View Full Version : Physics Q&A Game
Just like the Astronomy Q&A game in the Astronomy forum, let's play a physics Q&A game here. The astronomy game has certainly taught me some very neat facts -- maybe this one will do the same.
I'm going to post one question posed by marcus:
"A favorite version of the question chroot just answered is the one about the airplane flying the polar route. It is going 200 meters per second and its wingspan is 30 meters----what is the voltage difference between the two wingtips? I am not asking this question because it is a cousin of the one asked by Ivan, just recalling it. the plane is in a region where the earth's magn. field is roughly vertical and of such and such a strength etc etc."
It seems that there should be no voltage. Faraday's law dictates that
E = -N d[phi]/dt
Since [phi] is defined as the flux: [phi] = B * A, where B is the magnitude of the magnetic field and A is the area enclosed by the loop. In this problem, neither A nor B is changing -- so the induced voltage should be zero.
Am I right? Was it really just a trick question?
- Warren
Ivan Seeking
May25-03, 09:32 PM
-[30X200][50X10^-6][+-30]= +-9 volts
When the plane takes off a voltage will be induced across the wings, and it will be sustained in flight. But you can't measure it with a voltmeter across the wingtips because the very same induction will occur with the meter leads. However, if the plane can land with the charge on the wingtips remaining, it can be measured on the ground with an electrometer, which is a very sensitive voltmeter.
Ivan Seeking
May25-03, 10:00 PM
Originally posted by Tyger
When the plane takes off a voltage will be induced across the wings, and it will be sustained in flight. But you can't measure it with a voltmeter across the wingtips because the very same induction will occur with the meter leads. However, if the plane can land with the charge on the wingtips remaining, it can be measured on the ground with an electrometer, which is a very sensitive voltmeter.
I bought perfect test leads for my perfect voltmeter. They have them down at the gedunken supply house.[:D]
Re: "There are more things in heaven and earth, Horatio, Than are dreamt of in your philosophy."
I like you already! [a)]
Originally posted by Ivan Seeking
-[30X200][50X10^-6][+-30]= +-9 volts
30 meter wingspan x 200 meter/sec speed
x 50 microTesla (the vertical component of the geomagn. field)
let's see how that multiplies out
it comes to 0.3 volts
but you have an extra factor of 30 you multiplied in
which brings it to 9.0
You are permitted to edit your answer if it is a typo!
If it is not a typo then explain please the factor of "[+-30]"
You may certainly be right to have it. I am often forgetting
details and there may be some obvious thing I am missing.
I got this problem from an old edition of Halliday Resnick
but dont remember the exact numbers
Ivan Seeking
May26-03, 12:07 AM
Originally posted by marcus
30 meter wingspan x 200 meter/sec speed
x 50 microTesla (the vertical component of the geomagn. field)
let's see how that multiplies out
it comes to 0.3 volts
but you have an extra factor of 30 you multiplied in
which brings it to 9.0
You are permitted to edit your answer if it is a typo!
If it is not a typo then explain please the factor of "[+-30]"
You may certainly be right to have it. I am often forgetting
details and there may be some obvious thing I am missing.
I got this problem from an old edition of Halliday Resnick
but dont remember the exact numbers
Shoot no. I was integrating the emf over the length of the wings. [s(] twas thinking E and not ε
Ivan Seeking
May26-03, 12:28 AM
canni still go next? canni canni please please [6)]
Originally posted by Ivan Seeking
canni still go next? canni canni please please [6)]
Your turn!
sorry I got preoccupied with something else and just got
back
please proceed
(this game was partly your idea----the teaser problem---which chroot generalized, you and chroot make the rules I think)
It's the Hall effect, Warren. :) Actually a real-life voltmeter should work fine, I think -- just stretch the leads out, doesn't matter if you even touch the wing or not.
I'm still not sure how this answer was arrived at. [s(]
- Warren
Ohhh.. the Hall effect. Hmph.
[;)]
- Warren
Ivan Seeking
May26-03, 02:18 AM
Originally posted by marcus
Your turn!
sorry I got preoccupied with something else and just got
back
please proceed
(this game was partly your idea----the teaser problem---which chroot generalized, you and chroot make the rules I think)
yippie![:D]
The dA/dt was just the width of the wings multiplied by the speed of the plane.
Ok. I don't know if this will be really easy, or really hard, or somewhere in between. A True Story.
The answer to a very complex physics or mathematics problem [I don't remember the exact nature of the problems] was to be calculated on one of the world's top supercomputers of the time; circa 1975. Two approaches were considered. The first and preferred approach was to calculate an extremely precise answer, and next was to approximate the answer. The best answer would require about a year of CPU time. The approximation could be calculated in relatively short order: in a matter of days.
The cost of CPU time caused this issue to be scrutinized very closely. It was decided that in spite of the money, the approximate answer was the best choice. What is the physical motivation [the physics] for their choice?
