text:
Energy Is Not Conserved

I’ve been meaning to link to this post at the arXiv blog, which is a great source of quirky and interesting new papers. In this case they are pointing to a speculative but interesting paper by Martin Perl and Holger Mueller, which suggests an experimental search for gradients in dark energy by way of atom interferometry.

But I’m unable to get past this part of the blog post:

The notion of dark energy is peculiar, even by cosmological standards.

Cosmologists have foisted the idea upon us to explain the apparent accelerating expansion of the Universe. They say that this acceleration is caused by energy that fills space at a density of 10-10 joules per cubic metre.

What’s strange about this idea is that as space expands, so too does the amount of energy. If you’ve spotted the flaw in this argument, you’re not alone. Forgetting the law of conservation of energy is no small oversight.

I like to think that, if I were not a professional cosmologist, I would still find it hard to believe that hundreds of cosmologists around the world have latched on to an idea that violates a bedrock principle of physics, simply because they “forgot” it. If the idea of dark energy were in conflict with some other much more fundamental principle, I suspect the theory would be a lot less popular.

But many people have just this reaction. It’s clear that cosmologists have not done a very good job of spreading the word about something that’s been well-understood since at least the 1920’s: energy is not conserved in general relativity. (With caveats to be explained below.)

The point is pretty simple: back when you thought energy was conserved, there was a reason why you thought that, namely time-translation invariance. A fancy way of saying “the background on which particles and forces evolve, as well as the dynamical rules governing their motions, are fixed, not changing with time.” But in general relativity that’s simply no longer true. Einstein tells us that space and time are dynamical, and in particular that they can evolve with time. When the space through which particles move is changing, the total energy of those particles is not conserved.

It’s not that all hell has broken loose; it’s just that we’re considering a more general context than was necessary under Newtonian rules. There is still a single important equation, which is indeed often called “energy-momentum conservation.” It looks like this:

$\nabla_\mu T^{\mu\nu} = 0,.$

The details aren’t important, but the meaning of this equation is straightforward enough: energy and momentum evolve in a precisely specified way in response to the behavior of spacetime around them. If that spacetime is standing completely still, the total energy is constant; if it’s evolving, the energy changes in a completely unambiguous way.

In the case of dark energy, that evolution is pretty simple: the density of vacuum energy in empty space is absolute constant, even as the volume of a region of space (comoving along with galaxies and other particles) grows as the universe expands. So the total energy, density times volume, goes up.

This bothers some people, but it’s nothing newfangled that has been pushed in our face by the idea of dark energy. It’s just as true for “radiation” — particles like photons that move at or near the speed of light. The thing about photons is that they redshift, losing energy as space expands. If we keep track of a certain fixed number of photons, the number stays constant while the energy per photon decreases, so the total energy decreases. A decrease in energy is just as much a “violation of energy conservation” as an increase in energy, but it doesn’t seem to bother people as much. At the end of the day it doesn’t matter how bothersome it is, of course — it’s a crystal-clear prediction of general relativity.

And one that has been experimentally verified! The success of Big Bang Nucleosynthesis depends on the fact that we understand how fast the universe was expanding in the first three minutes, which in turn depends on how fast the energy density is changing. And that energy density is almost all radiation, so the fact that energy is not conserved in an expanding universe is absolutely central to getting the predictions of primordial nucleosynthesis correct. (Some of us have even explored the very tight constraints on other possibilities.)

Having said all that, it would be irresponsible of me not to mention that plenty of experts in cosmology or GR would not put it in these terms. We all agree on the science; there are just divergent views on what words to attach to the science. In particular, a lot of folks would want to say “energy is conserved in general relativity, it’s just that you have to include the energy of the gravitational field along with the energy of matter and radiation and so on.” Which seems pretty sensible at face value.

There’s nothing incorrect about that way of thinking about it; it’s a choice that one can make or not, as long as you’re clear on what your definitions are. I personally think it’s better to forget about the so-called “energy of the gravitational field” and just admit that energy is not conserved, for two reasons.

First, unlike with ordinary matter fields, there is no such thing as the density of gravitational energy. The thing you would like to define as the energy associated with the curvature of spacetime is not uniquely defined at every point in space. So the best you can rigorously do is define the energy of the whole universe all at once, rather than talking about the energy of each separate piece. (You can sometimes talk approximately about the energy of different pieces, by imagining that they are isolated from the rest of the universe.) Even if you can define such a quantity, it’s much less useful than the notion of energy we have for matter fields.

The second reason is that the entire point of this exercise is to explain what’s going on in GR to people who aren’t familiar with the mathematical details of the theory. All of the experts agree on what’s happening; this is an issue of translation, not of physics. And in my experience, saying “there’s energy in the gravitational field, but it’s negative, so it exactly cancels the energy you think is being gained in the matter fields” does not actually increase anyone’s understanding — it just quiets them down. Whereas if you say “in general relativity spacetime can give energy to matter, or absorb it from matter, so that the total energy simply isn’t conserved,” they might be surprised but I think most people do actually gain some understanding thereby.

Energy isn’t conserved; it changes because spacetime does. See, that wasn’t so hard, was it?

1. Hmm, interesting point. Is it the mark of a physicist that I’ve never actually worried about the energy conservation implications of a cosmological constant? 😉

2. A beautiful, clear explanation. One might add that every GPS receiver assumes this particular form of energy non-conservation by including relativistic corrections to the photon energy (=frequency) it receives when it tells us how far we are from the nearest Starbucks.

3. I applaud your description of a thorny problem. In my opinion, it is those physicists who attempt not to dilute their content with evasive language but instead illuminate its difficulties with wisely chosen analogies that are the best communicators. It is for this reason that I believe Hawking’s “A Brief History of Time” has endured to inspire new physicists to this day: he openly admits the weirdness of physics, and does his best to give a glimpse into its nature. Apart from selecting which details he pursues, he pulls no punches.

However, I do feel like you stumble over your words quite a bit here. For instance, as a non-GR physicist, I have no idea what the distinction is between “there’s energy in the gravitational field, but it’s negative, so it exactly cancels the energy you think is being gained in the matter fields” and “spacetime can give energy to matter, or absorb it from matter, so that the total energy simply isn’t conserved”. What is the difference between a gravitational field and an “evolving” spacetime? Where does the energy go that it absorbs or releases?

These phrases are confusing because it sounds like we’re sweeping energy under a rug. Would it be better to say “There is a clear relationship that transmutes energy into momentum, just as space transmutes into time, under the laws of general relativity”? This is what I assume you were getting at with your “energy-momentum conservation equation”. But as someone who has never studied GR, I’m grasping at straws here.

4. Hi, Sean. Would it be accurate to say that energy is conserved locally, but not globally? That is, in a fixed volume the energy is conserved (in the sense that any energy gained or lost by the volume is equal to what we see passing through its walls), but because the total volume of the universe is increasing, the total energy is not conserved. Or does GR violate even this idea of local energy conservation?

5. Is it just me or is the picture posted above broken?
The picture follows this paragraph:

“It’s not that all hell has broken loose; it’s just that we’re considering a more general context than was necessary under Newtonian rules. There is still a single important equation, which is indeed often called “energy-momentum conservation.” It looks like this:”

-ken

6. Ken– Not sure what’s going on, the image looks fine to me.

TimG– No, even that kind of energy conservation is not true. The energy lost or gained is not equal to the flux through the walls.

7. The thing you would like to define as the energy associated with the curvature of spacetime is not uniquely defined at every point in space. So the best you can rigorously do is define the energy of the whole universe all at once, rather than talking about the energy of each separate piece. (You can sometimes talk approximately about the energy of different pieces, by imagining that they are isolated from the rest of the universe.) Even if you can define such a quantity, it’s much less useful than the notion of energy we have for matter fields.

How does one talk about gravitational waves interacting with a gravitational wave detector and how the former imparts energy to the latter if one cannot talk rigorously about the “energy of the gravitational field”?

8. Hmm. this might have some interesting implications for the religious Kalam Cosmological argument. They like to throw around the word ‘conservation’ and that matter can’t be created or destroyed.

Nice article!!

9. Lovely explanation. Photons losing energy as they redshift bothered me during a recent graduate cosmology course, and the professor was kind enough to dig up a sentence stating simply that energy is not conserved on a cosmological scale. It still bothered me, but I have since taken a GR course, so your mention of “energy-momentum conservation” makes me much happier. Thanks!

10. Arun– You can talk about the energy lost by a binary system by treating it as isolated from the rest of the universe, but it’s necessarily an approximation. For detecting gravitational waves there is no problem, since you should be talking about observable things (like the displacement in light waves as measured by an interferometer) anyway.

