6 added a disclaimer of Gregor's objection to my answer edited Jun 23 '18 at 18:40 Jason Arthur Taylor 34122 silver badges99 bronze badges MyGregor Michalicek has argued (see comments below) that one should overlook this issue, and answer a different version of the question in which the simple clause was removed. I disagree. I think there are many examples in history of science when bad science was done because people glossed over issues or ignored implied or hidden assumptions in their research. At any rate, my answer of "no" being the best answer is probably impossible to prove. But, to attempt to make my case, what I am going to do is see if the other answer here is both correct and simple. If I am right, it will either be wrong or not simple. Then NoEigenvalue's answer claims, "First you must know that only electrons with an energy close to the Fermi energy can participate to the conduction process...At $$T>0$$...you are [only] now able to excite electrons in higher energy states, which you need to do if you want to accelerate them in a direction; i.e., if you want to do work [i.e, produce current] on them." (emphasis supplied) This argument is not explained. It is actually backwards as well. I debunked this misconception here. It is obviously false when you consider either of these metals at low temperatures, where according to the argument there would be no electrons to carry current, and conduction would be zero. In reality this is where conduction is infinite/maximizedmaximized and resistivity (the inverse) is zeroapproaches a minimum:$T=0K$"> At the end of the day, I think nature isn’t always so trivial as the fictional version people sometimes come up with due to their own arrogance in wanting to think they personally know what's going on and, apparently, the version of reality you seek via your question. Humility is an important partner of curiosity and good research, and one should not leave it behind in a selfish desire to trick oneself into moving forward. My answer of "no" being the best answer is probably impossible to prove. But, to attempt to make my case, what I am going to do is see if the other answer here is both correct and simple. If I am right, it will either be wrong or not simple. Then NoEigenvalue's answer claims, "First you must know that only electrons with an energy close to the Fermi energy can participate to the conduction process...At $$T>0$$...you are [only] now able to excite electrons in higher energy states, which you need to do if you want to accelerate them in a direction; i.e., if you want to do work [i.e, produce current] on them." (emphasis supplied) This argument is not explained. It is actually backwards as well. I debunked this misconception here. It is obviously false when you consider either of these metals at low temperatures, where according to the argument there would be no electrons to carry current, and conduction would be zero. In reality this is where conduction is infinite/maximized and resistivity (the inverse) is zero:$T=0K$"> At the end of the day, nature isn’t always so trivial as the fictional version people sometimes come up with due to their own arrogance in wanting to think they personally know what's going on and, apparently, the version of reality you seek via your question. Gregor Michalicek has argued (see comments below) that one should overlook this issue, and answer a different version of the question in which the simple clause was removed. I disagree. I think there are many examples in history of science when bad science was done because people glossed over issues or ignored implied or hidden assumptions in their research. At any rate, my answer of "no" being the best answer is probably impossible to prove. But, to attempt to make my case, what I am going to do is see if the other answer here is both correct and simple. If I am right, it will either be wrong or not simple. Then NoEigenvalue's answer claims, "First you must know that only electrons with an energy close to the Fermi energy can participate to the conduction process...At $$T>0$$...you are [only] now able to excite electrons in higher energy states, which you need to do if you want to accelerate them in a direction; i.e., if you want to do work [i.e, produce current] on them." (emphasis supplied) This argument is not explained. It is actually backwards as well. I debunked this misconception here. It is obviously false when you consider either of these metals at low temperatures, where according to the argument there would be no electrons to carry current, and conduction would be zero. In reality this is where conduction is maximized and resistivity (the inverse) approaches a minimum:$T=0K$"> At the end of the day, I think nature isn’t always so trivial as the fictional version people sometimes come up with due to their own arrogance in wanting to think they personally know what's going on and, apparently, the version of reality you seek via your question. Humility is an important partner of curiosity and good research, and one should not leave it behind in a selfish desire to trick oneself into moving forward. 