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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJun 15th 2010
• (edited Jun 15th 2010)
Has anyone been following the long series of articles by Nishimura on generalized smooth spaces?

He relates Frölicher-space technology with synthetic differential geometry. My impression from a quick glance is that he is following the basic approach of

- Anders Kock, _Convenient vector spaces embed into the Cahiers Topos_

In fact, looking at that I see that I missed all along that I had been re-inventing the wheel. The site ThCartSp (infinitesimally thickened Cartesian spaces) that I have been so happily studying oo-shaves on is the site for the Cahiers-topos and well-studied. I had not known that!
• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeJun 15th 2010

Hm, Nishimura has a special way of writing. In various articles he proceeds from conjectures. Not sure yet what to make of all this.

• CommentRowNumber3.
• CommentAuthorAndrew Stacey
• CommentTimeJun 15th 2010

The link to the articles got mangled somehow and didn’t make it through the parser. Could you repost it, please? If you’re cut-and-pasting from somewhere, use the MarkdownItex formatter as it handles urls better (you appear to have used the Html formatter in the above). The native Markdown link syntax is [text](url).

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeJun 15th 2010
• (edited Jun 15th 2010)

use the MarkdownItex formatter as it handles urls better (you appear to have used the Html formatter in the above).

I switched to HTML after Mardown didn’t work! But HTML didn’t work either.

This is from a general experience I have on the nLab: if I link to long page names with extra symbols and anchors, then the Markdown syntax gets mixed up, while HTML still works.

Anyway, the link was simply to the list of articles that one gets when going to arXiv and searching for author name “Nishimura”.

• CommentRowNumber5.
• CommentAuthorAndrew Stacey
• CommentTimeJun 15th 2010

Do you paste the links in literally, as text? If so, all the filters are going to complain. The absolute best way is to use the native Markdown syntax that I outlined above: [text](link) since then Markdown intelligently replaces the ampersands and stuff. For example, this discussion has an ampersand in it.

• CommentRowNumber6.
• CommentAuthorAndrew Stacey
• CommentTimeJun 15th 2010

Back on track, thanks for reminding me of those papers - I’ve been bookmarking them to read later but haven’t gotten round to it. I’ll bump them up the list!

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeJun 15th 2010

Andrew,

when I put links into nLab pages to subsections of other pages, very frequently the method that you recommend fails, while the method that you say will fail works fine for me! :-)

It’s interesting that you suggest the opposite. Now I am wondering what’s going on.

• CommentRowNumber8.
• CommentAuthorAndrew Stacey
• CommentTimeJun 16th 2010


text:
+0

# Inequaity

0
153
1

What interval consists of all u such that neither 2u nor -20u is in the interval $$(-10,10)$$?

May 15, 2022

#1
+9462
0

If 2u is not in the interval, then $$2u \leq -10 \text{ or }2u \geq 10$$. That means $$u \leq -5 \text{ or }u \geq 5$$.

Moreover, -20u is not in the interval, so $$-20u \leq -10 \text{ or }-20u \geq 10$$. That means $$u \leq -\dfrac12 \text{ or }u \geq \dfrac12$$.

If $$u \leq -5 \text{ or }u \geq 5$$, then it is true that $$u \leq -\dfrac12 \text{ or }u \geq \dfrac12$$. So $$u \leq -\dfrac12 \text{ or }u \geq \dfrac12$$ is extraneous. The range of values of u is $$u \leq -5 \text{ or }u \geq 5$$. Expressed in interval notation, $$u \in (-\infty, -5]\cup [5, \infty)$$.

May 15, 2022

format: plain


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# Basic Special Relativity Question

Discussion in 'Physics & Math' started by Fednis48, Apr 22, 2013.

1. ### Fednis48Registered Senior Member

Messages:
725
This problem has been extracted from this thread because Tach refuses to address it unless it's in a new topic. The question is thus:

In a relativistically moving train, a metal rod is dropped from the ceiling, parallel to the floor. To stay in the realm of special relativity, let's say it's moving down at constant velocity rather than falling under gravity. In the reference frame of a stationary platform watching the train zoom by, does the rod stay parallel to the ground as it falls? The following calculation is Pete's attempt at a solution, and it concludes that the rod is not parallel to the ground in the platform frame.

$S =$ rest frame of the rod
$S' =$ rest frame of the train
$S'' =$ rest frame of the platform

$v = (0, u)$ is the velocity of $S'$ relative to $S$
$v' = (V, 0)$ is the velocity of $S''$ relative to $S'$
$\gamma = 1/\sqrt{1-u^2/c^2}$
$\gamma' = 1/\sqrt{1-V^2/c^2}$

The rod has two ends, A and B. In the rod frame, we place them at (x,y) = (0,0) and (1,0) respectively.

In $S$:
\begin{align} A &= (x_A, \ y_A) \\ &= (0, \ 0) \end{align}
\begin{align} B &= (x_B, \ y_B) \\ &= (1, \ 0) \end{align}
The angle of the rod with the x-axis at time t is:
\begin{align} \tan(\theta) &= \frac{y_B-y_A}{x_B-x_A} \\ &= 0 \end{align}

In $S'$:
\begin{align} A' &= (x_A'(t'), \ y_A'(t')) \\ &= (x_A, \ \frac{y_A}{\gamma} - ut') \\ &= (0, \ -ut') \end{align}
\begin{align} B' &= (x_B'(t'), \ y_B'(t')) \\ &= (x_B, \ \frac{y_B}{\gamma} - ut') \\ &= (1, \ -ut') \end{align}
The angle of the rod with the x-axis at time t' is:
\begin{align} \tan(\theta') &= \frac{y_B'(t') - y_A'(t')}{x_B'(t')-x_A'(t')} \\ &= 0 \end{align}

In $S''$:
\begin{align} A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\ &= \left(\frac{x_A'}{\gamma'} - Vt'', \ -ut' \right) \\ &= \left(-Vt'', \ -u\gamma'(t'' + \frac{Vx_A''}{c^2}) \right ) \\ &= \left(-Vt'', \ \frac{-ut''}{\gamma'} \right) \end{align}
\begin{align} B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\ &= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \right) \end{align}
The angle of the rod with the x-axis at time t'' is:
\begin{align} \tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\ &= \frac{uV\gamma' / \gamma c^2}{1 / \gamma'} \\ &= \frac{uV\gamma'^2}{\gamma c^2} \end{align}

3. ### PeteIt's not rocket surgeryModerator

Messages:
10,166
Thanks Fednis48.
Tach, I'd particularly like to know whether you think there are any mistakes with the transformation of $A'$ to $A''$ and $B'$ to $B''$.

5. ### Neddy BateValued Senior Member

Messages:
1,403
Another way to look at it is to consider that if the rod is parallel to the floor, then both ends of the rod must impact the floor simultaneously. And conversely, that if the ends of the rod do not impact the floor simultaneously, then the rod must not have been parallel to the floor.

In the train frame, let both ends of the rod impact the floor simultaneously at time t'. Let one end of the rod impact the floor at location x', and let the other end of the rod impact the floor at location x'+L where L is the proper length of the rod.

