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 Please use this identifier to cite or link to this item: http://hdl.handle.net/1807/31333

 Title: Automated Target Detection in Diagnostic Ultrasound based on the CLEAN Algorithm Authors: Masoom, Hassan Advisor: Adve, RavirajCobbold, Richard Department: Electrical and Computer Engineering Keywords: UltrasoundMedical ImagingSignal ProcessingDetectionComputer Aided DetectionDiagnostics Issue Date: 14-Dec-2011 Abstract: In this thesis, we present an algorithm for the automated detection of abnormalities (targets) in ultrasound images. The algorithm uses little a priori information and does not require training data. The proposed scheme is a combination of the CLEAN algorithm, originally proposed for radio astronomy, and constant false alarm rate (CFAR) processing, developed for use in radar systems. Neither of these algorithms appears to have been previously used for target detection in ultrasound images. The CLEAN algorithm identifies areas in the ultrasound image that stand out above a threshold in relation to the background; CFAR techniques allow for an automated and adaptive selection of the threshold. The algorithm was tested on simulated B-mode images. Using a contrast-detail analysis, probability of detection curves indicate that, depending on the contrast, the method has considerable promise for the automated detection of abnormalities with diameters greater than a few millimetres. URI: http://hdl.handle.net/1807/31333 Appears in Collections: Master

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# Planar Flow in Normalizing Flows

While I've read "Variational Inference with Normalizing Flows" (abstract), I don't understand about an intuition of Planar Flow.

The author defined Planar Flow as below

Let $$\boldsymbol{w} \in \mathbb{R}^D, \boldsymbol{u} \in \mathbb{R}^D, > b \in \mathbb{R}$$ and $$h(\cdot)$$ be a smooth element-wise non-linearity.
Then the following formula is Planar Flow

$$\begin{array}{c} f(\mathbf{z}) = \mathbf{z} + \mathbf{u}h(\mathbf{w}^T\mathbf{z}+b) \\ \psi(\mathbf{z})=h^{\prime}\left(\mathbf{w}^{\top} \mathbf{z}+b\right) \mathbf{w} \\ |\operatorname{det} \frac{\partial f}{\partial \mathbf{z}}|=| \operatorname{det}\left(\mathbf{I}+\mathbf{u} \psi(\mathbf{z})^{\top}\right)|=| 1+\mathbf{u}^{\top} \psi(\mathbf{z}) | \quad (1)\end{array}$$

The author said that

The flow defined by the transformation (1) modified the initial density $$q_0$$ by applying a series of contractions and expansions in the direction perpendicular to the hyperplane $$\mathbf{w}^T\mathbf{z}+b=0$$.

I couldn't understand that why the transformation (1) move the vector $$\mathbf{z}$$ along the direction perpendicular to the hyperplane $$\mathbf{w}^T\mathbf{z}+b=0$$.

Would anybody elaborate this?

For every $$z,$$ notice that the displacement from $$z$$ to its destination $$f(z),$$ given by $$f(z)-z,$$ is a multiple of the fixed vector $$u.$$ Thus, if you were to diagram the effect of $$f$$ by drawing arrows from a selected set of original values $$z_i$$ to their destinations $$f(z_i),$$ all the arrows would be parallel. See the right hand plot in the figure below.

Next, notice that each level set of $$f$$ is a union of level sets of the function

$$z \to w^\top z,$$

which are parallel hyperplanes. On any such hyperplane given by $$w^\top z = c,$$ for some constant real number $$c,$$ all the arrows equal

$$f(z) - z = u\,h(w^\top z + b) = u\, h(c + b).$$

That shows they all have common length $$|h(c+b)|\,||u||$$ for every $$z$$ on that hyperplane.

Why one might call these characteristics "planar" is inscrutable.

Here at the left is an example of a generic $$f:\mathbb{R}^2\to\mathbb{R}^2$$ from Analysis with complex data, anything different?:

On the right is a "planar flow" transformation. The arrows are colored according to the value of $$h.$$ The common direction of displacement is $$u = (2,-1)$$ and the amount of displacement varies in the direction $$w = (10,-1).$$

The equation $$\mathbf{w}^T\mathbf{z_1}+b=0$$ defines a (hyper)plane. The vector $$\mathbf{w}$$ is the normal vector. For a refresher on multivariable calculus, see here.

So what happens if you have a fixed vector $$\mathbf{w}$$, a fixed scalar $$b$$, and you plug in a different point $$\mathbf{z_2}$$ into the above equation, and get

$$\mathbf{w}^T\mathbf{z_z}+b= 1?$$ $$1$$ isn't $$0$$, so obviously this new point $$\mathbf{z}_2$$ isn't in the same plane. But what does $$1$$ represent?

$$\mathbf{z}_2$$ is in a different plane. This new plane has the same normal vector, $$\mathbf{w}$$, so this new plane is parallel to the old one. It's just shifted.

The function $$\mathbf{w}^T\mathbf{z}+b$$ operates on all of $$\mathbb{R}^d$$, so you can plug in any vector $$\mathbf{z}$$. The output represents perpendicular distance from some prototypical plane.

Then $$h$$, the "smooth element-wise non-linearity" will take this scalar output and map it into another, less interpretable scalar.

Then that scalar gets multiplied to a vector. This product is added to the original input $$\mathbf{z} \in \mathbb{R}^d$$. If the original point was on the plane, and if $$h$$ maps $$0$$ to $$0$$, then nothing gets added to the original vector $$\mathbf{z}$$.

On the other hand, if $$\mathbf{z}$$ was far away from the original plane, then a significant amount $$\mathbf{u}h(\mathbf{w}^T\mathbf{z}+b)$$ gets added to the original input vector.

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Algebra and Trigonometry 10th Edition

(f + g)(t - 2) = $t^{2}$ - 3t + 3
The notation (f + g)(t - 2) can be rewritten as f(t - 2) + g(t - 2). Solving by adding the functions then evaluating at t - 2 will yield the same result as evaluating each function and then adding together. In this question: f(x) = x + 3 g(x) = $x^{2}$ - 2 Method 1: Add the functions then evaluate First we want to add the two functions. The new function will be called h(x). h(x) = f(x) + g(x) h(x) = (x + 3) + ($x^{2}$ - 2) h(x) = x + 3 + $x^{2}$ - 2 Combine like variables to get: h(x) = $x^{2}$ + x + 1 Evaluate at x = t - 2: h(t - 2) = $(t - 2)^{2}$ + (t - 2) + 1 = $t^{2}$ - 4t + 4 + t - 2 + 1 = $t^{2}$ - 3t + 3 Method 2: Evaluate the functions and then add First we want to evaluate f(x) at x = t - 2: f(t - 2) = t - 2 + 3 = t + 1 Then we want to evaluate g(x) at x = t - 2: g(t - 2) = $( t - 2)^{2}$ - 2 = $t^{2}$ - 4t + 4 - 2 = $t^{2}$ - 4t + 2 Add the numbers together: (f + g)(t - 2) = t + 1 + $t^{2}$ - 4t + 2 = $t^{2}$ - 3t + 3 Using both methods (f + g)(t - 2) = $t^{2}$ - 3t + 3


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# Argonne National Laboratory

Press Release | Argonne National Laboratory

# Argonne awarded $6 million to develop technologies for recycling nuclear fuel Two projects will demonstrate improved separation technologies to help the U.S. get even more energy out of safe, clean nuclear power. A very small fraction of the energy available in nuclear fuel is used during the production of nuclear energy. The amount that remains can be recovered and recycled to create a lot of electricity. The U.S. Department of Energy’s (DOE) Advanced Research Projects Agency-Energy (ARPA-E) estimates that the untapped energy remaining in used nuclear fuel, sometimes referred to as spent nuclear fuel, can generate enough clean energy to power more than 70 million homes. Recently, ARPA-E awarded more than$6 million to DOE’s Argonne National Laboratory to develop and demonstrate innovative processes and technologies that focus on recycling used nuclear fuel (UNF). The funding is awarded through the Converting UNF Radioisotopes Into Energy (CURIE) program.

Nuclear energy is the source of approximately one-fifth of total electricity used in the U.S., and it provided half of the nation’s carbon-free electricity in 2021. Improving electrochemical separation and other UNF recovery and recycling technologies could help nuclear energy do more. More efficient use of nuclear energy would significantly decrease domestic reliance on fossil fuels and the related greenhouse gases that contribute to climate change. It would reduce waste and produce new fuel for advanced reactors. This last benefit would lead to deployment of new advanced reactors, which would provide even more safe, clean, reliable electricity.

Our partnerships with ARPA-E and industry are incredibly important to closing remaining technology gaps.” — Krista Hawthorne, section manager, pyroprocess engineering

Argonne will use ARPA-E’s funding to advance two projects. The larger project will combine Argonne’s leadership in nuclear research and development with the experience of nuclear industry partners Oklo Inc. and Deep Isolation. Argonne will collaborate with Case Western Reserve University on both projects.

The larger project was awarded $4.9 million. It focuses on increasing the efficiency and, therefore, the economics of nuclear fuel recycling processes. Argonne, Oklo Inc. and Deep Isolation are working to improve a process that converts UNF oxide fuel to a metal that can then be processed to provide usable materials for advanced reactor fuel. The project will combine improvements in stable, next-generation anode materials and sensors used to control the process to maximize the amount of material that can ultimately be recovered. Krista Hawthorne, the pyroprocess engineering section manager in Argonne’s Chemical and Fuel Cycle Technologies division, leads the effort. Improving the efficiency of fuel recycling processes will help recover and recycle valuable resource materials remaining in used nuclear fuel,” said Hawthorne. Our partnerships with ARPA-E and industry are incredibly important to closing remaining technology gaps.” The second project was awarded$1.52 million. With this project, Argonne researchers will develop, produce and test more efficient industrial chemical processing equipment to reprocess UNF. The project lead is Argonne radiochemist Anna Servis.

State-of-the-art facilities that reprocess light water nuclear reactor fuels use equipment known as packed bed columns. Packed bed columns are tall canisters filled with a packing material that mixes the fluids flowing from one end to the other during chemical separations.

Servis’ team’s design instead uses PACERs (PAcked Centrifugal Equipment for Radiochemical separations). PACERs are rotating packed bed contactors that are considerably more efficient than the columns. They use centrifugal forces much greater than the pull of gravity to boost the speed and efficiency of chemical separations. They are also notably smaller. For scale, imagine replacing a water tower-sized packed bed column with a shed-sized PACER.

Used nuclear fuel contains a treasure trove of raw materials, including energy-rich uranium that can be recycled into new fuel, precious metals and versatile radioisotopes,” said Servis. Mining our existing UNF stockpiles reduces demand for non-renewable minerals and conserves resources for future generations.”

