Do “fields” always combine by addition?
$begingroup$
"Field" is a fun word which clearly has several meanings.
In all fields I can think of in my learning career, the fields obey superposition. I can calculate the fields generated by each object independently, and then sum them to determine the total field. But all the fields I can think of are relatively simple.
Are there fields for which this superposition principle does not apply? In other words, if I have a system where two mathematical vector spaces do not add (perhaps they saturate due to nonlinear effects), would a a physicist say "that's not a field because it doesn't admit the superposition principle?" Is there another name which is used in such circumstances instead?
field-theory superposition
asked Feb 27 at 1:59
Cort AmmonCort Ammon
23.4k34778
$endgroup$
add a comment |
$begingroup$
"Field" is a fun word which clearly has several meanings.
In all fields I can think of in my learning career, the fields obey superposition. I can calculate the fields generated by each object independently, and then sum them to determine the total field. But all the fields I can think of are relatively simple.
Are there fields for which this superposition principle does not apply? In other words, if I have a system where two mathematical vector spaces do not add (perhaps they saturate due to nonlinear effects), would a a physicist say "that's not a field because it doesn't admit the superposition principle?" Is there another name which is used in such circumstances instead?
field-theory superposition
asked Feb 27 at 1:59
Cort AmmonCort Ammon
23.4k34778
$endgroup$
$begingroup$
Consider the metric field in general relativity. A linear combination of legitimate metric fields is not necessarily a legitimate metric field. A simple example is $g_{ab}$ minus $g_{ab}$, which is zero -- definitely not a legitimate metric field. But we still call the metric field a field, specifically a particular type of tensor field. "Field" usually just means something like "a dynamic entity that is a smooth function of all the space coordinates."
$endgroup$
– Dan Yand
Feb 27 at 2:28
add a comment |
$begingroup$
"Field" is a fun word which clearly has several meanings.
In all fields I can think of in my learning career, the fields obey superposition. I can calculate the fields generated by each object independently, and then sum them to determine the total field. But all the fields I can think of are relatively simple.
Are there fields for which this superposition principle does not apply? In other words, if I have a system where two mathematical vector spaces do not add (perhaps they saturate due to nonlinear effects), would a a physicist say "that's not a field because it doesn't admit the superposition principle?" Is there another name which is used in such circumstances instead?
field-theory superposition
asked Feb 27 at 1:59
Cort AmmonCort Ammon
23.4k34778
$endgroup$
"Field" is a fun word which clearly has several meanings.
In all fields I can think of in my learning career, the fields obey superposition. I can calculate the fields generated by each object independently, and then sum them to determine the total field. But all the fields I can think of are relatively simple.
Are there fields for which this superposition principle does not apply? In other words, if I have a system where two mathematical vector spaces do not add (perhaps they saturate due to nonlinear effects), would a a physicist say "that's not a field because it doesn't admit the superposition principle?" Is there another name which is used in such circumstances instead?
field-theory superposition
field-theory superposition
asked Feb 27 at 1:59
Cort AmmonCort Ammon
23.4k34778
asked Feb 27 at 1:59
Cort AmmonCort Ammon
23.4k34778
asked Feb 27 at 1:59
Cort AmmonCort Ammon
23.4k34778
asked Feb 27 at 1:59
Cort AmmonCort Ammon
23.4k34778
asked Feb 27 at 1:59
Cort AmmonCort Ammon
23.4k34778
23.4k34778
$begingroup$
Consider the metric field in general relativity. A linear combination of legitimate metric fields is not necessarily a legitimate metric field. A simple example is $g_{ab}$ minus $g_{ab}$, which is zero -- definitely not a legitimate metric field. But we still call the metric field a field, specifically a particular type of tensor field. "Field" usually just means something like "a dynamic entity that is a smooth function of all the space coordinates."
