Mixing
Definition of inner product as in the case of work
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According to the mathematical definition of "vectors", vectors are simply the elements of a set $V$ which forms a vector space structure $(V,F,+,*)$. The definition of inner product states that it is a function from $langle ,rangle :Vtimes V rightarrow F$ with the properties of inner products being satisfied as such. My questions is: How do we then define the dot product like work $dW=F.dx$ where $F$ and $dx$ are both vectors as per the mathematical definition? Because both the vector $F$ and $dx$ belong to two different vector spaces, so how do we define "inner product" between them?
newtonian-mechanics work vectors
asked 1 hour ago
Naman Agarwal
350113
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up vote
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According to the mathematical definition of "vectors", vectors are simply the elements of a set $V$ which forms a vector space structure $(V,F,+,*)$. The definition of inner product states that it is a function from $langle ,rangle :Vtimes V rightarrow F$ with the properties of inner products being satisfied as such. My questions is: How do we then define the dot product like work $dW=F.dx$ where $F$ and $dx$ are both vectors as per the mathematical definition? Because both the vector $F$ and $dx$ belong to two different vector spaces, so how do we define "inner product" between them?
newtonian-mechanics work vectors
asked 1 hour ago
Naman Agarwal
350113
1
Don't both vectors live in $mathbbR^3$?
– ZeroTheHero
1 hour ago
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up vote
4
down vote
favorite
up vote
4
down vote
favorite
According to the mathematical definition of "vectors", vectors are simply the elements of a set $V$ which forms a vector space structure $(V,F,+,*)$. The definition of inner product states that it is a function from $langle ,rangle :Vtimes V rightarrow F$ with the properties of inner products being satisfied as such. My questions is: How do we then define the dot product like work $dW=F.dx$ where $F$ and $dx$ are both vectors as per the mathematical definition? Because both the vector $F$ and $dx$ belong to two different vector spaces, so how do we define "inner product" between them?
newtonian-mechanics work vectors
asked 1 hour ago
Naman Agarwal
350113
According to the mathematical definition of "vectors", vectors are simply the elements of a set $V$ which forms a vector space structure $(V,F,+,*)$. The definition of inner product states that it is a function from $langle ,rangle :Vtimes V rightarrow F$ with the properties of inner products being satisfied as such. My questions is: How do we then define the dot product like work $dW=F.dx$ where $F$ and $dx$ are both vectors as per the mathematical definition? Because both the vector $F$ and $dx$ belong to two different vector spaces, so how do we define "inner product" between them?
newtonian-mechanics work vectors
newtonian-mechanics work vectors
asked 1 hour ago
Naman Agarwal
350113
asked 1 hour ago
Naman Agarwal
350113
asked 1 hour ago
Naman Agarwal
350113
asked 1 hour ago
Naman Agarwal
350113
asked 1 hour ago
Naman Agarwal
350113
350113
1
Don't both vectors live in $mathbbR^3$?
– ZeroTheHero
1 hour ago
add a comment |Â
1
Don't both vectors live in $mathbbR^3$?
– ZeroTheHero
1 hour ago
1
1
Don't both vectors live in $mathbbR^3$?
– ZeroTheHero
1 hour ago
Don't both vectors live in $mathbbR^3$?
– ZeroTheHero
1 hour ago
add a comment |Â
4 Answers
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I'm not sure if your claim that $F$ and $dx$ lie in different vector spaces stems from the fact that one is infinitesimal and the other isn't, or that they have different physical units and so can't be added together. If it's the former, then we can consider a line integral as a limit of sums of dot products of non-infinitesimal vectors.
If it's the latter, then we can get at the same issue without needing to consider the complication of vector spaces by asking why we can multiply together two scalar quantities with different units, even though they can't be added together and so don't (obviously) lie in the same field (in the abstract algebra sense of the word). Mathematically formalizing elementary dimensional analysis is actually surprisingly nontrivial: Terry Tao has a nice treatment here.