Argonne National Laboratory seeks solutions to pressing national problems in science and technology. The nation’s first national laboratory, Argonne conducts leading-edge basic and applied scientific research in virtually every scientific discipline. Argonne researchers work closely with researchers from hundreds of companies, universities, and federal, state and municipal agencies to help them solve their specific problems, advance America’s scientific leadership and prepare the nation for a better future. With employees from more than 60 nations, Argonne is managed by UChicago Argonne, LLC for the U.S. Department of Energy’s Office of Science.

The U.S. Department of Energy’s Office of Science is the single largest supporter of basic research in the physical sciences in the United States and is working to address some of the most pressing challenges of our time. For more information, visit https://energy.gov/science.


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[R] Transferring a dataset from WINXP to Mac - problems with umlauts

David Winsemius dwinsemius at comcast.net
Wed Mar 28 14:27:21 CEST 2012

```On a Mac running with a US locale, when I first type <option>-u (in
Mail.app that shows a highlighted "naked umlaut)  and then type 'u' I
get:

ü
Similarly with opt-u, 'a'

ä

--
David.
On Mar 28, 2012, at 7:15 AM, Frank Bloos wrote:

> Dear R-Help,
>
> is it possible to correctly display umlauts in a dataset, which has
> been created under WINXP, on a Mac-system?
>
> This is the problem:
> I have imported a German MS-Access database into R 2.14.1 by using
> RODBC on a WINXP-System. It runs with a Latin-1 encoding (codepage
> 1252). However, most of the work is done with a Mac (UTF-8 encoding)
> and the R-dataset has been transferred to the MAC.  Here, umlauts
> are displayed as \x.. (i.e. \xe4 instead of ä). If I set the locale
> to Sys.setlocale(locale="de_DE.ISO8859-15"), umlauts are displayed
> as black diamonds.
>
> Since it is awkward to enter \x-codes when I work with the dataset,
> I would rather use the umlauts. Is there a way?
>
> Thanks, FB
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help


text:
# Prove that the set of all infinite subsets of $\mathbb{Q}$ has cardinality $\mathfrak{c}$ [closed]

How to prove the set of all infinite subsets of $$\mathbb{Q} = \mathfrak{c}$$, if you first prove that the cardinality of the set of all finite subsets of $$\mathbb{Q}$$ is $$\aleph_{0}$$.

I know how to prove the finite subset of $$\mathbb{Q}$$ is $$\aleph_{0}$$ but hard to reach this problem

New contributor
RLOUIS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• Do you know that the set of all subsets of $\Bbb N$ has cardinality $\mathfrak{c}$? – Brian M. Scott 2 days ago
• I know the cardinality of the set of all real numbers is 2^N0 – RLOUIS 2 days ago
• Is N you way of writing $\aleph$? – Hagen von Eitzen 2 days ago
• Yes, N is ℵ. I just write n simply. – RLOUIS 2 days ago
• Ok I got your point. Thank you for your help. – RLOUIS 2 days ago


text:
A Geometrical View of Time Dilation and the Twin Paradox

Based on the number of questions we receive on the topic at Physics Forums, there is a lot of confusion in the general public about how time dilation works and the resolution of the twin paradox. More often than not, the confusion is based upon misunderstandings about how time dilation works and an unawareness of the relativity of simultaneity and how time is treated within special relativity. The aim of this Insight is to make an attempt at explaining these issues from a geometrical viewpoint and to see that we have an analogous effect in the geometry which we are used to, although we do not find it strange.

Table of Contents

A geometrical analogue

Consider the following two lines in normal two-dimensional space:

In order to describe the lines, we can introduce a coordinate system ##S##:

In this coordinate system, we may use the coordinate ##x## to parametrise the lines, we find that

$$y_1(x) = 0, \quad y_2(x) = kx,$$

where ##k## is some constant. The lengths of each line between ##x = 0## and ##x = x_0## are simply given by Pythagoras’ theorem ##\ell^2 = \Delta x^2 + \Delta y^2## and we find

$$\ell_1 = x_0, \quad \ell_2 = x_0\sqrt{1+k^2}.$$

Consequently we find that the ratio between these two lengths is ##\ell_2/\ell_1 = \sqrt{1+k^2}##. This is shown in the following figure:

However, there is nothing special about the coordinate system ##S##. In particular, the length of the lines between any two given points on the lines does not depend on the coordinate system chosen and we could just as well have chosen to introduce the coordinate system ##S’## according to:

In this new coordinate system, we can parametrise the lines using the ##x’## coordinate

$$y_1′(x’) = -kx’, \quad y_2′(x’) = 0.$$

For the distance between the points corresponding to ##x’ = 0## and ##x’=x_0’##, Pythagoras’ theorem now results in

$$\ell_1′ = x_0′ \sqrt{1+k^2}, \quad \ell_2′ = x_0’$$

and so the line ##\gamma_1## is longer than ##\gamma_2## for the same difference in the parameter ##x_0’##, unlike in the case when the lines were parametrised by ##x##, where we obtained the opposite result. Yet there is nothing strange going on here. If we select a point on ##\gamma_2## and draw the lines which have the same ##x## and ##x’## values as that point, it becomes clear that the points on ##\gamma_1## which share the same parameter value differs depending on whether we used ##x## or ##x’## for the parametrisation. We should therefore not be surprised to find that the length to these two different points is different. Let us call this effect the “relativity of the same ##x##-coordinate”. Naturally, this effect does not affect the actual length of any of the lines, but only what lengths are being compared. This is illustrated in the following figure:

The points ##B## and ##B’## have the same ##x## and ##x’## coordinate as ##A##, respectively, yet correspond to different points on the curve ##\gamma_1## and therefore the distance from the intersection along this curve is different.

Curves intersecting at two points

Now consider the following two curves:

Just from the figure, it should be clear that ##\gamma_2## is longer and we can verify this by introducing the coordinate system ##S## as above and computing the lengths of each of the curves, with the length of ##\gamma_2## being split into two equal contributions, one from each line segment.

We find that

$$\ell_1 = 2\ell, \quad \ell_2 = 2\ell \sqrt{1+k^2}.$$

We could also compute the length of the curves by looking at the ##S’## coordinate system:

By our previous discussion, the length of ##\gamma_2## up to the half-way point is given by

$$\ell_2’/2 = \ell’,$$

while the length of ##\gamma_1## to the same ##x’## coordinate is given by

$$\ell_1’/2 = \ell’ \sqrt{1+k^2}.$$

If we were not careful about the relativity of the same ##x##-coordinate, we might do the error of thinking that ##\ell_1’/2## would be half the length of the curve ##\gamma_1##. However, since we are aware of this, we know better than that and realise that if we multiplied the result by two, we would obtain a result that was too large.

Application to relativity

So how does this connect to relativity and time dilation? In its foundation, relativity is a theory about the geometry of space-time, which may be described using a coordinate system with four coordinates ##(t,x,y,z)##, where the first coordinate is the time coordinate and the last three are spatial coordinates. It is important to realise that, just as in the case we just described, these are just coordinated that, by themselves, do not have a direct physical interpretation. For the sake of brevity, let us consider only one spatial coordinate ##x## as the generalisation is straightforward.

In relativistic space-time, geometry works a bit differently as compared to the geometry with which we are used to working with. In particular, this manifests itself by a change in Pythagoras’ theorem, which now comes with a minus sign

$$c^2 \tau^2 = c^2 \Delta t^2 – \Delta x^2$$

as a direct consequence of the speed of light is equal in all frames. Here, ##\tau## is the proper time along a straight line in space-time with ##\Delta t## being the difference between the endpoint time-coordinates and ##\Delta x## being the difference between the endpoint space-coordinates. This also changes what kind of coordinate transformations we can do in order for the proper time to be the same regardless of the coordinates. Instead of rotating both coordinate axes in the same direction as happens for rotations, the coordinate axes are rotated by the same angle in opposite directions.

In relativity, the proper time defined above is the only time that will be measured by any observer. A curve in space-time is called a world line and will generally describe the position of the observer as a function of time. The proper time of a world line, i.e., the “length” of the world line with the modified Pythagorean theorem, is the time a clock following that world line will measure, the time coordinate ##t## has no particular meaning other than that it happens to numerically be the same as the time measured by an observer for which the spatial coordinates are not changing. This is completely analogous with our previous example, where the value of the ##x##-coordinate was equal to the length of a line for which the ##y##-coordinate was not changing.

We can now do the exact same exercise as we did use regular geometry. Consider the following two world lines described in the coordinate system ##S##:

The first world line ##\gamma_1## has a constant ##x##-component while the second world line describes an observer travelling with velocity ##v##. The world lines may be parametrised as

$$x_1(t) = 0, \quad x_2(t) = vt.$$

We can now compute the proper time from the time coordinate taking values ##t=0## to ##t=t_0## for both world lines using the modified Pythagorean theorem and find that

$$\tau_1 = t_0, \quad \tau_2 = t_0\sqrt{1-\frac{v^2}{c^2}}$$

or, equivalently,

$$\frac{\tau_1}{\tau_2} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \equiv \gamma,$$

where ##\gamma## is the famous gamma-factor of special relativity. In other words, the proper time elapsed for an observer at rest between two time coordinates will differ from the proper time elapsed for a moving observer between the same time coordinates by a factor ##\gamma##. This is the essence of time-dilation but, just as for the example we started with, it crucially depends on the coordinate system.

Taking a different coordinate system ##S’## where ##\gamma_2## describes a world line with constant ##x’##

the exact same argumentation will now lead us to conclude

$$\frac{\tau_1′}{\tau_2′} = \sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\gamma}.$$

This is the exact opposite relation as that we found in the coordinate system ##S##, just as was the case in our first example! The resolution of this apparent mismatch is also completely analogous. We should not be surprised to find a different relation between the times elapsed between different ##t## coordinates and that elapsed between different ##t’## coordinates simply because the point on the curve ##\gamma_1## with the same ##t## coordinate as the point ##A## on ##\gamma_2## will not be the same as the point on ##\gamma_1## with the same ##t’## coordinate as ##A##:

This effect is called the “relativity of simultaneity” and simply states that two events which are simultaneous in one set of coordinates are generally not simultaneous in a different set of coordinates (simultaneous simply refers to having the same time-coordinate). As such, there is really nothing strange about the proper times to ##B## and ##C## along with ##\gamma_1## being different since they are simply not the same point.

The twin paradox

The twin paradox is the analogue of our second example with two curves starting and ending at the same points. In the case of two twins, where one leaves to travel to a faraway land and later returns, the world lines in the coordinate system ##S## would look like:

Just applying the modified Pythagorean theorem it becomes clear that the proper times along the world lines are given by

$$\tau_1 = 2\tau, \quad \tau_2 = 2\tau \sqrt{1-\frac{v^2}{c^2}} \quad \Longrightarrow \quad \frac{\tau_1}{\tau_2} = \gamma$$