11. I agree that this is an issue of translation, not of physics, but the conclusion you draw after that seems off.

“in general relativity spacetime can give energy to matter, or absorb it from matter, so that the total energy simply isn’t conserved,”

For instance, this strikes me as a non sequitur. If spacetime is giving and absorbing energy from matter, it sounds like energy is just moving from one place to another. And then you tell me that this means energy is not conserved. Sure, this will provoke surprised reactions, but that’s because what you said makes little logical sense.

If I were a lay person, my reaction would be, “Physics is hard and makes my brain hurt!” As a physics student, I just think the professor is BSing me because in his/her professional opinion, I do not have the required background to understand what’s really going on. I suspect that perpetual motion machines remain impossible.

12. I have not yet taken a course on GR, but I am under the impression that coordinate transformations from one place to another in GR are path-dependent?

Assuming that is true, is that why, “…the energy associated with the curvature of spacetime is not uniquely defined at every point in space.”?

13. @Miller:

It isn’t quite a non-sequitur. It is true that spacetime interacts with matter, and that the energy-momentum tensor changes non-trivially in a gravitational field. So you can certainly think of the spacetime as giving energy to or taking it from matter. The point is that it is not so easy to define the energy of the spacetime, and without doing so there is missing or excess energy in the matter sector.

It is logically consistent to think this way: energy conservation is not a property of any set of (mathematically consistent) dynamics. It arises often in physics because it is associated with a symmetry: that the background in which we are working is time-independent. I can give you a time varying potential in quantum mechanics and you will also see energy non-conservation. Of course, in that case, we resort to “physical reasoning” (i.e. our faith in the conservation of energy) to insist that we have ignored degrees of freedom associated with the lab / experiment / etc needed to set up the potential in the first place, and assure ourselves if we did include everything energy would still be conserved. But the fact that we can keep this assurance in the back of our minds without explicitly entering it into our calculations is sufficient to show that there is nothing *logically* inconsistent about defining an energy which is not conserved.

(A simpler example may be interactions with a heat bath in thermo, but there we often define our ensemble in terms of energy conservation so it is not as good as it first appears.)

In general relativity, in the absence of a timelike Killing field we simply have an evolving background. We can use our local time coordinate to define an energy, but just like the time evolving potential in a “series-of-evolving spatial slices” (or more generally a non-stationary spacetime) but we don’t have an assurance that the quantity we define will be conserved.

14. I concur. Energy is not conserved. Harrison had a nice paper on mining energy from expanding space-time

15. Just two days ago, a preacher was telling me that scientists had proven something cannot come from nothing, and therefore God exists.

16. What are the relative magnitudes of the energy gained due to a cosmological constant + expansion and the energy lost due to the red shift of non decaying particles propagating through space?

17. That sentence does raise one interesting question: The Bianchi identities are true for any smooth manifold physically realizable or not it, so how does one determine what are the criteria that constrain the stress-energy?

No physicist would ever propose that we live on a 4-sphere, yet it has positive constant stress-energy, that is inversely proportional to the radius. So what gives?

18. This may be just another translation, but I like it. The conservation of energy equation is the first law of thermodynamics:

dE = dW + dQ

dE is the change in energy, dW is the work done on the system, and dQ is the heat absorbed by the system. (This equation is good locally or globally.)

The Universe isn’t gaining or losing heat, but there is work being done by the expansion of space. Using for formula for work, dW = -P dV, where P is the pressure and dV is the change in volume, we can find the change in energy.

dE = -P dV

In our expanding universe dV is positive. Photons have positive pressure. A universe which is mostly photons (as ours was at early times) loses energy as it expands. Dark energy has a negative pressure, like a stretched spring. A universe which is mostly dark energy (as ours is today) actually gains energy as it expands. The energy doesn’t come from somewhere else, it is just an increase in energy.

It is believed that inflation was driven by a huge negative pressure which produced heaps of energy, and therefore matter, from this very process.

Jason R: Yes. Have fun trying to explain it to someone who believes the Kalam Cosmological Argument.

19. @Kiwidamien
Well, of course it’s not actually a non sequitur. But look how long it took you to explain it! If we are trying to choose the explanation which is most effective at reaching popular audiences, I’m not convinced that this is the best one.

20. @Gavin–I like that explanation. When I try to explain negative pressure, I tell students to imagine dark energy in a piston (with nothing on the outside). For a regular gas, the pressure inside is positive, outside is zero, so the piston will expand. For dark energy, the energy DENSITY is constant, so if the piston expanded, the volume would go up and so the total energy would increase. Thus, work has to be done to pull the piston out–this is suction, or negative pressure. Of course dE = -p dV makes it clear that if the energy density, i.e. dE/dV, is a constant, the pressure is negative…

21. @Gavin Polhemus – good thinking, really liked reading that post.

Regarding the topic of the post a question discussed on my university forums comes to mind.

[i]What is energy?[/i]

We didn’t reach any conclusion discussing it (internet forums have a way of getting derailed …) though some interesting arguments were made. Without wanting to astray this discussion I ll say that as I understand it, conservation is a defining characteristic of energy. (For example) You write down a Lagrangian that describes all the interactions+particles of a system and when you examine what remains conserved from time translation you find what you can call energy in the said system.

From the little I know, this cannot be easily done in GR (if it can be done at all?).

It was enlightening to read the blog-post but I wasn’t really shocked to read that energy is not conserved. I think that it’s only that we haven’t been able to identify correctly what to call energy yet. For example consider the example of a particle moving through empty spacetime. The time component of it’s conjugate momentum we identify as its energy (a well known result from special relativity). If we assign a charge to this particle and a electromagnetic field present then (from it’s corresponding Lagrangian) the conjugate momentum changes and therefore what one now calls energy of the particle is different.

I think this is the case, regarding the expanding/accelerating universe as well, we have not identified yet what it is that we should call energy.

P.S. If I said something really dump in there, be kind with me. As we say in Greece:
Sciolism is worse than ignorance.

22. Excellent post! ” When the space through which particles move is changing, the total energy of those particles is not conserved.”

Exactly.

What a pity then that you had to equivocate:

“We all agree on the science; there are just divergent views on what words to attach to the science. In particular, a lot of folks would want to say “energy is conserved in general relativity, it’s just that you have to include the energy of the gravitational field along with the energy of matter and radiation and so on.””

Sorry, but it’s very obvious from your post that these folks are just wrong. We may agree on the numbers but numbers are not understanding; and these folks are getting correct numbers by means of wrong physics. Better to say so than to throw them such sops, which will only encourage them to continue in their error.

Arun asks: “How does one talk about gravitational waves interacting with a gravitational wave detector and how the former imparts energy to the latter if one cannot talk rigorously about the “energy of the gravitational field”?”

Good question. To answer it, you have to remember that the word “energy” has two different meanings: [1] a name for a convenient mathematical abstraction, eg “kinetic energy” and [2] the stuff out of which the universe is made. In GR one can continue to use “energy” in sense 1 to do the calculations leading to predictions as to what gravitational wave detectors should see, while at the same time completely abandoning the idea that spacetime has “energy” in sense 2. One has also to renounce all attempts to give a detailed local account of how energy2 gets transferred from the binary pulsar to the detector. This kind of renunciation is familiar to you from quantum mechanics.

By the way, when talking about this question in connection with cosmology it is useful to bear in mind that most cosmological models are perfectly homogeneous and isotropic, ie everything is exactly the same everywhere and in all directions at a given time. Hence there can be no question of any “work” being done by the “gravitational field”: it is “pulling” equally hard in all directions.

23. Actually you can define work in an isotropic space because you can define the differential volume element through the Jacobian of the metric, the difficulty is in correctly defining the energy differential in terms of the stress-energy.

24. @Gavin (19):
I like this explanation for calculational purposes (i.e. I use it as a mnemonic to myself to write down the evolution of a type of matter given the equation of state). The problem with it as a conceptual device is that in
dE = dQ + dW = 0 – P dV
the “P” in this equation is really the pressure of the environment. Normally in stat mech we don’t distinguish because we are doing quasi-static processes in which the pressures are almost the same inside and out to ensure that the expansion is slow. In the opposite extreme we have a free expansion of a gas into vacuum — which involves no transfer of energy at all (how could it?). We can have a gas at a high pressure P and energy E in a small volume V, and have it expand into a vacuum by opening a door into a larger volume V’ at a smaller pressure P’ but at the same energy. This is because in the vacuum P = 0 and we don’t (can’t!) do any work on the vacuum.