By the Lorentz transformation, the platform frame measures the time of impact of one end of the rod to be t_1:
$t_1 = \gamma(t' + \frac{vx'}{c^2})$

and the platform frame also measures the time of impact of the other end of the rod to be t_2:
$t_2 = \gamma(t' + \frac{v(x'+L)}{c^2})$

In order for the rod to be parallel to the floor, both ends of the rod must rod impact the floor simultaneously, thus t_1 must equal t_2:
$t_1 = t_2$

which lets us set the right hand sides of the first wo equation equal to each other:
$\gamma(t' + \frac{vx'}{c^2}) = \gamma(t' + \frac{v(x'+L)}{c^2})$

which simplifies to:
$x' = x'+L$

which simplifies to:
$L = 0$

Demonstrating that, according to the platform frame, the ends of the rod can only impact the floor simultaneously if the length of the rod is zero L=0. Thus, for any rod of non-zero length, the platform frame says the ends of the rod do not impact the floor simultaneously, and therefore such a rod cannot be parallel to the floor, according to the platform frame.

7. ### PeteIt's not rocket surgeryModerator

Messages:
10,166
Right, the essence of the problem is in two questions Tach refused to answer in the other thread:

Do the ends of the rod hit the floor simultaneously in the train frame?
Do the ends of the rod hit the floor simultaneously in the platform frame?

8. ### Fednis48Registered Senior Member

Messages:
725
I think it looks solid. The step where you write y'' as $-ut'$ is a little odd, since t' isn't well-defined in the platform frame. So there's probably a more precise way to write it. But the conclusion looks right.

9. ### PeteIt's not rocket surgeryModerator

Messages:
10,166
Yeah, I wasn't really happy with that either. I considered writing t'(x'', t'') to indicate that in S'', t' depends on x'' and t'', but it looked cluttered.

10. ### Fednis48Registered Senior Member

Messages:
725
Ahh, gotcha. That would look cluttered; the mere fact that you went through it in that more precise way satisfies my puzzlement.

11. ### TachBannedBanned

Messages:
5,265
Basic algebra would give the result $-\frac{uV\gamma'}{c^2}$. (not that the physics treatment is correct, either).
You may want to check your math. The fact that Neddy Bate and Fednis48 agree with you result speaks volumes.
Once you figure out where your new error is , we can talk about the rest.

Messages:
5,265

13. ### FarsightValued Senior Member

Messages:
3,475
You can maybe simplify matters somewhat by imagining two separated light sources A and B which are turned on at the same time according to the stationary observer in the middle. However the moving observer who is instantaneously in the middle heading from A to B will claim that B was turned on before A. Put A and B on the ends of a falling rod, and the moving observer will claim that the B end hits the ground first. Flipping it around and putting the rod on the moving train then tells you that the rod is not parallel to the ground in the platform frame.

14. ### Neddy BateValued Senior Member

Messages:
1,403
I'm very happy that you are now on record calling the Lorentz transformations "garbage." Speaks volumes about you.

The only thing I "conceded" in the other thread was that I was neglected gravity. I'm glad you popped into this thread to go on record that you still think gravity must be considered, even though gravity is not part of the problem in this thread.

Yeah! Go Tach!

15. ### eramSciengineerValued Senior Member

Messages:
1,875
Careful...Hell hath no fury like a Tach scorned.

16. ### Fednis48Registered Senior Member

Messages:
725
Edit: My bad, that's not right! There is a $\gamma'^2$ in the numerator and a $\gamma$ in the denominator. One is primed and the other is not, so they don't cancel. Pete's algebra is right, as far as I can tell. We specifically spun this thread off at your request to get you to point out errors, so if you see any, please identify them precisely ("basic algebra" is not precise enough, because it doesn't indicate which step(s) in the math are mistaken).

Original Post: Good point. Since the question of this thread is basically "does $\theta=0$", I was just focused on the numerator of the fraction, but the denominator turns into $1/\gamma$, which Pete failed to include. Dunno where you got that overall minus sign, though; I find the result to be $\frac{uV\gamma'}{c^2}$. Of course, both of our expressions are still nonzero.

Last edited: Apr 22, 2013
17. ### Neddy BateValued Senior Member

Messages:
1,403
LOL, his standard is to spew venom with every post, so how much worse can it get?

18. ### eramSciengineerValued Senior Member

Messages:
1,875
I don't wanna find out lol

19. ### TachBannedBanned

Messages:
5,265
I understand that you desperately want to debate but your basic algebra skills get in the way. This is very basic math.....

As far you can tell...you are wrong. Want to try again?

20. ### TachBannedBanned

Messages:
5,265
how about you helped the other two guys, they don't get it. Neither do you.

21. ### TachBannedBanned

Messages:
5,265
No, I called your reasoning, garbage, let's be precise.

22. ### Neddy BateValued Senior Member

Messages:
1,403
See his two posts above this. Not too bad, because he doesn't even bother to demonstrate his claims.

23. ### PeteIt's not rocket surgeryModerator

Messages:
10,166
Hi Tach,
Please point out any errors you see, so we can fix them. I'd also like to know what problems you see with the physical treatment.
I don't want to waste time, I just want to find the right answers.


text:
# Bernard Sugar Free Biscuit Mix

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This low-sodium, sugar free, low-cholesterol Biscuit Mix was especially developed and is manufactured for use in all types of dietary feeding, particularly low-sodium feeding. While this fine mix can be used for numerous restricted diets, it is interesting to note that if only one type of restriction is necessary, the fine balance of the finished product can be utilized for any other non-restricted diet. In fact, this is a satisfactory mix for every day non-restricted use.


text:
## EPA updates air pollution standards for petroleum refineries; fenceline monitoring

##### 30 September 2015

The US Environmental Protection Agency (EPA) has updated air pollution standards to further control toxic air emissions from petroleum refineries. The rule requires first-of-its-kind fenceline monitoring to better protect and inform nearby communities, while also strengthening emission controls for flares, pressure relief devices, storage tanks, and delayed coker operations that will reduce thousands of tons of hazardous air pollutants.

When fully implemented, the rule will result in a reduction of 5,200 tons per year of toxic air pollutants, and 50,000 tons per year of volatile organic compounds (VOCs). Also, as a co-benefit of these final standards, EPA projects that these standards will eliminate emissions of greenhouse gases equivalent to approximately 660,000 tons per year of CO2.

The EPA estimates the capital cost of this final rule to be approximately $283 million, with an annualized cost of approximately$63 million. The EPA estimates that these final standards will have a negligible impact on the costs of petroleum products.

The action requires continuous monitoring of benzene concentrations at the fenceline of refinery facilities to ensure that refineries appropriately manage toxic emissions. The rule requires corrective action to protect neighboring communities from being exposed to harmful levels of emissions if the established standard level is exceeded. The new fenceline monitors must encircle the facility to detect benzene at very low levels, and the monitoring data will be posted on EPA’s website. In addition, in response to public feedback, the rule provides room for alternative monitoring methods in the future as technology advances.

Other specific requirements in this rule will virtually eliminate visible flare emissions and releases by pressure release devices by requiring a comprehensive program of process changes and pollution prevention measures for these emission sources. It will also require additional emission reductions from storage tanks and delayed coking units, some of which had no previous required controls.

This final rule is based on the risk and technology review of two emissions standards already in place at refineries: the National Emission Standards for Hazardous Air Pollutants From Petroleum Refineries (Refinery MACT 1) and the National Emission Standards for Hazardous Air Pollutants for Petroleum Refineries: Catalytic Cracking Units, Catalytic Reforming Units, and Sulfur Recovery Units (Refinery MACT 2).