The "quick and dirty" answer is that when you think of the formal mathematical/abstract-algebraic structure of the space of possible values for a physical quantity, you forget about units and treat everything as dimensionless. (E.g. you think of all three-vectors as living in $mathbbR^3$, regardless of their physical units.) Then afterward, you declare operations that don't make dimensional sense to be "physically invalid" even though they're perfectly legitimate from a formal mathematical perspective. "Hard-coding" the dimensional restrictions on the set of allowed operations directly into the formal mathematical structure is usually way more effort than it's worth.
answered 51 mins ago
tparker
21.5k143114
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up vote
0
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Thinking about this in the finite case, without dealing with infinitesimals, you have $Fcdot Delta x$. Both vectors in this inner product do live in the same space, namely $mathbbR^3$ in usual 3-d space, at least up to an isomorphism.
answered 1 hour ago
Hugo V
2868
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If you consider $F$ and $dx$ as in different Vector spaces, you are entering in the region of Differential Geometry. Here the things are more complicated, since vectors are defined as the element of the tangent space defined in each point of the manifold, and in order to compare vectors in different points you have to define a rule (which is the parallel transport), which in turn is defined by a covariant derivative.
In your example though, $vecF=(F_x,F_y,F_z)$ and $dvecx=(dx,dy,dz)$ so that they can be thought of belonging to $mathbbR^3$, and the '$cdot$' (dot) product is just the usual one.
answered 53 mins ago
Kevin De Notariis
664
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I assume you're referring to the fact that they belong to different vector spaces because they have different units. The difficulty you're running into is just an indication of an inconvenience in the set of definitions and foundations for mathematics that currently happens to be popular. To understand this, it may be helpful to look at the history.
Something like what we today define as a vector space was first defined by Peano in 1888.
Vectors, and including the words "vector" and "scalar," were defined by Gibbs around 1888, as a way of simplifying the quaternion system for his students at Yale.
Physicists' modern definition of a vector, which involves its transformation properties, was standardized ca. 1930-1950.
Set theory was cast into something like its present form by by ZFC in 1908-1922.
Keep in mind that all of these developments really occurred over the period of many years, not on specific dates, so what we really have is an overlapping set of time periods, with people working independently of one another and not necessarily producing consistent systems.
The basic notions that we really need in order to do linear algebra are algebraic, i.e., essentially syntactical facts, e.g., the property $(atextbfu)cdottextbfv=a(textbfucdottextbfv)$. When you look at an axiom like this one and then apply it, it makes no difference whether or not the objects $a$, $textbfu$, and $textbfv$ belong to certain sets. The notion of a set came later than the notion of a vector space. The fact that people today usually use ZFC as a foundation and define things like vector spaces in terms of operations on some set does not mean that it has to be done that way, was historically done that way, or is always convenient to do that way.
So for example, suppose an object starts at rest and is accelerated by a constant force $textbfF$, so that $textbfF=(2m/t^2)textbfx$. The identities that define the properties of the inner product have the same syntactical form, and therefore lead to the same results, regardless of the fact that it wouldn't make sense to talk about $textbfF+textbfx$. You can find the work done by this force, and while you're doing that, you can freely use the identity $(atextbfu)cdottextbfv=a(textbfucdottextbfv)$.
If you like, you can describe forces and displacements as different vector spaces with some machinery to connect them so that you can do dot products, or if you like, you can think of them as both belonging to some sort of space that has some forbidden operations, such as not being able to some additions. It doesn't matter which of these you do, and in practice nobody does anything like this formally.
Keep in mind also that the physicist's definition of a vector is more restrictive than a mathematician's, e.g., if you form an ordered pair consisting of $(S,q)$, where $S$ is the current value of the S&P 500 stock market index, and $q$ is an electric charge, then to a mathematician, this is a vector that lives in some vector space, but to a physicist this is not a vector at all, because it doesn't transform as a vector.
answered 18 mins ago
Ben Crowell
45.2k3147275
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I'm not sure if your claim that $F$ and $dx$ lie in different vector spaces stems from the fact that one is infinitesimal and the other isn't, or that they have different physical units and so can't be added together. If it's the former, then we can consider a line integral as a limit of sums of dot products of non-infinitesimal vectors.