and therefore more proper time will pass for the twin staying behind, i.e., the red curve. Going to the coordinate system ##S’##, we can compute the relation between the proper time ##\tau’## taken for the travelling twin to reach the turnaround point and the proper time elapsed for the staying twin during the same ##t’## interval as

$$\frac{\tau_1′}2 = \tau’ \sqrt{1-\frac{v^2}{c^2}}, \quad \frac{\tau_2′}2 = \tau’.$$

However, just as in our example above, the relativity of simultaneity tells us that ##\tau_1’## is not equal to half the proper time elapsed for the staying twin until the reunion:

Instead, the change of geometry indicates that we are not counting part of the world line if we make this assumption and therefore underestimate the elapsed proper time. If taken properly into account, we would find that the time elapsed for the staying twin is longer also as computed by the travelling observer. This resolves the twin paradox and shows that it only appears due to not taking the relativity of simultaneity properly into account.

Click for Forum Comments

Tags:
90 replies
1. whatif says:

Your experience is much more closely related to your past light cone than your frame.Rest frame was, perhaps, a bad choice of wording. What I meant was all frames in which I was stationary at the time, including non inertial frames of acceleration, which I guess correlates to the light cone. Yes/no?
To be more precise about the twin paradox. So long as each twin occupies a different inertial frame continuing to separate apart from the other's frame then each continues to age faster than the other. That, I think, exemplifies what requires insight; or just plane acceptance because it seems to be implicitly verified by experiment.

2. Dale says:
whatif

Only for the spacetime interval.The choice of reference frame or coordinates is not only arbitrary for the spacetime interval. It is also arbitrary for the expression of any physical law and for the prediction of the outcome of any physical measurement for any experiment. That is the meaning of the first postulate.

whatif

Except that my rest frame is my experience.Not really. Your experience is much more closely related to your past light cone than your frame. And your past light cone is the same in all reference frames.

3. whatif says:

Of course the choice of axes is arbitrary.A choice of inertial system is exactly analogous to an arbitrary choice of (mutually orthogonal) coordinate axes.It is completely arbitrary.Only for the spacetime interval. As I see it, time is not independent of the location in space, but a restriction on the 3D location that can be occupied. In the 2D analogue there is no such interdependence/restriction. The article introduces the rotation of axes and ‘modified Pythagorean theorem’ without explanation, which, to me, reduces the idea to an arduous mathematical how to, rather than an insight.

The twin paradox has a resolution when the twins reunite. However, so long as the twins separate in different inertial frames each is older than the other, forever in each's own frame. That, I imagine, is time dilation.

It is completely arbitrary. There is no requirement that I do my calculations using the frame in which I am rest.Except that my rest frame is my experience.

4. whatif says:
Mister T

Are you talking about the lengths of the line segments that are drawn on the spacetime diagrams?I was talking about the lengths of the projections of the line (object) onto the axes in the analogue, which was just about 2 D space.

5. Nugatory says:
whatif

it still seems to me that a significant flaw in the analogue is that the different coordinate systems come down to an arbitrary choice of axes but the transformations between the different moving frames is not about an arbitrary choice of axes; I think.It is completely arbitrary. There is no requirement that I do my calculations using the frame in which I am rest, and no matter which frame I use I will get the same results for anything and everything that happens – for example, the time on my wristwatch when light from some event reaches my eyes.

Often I choose to use the frame in which I am at rest, but that's not because that frame is special, it's because the problem at hand is most easily solved in that frame.

6. Orodruin says:
whatif

but it still seems to me that a significant flaw in the analogue is that the different coordinate systems come down to an arbitrary choice of axes but the transformations between the different moving frames is not about an arbitrary choice of axes; I think.Then this is where your intuition fails you. It is exactly what it comes down to. A choice of inertial system is exactly analogous to an arbitrary choice of (mutually orthogonal) coordinate axes. The geometry of Minkowski space just implies that the corresponding transformation is a hyperbolic rotation rather than a normal rotation.

7. Ibix says:
whatif

I think that I understand the resolution of the twin paradox using a Minkowski diagram, but this geometrical analogue and explanation does not help me better understand time dilation. I would say that maybe it was not intended to if that was not stated as an aim.The ticks of a clock are regular marks on the timelike coordinate axis. If you are in motion with respect to me, your timelike coordinate axis is at an angle to mine. The spacing between your regular ticks, as measured by me, is the projection of the spacing you measure onto my coordinate axis. That's time dilation. The Euclidean analogy is two rulers at an angle to each other. Assuming the rulers cross at their zero mark, both rulers agree that the 1m mark on the other ruler is level with their 95cm mark (or whatever – depends on the angle). Minkowski geometry is just more mathematically interesting.

whatif

Also, no analogue is perfect but it still seems to me that a significant flaw in the analogue is that the different coordinate systems come down to an arbitrary choice of axes but the transformations between the different moving frames is not about an arbitrary choice of axes; I think.Of course the choice of axes is arbitrary. How is an experiment affected if I observe it sutting still, or walking past, or flying by in a spaceship at 0.95c? Each of those options would naturally choose a different frame. The transform between any pair of frames is well defined, but no frame is right or wrong – the choice is completely arbitrary.

8. Mister T says:
whatif

I think the lengths in the different coordinate systems of the geometric analogue is just a matter of linear scaling with different base vectors.Are you talking about the lengths of the line segments that are drawn on the spacetime diagrams?

9. whatif says:

You need to distinguish carefully between the two cases Orodruin was talking about in the article, and I don't think you are doing so.I think that I understand the resolution of the twin paradox using a Minkowski diagram, but this geometrical analogue and explanation does not help me better understand time dilation. I would say that maybe it was not intended to if that was not stated as an aim.

Also, no analogue is perfect but it still seems to me that a significant flaw in the analogue is that the different coordinate systems come down to an arbitrary choice of axes but the transformations between the different moving frames is not about an arbitrary choice of axes; I think.

10. Ibix says:
whatif

That was my point, except that that it applies in the same inertial reference frame. While the transformations are mathematically linear I understood that length varied according to relative motion of the frame in which it is measured.You need to distinguish carefully between the two cases Orodruin was talking about in the article, and I don't think you are doing so. One case is Euclidean geometry, in which length is the same in all coordinate systems and time is no part of geometry. The other case is the Minkowski geometry that underlies relativity, which includes time as a dimension. In this case, a concept called "interval" is analogous to length in Euclidean geometry while the spatial length of an object is analogous to the difference in x coordinates between the ends of a rod in Euclidean geometry.

The interval is invariant between frames in relativity, but length is not. In Euclidean geometry, length is invariant and interval isn't a concept. The coordinate transforms between frames are linear in both cases.

11. whatif says:
the Lorentz transformations are linear transformations.Sorry, yes they are.

Length has nothing to do with base vectors. It is invariant and does not depend on the coordinate system.That was my point, except that that it applies in the same inertial reference frame. While the transformations are mathematically linear I understood that length varied according to relative motion of the frame in which it is measured.

12. Orodruin says:
whatif

I think the lengths in the different coordinate systems of the geometric analogue is just a matter of linear scaling with different base vectors.Length has nothing to do with base vectors. It is invariant and does not depend on the coordinate system.

whatif

As I understand it, the lengths and time intervals for special relativity are different, not as a result of arbitrary scaling but as a result of relative motion between of the inertial frames and the transformations are nonlinear.No, the Lorentz transformations are linear transformations. The entire point is that spacetime intervals do not depend on the coordinate system, just as little as the distance between you and your neighbour does not depend on the coordinate system.

13. whatif says:

I am a novice about relativity, but I found this convoluted and very difficult to follow. I think the lengths in the different coordinate systems of the geometric analogue is just a matter of linear scaling with different base vectors. The lengths are identical in each frame when accounting for the scaling. As I understand it, the lengths and time intervals for special relativity are different, not as a result of arbitrary scaling but as a result of relative motion between of the inertial frames and the transformations are nonlinear.

14. greswd says:

[QUOTE=”Orodruin, post: 5447359, member: 510075″]But exactly here lies the problem. I am not arguing that the coordinates are abad idea, I just question the pedagogical value to people who are just learning relativity. I do not see the point of introducing non-inertial coordinate systems on top of the struggles they already have.[/QUOTE]
That’s true. But the moment they understand time dilation, they will demand to know things from the travelling twin’s perspective.

15. Orodruin says:

[QUOTE=”the_emi_guy, post: 5448229, member: 415127″]Again, I am sharing that there is a way to introduce the clock skew phenomena in a way that does not require proper time, Minkowsky diagrams, etc.[/QUOTE]
This to me is counter productive for the real understanding of SR, which should be the actual goal. You will still require the use of the Lorentz transformations, the application of the relativity of simultaneity, etc. This completely defeats the pedagogical purpose as it is essentially just giving the formulas and letting the student figure out their meaning. It is like giving a student the formula for the coordinate change due to a rotation and expect them to figure out the rest.

Proper time and Minkowski diagrams are fundamental tools in understanding SR. You should teach them, not try to sweep them under the carpet.

16. Dale says:

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]On the other hand, the “traveling” twin is undergoing acceleration bringing him into a different inertial frame. This is outside the scope of SR unless we assume a non-physical observer[/QUOTE]I disagree with both of these sentences.

In the first, no object goes into or out of any frame. All objects or observers are in every frame at all points of the journey. An object may be at rest in one frame and moving in another, but it is in both frames.

In the second, SR applies as long as gravity is negligible (spacetime is flat). An accelerating observer is certainly within the scope of SR.

17. PeterDonis says:

[QUOTE=”the_emi_guy, post: 5448229, member: 415127″]but the explanation seems to be a stumbling block for many (thus the existence of this very thread)[/QUOTE]

This thread is about Orodruin’s post describing the geometric method, which, as he says, he uses to teach the subject.

[QUOTE=”the_emi_guy, post: 5448229, member: 415127″]You have never heard of this approach? I thought it was fairly well known.[/QUOTE]

