Can two atoms be a crystal?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
In the physics literature, you can often find the term "two-ion crystal", when talking about two ions that are confined in a e.g. Paul trap. How is this possible? Shouldn't a crystal be a structure which repeats in space multiple (>2) times? Otherwise, what are the necessary requirements to define something as a crystal?
EDIT: one of the first ≈5k results found by Googling "two-ion crystal" https://arxiv.org/abs/1202.2730
solid-state-physics atomic-physics terminology crystals ion-traps
edited Jan 22 at 17:54
Qmechanic♦
104k121881195
asked Jan 22 at 13:47
m137m137
1829
$endgroup$
|
show 1 more comment
$begingroup$
In the physics literature, you can often find the term "two-ion crystal", when talking about two ions that are confined in a e.g. Paul trap. How is this possible? Shouldn't a crystal be a structure which repeats in space multiple (>2) times? Otherwise, what are the necessary requirements to define something as a crystal?
EDIT: one of the first ≈5k results found by Googling "two-ion crystal" https://arxiv.org/abs/1202.2730
solid-state-physics atomic-physics terminology crystals ion-traps
edited Jan 22 at 17:54
Qmechanic♦
104k121881195
asked Jan 22 at 13:47
m137m137
1829
$endgroup$
1
$begingroup$
Can you give a link in your question to an example of what you are talking about?
$endgroup$
– Aaron Stevens
Jan 22 at 14:35
2
$begingroup$
I mean, is a taco a sandwich? It seems they're using "crystal" in a different way than most do, but there's no law against that.
$endgroup$
– knzhou
Jan 22 at 14:43
2
$begingroup$
The necessary requirenment for a crystal is that it is periodic along it's lattice vectors. I.e. there exists at least one vector $v$ so that a translation along it conserves the system (btw. the number of such lineraly independent vektors defines the dimension of the crystal), by which i mean the observables, for instance the electron density within a solid in equilibrium: $rho(x + n v) = rho(x)$, $n inmathbfN$. This is by no means true for two isolated ions.
$endgroup$
– denklo
Jan 22 at 15:01
1
$begingroup$
@denklo That's an answer
$endgroup$
– FGSUZ
Jan 22 at 15:10
2
$begingroup$
@denklo That's an interesting definition; it's particularly notable in that it leaves out all real materials, so that (under that definition) crystals don't exist in the real world. (Among other shortcomings, such as leaving out quasicrystals, which are accepted as crystals by, say, the American Crystallographic Association, and other institutions whose opinions are generally regarded as important in this area.)
$endgroup$
– Emilio Pisanty
Jan 22 at 15:10
|
show 1 more comment
$begingroup$
In the physics literature, you can often find the term "two-ion crystal", when talking about two ions that are confined in a e.g. Paul trap. How is this possible? Shouldn't a crystal be a structure which repeats in space multiple (>2) times? Otherwise, what are the necessary requirements to define something as a crystal?
EDIT: one of the first ≈5k results found by Googling "two-ion crystal" https://arxiv.org/abs/1202.2730
solid-state-physics atomic-physics terminology crystals ion-traps
edited Jan 22 at 17:54
Qmechanic♦
104k121881195
asked Jan 22 at 13:47
m137m137
1829
$endgroup$
In the physics literature, you can often find the term "two-ion crystal", when talking about two ions that are confined in a e.g. Paul trap. How is this possible? Shouldn't a crystal be a structure which repeats in space multiple (>2) times? Otherwise, what are the necessary requirements to define something as a crystal?