If it's the latter, then we can get at the same issue without needing to consider the complication of vector spaces by asking why we can multiply together two scalar quantities with different units, even though they can't be added together and so don't (obviously) lie in the same field (in the abstract algebra sense of the word). Mathematically formalizing elementary dimensional analysis is actually surprisingly nontrivial: Terry Tao has a nice treatment here.
The "quick and dirty" answer is that when you think of the formal mathematical/abstract-algebraic structure of the space of possible values for a physical quantity, you forget about units and treat everything as dimensionless. (E.g. you think of all three-vectors as living in $mathbbR^3$, regardless of their physical units.) Then afterward, you declare operations that don't make dimensional sense to be "physically invalid" even though they're perfectly legitimate from a formal mathematical perspective. "Hard-coding" the dimensional restrictions on the set of allowed operations directly into the formal mathematical structure is usually way more effort than it's worth.
answered 51 mins ago
tparker
21.5k143114
add a comment |Â
up vote
3
down vote
I'm not sure if your claim that $F$ and $dx$ lie in different vector spaces stems from the fact that one is infinitesimal and the other isn't, or that they have different physical units and so can't be added together. If it's the former, then we can consider a line integral as a limit of sums of dot products of non-infinitesimal vectors.
If it's the latter, then we can get at the same issue without needing to consider the complication of vector spaces by asking why we can multiply together two scalar quantities with different units, even though they can't be added together and so don't (obviously) lie in the same field (in the abstract algebra sense of the word). Mathematically formalizing elementary dimensional analysis is actually surprisingly nontrivial: Terry Tao has a nice treatment here.
The "quick and dirty" answer is that when you think of the formal mathematical/abstract-algebraic structure of the space of possible values for a physical quantity, you forget about units and treat everything as dimensionless. (E.g. you think of all three-vectors as living in $mathbbR^3$, regardless of their physical units.) Then afterward, you declare operations that don't make dimensional sense to be "physically invalid" even though they're perfectly legitimate from a formal mathematical perspective. "Hard-coding" the dimensional restrictions on the set of allowed operations directly into the formal mathematical structure is usually way more effort than it's worth.
answered 51 mins ago
tparker
21.5k143114
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I'm not sure if your claim that $F$ and $dx$ lie in different vector spaces stems from the fact that one is infinitesimal and the other isn't, or that they have different physical units and so can't be added together. If it's the former, then we can consider a line integral as a limit of sums of dot products of non-infinitesimal vectors.
If it's the latter, then we can get at the same issue without needing to consider the complication of vector spaces by asking why we can multiply together two scalar quantities with different units, even though they can't be added together and so don't (obviously) lie in the same field (in the abstract algebra sense of the word). Mathematically formalizing elementary dimensional analysis is actually surprisingly nontrivial: Terry Tao has a nice treatment here.
The "quick and dirty" answer is that when you think of the formal mathematical/abstract-algebraic structure of the space of possible values for a physical quantity, you forget about units and treat everything as dimensionless. (E.g. you think of all three-vectors as living in $mathbbR^3$, regardless of their physical units.) Then afterward, you declare operations that don't make dimensional sense to be "physically invalid" even though they're perfectly legitimate from a formal mathematical perspective. "Hard-coding" the dimensional restrictions on the set of allowed operations directly into the formal mathematical structure is usually way more effort than it's worth.
answered 51 mins ago
tparker
21.5k143114
I'm not sure if your claim that $F$ and $dx$ lie in different vector spaces stems from the fact that one is infinitesimal and the other isn't, or that they have different physical units and so can't be added together. If it's the former, then we can consider a line integral as a limit of sums of dot products of non-infinitesimal vectors.