$endgroup$
– Dan Yand
Feb 27 at 2:28
add a comment |
$begingroup$
Consider the metric field in general relativity. A linear combination of legitimate metric fields is not necessarily a legitimate metric field. A simple example is $g_{ab}$ minus $g_{ab}$, which is zero -- definitely not a legitimate metric field. But we still call the metric field a field, specifically a particular type of tensor field. "Field" usually just means something like "a dynamic entity that is a smooth function of all the space coordinates."
$endgroup$
– Dan Yand
Feb 27 at 2:28
$begingroup$
Consider the metric field in general relativity. A linear combination of legitimate metric fields is not necessarily a legitimate metric field. A simple example is $g_{ab}$ minus $g_{ab}$, which is zero -- definitely not a legitimate metric field. But we still call the metric field a field, specifically a particular type of tensor field. "Field" usually just means something like "a dynamic entity that is a smooth function of all the space coordinates."
$endgroup$
– Dan Yand
Feb 27 at 2:28
$begingroup$
Consider the metric field in general relativity. A linear combination of legitimate metric fields is not necessarily a legitimate metric field. A simple example is $g_{ab}$ minus $g_{ab}$, which is zero -- definitely not a legitimate metric field. But we still call the metric field a field, specifically a particular type of tensor field. "Field" usually just means something like "a dynamic entity that is a smooth function of all the space coordinates."
$endgroup$
– Dan Yand
Feb 27 at 2:28
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There are really two parts to your question: first, given two field configurations $phi_A$ and $phi_B$, does it make sense to think of a field configuration $phi_C = phi_A + phi_B$? Second, is the time evolution of $phi_C$ the same as the sum of the time evolutions of $phi_A$ and $phi_B$? If it isn't, there's not much point in writing $phi_C$ as a sum in the first place.
To answer the first question: not always. Essentially by definition, field combinations can be added if the space of possible field values is a vector space. This is the simplest option, but not the only one. For example, for a permanent magnet at low temperature, the local magnetization field has a constant magnitude but can vary its direction; it can take on values in a sphere. But the sum of two vectors on a sphere doesn't necessarily lie on the same sphere, so taking sums doesn't make sense. For a more sophisticated example, the Higgs field does something quite similar.
Sometimes one refers to theories with fields of this type as nonlinear sigma models. We still call these entities fields; my impression is that any function either from or to spacetime can be called a field.
Even in cases like this, you can still add field configurations if you think of them as small deviations from a uniform background configuration. Geometrically, this is just the fact that when you zoom in around a point on a sphere, it looks like a plane, which is a vector space. That's part of why you haven't seen examples of fields that aren't additive. The zoomed-in perspective can do a lot, but it can't describe, for example, topological field configurations which wrap around the sphere.
To answer the second question: not always. Time evolution can be calculated using the superposition principle if the equations of motion are linear, which happens if the Lagrangian is quadratic in the fields. There is nothing stopping you from adding higher-order terms, and any interesting field theory is full of them; otherwise particles would just pass right through each other.
The fact that most fields you've learned about are free can be understood in the light of effective field theory. For example, for the electromagnetic field, effective field theory tells us that at low energies, almost all contributions to the Lagrangian are strongly suppressed, with the suppression higher the higher-order the term. Thanks to other symmetries at play, the only terms that aren't negligible are the quadratic ones, which are why they were understood a century before the rest. For QED, the full Lagrangian for the electromagnetic field is given by the Euler-Heisenberg Lagrangian and includes, e.g. light-by-light scattering, a nonlinear effect.
answered Feb 27 at 2:21
knzhouknzhou
44.7k11122216
$endgroup$
add a comment |
$begingroup$
No, not at all. You would just classify them as non-interacting.