Alexander
May26-03, 08:40 AM
Originally posted by Tyger
... However, if the plane can land with the charge on the wingtips remaining, it can be measured on the ground with an electrometer, which is a very sensitive voltmeter.
Charge can't remain (separated when plane is on the ground) because plane is made of metal.
Alexander
May26-03, 09:01 AM
Originally posted by damgo
It's the Hall effect, Warren. :) Actually a real-life voltmeter should work fine, I think -- just stretch the leads out, doesn't matter if you even touch the wing or not.
No, voltmeter won't show anything (because emf [vB]l is evenly distributed along a wire l).
Originally posted by Alexander
No, voltmeter won't show anything (because emf [vB]l is evenly distributed along a wire l).
I often disagree with you Alexander or else am put off by your argumentatitive tone of voice (I do not remember which it is) but here I must commend you and voice agreement.
This is not normally considered an example of the Hall effect.
It is a standard textbook example that comes in right after the Faraday law. Halliday Resnick has it as an antenna of a car driving where the field is horizontal and the vertical antenna is cutting across field lines.
Giancoli has it with airplane wings cutting across vertical field lines.
The basic law (as in so many things) is the Lorentz force.
(Which also explains the classical Hall effect! but the problem is elementary and appears long before the Hall effect is discussed.)
A voltmeter could not be used, I reckon, to measure the voltage difference between the wingtips in flight (for the reason Alexander gave----no current would flow thru the meter altho the voltage difference would be there)
A conductor is not necessarily a constant potential thing when it is moving thru a B field as someone said, maybe Alexander.
Lorentz force is so basic, almost the F=ma of E/M
Lorentz force law is really the definition of the E and B fields, it is part of the groundwork.
dropping out the electrostatic part of the law it just says
F = qv X B
The B arrows near the north geomagnetic pole, which is a "south" pole---paradoxically---point DOWN.
So conduction electrons in the wing feel a force to the right
The voltage is higher on the left wingtip, where positive charge carriers would concentrate if there were some.
In Giancoli the problem comes right after the example of the bar sliding along two parallel rails, in a uniform vertical field
The rails merely permit the voltage difference between the ends of the bar to be measured (it is a Faraday law loop including the rails)
The airplane wing is the bar with the two rails removed.
You could put rails in the sky and have the airplane swoop down and fly along them with its wingtips touching them so that the voltage could be measured but the FAA and the Airline's Pilots Association would not allow this so one must trust in Faraday and believe on grounds of faith.
^^^ Well, you would need 'leads' with a great deal of capacitance; then charges will build up at the ends and cause a voltage difference across the voltmeter inputs. emfs in 'series' along each lead act just like one big emf, remember.
Originally posted by damgo
^^^ Well, you would need 'leads' with a great deal of capacitance; then charges will build up at the ends and cause a voltage difference across the voltmeter inputs. emfs in 'series' along each lead act just like one big emf, remember.
not a bad idea. As I picture your idea it involves remote switches that can be used to isolate parts of a conductor running the length of the wing
one might even "freeze" the voltage differerence in carefully insulated elements inside the airplane wing and keep
the difference "on hold" so to speak until the plane could
land and have it measured.
the humble textbook writers did not bother to discuss how the voltage could be measured!
Originally posted by Alexander
Charge can't remain (separated when plane is on the ground) because plane is made of metal.
So lets make the plane out of fiberglass with metal wingtips and have a wire and a switch between them. We can close the switch when the plane takes off and open it before it lands, keeping the charge on the wingtips. And I see marcus thought of this too.
Alexander
May26-03, 03:56 PM
Originally posted by marcus
I often disagree with you Alexander or else am put off by your argumentatitive tone of voice (I do not remember which it is) but here I must commend you and voice agreement... (blah blah blah)...
Markus, why this long post? Any relevance to a voltmeter reading? Looking across this long post I could not find what you state about voltmeter. Can you be concrete in you posts, without mentioning all what textbooks have cover-to-cover?
Voltmeter with its leads span as airplane wings will show zero reading despite that there may be quite high potential between tips of leads. (And I can prove that if needs be.)
From you post I understood that you claim opposite. Well, then prove it.
Originally posted by Alexander
Voltmeter with its leads span as airplane wings will show zero reading despite that there may be quite high potential between tips of leads. (And I can prove that if needs be.)
.
Me too, this is what I was saying. You did not understand.
Originally posted by Alexander
From you post I understood that you claim opposite. Well, then prove it.
Then you REALLY misunderstood me---180 degrees. Well that happens. Better luck next time.
My post was opposed to using the Hall effect (which damgo referred to) as an explanation. Sorry it confused you!