This rulemaking was informed by a Petroleum Refinery Risk and Technology Review (RTR). During this scientific and technical review, EPA also engaged with key communities in the two years prior to proposing a regulation in 2014.


text:
# Composition of Lipschitz and differentiable functions is also differentiable

The proposition that I'm trying to prove is:

Let $f:U\rightarrow \mathbb R^n$ be Lipschitz defined on $U\subset \mathbb R^m$ an open set and $a\in U$. Let $g:V\rightarrow \mathbb R^p$ differentiable in the open set $V\subset \mathbb R^m$ with $f(U)\subset V$, and $b=f(a)$. If $g'(b)=0$, then $g\circ f:U\rightarrow \mathbb R^p$ is differentiable in $a$, and $(g\circ f(a))'=0$.

I've tried to prove by the definition, i.e., that $g\circ f(a+v)=g\circ f(a)+(g\circ f(a))'v+r(v)$, with $\lim_{v\rightarrow0}\frac{r(v)}{|v|}=0$.

So what I got is $\frac{r(v)}{|v|}=\frac{g( f(a+v))-g( f(a))}{|v|}$, I don't know how to solve this limit, I've tried to use the Lipschitz condition to change $|v|$ to $|f(a+v)-f(a)|$, and get $k\frac{g( f(a+v))-g( f(a))}{|f(a+v)-f(a)|}$, $k$ the Lipschitz constant, but I can't go further. Any tips?

If $f(a+v)=f(a)$ then $r(v)=0$. If $f(a+v)\ne f(a)$, then $$\left\vert\frac{r(v)}{|v|}\right\vert=\left\vert\frac{g(f(a+v)-g(f(a))}{|f(a+v)-f(a)|}\right\vert\frac{|f(a+v)-f(a)|}{|v|}\le k \frac{\vert g(f(a+v)-g(f(a))\vert}{|f(a+v)-f(a)|}.$$ Using the fact that $g'(b)=0$ you have that $\frac{\vert g(y)-g(b)\vert}{|y-b|}\le\varepsilon$ whenever $|y-b|\le \delta$. Now, since $f$ is continuous, $f(a+v)\to f(a)=b$ when $v\to 0$ and so you can find $\eta>0$ such that if $|v|\le\eta$, then $|f(a+v)-b|\le \delta$. In turn, $$\frac{\vert g(f(a+v)-g(f(a))\vert}{|f(a+v)-f(a)|}\le \varepsilon.$$


text:
#### Still need help?

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## Workaround

Note that if you try to open an attachment in a secure site, for example a PDF document, the relevant application will try to open it but will report the file cannot be found. However, if you right click the document and choose "Save as" this will work. To fix this, in IE go to the "Tools>Internet Options" menu and click on the Advanced Tab, scroll down to the bottom and look for "Do not save encrypted pages to disk" option - make sure that you un-tick this. This problem is known to affect IE6 and has been reported in Microsoft's knowledge base.

## Cause

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text:
# Decrypting an Affine Cipher with known characters

I am trying to cryptanalyse multiple cipher–texts that I know are encrypted by different Affine ciphers. I have already analysed the frequency that each character occurs, and compared it to a frequency table of the languages the plain-texts could be in and as a result have a pretty decent idea of what a few characters could correspond to. For example I know in one cipher-text:

X = e, S = t, V = s

Converting those into their numerical correspondences yields

23 = 4, 18 = 19, 21 = 18

I have therefore deduced that:

$4a+b=23 \pmod{26}$

$19a+b=18 \pmod{26}$

$18a+b=21 \pmod{26}$

I figure I could probably calculate the values of a and b by manually trying every iteration of a (which wouldn't be terrible since I can rule out all even values) and hoping once they all match with the same b it works for all characters, but I was just wondering if there was a faster/more reliable way of doing it Mathematically. Is there some way I can solve for $a$ and $b$ that I am unaware of?

If you subtract equation 3 from equation 2, you get $a=-3$, and can solve for $b=75$. This fits equation 2 and 3, but not equation 1. So, no solution exists.


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# American Institute of Mathematical Sciences

2015, 12(4): 739-760. doi: 10.3934/mbe.2015.12.739

## Optimal design for parameter estimation in EEG problems in a 3D multilayered domain

 1 Center for Research in Scientific Computation, North Carolina State University, Raleigh, NC 27695-8212 2 Centro de Matemática Aplicada, Universidad de San Martín, Buenos Aires, Argentina 3 Instituto de Ciencias, Universidad Nacional Gral. Sarmiento, Buenos Aires, Argentina 4 Dep. de Matemática, Facultad de Ingeniería, Universidad de Buenos Aires, Argentina

Received  April 2014 Revised  December 2014 Published  April 2015

The fundamental problem of collecting data in the best way'' in order to assure statistically efficient estimation of parameters is known as Optimal Experimental Design. Many inverse problems consist in selecting best parameter values of a given mathematical model based on fits to measured data. These are usually formulated as optimization problems and the accuracy of their solutions depends not only on the chosen optimization scheme but also on the given data. We consider an electromagnetic interrogation problem, specifically one arising in an electroencephalography (EEG) problem, of finding optimal number and locations for sensors for source identification in a 3D unit sphere from data on its boundary. In this effort we compare the use of the classical $D$-optimal criterion for observation points as opposed to that for a uniform observation mesh. We consider location and best number of sensors and report results based on statistical uncertainty analysis of the resulting estimated parameters.
Citation: H. T. Banks, D. Rubio, N. Saintier, M. I. Troparevsky. Optimal design for parameter estimation in EEG problems in a 3D multilayered domain. Mathematical Biosciences & Engineering, 2015, 12 (4) : 739-760. doi: 10.3934/mbe.2015.12.739
##### References:

show all references

##### References:
 [1] Xuhui Peng, Rangrang Zhang. Approximations of stochastic 3D tamed Navier-Stokes equations. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5337-5365. doi: 10.3934/cpaa.2020241 [2] Yi An, Bo Li, Lei Wang, Chao Zhang, Xiaoli Zhou. Calibration of a 3D laser rangefinder and a camera based on optimization solution. Journal of Industrial & Management Optimization, 2021, 17 (1) : 427-445. doi: 10.3934/jimo.2019119 [3] Xin-Guang Yang, Lu Li, Xingjie Yan, Ling Ding. The structure and stability of pullback attractors for 3D Brinkman-Forchheimer equation with delay. Electronic Research Archive, 2020, 28 (4) : 1395-1418. doi: 10.3934/era.2020074 [4] Vieri Benci, Sunra Mosconi, Marco Squassina. Preface: Applications of mathematical analysis to problems in theoretical physics. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020446 [5] Chongyang Liu, Meijia Han, Zhaohua Gong, Kok Lay Teo. Robust parameter estimation for constrained time-delay systems with inexact measurements. Journal of Industrial & Management Optimization, 2021, 17 (1) : 317-337. doi: 10.3934/jimo.2019113 [6] George W. Patrick. The geometry of convergence in numerical analysis. Journal of Computational Dynamics, 2021, 8 (1) : 33-58. doi: 10.3934/jcd.2021003 [7] Gunther Uhlmann, Jian Zhai. Inverse problems for nonlinear hyperbolic equations. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 455-469. doi: 10.3934/dcds.2020380 [8] Min Chen, Olivier Goubet, Shenghao Li. Mathematical analysis of bump to bucket problem. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5567-5580. doi: 10.3934/cpaa.2020251 [9] Qianqian Han, Xiao-Song Yang. Qualitative analysis of a generalized Nosé-Hoover oscillator. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020346 [10] Laurence Cherfils, Stefania Gatti, Alain Miranville, Rémy Guillevin. Analysis of a model for tumor growth and lactate exchanges in a glioma. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020457 [11] Yi-Hsuan Lin, Gen Nakamura, Roland Potthast, Haibing Wang. Duality between range and no-response tests and its application for inverse problems. Inverse Problems & Imaging, , () : -. doi: 10.3934/ipi.2020072 [12] Kha Van Huynh, Barbara Kaltenbacher. Some application examples of minimization based formulations of inverse problems and their regularization. Inverse Problems & Imaging, , () : -. doi: 10.3934/ipi.2020074 [13] Yining Cao, Chuck Jia, Roger Temam, Joseph Tribbia. Mathematical analysis of a cloud resolving model including the ice microphysics. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 131-167. doi: 10.3934/dcds.2020219 [14] Xin Guo, Lei Shi. Preface of the special issue on analysis in data science: Methods and applications. Mathematical Foundations of Computing, 2020, 3 (4) : i-ii. doi: 10.3934/mfc.2020026 [15] Martin Kalousek, Joshua Kortum, Anja Schlömerkemper. Mathematical analysis of weak and strong solutions to an evolutionary model for magnetoviscoelasticity. Discrete & Continuous Dynamical Systems - S, 2021, 14 (1) : 17-39. doi: 10.3934/dcdss.2020331 [16] Feifei Cheng, Ji Li. Geometric singular perturbation analysis of Degasperis-Procesi equation with distributed delay. Discrete & Continuous Dynamical Systems - A, 2021, 41 (2) : 967-985. doi: 10.3934/dcds.2020305 [17] Illés Horváth, Kristóf Attila Horváth, Péter Kovács, Miklós Telek. Mean-field analysis of a scaling MAC radio protocol. Journal of Industrial & Management Optimization, 2021, 17 (1) : 279-297. doi: 10.3934/jimo.2019111 [18] Wen Li, Wei-Hui Liu, Seak Weng Vong. Perron vector analysis for irreducible nonnegative tensors and its applications. Journal of Industrial & Management Optimization, 2021, 17 (1) : 29-50. doi: 10.3934/jimo.2019097 [19] Lars Grüne, Matthias A. Müller, Christopher M. Kellett, Steven R. Weller. Strict dissipativity for discrete time discounted optimal control problems. Mathematical Control & Related Fields, 2020  doi: 10.3934/mcrf.2020046 [20] Hai Huang, Xianlong Fu. Optimal control problems for a neutral integro-differential system with infinite delay. Evolution Equations & Control Theory, 2020  doi: 10.3934/eect.2020107

2018 Impact Factor: 1.313


text:
## CryptoDB

### Paper: Multiplicative Masking and Power Analysis of AES

Authors: Jovan Dj. Goli\'{c} URL: http://eprint.iacr.org/2002/091 Search ePrint Search Google The recently proposed multiplicative masking countermeasure against power analysis attacks on AES is interesting as it does not require the costly recomputation and RAM storage of S-boxes for every run of AES. This is important for applications where the available space is very limited such as the smart card applications. Unfortunately, it is here shown that this method is in fact inherently vulnerable to differential power analysis. Other possible random masking methods are also discussed.
##### BibTeX
@misc{eprint-2002-11614,
title={Multiplicative Masking and Power Analysis of AES},
booktitle={IACR Eprint archive},
keywords={secret-key cryptography / AES, differential power analysis, countermeasures, multiplicative masking},
url={http://eprint.iacr.org/2002/091},
note={Presented at (internal) Gemplus Quarterly meeting, La Ciotat, France, October 30-31, 2001. jovan.golic@gemplus.com 11876 received 8 Jul 2002},
author={Jovan Dj. Goli\'{c}},
year=2002
}


text:
# simnibs.Elements¶

class simnibs.Elements(triangles=None, tetrahedra=None)

Mesh elements.

Can only handle triangles and tetrahedra!

Parameters
• (optional) (triangles) – List of nodes composing each triangle

• tetrahedra(optional) ((Nx3) ndarray) – List of nodes composing each tetrahedra

elm_number

element ID (u from 1 till nr)

Type

(Nx1) ndarray

elm_type

elm-type (2=triangle, 4=tetrahedron, etc)

Type

(Nx1) ndarray

tag1

first tag for each element

Type

(Nx1) ndarray

tag2

second tag for each elenent

Type

(Nx1) ndarray

node_number_list

4xnumber_of_element matrix of the nodes that constitute the element. For the triangles, the fourth element = -1

Type

(Nx4) ndarray

nr

Number or elemets

Type

int

Notes

Node and element count starts at 1!

__init__(triangles=None, tetrahedra=None)

Initialize self. See help(type(self)) for accurate signature.

Methods

 __init__([triangles, tetrahedra]) Initialize self. add_triangles(triangles, tag) Add triangles to mesh in-place connected_components([element_indexes]) Finds connected components find_adjacent_tetrahedra() Find the tetrahedra adjacent (sharing a face) to each element find_all_elements_with_node(node_nr) Finds all elements that have a given node find_neighbouring_nodes(node_nr) Finds the nodes that share an element with the specified node get_faces([tetrahedra_indexes]) Creates a list of nodes in each face and a list of faces in each tetrahedra get_outside_faces([tetrahedra_indexes]) Creates a list of nodes in each face that are in the outer volume node_elm_adjacency() Generates a sparse matrix with element indexes adjacent to each node nodes_with_tag(tags) Gets all nodes indexes that are part of at least one element with the given tags

Attributes

 elm_number Element numbers (1, …, nr) nr Number of elements tetrahedra Tetrahedra element numbers triangles Triangle element numbers


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# Beam Reaction Problem With Hinged Joints

The problem was assigned in my Structural Analysis class, and I am having some difficulty with it.

My professor said the problem is "very easy" if we just think about what the fixed end moment on a cantilever is. However, I don't see how this is relevant considering the member ACE is not a cantilever...

The only other way I see to do this is using the moment distribution method, but I'm uncertain as to what happens with the internal hinges; we've never had a problem like this in class, and there is no relevant example in my textbook. Any suggestions?