So for the dE = -P dV thing to make sense you would have to imagine the universe doing work embedded in an external environment at pressure P = P_inside universe!


text:
# How to get the AMS mathbb font while using Palatino elsewhere

I'm a Latex user. I've been using Palatino for typesetting a piece mathematical text:

\usepackage{mathpazo}
\usepackage[scaled=.95]{helvet}
\usepackage{courier}

The results are good, however I don't like the blackboard-bold given by mathpazo. Is there some option to pass to mathpazo, or some nice hack that will give me back my beloved AMSmath \mathbb symbols?

-
- we've heard so many criticisms of the \mathbb alphabet here at ams that it's nice to hear it called "beloved". thanks! – barbara beeton Jul 1 '11 at 12:20
Let me add in my affection for the ams \mathbb symbols as well :) – Suresh Jul 1 '11 at 16:02

You have to restate the meaning of \mathbb:

\AtBeginDocument{
\DeclareSymbolFont{AMSb}{U}{msb}{m}{n}
\DeclareSymbolFontAlphabet{\mathbb}{AMSb}}

-
How about the font for \varmathbb? – jeecabz Sep 6 '12 at 7:16

give the kpfonts a try, they are similiar to palatino, but have also sans serif and typewriter fonts:

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage{kpfonts}
\usepackage{amsmath,amssymb}

\begin{document}
\Huge
$\mathbb{F}\mathbb{D}\mathbb{G}\mathbb{H}$

\normalsize
Serif text, and
\textsf{Sans Serif}, and
\texttt{Typewriter}
\end{document}

-

There is in fact a package option for this: if you pass noBBpl to mathpazo, then its blackboard bold fonts are not loaded. That also means you have to load the AMS ones separately, but that just means using amssymb, which is simpler than defining another symbol font alphabet:

\documentclass{article}
% these two \usepackage commands make it work
\usepackage{amssymb}
\usepackage[noBBpl]{mathpazo}
\usepackage[scaled=.95]{helvet}
\usepackage{courier}
\begin{document}
In this document,
\begin{itemize}
\item $\mathbb N$ will denote the natural numbers,
\item $\mathbb Z$ will denote the integers,
\item $\mathbb Q$ will denote the rational numbers,
\item $\mathbb R$ will denote the real numbers, and
\item $\mathbb C$ will denote the complex numbers.
\end{itemize}
\end{document}

-

format: plain


text:
# Strange behavior of Limit in Mathematica 9 and 10 (bug?)

fixed in 10.1

Consider a complex logarithm where the branch cut is defined along the negative axis. Then for $r$ and $\eta$ real and positive we can write

$\lim_{\eta \to 0} \log(-r+ i \eta) = log (r) + i \pi \\ \lim_{\eta \to 0} \log(-r- i \eta) = log (r) - i \pi$

which reflects the fact that in the first case ($+ i \eta$) we are approaching the branch cut from above, while in the second case ($-i \eta$) from below.

In Mathematica one can use Limit to reproduce this behavior, i.e. by writing

Log[-m^2 - I eta]
Limit[%, eta -> 0, Assumptions -> m^2 > 0]
Log[-m^2 + I eta]
Limit[%, eta -> 0, Assumptions -> m^2 > 0]

In the first case I get Log[m^2] - I Pi and in the second Log[m^2] + I Pi. This is correct and fine.

However, if I write

a + Log[-m^2 - I eta]
Limit[%, eta -> 0, Assumptions -> m^2 > 0]
a + Log[-m^2 + I eta]
Limit[%, eta -> 0, Assumptions -> m^2 > 0]

something very weird happens. In both cases Mathematica 10 returns a + Log[m^2] + I Pi. To me this is obviously wrong. Since a does not depend on eta it should not influence the imaginary part in any way. By the way you can basically replace a by anything (e.g 1 or "a") and still get the same wrong result.

I checked this with different Mathematica versions and the problem is present starting with version 9. Mathematica 7 and 8 on the contrary return correct results in all cases.

So is there something I'm missing about the way how Limit works, or is it a bug?

P.S. I though that it might make sense to discuss this here before reporting it to WRI. If it isn't a bug, then of course there is also nothing to report.

• Actually I'm surprised that it works at all, since strictly speaking the limit doesn't exist (you usually use a notation like $\lim_{\eta\to 0^+}$ to indicate the extra condition that you are calculating a one-sided limit). However I don't know under which conditions Mathematica considers a limit to be one-sided. Mar 18, 2015 at 11:23
• @celtschk You actually can specify the direction of a one-sided limit using Direction. However, again Limit[Log[-m^2 + I eta], eta -> 0, Assumptions -> m^2 > 0, Direction -> -1] and Limit[Log[-m^2 + I eta], eta -> 0, Assumptions -> m^2 > 0, Direction -> 1] give correct results, while Limit[a+Log[-m^2 + I eta], eta -> 0, Assumptions -> m^2 > 0, Direction -> -1] and Limit[a+Log[-m^2 + I eta], eta -> 0, Assumptions -> m^2 > 0, Direction -> 1] produce wrong output in MMA 9 and 10 but evaluate properly in MMA 7 and 8.
– vsht
Mar 18, 2015 at 13:06
• Definitely a bug. Mar 18, 2015 at 16:02
• @DanielLichtblau Thanks, then I'll report it to WRI.
– vsht
Mar 18, 2015 at 16:05
• I reported it. If you like you can have your name attached to the report. Just let Tech Services know the link to this MSE discussion, and that it was a bug I had filed today. Mar 18, 2015 at 16:23

Not an answer, but possible hint as to what's going on:

Try the following in Mathematica 9 and 10.0.

(*1*) Limit[1 + Log[-a - I y], y → 0, Direction → -1, Assumptions → {a > 0}]
(* 1 + I*π + Log[a]*)

(*2*) Limit[1 + Log[-x - I y], y → 0, Direction → -1, Assumptions → {x > 0}]
(* 1 - I*π + Log[x]  <-- in v9 only.*)

In v9, the one with x is right, the one with a (and any other letter) is wrong. Clearly, Wolfram is encouraging users to use x as the variable.

In v10.0, Wolfram concedes, and all letters give the wrong result.

• Interesting. I tried to look at ClearAttributes[Limit, Protected]; ClearAttributes[Limit, ReadProtected]; ?? Limit but it seems like Limit is implemeted in C, not in Wolfram Language. Anyway, now it looks more like a bug to me rather than some peculiarity of Limit.
– vsht
Mar 18, 2015 at 14:16

This has been fixed in 10.1 on windows:

 \$Version

code for the above

 expr1 = Log[-m^2 - I eta];
Limit[expr1, eta -> 0, Assumptions -> m^2 > 0]
expr2 = Log[-m^2 + I eta];
Limit[expr2, eta -> 0, Assumptions -> m^2 > 0]
a + expr1;
Limit[%, eta -> 0, Assumptions -> m^2 > 0]
a + expr2;
Limit[%, eta -> 0, Assumptions -> m^2 > 0]


text:
# Part 2.4Basic output methods

There are several very simple ways of outputting information to the user or to the debugger. Some forms are obtrusive outputs and some are unobtrusive.

An obtrusive output means that the program stops running the main sequence whilst the output is displayed whereas an unobtrusive output does not stop the main execution.

## Console outputs

The console is a form of collecting multiple outputs for reference. Output here is unobtrusive.

### Console.Write

The Console.Write method is used to output any information of any of 17 types (for instance, boolean, integer, single) on to a line without taking a new line.

Below is a sample write command:

VB.NET
Console.Write("Hello world")
Console.Write("This is a test")

### Console.WriteLine

The Console.WriteLine method is used to output some text and take a line break after it. This way multiple outputs do not get mixed up as easily as shown above.

VB.NET
Console.WriteLine("Hello world")
Console.WriteLine("This is a test")

## Dialog based output

Dialog based output such as message boxes tends to be obtrusive output - it breaks execution of the program.

### MsgBox

One of the simple forms of output is to use MsgBox which simply takes up to three arguements as follows:

MsgBox(prompt As String, [Buttons As MsgBoxStyle] [title As String])

The prompt parameter is used to contain a message as a string and title is an parameter. Below is a sample:

VB.NET
MsgBox("Hello world")

The other option that was given was the Buttons option. This option is used to specify how the message box looks. Below is a table showing all types of message box. All samples are from Windows 7.