5 added graph and structure to my arugment as well as a summary sentence, fixed some typos edited Jun 23 '18 at 13:08 Jason Arthur Taylor 34122 silver badges99 bronze badges Next we see $$\sigma=ne^2l/mv_F$$ and an argument about how if $$v_F$$ is lower in copper than in aluminum, then the current will be higher. Not stated is that this relies upon $$n$$ and $$l$$ being the same in both metals, something that is unlikely. Indeed, this was my point made above, that a derivation of $$E_F(n)$$ was missing. It seems rather convenient given that it changes the later argument; aluminum is triply ionic instead of singly ionic, so $$n$$ gets a factor of three from that, and it ends up having a free electron density that is a factor of $$21.1\times 10^{22} {\rm cm}^{-3}/8.49\times 10^{22} {\rm cm}^{-3}\simeq 2.5$$ that of copper. This relatively large factor overtakes the energy/velocity one created by NoEigenvalue in an apparent attempt to make the final argument to answer a question in a simple way because that factor of a mere $$11/7\simeq 1.6$$ is further softened to 1.3 due to the square root function in converting from energies to velocities. So, at face value, if you include the variance of $$n$$ in NoEigenvalue's argument it actually would say aluminum has a greater conductivity. This clearly is a problem with the answer given that the question is why does copper have a greater conductivity. Next we see $$\sigma=ne^2l/mv_F$$ and an argument about how if $$v_F$$ is lower in copper than in aluminum, then the current will be higher. Not stated is that this relies upon $$n$$ and $$l$$ being the same in both metals, something that is unlikely. Indeed, this was my point made above, that a derivation of $$E_F(n)$$ was missing. It seems rather convenient given that it changes the later argument; aluminum is triply ionic instead of singly ionic, so $$n$$ gets a factor of three from that, and it ends up having a free electron density that is a factor of $$21.1\times 10^{22} {\rm cm}^{-3}/8.49\times 10^{22} {\rm cm}^{-3}\simeq 2.5$$ that of copper. This relatively large factor overtakes the energy/velocity one created by NoEigenvalue in an apparent attempt to make the final argument to answer a question in a simple way because that factor of a mere $$11/7\simeq 1.6$$ is further softened to 1.3 due to the square root function in converting from energies to velocities. Next we see $$\sigma=ne^2l/mv_F$$ and an argument about how if $$v_F$$ is lower in copper than in aluminum, then the current will be higher. Not stated is that this relies upon $$n$$ and $$l$$ being the same in both metals, something that is unlikely. Indeed, this was my point made above, that a derivation of $$E_F(n)$$ was missing. It seems rather convenient given that it changes the later argument; aluminum is triply ionic instead of singly ionic, so $$n$$ gets a factor of three from that, and it ends up having a free electron density that is a factor of $$21.1\times 10^{22} {\rm cm}^{-3}/8.49\times 10^{22} {\rm cm}^{-3}\simeq 2.5$$ that of copper. This relatively large factor overtakes the energy/velocity one created by NoEigenvalue in an apparent attempt to make the final argument to answer a question in a simple way because that factor of a mere $$11/7\simeq 1.6$$ is further softened to 1.3 due to the square root function in converting from energies to velocities. So, at face value, if you include the variance of $$n$$ in NoEigenvalue's argument it actually would say aluminum has a greater conductivity. This clearly is a problem with the answer given that the question is why does copper have a greater conductivity. 4 added graph and structure to my arugment as well as a summary sentence, fixed some typos edited Jun 23 '18 at 13:02 Jason Arthur Taylor 34122 silver badges99 bronze badges My answer of no"no" being the best answer is probably impossible to prove. But, to attempt to make my case, what I am going to do is see if the other answer here is both correct and simple. If I am right, it will either be wrong or not simple. As a preliminary remark, the other answer here by NoEigenvalue here currently has a green checkbox, 5 upvotes, an edit by a reviewer, and a favorable comment. So, at first glance, one would think my answer is wrong, since if I am right it suggests that not just one person here but the community as well is wrong also. The answer starts out attempting to explain what the Fermi energy is using the Bohr model of an atom. However, it then skips to saying what Fermi energies are for some metals. It does withthis by referencing external tables. It entirely skips the derivation based on free electron densities, and makes no attempt to connect them. I will show that this poses a problem later, because thean unstated dependence affects NoEigenvalue's argument. Then NoEigenvalue's answer claims, "First you must