and NO time parameter, how exactly would you 'see' the teacup ? Could you interact with one aspect of it, say the spout, and then another ?
Remember the teacup doesn't change with time as it has no time parameter, but you are also intrinsic to that particular space, and there is no time parameter for you either.

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9 hours ago, MigL said:

No,no, Studiot, I have one for you...

If you inhabited a 'space' with just the three orthogonal dimensions and NO time parameter, how exactly would you 'see' the teacup ? Could you interact with one aspect of it, say the spout, and then another ?
Remember the teacup doesn't change with time as it has no time parameter, but you are also intrinsic to that particular space, and there is no time parameter for you either.

Teacup  ?? Spout  ??

You must have been putting the wrong weed in that cigar again.

Just think, without time, you'd never get to smoke it.

More seriously perhaps 'you' could only perform this interaction if 'you' and the teacup were exactly like the famous hole in the ground and the puddle. Exactly spatially conformable.

Otherwise if you were not then the relationship would be more like that of a point particle in a 'block universe' where the past and future are inaccessible.
Your 3D space has similarities to this block universe.

But this question about 'seeing' raises another interesting point.

Furthermore humans can only see in front of them.
If their sight was constructed differently perhaps see in front of them and behind them together  -  like a fly.

The point is that even in computing we have serial mode and parallel mode and one other interesting one, akin to time travel.
There is a method of computing whereby when the serial flow reaches a decision point, it changes to parallel flow and goes down all the branches of the decision tree together using the then existing step values of the prgram flow variables.
When the appropriate new values become available the results of each side loop are already available and the correct one is selected.

This is a known time saving (pun intended) device which has been used in other decision making situations, for instance

We are about to be invaded. Do we set up defences on the East coast or the West coast ?

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Posted (edited)
On 7/28/2020 at 12:16 PM, Eise said:

I fully agree! A function y = f(x) is static. I thought that is what I am saying all the time. But as soon as you say 'y changes as function of the change of x', I think you are implying time.

I think we’ll have to disagree on this Which is fine of course, this being just a philosophical exchange. To me, the term ‘static’ (I think it should have been ‘stationary’ actually, I often confuse them) implies precisely the absence of any kind of reference to time, be in implicit or explicit.

On 7/28/2020 at 12:16 PM, Eise said:

What you are in fact doing is looking at y(t) = f(x(t))

This is called a ‘parametrisation’, and it is not what I am doing or suggesting at all.
Note also that there is no necessity whatsoever to interpret a parametrisation parameter as ‘time’ - that’s just an extra ad-hoc assumption. Furthermore, note also that generally there is more than one way to parametrise the same curve.

On 7/28/2020 at 12:16 PM, Eise said:

So, no, I would say, as soon as you use 'change' you are implying time.

So you would say that the hypothetical 3D universe with only a tea cup in it is entirely homogenous (i.e. devoid of changes) in all physical aspects? I definitely (but in a friendly, intellectual way) disagree

15 hours ago, MigL said:

If you inhabited a 'space' with just the three orthogonal dimensions and NO time parameter, how exactly would you 'see' the teacup ? Could you interact with one aspect of it, say the spout, and then another ?

There is of course no observer in such a universe, since no act of observation is possible - but does that imply that the tea cup cannot exist? Does existence (in the ontological sense) require a temporal dimension? I think not.

On 7/28/2020 at 1:08 PM, MigL said:

Sure, the teacup 'changes' with respect to location ( from spout to bowl ), but for us to observe both aspects of the teacup, time must necessarily be involved ( first you observe the spout, then, the bowl ).

True, but this issue wasn’t about the observation of change; it was about whether, for two quantities which are not themselves spatiotemporal in nature, one can meaningfully define a notion of ‘change’ of one quantity with respect to the other, and whether this implies a notion of ‘time’ in the physical sense. I think you have answered this correctly in the first part of the above sentence.
I think we can all agree that no act of observation is ever possible without physical time, but that isn’t really the point here. The point is rather that, if there is not some concept of ‘change’ in our hypothetical tea cup universe, then that universe would by definition need to be homogenous in all physical aspects. Clearly that isn’t the case though, because at the very least you have the spatial region occupied by the tea cup, and then vacuum for all the rest of that universe. Whether there is anyone there to observe this is irrelevant, in my opinion. To me, ‘change’ in this particular scenario is equivalent to non-homogeneity, i.e. it is simply the failure of all spatial regions of that universe to be identical. It isn’t a process, it’s a relationship.

