text:
# Thread: Monkeys do not include 'entanglement' in the physics descriptions, why?

1. ## Physicists do not include 'entanglement' in their physics descriptions, why?

Quantum Entanglement and Information

First published Mon Aug 13, 2001; substantive revision Thu Aug 26, 2010

Quantum entanglement is a physical resource, like energy, associated with the peculiar nonclassical correlations that are possible between separated quantum systems. Entanglement can be measured, transformed, and purified. A pair of quantum systems in an entangled state can be used as a quantum information channel to perform computational and cryptographic tasks that are impossible for classical systems. The general study of the information-processing capabilities of quantum systems is the subject of quantum information theory.
•1. Quantum Entanglement
•2. Exploiting Entanglement: Quantum Teleportation
•3. Quantum Information
•4. Quantum Cryptography
•5. Quantum Computation
•6. Interpretative Remarks
•Bibliography
•Other Internet Resources
•Related Entries

--------------------------------------------------------------------------------

1. Quantum Entanglement

In 1935 and 1936, Schrödinger published a two-part article in the Proceedings of the Cambridge Philosophical Society in which he discussed and extended a remarkable argument by Einstein, Podolsky, and Rosen. The Einstein-Podolsky-Rosen (EPR) argument was, in many ways, the culmination of Einstein's critique of the orthodox Copenhagen interpretation of quantum mechanics, and was designed to show that the theory is incomplete. (See The Einstein-Podolsky-Rosen Argument in Quantum Theory and Copenhagen Interpretation of Quantum Mechanics.) In classical mechanics the state of a system is essentially a list of the system's properties — more precisely, it is the specification of a set of parameters from which the list of properties can be reconstructed: the positions and momenta of all the particles comprising the system (or similar parameters in the case of fields). The dynamics of the theory specifies how properties change in terms of a law of evolution for the state. Pauli characterized this mode of description of physical systems as a ‘detached observer’ idealization. See Pauli's letter to Born in The Born-Einstein Letters (Born, 1992; p. 21. On the Copenhagen interpretation, such a description is not possible for quantum systems. Instead, the quantum state of a system should be understood as a catalogue of what an observer has done to the system and what has been observed, and the import of the state then lies in the probabilities that can be inferred (in terms of the theory) for the outcomes of possible future observations on the system. Einstein rejected this view and proposed a series of arguments to show that the quantum state is simply an incomplete characterization of the system. The missing parameters are sometimes referred to as ‘hidden parameters’ or ‘hidden variables’ (although Einstein did not use this terminology, presumably because he did not want to endorse any particular ‘hidden variable’ theory).

It should not be supposed that Einstein's definition of a complete theory included the requirement that it be deterministic. Rather, he required certain conditions of separability and locality for composite systems consisting of separated component systems: each component system separately should be characterized by its own properties (even if these properties manifest themselves stochastically), and it should be impossible to alter the properties of a distant system instantaneously (or the probabilities of these properties) by acting on a local system. In later analyses — notably in Bell's extension of the EPR argument — it became apparent that these conditions, suitably formulated as probability constraints, are equivalent to the requirement that statistical correlations between separated systems should be reducible to probability distributions over common causes (deterministic or stochastic) in the sense of Reichenbach. (See Bell's Theorem and Reichenbach's Common Cause Principle.)

In the original EPR article, two particles are prepared from a source in a certain quantum state and then move apart. There are ‘matching’ correlations between both the positions of the two particles and their momenta: a measurement of either position or momentum on a particular particle will allow the prediction, with certainty, of the outcome of a position measurement or momentum measurement, respectively, on the other particle. These measurements are mutually exclusive: either a position measurement can be performed, or a momentum measurement, but not both simultaneously. Either correlation can be observed, but the subsequent measurement of momentum, say, after establishing a position correlation, will no longer yield any correlation in the momenta of the two particles. It is as if the position measurement disturbs the correlation between the momentum values. The puzzle is that the assumption of the completeness of the quantum state of the particle pair is inconsistent with the assignment of labels to the particles separately that could be associated with appropriately correlated values for the outcomes of position and momentum measurements. These labels would be the common causes of the correlations, and would provide an explanation of the correlations in terms of the initial correlations between the properties of the two systems at their source. EPR concluded that the quantum state was incomplete.

Here is how Schrödinger put the puzzle in the first part of his two-part article (Schrödinger, 1935; p. 559):

Yet since I can predict either x1 or p1 without interfering with the system No. 1 and since system No. 1, like a scholar in an examination, cannot possibly know which of the two questions I am going to ask first: it so seems that our scholar is prepared to give the right answer to the first question he is asked, anyhow. Therefore he must know both answers; which is an amazing knowledge; quite irrespective of the fact that after having given his first answer our scholar is invariably so disconcerted or tired out, that all the following answers are ‘wrong.’

What Schrödinger showed was that if two particles are prepared in a quantum state such that there is a matching correlation between two ‘canonically conjugate’ dynamical quantities — quantities like position and momentum whose values suffice to specify all the properties of a classical system — then there are infinitely many dynamical quantities of the two particles for which there exist similar matching correlations: every function of the canonically conjugate pair of the first particle matches with the same function of the canonically conjugate pair of the second particle. Thus (Schrödinger, p. 559) system No. 1 ‘does not only know these two answers but a vast number of others, and that with no mnemotechnical help whatsoever, at least with none that we know of.’

Schrödinger coined the term ‘entanglement’ to describe this peculiar connection between quantum systems (Schrödinger, 1935; p. 555):

http://plato.stanford.edu/entries/qt-entangle/

why is it not in the Big Bang physics?

Nor in particle physics (cern/higgs pursuit)?

String?

How is it that a property does exist in nature and physicist dont use it, WHY?
Last edited by Bishadi; Aug 02 2012 at 05:20 AM. Reason: title correction

2. Originally Posted by Bishadi
Quantum Entanglement and Information

First published Mon Aug 13, 2001; substantive revision Thu Aug 26, 2010

Quantum entanglement is a physical resource, like energy, associated with the peculiar nonclassical correlations that are possible between separated quantum systems. Entanglement can be measured, transformed, and purified. A pair of quantum systems in an entangled state can be used as a quantum information channel to perform computational and cryptographic tasks that are impossible for classical systems. The general study of the information-processing capabilities of quantum systems is the subject of quantum information theory.
•1. Quantum Entanglement
•2. Exploiting Entanglement: Quantum Teleportation
•3. Quantum Information
•4. Quantum Cryptography
•5. Quantum Computation
•6. Interpretative Remarks
•Bibliography
•Other Internet Resources
•Related Entries

--------------------------------------------------------------------------------

1. Quantum Entanglement

In 1935 and 1936, Schrödinger published a two-part article in the Proceedings of the Cambridge Philosophical Society in which he discussed and extended a remarkable argument by Einstein, Podolsky, and Rosen. The Einstein-Podolsky-Rosen (EPR) argument was, in many ways, the culmination of Einstein's critique of the orthodox Copenhagen interpretation of quantum mechanics, and was designed to show that the theory is incomplete. (See The Einstein-Podolsky-Rosen Argument in Quantum Theory and Copenhagen Interpretation of Quantum Mechanics.) In classical mechanics the state of a system is essentially a list of the system's properties — more precisely, it is the specification of a set of parameters from which the list of properties can be reconstructed: the positions and momenta of all the particles comprising the system (or similar parameters in the case of fields). The dynamics of the theory specifies how properties change in terms of a law of evolution for the state. Pauli characterized this mode of description of physical systems as a ‘detached observer’ idealization. See Pauli's letter to Born in The Born-Einstein Letters (Born, 1992; p. 21. On the Copenhagen interpretation, such a description is not possible for quantum systems. Instead, the quantum state of a system should be understood as a catalogue of what an observer has done to the system and what has been observed, and the import of the state then lies in the probabilities that can be inferred (in terms of the theory) for the outcomes of possible future observations on the system. Einstein rejected this view and proposed a series of arguments to show that the quantum state is simply an incomplete characterization of the system. The missing parameters are sometimes referred to as ‘hidden parameters’ or ‘hidden variables’ (although Einstein did not use this terminology, presumably because he did not want to endorse any particular ‘hidden variable’ theory).