EDIT: one of the first ≈5k results found by Googling "two-ion crystal" https://arxiv.org/abs/1202.2730
solid-state-physics atomic-physics terminology crystals ion-traps
solid-state-physics atomic-physics terminology crystals ion-traps
edited Jan 22 at 17:54
Qmechanic♦
104k121881195
asked Jan 22 at 13:47
m137m137
1829
edited Jan 22 at 17:54
Qmechanic♦
104k121881195
asked Jan 22 at 13:47
m137m137
1829
edited Jan 22 at 17:54
Qmechanic♦
104k121881195
edited Jan 22 at 17:54
Qmechanic♦
104k121881195
edited Jan 22 at 17:54
Qmechanic♦
104k121881195
104k121881195
asked Jan 22 at 13:47
m137m137
1829
asked Jan 22 at 13:47
m137m137
1829
asked Jan 22 at 13:47
m137m137
1829
1829
1
$begingroup$
Can you give a link in your question to an example of what you are talking about?
$endgroup$
– Aaron Stevens
Jan 22 at 14:35
2
$begingroup$
I mean, is a taco a sandwich? It seems they're using "crystal" in a different way than most do, but there's no law against that.
$endgroup$
– knzhou
Jan 22 at 14:43
2
$begingroup$
The necessary requirenment for a crystal is that it is periodic along it's lattice vectors. I.e. there exists at least one vector $v$ so that a translation along it conserves the system (btw. the number of such lineraly independent vektors defines the dimension of the crystal), by which i mean the observables, for instance the electron density within a solid in equilibrium: $rho(x + n v) = rho(x)$, $n inmathbfN$. This is by no means true for two isolated ions.
$endgroup$
– denklo
Jan 22 at 15:01
1
$begingroup$
@denklo That's an answer
$endgroup$
– FGSUZ
Jan 22 at 15:10
2
$begingroup$
@denklo That's an interesting definition; it's particularly notable in that it leaves out all real materials, so that (under that definition) crystals don't exist in the real world. (Among other shortcomings, such as leaving out quasicrystals, which are accepted as crystals by, say, the American Crystallographic Association, and other institutions whose opinions are generally regarded as important in this area.)
$endgroup$
– Emilio Pisanty
Jan 22 at 15:10
|
show 1 more comment
1
$begingroup$
Can you give a link in your question to an example of what you are talking about?
$endgroup$
– Aaron Stevens
Jan 22 at 14:35
2
$begingroup$
I mean, is a taco a sandwich? It seems they're using "crystal" in a different way than most do, but there's no law against that.
$endgroup$
– knzhou
Jan 22 at 14:43
2
$begingroup$
The necessary requirenment for a crystal is that it is periodic along it's lattice vectors. I.e. there exists at least one vector $v$ so that a translation along it conserves the system (btw. the number of such lineraly independent vektors defines the dimension of the crystal), by which i mean the observables, for instance the electron density within a solid in equilibrium: $rho(x + n v) = rho(x)$, $n inmathbfN$. This is by no means true for two isolated ions.
$endgroup$
– denklo
Jan 22 at 15:01
1
$begingroup$
@denklo That's an answer
$endgroup$
– FGSUZ
Jan 22 at 15:10
2
$begingroup$
@denklo That's an interesting definition; it's particularly notable in that it leaves out all real materials, so that (under that definition) crystals don't exist in the real world. (Among other shortcomings, such as leaving out quasicrystals, which are accepted as crystals by, say, the American Crystallographic Association, and other institutions whose opinions are generally regarded as important in this area.)
$endgroup$
– Emilio Pisanty
Jan 22 at 15:10
1
1
$begingroup$
Can you give a link in your question to an example of what you are talking about?
$endgroup$
– Aaron Stevens
Jan 22 at 14:35
$begingroup$
Can you give a link in your question to an example of what you are talking about?
$endgroup$
– Aaron Stevens
Jan 22 at 14:35
2
2
$begingroup$
I mean, is a taco a sandwich? It seems they're using "crystal" in a different way than most do, but there's no law against that.
$endgroup$
– knzhou
Jan 22 at 14:43
$begingroup$
I mean, is a taco a sandwich? It seems they're using "crystal" in a different way than most do, but there's no law against that.