Ivan Seeking
May26-03, 05:57 PM
Well, I see we've all been very busy while I was away. [:D]
If you are all ready to get back to the next question? I'm asking not pushing:[:)]
The answer to a very complex physics or mathematics problem [I don't remember the exact nature of the problems] was to be calculated on one of the world's top supercomputers of the time; circa 1975. Two approaches were considered. The first and preferred approach was to calculate an extremely precise answer, and next was to approximate the answer. The best answer would require about a year of CPU time. The approximation could be calculated in relatively short order: in a matter of days.
The cost of CPU time caused this issue to be scrutinized very closely. It was decided that in spite of the money, the approximate answer was the best choice. What is the physical motivation [the physics] for their choice?
Too vague or have you just been too busy?
Machine reliability. The machine would have a high chance of breaking down and losing all the data in a year.
Originally posted by Ivan Seeking
It was decided that in spite of the money, the approximate answer was the best choice.
I'm not sure I understand. The approximate answer is cheap, right? Why then would they decide to do it "in spite of the money," if it's the cheap solution?
- Warren
Ivan Seeking
May26-03, 08:41 PM
Originally posted by chroot
I'm not sure I understand. The approximate answer is cheap, right? Why then would they decide to do it "in spite of the money," if it's the cheap solution?
- Warren
Another motivation was realized.
Ivan Seeking
May26-03, 08:44 PM
Originally posted by Tyger
Machine reliability. The machine would have a high chance of breaking down and losing all the data in a year.
No one was concerned about the machine breaking down. But I did respond to your post. [:D]
Alexander
May26-03, 08:59 PM
Originally posted by marcus
...
Lorentz force law is really the definition of the E and B fields, it is part of the groundwork.
No, Lorents force is not definition of electric nor nagnetic field. Electric field is defined as a force on a static charge (which, is derived in QED to actually be a result of creation- of virtual photons around one electric charge and disappearance of virtual photon around another charge), and magnetic field is defined as the specific term arizing from Lorents transformation of electric field from one reference system to another (namely - B is just the vector product of electric field with velocity of reference system [vxE]/c[sup]2[/[sup]).
Lorentz force is so basic, almost the F=ma of E/M
Really?
Here is a simple problem on understanding of Lorents force. Two protons move in perpendicular direction. At some moment of time vector of velocity of one of them is directed toward another one. Imagine L letter, first proton is in upper point and its velocity is directed down, second is in lower point right under it, and its velocity is directed to the right). What is the Lorents force F = q[vxB] of the lower proton on the upper one? Well, lower proton creates magnetic field B directed (in the vicinity of the upper electron) "out of the screen", which when vectorly multiplied by the velocity vector of upper proton gives the Lorents force on upper proton directed to the left.
Now, upper proton also creates a magnetic field. The magnitude of this magnetic field is non-zero everywhere except on the axis of motion of the upper proton (= on the line of its velocity vector). Therefore, lower proton happens to be in ZERO megnetic field (because this proton is located directly on the path of the upper proton) and therefore the magnitude of Lorents force on it is ZERO.
But we have a contradiction here: One objects (lower proton) exert force on another one, but gets no reaction force back.
How so? Momentum can't conserve if one objects exerts force on another and is not subjected by the same but oppositely directed force.
Can you resolve this apparent paradox (taking into consideration that Lorents force is "so basic" for you)?
Originally posted by Ivan Seeking
No one was concerned about the machine breaking down. But I did respond to your post. [:D]
Uh maybe because they wanted the answer fast so they could publish their papers before the other guys and get their PhD's...
- Warren
Ivan Seeking
May26-03, 09:34 PM
Originally posted by chroot
Uh maybe because they wanted the answer fast so they could publish their papers before the other guys and get their PhD's...
- Warren
Hee hee. No. There was a physical reason for their choice.[:D]
No offense, Ivan, but there's really not enough information here to make a guess. And this sort of "riddle" really isn't the sort of question I was hoping to see here.
- Warren
Ivan Seeking
May26-03, 11:33 PM
Originally posted by chroot
No offense, Ivan, but there's really not enough information here to make a guess. And this sort of "riddle" really isn't the sort of question I was hoping to see here.
- Warren
Yes. Perhaps I failed at formulate a good question out of this. I really didn't mean to make a riddle as much as to provide hints. Oh well, I will fess up and punt to you chroot. The answer is:
It was found that over the period of one year, the chance of an error due to bombardment by cosmic rays was nearly 1. As a result, this error would have made the exact answer less reliable than the approximation. Actually Tyger was getting close. Also, I would think that the chances for other errors could also be significant.
Here's to you chroot.
Well, hell, Ivan, machine reliability clearly includes this.
In fact, cosmic ray strikes are the most common cause of single-bit errors in modern computer memories -- hence the common SECDED technology -- "single error correction, double error detection." I was thinking "cosmic rays" the entire time, but thought that "machine reliability" included it!
The advance of satellite technology, I thought, had made the cosmic ray argument well-known.
When you said a "physical reason," I thought you meant something about the problem being solved -- not something about the physical machine. :)
In any event, it is certainly an interesting thing!