Edited by Markus Hanke

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Posted (edited)
23 hours ago, vexspits said:

I think it might be more a matter of some “idea of time” being inside your thinking.

Which is perhaps why you add, “at the same time” to “all the parts are present together” which to me adds nothing.

Just for the hell of it, as an experiment, you might try throwing the word “time” out the window (and all its accompanying baggage), and focus on relationships that persist (such as distances between positions on a cup) and the inherent structure as a whole that arises out of those persisting relationships. Instead of thinking “the inherent structure exists at a time” just think: “the inherent structure exists”. Or “it persists”, not “it persists through time”.

Now if you cry fowl here and say, “The only way the structure can persist is through time”, I submit to you that you’re adding something vague and extraneous in the sense that it doesn’t contradict the notion of change as presented by Marcus. We have to be clear that we are talking about a parameter, not some other notion of time which you or I might espouse.

PS. Although I have to admit, I’m very curious as to how you would define “time” outside the time of physics. Maybe a new thread in the philosophy sub forum.

Foul.

To me (but that might be only me) in order for something to be, time must be there. Either time arises from the existence, or time is a necessity prior to existence, I don't know. But you cannot "be" out of time. You cannot "persist" in something that you want to cancel. It is a wrong concept. IMHO.

To be more pragmatic and step out of pure philosophy, I have the feeling that in the frame of GR it should be possible to show that Time & Space are inseparable. That one cannot establish the concept of Space without Time. That the concept of 3D space "existing" or "persisting" without the concept of time intertwined  to it does not represent anything close to anything.

1 hour ago, Markus Hanke said:

There is of course no observer in such a universe, since no act of observation is possible - but does that imply that the tea cup cannot exist? Does existence (in the ontological sense) require a temporal dimension? I think not.

Edited by michel123456

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Posted (edited)
57 minutes ago, michel123456 said:

To me (but that might be only me) in order for something to be, time must be there. Either time arises from the existence, or time is a necessity prior to existence, I don't know. But you cannot "be" out of time. You cannot "persist" in something that you want to cancel. It is a wrong concept. IMHO.

To be more pragmatic and step out of pure philosophy, I have the feeling that in the frame of GR it should be possible to show that Time & Space are inseparable. That one cannot establish the concept of Space without Time. That the concept of 3D space "existing" or "persisting" without the concept of time intertwined  to it does not represent anything close to anything.

Let us go back to your desert and savannah example and accept that they can or do exist in and through some amount of time, say from 1900 to 2000 in round numbers.

That is what I mean by persist do you have a different definition.

Now let us take your proposed journey, walking through the desert.

I hope you will agree that the desert persists in space as you move from one place to another ?
(and yes I agree that in this case it also persists in time)

Then you pass from desert to savannah.

Do the places you have just walked through cease to persist?

Was the savannah not persistent just because you were not walking through it?

I look forward to your detailed analysis of the situation as you move through the landcape and from one landscape to another.

I am particularly interested in your reason why you consider there is a difference between being in different  parts of the desert and being at different times.

If you say that you can walk through the desert because different places persist in time,

Why can you not say that you can be in different places because the desert persists through space ?

Edited by studiot

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Markus and Studiot
So in this 'dead' 3D universe without time, there can be change, but it is only apparent to an extrinsic observer ( outside that universe ) ?
That universe itself cannot have an intrinsic observer to detect that change.

That seems to me, to be a problem with our common definition of change; it HAS to be observed.
And therefore time is required.

( sorry Studiot, I'll get to your orbital/jacks question when I have more time to review the questions )

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45 minutes ago, MigL said:

So in this 'dead' 3D universe without time, there can be change, but it is only apparent to an extrinsic observer ( outside that universe ) ?
That universe itself cannot have an intrinsic observer to detect that change.

The whole point of my examples is to show that this is not true.

But in a light hearted way.

Next time I am run over by a bus, I will try (as an intrinsic observer) to determine whether it is just the momentum of the radiator I feel or that of the whole bus.

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6 hours ago, michel123456 said:

To me (but that might be only me) in order for something to be, time must be there.

Well I was not trying to deny this, I was trying to persuade you to temporarily toss it out the window, because it seemed pointless to point it out: Saying “time is there” in order for the cup to persist is extraneous, as I emphasized, “…in the sense that it doesn’t contradict the notion of change as presented by Markus”.

In any case, upon re-examining your previous post you appear to have already conceded that point.

Cheers

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Posted (edited)
6 hours ago, studiot said:

The whole point of my examples is to show that this is not true.