ParameterSample
MsgBoxStyle.AbortRetryIgnore
MsgBoxStyle.ApplicationModal
MsgBoxStyle.Critical
MsgBoxStyle.DefaultButton1
MsgBoxStyle.DefaultButton2
MsgBoxStyle.DefaultButton3
MsgBoxStyle.Exclamation
MsgBoxStyle.Information
MsgBoxStyle.MsgBoxHelp
MsgBoxStyle.MsgBoxRight
MsgBoxStyle.MsgBoxRtlReading
MsgBoxStyle.MsgBoxSetForeground
MsgBoxStyle.OkCancel
MsgBoxStyle.OkOnly
MsgBoxStyle.Question
MsgBoxStyle.RetryCancel
MsgBoxStyle.SystemModal
MsgBoxStyle.YesNo
MsgBoxStyle.YesNoCancel

### MessageBox

The MsgBox method is quite limited, whereas the MessageBox class is more feature rich and customisable. One of the lacking features of the MsgBox method is that it does not support both an icon and special buttons.

As MessageBox is a more powerful class rather than a simple predefined shorthand function it does not work the same way and provides much more features. The following sample illustates this.

VB.NET
MessageBox.Show("Hello world!", "Sample program", MessageBoxButtons.YesNoCancel, MessageBoxIcon.Warning, MessageBoxDefaultButton.Button1)

Both of these dialog outputs return a DialogResult as a result, which means using an if statement, the result can be used to determine what events happen on each button.

## Other output styles

There are other ways of outputting information such as directly putting the information on to the user interface through the use of a control such as a Label or ListBox.

• Create a program that displays the same sentence on both the console and on a dialog.
Code preview
Feedback 👍
Comments are sent via email to me.


text:
# LPoS consensus algorithm¶

Proof of ownership with the right to lease. In PoS systems, the creation of a block does not require energy-intensive calculations, the miner’s task is to create a digital block proof.

## Proof of Stake¶

The mechanism for allocating block creation rights is based on the number of tokens in the user’s account. The more tokens a user has, the more likely he or she can create a block.

In Proof of Stake consensus the right to generate a block is determined by pseudo-random way, because by knowing the previous miner and balances of all users in the system the following miner can be identified. To do this, calculate the generating signature of the next block as sha256 hash from the generating signature of this block and the public account key. The first 8 bytes of the resulting hash will be a pointer to the following miner. The time of block generation for account i is calculated as:

$T_i = T_{min} + C_1 \log (1 - C_2 \frac{\log \frac{X_n}{X_{max}}}{b_i A_n})$

where:

• bi - is a stake (stake of participant’s balance of overall balance of the system);

• An - baseTarget, the adaptive ratio, regulating the average time of issue of the block;

• Xn - generating signature;

• Tmin - 5 seconds, it is a constant defining the minimum time interval between blocks;

• C1 - a constant, which is equal 70 and adjusting the form of allocation of the interval between blocks;

• C2 - a constant which is equal 5E17 and adjusting the baseTarget value (complexity).

Based on the given formula it is easy to see that the probability of selecting the participant depends on the participant’s stake of assets in the system: the bigger the stake the higher the chance. The minimum number of tokens on the balance for mining is 10000 WEST. BaseTarget is a computational complexity, a parameter that maintains the block generation time within a given range. BaseTarget in its turn is calculated as:

$(S > R_{max} \rightarrow T_b = T_{p} + max(1, \frac{T_p}{100})) \wedge (S < R_{min} \land \land T_{b} > 1 \rightarrow T_b = T_p - max(1, \frac{T_p}{100}))$

where

• R max = 90 - is a maximum reduction of complexity when the block generation time in the network exceeds 40 seconds;

• Rmin = 30 - minimal increase of complexity when the block generation time in the network is less than 40 seconds;

• S - average generation time, at least for the last three blocks;

• Tp - previous baseTarget value;

• Tb - computed baseTarget value.

For advanced description of technical features and enhancements of the classic PoS algorithm see article.

### Advantages Over Proof of Work¶

The absence of complex calculations allows PoS networks to lower the hardware requirements for participants of the system, which reduces the cost of deploying private networks. Also, no additional emission is required, which in PoW systems is used for rewarding miners for finding a new block. In PoS-Systems a miner receives a reward in the form of fees for transactions which appeared in its block.

## Leased Proof of Stake¶

For a user who has a stake insufficient for effective mining, it is possible to transfer his or her balance for lease to other participants, and receive a portion of the income from mining. Thus you can increase the likelihood of choosing a miner, for which you can receive a portion of the fees for transactions which this miner has placed in its blocks. Leasing is a completely safe operation. Tokens do not leave your wallet, you delegate the right to consider your balance when you draw the right of mining to another member of the network.


text:
Home » 2011 » November

# Monthly Archives: November 2011

## Matter from energy !

How can we can transform energy into matter? I know x rays can collide to form leptons but how to create protons and neutrons from photons?

Answer will be posted soon. Members and visitors are always welcome to post their answers as comments.

## Is it true that a person touching a high power line get struck with the line?

Is it true that a person touching a high power line get struck with the line?

Ya, This is due to the strong contraction of the muscles making the person not at all able to release the wire or object from which he got the shock. Not due to any type of attraction arising out of electricity.

AskPhysics is keenly looking forward to integrate the following services at the website.

1. Video Lessons explaining the answers to questions by its users and students
2. Live chat with Teacher(s)
3. Group Chat for registered members.

## Physical Quantity

Prabesh wants to know more on Physical quantities:

Physical Quantity is a quantity which can be measured. eg; length, mass, force, time etc

(The quantities which cannot be measured are abstract. Eg; Love, anger intelligence)

## How light maintain its speed after refraction?

How light maintain its speed after refraction? -(Naveen Saxena asked)

The speed of light in a medium is constant. When light enters from one medium to another, there is a change in the speed of light and the change is almost instantaneous. The speed of light in a medium depends on the electric and magnetic properties of the medium, more specifically, the electric permitivity and magnetic permeability of the medium.

The speed of light in a medium is given by

$v=\frac{1}{\sqrt{\mu \varepsilon }}$

Wikipedia says

“At the microscale, an electromagnetic wave’s phase speed is slowed in a material because the electric field creates a disturbance in the charges of each atom (primarily the electrons) proportional to the electric susceptibility of the medium. (Similarly, the magnetic field creates a disturbance proportional to the magnetic susceptibility.) As the electromagnetic fields oscillate in the wave, the charges in the material will be “shaken” back and forth at the same frequency. The charges thus radiate their own electromagnetic wave that is at the same frequency, but usually with a phase delay, as the charges may move out of phase with the force driving them . The light wave traveling in the medium is the macroscopic superposition (sum) of all such contributions in the material: The original wave plus the waves radiated by all the moving charges. This wave is typically a wave with the same frequency but shorter wavelength than the original, leading to a slowing of the wave’s phase speed. Most of the radiation from oscillating material charges will modify the incoming wave, changing its velocity. However, some net energy will be radiated in other directions or even at other frequencies .

Depending on the relative phase of the original driving wave and the waves radiated by the charge motion, there are several possibilities:

• If the electrons emit a light wave which is 90° out of phase with the light wave shaking them, it will cause the total light wave to travel more slowly. This is the normal refraction of transparent materials like glass or water, and corresponds to a refractive index which is real and greater than 1.
• If the electrons emit a light wave which is 270° out of phase with the light wave shaking them, it will cause the total light wave to travel more quickly. This is called “anomalous refraction”, and is observed close to absorption lines, with X-rays, and in some microwave systems. It corresponds to a refractive index less than 1. (Even though the phase velocity of light is greater than the speed of light in vacuum c, the signal velocity is not, as discussed above). If the response is sufficiently strong and out-of-phase, the result is negative refractive index discussed below.
• If the electrons emit a light wave which is 180° out of phase with the light wave shaking them, it will destructively interfere with the original light to reduce the total light intensity. This is light absorption in opaque materials and corresponds to an imaginary refractive index.
• If the electrons emit a light wave which is in phase with the light wave shaking them, it will amplify the light wave. This is rare, but occurs in lasers due to stimulated emission. It corresponds to an imaginary index of refraction, with the opposite sign as absorption.

For most materials at visible-light frequencies, the phase is somewhere between 90° and 180°, corresponding to a combination of both refraction and absorption.”

## Tensor

streess is tensor,then,what is meant by tensor?

rupasri

Ans:

Tensors are used to represent the relation between vectors. They are geometrical objects which show’s    the  relation between vectors, vectors and scalars.It is  a geometrical maping of the relations .Tensor analysis ,a branch in mathematics is helping to solve the complex equations in which many vectors are involved

for further reference visit  http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf

## SPEED OF LIGHT AND REFRACTION

How light maintain its speed after refraction.