Why don't you ask another question?
- Warren
Ivan Seeking
May27-03, 12:01 AM
Originally posted by chroot
Well, hell, Ivan, machine reliability clearly includes this.
In fact, cosmic ray strikes are the most common cause of single-bit errors in modern computer memories -- hence the common SECDED technology -- "single error correction, double error detection." I was thinking "cosmic rays" the entire time, but thought that "machine reliability" included it!
The advance of satellite technology, I thought, had made the cosmic ray argument well-known.
When you said a "physical reason," I thought you meant something about the problem being solved -- not something about the physical machine. :)
In any event, it is certainly an interesting thing!
Why don't you ask another question?
- Warren
Since Tyger included the specific reference to breakdowns, I tried to point to his answer but felt that he was otherwise off course.
Darn. That may be the only other interesting thing I have ever learned. [:D] O.K. How about this, I should say up front that I have an answer, but my confidence in this answer is not quite 100% [comes from a so-so professor]. So I will play this one by ear. At some appropriate point I will interject my two cents worth.
The total force due to friction between two materials is calculated as the normal force times the coefficient of friction between the materials. Why do racing cars use wide tires?
Ivan Seeking
May27-03, 12:22 AM
Originally posted by chroot
The advance of satellite technology, I thought, had made the cosmic ray argument well-known.
This is what I was not sure about. Since I have heard little discussion about this topic outside of NASA, I feared this could be an extremely obscure argument, or that this proof may have been a cosmic error! [:D] I was waiting to get run up the flag pole after giving my answer!
On Radioactive Waves
May27-03, 02:01 AM
Why do racing cars use wide tires?
that easy!? greater surface area ensures that it is harder to reach the static coefficiant of friction.
am i right?
Ivan Seeking
May27-03, 02:07 AM
Originally posted by On Radioactive Waves
that easy!? greater surface area ensures that it is harder to reach the static coefficiant of friction.
am i right?
I don't think I understand your answer. Can you splain. [:D]
On Radioactive Waves
May27-03, 02:24 AM
pressure of the tire to the surface is force exerted per unit area. the normal force is not a point source force but one distributed evenly over the area of interaction. greater area means more friction // less slipping.
Ivan Seeking
May27-03, 02:55 AM
Originally posted by On Radioactive Waves
pressure of the tire to the surface is force exerted per unit area. the normal force is not a point source force but one distributed evenly over the area of interaction. greater area means more friction // less slipping.
The frictional force f is calculated as μN. Where N is the normal force [the weight of the car], and μ is the coefficient of friction. Sorry.[:)]
On Radioactive Waves
May27-03, 03:27 AM
are you talking about the static or kinetic coefficient of friction?
ffk = [mu]kN
ffs is less than or equal to [mu]sN
sorry i dont know how to do subscripts
Ivan Seeking
May27-03, 03:40 AM
Originally posted by On Radioactive Waves
are you talking about the static or kinetic coefficient of friction?
ffk = [mu]kN
ffs is less than or equal to [mu]sN
sorry i dont know how to do subscripts
Both apply but it doesn't matter. In either case here, we only need to total weight of the car, and the applicable coefficient of friction in order to calculate the frictional forces. [:)]
On Radioactive Waves
May27-03, 03:50 AM
so the answer is not greater traction? sorry i didnt explain my answer well. peace
Ivan Seeking
May27-03, 04:04 AM
Originally posted by On Radioactive Waves
so the answer is not greater traction? sorry i didnt explain my answer well. peace
How does the perfomance of the vehicle benefit from wide tires? Clearly we do get some traction advantage but why; by what physical mechanism or process? A simple application of the fritional force theory seems to yield no advantage. Peace on you too. [:D]
On Radioactive Waves
May27-03, 04:10 AM
hmm is it someting to do with torque?
Originally posted by Ivan Seeking
How does the perfomance of the vehicle benefit from wide tires? Clearly we do get some traction advantage but why; by what physical mechanism or process? A simple application of the fritional force theory seems to yield no advantage. Peace on you too. [:D]
wider tires allow one to inflate the tires to a lower pressure
and still support the car
the weight of the car is equal to the footprint area of the tires
multiplied by the air overpressure
with wider tires, it is more practical to have a large footprint area and thus a lower pressure (for the same weight of car)
intuitively lower pressure will contribute to better adherence to the road (small irregularities wont cause vibration that bounces the tire out of contact)
Ivan Seeking
May27-03, 03:25 PM
Originally posted by marcus
intuitively lower pressure will contribute to better adherence to the road (small irregularities wont cause vibration that bounces the tire out of contact)
Can you elaborate on this direction of thought?