$endgroup$
– knzhou
Jan 22 at 14:43
2
2
$begingroup$
The necessary requirenment for a crystal is that it is periodic along it's lattice vectors. I.e. there exists at least one vector $v$ so that a translation along it conserves the system (btw. the number of such lineraly independent vektors defines the dimension of the crystal), by which i mean the observables, for instance the electron density within a solid in equilibrium: $rho(x + n v) = rho(x)$, $n inmathbfN$. This is by no means true for two isolated ions.
$endgroup$
– denklo
Jan 22 at 15:01
$begingroup$
The necessary requirenment for a crystal is that it is periodic along it's lattice vectors. I.e. there exists at least one vector $v$ so that a translation along it conserves the system (btw. the number of such lineraly independent vektors defines the dimension of the crystal), by which i mean the observables, for instance the electron density within a solid in equilibrium: $rho(x + n v) = rho(x)$, $n inmathbfN$. This is by no means true for two isolated ions.
$endgroup$
– denklo
Jan 22 at 15:01
1
1
$begingroup$
@denklo That's an answer
$endgroup$
– FGSUZ
Jan 22 at 15:10
$begingroup$
@denklo That's an answer
$endgroup$
– FGSUZ
Jan 22 at 15:10
2
2
$begingroup$
@denklo That's an interesting definition; it's particularly notable in that it leaves out all real materials, so that (under that definition) crystals don't exist in the real world. (Among other shortcomings, such as leaving out quasicrystals, which are accepted as crystals by, say, the American Crystallographic Association, and other institutions whose opinions are generally regarded as important in this area.)
$endgroup$
– Emilio Pisanty
Jan 22 at 15:10
$begingroup$
@denklo That's an interesting definition; it's particularly notable in that it leaves out all real materials, so that (under that definition) crystals don't exist in the real world. (Among other shortcomings, such as leaving out quasicrystals, which are accepted as crystals by, say, the American Crystallographic Association, and other institutions whose opinions are generally regarded as important in this area.)
$endgroup$
– Emilio Pisanty
Jan 22 at 15:10
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Coulomb crystals are the structures formed by ions in a trap when they are sufficiently cold: once they stop jiggling around, they come down to equilibrium positions which need to balance the need to get down to the center of the trap, where the trapping potential is at its minimum, with the mutual repulsion between the ions.
This usually results in an orderly stacking of the ions, often with very clear local symmetries in a bunch of places. Here's one example, formed in an elongated ion trap (with experiment on the left and a simulation on the right; the lines are blurry because the whole thing is rigidly rotating about its vertical axis):
Image source
Within an ion-trapping context, the phrase "two-ion crystal" is a perfectly natural phrase to use for the case where you have coulomb-crystal dynamics, with a trapping potential and a Coulomb repulsion balancing out to give the equilibrium positions, and you have $N=2$ ions in the structure. If the phrase doesn't make sense to you, then that's just an indication that you're not within that text's intended audience.
Now, is the word "crystal" being used correctly here? The real answer is that it doesn't matter, at all: this is unambiguous notation, and lack of ambiguity is the single requirement that we make of notation.
answered Jan 22 at 15:12
Emilio PisantyEmilio Pisanty
83.1k22203417
$endgroup$
add a comment |
$begingroup$
Why can't it be? This is a matter of definition and whether one is comfortable or not with such a definition, which is determined by one's intuitive feel. Certainly, the papers authors are comfortable with it, otherwise they wouldn't use it.
To me, perhaps because of my affinity for theoretical mathematics and also computer programming, I see no problem at all with a crystal as having only one repeating unit. I tend to think of such a thing as being most similar to a mathematical set (though perhaps given the identical nature of the repeating units, a "multiset" might be the better choice if one wants to get strict), and a set can have one or even zero elements within it. That said, a "crystal with zero repeating units" is perhaps uninteresting physically, but mathematically even still makes sense, and moreover, a crystal can be abstracted into its set of lattice points which, as just that, a set, is entirely mathematically reasonable to consider as having 1 or even 0 points. In fact, from such a standpoint, the definition that admits 0 points can be "simpler" when one looks at how it is formulated in formal logic.