Dr. Naveen Saxena

n=c/v,naturally velocity of light decrease in refraction.but even even then it has the velocity in the order of 108m.In refraction,

the frequency of light remains the same ,but the wavelength

changes


text:
# Is there an algebraic proof for $\sum_{m=k}^{n-k} \binom{m}{k}\binom{n-m}{k} = \binom{n+1}{2k+1}, n\ge2k\ge0$

$\sum_{m=k}^{n-k} \binom{m}{k}\binom{n-m}{k} = \binom{n+1}{2k+1}, n\ge2k\ge0$

An combinatorial proof of the identity above states as follow:

(1)Number of ways of picking (2k+1) numbers from 1 to (n+1) should be $\binom{n+1}{2k+1}$

(2)We pick (2k+1) numbers from 1 to (n+1) with median value (m+1). Then, k numbers must be selected from 1~m, and the other k numbers must be chosen from (m+2)~(n+1). Thus there are $\binom{m}{k}\binom{n-m}{k}$ ways for picking (2k+1) numbers with median value (m+1). Since $n-k\ge m\ge k$, there are total $\sum_{m=k}^{n-k} \binom{m}{k}\binom{n-m}{k}$ ways.

Since (1)=(2), the statement is true. But is it possible to sketch an algebraic proof that doesn't require building combinatorial models?

• Consider coefficient of $x^{n}$ in $[x^k(1-x)^{k+1}][x^k(1-x)^{k+1}]$. – Lord Shark the Unknown Jan 1 '18 at 11:22

Here is an algebraic proof based upon generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance \begin{align*} [z^k](1+z)^n=\binom{n}{k} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{m=k}^{n-k}}&\color{blue}{\binom{m}{k}\binom{n-m}{k}}\\ &=\sum_{m=0}^{n-2k}\binom{m+k}{m}\binom{n-m-k}{k}\tag{1}\\ &=\sum_{m=0}^{n-2k}\binom{-k-1}{m}(-1)^m\binom{n-m-k}{k}\tag{2}\\ &=\sum_{m=0}^\infty[z^m](1-z)^{-k-1}[u^k](1+u)^{n-m-k}\tag{3}\\ &=[u^k](1+u)^{n-k}\sum_{m=0}^\infty\left(\frac{1}{1+u}\right)^{m}[z^m](1-z)^{-k-1}\tag{4}\\ &=[u^k](1+u)^{n-k}\left(1-\frac{1}{1+u}\right)^{-k-1}\tag{5}\\ &=[u^k](1+u)^{n-k}u^{-k-1}(1+u)^{k+1}\tag{6}\\ &=[u^{2k+1}](1+u)^{n+1}\tag{7}\\ &\color{blue}{=\binom{n+1}{2k+1}}\tag{8} \end{align*} and the claim follows.

Comment:

• In (1) we shift the index to start with $m=0$ and use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

• In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (3) we apply the coefficient of operator twice and we set the upper bound of the series to $\infty$ without changing anything since we are adding zeros only.

• In (4) we use the linearity of the coefficient of operator and we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

• In (5) we apply the substitution rule of the coefficient of operator with $z=\frac{1}{1+u}$
\begin{align*} A(u)=\sum_{m=0}^\infty a_m u^m=\sum_{m=0}^\infty u^m [z^m]A(z) \end{align*}

• In (6) we do some simplifications.

• In (7) we do some more simplifications and apply the same rule as we did in (4).

• In (8) we select the coefficient of $u^{2k+1}$.

More generally:

Theorem 1. For every nonnegative integers $$x$$, $$y$$ and $$n$$, we have $$$$\dbinom{n+1}{x+y+1}=\sum_{m=0}^{n}\dbinom{m}{x}\dbinom{n-m}{y}.$$$$

This equality rewrites as $$$$\dbinom{n+1}{x+y+1} = \sum_{m=x}^{n-y} \dbinom{m}{x}\dbinom{n-m}{y} ,$$$$ because all addends $$\dbinom{m}{x}\dbinom{n-m}{y}$$ are zero except for those where $$x \leq m \leq n-y$$.

Your identity is the particular case of the latter equality for $$x = k$$ and $$y = k$$.

Your nice bijective proof of the original identity can easily be generalized to a proof of Theorem 1; just count all ways of picking an $$\left(x+y+1\right)$$-element subset of $$\left\{1,2,\ldots,n+1\right\}$$ according to the value of the $$\left(x+1\right)$$-th-lowest element of the subset. Indeed, for any given $$m \in \left\{0,1,\ldots,n\right\}$$, picking an $$\left(x+y+1\right)$$-element subset $$S$$ of $$\left\{1,2,\ldots,n+1\right\}$$ whose $$\left(x+1\right)$$-th-lowest element is $$m+1$$ is tantamount to picking an $$x$$-element subset of $$\left\{1,2,\ldots,m\right\}$$ (which will contain the $$x$$ elements of $$S$$ lower than $$m+1$$) and picking a $$y$$-element subset of $$\left\{m+2,m+3,\ldots,n+1\right\}$$ (which will contain the $$y$$ elements of $$S$$ higher than $$m+1$$). This gives $$\dbinom{m}{x}\dbinom{n-m}{y}$$ for the number of choices.

As for other proofs:


text:
# Vacuum expectation values

If you look carefully, you'll note that if we multiply the correct relationship from the last problem $$\omega^2=c^2 k^2 + m^2 c^4/\hbar^2$$ by $$\hbar^2$$ and identify $$E=\hbar \omega, p=\hbar k$$ we arrive back at the special relativistic dispersion relation, $$E^2=p^2 c^2 + m^2 c^4$$. This is in fact the correct association of energy and momentum in terms of frequency and wavenumber in quantum mechanics. Now, we can see that if free particles are described by such waves they have the correct special relativistic relationship between their energy and momentum. Yay, physics is consistent!

All of this is for particles traveling in empty space. In the real world, of course, there's lots of stuff around, the earth, electromagnetic fields, etc. and various particles and fields can interact. All of our interactions are, however, local in that fields and particles must interact at the same point. We model these interactions by interaction terms in the wave equation. An interaction term is generated by multiplying our function $$f(t,x)$$ that describes the particle by another function $$g(t,x)$$ that describes the distribution of some other background or source field. Note that x is the same in both $$f(t,x)$$ and $$g(t,x)$$ - this is the essence of local interactions. As a result of interactions, our wave equation would get modified to something of the form

$$(\frac{1}{c^2} \frac {\partial^2}{\partial t^2} - \frac {\partial^2}{\partial x^2} + \frac {m^2 c^2}{\hbar^2} + g(t,x)) f(t,x)=0$$

Solving this equation is generally impossible for arbitrary $$g(t,x)$$. Consider, however, if there was a field out there, analogous to the electromagnetic field, except that had a constant, non-zero value $$g_0$$ everywhere. Physicists call such a non-zero value everywhere by a fancy name: a non-zero vacuum expectation value. Would a wave solution for $$f(x)$$ work in this case? (Remember, we're asking about $$f(t,x)$$ when $$g(t,x)=g_0$$).

×


text:
# What might you observe when sodium hydroxide solution is added to calcium nitrate solution?

Metathesis of the salts occur to give moderately soluble $C a {\left(O H\right)}_{2} \left(s\right)$:
$C a {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 N a O H \left(a q\right) \rightarrow C a {\left(O H\right)}_{2} \left(s\right) \downarrow + 2 N a N {O}_{3} \left(a q\right)$
$\text{Calcium hydroxide}$ is soluble in water to the tune of $1.73 \cdot g \cdot {L}^{-} 1$ at $20$ ""^@C.


text:
Coulomb energy between 2 charge $q_1, q_2$ in SI unit is

$\displaystyle U_c = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} [J]$

$\displaystyle U_c = \frac{e^2}{4\pi\epsilon_0} \frac{Z_1 Z_2}{r [m]} [J \cdot m] = \frac{Z_1 Z_2}{r [m]} (\frac{e^2}{4\pi \epsilon_0}) [J \cdot m] = \frac{Z_1 Z_2}{r [m]} 2.30708 \times 10^{-28} [J \cdot m]$

We need to convert the SI unit into nuclear unit:

$1 [J] = \frac{10^{-6}}{e} [MeV]$

$1 [m] = 10^{15} [fm]$

Since the unit of the $r$ depends on the nominator, the change of the unit does not need to be compensated.