Ivan Seeking
May27-03, 03:34 PM
Originally posted by marcus
(small irregularities wont cause vibration that bounces the tire out of contact)
What does this matter if the frictional force calculated is independent of surface area? [:D]
Originally posted by Ivan Seeking
Can you elaborate on this direction of thought?
hi Ivan, I dont really believe the normal force is constant
I think irregularities in the road cause the wheel to bounce up and down and at times nearly leave the road
If I were designing a racing vehicle I would want low pressure in the tires so I could have a tight suspension (with little or no play). And so you could pick up the tire a ways before it would actually lose contact with the road.
i would probably want a large footprint to even out little irregularities in the road and to retain contact even with a lot of up/down motion
this is guessing, you asked me to elaborate.
Ivan Seeking
May27-03, 05:04 PM
Originally posted by marcus
lower pressure will contribute to better adherence to the road (small irregularities wont cause vibration that bounces the tire out of contact)
Well, I think I'll call this good enough. I was really looking for any of the following the references:
1).The coefficient of friction is higher for a soft tire:
Buried in this argument are any number of ideas that ultimately lead to the final value of the coefficient of friction. For example, the tire deforms and can then push against vertical surfaces. But this is all that happens at the microscopic level. So really we end up with a number of different levels of "contact" characteristics between the surfaces. After rationalizing this a bit we realize that all arguments boils down to the measured coefficient of friction for a particular tire configuration; e.g. pressure, temperature, stickiness, surface relaxation times, etc. etc. etc. The more one thinks about this, the more variables seem to crop up.
2).There is more rubber to remove:
To insure maximum friction, we typically [historically] want to remove as much rubber from the tire as possible. Obviously, as "On Radioactive Waves" almost started to touch on, if we add rubber by making a thicker tire, we create greater angular inertia than with a wider tire. This then affects the acceleration performance of the car, in addition perhaps to making for an impractical tire design.
I am sure other variables exist that I have never even considered. In any event, good enough, here's to you marcus. [:D]
Originally posted by Ivan Seeking
good enough, here's to you marcus. [:D]
In or around 1990 the BIPM, under the international authority of the General Conference on Weights and Measures (CGPM), established new Electrical Standards known as the 1990 volt and the 1990 ampere.
These are defined independently of the metal kilogram prototype in Paris and are in fact based solely on the atomic clock together with two quantum effects.
What are the two (low temperature) quantum mechanical effects used to define V90 and A90 ?
Or, equivalently if you prefer to look at it that way, the effects used to define V90 and Ω90 .
The definitions are based on ADOPTING exact values for two fundamental physical constants. What are they?
PS: it is sort of analogous to defining the meter by adopting an exact value for the speed of light---the official way since 1983.
You cant measure the speed of light in vacuo any more because it has a decreed exact value---in the 1990 electrical context there are two more constants which you cant measure.
'These are the electrical standards used in practice, though not yet finally made official.
The adopted values are among those listed at the NIST fundamental constants website in the "adopted values" section.
To sum up----the 1990 electrical standards have no logical connection to the metal kilogram and the newton force based on it---or on the old force between infinite parallel wires definition of the ampere. How are they defined?
The quantum magentic flux unit is used, and I believe that is defined with h-bar and e.
JMD
Originally posted by nbo10
The quantum magentic flux unit is used, and I believe that is defined with h-bar and e.
JMD
Yes! but that is only one fundamental constant and only enough to define one unit-----the 1990 ampere.
be more complete and specific.
How, for example, is the 1990 volt defined? Using the atomic clock plus what quantum device?
good, but not yet complete
BTW if you do dimensional analysis of hbar/e2
you will find that (if you carefully do it in SI units only and
follow SI conventions) you will get a ratio of voltage to current.
But that sort of ratio is called a resistance.
check out the adopted constants at the NIST website
nist.gov, if I remember right---the world's premier "fundamental
constants" website
They mostly likely use a squid or josephson junction. With the use of the quantum hall effect both a volt and amp and be "defined".
JMD
Originally posted by nbo10
They mostly likely use a squid or josephson junction. With the use of the quantum hall effect both a volt and amp and be "defined".
JMD
http://physics.nist.gov/cuu/Constants/
these adopted values define the current (1990) electrical standards. I'm quoting from this section of the site
http://physics.nist.gov/cgi-bin/cuu/Category?view=html&Adopted+values.x=102&Adopted+values.y=11
conventional value of Josephson constant
Value 483 597.9 GHz V-1
Standard uncertainty (exact)
conventional value of von Klitzing constant
Value 25 812.807 Ω
Standard uncertainty (exact)
Specifying exact values of these two is equivalent to specifying exact values of hbar and e.
Your turn JMD---you pose a question that you know the answer to
and whoever gets it right has the next turn.
Lets try an easy one,
What is the diference in first and second order phase tranistions?
JMD
Alexander
May27-03, 11:11 PM
Originally posted by Ivan Seeking
Why do racing cars use wide tires?
Simply because wide tires wear less than narrow ones - thus they last longer.
Originally posted by nbo10
Lets try an easy one,
What is the diference in first and second order phase tranistions?