Moreover, two atoms constitute only a minimal repeating unit where and when the two atoms are different. If they are the same, then you have actually do have two repeating units in the "lattice".
answered Jan 23 at 9:49
The_SympathizerThe_Sympathizer
3,894923
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Coulomb crystals are the structures formed by ions in a trap when they are sufficiently cold: once they stop jiggling around, they come down to equilibrium positions which need to balance the need to get down to the center of the trap, where the trapping potential is at its minimum, with the mutual repulsion between the ions.
This usually results in an orderly stacking of the ions, often with very clear local symmetries in a bunch of places. Here's one example, formed in an elongated ion trap (with experiment on the left and a simulation on the right; the lines are blurry because the whole thing is rigidly rotating about its vertical axis):
Image source
Within an ion-trapping context, the phrase "two-ion crystal" is a perfectly natural phrase to use for the case where you have coulomb-crystal dynamics, with a trapping potential and a Coulomb repulsion balancing out to give the equilibrium positions, and you have $N=2$ ions in the structure. If the phrase doesn't make sense to you, then that's just an indication that you're not within that text's intended audience.
Now, is the word "crystal" being used correctly here? The real answer is that it doesn't matter, at all: this is unambiguous notation, and lack of ambiguity is the single requirement that we make of notation.
answered Jan 22 at 15:12
Emilio PisantyEmilio Pisanty
83.1k22203417
$endgroup$
add a comment |
$begingroup$
Coulomb crystals are the structures formed by ions in a trap when they are sufficiently cold: once they stop jiggling around, they come down to equilibrium positions which need to balance the need to get down to the center of the trap, where the trapping potential is at its minimum, with the mutual repulsion between the ions.
This usually results in an orderly stacking of the ions, often with very clear local symmetries in a bunch of places. Here's one example, formed in an elongated ion trap (with experiment on the left and a simulation on the right; the lines are blurry because the whole thing is rigidly rotating about its vertical axis):
Image source
Within an ion-trapping context, the phrase "two-ion crystal" is a perfectly natural phrase to use for the case where you have coulomb-crystal dynamics, with a trapping potential and a Coulomb repulsion balancing out to give the equilibrium positions, and you have $N=2$ ions in the structure. If the phrase doesn't make sense to you, then that's just an indication that you're not within that text's intended audience.
Now, is the word "crystal" being used correctly here? The real answer is that it doesn't matter, at all: this is unambiguous notation, and lack of ambiguity is the single requirement that we make of notation.
answered Jan 22 at 15:12
Emilio PisantyEmilio Pisanty
83.1k22203417
$endgroup$
add a comment |
$begingroup$
Coulomb crystals are the structures formed by ions in a trap when they are sufficiently cold: once they stop jiggling around, they come down to equilibrium positions which need to balance the need to get down to the center of the trap, where the trapping potential is at its minimum, with the mutual repulsion between the ions.
This usually results in an orderly stacking of the ions, often with very clear local symmetries in a bunch of places. Here's one example, formed in an elongated ion trap (with experiment on the left and a simulation on the right; the lines are blurry because the whole thing is rigidly rotating about its vertical axis):
Image source
Within an ion-trapping context, the phrase "two-ion crystal" is a perfectly natural phrase to use for the case where you have coulomb-crystal dynamics, with a trapping potential and a Coulomb repulsion balancing out to give the equilibrium positions, and you have $N=2$ ions in the structure. If the phrase doesn't make sense to you, then that's just an indication that you're not within that text's intended audience.