For instance, in classical field theory the electric field $mathbf{E}$ and the the gravitational field $mathbf{g}$ are all perfectly well defined vector fields throughout all space, but don't add at all.
answered Feb 27 at 2:18
InertialObserverInertialObserver
3,0201027
$endgroup$
$begingroup$
And indeed from a dimensional analysis point of view, $mathbf{E}$ and $mathbf{g}$ cannot add, unless at least one of them is multiplied by an appropriate dimensional constant. And if such a physically significant constant did exist, that would be related to the interaction.
$endgroup$
– lastresort
Feb 27 at 4:20
add a comment |
$begingroup$
A field is a mathematical structure with addition/subtraction and multiplication/division. So yes every field (combines) additively at least internally. Two different fields aren't going to add unless you can define a mapping between the fields (see InertialObserver's answer).
Addition; however, may look different than what you expect. There exists finite fields where addition may be modulo a certain number. E.g. a 8 bit register in computing (modulo 256), or the set of rotation (modulo 2 pi).
Nonlinear effects do exist, but they are nonlinear functions of an underlying field that does obey superposition.
answered Feb 27 at 2:43
Paul ChildsPaul Childs
2305
$endgroup$
6
$begingroup$
You are giving the mathematical definition of a field. This is not the definition used in physics. In physics "field" is a term for a function of several continuous variables (most commonly the three or four dimensions of space(-time)). Its values don't even have to live in a mathematical field E.g., the electrical field is a vector field with three real components and as such doesn't allow for division.
$endgroup$
– tobi_s
Feb 27 at 7:24
$begingroup$
Not true. The physics definition is the mathematical definition, it is just in general used for a specific kind. A vector field in physics meets the definition of a mathematical field. Division is simply just the inverse of multiplication (cross product - but this looks different due to lack of an identity).
$endgroup$
– Paul Childs
Feb 28 at 1:37
$begingroup$
A field in physics can be composed of arbitrary tensors, spinors, even twistors. You could also have discrete-valued fields. Do you have multiplication laws for all of these classes of objects that satisfy the axioms for a mathematical field?
$endgroup$
– tobi_s
Feb 28 at 2:58
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are really two parts to your question: first, given two field configurations $phi_A$ and $phi_B$, does it make sense to think of a field configuration $phi_C = phi_A + phi_B$? Second, is the time evolution of $phi_C$ the same as the sum of the time evolutions of $phi_A$ and $phi_B$? If it isn't, there's not much point in writing $phi_C$ as a sum in the first place.
To answer the first question: not always. Essentially by definition, field combinations can be added if the space of possible field values is a vector space. This is the simplest option, but not the only one. For example, for a permanent magnet at low temperature, the local magnetization field has a constant magnitude but can vary its direction; it can take on values in a sphere. But the sum of two vectors on a sphere doesn't necessarily lie on the same sphere, so taking sums doesn't make sense. For a more sophisticated example, the Higgs field does something quite similar.
Sometimes one refers to theories with fields of this type as nonlinear sigma models. We still call these entities fields; my impression is that any function either from or to spacetime can be called a field.
Even in cases like this, you can still add field configurations if you think of them as small deviations from a uniform background configuration. Geometrically, this is just the fact that when you zoom in around a point on a sphere, it looks like a plane, which is a vector space. That's part of why you haven't seen examples of fields that aren't additive. The zoomed-in perspective can do a lot, but it can't describe, for example, topological field configurations which wrap around the sphere.
To answer the second question: not always. Time evolution can be calculated using the superposition principle if the equations of motion are linear, which happens if the Lagrangian is quadratic in the fields. There is nothing stopping you from adding higher-order terms, and any interesting field theory is full of them; otherwise particles would just pass right through each other.