JMD
For starters here are two examples-----the melting of ice is a first order transition.
The change in magnetic state at a critical temperature is a second order transition.
there are different definitions around----one is that first order involves a substantial supply of heat or change in free energy content----and second order just happens without a big change in heat content either way.
In a second order transition the specific heat may change and various other properties (like magnetisation). The change may be less obvious.
An online dictionary I gave a slightly different take on it: first order is abrupt and involves a gross change in properties like from solid to liquid
Second order (this dictionary indicated) was continuous as the critical temperature was approached----rather than abrupt---and likely to involve more subtle properties of the material.
the dictionary practically equated first order with discontinuous transition and second order with continuous.
I think there must be several alternative ways of classifying phase transitions by first or second order, and perhaps higher orders than two.
******for reference here are the dictionary entries*****
Phase transition
A change of state such as occurs in the boiling or freezing of a liquid, or in the change between ferromagnetic and paramagnetic states of a magnetic solid. An abrupt change, characterised by a jump in an order parameter is known as first order; a change in which the order parameter evolves smoothly to or from zero is called continuous or second order.
Order parameter
A variable such as ..... the magnetisation in an Ising model... used to describe the degree of order in a phase above (below) its critical point. In a continuous phase transition (second order phase transition), the order parameter goes continuously to zero as the critical point is approached from above (below).
Almost what I'm getting at.
What happenes to various properties, during a phase change. ie volume, entropy, compressibility, heat capacity. All of which can be expressed from the expression of the free energy. There is a distinct behavior for first and second order tranistions.
JMD
Originally posted by nbo10
Almost what I'm getting at.
What happenes to various properties, during a phase change. ie volume, entropy, compressibility, heat capacity. All of which can be expressed from the expression of the free energy. There is a distinct behavior for first and second order tranistions.
JMD
In Type I there is an abrupt, discontinuous change in these properties (vol, entropy, heat capacity)
In Type II the change is continuous.
www.mit.edu/~levitov/8.334/lec2.pdf
Some MIT lecture notes on statistical mechanics, by Levitov.
Discussing the Landau model of phase transition and
Type I/Type II also called first order/second order transition.
Has a concise summary of the difference at the end of lecture 2, which I will try to copy for extra clarity. Nice question "nbo10".
BTW symmetry is an important feature of Type II. In type I (most common type) it can happen that there is no significant change in symmetry. But in type II (in magnetization for example) there can be an abrupt change in the amount of organization or symmetry.
The notes talk about this.
want to copy in the end-of-lecture summary from those notes
www.mit.edu/~levitov/8.334/lec3.pdf
Type I phase transitons are discontinuous, accompanied by jumps in physical quantities, such as density or entropy, latent heat, and volume change. The symmetry is typically not changed in such transitions.
Type II phase transitions are very different. In such transitions, physical quantities vary continuously, while symmetry changes abruptly. The phase transformation in a type II transition is described by spontaneous symmetry breaking.
In first order phase tranistions, there is a discontiniuty in the first derivative in the free energy.
second order phase tranistions have a disconitiuty in the second derivative in the free energy.
In ice the volume changes as the sturcture changes, First order PT
The tranistion from normal metal to a superconducting state there is a jump in the specific heat, a second order PT
There is no gradual change in symtmetry, there is an abrupt change in going from one state to another.
JMD
Originally posted by nbo10
In first order phase tranistions, there is a discontiniuty in the first derivative in the free energy.
second order phase tranistions have a disconitiuty in the second derivative in the free energy.
In ice the volume changes as the sturcture changes, First order PT
The tranistion from normal metal to a superconducting state there is a jump in the specific heat, a second order PT
There is no gradual change in symtmetry, there is an abrupt change in going from one state to another.
JMD
Well I can't say scored a complete hit---in terms of what you expected. Would you like to have a second turn?
(You have to receive and acknowledge a right answer for the turn to move on to another person.)
It was close.
I have one more that is semi-interesting, a engineering proferssor asked this to my class. By the way his solution wasn't even close to being correct.
When you clap your hands, what keeps your hands from passing through each other? Or when anything touches anything, what keeps it from mergeing together?
Bonus if you can guess how the professor answered this question.
JMD
I would say the Exclusion Principle. It prevents matter from being any more "crowded together" than it is.
The professor's answere might be electrical repulsion.
The professors responce was Pauli's Exclusion Principle. but that is not correct. Pauli's Exclusion Principle imposses symmetry on the wavefunctions.
"electrical repulsion" is also incorrect. You might be thinking along the correct path.
Another bonus, restate the question in a scientific way.
JMD
Originally posted by nbo10
It was close.
.....
When you clap your hands, what keeps your hands from passing through each other? Or when anything touches anything, what keeps it from merging together?