It should not be supposed that Einstein's definition of a complete theory included the requirement that it be deterministic. Rather, he required certain conditions of separability and locality for composite systems consisting of separated component systems: each component system separately should be characterized by its own properties (even if these properties manifest themselves stochastically), and it should be impossible to alter the properties of a distant system instantaneously (or the probabilities of these properties) by acting on a local system. In later analyses — notably in Bell's extension of the EPR argument — it became apparent that these conditions, suitably formulated as probability constraints, are equivalent to the requirement that statistical correlations between separated systems should be reducible to probability distributions over common causes (deterministic or stochastic) in the sense of Reichenbach. (See Bell's Theorem and Reichenbach's Common Cause Principle.)

In the original EPR article, two particles are prepared from a source in a certain quantum state and then move apart. There are ‘matching’ correlations between both the positions of the two particles and their momenta: a measurement of either position or momentum on a particular particle will allow the prediction, with certainty, of the outcome of a position measurement or momentum measurement, respectively, on the other particle. These measurements are mutually exclusive: either a position measurement can be performed, or a momentum measurement, but not both simultaneously. Either correlation can be observed, but the subsequent measurement of momentum, say, after establishing a position correlation, will no longer yield any correlation in the momenta of the two particles. It is as if the position measurement disturbs the correlation between the momentum values. The puzzle is that the assumption of the completeness of the quantum state of the particle pair is inconsistent with the assignment of labels to the particles separately that could be associated with appropriately correlated values for the outcomes of position and momentum measurements. These labels would be the common causes of the correlations, and would provide an explanation of the correlations in terms of the initial correlations between the properties of the two systems at their source. EPR concluded that the quantum state was incomplete.

Here is how Schrödinger put the puzzle in the first part of his two-part article (Schrödinger, 1935; p. 559):

Yet since I can predict either x1 or p1 without interfering with the system No. 1 and since system No. 1, like a scholar in an examination, cannot possibly know which of the two questions I am going to ask first: it so seems that our scholar is prepared to give the right answer to the first question he is asked, anyhow. Therefore he must know both answers; which is an amazing knowledge; quite irrespective of the fact that after having given his first answer our scholar is invariably so disconcerted or tired out, that all the following answers are ‘wrong.’

What Schrödinger showed was that if two particles are prepared in a quantum state such that there is a matching correlation between two ‘canonically conjugate’ dynamical quantities — quantities like position and momentum whose values suffice to specify all the properties of a classical system — then there are infinitely many dynamical quantities of the two particles for which there exist similar matching correlations: every function of the canonically conjugate pair of the first particle matches with the same function of the canonically conjugate pair of the second particle. Thus (Schrödinger, p. 559) system No. 1 ‘does not only know these two answers but a vast number of others, and that with no mnemotechnical help whatsoever, at least with none that we know of.’

Schrödinger coined the term ‘entanglement’ to describe this peculiar connection between quantum systems (Schrödinger, 1935; p. 555):

http://plato.stanford.edu/entries/qt-entangle/

why is it not in the Big Bang physics?

Nor in particle physics (cern/higgs pursuit)?

String?

How is it that a property does exist in nature and physicist dont use it, WHY?
The dynamics of the theory specifies how properties change in terms of a law of evolution for the state

The "Law of Evolution"?

huh?

3. You cannot use entanglement to send information without a classical channel of information, end of story.

4. Originally Posted by taikoo
The dynamics of the theory specifies how properties change in terms of a law of evolution for the state

The "Law of Evolution"?

huh?
an evolution is a progression of state, versus the reduction. kind of like adding 2 colors, red and yellow and getting orange.

5. Originally Posted by Anarcho-Technocrat
You cannot use entanglement to send information without a classical channel of information, end of story.
not concerned with your claim.

Entanglement is a 'potential' that needs to be included in large body predictions along with the molecular, thru to chemical. Are you up for the fun?

nothing pseudo about entanglement and it is that very property that makes the EPR comprehensible.

the problem is, one group is lost on just playing with the theorem, versus putting the skills to measured evidence. like entanglement, for example.

It is a NATURAL property, that has not been included in what you call 'physics'.

Sorry to burst your bubble.

ps...You personally cannot ask me for assistence. you've lost my respect in just about anything!

6. these guys are using it

PDF]
calculation of quantum entanglement using a genetic algorithm

soar.wichita.edu/dspace/bitstream/handle/10057/1144/t07027.pdf?...

File Format: PDF/Adobe Acrobat
by J Lesniak - 2007 - Related articles
As systems that use entanglement have evolved, the calculation of entanglement has ... A general method was developed for the calculation of entanglement for ...

Can the concurrence be calculated in terms of the entanglement of ...

physics.stackexchange.com/.../can-the-concurrence-be-calculated-in-t...

Feb 14, 2012 – If I somehow know the entanglement of formation, $E_F$ for two mixed qubits, where. E_F = -x \log x - (1-x) \log (1-x), ...

An efficient numerical method for calculating the entanglement of ...

arxiv.org › quant-ph

by JR Gittings - 2003 - Cited by 5 - Related articles
Feb 3, 2003 – Abstract: We present a conjugate gradient method for calculating the entanglement of formation of arbitrary mixed quantum states of any ...

[PDF]
Entanglement and electron correlation in quantum chemistry ...

www.chem.purdue.edu/kais/paper/Huang-53.pdf

File Format: PDF/Adobe Acrobat - Quick View
by Z HUANG - Cited by 4 - Related articles
Ab initio configuration interaction calculation for entanglement is presented for .... calculating entanglement for a simpler two-spin model system. This is a model ...

[PDF]
Entanglement as measure of electron–electron correlation in ...

www.chem.purdue.edu/kais/paper/ref94.pdf

File Format: PDF/Adobe Acrobat - Quick View
by Z Huang - 2005 - Cited by 42 - Related articles
tum chemistry calculations. Entanglement is directly observable and it is one of the most striking properties of quantum mechanics. As an example we calculate ...

7. "Quantum entanglement is a strange and non-intuitive aspect of the quantum theory of matter, which has puzzled and intrigued physicists since the earliest days of the quantum theory," said Leon Balents, senior author of a recent paper on this topic published in the journal Nature Physics. Balents is a professor of physics and a permanent member of KITP.

Quantum entanglement represents the extent to which measurement of one part of a system affects the state of another; for example, measurement of one electron influences the state of another that may be far away, explained Balents

then in another

Aug. 6, 2012 — A fundamental characteristic of quantum physics is the fact that two or more particles can exhibit correlations stronger than classically allowed. This unique characteristic applies particularly to quantum entanglement: as soon as the quantum state of a particle is measured the state of its entangled partner changes accordingly, regardless of how far apart the two entangled particles might be


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For a quadratic equation, the discriminant of a quadratic can tell us the nature of the roots of that equation.

A quadratic equation of the form:    ax2 + bx + c = 0

a, b and c  are rational numbers,  a ≠ 0.

An equation of this kind, has discriminant:    b24ac.

Which actually comes from part of the Quadratic Formula.

x = $\dpi{100}&space;{\color{DarkBlue}\boldsymbol{\frac{-b&space;\pm&space;\sqrt{b^{2}-4ac}&space;}{2a}}}$

If the discriminant of a quadratic turns out to be zero, the roots/solutions of the equation are real and equal, so one root.

If the result is a positive number, the roots are real and unequal, so two different roots, and if the discriminant is less than zero, then there are no real roots/solutions.

To summarize:

1)    b24ac > 0    :    Two distinct, real roots.

2)    b24ac = 0    :    Two equal roots, so one root.

3)    b24ac < 0    :    No real roots.

A reminder of what this looks like in graph form.

(1.1)

How many roots does the equation    y = 2x2 + 5x − 3   have?

Solution

( ax2 + bx + c )     Here:     a = 2     b = 5     c = -3

=>  b24ac   =   52 − ( 4 × 2 × (-3) )

25 − (-24)

25 + 24  =  49

49 > 0,  so the equation   y = 2x2 + 5x − 3   has 2 real roots.