Now, is the word "crystal" being used correctly here? The real answer is that it doesn't matter, at all: this is unambiguous notation, and lack of ambiguity is the single requirement that we make of notation.
answered Jan 22 at 15:12
Emilio PisantyEmilio Pisanty
83.1k22203417
$endgroup$
Coulomb crystals are the structures formed by ions in a trap when they are sufficiently cold: once they stop jiggling around, they come down to equilibrium positions which need to balance the need to get down to the center of the trap, where the trapping potential is at its minimum, with the mutual repulsion between the ions.
This usually results in an orderly stacking of the ions, often with very clear local symmetries in a bunch of places. Here's one example, formed in an elongated ion trap (with experiment on the left and a simulation on the right; the lines are blurry because the whole thing is rigidly rotating about its vertical axis):
Image source
Within an ion-trapping context, the phrase "two-ion crystal" is a perfectly natural phrase to use for the case where you have coulomb-crystal dynamics, with a trapping potential and a Coulomb repulsion balancing out to give the equilibrium positions, and you have $N=2$ ions in the structure. If the phrase doesn't make sense to you, then that's just an indication that you're not within that text's intended audience.
Now, is the word "crystal" being used correctly here? The real answer is that it doesn't matter, at all: this is unambiguous notation, and lack of ambiguity is the single requirement that we make of notation.
answered Jan 22 at 15:12
Emilio PisantyEmilio Pisanty
83.1k22203417
answered Jan 22 at 15:12
Emilio PisantyEmilio Pisanty
83.1k22203417
answered Jan 22 at 15:12
Emilio PisantyEmilio Pisanty
83.1k22203417
answered Jan 22 at 15:12
Emilio PisantyEmilio Pisanty
83.1k22203417
83.1k22203417
add a comment |
add a comment |
$begingroup$
Why can't it be? This is a matter of definition and whether one is comfortable or not with such a definition, which is determined by one's intuitive feel. Certainly, the papers authors are comfortable with it, otherwise they wouldn't use it.
To me, perhaps because of my affinity for theoretical mathematics and also computer programming, I see no problem at all with a crystal as having only one repeating unit. I tend to think of such a thing as being most similar to a mathematical set (though perhaps given the identical nature of the repeating units, a "multiset" might be the better choice if one wants to get strict), and a set can have one or even zero elements within it. That said, a "crystal with zero repeating units" is perhaps uninteresting physically, but mathematically even still makes sense, and moreover, a crystal can be abstracted into its set of lattice points which, as just that, a set, is entirely mathematically reasonable to consider as having 1 or even 0 points. In fact, from such a standpoint, the definition that admits 0 points can be "simpler" when one looks at how it is formulated in formal logic.
Moreover, two atoms constitute only a minimal repeating unit where and when the two atoms are different. If they are the same, then you have actually do have two repeating units in the "lattice".
answered Jan 23 at 9:49
The_SympathizerThe_Sympathizer
3,894923
$endgroup$
add a comment |
$begingroup$
Why can't it be? This is a matter of definition and whether one is comfortable or not with such a definition, which is determined by one's intuitive feel. Certainly, the papers authors are comfortable with it, otherwise they wouldn't use it.
To me, perhaps because of my affinity for theoretical mathematics and also computer programming, I see no problem at all with a crystal as having only one repeating unit. I tend to think of such a thing as being most similar to a mathematical set (though perhaps given the identical nature of the repeating units, a "multiset" might be the better choice if one wants to get strict), and a set can have one or even zero elements within it. That said, a "crystal with zero repeating units" is perhaps uninteresting physically, but mathematically even still makes sense, and moreover, a crystal can be abstracted into its set of lattice points which, as just that, a set, is entirely mathematically reasonable to consider as having 1 or even 0 points. In fact, from such a standpoint, the definition that admits 0 points can be "simpler" when one looks at how it is formulated in formal logic.
Moreover, two atoms constitute only a minimal repeating unit where and when the two atoms are different. If they are the same, then you have actually do have two repeating units in the "lattice".
answered Jan 23 at 9:49
The_SympathizerThe_Sympathizer
3,894923
$endgroup$
add a comment |
$begingroup$
Why can't it be? This is a matter of definition and whether one is comfortable or not with such a definition, which is determined by one's intuitive feel. Certainly, the papers authors are comfortable with it, otherwise they wouldn't use it.