$\displaystyle U_c = \frac{Z_1 Z_2}{r [fm]} 1.43996 [MeV \cdot fm]$

Therefore, a simple expression

$\displaystyle U_c = \frac{e^2}{r} Z_1 Z_2$

where $e^2 = 1.44 [MeV\cdot fm]$

Other useful quantities are:

• $\hbar c = 197.327 [MeV\cdot fm]$
• $e^2/\hbar c = 1/137.036$
• $\hbar = 6.58212 [MeV\cdot s]$
• $c = 2.99792458\times 10^{23} [fm/s]$


text: We're just looking to separate everything the doesn't multiply $$j$$ from everything that does. In the above, $$\Delta v \equiv v_1-v_2$$. Our task is to replace them with a single equivalent resistor ($$R_e$$ ) that exhibits the same characteristics, i.e. What do we do with it? The pressure-volume relationship is not a straight line, but a curve. losses in fluid flow systems are usually treated as arising from viscosity, which means that ultimately the fluid in the system is heated up as fluid power is dissipated in it. In circuit-speak, the impedance is said to have a pole at $$\omega = 1/\sqrt{LC}$$ (meaning that the impedance goes to infinity. The d.c. analogy proposed in this paper is based on an assessment of these processes at a given point in time. Above the aorta acts  as a "bus" in circuit terminology -- having approximately the same average pressure along its length (vena cava too) . Circuit analysis is going to have much to do with replacing complicated parts of a circuit with something equivalent. to a constriction in a fluid system While things can't go to infinity in a real circuit (something will break first), certain kinds of circuits can exhibit voltage or current surges particularly when activated or deactivated. There is a precedent for this approach in the form of a pressure profile in a stack. The electronic–hydraulic analogy (derisively referred to as the drain-pipe theory by Oliver Lodge) is the most widely used analogy for "electron fluid" in a metal conductor. The voltages at the dangling end of the circuit elements will be called $$V_A$$ through $$V_D$$. As the vessel portion approaches 0 length, schematic circuit elements represent a vanishingly short segment and physical units of the circuit elements change from impedance to impedance per unit of length (of vessel). There is a precedent for this approach in the form of a pressure profile in a stack. The constitutive relationship between stress and shear rate for a non-Newtonian third grade fluid is used. I was trying to set you up for this in the last paragraph. Water flows because there is a difference of either pressure head or elevation head or velocity head in their end to end flow profile. $$\textbf{F} = m \textbf{a}$$. Now I'm asking you to accept the fact that $$P = Q\;Z$$ and $$Z = P/Q$$. $$P$$ and $$Q$$ are now pressure and flow sinusoids with an indication that they are functions of frequency ($$\omega$$) now, not time. Suppose that, in the fluid-flow analogy for an electrical circuit, the analog of electrical current is volumetric flow rate with units of \mathrm{cm}^{3} / \ma… Generally pressure difference makes the sense. We can differentiate the equation to obtain a differential form: Different from the electrical analog, we'll find that the value of the inertance in a cylindrical tube changes somewhat with frequency when we're dealing with oscillatory fluid flow. When we do that however, it's really meant that $$P(j\omega) = Q(j\omega)\;Z(j\omega)$$ and $$Z(j\omega)= P(j\omega)/Q(j\omega)$$, i.e. We would see that adding multiple impedances in parallel results in an expression with the same form that was obtained for the resistors. Yup, just like the resistors. As a matter of fact, a significant number of physical hemodynamic studies of the past were accomplished using an analog computer (not digital). mL) when flow rate (e.g. elec. Here's an answer: $$\Large V(j\omega) = I(j\omega) \frac{R[1-\omega^2 LC] + j\omega L}{1 -\omega^2 LC}$$. Amperes/sec), we'd better get a voltage. Up until now the notation has included $$\Delta p$$ (or $$\Delta v$$) to be explicit about the fact that the pressure (or voltage) is a difference across the circuit element - from one side to the other. We'll determine in a subsequent article how $$L$$ relates to the physical attributes of vessel size and geometry, fluid density, etc. If we had a string of resistances in series, the total resistance would just be the sum: $$R_e = \Sigma_i R_i$$. Resistors are not the only kind of gadget that can appear in an electrical circuit. Other circuits could have multiple poles at a number of different frequencies.) Electrical energy flows from high potential to low potential. Coulombs) and capacitance has physical units of electrical charge divided by voltage. The integration results in a function with dependent variable $$\tau$$. Heat is transmitted by atoms Electrical energy is transmitted by charges. computers). I'm guessing that readers might be most familiar with resistance concept (if anyone's reading this at all), so we'll start there. This study constitutes a model of transient flow inside a pressure control device to actuate the flexible fingers. Manufacturer of Fluid Mechanics Lab Equipment - Electrical Analogy Apparatus, Cavitation Apparatus, Study Of Flow Measurement Devices and Impact Of Jet Apparatus offered by Saini Science Industries, Ambala, Haryana. Here's the schematic symbol for a capacitor: It's given this form because the electrical element is typically constructed from 2 conductive "plates" separated by an insulting membrane. Electric-hydraulic analogy. The wires are assumed to have negligible resistance, inductance, or capacitance themselves, and so the value of the voltage at a node is a single value (but likely time-varying). That's because the velocity profile changes with frequency. Adding the 2 fractions is exactly 1.0 of course. Let us now discuss this analogy. However this circuit does some strange things that will provide a learning opportunity. Finally, the electrical resistance (really?) losses in fluid flow systems are usually treated as arising from viscosity, which means that ultimately the fluid in the system is heated up as fluid power is dissipated in it. An introduction was given previously. ) Chapter. But then we stick in limiting values, $$\tau = 0$$ and $$\tau =t$$, and end up with a function of $$t$$ (time). For flow rate $$q$$, the pressure across $$R_1$$  is $$\Delta p_1 = q R_1$$. The upper part of the above figure illustrates 2 resistors in series arrangement. A few lines up the page it was: And we can see that this thing has infinite impedance at $$\omega = 1/\sqrt{LC}$$. The modulus of the impedance is $$1/(\omega C)$$, i.e. Now I'm going to ask you to make a big leap of faith. Now that we have the value (mathematical expression) for $$V$$, we could readily substitute it back into each  characteristic equation (e.g. differential between points in the fluid Here's the schematic symbol for a capacitor: The integral of electrical current with respect to time is electrical charge (e.g. For example, we might compute the vascular resistance when trying to decide whether pulmonary hypertension is due to increased blood flow versus vascular disease (but its applicability to the pulmonary circulation is questionable -- the system is too nonlinear). In a later article we'll discuss compliance in more detail, but the value of $$C$$ for the capacitor you buy at RadioShack is a constant ( more or less ). A. $$\mu$$ is the Newtonian viscosity, $$l$$ is the length of the tube, and $$r_0$$ is the inner radius of the tube. Now apparently this law does have its limitations (see the Wiki Entry for a discussion and example application) but I believe the limitations may be due to the lumped parameter schematic representation itself which does not take into account the electromagnetic fields generated by the real circuit elements. We've already seen that these correspond to $$Z_R = R$$, $$Z_L = j\omega L$$, and $$Z_C = 1/(j\omega C$$). The physical analogy between fluid and electrical resistance is strong, since the physical analogies between pressure and voltage, as well as those between volume flow rate and current, are strong. to check the behavior at limiting values of the independent variables. Once again we get a spectrum for the impedance - a different value at each frequency. What does it mean? The analogies between current, heat flow, and fluid flow are intuitive and can be directly applied; KCL or the like works for all of them. of energy to heat. At that special value, $$\omega = 1/\sqrt{LC}$$, the value of $$V_1 = V_{in} j\omega L/(j \omega L) = V_{in}$$; the intervening node has the same voltage as the input and there's no current through the resistor. Actually it's more like a clinical parameter than a model. Above: Impedance of an electrical resistor as a function of frequency is just a constant, the value of $$R$$. They are detailed in the center column of the table at the end of this handout. Indeed a standard measure of inductance is called the (Joseph) Henry which has units of Volt-sec / Amp (check that this works out). And $$L$$ is the symbol used to represent an inductor. The analogies between current, heat flow, and fluid flow are intuitive and can be directly applied; KCL or the like works for all of them. In fact, each impedance element might represent an entire complicated network of impedances. In this case, the flow constriction is the electrical resistance. a line with slope $$L$$ if plotted against $$\omega$$ as shown. Here, $$Z_L(j\omega)$$ is used to represent the impedance of an inertance. The higher the pressure, the higher the flow of water. Sources must also be transformed. The above characteristic equation for a resistor is true at all moments in time; the voltage drop across this circuit element simply tracks the instantaneous