I don’t mean solving the problem yourself; of course that can be done. I mean using it to teach someone else. That’s what you’re claiming: that your approach will work better for teaching. Have you tried that? Has it worked?

18. the_emi_guy says:

[QUOTE=”PeterDonis, post: 5448209, member: 197831″]But the single traveling twin doesn’t do this. He just goes out, comes back, and never adjusts his clock at all during the whole trip–yet when he comes back, his clock has less elapsed time than his stay-at-home twin. So it seems to me that your “explanation” is introducing extraneous factors that aren’t there in the original scenario. That seems more likely to confuse than to enlighten a person who is struggling to understand the scenario.[/QUOTE]

Sure, the two observer problem statement is simpler (2 < 3), but the explanation seems to be a stumbling block for many (thus the existence of this very thread). And even if they learn how to draw the diagrams, they may not have an intuitive feel for what is going on. Again, I am sharing that there is a way to introduce the clock skew phenomena in a way that does not require proper time, Minkowsky diagrams, etc. As soon as a student gets time dilation he/she can discover and understand clock skew behavior with three observers.[QUOTE="PeterDonis, post: 5448209, member: 197831"]Have you ever actually tried this? Has it worked?[/QUOTE]You have never heard of this approach? I thought it was fairly well known. Section 4 of the attached journal is one reference and begins:"In this section we will demonstrate a method that can be used in the “clock paradox” problem without the need to consider the effects of various accelerations and decelerations, at least in principle. Such a method was first proposed by Lord Halsbury in 1957 [17], as a “triplet” or a “three clocks” problem, and can be briefly stated as follows..."

19. PeterDonis says:

[QUOTE=”the_emi_guy, post: 5448196, member: 415127″]I think that there is unrecognized value in seeing that an outbound (non-returning) observer sees Earth clock running slow, later synchronizes his “ahead” clock with an inbound clock, then watches the inbound clock go from being ahead to being behind Earth do to the greater inbound vs. outbound relative velocity.[/QUOTE]

But the single traveling twin doesn’t do this. He just goes out, comes back, and never adjusts his clock at all during the whole trip–yet when he comes back, his clock has less elapsed time than his stay-at-home twin. So it seems to me that your “explanation” is introducing extraneous factors that aren’t there in the original scenario. That seems more likely to confuse than to enlighten a person who is struggling to understand the scenario.

[QUOTE=”the_emi_guy, post: 5448196, member: 415127″]Doing this in all three observer rest frames and getting the same clock skew I think is compelling to a beginner, and the concept of proper time was not even needed or required.[/QUOTE]

Have you ever actually tried this? Has it worked?

20. the_emi_guy says:

Peter, thanks for the thoughtful response, I appreciate it.

I think that there is unrecognized value in seeing that an outbound (non-returning) observer sees Earth clock running slow, later synchronizes his “ahead” clock with an inbound clock, then watches the inbound clock go from being ahead to being behind Earth do to the greater inbound vs. outbound relative velocity. Note that this clock skew is relative to outbound’s local clock, Earth’s clock ended up being behind, and inbound’s clock even more behind.

Doing this in all three observer rest frames and getting the same clock skew I think is compelling to a beginner, and the concept of proper time was not even needed or required.

21. PeterDonis says:

[QUOTE=”the_emi_guy, post: 5448087, member: 415127″]These folks are better served by showing them how simple application of time dilation to all three inertial frames (see post #18) shows this in a trivial and intuitive manner (and I do mean “inertial frames” here, with each observer at rest in his frame).[/QUOTE]

Whether this is “trivial and intuitive” is debatable (see below), but it seems to me that these folks would be served better still by giving them the most general method of all: spacetime geometry and lengths of curves. What we have here is a triangle in spacetime: one side is the stay-at-home twin’s worldline, and the other two sides are the traveling twin’s outbound and inbound worldlines. Then all we need is the Minkowski spacetime analogue to the Pythagorean theorem–of which the time dilation equation you are implicitly using in your method is just a special case.

[QUOTE=”the_emi_guy, post: 5448087, member: 415127″]As a [I]follow on[/I] we could consider “what if there were only two observers?”[/QUOTE]

That’s not a follow-on; it’s how the original scenario was posed. When you formulate it in terms of three observers, you are, on the face of it, talking about a [I]different[/I] scenario. You then have to explain how the result you derive with three observers gives the right answer for the case of only two, which was the actual scenario posed. This is certainly doable, but I question whether it is “trivial and intuitive” to a person who is struggling with time dilation and differential aging.

[QUOTE=”the_emi_guy, post: 5448087, member: 415127″]what did the second observer do? Did he turn around instantaneously experiencing infinite acceleration (a non-physical physics problem)? What if he accelerates and decelerates gradually over time? What if he slingshots himself around another planet experiencing weightlessness the whole way? It may very well turn out that all of these produce the same result because spacetime does not get curved and SR still applies (or maybe not).[/QUOTE]

The first case is the one that is equivalent to your analysis–although I would prefer to phrase it in terms of taking the limit as the turnaround time (the time elapsed on the traveling twin’s clock during the period when he is accelerating to turn around) goes to zero. The key point is that, if the answer we are interested in is elapsed proper time for the two twins, we don’t really care about the details of how the traveling twin turns around, as long as the time elapsed on his clock while doing so is negligible compared to the time elapsed on his clock during the inbound and outbound legs.

The second case, accelerating and decelerating gradually over time, is [I]not[/I] equivalent to your analysis. And the simplest way to state the reason why is by using spacetime geometry and worldlines: the worldline of the traveling twin is [I]different[/I] in this case than it is in the idealized case you are analyzing (which is equivalent to the idealized case of instantaneous turnaround). So you would not expect the traveling twin’s elapsed time to be the same. It will still be shorter than the stay-at-home twin’s elapsed time when they meet up again, but by a different amount (which will depend on the details of how the twin accelerates and decelerates gradually). The spacetime geometry method adapts easily to this case; it just amounts to evaluating arc length along a curve instead of a pair of straight lines.

The two cases above can both be formulated in flat spacetime. The third, by contrast, requires curved spacetime, because in flat spacetime there is no way for the traveling twin to turn around without feeling acceleration, whereas in this third “slingshot” case, the traveling twin is weightless, feeling no acceleration, the whole way. The spacetime geometry method carries over to this case with no problem at all, since, as I noted, it is completely general.

22. the_emi_guy says:

[QUOTE=”PAllen, post: 5447959, member: 275028″]
What if there are only two observers? How is it useful, pedogogically, to say we must introduce a third to explain what happens? Especially, if the spacetime geometry viewpoint is very simple and requires only one frame (not necessarily one in which any observer is at rest).[/QUOTE]

Thanks for asking this, it gets right to my point.

Many folks who visit PF have learned about time dilation between observers who are in relative motion but want to get a handle on how perfectly good clocks can wind up sitting side-by-side but out of sync due to some motion that occurred. These folks are better served by showing them how simple application of time dilation to all three inertial frames (see post #18) shows this in a trivial and intuitive manner (and I do mean “inertial frames” here, with each observer at rest in his frame).

As a [I]follow on[/I] we could consider “what if there were only two observers?”.
Well, what did the second observer do? Did he turn around instantaneously experiencing infinite acceleration (a non-physical physics problem)? What if he accelerates and decelerates gradually over time? What if he slingshots himself around another planet experiencing weightlessness the whole way? It may very well turn out that all of these produce the same result because spacetime does not get curved and SR still applies (or maybe not). Volumes have been written about all these various scenarios.

23. PeterDonis says:

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]I can have an observer (real human observer) in an single inertial frame[/QUOTE]

You are misusing the term “inertial frame”. The correct way of stating what you are trying to state here is: I can have an observer in the same inertial [I]state of motion[/I]… The observer’s state of motion is the physical thing. The “inertial frame” is a mathematical construct that we use to help us describe what is going on. And, as others have remarked, the same observer’s worldline, with the same state of motion, can be described using as many different inertial frames–or non-inertial frames, for that matter–as we like.

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]the “traveling” twin is undergoing acceleration bringing him into a different inertial frame.[/QUOTE]

No. The traveling twin undergoes acceleration which [I]changes his state of motion[/I]. (Even this terminology has limitations, but it will do for this discussion.) But there is no need to switch inertial frames to describe this; you can describe it all perfectly well using a single inertial frame.

[QUOTE=”the_emi_guy, post: 5447914, member: 415127″]So acceleration (and presumably gravity by equivalence property)[/QUOTE]

You are misstating this as well. “Gravity” is an ambiguous term. The correct statement, which has already been given in this thread, is that SR deals with any scenario in which spacetime is flat. GR deals with scenarios in which spacetime can be curved.

24. PAllen says:

[QUOTE=”the_emi_guy, post: 5447914, member: 415127″]So acceleration (and presumably gravity by equivalence property) is withing the scope of SR, I’ll have to take your word for this. Still, I’ll repeat from earlier:
[/QUOTE]
No, per modern understanding, gravity involves curved spacetime, while SR involves flat spacetime. The equivalence principle is local,and is analogous to the statement that a small region of sphere is well approximated by a plane. Tidal gravity effects distinguish gravity from acceleration, but they become very small in a small region.
[QUOTE=”the_emi_guy, post: 5447914, member: 415127″]
Why bother with all of this complexity when formulating the problem with three observers (first proposed by Lord Halsbury) gets the expected result with simple plug ‘n chug application of time dilation applied twice to of the three observers.[/QUOTE]

What if there are only two observers? How is it useful, pedogogically, to say we must introduce a third to explain what happens? Especially, if the spacetime geometry viewpoint is very simple and requires only one frame (not necessarily one in which any observer is at rest).

25. the_emi_guy says:

So acceleration (and presumably gravity by equivalence property) is withing the scope of SR, I’ll have to take your word for this. Still, I’ll repeat from earlier:

Why bother with all of this complexity when formulating the problem with three observers (first proposed by Lord Halsbury) gets the expected result with simple plug ‘n chug application of time dilation applied twice to of the three observers.

26. Orodruin says:

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]If we are not talking about actual masses traveling in these inertial frames then we are not talking about physics.[/QUOTE]