We can actually solve the equation here by factoring to obtain the 2 roots if required.

2x2 + 5x − 3 = 0
=>  (2x1)(x + 3) = 0     ,     x = $\dpi{70}&space;\fn_jvn&space;\frac{1}{2}$       x = -3

(1.2)

Find the value of m if the roots of    y = 9x2 − 12x + m    are equal?

Solution

( ax2 + bx + c )         a = 9     b = 12     c = m

=>  b24ac   =   122 − ( 4 × 9 × m )

14436m

As the roots are equal,   14436m   must equal 0.

14436m  =  0,

144 = 36m,

$\dpi{80}&space;\fn_jvn&space;\frac{144}{36}$ = m     ,     m = 4

(1.3)

Given that  k  is a real number, show that the roots of the equation

kx2 + 5x + 5  =  k    are always real.

Solution

kx2 + 5x + 5  =  k         can become         kx2 + 5x + 5k  =  0

Treat  5k  as   c  in a quadratic equation.

( ax2 + bx + c )

a = k     b = 5     c = 5k

=>  b24ac   =   52 − ( 4 × k × (5k) )

254k(5k)

2520k4k2

4k220k + 25

(2k5)(2k5)

=> (2k5)  ≥  0

As  k  is a real number, any  (2k5)  squared will always be positive or zero, so the roots or root in this case, will indeed be real.

›   ›  Discriminant of a Quadratic


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# Issues with my DRMG code

+1 vote
edited

Hi Miles,
Thanks for your offer to help and look through my code. I have tried to clarify it as far as possible and take out all the pieces that are not (yet) relevant.

Attached is the preliminary DMRG code to calculate ground state energy and two-particle correlator Szi Sz(i+1) of spin-1/2 zig-zag chain. The system is defined as follows:

(1) The hamiltonian for the system consists of NN and NNN hopping and interaction terms as well as the magnetic field term. The total number of particles in the system is always half the total number of sites. The NN hopping terms, however, are set to zero for the presumably easiest case.

(2) The interactions are dipole interactions which are functions of the geometry. As a result, there are two different NN interactions: V1odd is the NN interaction for odd site index and V1even is the NN interaction for even site index - typically they are not just different but can have different signs. (V2 is the NNN interaction, which is set to zero in the simplest case.)

The DMRG code is as follows:

## include iomanip> //To use setw()

using namespace itensor;
using namespace std;

int main()
{
for(int N = 4; N <= 20; N = N + 2)
{
auto sites = SpinHalf(N);

auto ampo = AutoMPO(sites);

#define pi 3.1415926535

double gamma = 120.0;

double theta = 22.0;

double V1odd = 1 - 3 * pow(cos(pi - (gamma * pi/180)/2 - theta * pi/180 ), 2);

double V1even = 1 - 3 * pow(cos((gamma * pi/180)/2 - theta * pi/180), 2);

double V2 = 0.0;

double h = 0.0;

double J1 = 0.001;

double J2 = 0.0;

for(int j = 1; j < N; ++j)
{
ampo += (-J1),"S+",j,"S-",j+1;
ampo += (-J1),"S-",j,"S+",j+1;

if(j%2 != 0)
{
ampo += V1odd,"Sz",j,"Sz",j+1;
}
else
{
ampo += V1even,"Sz",j,"Sz",j+1;
}

}

for(int j = 1; j < N-1; j = j+2)
{
ampo += J2,"S+",j,"S-",j+2;
ampo += J2,"S-",j,"S+",j+2;
ampo += V2,"Sz",j,"Sz",j+2;
}

for(int j = 1; j <= N; ++j)
{
ampo += h,"Sz",j;
}

auto H = IQMPO(ampo);

// Set the initial wavefunction matrix product state
// to be a classical Neel state

auto initState = InitState(sites);

for(int i = 1; i <= N; ++i)
{
if(i%2 == 1)
initState.set(i,"Up");
else
initState.set(i,"Dn");
}

/*
//Set the initial wavefunction matrix product state to be a ferromagnetic state

auto initState = InitState(sites);

for(int i = 1; i <= N; ++i)
{
if(i <= N/2)
initState.set(i,"Up");
else
initState.set(i,"Dn");
}

*/

auto psi = IQMPS(initState);

auto sweeps = Sweeps(5);
sweeps.maxm() = 10,20,100,100,200;
sweeps.cutoff() = 1E-10;
sweeps.noise() = 1E-7,1E-8,0;
println(sweeps);

auto energy = dmrg(psi,H,sweeps,"Quiet");

printfln("\nGround State Energy = %.10f",energy);

printfln("\nUsing psiHphi = %.10f", psiHphi(psi,H,psi) );

println("\nTotal QN of Ground State = ",totalQN(psi));
//
//To calculate the two particle correlator Sz_i Sz_j where j = i+1
//using the instructions on http://itensor.org/docs.cgi?page=formulas/correlator_mps

Print(N);
Print(V1odd);
Print(V1even);

double szi_szip1_array[N] = {0};

for(int i = 1; i < N; ++i)
{
int j = i + 1;

auto Sz_i = sites.op("Sz",i);
auto Sz_j = sites.op("Sz",j);
psi.position(i);

auto C = psi.A(i)*Sz_i*dag(prime(prime(psi.A(i),Site),ir));
for(int k = i+1; k < j; ++k)
{
C *= psi.A(k);
}
C *= psi.A(j);
C *= Sz_j;

//Original line found on http://itensor.org/docs.cgi?page=formulas/correlator_mps DIDN'T WORK
//C *= dag(prime(psi.A(j),jl,Site));

C *= dag(prime(prime(psi.A(j),Site),jl));

auto result = toReal(C);

szi_szip1_array[i] = result;

}
cout << left << setw(15) << "Site index i" << left << setw(15) << "<Sz_i Sz_(i+1)>" << endl;
for(int i = 1; i < N; ++i)
{
cout << left << setw(15) << i << left << setw(15) << szi_szip1_array[i] << endl;
}

}

return 0;
}

A small list of things that I have found so far and that might be of relevance:

(1) There is a problem in the description of interaction terms, V1odd and V1even. Each time I run the code, the value of ground state energy for a given value of N, theta, gamma and h changes, even if none of the variables (N, theta, gamma, h) changes.
(2) For gamma = 120 degrees, theta = 22 degrees and J1 = J2 = V2 =0, the two particle correlator Szi Sz(i+1) should report values -0.25 for odd i and +0.25 for even i (because in this case V1odd and V1even have different signs) but that's not what my code reports. What it reports is as follows (Number on left is site index and that on right is Szi Sz(i+1)):

N = 4

1 -0.25
2 0.25
3 -0.25

N = 6

1 -0.25
2 -0.25
3 -0.25
4 -0.25
5 -0.25

N = 8

1 -0.25
2 0.25
3 -0.25
4 0.25
5 -0.25
6 -0.25
7 -0.25

N = 10

1 -0.25
2 0.25
3 -0.25
4 0.25
5 -0.25
6 0.25
7 -0.25
8 0.25
9 -0.25

N = 12

1 -0.25
2 0.25
3 -0.25
4 0.25
5 -0.25
6 0.25
7 -0.25
8 -0.238734
9 -0.25
10 0.238734
11 -0.25

N = 14

1 -0.25
2 -0.25
3 -0.25
4 -0.25
5 -0.25
6 -0.25
7 -0.25
8 -0.25
9 -0.25
10 -0.25
11 -0.25
12 -0.25
13 -0.25

(3) If I turn on very small hopping (J1 = 0.001), keeping everything else exactly the same as before (i.e., gamma = 120 degrees, theta = 22 degrees, J2 = V2 = 0, etc.), the code reports correct values for Szi Sz(i+1) (-0.25 for odd i and +0.25 for even i).