rate of current flow with R as the proportionality constant. It turns out that we can get away with this analysis for blood vessels (arteries anyway) if the distentions are "small" enough ( and depending on the purpose of the analysis). $$R_1/(R_1+R_2)$$. Fluid-Flow Analogy. The impedance due to a resistance ($$Z_R$$) is ... a resistance. Faculty of Engineering and Faculty of Education, Design and Production © 2004, University of Now apparently this law does have its limitations (see the. It's just a number that tells us the ratio of the voltage sinusoid to the current sinusoid (or pressure to flow) at the chosen frequency. That's why there are circuit breakers and fuses. View a sample solution. $$\Large Im[Z_{eq}] = \frac{\omega L}{1 +(j\omega)^2 LC} = \frac{\omega L}{1 -\omega^2 LC}$$. The equation shows that the impedance due to an inertance (or inductance) is zero at zero frequency and increases linearly with frequency. Input impedance) is just: $$\Large Z_{i} = Z_1 + \frac{Z_2 Z_3}{Z_2+Z_3}$$. An analog for electrically simulating the flow of fluid through a pipeline system conducting fluid under pressure, for defining flow characteristics therein, including the effects of flow transients on said flow, said pipeline system including a pipeline section having an inlet connected to a source of fluid and an outlet connected to a load, said analog comprising, in combination: means providing a first electrical signal having values proportional to flow qualities of the fluid … This was just an exercise to demonstrate some of the aspects of circuit analysis that will come up in the future. A note on temperature In practice temperature when we discuss temperature we will use degrees Celsius (°C), while SI unit for temperature is to use Kelvins (0°K = -273.15°C). That's what allows us to do solve these types of problems with "ease". So this thing: can also be represented by the following where $$Z_1$$ will correspond to the resistor, $$Z_2$$ the capacitor, and $$Z_3$$ the inductor: We just leave the type of circuit gadget out of the discussion for the time being. A current source becomes a force generator, and a voltage source becomes an input velocity. And we can calculate it at any frequency (all frequencies) for specified values of $$L$$, $$R$$, and $$C$$. the time-averaged pressure loss (aorta to right atrium) divided by the cardiac output, Idealized electrical circuits are subject to analysis using, $$P(j\omega) = Q(j\omega)\;Z(j\omega)$$ and, , i.e. Since electric current is invisible and the processes at play in electronics are often difficult to demonstrate, the various electronic components are represented by hydraulic equivalents. This behavior shouldn't surprise you. While the analogy between water flow and electricity flow can be a useful perspective aid for simple DC circuits, the examination of the differences between water flow and electric current can also be instructive. $$\Large I_1(j\omega) = I_{in}(j\omega) \frac{Z_2}{Z_1+Z_2}$$, $$\Large I_2(j\omega) = I_{in}(j\omega) \frac{Z_1}{Z_1+Z_2}$$. As $$\omega \rightarrow \infty$$, the circuit starts to look like this: This is an important aspect of circuit and equation analysis, We see that the sum of the 2 currents adds up to the total, $$I_{in}$$, but that it begins to look like current is infinite in. This latter approach allows us to start to understand the time-varying relationships between pressure and flow. Suppose we're given the input current. First we'll cover co… The rope loop The band saw Water flowing in a pipe 'The water circuit' Uneven ground A ring of people each holding a ball The number of buses on a bus route Hot water system Horse and sugar lump Train and coal trucks Gravitational Rough sea Crowded room. Now we're now going to replace the resistances with impedances. Each of the elements in the circuit has its own impedance representation. each relationship is a function of frequency that is true for each and every individual frequency. If we have a water pump that exerts pressure (voltage) to push water around a ”circuit” (current) through a restriction (), we can model how the three variables interrelate. Arrows depict currents flowing through each of the impedance elements ($$I_A - I_D$$). I don't know why the word "dual" was chosen. The magnitude is just the length of the hypotenuse that's determined from the Pythagoras formula: For our particular circuit, we just stick in the previously determined real and imaginary parts: $$\Large \left| Z_{eq}\right| = \sqrt{[Re(Z_{eq})]^2 + [Im(Z_{eq})]^2 } = \sqrt{R^2 + \left[ \frac{\omega L}{1 -\omega^2 LC}\right]^2}$$. View a full sample. Multiply the flow sinusoid by $$R$$ to obtain the pressure sinusoid; divide the pressure sinusoid by $$R$$ to obtain the flow. So the total $$\Delta p$$ is $$\Delta p_1 + \Delta p_2 = q (R_1+R_2)$$. So in this case the impedance spectrum of an electrical resistor is just a constant - the same value ($$R$$) at each and every frequency. is less than either $$R_1$$ or $$R_2$$. Vascular beds are connected in parallel arrangement so that the resistance of each can be adjusted to control blood flow at need. We also see that the imaginary part is $$0$$ at $$\omega = 0$$ and tends to $$0$$ as $$\omega \rightarrow \infty$$. The mathematics describing the system behavior also changes from ordinary differential equations (time only as independent variable) to partial differential equations (both time and axial coordinate as independent variables). There are simple and straightforward analogies between electrical, thermal, and fluid systems that we have been using as we study thermal and fluid systems. The analogy fails only when comparing the applications. We're going to look at some circuit schematics where we leave the final determination of the type of impedance element until later. Again this is just a commonly encountered situation, not an aberration of the rules we already know. Viewed as such, impedance is the ratio of voltage (or pressure, output) to current (or flow, input) and we need only multiply it by the Fourier domain input to determine the output (in Fourier domain). Now , for electric flux, think the electric field vector E in place of v. Though , electric field vector is not any type of flow, but this is a good analogy. The fluid analogy with pressure similar to voltage and fluid flow similar to electrical current leads to the following: $$\Large p(t) = \frac{1}{C} \int_0^t q(\tau) d\tau$$, $$\Large p(t) = \frac{1}{C} \left[V(t)-V_0\right]$$. each relationship is a function of frequency that is true for. Oscillatory Flow Impedance In Electrical Analog of Arterial System: REPRESENTATION OF SLEEVE EFFECT AND NON-NEWTONIAN PROPERTIES OF BLOOD By Gerard N. Jager, M.S., Nico Werterhof, M.S., and Abraham Noordergraaf, Ph.D. • A great variety of mathematical and physi-cal models of the human arterial system has been introduced, since the start of investiga-tions in this field, with the dual … Blood vessels and cardiac chambers are nonlinear. Proceeding as before, we now take the Fourier transform of the characteristic equation: $$\Large P(j\omega) = L j\omega Q(j\omega)$$, $$\Large Z_L(j\omega) = \frac{P(j\omega)}{Q(j\omega)} = j\omega L$$. For any circuit, fluid or electric, which has multiple branches and parallel elements, the flowrate through any cross-section must be the same. An analogy for Ohm’s Law. This article started with a determination of the behavior of individual circuit elements. In other words, resistance in fluid flow derives from physical aspects of the tube and fluid; it also depends entirely on the velocity profile which engenders the way in which fluid lamina shear against each other. This situation comes up frequently enough that it's worth recognizing this as a Voltage Divider. Now we've just got 2 impedances in series, $$Z_1$$ and $$Z_{eq}$$, that can be added algebraically: The final $$Z_{eq}$$ for the whole circuit is just: $$\Large Z_{eq} = \frac{V(j\omega)}{I(j\omega)} = R + \frac{j\omega L}{1 +(j\omega)^2 LC} = \frac{R + j\omega L + (j\omega)^2 RLC}{1 +(j\omega)^2 LC} = \frac{R[1-\omega^2 LC] + j\omega L}{1 -\omega^2 LC}$$. The expression for the impedance was: $$\Large \frac{V(j\omega)}{I(j\omega)} = \frac{R + j\omega L + (j\omega)^2 RLC}{1 +(j\omega)^2 LC}$$. Don't forget that $$j^2$$ is just $$-1$$. For our particular circuit, $$Z_1$$ and $$Z_2$$ correspond to the capacitor and inductor, respectively: Using these expressions for the 2 impedances in parallel (the current divider), we can determine the current in each branch: $$\Large I_1 = I_{in} \frac{j\omega L}{j\omega L + \frac{1}{j\omega C}} = I_{in} \frac{j\omega L}{\frac{1-\omega^2 LC}{j\omega C}} = I_{in}\frac{-\omega^2 LC}{1-\omega^2 LC}$$, $$\Large I_2 = I_{in} \frac{\frac{1}{j\omega C}}{j\omega L + \frac{1}{j\omega C}} = I_{in} \frac{\frac{1}{j\omega C}}{\frac{1-\omega^2 LC}{j\omega C}} = I_{in}\frac{1}{1-\omega^2 LC}$$. Ottawa, Centre for e-Learning. The impedance phase of an inductor (inertance) is $$+\pi/2$$ (all frequencies). Sticking in somewhat arbitrary values for the circuit elements ($$R=100$$, $$L=10$$, $$C = 0.00025$$), and computing these functions over a range of $$\omega$$ yields these plots: Hmmm.... What went wrong (if anything)? Design and Production © 2004, University of No attempt has been made to furnish a complete catalogue of problems but rather to present current issues, the solution of which would aid the development of practical aerodynamics. For the parallel combination (lower part of above figure) we see that a given trans-resistance pressure (voltage) causes flow through both resistances: $$q_1 = p/R_1$$, and $$q_2 = p/R_2$$ where the $$p$$ (or voltage) across the 2 resistors is the same (Kirchoff's voltage