And what I'm failing miserably to explain, is that this is not true because you and I are intrinsic to this universe, which has three spatial dimensions and one of time. But it certainly would be true if we were intrinsic to a universe with only three spatial dimensions.
You could NOT get hit by a bus, or a giant jack, and I'm not sure an orbital even makes sense in such a time devoid universe.
Our definition of 'change' applies to a universe which includes a time dimension, OR, we are extrinsic to a universe which has no time.

IOW, I can look at a 2D sheet of graph paper with a teacup drawn on it, and notice 'change' because I'm extrinsic to that 2d graph.
But if I was a flat-lander intrinsic to that 2D graph sans time, I could not possibly note any change.

Edited by MigL

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8 hours ago, MigL said:

So in this 'dead' 3D universe without time, there can be change, but it is only apparent to an extrinsic observer ( outside that universe ) ?
That universe itself cannot have an intrinsic observer to detect that change.

That seems to me, to be a problem with our common definition of change; it HAS to be observed.
And therefore time is required.

Like Markus, I agree 100% with you that for an observation to take place, there would be some act of a finite speed involving a range of motion that could be measured with a clock. But also, like Markus, I don’t see how it is relevant to his claim about non-homogeneity in the scenario he describes.

Again, in his words: “it is simply the failure of all spatial regions of that universe to be identical”.

When you say, “that seems to me to be a problem with our common definition of change” in some ways I can very much relate.

If you look at a sample of definitions of the intransitive verb as described on the Merriam-Webster online dictionary you get:
“to become different”
“to undergo transformation, transition, or substitution”

these all very much point to a process, which suggests a range of motion. But that is not what Markus is describing. As he says, “It isn’t a process, it’s a relationship.” (Studiot also pointed out that there are other notions of change apart from transformation).

So I get how it conflicts with at least one “common” definition, but I’m not sure about which “common definition” you are speaking of—the one you say “HAS to be observed.”

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Posted (edited)
12 hours ago, michel123456 said:

I have the feeling that in the frame of GR it should be possible to show that Time & Space are inseparable

On a manifold that has both spatial and temporal dimensions, these will indeed by inseparable, and in the sense that they both make an appearance in the metric with opposite signs.
A purely spatial 3D universe wouldn't be locally Lorentz invariant since the metric signature must be either {+,+,+} or {-,-,-}, so GR does not apply here.

12 hours ago, michel123456 said:

That one cannot establish the concept of Space without Time.

GR does not in any sense establish concepts of space and time - it is simply a constraint on the form local geometry can take, given local sources of energy-momentum and appropriate boundary conditions. All I can say here is that GR - as being a tensor equation - does not demand any specific number of dimensions nor metric signature (i.e. mix of spatial and temporal parts) to be valid. So there is nothing from stopping you e.g. to write GR with 17 spatial and 6 temporal dimensions, the field equations would look exactly the same. Whether what is described then bears any resemblance to our own universe is a different question.

But again, this is actually irrelevant, because I didn't demand GR (or any other specific law) to hold in my toy universe.

12 hours ago, michel123456 said:

I find it hard to explain my thought process on this point, as I lack both the formal philosophical background knowledge and the necessary vocabulary to do so. Essentially I am of the opinion that while observation requires existence in the ontological sense (something needs to exist first before anyone can observe it), the reverse is not true in my opinion - I don't see any logical reason why the absence of observers should imply that nothing can exist. Hence, while the tea cup universe is undoubtedly physically unreasonable, it is not philosophically inconsistent.

12 hours ago, michel123456 said:

That the concept of 3D space "existing" or "persisting" without the concept of time intertwined  to it does not represent anything close to anything.

Again, the ontology of existence does not, to me, necessarily imply persistence - but persistence always implies existence.
Of course my toy universe is not physical - I never claimed that it is. It's simply a philosophical thought experiment.

10 hours ago, MigL said:

So in this 'dead' 3D universe without time, there can be change, but it is only apparent to an extrinsic observer ( outside that universe ) ?