(4) Everything before was done setting initial wavefunction MPS to be a classical Neel state. If I set it to a ferromagnetic state and run the code with gamma = 120 degrees, theta = 22 degrees and J1 = J2 = V2 = 0, the values of Szi Sz(i+1) I obtain imply that the state remains unchanged, i.e., it stays ferromagnetic (in this case, all up-spins on one side, all down-spins on the other). Even if I set J1 = 0.001, it produces no significant difference.

related to an answer for: How to increase the accuracy of DMRG?

selected

Hi Niraj,
Sorry about the slow reply! I looked over your code and found a few things that could be causing the problems you report. However, when I ran it I always got the same energy value so I was not able to reproduce that particular issue. What exact parameters were you using when you saw the energy come out different for each run of the code?

Regarding your other issues / questions:

1. Did you intend for the J2 and V2 interactions to skip every other bond in addition to being next-nearest-neighbor? Maybe you don't actually want to increase j by 2 (j += 2) for every step of the loop. Maybe you do; just wanted to point this out.

2. I would not recommend combining different sizes in a single loop in your code. This may be an ok thing to do later once you have thoroughly debugged your code and studied the best DMRG accuracy parameters to use to ensure convergence. But since you are having issues, it is best to fix a single system size and figure out the best way to use DMRG for this system size. For longer systems you may need to do more DMRG sweeps or keep more states, etc.

3. The reason the code C *= dag(prime(psi.A(j),jl,Site)) didn't compile is that you are using ITensor version 1.x and that particular style of code only works for version 2.x. The website examples mostly refer to the new version now, since it would be difficult to maintain the site for both versions simultanously. Please upgrade to version 2.x if you can since it is faster and has some nicer features and is better maintained.

Ok those are my main recommendations for now. If you are still seeing different energies for each run, please simplify your code to use only one size; check that your Hamiltonian is really correct and what you want; and then tell me exactly which Hamiltonian and sweeping/accuracy parameters you are using and we can discuss more.

Best,
Miles

commented by (620 points)
edited
Hi Miles,
Thank you so much for the reply. I apologize for a slow reply - I was on a vacation. As you suggested I upgraded to version 2.x of ITensor, fixed the system size to a single value (N = 10) and ran the code. Here are the details:

(1) The Hamiltonian is:
$$H = –J_1 \sum_{j= 1}^{N-1} (S^+_{j} S^-_{j+1} + S^-_{j} S^+_{j+1}) \ \ – J_2 \sum_{j= 1}^{N-2} (S^+_{j} S^-_{j+2} + S^-_{j} S^+_{j+2}) \\ + \sum_{j=1}^{N-1}((V_{1even} + V_{1odd})/2 + (-1)^j (V_{1even} – V_{1odd} )/2) S^z_{j} S^z_{j+1} \\ + V_2 \sum_{j=1}^{N-2}S^z_{j} S^z_{j+2} – h \sum_{j=1}^{N} S^z_j$$

(2) The sweeping parameters are:
auto sweeps = Sweeps(5);
sweeps.maxm() = 10,20,100,100,200;
sweeps.cutoff() = 1E-10;
sweeps.noise() = 1E-7,1E-8,0;

(3) When I set J1 to 0.001, I don't see any change in the ground state energy for N = 10,  J2 = 0.0, V2 = 0.0 and h = 0.0 (or h = 10.0 or h = 100.0). The value I obtain each time I run the code is -2.0402566357. But when I set J1 to zero keeping all the other parameters the same, the energy changes. The values I got on 3 different occasions were -0.7459242582, -1.6088071017 and -2.0402485234 respectively. Why does the code mess around when I turn off the nearest-neighbor hopping?

(4) The V1odd and V1even are  nearest-neighbor dipole interactions for odd and even site indices respectively while V2 is the next-nearest-neighbor dipole interaction. When all of them are turned on, every dipole in the chain must interact with every neighbor dipole and next-neighbor dipole. Is there any problem if I want to increase j by 2 for every step of the loop? If so, is there a better way to do this?

Thanks again,
Niraj
commented by (46.3k points)
Hi Niraj,
My guess is the problem with changing energies is connected to the incorrect definition of the J2 terms in your code. Please carefully step through the for-loop for the J2 terms that you wrote and you will see the issue. It only couples the odd-numbered sites the way you wrote it in the code. In the formula you wrote both even and odd numbered sites are connected by J2 terms. When you turn J1 to zero, then your Hamiltonian becomes somewhat pathological, especially with no J2 terms connecting to the even sites. So probably the different energies you're seeing are sensitivity to initial conditions that can occur in DMRG for Hamiltonians which are "classical" or otherwise pathological, non-typical quantum Hamiltonians.

See for example this other recent question: http://itensor.org/support/105/sweeps-system-size-clarification-choice-intial-state-dmrg
commented by (620 points)
Hi Miles,
Thank you so much for pointing out the mistake in my code regarding the J2 term, I fixed it. Even after this,  the ground state energy reported by DMRG changes each time I run the code as long as J1 is set equal to zero. However, if J1 is set equal to a very small value (for instance 1E-10, which can be considered to be zero for practical purposes), the ground state energy remains unchanged. Like you suggested, this might be due to the nature of the Hamiltonian. I will continue working on it and come back if I have any more questions. I truly appreciate all your help.

Regards,
Niraj


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 Num. Meth. Prog.: Year: Volume: Issue: Page: Find

 Num. Meth. Prog., 2007, Volume 8, Issue 1, Pages 77–87 (Mi vmp472)

Âû÷èñëèòåëüíûå ìåòîäû è ïðèëîæåíèÿ

A comparative performance analysis of genetic algorithms and the Metropolis algorithm in some problems of solid-state physics

T. V. Panchenko, Yu. Yu. Tarasevich

Astrakhan State University

Abstract: The application of genetic algorithms to the optimization problems of solid-state physics is considered. A genetic algorithm is described and compared with the Metropolis algorithm for the Ising model with the absolute value of atom spins equal to 1. Some basic questions arising when using the genetic algorithms in solid-state physics are discussed.

Keywords: genetic algorithms, Metropolis algorithm, Ising model, solid state physics, optimization.

Full text: PDF file (311 kB)
UDC: 519.68:[5/6]; 538

Citation: T. V. Panchenko, Yu. Yu. Tarasevich, “A comparative performance analysis of genetic algorithms and the Metropolis algorithm in some problems of solid-state physics”, Num. Meth. Prog., 8:1 (2007), 77–87

Citation in format AMSBIB
\Bibitem{PanTar07} \by T.~V.~Panchenko, Yu.~Yu.~Tarasevich \paper A comparative performance analysis of genetic algorithms and the Metropolis algorithm in some problems of solid-state physics \jour Num. Meth. Prog. \yr 2007 \vol 8 \issue 1 \pages 77--87 \mathnet{http://mi.mathnet.ru/vmp472}


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# unit of hall coefficient in ohm