• I'm not sure if this qualifies as a duplicate, but this question (and the requisite solution) is very similar to a previous one. – Wasabi Dec 5 '15 at 20:09
• It actually seems pretty different. That problem was statically determinate I believe. – WGD Dec 5 '15 at 22:13
• @WGD, yes, but the essential trick to solving it can be seen in its answer (which is the same as suggested by CableStay in their answer below). You can "split" the structure into three segments: from the fixed end to the first hinge (which is equivalent to an indeterminate fixed-and-pinned beam with a concentrated load at its end), an isostatic beam between the hinges and a simply supported beam as of the second hinge (which you don't even need to solve in this case). – Wasabi Dec 5 '15 at 23:58


text:
Properties

 Label 256.4.b.b Level $256$ Weight $4$ Character orbit 256.b Analytic conductor $15.104$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$

Related objects

Newspace parameters

 Level: $$N$$ $$=$$ $$256 = 2^{8}$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 256.b (of order $$2$$, degree $$1$$, not minimal)

Newform invariants

 Self dual: no Analytic conductor: $$15.1044889615$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-1})$$ Defining polynomial: $$x^{2} + 1$$ Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$2$$ Twist minimal: no (minimal twist has level 128) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$

$q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of $$i = \sqrt{-1}$$. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + 2 i q^{3} -6 i q^{5} -20 q^{7} + 23 q^{9} +O(q^{10})$$ $$q + 2 i q^{3} -6 i q^{5} -20 q^{7} + 23 q^{9} -14 i q^{11} + 54 i q^{13} + 12 q^{15} -66 q^{17} + 162 i q^{19} -40 i q^{21} -172 q^{23} + 89 q^{25} + 100 i q^{27} -2 i q^{29} -128 q^{31} + 28 q^{33} + 120 i q^{35} -158 i q^{37} -108 q^{39} -202 q^{41} + 298 i q^{43} -138 i q^{45} -408 q^{47} + 57 q^{49} -132 i q^{51} + 690 i q^{53} -84 q^{55} -324 q^{57} + 322 i q^{59} -298 i q^{61} -460 q^{63} + 324 q^{65} + 202 i q^{67} -344 i q^{69} + 700 q^{71} + 418 q^{73} + 178 i q^{75} + 280 i q^{77} + 744 q^{79} + 421 q^{81} -678 i q^{83} + 396 i q^{85} + 4 q^{87} + 82 q^{89} -1080 i q^{91} -256 i q^{93} + 972 q^{95} -1122 q^{97} -322 i q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q - 40q^{7} + 46q^{9} + O(q^{10})$$ $$2q - 40q^{7} + 46q^{9} + 24q^{15} - 132q^{17} - 344q^{23} + 178q^{25} - 256q^{31} + 56q^{33} - 216q^{39} - 404q^{41} - 816q^{47} + 114q^{49} - 168q^{55} - 648q^{57} - 920q^{63} + 648q^{65} + 1400q^{71} + 836q^{73} + 1488q^{79} + 842q^{81} + 8q^{87} + 164q^{89} + 1944q^{95} - 2244q^{97} + O(q^{100})$$

Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/256\mathbb{Z}\right)^\times$$.

 $$n$$ $$5$$ $$255$$ $$\chi(n)$$ $$-1$$ $$1$$

Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
129.1
 − 1.00000i 1.00000i
0 2.00000i 0 6.00000i 0 −20.0000 0 23.0000 0
129.2 0 2.00000i 0 6.00000i 0 −20.0000 0 23.0000 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
8.b even 2 1 inner

Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 256.4.b.b 2
4.b odd 2 1 256.4.b.f 2
8.b even 2 1 inner 256.4.b.b 2
8.d odd 2 1 256.4.b.f 2
16.e even 4 1 128.4.a.a 1
16.e even 4 1 128.4.a.d yes 1
16.f odd 4 1 128.4.a.b yes 1
16.f odd 4 1 128.4.a.c yes 1
48.i odd 4 1 1152.4.a.d 1
48.i odd 4 1 1152.4.a.j 1
48.k even 4 1 1152.4.a.c 1
48.k even 4 1 1152.4.a.i 1

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
128.4.a.a 1 16.e even 4 1
128.4.a.b yes 1 16.f odd 4 1
128.4.a.c yes 1 16.f odd 4 1
128.4.a.d yes 1 16.e even 4 1
256.4.b.b 2 1.a even 1 1 trivial
256.4.b.b 2 8.b even 2 1 inner
256.4.b.f 2 4.b odd 2 1
256.4.b.f 2 8.d odd 2 1
1152.4.a.c 1 48.k even 4 1
1152.4.a.d 1 48.i odd 4 1
1152.4.a.i 1 48.k even 4 1
1152.4.a.j 1 48.i odd 4 1

Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{4}^{\mathrm{new}}(256, [\chi])$$:

 $$T_{3}^{2} + 4$$ $$T_{7} + 20$$

Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$4 + T^{2}$$
$5$ $$36 + T^{2}$$
$7$ $$( 20 + T )^{2}$$
$11$ $$196 + T^{2}$$
$13$ $$2916 + T^{2}$$
$17$ $$( 66 + T )^{2}$$
$19$ $$26244 + T^{2}$$
$23$ $$( 172 + T )^{2}$$
$29$ $$4 + T^{2}$$
$31$ $$( 128 + T )^{2}$$
$37$ $$24964 + T^{2}$$
$41$ $$( 202 + T )^{2}$$
$43$ $$88804 + T^{2}$$
$47$ $$( 408 + T )^{2}$$
$53$ $$476100 + T^{2}$$
$59$ $$103684 + T^{2}$$
$61$ $$88804 + T^{2}$$
$67$ $$40804 + T^{2}$$
$71$ $$( -700 + T )^{2}$$
$73$ $$( -418 + T )^{2}$$
$79$ $$( -744 + T )^{2}$$
$83$ $$459684 + T^{2}$$
$89$ $$( -82 + T )^{2}$$
$97$ $$( 1122 + T )^{2}$$


text:
# How do you write in standard form 6.36 x 10^-3?

$6.36 \cdot {10}^{-} 3 = \frac{6.36}{10} ^ 3 = \frac{6.36}{1000} = 0.00636$


text:
Definition:Topological Ring

Definition

Let $\struct{R, +, \circ}$ be a ring.

Let $\tau$ be a topology over $R$.

Then $\struct{R,+,\circ,\tau}$ is a topological ring if and only if:

$(1): \quad \struct{R, +, \tau}$ is a topological group
$(2): \quad \struct{R, \circ,\tau}$ is a topological semigroup.

Also see

• Results about topological rings can be found here.


text:
# K Value

Written by Jerry Ratzlaff on . Posted in Civil Engineering

K value, abbreviated as K, is the horizontal distance required to achieve a 1% change in the slope of the vertical curve.

## K Value formula

 $$\large{ K = \frac{ L }{ A } }$$

### Where:

 Units English Metric $$\large{ K }$$ = rate of vertical curvature $$\large{ft}$$ $$\large{m}$$ $$\large{ L }$$ = length of vertical curve $$\large{ft}$$ $$\large{m}$$ $$\large{ A }$$ = algebric difference in grade (%) $$\large{dimensionless}$$


text:
# Relativistic Work-Kinetic Energy Theorem

[Total: 1    Average: 5/5]

I was bothered for a long time by the reasons for the relativistic validity of the work-kinetic energy relation $\Delta E=Fd$, which holds without any need for corrections. We’ve discussed this before here on PF,  but I think at this point I understand it better, so I thought I’d post a summary of my present understanding.

Einstein’s treatment

At one time I somehow got the idea that Einstein had never given any logical justification for the fact that $\Delta E=Fd$ was valid and exact relativistically. I think this was because I hadn’t carefully thought through section 10 of “On the electrodynamics…,” where he actually gives a rigorous treatment. He starts with two assumptions:

(1) We have $ma=eE$ for the motion of an electron in any instantaneously comoving frame, in an electric field E. (He explicitly uses two such frames.). There’s no way this can be subject to a correction factor such as $\gamma$, since the electron is at rest.