Wrong again. Please read my previous post.

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]Nope, I can have an observer (real human observer) in an single inertial frame and change coordinates all day. While traveling through New York my position can be rendered in Cartesian coordinates, then when I cross the border into New Jersey we decide to render my position in some polar coordinates. Nothing about this change in coordinates implies a change in inertial frame. On the other hand, the “traveling” twin is undergoing acceleration bringing him into a different inertial frame. This is outside the scope of SR unless we assume a non-physical observer.[/QUOTE]
Again, a very common misconception. What you fail to realise is that the changes from Cartesian to polar coordinates is just the same as applied to space as a Lorentz transformation is to space-time (not completely true, a Lorentz transformation is the equivalent of a change between different [I]Cartesian[/I] coordinate systems in space). You can easily use a curvilinear coordinate system in SR as well and there are many examples. Unless you accept that SR is not a theory about observers but a theory about the geometry of space-time, you will be very hard pressed to try to learn GR.

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]My question is why bother? With three observers, each in his own unchanging inertial frame, we get the expected result with simple plug ‘n chug application of time dilation applied twice to of the three observers, all comfortably within SR.[/QUOTE]
Everything in this thread is SR. That SR cannot deal with acceleration is a myth.

27. PAllen says:

[QUOTE=”the_emi_guy, post: 5447872, member: 415127″]If we are not talking about actual masses traveling in these inertial frames then we are not talking about physics.

[/QUOTE]
This is a nonsensical claim, and the rest of your claims follow from it. So far as I know, there is no reputable modern theoretical physicist who would agree with the claim that acceleration and accelerated observers are outside the scope of SR.

28. the_emi_guy says:

[QUOTE=”Orodruin, post: 5447808, member: 510075″]No, this is not true. You can use any actual observer to define a reference frame, but it is not necessary for the theory to make sense. [/QUOTE]
If we are not talking about actual masses traveling in these inertial frames then we are not talking about physics.

[QUOTE=”Orodruin, post: 5447808, member: 510075″]Changes of inertial frames are just changes of how you assign coordinates. What you are claiming is essentially the equivalent of saying that you cannot use spherical coordinates unless you have an actual sphere, which is obviously absurd. .[/QUOTE]
Nope, I can have an observer (real human observer) in an single inertial frame and change coordinates all day. While traveling through New York my position can be rendered in Cartesian coordinates, then when I cross the border into New Jersey we decide to render my position in some polar coordinates. Nothing about this change in coordinates implies a change in inertial frame. On the other hand, the “traveling” twin is undergoing acceleration bringing him into a different inertial frame. This is outside the scope of SR unless we assume a non-physical observer.

I am not just a “novice” making this up. I have seen 30 page dissertations on the subject of how to properly handle the acceleration problem inherent in the two observer twin paradox. Just go to Wikipedia under “twin paradox” and search for the word “acceleration”.

My question is why bother? With three observers, each in his own unchanging inertial frame, we get the expected result with simple plug ‘n chug application of time dilation applied twice to of the three observers, all comfortably within SR.

29. Orodruin says:

[QUOTE=”the_emi_guy, post: 5447761, member: 415127″]We have to allow the observers to have physical bodies, and the clocks to have actual mass, otherwise this is not a physics problem but a mathematical exercise.[/QUOTE]
No, this is not true. You can use any actual observer to define a reference frame, but it is not necessary for the theory to make sense. Changes of inertial frames are just changes of how you assign coordinates. What you are claiming is essentially the equivalent of saying that you cannot use spherical coordinates unless you have an actual sphere, which is obviously absurd. It may not be the most convenient coordinate system unless you have some sort of spherical symmetry, but you certainly can use it.

A physical exercise is any exercise which makes experimentally verifiable predictions. You can use any coordinate system which might be convenient to make those predictions. Like many novices, you are putting way too much relevance into coordinates instead of focusing on the actual issue in the twin paradox problem – the geometry of space-time. You can use any coordinate system you wish, the geometry will not change.

[QUOTE=”the_emi_guy, post: 5447761, member: 415127″]More importantly, why would we take a problem that when formulated with three real human observers is trivial to analyze, and formulate it in a way that is very complicated (look at how many questions PF gets on this topic), and requires hypothetical massless observers?[/QUOTE]
There are no massless hypothetical observers in this example. Just different coordinate systems on the same geometrical object.

30. the_emi_guy says:

We have to allow the observers to have physical bodies, and the clocks to have actual mass, otherwise this is not a physics problem but a mathematical exercise.

More importantly, why would we take a problem that when formulated with three real human observers is trivial to analyze, and formulate it in a way that is very complicated (look at how many questions PF gets on this topic), and requires hypothetical massless observers?

31. Orodruin says:

[QUOTE=”the_emi_guy, post: 5447701, member: 415127″]Seems to me that SR cannot be applied to a problem that involves three inertial frames and only two
observers. Isn’t an observer moving from one inertial frame to another is out of the scope of SR?[/QUOTE]

No. It is perfectly within the scope of SR. You do not need an actual observer to define an inertial frame in SR.

32. the_emi_guy says:

Seems to me that SR cannot be applied to a problem that involves three inertial frames and only two
observers. Isn’t an observer moving from one inertial frame to another is out of the scope of SR?
Seems that any solution would require a footnote indicating that
the presumably infinite acceleration experienced by the “travelling” twin is assumed to have no implication.

On the other hand, allowing three observers, one in each frame, is a valid SR problem, and its solution is a trivial
application of time dilation and removes all of the complexity of the two observer problem.

Observers:
Observer1 (Earthbound observer) is on earth.
Observer2 (outbound observer) passes earth traveling toward Alpha Centauri (and never comes back)
Observer3 (inbound observer) is traveling from Alpha Centauri and first passes outbound observer, then later passes earth.

Events:
Earth passes his time to outbound observer when they meet.
Outbound observer passes his time to inbound observer when they meet.

Analysis:
Outbound observer’s point of view (exclusively): Earthbound clock is running slow (simple time dilation)
thus inbound observer’s clock is set ahead of Earth’s clock when outbound and inbound meet.

Now continuing along with outbound observer (*not* switching inertial frames). Earth’s clock is still running slow, but the
inbound observer’s clock is running even slower since this relative velocity is larger.
Thus, by the time that the outbound observer sees the inbound observer reach earth, he will see that the inbound observer’s clock will have gone from being ahead of Earth’s to being behind Earth’s by simple application of time dilation applied to this larger relative velocity.

This can be worked using simple SR time dilation from any of the three inertial frames to obtain identical results.

33. Dale says:

[QUOTE=”Orodruin, post: 5447359, member: 510075″]But exactly here lies the problem. I am not arguing that the coordinates are abad idea, I just question the pedagogical value to people who are just learning relativity. I do not see the point of introducing non-inertial coordinate systems on top of the struggles they already have.[/QUOTE]I agree. Pedagogically, for a novice I would encourage them to stick with inertial frames. Then if they persist use the paper in the spirit of “your question does have an answer, here it is, come back to it when you are ready”.

34. Orodruin says:

[QUOTE=”Dale, post: 5447343, member: 43978″]I like it. It is one of my favorites.

That said, I wouldn’t worry too much about details like the part of the trip where the distance is constant. It is a non-inertial frame so distance and speed don’t have their ordinary meanings anyway. The point is mostly that the travelling twin’s “perspective” is not symmetrical to the home twin.[/QUOTE]
But exactly here lies the problem. I am not arguing that the coordinates are abad idea, I just question the pedagogical value to people who are just learning relativity. I do not see the point of introducing non-inertial coordinate systems on top of the struggles they already have.

35. Dale says:

[QUOTE=”greswd, post: 5446549, member: 445614″]What do you think of this derivation of the space-time diagram from the travelling twin’s perspective?

As described in this paper: [URL]http://arxiv.org/abs/gr-qc/0104077v2[/URL][/QUOTE]I like it. It is one of my favorites.

That said, I wouldn’t worry too much about details like the part of the trip where the distance is constant. It is a non-inertial frame so distance and speed don’t have their ordinary meanings anyway. The point is mostly that the travelling twin’s “perspective” is not symmetrical to the home twin.

36. PAllen says:

[QUOTE=”Orodruin, post: 5446561, member: 510075″]This is full of misconceptions. First of all, all objects exist in all frames. Frames are just different ways of assigning coordinates to events. You can use any viable frame to describe any situation as long as you do it properly. Some frames are better suited for the analysis, but that does not imply a special status. The “radar coordinates” introduced in the paper is a non-minkowski coordinate system. You should be as little surprised about having different proper time per radar time as you should be that changing a polar coordinate results in different displacements for different distances from the origin in normal Euclidean space.

Apart from that, I do not see anything particularly new or of additional pedagogical value here. Just the added complexity of not using a minkowski coordinate chart.[/QUOTE]
Note that radar coordinates are exactly Minkowski coordinates if the defining world line is eternally inertial. They are asymptotically the same as Fermi-Normal coordinates near the defining world line. They have the specific feature that as long as the defining world line is inertial for all proper time before some event, and also inertial for all proper time after some event, no matter how complex the motion in between, and however ‘long’ such complex motion occurs, they give a complete coordinate chart of Minkowski space (unlike Fermi-Normal coordinates). Thus, they can be seen as a generalization of Minkowski coordinates to describe the ‘experience’ of a non-inertial observer, having several nice properties (including also they are built from more achievable measurements than positing very long rulers).

All the same, one must never attach ‘more reality’ to a description in radar coordinates than in some other coordinates.

37. Nugatory says:

[QUOTE=”greswd, post: 5446574, member: 445614″]This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time[/quote]
[quote]By “same frame”, I mean same inertial reference frame. Both being at rest w.r.t. each other.[/QUOTE]
Whether the twins are at rest relative to one another has nothing to do with whether they are in the same frame. Both twins are always in all frames all the time.

38. greswd says:

[QUOTE=”Orodruin, post: 5446561, member: 510075″]This is full of misconceptions. First of all, all objects exist in all frames. Frames are just different ways of assigning coordinates to events. You can use any viable frame to describe any situation as long as you do it properly. Some frames are better suited for the analysis, but that does not imply a special status. The “radar coordinates” introduced in the paper is a non-minkowski coordinate system. You should be as little surprised about having different proper time per radar time as you should be that changing a polar coordinate results in different displacements for different distances from the origin in normal Euclidean space.

Apart from that, I do not see anything particularly new or of additional pedagogical value here. Just the added complexity of not using a minkowski coordinate chart.[/QUOTE]
By “same frame”, I mean same inertial reference frame. Both being at rest w.r.t. each other.

I’m not sure what you mean by “special status”, I don’t think I mentioned that any frame had a special status.

The “time compression” may be viewed as a form of gravitational blueshift. This and the “same frame” effect are described as being due to “shock like scale discontinuity”.

Speaking of additional pedagogical value, I was just curious about the travelling twin’s perspective. I think many people have never seen such a diagram before too.

39. Orodruin says:

[QUOTE=”greswd, post: 5446557, member: 445614″]I think many people are interested in the travelling twin’s perspective.

This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time, this is not the case in the staying twin’s frame.

And despite both being in the same frame, the travelling twin sees the other twin undergo time compression (the opposite of time dilation).[/QUOTE]
This is full of misconceptions. First of all, all objects exist in all frames. Frames are just different ways of assigning coordinates to events. You can use any viable frame to describe any situation as long as you do it properly. Some frames are better suited for the analysis, but that does not imply a special status. The “radar coordinates” introduced in the paper is a non-minkowski coordinate system. You should be as little surprised about having different proper time per radar time as you should be that changing a polar coordinate results in different displacements for different distances from the origin in normal Euclidean space.

Apart from that, I do not see anything particularly new or of additional pedagogical value here. Just the added complexity of not using a minkowski coordinate chart.

40. greswd says:

[quote=”greswd, post: 5446549″]What do you think of this derivation of the space-time diagram from the travelling twin’s perspective?

[img]https://www.physicsforums.com/attachments/triplets10-png.55802/[/img]

As described in this paper: http://arxiv.org/pdf/gr-qc/0104077v2.pdf%5B/quote%5DI think many people are interested in the travelling twin’s perspective.

This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time, this is not the case in the staying twin’s frame.

And despite both being in the same frame, the travelling twin sees the other twin undergo time compression (the opposite of time dilation).

41. greswd says:

What do you think of this derivation of the space-time diagram from the travelling twin’s perspective?

[img]https://www.physicsforums.com/attachments/triplets10-png.55802/[/img]

As described in this paper: [URL]http://arxiv.org/abs/gr-qc/0104077v2[/URL]

42. Smattering says:

[QUOTE=”Orodruin, post: 5287000, member: 510075″]Orodruin submitted a new PF Insights post[/QUOTE]

This is indeed a very good read. Thanks a lot for the effort.

43. Orodruin says:

[QUOTE=”SlowThinker, post: 5287442, member: 572662″]Perhaps you’re going a bit fast near the end.

It seems the twin staying behind should have more proper time, since ##frac{tau_1}{tau_2}=gammageq1##.[/QUOTE]
Youre right of course, fixed.

44. SlowThinker says:

Perhaps you’re going a bit fast near the end.
[QUOTE]Just applying the modified Pythagorean theorem it becomes clear that the proper times along the world lines are given by
$$tau_1=2tau, tau_2=2tausqrt{1-frac{v^2}{c^2}} implies frac{tau_1}{tau_2}=gamma$$
and therefore [B]less[/B] proper time will pass for the twin staying behind, i.e., the red curve.[/QUOTE]
It seems the twin staying behind should have more proper time, since ##frac{tau_1}{tau_2}=gammageq1##.

45. Orodruin says:

[QUOTE=”DaleSpam, post: 5287092, member: 43978″]The geometrical view does help clear up a lot of the confusing aspects of SR, but it seems to be a big leap for new students.[/QUOTE]
I find the geometrical comparison to Euclidean space helps students and it is how I start the presentation of the subject in my SR class, by introducing Lorentz transforms as the hyperbolic rotations leaving the Minkowski line element invariant. Admittedly, these are students who are in their fourth year of university and they may be more ready for it. I had the idea of writing this Insight three days ago when I was giving the corresponding lecture, although I have tried to make it slightly more accessible than the level I present it on in class.

46. Dale says:

Thanks Orodruin. This is definitely worth referencing.

The geometrical view does help clear up a lot of the confusing aspects of SR, but it seems to be a big leap for new students. I wonder which is more difficult for most students: understanding the geometrical view or understanding the relativity of simultaneity.

47. greswd says:

I think many people are interested in the travelling twin's perspective.This diagram is bizarre because both twins are in the SAME FRAME for a finite duration of time, this is not the case in the staying twin's frame.And despite both being in the same frame, the travelling twin sees the other twin undergo time compression (the opposite of time dilation).

48. RJLiberator says:

Excellent write up. Well-written! Thanks.

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# Give an orthonormal basis for null(T), for $T \in \mathbb{L(\mathbb{C^4)}}$

Question:

Give an orthonormal basis for null(T), for $T \in \mathbb{L(\mathbb{C^4)}}$ is the map with canonical matrix: $$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ \end{bmatrix}$$

My Steps:

First I put the matrix in RREF to find the dimension of the null space: $$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ Resulting in a null space dimension 3. With this I solved the matrix for the null space getting $$x_1 = -a -b -c$$ $$x_2 = a$$ $$x_3 = b$$ $$x_4 = c$$ Solving for the basis of the null space, I got $$a \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix} + b \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \\ \end{bmatrix} + c \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \\ \end{bmatrix}$$ Was this the proper way to solve for the basis of the null space given this matrix and if so, since a, b, and c are arbitrary scalars, could I apply the Gram-Schmidt procedure on the three vectors to find the orthonormal basis?

• Yes, go ahead with Gram-Schmidt. – Friedrich Philipp Nov 11 '17 at 23:41

$$H_4H_4^T = 4I$$
where $$H_4 = \begin{bmatrix} 1 & 1 & 1 & 1 \\ -1 & 1 & -1 & 1 \\ -1 & -1 & 1 & 1 \\ 1 & -1 & -1 & 1 \\ \end{bmatrix}$$


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# Need help checking working (limit at infinity)

Need help checking my working with this limit at infinity.

$$\lim_{x\to \infty} (1+x^2)^{1/\ln x}$$

All help is appreciated

My working was-

$$\ln(1+x^2)^{1/\ln(x)} = \frac {\ln(1+x^2)}{\ln(x)}$$

Using L'Hopital

$$\frac {\frac {2x}{1+x^2}}{\frac {1}{x}}$$

$$\frac {2x}{1+x^2}$$

My question is does the 1 here simply get cancelled as x approaches infinity. That would give me the correct answer but I want to know that I didn't accidentally get to it in which case my working would be wrong. $$2=\lim_{x\to \infty} \ln(1+x^2)^{1/\ln(x)}$$

$$e^2=\lim_{x\to \infty} (1+x^2)^{1/\ln(x)}$$

• You get ${2x^2\over 1+x^2}$ after using L'Hopital. – Edu Aug 7 '17 at 13:19
• what is $n$ here in your formula? – Dr. Sonnhard Graubner Aug 7 '17 at 13:20
• So you really mean $\ln\Big((1+x^2)^{1/\ln(x)}\Big)$ and not $\Big(\ln(1+x^2)\Big)^{1/\ln(x)}$ right ? – Surb Aug 7 '17 at 13:29
• Please get rid of $n$ in your question. – zhw. Aug 7 '17 at 15:23
• @zhw. sorry. Thank you for pointing that out – Virens Aug 7 '17 at 18:08

Simplifying the fraction after applying L'Hopital, you should get $$\frac{2x^2}{1+x^2}$$ instead of $$\frac{2x}{1+x^2}$$
Then to get $$\lim_{x \to \infty} \frac{2x^2}{1+x^2}$$ you can again apply L'Hopital and get $$\lim_{x \to \infty} \frac{2x^2}{1+x^2}=\lim_{x \to \infty} \frac{4x}{2x}=2$$


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## April 2, 2007

### Rota on Combinatorics

#### Posted by David Corfield

From an interview with Gian-Carlo Rota and David Sharp:

Combinatorics is an honest subject. No adèles, no sigma-algebras. You count balls in a box, and you either have the right number or you haven’t. You get the feeling that the result you have discovered is forever, because it’s concrete. Other branches of mathematics are not so clear-cut. Functional analysis of infinite-dimensional spaces is never fully convincing; you don’t get a feeling of having done an honest day’s work. Don’t get the wrong idea - combinatorics is not just putting balls into boxes. Counting finite sets can be a highbrow undertaking, with sophisticated techniques.

I remember somewhere else he spoke of his mathematical ‘bottom line’ as concerned with putting balls in boxes.

Much combinatorics of our day came out of an extraordinary coincidence. Disparate problems in combinatorics, ranging from problems in statistical mechanics to the problem of coloring a map, seem to bear no common features. However, they do have at least one common feature: their solution can be reduced to the problem of finding the roots of some polynomial or analytic function. The minimum number of colors required to properly color a map is given by the roots of a polynomial, called the chromatic polynomial; its value at $N$ tells you in how many ways you can color the map with $N$ colors. Similarly, the singularities of some complicated analytic function tell you the temperature at which a phase transition occurs in matter. The great insight, which is a long way from being understood, was to realize that the roots of the polynomials and analytic functions arising in a lot of combinatorial problems are the Betti numbers of certain surfaces related to the problem. Roughly speaking, the Betti numbers of a surface describe the number of different ways you can go around it. We are now trying to understand how this extraordinary coincidence comes about. If we do, we will have found a notable unification in mathematics.

Does anyone know whether progress has been made in explaining the ‘extraordinary coincidence’?

Posted at April 2, 2007 2:43 PM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1224

### Re: Rota on Combinatorics

The minimum number of colors required to properly color a map is given by the roots of a polynomial, called the chromatic polynomial; its value at N tells you in how many ways you can color the map with N colors.

Perhaps I’m only showing my naivete, but I’m not sure what the roots have to do with anything. As he says only a moment later, the natural interest is in the values at natural numbers. There are a lot of polynomials that it has never occurred to me to seek the roots of – Kazhdan-Lusztig polynomials, for example.

The idea that polynomials with positive coefficients (as combinatorial ones usually have, of course) should be Poincare polynomials of some space has made great advances. My favorite example is the Geometric Satake Correspondence, in which the weight multiplicities in a G-representation (w.r.t. the circle in a principal SL_2) are shown to be the coefficients of the intersection homology Poincare polynomial of a Schubert variety in the loop Grassmannian of the Langlands dual of G.

If a polynomial arises as the q-analogue of some number, then the evidence is especially compelling that there should be a space whose F_q-points are being counted, and the Weil conjectures say that one could equivalently (to some extent) compute the Poincare polynomial of that space.

Posted by: Allen Knutson on April 2, 2007 5:15 PM | Permalink | Reply to this

### Re: Rota on Combinatorics

Allen wrote:

Rota wrote:

The minimum number of colors required to properly color a map is given by the roots of a polynomial, called the chromatic polynomial; its value at N tells you in how many ways you can color the map with N colors.