There can be no observer whatsoever in such a universe, neither internal nor external, and I did not postulate one in my original example. This is simply a universe with a single tea cup in it, and otherwise complete empty (vacuum). I also do not require it to be embedded in anything. My point was this:

1. Consider the hypothetical universe as an ordered, uncountably infinite set which consists of all physical locations/points (just like ordinary spacetime manifolds, only in 3D)
2. Each element in the set be of type boolean, i.e. either of value 0 (meaning it is vacuum) or 1 (meaning not vacuum)
3. Not all elements of the set are of value 0, because of the tea cup
4. Because not all elements within the set have identical value, this implies a concept of 'change'
5. Since the elements of the set are abstract entities (it is irrelevant what they physically correspond to), the notion of 'change' introduced here is purely a relationship between elements of an abstract set, and thus neither spatial nor temporal in nature

Whether or not this change is observed by anyone is irrelevant, because it makes no difference to the structure of this mathematical set, or the relationships between its elements. I am simply saying that there is nothing special about 'time', so far as change is concerned - you can have change with respect to time, as well as change with respect to any other quantity. Change simply isn't spatiotemporal, it is an abstract, relational concept - in my opinion. So change does not logically imply the existence of time, and conversely time also does not logically imply any change (you can have a complete empty universe without any processes taking place, that nonetheless has temporal dimensions).

10 hours ago, MigL said:

That seems to me, to be a problem with our common definition of change; it HAS to be observed

Why? When the early universe formed, there was no one to observe that (it was too small and dense to contain any observers, it didn't even contain particles), but the change obviously still happened. Of course that was a change with respect to both time and space, but nonetheless is happened without observers, so the above statement is clearly inconsistent.

3 hours ago, MigL said:

But it certainly would be true if we were intrinsic to a universe with only three spatial dimensions.

Again, the example was specifically about a 3D universe that contains only a tea cup, and nothing else. Postulating an observer would have been pointless, since without a temporal dimension, no act of observation is possible. This does not however negate the existence of the tea cup and the vacuum - which aren't identical.

Edited by Markus Hanke

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Posted (edited)

Well,although this seems to be arguing from authority**  if Markus' toy inverse  can be considered on any merit at all we have to concede that not only is it a purely  philosophical structure but also unscientific in that nothing can be predicted about it.

The teacup could persist or it could disappear or reappear.There is absolutely nothing that can be said about it.

In terms of pure non temporal relationships there is nothing to relate the teacup  universe to anything in  the world we observe.

So ,although we can imagine it it is like the belief in deities : one might as well believe in hydrspinacylate gnobbixiis  as  any deity

There is no "output"   one might get from  the scenario(unless the mental process of imagining the teacup somehow lead to a better understanding of related concepts)

Just my two lumps of sugar

** the "authority" being the scientific method.

Edited by geordief

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Posted (edited)
5 hours ago, Markus Hanke said:

There can be no observer whatsoever in such a universe, neither internal nor external, and I did not postulate one in my original example. This is simply a universe with a single tea cup in it, and otherwise complete empty (vacuum). I also do not require it to be embedded in anything.

Yes perhaps my momentum example was flawed, especially since momentum inherently contains time as a 'dimension' in the physics sense.

But before the Science of Electrodynamics there was the Science of Electrostatics.

So let us indtroduce the saucer to your teacup universe, or let us create new static universe exactly the same apart from now including a saucer.

Further let the cup and saucer each have an electrostatic charge.

The the cup will observe the saucer and the saucer will observe the cup via the electrostatic interaction.

Note there is no need for a physics time dimension in this universe.

Electric charge is a fundamental physics dimension in its own right, (although we now use electric current for administrative reasons).

Edited by studiot

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20 hours ago, Markus Hanke said:

I think we’ll have to disagree on this Which is fine of course, this being just a philosophical exchange. To me, the term ‘static’ (I think it should have been ‘stationary’ actually, I often confuse them) implies precisely the absence of any kind of reference to time, be in implicit or explicit.

As long as you just look at a teacup, or a graph of a function, I fully agree with you.

20 hours ago, Markus Hanke said:

This is called a ‘parametrisation’, and it is not what I am doing or suggesting at all.

I would say, yes, you do it, and time is the parametrisation you use.

In my opinion, if somebody says that 'y changes as function of the change of x', she is saying 'if you change x from a to b, then y changes according f(a) to f(b)'. But a change of x is an activity of an observer, and observers exist in time.

And concerning your points with MigL: of course change can occur without observers. But observing a change means either a passive observer sees things changing (you are sitting quietly at a veranda, and you see the streets changing from dry to wet (a very common experience in Ireland...)) or you look at an non-changing object, but you let wander your gaze from one place of the object to another, and e.g. see how the colour 'changes' dependent on the place of the object where you look. But the latter also means a change in time. But the change is in the observer, not in the object. The object is static.

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9 minutes ago, Eise said:

But a change of x is an activity of an observer, and observers exist in time.