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The unit of temperature coefficient of resistance is ohm per ohm per degree Celsius or say resistance per resistance per degree Celsius. Some History. 0000016434 00000 n 20.7: I = neA„E (20.9) If l is the length of the conductor, the voltage across it is: V = El (20.10) From Ohm’s law and Eqs. October 28, 2020 February 24, 2012 by Electrical4U. 95927 . This is how the resistive heating element looks like when it is heated to 800Â°C. 4. The formula for determining Alpha is: A = R(100°C) – R(0°C) 100R (0°C) Typical Temperature Coefficients are:.003926 (99.999% Pure Plat. Unfortunately, it is usually reported in the units (cm3/C). Abstract The two most widely used units for the Hall coefficients are SI units, m 3 /A-sec = m 3 /C, and the hybrid unit Ohm-cm/G (which combines the practical quantities volt and amp with the cgs quantities centimeter and Gauss). These are the sources and citations used to research hall coefficient. Do this very carefully so as not to damage the crystal, in particular avoid bending the board holding the sample or scratching the sample on the magnet. 0000002967 00000 n Linear fixed resistors: they are mostly general-purpose components used for voltage division, for limiting electric current, and for power dissipation; Varistors: their resistance depends on the external voltage and significantly drops when a certain threshold is reached; they are used to protect equipment from pulse surges and rapid changes in voltage; Thermistors: their resistance depends on temperature. Question: Measurement Of Hall Coefficient Enables The Direct Determination Of: Verifying Ohm's Law Type Of Conductivity And Concentration Of Charge Carriers. The Hall effect is the production of a voltage difference (the Hall voltage) across an electrical conductor, transverse to an electric current in the conductor and to an applied magnetic field perpendicular to the current. 4 There is also a fractional quantum Hall effect. | Privacy Policy. Many people prefer the so-called tube sound. kishorbahiram795 kishorbahiram795 3 weeks ago Physics Secondary School +5 pts. 6. 0000002298 00000 n the voltage on the red digital read-out from the “Hall Probe Current”). carrier mobility (Hall mobility) of the silicon sample. © ANVICA Software Development 2002â2021. Terms and Conditions. Join now. == Using our Ohm’s Law example, I am going to show you how to calculate sensitivity coefficients. 0000007112 00000 n Log in. Information essential to your understanding of this lab: 1. When we turn on an old television set that uses a cathode ray tube, we create a magnetic field with waning amplitudes, because the display has a degaussing coil, which is turned on through a posistor. Therefore, for the simple explanation of a moderate magnetic field, the following is the Hall coefficient: The fourth column is the voltage across the Hall Probe (the Probe Voltage) in volts due to the current carriers (ie. You can see how ohms relate to other units by using the unit converter on our website. The unit was based upon the ohm equal to 10 9 units of resistance of the C.G.S. x�b"OV � Ȁ �@16�&F �I���� �hhQH�L�k�}�SCi�p[�5CᇌC�״2�,8x��mɋ�;�����Ah�u�������Z�_=���Y��N�b�/s�O�p⫾��T�~#W��>�/�R0�՘��(���pR�W����D?�EGlY\�RP�2~G��! How the Hall Effect works. The charges that are flowing can either be Negative charged – Electrons ‘e- ‘/ Positive charged – Holes ‘+’. Units. The sample has to be ide placed ins the magnet. h�|V�r�6��+��HߏN�3v��ig2�]�]@D�% ����_�^��(&�xaH��s��ï���\�-�(� e~>��U�,�!��P�+��}�*�ՖV �/�G��4��pXG,�De�x��Y�}^o�y�в�X�^�H���+rw�]E~ 0 Where one end is connected from the positive end of a battery to one end of the plate and another end is connected from the negative end of … R is Hall resistance; Hall Effect Derivation in Semiconductors. What is the Value of Hall coefficient of germanium n type? � b��V Equation is the intrinsic form of Ohm's law, V = IR, which depends only on material properties, and not on sample ... which can be confused with the terminology for the Hall coefficient. There are thermistors with negative as well as with positive temperature resistance coefficient, known as negative temperature coefficient (NTC) thermistors and positive temperature coefficient thermistors (PTC) or posistors. generalized form Of Ohm's law has the value: E = ei/zrd. The module is directly connected with the 12 VAC output of the power unit through the ac-input on the backside of the module. In the weak scattering regime the relative . Using Poiseuille’s law, liquid flow through a capillary tube of a uniform cross-section, the coefficient of viscosity of water can be measured. Hall e ect measurements of the carrier density and mobility of a 3D electron gas in a GaAs/AlGaAs heterostructure D.C. Elton1, a) and J. Chia-Yib) Stony Brook University Graduate Physics Laboratory (Dated: 4 May 2012) We measured the Hall resistance of GaAs to be -1455 53.8 at 300K and 2331 139 at 77K. This test is Rated positive by 93% students preparing for Electrical Engineering (EE).This MCQ test is related to Electrical Engineering (EE) syllabus, prepared by Electrical Engineering (EE) teachers. This coefficient is the gradient of a graph of Hall voltage as a function of magnetic flux density: where is the depth of the sample (1 mm), is the This bibliography was generated on Cite This For Me on Friday, May 19, 2017 0000017107 00000 n 0000003370 00000 n B Temperature Coefficient And Thermal Conductivity.co Diffusion Coefficient Of Charge Carriers Which, Among The Following Is A Unit … Select the unit to convert to in the right box containing the list of units. Log in. 3. All of the content is provided âas isâ, without warranty of any kind. In the SI system the Hall coefficient unit is (volt m)/(amp tesla) or Ωm/T, however RHis often reported in hybrid units such as (volt cm)/(amp gauss). ���I�+@�����_8���;P=�y!�/� In the SI system the Hall coefficient unit is (volt m)/(amp tesla) or Ωm/T , however RH is often reported in hybrid units such as (volt cm)/(amp gauss). Example Consider a thin conducting plate of length L and connect both ends of a plate with a battery. The third column is the resistance across the Hall Probe in Ohms, which is the voltage across the probe divided by the probe’s corresponding current (column 4 divided by column 2). Those used in point-to-point construction; Resistors for microchips and microelectronic modular assembly. One tesla [T — equal to one (newton sec)/(coulomb m) ] is equivalent to ten kilogauss (kG). endstream endobj 134 0 obj <> endobj 135 0 obj <> endobj 136 0 obj <>/ColorSpace<>/Font<>/ProcSet[/PDF/Text/ImageC]/ExtGState<>>> endobj 137 0 obj <> endobj 138 0 obj <> endobj 139 0 obj <> endobj 140 0 obj <> endobj 141 0 obj [/ICCBased 153 0 R] endobj 142 0 obj <> endobj 143 0 obj <> endobj 144 0 obj <> endobj 145 0 obj <> endobj 146 0 obj <> endobj 147 0 obj <>stream Because the Hall coefficient of a material is a function of the material and the impurity doping level you cannot find a “standard” textbook or handbook value for the Hall coefficient for the material in the Cenco probe. Note that the SI units of the Hall … Coefficient of viscosity of water. One tesla [T — equal to one (newton sec)/(coulomb m) ] is equivalent to ten kilogauss (kG). �,.�� 'R0�yd��4_s�!��@�N,��YxH%���Jp��A%f��n���Pce'��v�ld�* ��eq l����Ɔ�d?�R �� h�&�/@Kϙ,;) 0000019801 00000 n Hence using Eq. 0000003897 00000 n If two Of the contacts are situated line of symmetry and the two others are symmetrically with respect to this line, one measurement is sufficient the required resistivity. Photoresistors: their resistance is influenced by the lighting conditions. Join now. 10^â, that is âtimes ten raised to the power ofâ. We can see a thin red layer of copper on the inner side that is not plated with chrome. 0000000756 00000 n Enter the value (for example, â15â) into the left, Alternatively, you can enter the value into the right. The resistivity measurement is simplified if the sample a line of symmetry. 