To me, perhaps because of my affinity for theoretical mathematics and also computer programming, I see no problem at all with a crystal as having only one repeating unit. I tend to think of such a thing as being most similar to a mathematical set (though perhaps given the identical nature of the repeating units, a "multiset" might be the better choice if one wants to get strict), and a set can have one or even zero elements within it. That said, a "crystal with zero repeating units" is perhaps uninteresting physically, but mathematically even still makes sense, and moreover, a crystal can be abstracted into its set of lattice points which, as just that, a set, is entirely mathematically reasonable to consider as having 1 or even 0 points. In fact, from such a standpoint, the definition that admits 0 points can be "simpler" when one looks at how it is formulated in formal logic.
Moreover, two atoms constitute only a minimal repeating unit where and when the two atoms are different. If they are the same, then you have actually do have two repeating units in the "lattice".
answered Jan 23 at 9:49
The_SympathizerThe_Sympathizer
3,894923
$endgroup$
Why can't it be? This is a matter of definition and whether one is comfortable or not with such a definition, which is determined by one's intuitive feel. Certainly, the papers authors are comfortable with it, otherwise they wouldn't use it.
To me, perhaps because of my affinity for theoretical mathematics and also computer programming, I see no problem at all with a crystal as having only one repeating unit. I tend to think of such a thing as being most similar to a mathematical set (though perhaps given the identical nature of the repeating units, a "multiset" might be the better choice if one wants to get strict), and a set can have one or even zero elements within it. That said, a "crystal with zero repeating units" is perhaps uninteresting physically, but mathematically even still makes sense, and moreover, a crystal can be abstracted into its set of lattice points which, as just that, a set, is entirely mathematically reasonable to consider as having 1 or even 0 points. In fact, from such a standpoint, the definition that admits 0 points can be "simpler" when one looks at how it is formulated in formal logic.
Moreover, two atoms constitute only a minimal repeating unit where and when the two atoms are different. If they are the same, then you have actually do have two repeating units in the "lattice".
answered Jan 23 at 9:49
The_SympathizerThe_Sympathizer
3,894923
answered Jan 23 at 9:49
The_SympathizerThe_Sympathizer
3,894923
answered Jan 23 at 9:49
The_SympathizerThe_Sympathizer
3,894923
answered Jan 23 at 9:49
The_SympathizerThe_Sympathizer
3,894923
3,894923
add a comment |
add a comment |
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1
$begingroup$
Can you give a link in your question to an example of what you are talking about?
$endgroup$
– Aaron Stevens
Jan 22 at 14:35
2
$begingroup$
I mean, is a taco a sandwich? It seems they're using "crystal" in a different way than most do, but there's no law against that.
$endgroup$
– knzhou
Jan 22 at 14:43
2
$begingroup$
The necessary requirenment for a crystal is that it is periodic along it's lattice vectors. I.e. there exists at least one vector $v$ so that a translation along it conserves the system (btw. the number of such lineraly independent vektors defines the dimension of the crystal), by which i mean the observables, for instance the electron density within a solid in equilibrium: $rho(x + n v) = rho(x)$, $n inmathbfN$. This is by no means true for two isolated ions.
$endgroup$
– denklo
Jan 22 at 15:01
1
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@denklo That's an answer
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– FGSUZ
Jan 22 at 15:10
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@denklo That's an interesting definition; it's particularly notable in that it leaves out all real materials, so that (under that definition) crystals don't exist in the real world. (Among other shortcomings, such as leaving out quasicrystals, which are accepted as crystals by, say, the American Crystallographic Association, and other institutions whose opinions are generally regarded as important in this area.)
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– Emilio Pisanty
Jan 22 at 15:10