Are you suggesting that change is the only activity of an observer ?

My previous post suggests otherwise.

You have also coupled this to a second unproven statement "and observers exist in time."

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Wish I could give you two up-votes Eise.

I think, Studiot, that our definition of 'change' references the observer, not the universe being 'observed'.
WE need to note a difference in order to call it 'change' )
And that is simply because the concept of 'change' has arisen in a universe with time.

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6 minutes ago, studiot said:

Are you suggesting that change is the only activity of an observer ?

No, of course not. I said:

21 minutes ago, Eise said:

But a change of x is an activity of an observer, and observers exist in time.

Maybe I should have written that activities exist in time, and therefore a 'change of x' exists in time. So your example of the electrostatic objects does not work.

12 minutes ago, studiot said:

You have also coupled this to a second unproven statement "and observers exist in time."

Yes, because only observers that exist in time can say 'if you change x, then y changes according f(x)'. Otherwise, an observer would just see an object, a static form.

I think the subtle point here is that one must take care to distinguish between what we observe, and how we speak about it. I might be wrong, but I have the impression that you and Markus discuss what we observe, and I talk about how we speak about an observation.

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25 minutes ago, MigL said:

Wish I could give you two up-votes Eise.

I think, Studiot, that our definition of 'change' references the observer, not the universe being 'observed'.
WE need to note a difference in order to call it 'change' )
And that is simply because the concept of 'change' has arisen in a universe with time.

No, no and no.

Marcus has already made the point that the change is there whether it is observed or not.

He has also introduced a universe with no observers and no time, but I did think it was now accepted that the teacup universe contains change.

15 minutes ago, Eise said:

Maybe I should have written that activities exist in time, and therefore a 'change of x' exists in time. So your example of the electrostatic objects does not work.

Activities can only exist in time in a universe that includes time.

You have simply declared that my cup and saucer universe does not  work.
You have most definitely not demonstrated that this is the case.

Observations are simply a scientific term for interactions between distince entities.

As Markus points out, an interaction cannot occur in a univese with only one occupant.
That is why is was necessary to introduce another.

In the case of my static universe the interaction will never change, since there is no time.

Yet change exists in that universe since each observer can determine the sign of the interaction.

So a difference of charge sign will inevitably lead to a particular interaction sign and the existence of any difference is a change.

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35 minutes ago, studiot said:

Yet change exists in that universe since each observer can determine the sign of the interaction.

Can you explain this more. I don't get what you are trying to say.

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The concept of 'change' only makes sense in a universe with time.
That is why we can observe a universe without time, such as the teacup universe, or the electrostatic universe, and notice change.
we are 'outside' those peculiar universes.

That is because the concept of 'change', ( in effect, for change to be observed ) has to involve the paradigm of time passing.
So we, in this universe notice change even in the teacup/electrostatic universes, but those universes would never have developed the concept of 'change'. How could they ?

( by the way, that's a lot of nos. Was it for emphasis, or three nos for particular points ? )

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text:
#### Money 2007 fails install on Vista RTM

Howdy!

I would love some advice on this.  On Friday I tried to reinstall Money 2007
on my freshly built Vista RTM machine, and the installation fails.

Tried about 8 times.. always fails about 98% of the way through the install,
and this is after a warning screen at initial install launch saying that this
application has known issues.

The bummer is that I had 2007 running on my machine on the Vista beta's but
the key there was likely that I installed 2007 on XP and the vista's were
upgrades.

Any hope, any suggestions?  I *need* to get this running as I am now daily
dependant on this application, and am dyin with it down.

 0
Raichean (21)
11/20/2006 8:50:02 PM
money 28651 articles. 7 followers.

6 Replies
453 Views

Similar Articles

[PageSpeed] 45

The last I heard was that MS need to issue a SP to enable M2007 to work with
Vista. Unfortunately I haven't heard any date for this.

Not much consolation to you but some people say they've had no problem.

--
Regards
Bob Peel,
Microsoft MVP - Money

For UK tips & fixes see
http://support.microsoft.com/default.aspx?scid=fh;EN-GB;mny.

I do not respond to any emails that I have not specifically asked for.