know that only electrons with an energy close to the Fermi energy can participate to the conduction process...At $$T>0$$...you are [only] now able to excite electrons in higher energy states, which you need to do if you want to accelerate them in a direction; i.e., if you want to do work [i.e, produce current] on them." (emphasis supplied) This argument is not explained. It is actually backwards as well. I debunked this misconception here. It is obviously false when you consider either of these metals at low temperatures, where according to the argument there'sthere would be no electrons to carry current, and conduction would be zero. In reality this is where conduction is infinite/maximized and resistivity (the inverse) is zero: Next the author claims that $$\sigma$$ is defined by $$ne^2τ/m$$$$ne^2\tau/m$$ where $$n$$ is the number of electrons [per unit volume]... Wrong. The definition of conductivity is the ratio of current to field, or the ease of current flow when pushed. The cited expression largely just gives an illusion of confidence, introducing a free parameter, τ$$\tau$$, with one new equation, which doesn't necessarily give any new physical information. It thus really defines $$τ(\sigma)$$$$\tau(\sigma)$$, which can be argued to be a momentum exchange time scale of some electron cloud with an ionic lattice. Next we see $$σ=ne^2l/mv_F$$$$\sigma=ne^2l/mv_F$$ and an argument about how if $$v_F$$ is lower in copper than in aluminum, then the current will be higher. Not stated is that this relies upon $$n$$ and $$l$$ being the same in both metals, something that is unlikely. Indeed, this was my point made above, that a derivation of $$E_F(n)$$ was missing. It seems rather convenient given that it changes the later argument; aluminum is triply ionic instead of singly ionic, so $$n$$ gets a factor of three from that, and it ends up having a free electron density that is a factor of $$21.1\times 10^{22} {\rm cm}^{-3}/8.49\times 10^{22} {\rm cm}^{-3}\simeq 2.5$$ that of copper. This relatively large factor overtakes the energy/velocity one created by NoEigenvalue in an apparent attempt to make the final argument to answer a question in a simple way because that factor of a mere $$11/7\simeq 1.6$$ is further softened to 1.3 due to the square root function in converting from energies to velocities. To sum up, I think I've shown that the allegedly simple answer is both both too simple and also incorrect in many aspects. This is not a proof that the best answer is no, but it is probably a strong case that no"no" is a reasonable answer if not the best answer. My answer of no being the best answer is probably impossible to prove. But, to attempt to make my case, what I am going to do is see if the other answer here is both correct and simple. If I am right, it will either be wrong or not simple. As a preliminary remark, the other answer here by NoEigenvalue here currently has a green checkbox, 5 upvotes, an edit by a reviewer, and a favorable comment. So, at first glance, one would think my answer is wrong, since if I am right it suggests that not just one person here but the community as well is wrong also. The answer starts out attempting to explain what the Fermi energy is using the Bohr model of an atom. However, it then skips to saying what Fermi energies are for some metals. It does with by referencing external tables. It entirely skips the derivation based on free electron densities, and makes no attempt to connect them. I will show that this poses a problem later, because the unstated dependence affects NoEigenvalue's argument. Then NoEigenvalue's answer claims, "First you must know that only electrons with an energy close to the Fermi energy can participate to the conduction process...At $$T>0$$...you are [only] now able to excite electrons in higher energy states, which you need to do if you want to accelerate them in a direction; i.e., if you want to do work [i.e, produce current] on them." (emphasis supplied) This argument is not explained. It is actually backwards as well. I debunked this misconception here. It is obviously false when you consider either of these metals at low temperatures, where according to the argument there's be no electrons to carry current, conduction would be zero. In reality this is where conduction is infinite/maximized and resistivity (the inverse) is zero: Next the author claims that $$\sigma$$ is defined by $$ne^2τ/m$$ where $$n$$ is the number of electrons [per unit volume]... Wrong. The definition of conductivity is the ratio of current to field, or the ease of current flow when pushed. The cited expression largely just gives an illusion of confidence, introducing a free parameter, τ, with one new equation, which doesn't necessarily give any new physical information. It thus really defines $$τ(\sigma)$$, which can be argued to be a momentum exchange time scale of some electron cloud with an ionic lattice. Next we see $$σ=ne^2l/mv_F$$ and an argument about how if $$v_F$$ is lower in copper than in aluminum, then the current will be higher. Not stated is that this relies upon $$n$$ and $$l$$ being the same in both metals, something that is unlikely. Indeed, aluminum is triply ionic instead of singly ionic, so $$n$$ gets a factor of three from that, and it ends up having a free electron density that is a factor of $$21.1\times 10^{22} {\rm cm}^{-3}/8.49\times 10^{22} {\rm cm}^{-3}\simeq 2.5$$ that of copper. This relatively large factor overtakes the energy/velocity one created by NoEigenvalue in an apparent attempt to make the final argument to answer a question in a simple way because that factor of a mere $$11/7\simeq 1.6$$ is further softened to 1.3 due to the square root function in converting from energies to velocities. To sum up, I think I've shown that the allegedly simple answer is both both too simple and also incorrect in many aspects. This is not a proof that the best answer is no, but it is probably a strong case that no is a reasonable answer if not the best answer. My answer of "no" being the best answer is probably impossible to prove. But, to attempt to make my case, what I am going to do is see if the other answer here is both correct and simple. If I am right, it will either be wrong or not simple. As a preliminary remark, the other answer here by NoEigenvalue currently has a green checkbox, 5 upvotes, an edit by a reviewer, and a favorable comment. So, at first glance, one would think my answer is wrong, since if I am right it suggests that not just one person here but the community as well is wrong also. The answer starts out attempting to explain what the Fermi energy is using the Bohr model of an atom. However, it then skips to saying what Fermi energies are for some metals. It does this by referencing external tables. It entirely skips the derivation based on free electron densities, and makes no attempt to connect them. I will show that this poses a problem later, because an unstated dependence affects NoEigenvalue's argument. Then NoEigenvalue's answer claims, "First you must know that only electrons with an energy close to the Fermi energy can participate to the conduction process...At $$T>0$$...you are [only] now able to excite electrons in higher energy states, which you need to do if you want to accelerate them in a direction; i.e., if you want to do work [i.e, produce current] on them." (emphasis supplied) This argument is not explained. It is actually backwards as well. I debunked this misconception here. It is obviously false when you consider either of these metals at low temperatures, where according to the argument there would be no electrons to carry current, and conduction would be zero. In reality this is where conduction is infinite/maximized and resistivity (the inverse) is zero: Next the author claims that $$\sigma$$ is defined by $$ne^2\tau/m$$ where $$n$$ is the number of electrons [per unit volume]... Wrong. The definition of conductivity is the ratio of current to field, or the ease of current flow when pushed. The cited expression largely just gives an illusion of confidence, introducing a free parameter, $$\tau$$, with one new equation, which doesn't necessarily give any new physical information. It thus really defines $$\tau(\sigma)$$, which can be argued to be a momentum exchange time scale of some electron cloud with an ionic lattice. Next we see $$\sigma=ne^2l/mv_F$$ and an argument about how if $$v_F$$ is lower in copper than in aluminum, then the current will be higher. Not stated is that this relies upon $$n$$ and $$l$$ being the same in both metals, something that is unlikely. Indeed, this was my point made above, that a derivation of $$E_F(n)$$ was missing. It seems rather convenient given that it changes the later argument; aluminum is triply ionic instead of singly ionic, so $$n$$ gets a factor of three from that, and it ends up having a free electron density that is a factor of $$21.1\times 10^{22} {\rm cm}^{-3}/8.49\times 10^{22} {\rm cm}^{-3}\simeq 2.5$$ that of copper. This relatively large factor overtakes the energy/velocity one created by NoEigenvalue in an apparent attempt to make the final argument to answer a question in a simple way because that factor of a mere $$11/7\simeq 1.6$$ is further softened to 1.3 due to the square root function in converting from energies to velocities. To sum up, I think I've shown that the allegedly simple answer is both both too simple and also incorrect in many aspects. This is not a proof that the best answer is no, but it is probably a strong case that "no" is a reasonable answer if not the best answer. 3 added graph and structure to my arugment as well as a summary sentence. edited Jun 23 '18 at 12:44 Jason Arthur Taylor 34122 silver badges99 bronze badges 2 fixed eaten hyperlinks, added green checkbox clause edited Jun 23 '18 at 4:13 Jason Arthur Taylor 34122 silver badges99 bronze badges 1 answered Jun 23 '18 at 4:06 Jason Arthur Taylor 34122 silver badges99 bronze badges