Perhaps I’m only showing my naivete, but I’m not sure what the roots have to do with anything. As he says only a moment later, the natural interest is in the values at natural numbers.

Probably Rota just meant that there are no ways to N-color a map (or really a graph) if its chromatic polynomial vanishes at N. Using this, we can in principle figure out the minimum number of colors required.

Rota was probably talking this way just to be comprehensible by layfolk while still sounding really cool.

But, there are weirdly interesting questions about the non-integral roots of the chromatic polynomial! From week22:

Consider the problem of trying to color the vertices of a graph with $n$ colors in such a way that no two vertices at opposite ends of any given edge have the same color. Let $P(n)$ denote the number of such n-colorings. This turns out to be a polynomial in $n$ - it’s not hard to see using recursion relations similar to the skein relations above. It also turns out that the 4-color theorem is equivalent to saying that the vertices of any planar graph can be 4-colored. (To see this, just use the idea of the “dual graph” of a graph - the vertices of the one being in 1-1 correspondence with the edges of the other.) So, another way to state the 4-color theorem is that for no planar graph does the polynomial $P(n)$ have a root at $n = 4$.

$P(n)$ is called the “chromatic polynomial” and has been intensively investigated. One very curious thing is this. Remember the golden mean

$\Phi = (\sqrt{5}+ 1)/2 = 1.61803398874989484820458683437...?$

Well, $\Phi + 1$ is never a root of the chromatic polynomial of a graph! (Unless the polynomial vanishes identically, which happens just when the graph has loops.) The proof is not all that hard, and it’s in Saaty and Kainen’s book.

However — and here’s where things get really interesting — in 1965, Hall, Siry and Vanderslice figured out the chromatic polynomial of a truncated icosahedron. (This looks like a soccer ball or buckyball.) They found that of the four real roots that weren’t integers, one agreed with $\Phi + 1$ up to 8 decimal places! Of course, here one might think the 5-fold symmetry of the situation was secretly playing a role. But in 1966 Barri tabulated a bunch of chromatic polynomials in her thesis, and in 1969 Berman and Tutte noticed that most of them had a root that agreed with $\Phi + 1$ up to at least 5 decimal places.

This curious situation was at least partially explained by Tutte in 1970. He showed that for a triangular planar graph (that is, one all of whose faces are triangles) with $n$ vertices one has

$|P(\Phi + 1)| \le \Phi^{5-n}$

This is apparently not a complete explanation, though, because the truncated icosahedron is not triangular.

This is not an isolated freak curiosity, either! In 1974 Beraha suggested checking out the behavior of chromatic polynomials at what are now called the “Beraha numbers”

$B(n) = 4 cos^2(\pi/n).$

Note by the way that $B(n)$ approaces 4 as $n$ approaches infinity. It turns out that the roots of chromatic polynomials seem to cluster near Beraha numbers — as if graphs were trying to come really close to being impossible to 4-color!

For example, the four nonintegral real roots of the chromatic polynomial of the truncated icosahedron are awfully close to $B(5), B(7), B(8)$ and $B(9)$. Beraha made the following conjecture: let $P_i$ be a sequence of chromatic polynomials of graphs such whose number of vertices approaches infinity as $i \to \infty$. Suppose $r_i$ is a real root of $P_i$ and suppose the $r_i$ approach some number $x$. Then $x$ is a Beraha number. In work in the late 60’s and early 70’s, Tutte proved some results showing that there really was a deep connection between chromatic polynomials and the Beraha numbers.

Well, to make a long story short (I’m getting tired), the Beraha numbers also have a lot to do with the quantum group $SU(2)$. This actually goes back to some important work of Jones right before he discovered the first of the quantum group knot polynomials, the Jones polynomial. He found that — pardon the jargon burst — the Markov trace on the Temperley-Lieb algebra is only nonnegative when the Markov parameter is the reciprocal of a Beraha number or less than 1/4. When the relationship of all this stuff to quantum groups became clear, people realized that this was due to the special nature of quantum groups when $q$ is an $n$th root of unity (this winds up corresponding to the Beraha number $B(n)$.) This all leads up to a paper that, unfortunately, I have not yet read:

• Zeroes of chromatic polynomials: a new approach to the Beraha conjecture using quantum groups, by H. Saleur, Comm. Math. Phys. 132 (1990) 657.

This apparently gives a “physicist’s proof” of the Beraha conjecture, and makes use of conformal field theory, that is, quantum field theory in 2 dimensions that is invariant under conformal transformations.

This all sounds quite cool, and I wish I knew what became of these ideas later on…

However, while Xiao-Song Lin found fascinating patterns in the roots of the Jones polynomials of random links, Dan Christensen and Igor Khavkine’s work suggests that some of these patterns may simply be due to the polynomials having integer coefficients!

The chromatic polynomial also has integer coefficients. So, maybe some of the Beraha conjecture could be explained simply by that fact??? I have no idea.

Posted by: John Baez on April 3, 2007 12:35 AM | Permalink | Reply to this

### Re: Rota on Combinatorics

The chromatic polynomial with q colors is nothing but the partition function of the antiferromagnetic q-state Potts model at T = 0; this forces adjacent spins/colors to be different. At non-zero temperature the partition function becomes the dichromatic polynomial, which depends on two variables q and T. At the critical temperature Tc, it is described by a CFT. If the number of colors is given by

q = 4 cos2(π/m+1)

the CFT at criticality has central charge

c = 1 - 6/m(m+1),

with the same value of m. I.e., the Beraha numbers correspond to the discrete unitary series on the CFT side. In particular, the Ising model corresponds to m = 3.

I always found it remarkable that zeroes of the partition function at T = 0 correspond to unitarity at criticality. But of course, this stuff has a severe limitation: it only works in 2D.

### Re: Rota on Combinatorics

Also note that when m -> infinity, c -> 1 and q -> 4. This has something to do with the four-color theorem.

### Re: Rota on Combinatorics

Is this what John was talking about when discussing Bar-Natan’s paper on the 4 color theorem at the end of TWF96?

Posted by: David Corfield on April 4, 2007 4:34 PM | Permalink | Reply to this

### Re: Rota on Combinatorics

I don’t know about that. It is, however, exactly what I was talking about above in the quote from week22. As $m \to \infty$, the Beraha number $q = 4 \cos^2 (\pi/m+1)$ approaches 4, and we should get some information about the 4-color theorem.

Posted by: John Baez on April 11, 2007 2:47 AM | Permalink | Reply to this

### Re: Rota on Combinatorics

Rota wrote:

Combinatorics is an honest subject. No adèles, no sigma-algebras. You count balls in a box, and you either have the right number or you haven’t. You get the feeling that the result you have discovered is forever, because it’s concrete.

Rota is engaged in some clever salesmanship here, and demonstrating how he could never resist a rhetorical flourish, even if it required a little inconsistency in his stance.

Here he’s suggesting that it takes balls to do combinatorics — that it’s a concrete, manly sort of activity. No fancy “adèles” with their French accents here! But in other pieces, and in his own work, he emphasized the power of abstraction in combinatorics, like the use of Hopf algebras, and espèces de structure.

Indeed, later here he writes:

Don’t get the wrong idea — combinatorics is not just putting balls into boxes. Counting finite sets can be a highbrow undertaking, with sophisticated techniques.

So, he’s actually claiming both ends of the spectrum for combinatorics: the concrete and the abstract.

But this is not unique to combinatorics. To find all the integer solutions of a polynomial equation sounds quite concrete — but before you know it, you wind up using things like adèles.

Posted by: John Baez on April 3, 2007 1:15 AM | Permalink | Reply to this

### Re: Rota on Combinatorics

One of my pet peeves is when people say that the $p$-adic numbers are merely formal devices but that the real numbers are, well, real. So, it’s nice to see Rota label the adeles as formal. If he had only meant the $p$-adic part was formal and not the real part, he surely wouldn’t have bothered to mention the adeles!

Posted by: James on April 3, 2007 1:58 PM | Permalink | Reply to this
Read the post Toegepaste wiskunde volgens Rota en Sharp
Weblog: Filosofie voor elke dag
Excerpt: David Corfields blog wees me gisteren op een interessant interview met de wiskundigen Gian-Carlo Rota en David Sharp. Het dateert al uit 1985, maar bevat heel wat boeiende inzichten over wiskunde, filosofie en kunstmatige intelligentie. Zo heet het artike
Tracked: April 3, 2007 5:36 PM

### Re: Rota on Combinatorics

He’s certainly not averse to abstract constructions:

“What can you prove with exterior algebra that you cannot prove without it?” Whenever you hear this question raised about some new piece of mathematics, be assured that you are likely to be in the presence of something important. In my time, I have heard it repeated for random variables, Laurent Schwartz’ theory of distributions, ideles and Grothendieck’s schemes, to mention only a few. A proper retort might be: “You are right. There is nothing in yesterday’s mathematics that could not also be proved without it. Exterior algebra is not meant to prove old facts, it is meant to disclose a new world. Disclosing new worlds is as worthwhile a mathematical enterprise as proving old conjectures.” (Indiscrete Thoughts, 48)

And, as Founding Editor of Advances in Mathematics, we have Rota to thank for the publication of:

A. Joyal, Une théorie combinatoire des séries formelles, Advances in Mathematics 42, (1981) 1-82.

A. Joyal and R. Street, Braided tensor categories, Advances in Mathematics 102 (1993), no. 1, 20-78.

J. Baez and J. Dolan, Higher-dimensional algebra III: $n$-categories and the algebra of opetopes, Advances in Mathematics 135 (1998), 145-206.

Posted by: David Corfield on April 3, 2007 5:49 PM | Permalink | Reply to this

### Re: Rota on Combinatorics

Okay, now about that ‘extraordinary coincidence’…

(Maybe I missed something.)

Posted by: Chris W. on April 11, 2007 2:18 AM | Permalink | Reply to this

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# Expansion Greater than Light Speed?

1. Apr 26, 2004

### Antonio Lao

Does anyone know whether the expansion of the universe ever exceeded the speed of light taking the early accelerated phase into consideration?

If not then the early photons must have bounded back and forth within the boundary of the early universe and a few or maybe many of all photons we are seeing now are rebounded photons.

Last edited: Apr 26, 2004
2. Apr 26, 2004

### Staff: Mentor

Yes, some galaxies appear to be recedinig at >C.
Even a finite universe need not have boundaries. There wouldn't be any bouncing - no edge to bounce off of.

3. Apr 26, 2004

### marcus

confirming what russ says,
most of the galaxies in the observable universe have redshifts z
greater than 2
(indeed one has been observed with z = 10, much more distant!)

and when we look at a galaxy with z = 2 or bigger

we are looking at an object which was already receding from us at a speed greater
than c
at the very moment when it emitted the light which we are now receiving
so we witness it in the act
of receding faster than c

that might also be true for z = 1.9, or z = 1.8
I would have to use a calculator to be sure
but certainly it is true for z = 2 and bigger.

At first sight it seems paradoxical and some essays have been written
to help people assimilate this, but it is a standard fact, just mindboggling when you first encounter it

Last edited: Apr 26, 2004
4. Apr 26, 2004

### marcus

I just used Morgan's calculator
with Hubble parameter = 71
and darkenergy 0.73
and matter 0.27
and
it said z = 1.7 implies the object was
receding at speed 1.02 c
that is just a little faster than light
so to a good approximation the cutoff is z = 1.7

using the present best estimates for the cosmological parameters
anything we see with redshift greater than 1.7