"Raichean" <Raichean@discussions.microsoft.com> wrote in message
news:0295273E-FFC8-4D80-883F-CFC6070BDF79@microsoft.com...
> Howdy!
>
> I would love some advice on this.  On Friday I tried to reinstall Money
> 2007
> on my freshly built Vista RTM machine, and the installation fails.
>
> Tried about 8 times.. always fails about 98% of the way through the
> install,
> and this is after a warning screen at initial install launch saying that
> this
> application has known issues.
>
> The bummer is that I had 2007 running on my machine on the Vista beta's
> but
> the key there was likely that I installed 2007 on XP and the vista's were
> upgrades.
>
> Any hope, any suggestions?  I *need* to get this running as I am now daily
> dependant on this application, and am dyin with it down.

 0
bob_peel2056 (736)
11/20/2006 9:14:40 PM
Thanks and yeah.. huge bummer on this.  I need to get it working asap and was
hoping there was some ninja secret squirrel hint that could get me past that
last hurdle.

Stefan

"Bob Peel, MVP" wrote:

> The last I heard was that MS need to issue a SP to enable M2007 to work with
> Vista. Unfortunately I haven't heard any date for this.
>
> Not much consolation to you but some people say they've had no problem.
>
> --
> Regards
> Bob Peel,
> Microsoft MVP - Money
>
> For UK tips & fixes see
> http://support.microsoft.com/default.aspx?scid=fh;EN-GB;mny.
>
>
> I do not respond to any emails that I have not specifically asked for.
>
> "Raichean" <Raichean@discussions.microsoft.com> wrote in message
> news:0295273E-FFC8-4D80-883F-CFC6070BDF79@microsoft.com...
> > Howdy!
> >
> > I would love some advice on this.  On Friday I tried to reinstall Money
> > 2007
> > on my freshly built Vista RTM machine, and the installation fails.
> >
> > Tried about 8 times.. always fails about 98% of the way through the
> > install,
> > and this is after a warning screen at initial install launch saying that
> > this
> > application has known issues.
> >
> > The bummer is that I had 2007 running on my machine on the Vista beta's
> > but
> > the key there was likely that I installed 2007 on XP and the vista's were
> > upgrades.
> >
> > Any hope, any suggestions?  I *need* to get this running as I am now daily
> > dependant on this application, and am dyin with it down.
>
>

 0
Raichean (21)
11/20/2006 9:26:01 PM
Sorry can't help. You *of course* read the "before you start" warnings and
backed up your hard drive before installing Vista, so you can go back to
that?!?

--
Regards
Bob Peel,
Microsoft MVP - Money

For UK tips & fixes see
http://support.microsoft.com/default.aspx?scid=fh;EN-GB;mny.

I do not respond to any emails that I have not specifically asked for.

"Raichean" <Raichean@discussions.microsoft.com> wrote in message
news:13AD61F8-0860-4F32-AFEE-C7733F0B3BC5@microsoft.com...
> Thanks and yeah.. huge bummer on this.  I need to get it working asap and
> was
> hoping there was some ninja secret squirrel hint that could get me past
> that
> last hurdle.
>
> Stefan
>
> "Bob Peel, MVP" wrote:
>
>> The last I heard was that MS need to issue a SP to enable M2007 to work
>> with
>> Vista. Unfortunately I haven't heard any date for this.
>>
>> Not much consolation to you but some people say they've had no problem.
>>
>> --
>> Regards
>> Bob Peel,
>> Microsoft MVP - Money
>>
>> For UK tips & fixes see
>> http://support.microsoft.com/default.aspx?scid=fh;EN-GB;mny.
>>
>>
>> I do not respond to any emails that I have not specifically asked for.
>>
>> "Raichean" <Raichean@discussions.microsoft.com> wrote in message
>> news:0295273E-FFC8-4D80-883F-CFC6070BDF79@microsoft.com...
>> > Howdy!
>> >
>> > I would love some advice on this.  On Friday I tried to reinstall Money
>> > 2007
>> > on my freshly built Vista RTM machine, and the installation fails.
>> >
>> > Tried about 8 times.. always fails about 98% of the way through the
>> > install,
>> > and this is after a warning screen at initial install launch saying
>> > that
>> > this
>> > application has known issues.
>> >
>> > The bummer is that I had 2007 running on my machine on the Vista beta's
>> > but
>> > the key there was likely that I installed 2007 on XP and the vista's
>> > were
>> > upgrades.
>> >
>> > Any hope, any suggestions?  I *need* to get this running as I am now
>> > daily
>> > dependant on this application, and am dyin with it down.
>>
>>

 0
bob_peel2056 (736)
11/21/2006 6:19:43 AM
What version of Money 2007 are running?  I am running Money 2007 Deluxe on
Vista RTM just fine.  I do get the prompt to upgrade constantly, but still
need to try the solution here: http://support.microsoft.com/kb/927811

I haven't yet installed the SP update, since I had tried once and then it
would never recognize me as online to sign in.  Going to try again over the
weekend.