(2) The electrical potential energy of an electron is given by $eV$. This is firmly established by experiment, and it doesn’t involve motion, so there is no reasonable way that we could imagine that it would need a correction factor like $\gamma$. In the example of a small spherical shell of charge between two capacitor plates, it’s derivable from the equation for the energy density of the field

He’s also worked out the transformations of the fields (E,B) and the coordinates, so starting from assumption 1 he’s able to find the dynamical laws in a frame where the electron is moving. He uses these dynamical laws, along with assumption 2, to find the kinetic energy of the electron at relativistic speeds.

Along the way, he also does the following. His dynamical laws give, in the parallel direction, $eE=m\gamma^3a$, and he calls this the force (in the parallel direction), which is equivalent to the common modern definition of the three-force as $dp/dt$. Because of assumption 2, it follows that $\Delta E=Fd$ holds exactly; he is *not* assuming $\Delta E=Fd$, he’s proving it (for this definition of force).

Using the relativistic energy-momentum four-vector

A more modern approach is as follows. Suppose that we’ve already determined the properties of the energy-momentum four-vector. This includes the relativistic definition of mass, $m^2=E^2-p^2$, and also the identity $p/E=v$. Then is becomes pretty straightforward to show that $W=Fd$ is relativistically exact. Consider one-dimensional motion, and let $p$ be the momentum component along our one spatial dimension. Define $F=dp/dt$. Then

$\frac{d E}{d x} = \frac{d E}{dp}\frac{d p}{d t} \frac{d t}{d x} = \frac{d E}{d p} \frac{F}{v}$

By implicit differentiation of the definition of mass, we find that $dE/dp=p/E$, and this in turn equals $v$ by the identity above. This leads to the claimed result, which is valid for both massless and material particles.

Using simple machines

I always felt that there should be some way of getting at the problem this way, but I don’t think it ends up working. Suppose we have a simple machine with mechanical advantage $A$. At least some simple machines (such as a lever) are capable of operating rigidly (based on the definition of Born-rigidity) for arbitrary relativistic motion. (Many others are not, because rigid motion would violate the Herglotz-Noether theorem. For example, a screw undergoing accelerated motion through a set of threads at relativistic speeds would bind.) For a machine that does work relativistically, the ratio of the input and output displacements is fixed to the exact value $1/A$, without relativistic correction.

The proportionality of work to $F$ and to $d$ follows from elementary considerations, independent of relativity. See the Feynman lectures, section I-4-2.

What seems more problematic in these arguments is the following. Suppose we have two things that could be called “force,” $F$ and $G$, and let them differ by $G=F h(v)$, where $h$ is some function and $v$ is velocity of object being acted upon. These could be, e.g., $F=dp/dt$ and $G=dp/d\tau$ (the spacelike component of the force four-vector). In this example, it turns out that the forces at the input and output of the machine are related by $F_1/F_2=A$, while $G_1/G_2\ne A$, but how do you show this? I don’t think you can show it without some other argument, as in the two sections above.

PhD in physics. I teach physics at Fullerton College, a community college in Southern California. I enjoy writing, playing viola, brewing beer, climbing and mountaineering.

Tags:
20 replies
1. Jano L. says:

I have trouble seeing what is the problem you're trying to solve. The relation $W=Fd$ is definition of work, not the work-energy theorem. The work-energy theorem says work equals change in kinetic energy of the particle. This follows mathematically from the equation of motion $md(\gamma v)/dt = F$ and Einstein's definition of energy $E=\gamma mc^2$.

2. Jano L. says:

I see the $way of writing formulae does not work well, so here's my comment in plaintext:I have trouble seeing what is the problem you're trying to solve. The relation W=Fd is definition of work, not the work-energy theorem. The work-energy theorem says work equals change in kinetic energy of the particle. This follows mathematically from the equation of motion md(γv)/dt=F and Einstein's definition of energy E=γmc^2. 3. pervect says: I’m not too sure my thoughts will be helpful for the intended purpose (which I assume ultimately involves a simple presentation of special relativity), but perhaps they’ll provide some insight. And they can be expressed pretty briefly. As long as special relativity has a non-trivial Hamiltonian, we can write Hamilton’s equations: $$frac{partial H}{partial q} = dot {p}$$ Now, if we can also identify H with energy, q with position, and$dot{p}$with force, then we have the work-energy theorem, the rate of change of the energy with position must be equal to the force. 4. bcrowell says: I see the$ way of writing formulae does not work well, so here’s my comment in plaintext:
I have trouble seeing what is the problem you’re trying to solve. The relation W=Fd is definition of work, not the work-energy theorem.

Right, I should have notated this as $Delta E=Fd$.

The work-energy theorem says work equals change in kinetic energy of the particle. This follows mathematically from the equation of motion md(γv)/dt=F and Einstein’s definition of energy E=γmc^2.

There are two disadvantages to your method. (1) In the 1905 paper, Einstein uses $Delta E=Fd$ to prove $E=mgamma c^2$, so he can’t use the latter to prove the former. (2) Your method doesn’t work for massless particles.

5. Jano L. says:

There are two disadvantages to your method. (1) In the 1905 paper, Einstein uses $Delta E=Fd$ to prove $E=mgamma c^2$, so he can’t use the latter to prove the former. (2) Your method doesn’t work for massless particles.

I do not know what Einstein was getting at there. I am not sure proving the relation $E=gamma mc^2$ makes sense. I think this is just a definition of energy, same as $frac{1}{2}mv^2$ is definition of kinetic energy in classical mechanics.

6. bcrowell says:

I’m not too sure my thoughts will be helpful for the intended purpose (which I assume ultimately involves a simple presentation of special relativity), but perhaps they’ll provide some insight. And they can be expressed pretty briefly. As long as special relativity has a non-trivial Hamiltonian, we can write Hamilton’s equations:

$$frac{partial H}{partial q} = dot {p}$$

Now, if we can also identify H with energy, q with position, and $dot{p}$ with force, then we have the work-energy theorem, the rate of change of the energy with position must be equal to the force.

This is an interesting approach, but if we examine it carefully I think it’s not quite as much of a slam-dunk as it might seem. And it does have the disadvantage that it won’t work as a presentation at the undergraduate level, since a physics major might not be exposed to Hamiltonian mechanics until after taking their upper-division SR course.

Spelling out the steps in more detail, I think we would have the following:

(1) The action has to be Lorentz-invariant and additive, and the only possibility that seems to present itself is $S=(ldots)mint_{t_1}^{t_2} dtau$, where … represents a constant.

(2) Working backward from this, infer that the Lagrangian for a free relativistic particle in one dimension is $L=(ldots)m/gamma$.

(3) Find the conjugate momentum, which is $p=(ldots)mgamma v$.

(4) Interpret $dp/dt$ as a force.

(5) Calculate the Hamiltonian.

(6) Identify the Hamiltonian with the energy.

(7) Use Hamilton’s equations to associate $partial H/partial x$ with minus the force.

The first thing to note is that this is much, much longer than the simple derivation I gave (the second of the three sections in my original post).

The next problem is that we have foundational issues in steps 1 and 4. Maybe there’s an explicit uniqueness theorem one could give at step 1? Otherwise it’s just a plausibility argument, with no a priori guarantee, for example, that the final result will come out consistent with Maxwell’s equations in the case of a charge moving in a field. In step 4, we have to decide what is the best definition of force. One could argue that this definition is good, because by the time we get to step 7 we will have shown that it preserves the form of the work-energy theorem without correction. But this is a weak justification if we aren’t at step 7 yet, and in fact we have a different definition of force, the four-force, which a priori would be preferable because it’s tensorial.