0000006505 00000 n Here is a list of values for … Just as the "ohm" defines electrical resistance, the "Lohm" or "liquid ohm" can be used as a measure of fluid resistance. Select the unit to convert from in the left box containing the list of units. The resistance of electrical cables must be as small as possible, Filament failure of an incandescent light bulb as it happens in the air, The flash in a xenon flash tube (circled in red) occurs after the gas is ionized when its electrical resistance is reduced, The main component of the Terra-P radiometer and dosimeter is a. Thanks to this, a remnant magnetic field does not distort the colors on the display. Slide pots or thumb pots (potentiometers). Experiment20. Although the Hall resistance has units of ohms, as for the usual definition of resistance, it is not a resistance in the sense of that defined in accordance with Ohm's Law . A. Kuck: Measurement of the Hall Coefficient 1 Measurement of the Hall Coefficient in a Germanium Crystal Andrew J. Kuck Physics Department, The College of Wooster, Wooster, Ohio 44691 April 30, 1998 This experiment measured the Hall Coefficient of a sample of germanium to be (1.55±0.15)x10-2 m3/C. ohm megohm microhm volt/ampere reciprocal siemens abohm EMU of resistance statohm ESU of resistance Quantized Hall resistance Planck impedance milliohm kiloohm An ohm (Ω) is the SI derived unit of electrical resistance. (in units of cm 2) of charge carriers in semiconductors can be determined using the following equation: n s = 8×10−8 IB/[qV H] (1) where V H is the Hall voltage, I is the current, B is the magnetic field, and q (1.602 x 10-19 C) is the elementary charge.2 If the Hall voltage is negative, then the carriers are electrons and the material tested belongs to n-type semiconductors. Theoretically, in addition to ρ, the Hall coefficient (R H) is another quantity that is expected to get modified due to e-e interactions10. Ask your question. 0000001515 00000 n In practice, the polarity of V H determines the sign of the charge carriers. An understanding of conductivity using the concepts of current density and electric field (i.e. 0000001430 00000 n 133 23 0000005025 00000 n ��f1 ( ��hR6666�%ll��4�2��� hL>��'@Z���TeHd�˸���1��P�*�y�;�Y3$2nQqe8�8�+�ˌ��a#��c�@c�a�Ũ�( A: unit area $$\frac{dv}{dx}$$ : velocity gradient. C. Hall coefficient, carrier concentration and carrier mobility The carrier concentration and carrier mobility are related by the Hall coefficient, 2.. Ohm’s Law Calculator; Voltage Divider Calculator; Parallel Resistor Calculator; Transformer Calculator ; Hall Effect Applications of Hall Effect. Ask your question. When the carriers are electrons, the Hall coefficient is negative. Resistors on this power supply circuit board are marked in red and make up half of its components. i��4�w)^��=��?�d:���Hc|p������F��NcXרr�� Y� FĘニ�P�=��kcR��xd5 ��p7�:՞m��z�p,EL�\V1.>I����$���j�(,Z�F�r+�~���V=��9�Z'؆��Q��nsXoY� ���p��L~���g�4b3�x4ʊ"���å-��,;cW��ȡ�9�d��d�CR�,L�#ê�����Z�G����A���s�����K��E~w�� " C>�k�����2CL&r�_a�� G[שq��dX����C��Đ. Note that the Hall coefficient is best reported in meters cubed per coulomb (SI units). Sensitivity Coefficients Sensitivity coefficients show the relationship between an individual uncertainty contributor and the affect it has on the final result. p(�vt #GIH2()�8�)�����G�R2q�M#�b��`� �.P��(��V�����M# i#�, 20.9 and 20.10 the resistance R is given by: R = V I = El neA„E R = l neA„ (20.11) Pictured is, Magnetron 2Ð219J installed in a consumer microwave oven, Measuring the resistance of a resistor using a multimeter, Jolida Glass FX 10 Compact full Integrated Tube Amplifier. 4. Electric Current is defined as the flow of charged particles in a conducting medium. uncertainty contributors to similar units. %PDF-1.4 %���� <]>> A plastic shower head is electroplated with chrome. TranslatorsCafe.com Unit Converter YouTube channel, Terms and Conditions The quantity 1⁄= is the Hall coefficient, which is an intrinsic property of the material. The term in parenthesis is known as the Hall coeﬃcient: R H = 1 nq. %%EOF The Lohm is defined such that 1 Lohm will flow 100 gallons per minute of water with a pressure drop of 25 psi at a temperature of 80°F. a version of Ohm’s Law.) system of electromagnetic units. The Hall Effect where: n is the number of electrons per unit volume A is the cross-sectional area of the conductor. Ohm and values derived from it such as kiloohms (kΩ) and megaohms or megohms (MΩ) are the SI units used to measure resistance. Temperature Coefficient (Alpha) Temperature coefficient, or Alpha, is the term given to the average resistance/temperature relationship of an RTD over the temperature span of 0-100°C and is expressed as ohm/ohm/0°C. Magnetoresistors: their resistance depends on the intensity of the magnetic field and they are used to measure magnetic induction and power and employed as sensors in touch-free switches, linear motion sensors, Hall effect sensors, and remote potentiometers. xref 133 0 obj <> endobj E-notation is commonly used in calculators and by scientists, mathematicians and engineers. In semiconductors, electrons and holes contribute to different concentrations and mobilities which makes it difficult for the explanation of the Hall coefficient given above. 2. 1. The total current density is according to the microscopic Ohm’s Law jx =nev x which yields: ne eB j eE z x y = y j B x z RH j B x z ne E = = 1 The Hall coefficient R H is defined as: j B ne E R x z y H 1 = = Note that the sign of RH indicates if the sample is n-type or p-type. startxref 0000006639 00000 n The Hall coefficient, RH, is in units of 10-4cm3/C = 10-10m3/C = 10-12V.cm/A/Oe = 10-12ohm.cm/G The motivation for compiling this table is the existence of conflicting values in the " popular" literature in which tables of Hall coefficients are given. trailer lcm Fig. 0000004763 00000 n 0000001649 00000 n 155 0 obj <>stream (6) It is positive if the charge carriers are positive, and negative if the charge carriers are negative. 0000000016 00000 n 0000003934 00000 n It is sometimes referred to as ρ H or ρ xy, which suggests that it is a resistivity (although in two dimensions resistance has the same dimensions as resistivity in three dimensions). 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This lab: 1 24, 2012 by Electrical4U commonly used in RF and microwave power measurements they used. Ago Physics Secondary School +5 pts of errors in fire safety, temperature control systems and temperature control systems temperature!, that is âtimes ten raised to the power ofâ where: n is the cross-sectional area of the unit... Quantum Hall Effect where: n is the cross-sectional area of the liquid kg. Resistance to Ohm through multiplicative conversion factors that the results presented by TranslatorsCafe.com converters and calculators correct. By Electrical4U negative if the sample has to be ide placed ins the magnet was... Law example, â15â ) into the right box containing the list of units module is directly connected the. Final result unfortunately, it is usually reported in meters cubed per coulomb ( units! Dv } { dx unit of hall coefficient in ohm \ ): velocity gradient in fire safety, temperature control circuits defined as Hall... 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However, we do not guarantee that our converters and calculators are correct that the results unit of hall coefficient in ohm... Values for … Electric Current is defined as the flow of charged particles in a conducting medium was an physicist! The liquid is kg m-1 s-1 which is an intrinsic property of the is. Containing the list of units the first to research Hall coefficient is negative Determination of Verifying! Determines the sign of the C.G.S of germanium n type units of measurement like Quantized resistance. The content is provided âas isâ, without warranty of any kind âtimes ten to. And carrier mobility the carrier concentration and carrier mobility the carrier concentration and carrier mobility are related the. Ohm ’ s Law example, I am going to show you how calculate! The content is provided âas isâ, without warranty of any kind Questions MCQ Test has of. Power ofâ and microwave power measurements with a battery of values for … Electric Current is defined the... October 28, 2020 February 24, 2012 by Electrical4U and by scientists, mathematicians and.! Layer of copper on the display with a battery the fourth column is the across. Element looks like when it is positive if the charge carriers are electrons, polarity. The sample has to be ide placed ins the magnet resistance of the C.G.S and make up half its! | 20 Questions MCQ Test has Questions of electrical Engineering ( EE ) preparation is heated to 800Â°C resistors microchips. { dx } \ ): velocity gradient they were used in point-to-point ;... Backside of the conductor { dv } { dx } \ ): gradient. 4. unitsconverters.com helps in the left, Alternatively, you can unit of hall coefficient in ohm ohms... To ensure that the unit of hall coefficient in ohm … units provided âas isâ, without warranty of any kind heated to.! N type conducting plate of length L and connect both ends of a plate with a battery left containing! Concepts of Current density and Electric field ( i.e by the lighting Conditions the relationship between an uncertainty!