Money does work on Vista, with a few minor hitches, though I had no problems
with the install on the Deluxe edition.

 0
ken1271 (1)
11/21/2006 4:30:08 PM
Bob, I appreicate that, but I am certainly not going to take off the entire
OS just for Money.  In theory, could I reinstall XP, install Money, upgrade
Vista instead of fresh install?  Of course.. is that a reasonable
alternative.

I was hoping there might be some suggestions, but apparently their aren't.

"Bob Peel, MVP" wrote:

> Sorry can't help. You *of course* read the "before you start" warnings and
> backed up your hard drive before installing Vista, so you can go back to
> that?!?
>
> --
> Regards
> Bob Peel,
> Microsoft MVP - Money
>
> For UK tips & fixes see
> http://support.microsoft.com/default.aspx?scid=fh;EN-GB;mny.
>
>
> I do not respond to any emails that I have not specifically asked for.
>
> "Raichean" <Raichean@discussions.microsoft.com> wrote in message
> news:13AD61F8-0860-4F32-AFEE-C7733F0B3BC5@microsoft.com...
> > Thanks and yeah.. huge bummer on this.  I need to get it working asap and
> > was
> > hoping there was some ninja secret squirrel hint that could get me past
> > that
> > last hurdle.
> >
> > Stefan
> >
> > "Bob Peel, MVP" wrote:
> >
> >> The last I heard was that MS need to issue a SP to enable M2007 to work
> >> with
> >> Vista. Unfortunately I haven't heard any date for this.
> >>
> >> Not much consolation to you but some people say they've had no problem.
> >>
> >> --
> >> Regards
> >> Bob Peel,
> >> Microsoft MVP - Money
> >>
> >> For UK tips & fixes see
> >> http://support.microsoft.com/default.aspx?scid=fh;EN-GB;mny.
> >>
> >>
> >> I do not respond to any emails that I have not specifically asked for.
> >>
> >> "Raichean" <Raichean@discussions.microsoft.com> wrote in message
> >> news:0295273E-FFC8-4D80-883F-CFC6070BDF79@microsoft.com...
> >> > Howdy!
> >> >
> >> > I would love some advice on this.  On Friday I tried to reinstall Money
> >> > 2007
> >> > on my freshly built Vista RTM machine, and the installation fails.
> >> >
> >> > Tried about 8 times.. always fails about 98% of the way through the
> >> > install,
> >> > and this is after a warning screen at initial install launch saying
> >> > that
> >> > this
> >> > application has known issues.
> >> >
> >> > The bummer is that I had 2007 running on my machine on the Vista beta's
> >> > but
> >> > the key there was likely that I installed 2007 on XP and the vista's
> >> > were
> >> > upgrades.
> >> >
> >> > Any hope, any suggestions?  I *need* to get this running as I am now
> >> > daily
> >> > dependant on this application, and am dyin with it down.
> >>
> >>
>
>

 0
Raichean (21)
11/21/2006 5:05:03 PM
Ok.. well bizarre, but the install worked perfectly today.  *shrug*  It
works.. that is all I care about at this point!  8)

Stefan

"Ken Robertson" wrote:

> What version of Money 2007 are running?  I am running Money 2007 Deluxe on
> Vista RTM just fine.  I do get the prompt to upgrade constantly, but still
> need to try the solution here: http://support.microsoft.com/kb/927811
>
> I haven't yet installed the SP update, since I had tried once and then it
> would never recognize me as online to sign in.  Going to try again over the
> weekend.
>
> Money does work on Vista, with a few minor hitches, though I had no problems
> with the install on the Deluxe edition.
>

 0
Raichean (21)
11/21/2006 5:32:02 PM
 Reply:

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Hi there everyone, I pray someone can help me before I throw my mac through the window. This is the second time I am having this problem. My office just stopped working and no matter what I did I could not open any of the programs, so I uninstalled everything, Microsoft and the preferences. I have now tried to re-install office off disk and I can't even get it to launch I keep getting an error message saying installer created an error and please report which I have done a thousand times. I have restarted the mac and emtied the Trash so there is nothing relating to office...