The fact that most fields you've learned about are free can be understood in the light of effective field theory. For example, for the electromagnetic field, effective field theory tells us that at low energies, almost all contributions to the Lagrangian are strongly suppressed, with the suppression higher the higher-order the term. Thanks to other symmetries at play, the only terms that aren't negligible are the quadratic ones, which are why they were understood a century before the rest. For QED, the full Lagrangian for the electromagnetic field is given by the Euler-Heisenberg Lagrangian and includes, e.g. light-by-light scattering, a nonlinear effect.
answered Feb 27 at 2:21
knzhouknzhou
44.7k11122216
$endgroup$
add a comment |
$begingroup$
There are really two parts to your question: first, given two field configurations $phi_A$ and $phi_B$, does it make sense to think of a field configuration $phi_C = phi_A + phi_B$? Second, is the time evolution of $phi_C$ the same as the sum of the time evolutions of $phi_A$ and $phi_B$? If it isn't, there's not much point in writing $phi_C$ as a sum in the first place.
To answer the first question: not always. Essentially by definition, field combinations can be added if the space of possible field values is a vector space. This is the simplest option, but not the only one. For example, for a permanent magnet at low temperature, the local magnetization field has a constant magnitude but can vary its direction; it can take on values in a sphere. But the sum of two vectors on a sphere doesn't necessarily lie on the same sphere, so taking sums doesn't make sense. For a more sophisticated example, the Higgs field does something quite similar.
Sometimes one refers to theories with fields of this type as nonlinear sigma models. We still call these entities fields; my impression is that any function either from or to spacetime can be called a field.
Even in cases like this, you can still add field configurations if you think of them as small deviations from a uniform background configuration. Geometrically, this is just the fact that when you zoom in around a point on a sphere, it looks like a plane, which is a vector space. That's part of why you haven't seen examples of fields that aren't additive. The zoomed-in perspective can do a lot, but it can't describe, for example, topological field configurations which wrap around the sphere.
To answer the second question: not always. Time evolution can be calculated using the superposition principle if the equations of motion are linear, which happens if the Lagrangian is quadratic in the fields. There is nothing stopping you from adding higher-order terms, and any interesting field theory is full of them; otherwise particles would just pass right through each other.
The fact that most fields you've learned about are free can be understood in the light of effective field theory. For example, for the electromagnetic field, effective field theory tells us that at low energies, almost all contributions to the Lagrangian are strongly suppressed, with the suppression higher the higher-order the term. Thanks to other symmetries at play, the only terms that aren't negligible are the quadratic ones, which are why they were understood a century before the rest. For QED, the full Lagrangian for the electromagnetic field is given by the Euler-Heisenberg Lagrangian and includes, e.g. light-by-light scattering, a nonlinear effect.
answered Feb 27 at 2:21
knzhouknzhou
44.7k11122216
$endgroup$
add a comment |
$begingroup$
There are really two parts to your question: first, given two field configurations $phi_A$ and $phi_B$, does it make sense to think of a field configuration $phi_C = phi_A + phi_B$? Second, is the time evolution of $phi_C$ the same as the sum of the time evolutions of $phi_A$ and $phi_B$? If it isn't, there's not much point in writing $phi_C$ as a sum in the first place.
To answer the first question: not always. Essentially by definition, field combinations can be added if the space of possible field values is a vector space. This is the simplest option, but not the only one. For example, for a permanent magnet at low temperature, the local magnetization field has a constant magnitude but can vary its direction; it can take on values in a sphere. But the sum of two vectors on a sphere doesn't necessarily lie on the same sphere, so taking sums doesn't make sense. For a more sophisticated example, the Higgs field does something quite similar.
Sometimes one refers to theories with fields of this type as nonlinear sigma models. We still call these entities fields; my impression is that any function either from or to spacetime can be called a field.
Even in cases like this, you can still add field configurations if you think of them as small deviations from a uniform background configuration. Geometrically, this is just the fact that when you zoom in around a point on a sphere, it looks like a plane, which is a vector space. That's part of why you haven't seen examples of fields that aren't additive. The zoomed-in perspective can do a lot, but it can't describe, for example, topological field configurations which wrap around the sphere.