At step 6, we have to check some technical criteria.

A general philosophical objection to the whole thing is that Hamiltonian mechanics lacks manifest Lorentz invariance at every step of the way. In particular, time is treated as a parameter rather than a coordinate.

7. bcrowell says:

I do not know what Einstein was getting at there. I am not sure proving the relation $E=gamma mc^2$ makes sense. I think this is just a definition of energy, same as $frac{1}{2}mv^2$ is definition of kinetic energy in classical mechanics.

Definitions are like babies. They have to come from somewhere. Einstein is proving that this equation is the only one for a massive particle that is consistent with Maxwell’s equations. In any case, this is not a definition of energy, since it doesn’t apply to massless particles.

8. vanhees71 says:

Inspired by this thread and bcrowells remark in #8 on “massless particles” (which don’t make too much sense in classical (i.e., non-quantum) relativistic physics anyway), I’ve extended the section on “naive dynamics” (i.e., dynamics not employing Hamilton’s principle of least action) in my SRT writeup by giving a derivation of the work-energy theorem and also a formulation of the dynamics of massless particles:

[URL]http://fias.uni-frankfurt.de/~hees/pf-faq/srt.pdf[/URL]

9. bcrowell says:

”massless particles” (which don’t make too much sense in classical (i.e., non-quantum) relativistic physics anyway)

Why?

10. vanhees71 says:

Very simple: There are no classical massless particles, at least I’ve never heard that one has observed them yet. This makes much sense from the point of view of quantum field theory. The only candidate that comes to mind within the standard model would be a photon, but this is the quantum (one-particle Fock state) of the electromagnetic field, which is a massless spin-1 field and thus necessarily a gauge field, and you cannot define a position observable in the strict sense. So it is hard to imagine, how to get a classical particle limit. The classical limit is that of classical electromagnetic fields, i.e., coherent states with a large average photon number rather than classical point particles.

The other massless fields in the Standard Model are the gluon fields, but the corresponding quanta, the gluons, are confined, so one cannot expect to find them as particles in the sense of a classical limit.

The very last possibility would be that one of the neutrinos is in fact massless. It’s hard to say at the present empirical status. If it would exist, it may be a candidate for something that could make sense as a classical approximation in the sense of a massless particle.

11. bcrowell says:

I don’t think #12 is the same as saying that massless particles don’t “make sense” classically. In any case, we deal with massless particles constantly in classical relativity, and we do it in practical contexts that relate directly to observation.

12. vanhees71 says:

Can you give an example where one does this? Anyway, there’s of course nothing wrong with it from a theoretical point of view. You can have perfectly valid equations of motion for massless particles.

13. bcrowell says:

Can you give an example where one does this? Anyway, there’s of course nothing wrong with it from a theoretical point of view. You can have perfectly valid equations of motion for massless particles.

Gravitational lensing would be a good example.

14. vanhees71 says:

In which sense has gravitational lensing to do with the motion of massless particles? Well, you might interpret ray optics (i.e., the eikonal approximation for classical electromagnetic waves) as massless-particle motion. In this sense the naive photon picture is not completely wrong…

15. bcrowell says:

In which sense has gravitational lensing to do with the motion of massless particles? Well, you might interpret ray optics (i.e., the eikonal approximation for classical electromagnetic waves) as massless-particle motion.

Yes.

16. vanhees71 says:

Ok, so that’s well understood by classical electromagnetism. No naive-photon picture is needed. What’s usually done is just the eikonal approximation to classical electromagnetism (i.e., ray optics derived from wave optics). Photons in a curved background spacetime are a quite tricky business!

17. bcrowell says:

Ok, so that’s well understood by classical electromagnetism. No naive-photon picture is needed.

I would put it the other way around. Ray optics is all you need for the application I mentioned, so there is no need for classical electromagnetism or quantum mechanics. Ray optics is not a “naive-photon picture.” In any case, I think this is of little or no relevance to the topic of this thread. If you want to start a separate thread on the status of massless particles in classical relativity, that might be more appropriate IMO.

18. vanhees71 says:

Sure, ray optics is all you need for this purpose. For things like the redshift-distance relation in cosmology you need its proper definition via the eikonal approximation of the Maxwell equations in a given background spacetime (here the FLRW pseudometric), usually for spherical waves.

It’s also historically interesting that in the 1930ies there was a debate about the correct interpretation of the Hubble redshift, where distinguished physicists like von Laue were involved. There they also had the debate, whether the already then favored derivation in the naive photon picture is correct. Finally Robertson showed, how to derive the relation within classical electromagnetics. It’s downloadable from NASA (I think not behind a paywall):

Robertson, H. The apparent luminosity of a receding nebula Zeitschrift für Astrophysik 15, 1938, 69


text:
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# Big Numbers

Take a number, say, 264, and write it in binary:

264 = 2^8 + 2^3.

Because we like binary so much, write the exponents in binary too:

\begin{aligned}
264 &= 2^{2^3} + 2^{2 + 1} \\
&= 2^{2^{2 + 1}} + 2^{2 + 1}
\end{aligned}

and so on, until every number is 2 or less. This is the hereditary base-2 representation of 264. Now, transform that number by increasing the base from 2 to 3, and subtracting one:

\begin{aligned}
G_2(264) &= 3^{3^{3 + 1}} + 3^{3 + 1} - 1 \\
&= 3^{3^{3 + 1}} + 2 \cdot 3^3 + 2 \cdot 3^2 + 2 \cdot 3 + 2 \\
&\approx 4.4 \cdot 10^{38}.
\end{aligned}

And again:

\begin{aligned}
G_3(G_2(264)) &= 4^{4^{4 + 1}} + 2 \cdot 4^4 + 2 \cdot 4^2 + 2 \cdot 4 + 2 \\
&\approx 3.2 \cdot 10^{616}.
\end{aligned}

Goodstein sequences are defined by these repeated applications of $G_n$. The Goodstein sequence starting at $m$ is $m_1 = m$, $m_2 = G_2(m_1)$, $m_3 = G_3(m_2)$, etc.

## Goodstein's theorem

Goodstein's theorem states that every Goodstein sequence eventually reaches zero. Despite the deceivingly explosive initial increase (the next value for $m = 264$ above is $m_4 \approx 2.5 \cdot 10^{10921}$), the surprising truth is that for all initial values of $m$, $m_k = 0$ for some finite $k$.