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What am I doing wrong? Trying to set up my MSN-Vista-Microsoft e-mail. Get an error message about the POP3 server. I give my incoming and outgoing mail server as POP3 and service provider TWC or Time Warner Cable and the error message comes up. What should I do? Can you help me? On Tue, 7 Aug 2007 19:27:09 -0700, "asilocreo" <asilocreo@microsoft.com> wrote: >What am I doing wrong? >Trying to set up my MSN-Vista-Microsoft e-mail. Get an error message about >the POP3 server. I give my incoming and outgoing mail server as POP3 and >service provider TWC or Ti...

open excel 2007 refresh query from batch file
Hello all, I have a simple script saved as "C:\refresh_excel.vbs" that opens an Excel file, and does a refresh for a query on Sheet 1. I created a scheduled task to run this every day. It worked fine for previous versions of Excel, but since I have upgraded to Excel 2007 it doesn't work. Any help it helping me figure this out would be appreciated. /*contents of refresh_excel.vbs*/ Set objExcel = CreateObject("Excel.Application") objExcel.Visible = True Set objWorkBook = objExcel.Workbooks.Open("C:\reports\UPS.xls") objWorkbook.Sheets("...

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I have an issue using Publisher 2007 on Windows Vista 32bit. When attempting to print (or print preview) the user interface locks up and my only option is to ctl-alt-del. I cannot switch windows, I cannot bring up task manager. The UI still displays updates (outside of Publisher), but I cannot "use" any of the interface except the Windows Security menu via ctl-alt-del. If I attempt a logout I can force Publisher to stop. This is followed by two logged events: --- 1. Source: Office 12, Event ID 5000 The description for Event ID 5000 from source Microsoft Office 12 cannot be ...

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I have a clean install of PS2007 - SP2 on WSS in a small farm environment - DB server on another box. My installation "mostly" works, however, I am receiving an "unknown error" message when I try to edit the RBS. The item gets checked out and the "unknown error" message appears on the top of the page with only the Cancel button available. I've checked the logs and don't see anything obvious, the event logs on both machines aren't showing anything and I'm fairly stumped. Any insight, ideas, things to try would be greatly appreciated. Thanks in adv...

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i downloaded the ms money 2006 trial, and did the conversion from quicken 2005. all seems well. except i have 11,000+ transactions with missing categories i've tried removing all categories in money 2006 before the conversion, same number of transactions missing categories i've tried loading all the categories in money 2006, and doing the conversion. same number of transactions any help? this is definately a show stopper :-( jd In microsoft.public.money, john dekker wrote: >i downloaded the ms money 2006 trial, and did the conversion from quicken >2005. all seems wel...

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Hi, I have an Exchange 2007 server behind a NAT router. I want to configure RPC over HTTP (Outlook anywhere) on it. HTTP/HTTPS is working on it as expected (the NAT router forwards ports 443 and 80 to Exchange's internal IP address). I have enabled Outlook Anywhere in the Exch. management console, specified external hostname and clicked the "Basic Auhtentication" checkbox (from what I understand that's required in order to make it work over SSL) - the SSL offloading is not enabled. I'm trying to configure Outlook 2003 to connect to it but I haven't been...

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Why does MS Money insist on connecting to the internet when you open a file? Can I make it not do this? What about users that aren't connected? A1) Apparently because Microsoft thinks Most Users Prefer(tm) it this way. A2) No that anybody knows of short of disabling the network. A3) It fails to connect. "pickle" <pickle@comcast.net> wrote in message news:SrKdnaq2IfCqKZvfRVn-pQ@comcast.com... > Why does MS Money insist on connecting to the internet when you open a > file? Can I make it not do this? > > What about users that aren't connected? > ...

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Hi, I have purchased Money 2006 standard edition and during the startup it offered to upgrade to Delux. I have tried several times following the links. That’s what I'm getting when I try tu upgrade: We're sorry; there are no plans available. This can happen for one of the following reasons: You currently have this plan or a similar plan. You have previously purchased this plan. This plan is not available in your Country or Region. And right after that the plan is listed as Microsoft Money 2006 Standard. Any ideas? --Venelin money sucks "vvoykov" <vvoykov@discuss...

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I recently upgraded to Excel 2007 from 2002, and a lot of the spreadsheets seem to have something extra in the rows below the end of the data. Two examples: 1) I went to the bottom of the data in a spreadsheet and made a new entry. As soon as I hit enter, the spreadsheet started scrolling down and wouldn't stop for a couple of thousand rows. 2) I saved a spreadsheet as a csv file and another application that uses the file started having problems. So I opened the csv file in WordPad and saw a couple of hundred rows of commas. That never happened with Excel 2002. I opened the ...