To answer the second question: not always. Time evolution can be calculated using the superposition principle if the equations of motion are linear, which happens if the Lagrangian is quadratic in the fields. There is nothing stopping you from adding higher-order terms, and any interesting field theory is full of them; otherwise particles would just pass right through each other.
The fact that most fields you've learned about are free can be understood in the light of effective field theory. For example, for the electromagnetic field, effective field theory tells us that at low energies, almost all contributions to the Lagrangian are strongly suppressed, with the suppression higher the higher-order the term. Thanks to other symmetries at play, the only terms that aren't negligible are the quadratic ones, which are why they were understood a century before the rest. For QED, the full Lagrangian for the electromagnetic field is given by the Euler-Heisenberg Lagrangian and includes, e.g. light-by-light scattering, a nonlinear effect.
answered Feb 27 at 2:21
knzhouknzhou
44.7k11122216
$endgroup$
There are really two parts to your question: first, given two field configurations $phi_A$ and $phi_B$, does it make sense to think of a field configuration $phi_C = phi_A + phi_B$? Second, is the time evolution of $phi_C$ the same as the sum of the time evolutions of $phi_A$ and $phi_B$? If it isn't, there's not much point in writing $phi_C$ as a sum in the first place.
To answer the first question: not always. Essentially by definition, field combinations can be added if the space of possible field values is a vector space. This is the simplest option, but not the only one. For example, for a permanent magnet at low temperature, the local magnetization field has a constant magnitude but can vary its direction; it can take on values in a sphere. But the sum of two vectors on a sphere doesn't necessarily lie on the same sphere, so taking sums doesn't make sense. For a more sophisticated example, the Higgs field does something quite similar.
Sometimes one refers to theories with fields of this type as nonlinear sigma models. We still call these entities fields; my impression is that any function either from or to spacetime can be called a field.
Even in cases like this, you can still add field configurations if you think of them as small deviations from a uniform background configuration. Geometrically, this is just the fact that when you zoom in around a point on a sphere, it looks like a plane, which is a vector space. That's part of why you haven't seen examples of fields that aren't additive. The zoomed-in perspective can do a lot, but it can't describe, for example, topological field configurations which wrap around the sphere.
To answer the second question: not always. Time evolution can be calculated using the superposition principle if the equations of motion are linear, which happens if the Lagrangian is quadratic in the fields. There is nothing stopping you from adding higher-order terms, and any interesting field theory is full of them; otherwise particles would just pass right through each other.
The fact that most fields you've learned about are free can be understood in the light of effective field theory. For example, for the electromagnetic field, effective field theory tells us that at low energies, almost all contributions to the Lagrangian are strongly suppressed, with the suppression higher the higher-order the term. Thanks to other symmetries at play, the only terms that aren't negligible are the quadratic ones, which are why they were understood a century before the rest. For QED, the full Lagrangian for the electromagnetic field is given by the Euler-Heisenberg Lagrangian and includes, e.g. light-by-light scattering, a nonlinear effect.
answered Feb 27 at 2:21
knzhouknzhou
44.7k11122216
answered Feb 27 at 2:21
knzhouknzhou
44.7k11122216
answered Feb 27 at 2:21
knzhouknzhou
44.7k11122216
answered Feb 27 at 2:21
knzhouknzhou
44.7k11122216
44.7k11122216
add a comment |
add a comment |
$begingroup$
No, not at all. You would just classify them as non-interacting.
For instance, in classical field theory the electric field $mathbf{E}$ and the the gravitational field $mathbf{g}$ are all perfectly well defined vector fields throughout all space, but don't add at all.
answered Feb 27 at 2:18
InertialObserverInertialObserver
3,0201027
$endgroup$
$begingroup$
And indeed from a dimensional analysis point of view, $mathbf{E}$ and $mathbf{g}$ cannot add, unless at least one of them is multiplied by an appropriate dimensional constant. And if such a physically significant constant did exist, that would be related to the interaction.