If it's the latter, then we can get at the same issue without needing to consider the complication of vector spaces by asking why we can multiply together two scalar quantities with different units, even though they can't be added together and so don't (obviously) lie in the same field (in the abstract algebra sense of the word). Mathematically formalizing elementary dimensional analysis is actually surprisingly nontrivial: Terry Tao has a nice treatment here.
The "quick and dirty" answer is that when you think of the formal mathematical/abstract-algebraic structure of the space of possible values for a physical quantity, you forget about units and treat everything as dimensionless. (E.g. you think of all three-vectors as living in $mathbbR^3$, regardless of their physical units.) Then afterward, you declare operations that don't make dimensional sense to be "physically invalid" even though they're perfectly legitimate from a formal mathematical perspective. "Hard-coding" the dimensional restrictions on the set of allowed operations directly into the formal mathematical structure is usually way more effort than it's worth.
answered 51 mins ago
tparker
21.5k143114
answered 51 mins ago
tparker
21.5k143114
answered 51 mins ago
tparker
21.5k143114
answered 51 mins ago
tparker
21.5k143114
21.5k143114
add a comment |Â
add a comment |Â
up vote
0
down vote
Thinking about this in the finite case, without dealing with infinitesimals, you have $Fcdot Delta x$. Both vectors in this inner product do live in the same space, namely $mathbbR^3$ in usual 3-d space, at least up to an isomorphism.
answered 1 hour ago
Hugo V
2868
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Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
Thinking about this in the finite case, without dealing with infinitesimals, you have $Fcdot Delta x$. Both vectors in this inner product do live in the same space, namely $mathbbR^3$ in usual 3-d space, at least up to an isomorphism.
answered 1 hour ago
Hugo V
2868
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Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
up vote
0
down vote
Thinking about this in the finite case, without dealing with infinitesimals, you have $Fcdot Delta x$. Both vectors in this inner product do live in the same space, namely $mathbbR^3$ in usual 3-d space, at least up to an isomorphism.
answered 1 hour ago
Hugo V
2868
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Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thinking about this in the finite case, without dealing with infinitesimals, you have $Fcdot Delta x$. Both vectors in this inner product do live in the same space, namely $mathbbR^3$ in usual 3-d space, at least up to an isomorphism.
answered 1 hour ago
Hugo V
2868
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Hugo V
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answered 1 hour ago
Hugo V
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answered 1 hour ago
Hugo V
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New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.
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If you consider $F$ and $dx$ as in different Vector spaces, you are entering in the region of Differential Geometry. Here the things are more complicated, since vectors are defined as the element of the tangent space defined in each point of the manifold, and in order to compare vectors in different points you have to define a rule (which is the parallel transport), which in turn is defined by a covariant derivative.
In your example though, $vecF=(F_x,F_y,F_z)$ and $dvecx=(dx,dy,dz)$ so that they can be thought of belonging to $mathbbR^3$, and the '$cdot$' (dot) product is just the usual one.
answered 53 mins ago
Kevin De Notariis
664
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Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
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If you consider $F$ and $dx$ as in different Vector spaces, you are entering in the region of Differential Geometry. Here the things are more complicated, since vectors are defined as the element of the tangent space defined in each point of the manifold, and in order to compare vectors in different points you have to define a rule (which is the parallel transport), which in turn is defined by a covariant derivative.
In your example though, $vecF=(F_x,F_y,F_z)$ and $dvecx=(dx,dy,dz)$ so that they can be thought of belonging to $mathbbR^3$, and the '$cdot$' (dot) product is just the usual one.
answered 53 mins ago
Kevin De Notariis
664
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Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
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up vote
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If you consider $F$ and $dx$ as in different Vector spaces, you are entering in the region of Differential Geometry. Here the things are more complicated, since vectors are defined as the element of the tangent space defined in each point of the manifold, and in order to compare vectors in different points you have to define a rule (which is the parallel transport), which in turn is defined by a covariant derivative.