was already receding from us faster than light
when it emitted the very light which we are now receiving

Morgan's calculator
http://www.earth.uni.edu/~morgan/ajjar/Cosmology/cosmos.html
is in the links in the reference sticky at the General Astronomy forum.

This thread could as well be in General Astronomy
instead of Theory Development. When I responded I didnt realize it was
in Theory Development. We've had similar questions there

Last edited by a moderator: Apr 20, 2017
5. Apr 26, 2004

### DrChinese

Last edited by a moderator: Apr 20, 2017
6. Apr 26, 2004

### LURCH

The second paragraph does not necessarily follow from the first. If the universe never expanded at a speed exceeding light-speed, it could still expand at light speed. Keep in mind that the "expansion of the universe" simply means, "how far stuff has gotten". So, for as long as there has been light, and that light has been traveling outward, the universe has been expanding at least at the speed of light.

7. Apr 26, 2004

### kurious

Inflation theory was thought up to explain, among other things, how regions separated
by distances greater than light can have possibly travelled, seem so alike- the idea being that if light couldn't bounce around and smooth things out , the faster than light stretching of space-time itself could.But what if there are particles out in space moving faster than light-most of the mass of the universe is still unaccounted for.

8. Apr 26, 2004

### Antonio Lao

Is this a stupid question? But if some of these galaxies are really receding beyond light would their relativistic mass be more than infinite?

9. Apr 26, 2004

### DrChinese

Not a stupid question at all. But read the Lineweaver reference first and save yourself a lot of grief.

10. Apr 26, 2004

### Staff: Mentor

...short answer: no, because expansion isn't motion. The galaxies aren't necessarily moving, but rather the space between them is expanding.

11. Apr 27, 2004

### kurious

...short answer: no, because expansion isn't motion. The galaxies aren't necessarily moving, but rather the space between them is expanding

But if that space is made of mass there is a relativistic mass problem!

12. Apr 27, 2004

### Antonio Lao

Thanks. It also dawn on me after I posted the "stupid" (not thinking) question that the galaxies were in a rest frame of their own and that it is the space that is expanding. But as kurious pointed out, if mass is attributed to space, then again the relativistic mass increase would be a problem.

Are modern theories trying to explain exotic matter, dark matter and dark energy to the mass property of space?

13. Apr 27, 2004

### kurious

Theorists are saying that dark energy 's density stays the same as space expands.
Normal energy like the cosmic microwave background gets less dense according to
1/ r^4. Dark matter is not thought to be a property of space but rather some exotic particles like axions or maybe even large accumulations of neutrinos.I have a problem with dark matter hypotheses: why doesn't it block our view of some stars and why don't we suddenly see new stars if dark matter moves around a bit, as it must surely do?

14. Apr 27, 2004

### Antonio Lao

What I am trying to do in my own research is to show that the mass we commonly used to know about can be splitted into two kinds of mass. The potential and the kinetic. The potential is the same as inertial and gravitational mass. And the kinetic mass has more to do with the true meaning of energy and momentum. There are logical distinctions between them. Their combination gave us the experimentally detectable mass.

When the potential mass and kinetic mass of a particle are equal then the experimental mass is zero (e.g. the mass of photon). With this idea in mind, we can say that the potential and kinetic mass of space are also equal so that the experimental (detectable) mass of space is also zero.

The W's and Z's bosons in the electroweak theory are all kinetic mass. And the mass responsible for gravity is the potential mass. The mass inside black holes and singularities are dominated by potential mass.

The universal expansion of space is the constant process of separation between potential and kinetic mass. All the potential mass are trying to get together, forming black holes, galaxies, stars, and planets while the kinetic mass spreadout uniformly causing the phenomenon of expansion in order to keep the detectable mass of space at the constant value of zero.

15. Apr 27, 2004

### Staff: Mentor

"space" itself is not a substance with mass.

16. Apr 27, 2004

### kurious

potential and kinetic energy

space" itself is not a substance with mass

Then what is it? What if the 10^120 Joules /m^3 of vacuum energy is Antonio's two cancelling mass types?
I thought of something similar a few weeks ago-the idea being that vacuum particles appear when positive energy moves away from a cloud of negative energy and then the particles disappear again by re-entering the cloud.

17. Apr 27, 2004

### Antonio Lao

Space is the seat where the potential and kinetic mass are exactly equal. I think, it is the same as saying the potential and kinetic energy are also equal. When the potential and kinetic energy are exactly equal, the Lagrangian at the point of spacetime is zero.

When the Lagrangian is exactly zero, the Hamiltonian is just the sum of the products of generalized space dimension (position coordinates) and the generalized momenta.

18. Apr 27, 2004

### Antonio Lao

The time independent Hamiltonian (Lagrangian=0) is equivalent to the quantum uncertainty condition of $\Delta \psi \Delta \phi \geq \frac{h}{2 \pi}$.

19. Apr 27, 2004

### AWolf

The zero mass of a photon is at rest mass. Since a photon cannot be at rest, it cannot be compared to the zero mass of space.

20. Apr 27, 2004

### Antonio Lao

if we take the point of view of the photon in vacuum, one photon can never overtake any other photons from the same source and along the same straight line (geodesic). So all photons from the same source appear motionless with respect to each other if the are along the same geodesic line. But can photons from a different source, or photon from same source but different direction, always appear to be traveling at light speed or less?


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# Semiconductor Doping Question

The question says:

Calculate the minority carrier life-time in Si having a donor concentration, NA = 1014 cm-3 if the diffusion length for minority carriers is 150 m.

Shouldnt it say "ND=1014"? because its a donor concentration and not an acceptor concentration?

Thanks

• I'm more curious about the 150m diffusion length – Dr Coconut Feb 4 '16 at 8:07

But yes, $N_D$ = donor concentration, $N_A$ = acceptor concentration.


text:
HAL will be down for maintenance from Friday, June 10 at 4pm through Monday, June 13 at 9am. More information

# Numerical simulation of a jet in crossflow. Application to GRID computing

1 SMASH - Simulation, modeling and analysis of heterogeneous systems
CRISAM - Inria Sophia Antipolis - Méditerranée , Université de Provence - Aix-Marseille 1
Abstract : This work describes the LES and hybrid RANS/LES simulations for a rounded jet issuing normally into a cross-flow, also known as jet in cross-flow (JICF). This test case is particularly suitable to study the capability of LES and hybrid approaches because of the presence of both separated and non-separated boundary layer regions. Simulations are performed for a jet to cross-flow velocity ratio $R = 2$ and a $Re = 82000$, based on the cross-flow velocity and the jet diameter. The simulations are carried out using a numerical solver of 3D compressible flows, named AERO. Large scales coherent structures observed in experimental flow visualisations are reproduced in both the simulations and the obtained mean and turbulent statistics are compared to experimental data. LES simulations seem to follow in a better way the experimental results than LNS. In the second case, there is not sufficiant mixing between jet and cross-flow and thus the jet bends less and a too large recirculation bubble forms behind the jet. This test case has been chosen also because we can split the domain in two different regions, the pipe and the cross-flow, with a limited interface. This is a good characteristic in order to test GRID computing, which is another goal of this work. GRID computing is an alternative way to carry out parallel computations with respect to the classical clusters or parallel computers. It consists in using computers allocated in different sites exchanging data between the sites through the web (Internet). In the present work, GRID computing is tested for both JICF and for the flow in a pipe.
Keywords :
Document type :
Reports
Domain :

Cited literature [33 references]

https://hal.inria.fr/inria-00070369
Contributor : Rapport de Recherche Inria Connect in order to contact the contributor
Submitted on : Friday, May 19, 2006 - 8:18:12 PM
Last modification on : Friday, February 4, 2022 - 3:13:09 AM
Long-term archiving on: : Monday, September 17, 2012 - 2:35:32 PM

### Identifiers

• HAL Id : inria-00070369, version 1

### Citation

Simone Camarri, Vanessa Mariotti, Maria Vittoria Salvetti, Bruno Koobus, Alain Dervieux, et al.. Numerical simulation of a jet in crossflow. Application to GRID computing. [Research Report] RR-5638, INRIA. 2006, pp.80. ⟨inria-00070369⟩

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text:
# Algorithmic description of implementing path integral molecular dynamics?

Suppose I have a functioning molecular dynamics package for classical MD. I want to extend this code to run path integral MD. I know that a surprisingly small number of new features need to be implemented, but the literature I am aware of is quite formal, where I'd be interested to look at a more algorithmic description of what steps need to be added to the usual integration of Newton's equations in classical MD.

Any references, books or articles, you've seen which fit the description, I'd love to look at. Alternatively, a simple reference implementation would be nice. It's quite easy to find reference implementations of path integral monte carlo, but not as much PIMD.


text:
# Java: Basic Movement AI not working

I am working on a basic AI system to get an entity to follow a player. I created both the player and the ghost entity, but it seems as if the ghost only wants to increment both x and y, it never wants to do the opposite, or choose only x or y to use. Here is where I implemented the ghost and player in my constructor:

public MansionAzazel() {
...
TileCoordinate ghostSpawn = new TileCoordinate(52, 50);
TileCoordinate playerSpawn = new TileCoordinate(54, 52);
player = new Player(playerSpawn.x(), playerSpawn.y(), key);
player.init(level);
ghost = new Ghost(ghostSpawn.x(), ghostSpawn.y(), player);
ghost.init(level);
...
}

Here is the update() function for the ghost:

public void update() {
int xa = 0, ya = 0;
if (anim < 1000000)
anim++;
else
anim = 0;

if (player.getX() > xa)
xa++;
if (player.getX() < xa)
xa--;
if (player.getY() > ya)
ya++;
if (player.getY() < ya)
ya--;

if (xa != 0 || ya != 0) {
move(xa, ya);
walking = true;
} else
walking = false;
}

In which to sum it all up, the ghost entity will only and always move in one direction because the coordinates are 0 and I don't know why.

• What is your question? – wes Aug 12 '13 at 18:50
• It looks like your ghost update function uses player.getX() when checking for Y coordinates. You might want to fix that. Also, you should output what player.getX() and player.getY() returns in your ghost update function to see what you get. – UnderscoreZero Aug 12 '13 at 22:01
• @UnderscoreZero I said that it returns -1 for both of them. I – Shzylo Aug 12 '13 at 22:04
• What I meant was change the line System.out.println(xa + ", " + ya); to System.out.println(xa + ", " + ya + " " + player.getX() + ", " + player.getY()); – UnderscoreZero Aug 12 '13 at 22:14
• @UnderscoreZero after I changed getX to getY from checking y coordinate they both print out positive 1 Also, I divided the player.getX() and player.getY() by 16 (tile size) it prints out the actual location, just not ghost. – Shzylo Aug 12 '13 at 22:14

• Okay, @JMB you were quite right about that. It does reset the variables each time. Instead of using the my variables xa and ya I used this.getX() and this.getY() which worked perfectly! – Shzylo Aug 13 '13 at 16:16