$endgroup$
– lastresort
Feb 27 at 4:20
add a comment |
$begingroup$
No, not at all. You would just classify them as non-interacting.
For instance, in classical field theory the electric field $mathbf{E}$ and the the gravitational field $mathbf{g}$ are all perfectly well defined vector fields throughout all space, but don't add at all.
answered Feb 27 at 2:18
InertialObserverInertialObserver
3,0201027
$endgroup$
$begingroup$
And indeed from a dimensional analysis point of view, $mathbf{E}$ and $mathbf{g}$ cannot add, unless at least one of them is multiplied by an appropriate dimensional constant. And if such a physically significant constant did exist, that would be related to the interaction.
$endgroup$
– lastresort
Feb 27 at 4:20
add a comment |
$begingroup$
No, not at all. You would just classify them as non-interacting.
For instance, in classical field theory the electric field $mathbf{E}$ and the the gravitational field $mathbf{g}$ are all perfectly well defined vector fields throughout all space, but don't add at all.
answered Feb 27 at 2:18
InertialObserverInertialObserver
3,0201027
$endgroup$
No, not at all. You would just classify them as non-interacting.
For instance, in classical field theory the electric field $mathbf{E}$ and the the gravitational field $mathbf{g}$ are all perfectly well defined vector fields throughout all space, but don't add at all.
answered Feb 27 at 2:18
InertialObserverInertialObserver
3,0201027
answered Feb 27 at 2:18
InertialObserverInertialObserver
3,0201027
answered Feb 27 at 2:18
InertialObserverInertialObserver
3,0201027
answered Feb 27 at 2:18
InertialObserverInertialObserver
3,0201027
3,0201027
$begingroup$
And indeed from a dimensional analysis point of view, $mathbf{E}$ and $mathbf{g}$ cannot add, unless at least one of them is multiplied by an appropriate dimensional constant. And if such a physically significant constant did exist, that would be related to the interaction.
$endgroup$
– lastresort
Feb 27 at 4:20
add a comment |
$begingroup$
And indeed from a dimensional analysis point of view, $mathbf{E}$ and $mathbf{g}$ cannot add, unless at least one of them is multiplied by an appropriate dimensional constant. And if such a physically significant constant did exist, that would be related to the interaction.
$endgroup$
– lastresort
Feb 27 at 4:20
$begingroup$
And indeed from a dimensional analysis point of view, $mathbf{E}$ and $mathbf{g}$ cannot add, unless at least one of them is multiplied by an appropriate dimensional constant. And if such a physically significant constant did exist, that would be related to the interaction.
$endgroup$
– lastresort
Feb 27 at 4:20
$begingroup$
And indeed from a dimensional analysis point of view, $mathbf{E}$ and $mathbf{g}$ cannot add, unless at least one of them is multiplied by an appropriate dimensional constant. And if such a physically significant constant did exist, that would be related to the interaction.
$endgroup$
– lastresort
Feb 27 at 4:20
add a comment |
$begingroup$
A field is a mathematical structure with addition/subtraction and multiplication/division. So yes every field (combines) additively at least internally. Two different fields aren't going to add unless you can define a mapping between the fields (see InertialObserver's answer).
Addition; however, may look different than what you expect. There exists finite fields where addition may be modulo a certain number. E.g. a 8 bit register in computing (modulo 256), or the set of rotation (modulo 2 pi).
Nonlinear effects do exist, but they are nonlinear functions of an underlying field that does obey superposition.
answered Feb 27 at 2:43
Paul ChildsPaul Childs
2305
$endgroup$
6
$begingroup$
You are giving the mathematical definition of a field. This is not the definition used in physics. In physics "field" is a term for a function of several continuous variables (most commonly the three or four dimensions of space(-time)). Its values don't even have to live in a mathematical field E.g., the electrical field is a vector field with three real components and as such doesn't allow for division.