This counterintuitive theorem is actually fairly easy to prove. We simply find an upper bound on the base $b$ used in the hereditary base-$b$ representations in the sequence. Finding the upper bound is actually the easiest part:

b < \infty

"But wait," you say, "that's obvious and unhelpful!" Well, by using a particular kind of infinity, this upper bound becomes useful. Precisely, we use $\omega$, the first transfinite ordinal number. For $m = 264$ as before, the bounds look like this:

\begin{array}{cclcl}
m_1 & = & 2^{2^{2 + 1}} + 2^{2 + 1} & \le & \omega^{\omega^{\omega + 1}} + \omega^{\omega + 1} \\
m_2 & = & 3^{3^{3 + 1}} + 2 \cdot 3^3 + 2 \cdot 3^2 + 2 \cdot 3 + 2 & \le & \omega^{\omega^{\omega + 1}} + \omega^\omega \cdot 2 + \omega^2 \cdot 2 + \omega \cdot 2 + 2 \\
m_3 & = & 4^{4^{4 + 1}} + 2 \cdot 4^4 + 2 \cdot 4^2 + 2 \cdot 4 + 1 & \le & \omega^{\omega^{\omega + 1}} + \omega^\omega \cdot 2 + \omega^2 \cdot 2 + \omega \cdot 2 + 1 \\
m_4 & = & 5^{5^{5 + 1}} + 2 \cdot 5^5 + 2 \cdot 5^2 + 2 \cdot 5 & \le & \omega^{\omega^{\omega + 1}} + \omega^\omega \cdot 2 + \omega^2 \cdot 2 + \omega \cdot 2 \\
m_5 & = & 6^{6^{6 + 1}} + 2 \cdot 6^6 + 2 \cdot 6^2 + 6 + 5 & \le & \omega^{\omega^{\omega + 1}} + \omega^\omega \cdot 2 + \omega^2 \cdot 2 + \omega + 5 \\
m_6 & = & 7^{7^{7 + 1}} + 2 \cdot 7^7 + 2 \cdot 7^2 + 7 + 4 & \le & \omega^{\omega^{\omega + 1}} + \omega^\omega \cdot 2 + \omega^2 \cdot 2 + \omega + 4 \\
\vdots
\end{array}

These upper bounds are well-defined using ordinal arithmetic. They are also strictly decreasing, and since the ordinals are well-ordered, every such strictly-decreasing sequence must reach zero in a finite number of steps. Note that we write $\omega \cdot 2$ rather than $2 \cdot \omega$ above because ordinal arithmetic is not commutative; in fact, $2 \cdot \omega = \omega < \omega \cdot 2$.

## Independence

Despite being a theorem about natural numbers, Goodstein's theorem is independent of the Peano axioms. It's a natural, concrete example of Gödel's incompleteness theorem for those axioms.

Kirby and Paris proved its independence by essentially showing that any proof of Goodstein's theorem requires induction over sets too large for Peano's axioms to handle. They showed that Goodstein's theorem implies Gentzen's theorem.

## The Goodstein function

How long are Goodstein sequences? It turns out they are extremely long. Let the Goodstein function $\mathcal{G}(m)$ be the length of the Goodstein sequence starting at $m_1 = m$.

$\mathcal{G}(12)$ is larger than Graham's number. $\mathcal{G}(m)$ grows faster than the Ackermann function $A(m, m)$. $\mathcal{G}(m)$ grows faster than the iterated Ackermann function $A^{A(m, m)}(m, m) = A(A(\cdots), A(\cdots))$. In fact, $\mathcal{G}(m)$ grows faster than any function that can be proven to be total in Peano arithmetic. That means $\mathcal{G}(m)$ grows faster than any function that can be shown to even exist by induction over the natural numbers.


text:
# Homework Help: Diver falling into swimming pool - forces!

1. Oct 24, 2005

### Punchlinegirl

A 88.0 kg diver falls from rest into a swimming pool from a height of 5.30 m. It takes 1.94 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
I found the acceleration to be 1.41 m/s^2 and then multiplied it by the mass to get 124 N. this wasn't right.
Can someone help me?
Thanks.

2. Oct 24, 2005

### whozum

I'm getting a larger acceleration, what is his speed as he hits the water?

3. Oct 24, 2005

### Punchlinegirl

I got the a to be 3.42 m/s^2.
5.30= (1/2)a(1.94^2)
I don't think I did it right though...

4. Oct 24, 2005

### whozum

He is falling under gravity's force at a rate of 9.8m/s^2 and it takes him 1.94 seconds from the moment he penetrates the water surface to the time he becomes stopped (under water), find his final velocity as he hits the water:

$$v_f^2 = v_i^2 + 2ad$$

Then You can find the force appliedo n him by multiplying his mass by the average acceleration you found.


text: A three-phase diode bridge rectifier is feeding a constant $DC$ current of $100 A$ to a highly inductive load. If three-phase, $415 V, 50 Hz AC$ source is supplying to this bridge rectifier then the rms value of the current in each diode, in ampere, is ___________.


text:
# HELIUM DROPLETS AS NANO-CRYOSTATS FOR MOLECULAR SPECTROSCOPY: AGGREGATION, STATE SELECTION AND ELECTRON SPIN RESONANCE

Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/38403

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 Title: HELIUM DROPLETS AS NANO-CRYOSTATS FOR MOLECULAR SPECTROSCOPY: AGGREGATION, STATE SELECTION AND ELECTRON SPIN RESONANCE Creators: Ernst, Wolfgang E. Issue Date: 2009 Publisher: Ohio State University Abstract: Droplets of about $10^{4}$ helium atoms generated in a supersonic expansion, represent a nanometer-sized superfluid medium of 0.4 K temperature and can be doped with one or several atoms or molecules that may form complexes in this cold environment. Using two-laser excitation schemes, we were able to identify the alkali trimers K$_3$, Rb$_3$, K$_2$Rb and KRb$_2$ in their lowest quartet states formed on helium droplets loaded with potassium and rubidium atoms and assign several excited states that underlie both Jahn-Teller and spin-orbit coupling. As helium provides a gentle and only weakly perturbing matrix, it appeared desirable to look for ways to measure fine and hyperfine structure directly in the microwave or radiofrequency regime. In preparation for experiments involving optical detection of electron spin transitions in cold molecules, we studied the electronic spin relaxation in alkali atoms and molecules that reside on the surface of a droplet. Measurements of the circular dichroism in the presence of a magnetic field showed that the populations of Zeeman sublevels in alkali atoms are not thermalized, while for dimers and trimers a temperature of 0.4 K was found, implicitly providing a first determination of the droplet s surface temperature. Optical detection of spin resonance is achieved in an optical pump-probe experiment with the electron spin transition induced in a microwave cavity in a magnetic field between the pump and probe regions. With the pump laser depleting a particular spin state by desorption of the species from the droplet beam or by optical pumping, the probe laser detects the successful spin flip induced by the microwave field. Examples will be presented showing up to 50 Rabi cycles of an electron spin transition on an alkali doped helium droplet during the flight time of 57 $\mu$s through the cavity. Description: J.~Nagl, G.~Aubock, A.~W.~Hauser, O.~Allard, C.~Callegari, and W.~E.~Ernst, Phys.~Rev.~Lett. 100, 063001(2008).G.~Aubock, J.~Nagl, C.~Callegari, and W.~E.~Ernst, J.~Chem.~Phys. 129, 114501(2008).J.~Nagl,G.~Aubock, C.~Callegari, and W.~E.~Ernst, Phys.~Rev.~Lett. 98, 075301 (2007).G.~Aubock, J.~Nagl, C.~Callegari, and W.~E.~Ernst, J.~Phys.~Chem. A 111, 7404(2007).G.~Aubock, J.~Nagl, C.~Callegari, and W.~E.~Ernst, Phys.~Rev.~Lett. 101, 035301(2008). Author Institution: Institute of Experimental Physics, Graz University of Technology; Petersgasse 16, A-8010 Graz, Austria URI: http://hdl.handle.net/1811/38403 Other Identifiers: 2009-MA-01