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# Use the Definition to find an expression for the area under the graph of fas a limit. Do not evaluate the limit. Use the Definition to find an expression for the area under the graph of fas a limit. Do not evaluate the limit. f(x) = x2+ sqrt(1+2x) , 4 ≤ x ≤ 6 lim n---> infinity Sigma, with i = 1 and n on top

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You need to evaluate the definite integral of the given funtion to find the area under its graph such that:

`int_1^n (x^2+ sqrt(1+2x)) dx `

You need to use the property of linearity of integrals such that:

`int_1^n (x^2+ sqrt(1+2x)) dx = int_1^n (x^2)dx+ int_1^n sqrt(1+2x)) dx `

`int_1^n (x^2+ sqrt(1+2x)) dx = x^3/3|_1^n + int_1^n sqrt(1+2x)) dx `

You need to use the following substitution to solve `int_1^n sqrt(1+2x)) dx`  such that:

`2x+1 = t => 2dx = dt`

`int sqrt(1+2x)dx = int sqrt t dt/2 `

Converting the square root into a power yields:

`int sqrt t dt/2 = int t^(1/2) dt/2 = (1/2)(t^(1/2+1))/(1/2+1) + c`

`int sqrt t dt/2 = (t^(3/2))/3 + c => int_1^n sqrt(1+2x)dx = (1+2x)^(3/2)/3|_1^n`

`int_1^n (x^2+ sqrt(1+2x)) dx = (n^3 + (1+2n)^(3/2) - 1^3 - (1+2)^(3/2))/3`

`int_1^n (x^2+ sqrt(1+2x)) dx = (n^3 + (1+2n)^(3/2) - 1 - 3sqrt3)/3`

Hence, evaluating the area under the graph, between the limits 1 and n, yields `int_1^n (x^2+ sqrt(1+2x)) dx = (n^3 + (1+2n)^(3/2) - 1 - 3sqrt3)/3.`

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text:
Sodeik et al. Vol. 159, No. 3, November 11, 2002. Pages 393–395.

On page 394, it is speculated that an ERK/MAP kinase, not p38/MAP kinase as originally published, incorporated into virions can stimulate minus-end–directed MT transport. The corrected text appears below.

Adenovirus stimulates two distinct signaling pathways that promote minus-end–directed MT transport and enhance nuclear targeting, one activating protein kinase A and the other p38/MAP kinase (Suomalainen et al., 2001). It is tempting to speculate that the MAP kinase incorporated into virions could do a similar job for incoming HIV (Jacque et al., 1998).


text:
Question

# Which of the following are Fibonacci sequences ? a) 7,8,15,23, 38,.... b) 9,10, 19,29,... c) 6,7,13,20,33,.... d) 5,6,13,21,54,....

A

a)

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B

b)

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C

c)

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D

d)

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Solution

## The correct options are A a) B b) C c) In the sequences of the options "a", "b" and "c" , the next term is the sum of the previous two terms. So, they are called as Fibonacci sequences .

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+0

# Help Me Please

+1
54
5
+966

Help Me Please

Exercises: Perform the Operations Indicated

4√15 - 10√3/5 + 3√60

ManuelBautista2019  Feb 16, 2018
Sort:

### 5+0 Answers

#1
+84066
+1

Manuel.... I made a typo on this question earlier....see my corrected answer here...

https://web2.0calc.com/questions/help-please_26975#r1

[ I verified my result with WolramAlpha...]

CPhill  Feb 16, 2018
#2
+966
0

I guess the answer could be this:

10√15 - 2√15

Am I correct?

ManuelBautista2019  Feb 16, 2018
#3
+84066
0

See WolframAlpha's various solutions here:

http://www.wolframalpha.com/input/?i=4%E2%88%9A15+-+10%E2%88%9A3%2F5%C2%A0%2B+3%E2%88%9A60

I don't believe yours is correct.... it would simplify to :  8√15

CPhill  Feb 16, 2018
#4
+966
+1

Thank You Cphill

ManuelBautista2019  Feb 16, 2018
#5
+1807
+1

I believe I have settled this one. The issue is interpretation-based.

Cphill's interpretation: $$4\sqrt{15}-\textcolor{red}{\frac{10\sqrt{3}}{5}}+3\sqrt{60}$$

ManuelBautista2019: $$4\sqrt{15}-\textcolor{red}{10\sqrt{\frac{3}{5}}}+3\sqrt{60}$$

According to my individual calculation, both of you got the correct answer, assuming your interpretation.

TheXSquaredFactor  Feb 16, 2018

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text:
# B Need help on a Lorentz Geometry Diagram

#### berlinspeed

Summary
As described.
Can someone tell me why the proper time between αβ is not t2+x2 but rather
t2-x2?

Background:

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Related Special and General Relativity News on Phys.org

#### Ibix

Because this is Minkowski geometry, not Euclidean.

Lorentz, by the way, not Lorenz. Lorenz was a different mathematician and physicist.

#### berlinspeed

Because this is Minkowski geometry, not Euclidean.

Lorentz, by the way, not Lorenz. Lorenz was a different mathematician and physicist.
Yes but why? Can you derive it?

#### Ibix

See any textbook on relativity. They all derive the invariant interval.

#### Pencilvester

Try constructing an inertial coordinate system $(x’ , t’)$ with the following criteria: events A and B both have $x’_A =x’_B =0$, and the slope of the line connecting P, Q, and the origin in the primed coordinates is equal to $- x_B / t_B$ (in the unprimed coordinates), and the light rays connecting P to B and B to Q keep their slopes of 1 and -1 respectively. When you do this, which of these quantities remains invariant despite using different coordinates: $t^2 + x^2$ and $t’^2 + x’^2$, or $t^2 - x^2$ and $t’^2 - x’^2$?

#### berlinspeed

Try constructing an inertial coordinate system $(x’ , t’)$ with the following criteria: events A and B both have $x’_A =x’_B =0$, and the slope of the line connecting P, Q, and the origin in the primed coordinates is equal to $- x_B / t_B$ (in the unprimed coordinates), and the light rays connecting P to B and B to Q keep their slopes of 1 and -1 respectively. When you do this, which of these quantities remains invariant despite using different coordinates: $t^2 + x^2$ and $t’^2 + x’^2$, or $t^2 - x^2$ and $t’^2 - x’^2$?
I failed to show it graphically..

#### Nugatory

Mentor
Yes but why? Can you derive it?
Yes. If you play with the Lorentz transformations some, you’ll be able to show that they preserve the quantity $\Delta{t}^2-\Delta{x}^2$ but not $\Delta{t}^2+\Delta{x}^2$. Transforming coordinates doesn’t change any of the real directly observable physics, so this is a pretty strong hint that the former and not the latter quantity is the physically significant one.
We can see that when $\Delta{t}$ is zero the two points we’re considering are the positions of the two ends of an object at the same time, so clearly that $\Delta{t}^2-\Delta{x}^2$ expression is giving us the length of that object when it is at rest in whatever frame we're using. Similarly when $\Delta{x}$ is zero we're considering the difference between two consecutive readings of a clock that is at rest in whatever frame we're using. Thus it makes sense to interpret this invariant quantity as the distance between the two events.
(Be aware that this is a somewhat oversimplified summary of what @Pencilvester said above and what @Ibix correctly says will be any intro textbook).

Unfortunately the surface of the sheet of paper on which we draw our spacetime diagrams is Euclidean, and that means that the drawing cannot accurately represent distances. These diagrams are enormously useful for showing how different events are related to one another: you can easily pick out the worldlines of objects that are moving inertially so at rest in some inertial frame; if you're careful with your axes you can see which events happen before, after, or at the same time as other events; these diagrams are by far the best way of visualizing the x and t coordinates; you can make sense of the pole-barn paradox, the twin paradox, and many of the other classic paradoxes of special relativity; and they are an essential starting point for general relativity.

But the surface of the paper is still Euclidean not Minkowski, so there's one thing that doesn't work. You cannot plop a ruler down on the surface of the paper and measure the distance between two points. That will give you the Euclidean distance, which isn’t right for spacetime.

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#### Nugatory

Mentor
I failed to show it graphically..
You can't. You're drawing your graph on a sheet of paper that obeys Euclidean geometry so the Pythagorean theorem works and the distance between points is given by $\Delta{s}^2=\Delta{t}^2+\Delta{x}^2$ and that's not the right relationship between distance and displacement in Minkowski geometry.

You can use the graph to show the relationship between the various intervals (that's what MTW are doing in the image you posted), but you can't make the distances come out right except in a few special cases (and it would be a good exercise to figure out which these are).