$endgroup$
– tobi_s
Feb 27 at 7:24
$begingroup$
Not true. The physics definition is the mathematical definition, it is just in general used for a specific kind. A vector field in physics meets the definition of a mathematical field. Division is simply just the inverse of multiplication (cross product - but this looks different due to lack of an identity).
$endgroup$
– Paul Childs
Feb 28 at 1:37
$begingroup$
A field in physics can be composed of arbitrary tensors, spinors, even twistors. You could also have discrete-valued fields. Do you have multiplication laws for all of these classes of objects that satisfy the axioms for a mathematical field?
$endgroup$
– tobi_s
Feb 28 at 2:58
add a comment |
$begingroup$
A field is a mathematical structure with addition/subtraction and multiplication/division. So yes every field (combines) additively at least internally. Two different fields aren't going to add unless you can define a mapping between the fields (see InertialObserver's answer).
Addition; however, may look different than what you expect. There exists finite fields where addition may be modulo a certain number. E.g. a 8 bit register in computing (modulo 256), or the set of rotation (modulo 2 pi).
Nonlinear effects do exist, but they are nonlinear functions of an underlying field that does obey superposition.
answered Feb 27 at 2:43
Paul ChildsPaul Childs
2305
$endgroup$
6
$begingroup$
You are giving the mathematical definition of a field. This is not the definition used in physics. In physics "field" is a term for a function of several continuous variables (most commonly the three or four dimensions of space(-time)). Its values don't even have to live in a mathematical field E.g., the electrical field is a vector field with three real components and as such doesn't allow for division.
$endgroup$
– tobi_s
Feb 27 at 7:24
$begingroup$
Not true. The physics definition is the mathematical definition, it is just in general used for a specific kind. A vector field in physics meets the definition of a mathematical field. Division is simply just the inverse of multiplication (cross product - but this looks different due to lack of an identity).
$endgroup$
– Paul Childs
Feb 28 at 1:37
$begingroup$
A field in physics can be composed of arbitrary tensors, spinors, even twistors. You could also have discrete-valued fields. Do you have multiplication laws for all of these classes of objects that satisfy the axioms for a mathematical field?
$endgroup$
– tobi_s
Feb 28 at 2:58
add a comment |
$begingroup$
A field is a mathematical structure with addition/subtraction and multiplication/division. So yes every field (combines) additively at least internally. Two different fields aren't going to add unless you can define a mapping between the fields (see InertialObserver's answer).
Addition; however, may look different than what you expect. There exists finite fields where addition may be modulo a certain number. E.g. a 8 bit register in computing (modulo 256), or the set of rotation (modulo 2 pi).
Nonlinear effects do exist, but they are nonlinear functions of an underlying field that does obey superposition.
answered Feb 27 at 2:43
Paul ChildsPaul Childs
2305
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A field is a mathematical structure with addition/subtraction and multiplication/division. So yes every field (combines) additively at least internally. Two different fields aren't going to add unless you can define a mapping between the fields (see InertialObserver's answer).
Addition; however, may look different than what you expect. There exists finite fields where addition may be modulo a certain number. E.g. a 8 bit register in computing (modulo 256), or the set of rotation (modulo 2 pi).
Nonlinear effects do exist, but they are nonlinear functions of an underlying field that does obey superposition.
answered Feb 27 at 2:43
Paul ChildsPaul Childs
2305
answered Feb 27 at 2:43
Paul ChildsPaul Childs
2305
answered Feb 27 at 2:43
Paul ChildsPaul Childs
2305
answered Feb 27 at 2:43
Paul ChildsPaul Childs
2305
2305
6
$begingroup$
You are giving the mathematical definition of a field. This is not the definition used in physics. In physics "field" is a term for a function of several continuous variables (most commonly the three or four dimensions of space(-time)). Its values don't even have to live in a mathematical field E.g., the electrical field is a vector field with three real components and as such doesn't allow for division.