JMD
Fluids do merge together so the idea is what makes a solid solid
and related issues
So most solids have some lattice of bonds
or an effective system of intermolecular forces
that gives them integrity
cells in your hands are bags of fluid enclosed in membrane
if you ask why dont my hands merge when they press together
it must have to do with the strength of membranes
that is like crystal lattice bond-work in two dimensions.
these are amusing questions. i wish you were prepared to
come up with some more
Where is nbo? I want to know if I got the answer right or not.
He asks why don't things merge on contact and the answer is that some things (fluids) do tend to mix on contact so it is a question about solids---how they maintain integrity.
Having one solid pass thru another would involve breaking a lot of bonds. So the bond energy is the answer.
but also, given that the lattice is going to hold, solids are still mostly empty space so there is a further question of why one atom cannot just pass thru another atom.
Even when gasses mix, atom does not pass thru atom and that (which is not what the question is about) may have to do with the exclusion principle. So I stick with my original answer:
Originally posted by marcus
Fluids do merge together so the idea is what makes a solid solid
and related issues
So most solids have some lattice of bonds
or an effective system of intermolecular forces
that gives them integrity
cells in your hands are bags of fluid enclosed in membrane
if you ask why dont my hands merge when they press together
it must have to do with the strength of membranes
that is like crystal lattice bond-work in two dimensions.
these are amusing questions. i wish you were prepared to
come up with some more
I have to sleep sometime
Thats good enough marcus. Your question
To restate the question
Why do solids have a bulk modulus?
Electrostatic potentials forming a lattice.
JMD
Originally posted by nbo10
Thats good enough marcus. Your question
It is more interesting for me when you ask the questions.
hope you answer this and return with a question.
One seldom sees the Einstein equation written in form
compatible with SI metric units, but suppose one did.
then the equation would be written this way:
Gmu,nu = 8pi (G/c4) Tmu,nu
In what SI units are the quantities on the left expressed?
In what SI unit would (c4/G) be expressed? I am intentionally looking at the reciprocal of what appears in the equation. Of course G is Newton's constant and c is the speed of light.
In what SI units are the Tmu,nu terms on the right expressed?
dont be offended if this is too trivial a question. the fact is that people usually see the equations written in units with c = G =1 so in my experience they dont think concretely about how things would look in familiar SI terms
Alexander
May29-03, 04:16 PM
Momentum-energy stress T is in N(ewtons), curvature metrics g is unitless.
G/c4 unit is N-1.
In some papers curvature g is expressed in m-2 and momentum-energy tensor T in N/m2=J/m3
Alexander
May29-03, 04:25 PM
Originally posted by nbo10
The professors responce was Pauli's Exclusion Principle. but that is not correct. Pauli's Exclusion Principle imposses symmetry on the wavefunctions.
JMD
Actually vise versa, the (anti)symmetry of wave function results in Pauli exclusion principle (repulsion of fermions) as a mathematical by-product.
Originally posted by Alexander
....
G/c4 unit is N-1.
I was asking about (c4/G) which is a force and the answer I was looking for is that in SI it is expressed in newtons, the SI force unit.
However you said something just as good! You said the reciprocal is expressed in reciprocal newtons, or as you say N-1.
Originally posted by Alexander
....
In some papers curvature [Gmu,nu, not g] is expressed in m-2 and momentum-energy tensor T in N/m2=J/m3
The curvature Gmu,nu is expressed in 1/area units----in other words---and the Tmu,nu is expressed in energy density units----joules per cubic meter.
As you point out, joules per cubic meter is also equivalent to the unit of pressure, newton per sq. meter.
And as you say this happens in some papers, while in other papers the curvature is made dimensionless, so it is not expressed in SI units or in any kind of units.
OK Alexander, your turn. Have a good one!
Alexander
May29-03, 11:46 PM
Origin of blue sky?
Alexander
May30-03, 06:22 AM
On what?
Originally posted by Alexander
On what?
That's really a second question. So I will put in a thought----it will still be nbo's turn because he said Rayleigh scattering.
I've always been assuming that the Rayleigh scattering that causes the blue of the sky was mostly from the outer electrons of the N2 molecule, just because nitrogen is the most common.
But it is not specific to nitrogen. Oxygen would work just as well and would look blue.
In fact if the Mars CO2 atmosphere were not so full of reddish dust it would look blue also.
So the question "off what?" seems as if it could be answered
"whatever light-element gases the air is made of"
or maybe one could just reply "the air".
Even air has an index of refractions, and it isn't homogenous. To my recall it's scattered off the irregularities in the atmosphere. It's also the cause of the Cyrillian blue we see when we look in the distance.
Alexander
May30-03, 04:45 PM
Originally posted by marcus
That's really a second question. So I will put in a thought----it will still be nbo's turn because he said Rayleigh scattering.
No, it is Tyger turn - he is correct.