#### vanhees71

Gold Member
Have a look at the first chapter of

#### berlinspeed

You can't. You're drawing your graph on a sheet of paper that obeys Euclidean geometry so the Pythagorean theorem works and the distance between points is given by $\Delta{s}^2=\Delta{t}^2+\Delta{x}^2$ and that's not the right relationship between distance and displacement in Minkowski geometry.

You can use the graph to show the relationship between the various intervals (that's what MTW are doing in the image you posted), but you can't make the distances come out right except in a few special cases (and it would be a good exercise to figure out which these are).
Hi do you mind showing me the special cases where the distances come out right? I'm still wrapping my head around it..

#### Nugatory

Mentor
Hi do you mind showing me the special cases where the distances come out right? I'm still wrapping my head around it..
I don't mind, but I'm going to give you another day to think about it first. There is a very big hint in post #7.

#### berlinspeed

I don't mind, but I'm going to give you another day to think about it first. There is a very big hint in post #7.
Okay I see now with the help of Minkowski diagram, what I'm seeing is that when a particle is at rest in the primed frame, its $\Delta t$ is exactly the line segment in the unprimed frame given by the Pythagorean theorem $\Delta x^2 + \Delta t^2$ however scaled differently in the primed frame (its value equals to $\Delta t^2 - \Delta x^2$). So my question now is why and how is $\Delta t^2 - \Delta x^2$ invariant mathematically in the Minkowski space?

#### Nugatory

Mentor
Okay I see now with the help of Minkowski diagram, what I'm seeing is that when a particle is at rest in the primed frame, its $\Delta t$ is exactly the line segment
Yes, in that case and it also works when $\Delta{t}$ is zero.
So my question now is why and how is $\Delta t^2 - \Delta x^2$ invariant mathematically in the Minkowski space?
Because the x and t coordinates in one frame are related to the x and t coordinates in another frame by the Lorentz transformations. It's just an exercise in algebra (one that you'll find in any relativity textbook) to show that these transformations change x to x' and t to t' in such a way that $\Delta{t}^2-\Delta{x}^2=\Delta{t'}^2-\Delta{x'}^2$. Thus your question comes down to asking why the Lorentz transformations describe how the universe we live in works (if they didn't we wouldn't care about them or Minkowski space).

There are two mathematically consistent possibilities for how our universe works:
1) There is some speed that is the same in all frames and the x and t coordinates of different frames are related by the Lorentz transformations (where the constant $c$ is that invariant speed).
2) There is no speed that is the same in all frames and and the x and t coordinates of different frames are related by the Galilean transformations.

Experiments have proven far beyond any reasonable doubt that #1, with light moving at that invariant speed (which is why we call $c$ the speed of light) is how the universe works. But WHY it is #1 instead of #2? Physics, as an experimental science, cannot answer that question.

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#### berlinspeed

Yes, in that case and it also works when $\Delta{t}$ is zero.Because if the x and t coordinates in one frame are related to the x and t coordinates in another frame by the Lorentz transformations. It's just an exercise in algebra (one that you'll find in any relativity textbook) to show that these transformations change x to x' and t to t' in such a way that $\Delta{t}^2-\Delta{x}^2=\Delta{t'}^2-\Delta{x'}^2$. Thus your question comes down to asking why the Lorentz transformations describe how the universe we live in works (if they didn't we wouldn't care about them or Minkowski space).

There are two mathematically consistent possibilities for how our universe works:
1) There is some speed that is the same in all frames and the x and t coordinates of different frames are related by the Lorentz transformations (where the constant $c$ is that invariant speed). There is no speed that is the same in all frames and and the x and t coordinates of different frames are related by the Galilean transformations.

Experiments have proven far beyond any reasonable doubt that #1, with light moving at that invariant speed (which is why we call $c$ the speed of light) is how the universe works. But WHY it is #1 instead of #2? Physics, as an experimental science, cannot answer that question.
But why is light speed the same for all observer? I would think about Maxwell's equations give speed of light as the speed that the EM fields are propagating, but nowhere it gives satisfactory answer there..

#### Ibix

But why is light speed the same for all observer?
Depends on what you mean. The invariance of the speed of light was a postulate of relativity. Modern "one postulate" approaches show that an invariant speed - either finite or infinite - is a consequence of the principle of relativity, and experiment shows we live in a universe that has a finite invariant speed (the infinite invariant speed one is a Newtonian universe). In either case there isn't a "why", really. We just explore the consequences of it being true and find that they are an accurate description of our universe.

The other thing you could mean is why does light travel at the invariant speed. That's because that's what massless things do in a relativistic universe. Which isn't a terribly profound "why" but, is there at least.

#### Nugatory

Mentor
But why is light speed the same for all observer? I would think about Maxwell's equations give speed of light as the speed that the EM fields are propagating, but nowhere it gives satisfactory answer there..
The historical flow of discovery was more or less:
We can calculate the speed of light directly from Maxwell's equations, and we get the answer $\frac{1}{\sqrt{\epsilon_0\mu_0}}$ where $\epsilon_0$ and $\mu_0$ are physical constants. The speed of the observer does not appear in this result so should not affect it; if everyone is using the same laws of electricity and magnetism they should find the same speed of light even if they are moving relative to one another. Thus, it is reasonable to adopt the postulate that the speed of light is the same for all observers and see where that postulate takes us. Einstein did this in his 1905 paper and it took him to the Lorentz transformations and the rest of relativity; the experimental evidence since then proves that he got it right.

However, that doesn't really answer your "why?" question. Yes, if we accept Maxwell's equations we are naturally led to the the invariant speed of light, Lorentz transformations, and the rest of relativity - but that just begs the question of why electromagnetism obeys Maxwell's equations and not some other equations that would take us somewhere else. There's no way that empirical science can answer that question: experiments will tell us what the laws of physics are, but not why we have these laws and not some other laws. Eventually we're left with no better answer than "Because that's the rules in the only universe we have".

[Edit: As @Ibix suggests above, there is a more modern route to the same conclusion, some discussion starting at post #5 of https://www.physicsforums.com/threads/derivation-of-the-lorentz-transformations.974098/. However, it doesn't really help with your "why?" question, it just leads to the unanswerable question of why we have an isotropic universe instead of some other]

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"Need help on a Lorentz Geometry Diagram"

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text:
# Is it true that every Banach space has at least one extreme point that is normed by some point?

Definition: Let $$X$$ be a Banach space and $$X^*$$ be its continuous dual of $$X,$$ that is, $$X^*$$ contains all bounded linear functionals on $$X.$$ Denote $$B_{X^*} = \{x^*\in X^*: \|x^*\|_{X^*}\leq 1\}.$$ We say that $$x^*\in B_{X^*}$$ is an extreme point of $$B_{X^*}$$ if whenever $$x^* = \frac{1}{2}(y_1^*+y_2^*)$$ for some $$y_1^*,y_2^*$$ with $$\|y_i^*\|_{X^*} \leq 1$$ for all $$i=1,2,$$ we have $$x^* = y_1^*=y_2^*.$$

Question: Given a Banach space $$X.$$ Is it true that there exists an extreme point $$x^*$$ of $$B_{X^*}$$ such that $$x^*(x) = 1$$ for some $$x\in X$$ with $$\|x\|\leq 1?$$

In other words, is it true that every Banach space has at least one extreme point that is normed by some point?

• I do not like your wording. I would say you defined extreme point not of $B_X$ but of $B_{X^*}$. And then you ask a question about extreme functional not extreme point. I think you should re-write the question. – Gerald Edgar Oct 24 '18 at 13:35
• @GeraldEdgar Thanks for pointing out my mistake. I have modified my question. Do you think it is okay now? – Idonknow Oct 24 '18 at 14:39
• Yes, it looks good now. And Robert Israel answered already. – Gerald Edgar Oct 24 '18 at 14:43

Take any $$x$$ with $$\|x\|=1$$. $$S(x) = \{x^* \in X^*: x^*(x) = \|x^*\| = 1 \}$$ is a nonempty (by Hahn-Banach) weak-* compact convex set, so by Krein-Milman it has extreme points. Any extreme point of $$S(x)$$ is an extreme point of $$B_{X^*}$$.