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– tobi_s
Feb 27 at 7:24
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Not true. The physics definition is the mathematical definition, it is just in general used for a specific kind. A vector field in physics meets the definition of a mathematical field. Division is simply just the inverse of multiplication (cross product - but this looks different due to lack of an identity).
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– Paul Childs
Feb 28 at 1:37
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A field in physics can be composed of arbitrary tensors, spinors, even twistors. You could also have discrete-valued fields. Do you have multiplication laws for all of these classes of objects that satisfy the axioms for a mathematical field?
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– tobi_s
Feb 28 at 2:58
add a comment |
6
$begingroup$
You are giving the mathematical definition of a field. This is not the definition used in physics. In physics "field" is a term for a function of several continuous variables (most commonly the three or four dimensions of space(-time)). Its values don't even have to live in a mathematical field E.g., the electrical field is a vector field with three real components and as such doesn't allow for division.
$endgroup$
– tobi_s
Feb 27 at 7:24
$begingroup$
Not true. The physics definition is the mathematical definition, it is just in general used for a specific kind. A vector field in physics meets the definition of a mathematical field. Division is simply just the inverse of multiplication (cross product - but this looks different due to lack of an identity).
$endgroup$
– Paul Childs
Feb 28 at 1:37
$begingroup$
A field in physics can be composed of arbitrary tensors, spinors, even twistors. You could also have discrete-valued fields. Do you have multiplication laws for all of these classes of objects that satisfy the axioms for a mathematical field?
$endgroup$
– tobi_s
Feb 28 at 2:58
6
6
$begingroup$
You are giving the mathematical definition of a field. This is not the definition used in physics. In physics "field" is a term for a function of several continuous variables (most commonly the three or four dimensions of space(-time)). Its values don't even have to live in a mathematical field E.g., the electrical field is a vector field with three real components and as such doesn't allow for division.
$endgroup$
– tobi_s
Feb 27 at 7:24
$begingroup$
You are giving the mathematical definition of a field. This is not the definition used in physics. In physics "field" is a term for a function of several continuous variables (most commonly the three or four dimensions of space(-time)). Its values don't even have to live in a mathematical field E.g., the electrical field is a vector field with three real components and as such doesn't allow for division.
$endgroup$
– tobi_s
Feb 27 at 7:24
$begingroup$
Not true. The physics definition is the mathematical definition, it is just in general used for a specific kind. A vector field in physics meets the definition of a mathematical field. Division is simply just the inverse of multiplication (cross product - but this looks different due to lack of an identity).
$endgroup$
– Paul Childs
Feb 28 at 1:37
$begingroup$
Not true. The physics definition is the mathematical definition, it is just in general used for a specific kind. A vector field in physics meets the definition of a mathematical field. Division is simply just the inverse of multiplication (cross product - but this looks different due to lack of an identity).
$endgroup$
– Paul Childs
Feb 28 at 1:37
$begingroup$
A field in physics can be composed of arbitrary tensors, spinors, even twistors. You could also have discrete-valued fields. Do you have multiplication laws for all of these classes of objects that satisfy the axioms for a mathematical field?
$endgroup$
– tobi_s
Feb 28 at 2:58
$begingroup$
A field in physics can be composed of arbitrary tensors, spinors, even twistors. You could also have discrete-valued fields. Do you have multiplication laws for all of these classes of objects that satisfy the axioms for a mathematical field?
$endgroup$
– tobi_s
Feb 28 at 2:58
add a comment |
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Consider the metric field in general relativity. A linear combination of legitimate metric fields is not necessarily a legitimate metric field. A simple example is $g_{ab}$ minus $g_{ab}$, which is zero -- definitely not a legitimate metric field. But we still call the metric field a field, specifically a particular type of tensor field. "Field" usually just means something like "a dynamic entity that is a smooth function of all the space coordinates."
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– Dan Yand
Feb 27 at 2:28