Molecule of N2 is too small (d~0.3 nm) to give any essential contribution into blue sky color. Recall that intensity of scattered light is proportional to (d/[lamb])4, so it drops dramatically with size. It would take about hundred times more air to get essential contribution into scattering from individual molecules. But the distribution of molecules in air is fluctuating, and it turns out that most contribution into intensity of scattered light comes from fluctuating pockets of air of order of 10-100 nm in size. These pockets can have dramatically different number of molecules per unit volume.
Density of larger pockets does not fluctuates essentially, and there are less of those pockets, so despite their larger cross section their contribution is less.
Kinda interesting how interplay of math factors creates tangible phenomenon (blue sky) from something which is not very tangible itself (fluctuating due to thermal motion voids/concentration of certain size range).
Alexander
May31-03, 11:53 PM
Tyger, your turn.
On Radioactive Waves
Jun21-03, 04:58 AM
Whats going on? This was too much fun to let the thread die. If Tyger dosnt want to go, someone else should pose a question.
Ivan Seeking
Jun21-03, 05:31 AM
Originally posted by On Radioactive Waves
Whats going on? This was too much fun to let the thread die. If Tyger dosnt want to go, someone else should pose a question.
Is that your question? [:D]
Why don't you go.
that I got this one. I'll come back with a question when I think of a good one, hopefully today.
Why are metals stronger and more ductile than non-metals?
electrostatic forces from the lattice of atoms that make of the metal
JMD
Originally posted by Tyger
Why are metals stronger and more ductile than non-metals?
this is not an answer exactly. I am thinking out loud: metals have comparatively small incomplete outer shells
and give up their outer electrons more easily than non-metals.
in a metallic lattice the outer electrons tend to be owned *collectively* by all the atoms---they are shared in common
such a lattice will be easier to deform (less rigid and brittle) and it will retain its integrity even while being deformed
in a non-metal lattice the ownership of the electrons is not widely shared but instead there is only sharing between nearby neighbors.
so deforming the lattice becomes a complex and difficult negotiation where specific bonds must be broken between old neighbors and reformed between new neighbors.
it is easy to see how cracks could develop and propagate in a non-metallic lattice
but "strong" is a word with various meanings and I suppose sometimes a carbon (non-metal) fiber could be strong---tho not ductile!
I yield to nbo, if there is any question about this, because he said "electrostatic forces" which covers all this in general tho it is not very specific---maybe I was just elaborating on nbo's earlier answer.
for a more specific answer. There is a more precise way of stating why.
KillaMarcilla
Jun22-03, 12:29 AM
Man, I'm clueless
Although this page was a real interestin' read:
http://matse1.mse.uiuc.edu/~tw/metals/prin.html
Ivan Seeking
Jun22-03, 01:24 AM
Originally posted by Tyger
for a more specific answer. There is a more precise way of stating why.
Band Theory.
The energy separation between successive levels is extremely small - referring to the electron orbitals for each atom - due to sharing of unfilled orbitals.
The answer I was looking for was Surface Tension. The electron gas provides a powerful surface tension that holds the faces of the crystal latice together so that they slide over each other rather than break apart. We can see this surface tension at work in Mercury where the droplets resist deformation despite the high density.
Originally posted by Tyger
The answer I was looking for was Surface Tension. The electron gas provides a powerful surface tension that holds the faces of the crystal latice together so that they slide over each other rather than break apart. We can see this surface tension at work in Mercury where the droplets resist deformation despite the high density.
nice explanation. would not have occurred to me to think of
strength and ductility of metals due to surface tension of electron gas
unless there is some objection, why don't you ask the next question Tyger
On Radioactive Waves
Jul18-03, 04:20 AM
I'll think of a question to post tommorow if someone dosn't beat me to it.
Once again, Tyger is lagging.
Originally posted by On Radioactive Waves
I'll think of a question to post tommorow if someone dosn't beat me to it....
that seems reasonable to me. I hope you do post a question.
The rules of the Q/A game are at the beginning of the "Astronomy" Q/A thread----I think "nicool" wrote them. The rules say something like it is up for grabs if the thread is cold for 3 days.
I forget exactly. But why stand on ceremony? Go for it.
It wouldn't just be raliegh, it's probably a little raman and raman-stokes. I would imagine the electrons would not fall back to their origonal energy levels everytime.
Pete
Integral
Jul20-03, 11:10 AM
SD645,
The rules of the thread are to post a single question, the one who answers correctly (first) earns the right to to post the next question. At this point, while I am tempted to simply delete the extra questions I will instead specify that your first question is the current one on the "table". It would be nice if you cleaned up the other questions.
If you can't do it I can.
sdeliver645
Jul20-03, 11:25 AM
Sorry to have disobeyed the rules of the thread (single question posting and waiting my turn). Frankly, the thread isn't very active so I didn't think people would mind if I posted some interesting questions.
Guess I was wrong.
P.S. I saved you the trouble of "cleaning up" my questions.
Integral
Jul20-03, 11:46 AM
Thanks,
Give it a bit more time. I am sure someone will step up to the plate soon. [:)]
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