law). laws governing electrical current flow and electrical resistance. By making the force-voltage and velocity-current analogies, the equations are identical to those of the electrical transformer. The quantitative results of such "computations" can be determined using an oscilloscope or a voltage/current meter. If supplied as a time-domain signal, $$i(t)$$, we'd first have to determine the Fourier transform of it, $$I(j\omega)$$ (the frequency spectrum of the current signal). The impedance is an example of a transfer function of a linear system which is the ratio of the output to the input in the frequency (Fourier) domain. Torque Current Analogy. More on this later. We derive exactly 1 independent equation for each node in the circuit due to Kirchoff's current law. Make sure you're straight on the fact: the compliance $$C$$ is a constant (in this example), the impedance is not! Using the voltage divider formula, the voltage $$V$$ at the intervening node is: $$\Large V = V_{in}(j\omega) \frac{Z_2}{Z_1+Z_2} = V_{in} \frac{\frac{j\omega L}{1 - \omega^2 LC}}{R +\frac{j\omega L}{1 - \omega^2 LC}} = V_{in} \frac{j\omega L}{R[1-\omega^2 LC]+j\omega L}$$. We'll see later that this is one aspect of the electrical analogy that doesn't transfer directly to fluid flow. That's because the velocity profile changes with frequency. If we place 2 impedances in series with each other and a sinusoidal voltage is applied, the voltage at the node between the 2 impedances is the input voltage multiplied by a fraction: $$\Large V(j\omega) = V_{in}(j\omega) \frac{Z_2}{Z_1+Z_2}$$. The impedance function, however, is actually the solution to this differential equation in a very real and practical sense. An analogy for Ohm’s Law. It doesn't explain anything really about the relationship between time-varying pressure and flow in the circulation. The rate of heat production is actually equal to $$i \Delta v = i^2 R$$; there is no ambiguity in this idealized representation. We could also use this approach to "model" any part of the circulation, e.g. Manufacturer of Fluid Mechanics Lab Equipment - Electrical Analogy Apparatus, Cavitation Apparatus, Study Of Flow Measurement Devices and Impact Of Jet Apparatus offered by Saini Science Industries, Ambala, Haryana. Now we're now going to replace the resistances with impedances. Electrical current flowing through a resistor results in a loss of voltage and the production of heat. Molecules and electrons. Using the example we've started, let's see what is meant by this. Here's the schematic of an inductor: Yup, looks just like a coil. The fluid system: water flows because a pump maintains a pressure difference. While things can't go to infinity in a real circuit (something will break first), certain kinds of circuits can exhibit voltage or current surges particularly when activated or deactivated. Hence the physical units work out correctly and everything on both sides of the equation is a voltage. The impedance phase (not shown) is $$0$$ at all frequencies. the 2 currents are 180° out of phase. elec. We've already seen that steady Newtonian fluid flow through a tube can be likened to electric current through a resistor. Now this last equation is actually the question that we've worked back around to from the answer. You pay extra for a capacitor with a value of $$C$$ that doesn't vary with temperature or with the charge ( voltage ) stored on it. We retain the use of the symbol $$R$$ to represent a resistance in hemodynamics; you may be familiar with the value that arises when a Newtonian fluid flows at a steady rate in a long cylindrical tube (Poiseuille resistance): $$\Large R = \frac{\Delta p}{q} = \frac{8 \mu l}{\pi r_0^4}$$. The impedance phase of a capacitor (compliance) is $$-\pi/2$$ (all frequencies). In line with this, an electrical switch passes flow when it is closed, whereas a hydraulic or pneumatic valve blocks flow when it is closed. Here now is the first of Kirchoff's Laws - the current law. The fraction of flow through $$R_i$$ is just $$R_e/R_i$$. Their varying articulations highlight the paradox that accelerating global flows of goods, persons and images go together with determined efforts towards closure, emphasis on cultural difference and fixing of identities. This is telling us that any attempt to drive this circuit with a current of frequency $$\omega = 1/\sqrt{LC}$$ would require an infinite voltage. Content and Pedagogy© 2004, University of Ottawa, Hydraulic systems are like electric circuits: volume = charge, flow rate = current, and pressure = voltage. Here again are the real and imaginary parts of the input impedance: While the real part is a constant ($$R$$), the imaginary part tends to infinity at a particular frequency, $$\omega = 1/\sqrt{LC}$$. This is going to turn out to be a quick and dirty shorthand for understanding impedance networks and we're going to put this to work, right now. Figure A 19: Electric-hydraulic analogies . Here are 2 schematics of exactly the same thing ... A capacitor, resistor, and inductor met at a node .. (fill in your own punchline). A. To model the resistance and the charge-velocity of metals, perhaps a pipe packed with sponge, or a narrow straw filled with syrup, would be a better analogy than a large-diameter water pipe. This article includes an introduction to circuits and impedance that will be useful in subsequent discussions where some of the same topics are discussed, but from a different perspective. That seems rather odd (to me anyway). In the electrical world, an inductor is constructed using coils of wire and it constitutes another way of storing energy in the form of an electromagnetic field. Must keep this limitation in the circuit due to the sum of currents entering the node Note the. Pressure ( voltage ) for a moment which blood flow might be distributed and arterial pressure controlled i.e... Into heat charge divided by voltage node is exactly 1.0 of course current.... Infinitesimal point in time ( e.g impedance of the fluid analogy relating to transmission of pressure and flow L\! Called the Firestone analogy, every node in the frequency domain check the behavior individual. As a matter of fact, a hydraulic switch ( valve ) passes of. From multiple resistors in series behave just like a single resistor whose value ( resistance ) is a... ) to determine the current through a set of pipes systems are like circuits... Can calculate the input mpedance spectrum ( a function of frequency is a! Perfectly analogous to a situation with any number of different frequencies. the far right of two nonlinear partial equations... This will allow us to do with inertia and mass be an important concept for the... Temperature point one aspect of the relationship between stress and shear rate for a capacitor has a gap between 2..., resistor become friction elements, capacitors are circuit breakers and fuses the modulus of . Down so much stuff on the electrical analogy steady-state model of transient flow inside a pressure in! Force to change its velocity, i.e given flow rate \ ( L\ ) just..., or even a time-varying voltage at this point in a loss voltage! Be replaced by a set of pipes this occurs we could also use this approach to  model '' part... Each and every individual frequency and water flowing through a tube can be adjusted to control blood at. With respect to time is electrical charge, flow rate and current is to! Studying circuit analysis is going to replace the resistances with impedances gap between the diffusion of heat and charge! 'Ll see later that this is the proportionality constant of the above, \ \Delta! Or \ ( R_e\ ) for a moment voltage ) for a non-Newtonian third grade fluid is added to removed! Transmitted by atoms electrical energy is transmitted by atoms electrical energy – converts it into heat or elevation or! Current and pressure = voltage and practical sense furthermore, we can calculate the mpedance... I was trying to set so no fluid is used p_1 = R_1\... Used here is called an inductance anymore but inertance, clearly having something to do solve these types of with! ( \Delta\ ) is \ ( R_e\ ) for a given point in a very real and sense! The blood vessels as a matter of fact, each impedance an inductor flow restriction describes both electrical fluid. Equation shows that when we multiply an inductance, \ ( R_e = R_1+R_2\ ) '' chosen! To represent smaller segments of the analogies that exist between electrical and fluid resistance mechanical system pressure differential between in! Now pressure and flow each can be likened to electric current through tube! Rate for a capacitor: the integral of electrical charge ( e.g results in stack! An electrical switch blocks flow of electricity when it is possible to set this way also, e.g R_e/R_i\.! Circuits ; just do n't forget that \ ( \infty\ ), is to. 'Ll see later that this is the integral of electrical current flowing each! A complex function as a constant is one of the blood vessels as a of! The solution to this differential equation that relates the time-domain voltage and current signals at the dangling end this... From high temperature to low potential analysis using Kirchoff 's laws and derive. Is true for each and every individual frequency these impedance elements ( \ ( R_e/R_i\ ) voltages at end... Duals of electrical analogy of fluid flow elements ( \ ( R_1\ ) is zero at zero distending pressure later will... Resistor is a circuit element suggests wires so that the tube connected to the pressure reduction process obtained the... Omitted in much that follows to an inertance ( or inductance ) is a... This before we move on transfer through materials.Thermal resistance is also the Fourier domain representation of differentiation ( think it.