In your example though, $vecF=(F_x,F_y,F_z)$ and $dvecx=(dx,dy,dz)$ so that they can be thought of belonging to $mathbbR^3$, and the '$cdot$' (dot) product is just the usual one.
answered 53 mins ago
Kevin De Notariis
664
New contributor
Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If you consider $F$ and $dx$ as in different Vector spaces, you are entering in the region of Differential Geometry. Here the things are more complicated, since vectors are defined as the element of the tangent space defined in each point of the manifold, and in order to compare vectors in different points you have to define a rule (which is the parallel transport), which in turn is defined by a covariant derivative.
In your example though, $vecF=(F_x,F_y,F_z)$ and $dvecx=(dx,dy,dz)$ so that they can be thought of belonging to $mathbbR^3$, and the '$cdot$' (dot) product is just the usual one.
answered 53 mins ago
Kevin De Notariis
664
New contributor
Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 53 mins ago
Kevin De Notariis
664
New contributor
Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 53 mins ago
Kevin De Notariis
664
answered 53 mins ago
Kevin De Notariis
664
664
New contributor
Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
0
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I assume you're referring to the fact that they belong to different vector spaces because they have different units. The difficulty you're running into is just an indication of an inconvenience in the set of definitions and foundations for mathematics that currently happens to be popular. To understand this, it may be helpful to look at the history.
Something like what we today define as a vector space was first defined by Peano in 1888.
Vectors, and including the words "vector" and "scalar," were defined by Gibbs around 1888, as a way of simplifying the quaternion system for his students at Yale.
Physicists' modern definition of a vector, which involves its transformation properties, was standardized ca. 1930-1950.
Set theory was cast into something like its present form by by ZFC in 1908-1922.
Keep in mind that all of these developments really occurred over the period of many years, not on specific dates, so what we really have is an overlapping set of time periods, with people working independently of one another and not necessarily producing consistent systems.
The basic notions that we really need in order to do linear algebra are algebraic, i.e., essentially syntactical facts, e.g., the property $(atextbfu)cdottextbfv=a(textbfucdottextbfv)$. When you look at an axiom like this one and then apply it, it makes no difference whether or not the objects $a$, $textbfu$, and $textbfv$ belong to certain sets. The notion of a set came later than the notion of a vector space. The fact that people today usually use ZFC as a foundation and define things like vector spaces in terms of operations on some set does not mean that it has to be done that way, was historically done that way, or is always convenient to do that way.
So for example, suppose an object starts at rest and is accelerated by a constant force $textbfF$, so that $textbfF=(2m/t^2)textbfx$. The identities that define the properties of the inner product have the same syntactical form, and therefore lead to the same results, regardless of the fact that it wouldn't make sense to talk about $textbfF+textbfx$. You can find the work done by this force, and while you're doing that, you can freely use the identity $(atextbfu)cdottextbfv=a(textbfucdottextbfv)$.
If you like, you can describe forces and displacements as different vector spaces with some machinery to connect them so that you can do dot products, or if you like, you can think of them as both belonging to some sort of space that has some forbidden operations, such as not being able to some additions. It doesn't matter which of these you do, and in practice nobody does anything like this formally.
Keep in mind also that the physicist's definition of a vector is more restrictive than a mathematician's, e.g., if you form an ordered pair consisting of $(S,q)$, where $S$ is the current value of the S&P 500 stock market index, and $q$ is an electric charge, then to a mathematician, this is a vector that lives in some vector space, but to a physicist this is not a vector at all, because it doesn't transform as a vector.
answered 18 mins ago
Ben Crowell
45.2k3147275
add a comment |Â
up vote
0
down vote
I assume you're referring to the fact that they belong to different vector spaces because they have different units. The difficulty you're running into is just an indication of an inconvenience in the set of definitions and foundations for mathematics that currently happens to be popular. To understand this, it may be helpful to look at the history.
Something like what we today define as a vector space was first defined by Peano in 1888.
Vectors, and including the words "vector" and "scalar," were defined by Gibbs around 1888, as a way of simplifying the quaternion system for his students at Yale.
Physicists' modern definition of a vector, which involves its transformation properties, was standardized ca. 1930-1950.
Set theory was cast into something like its present form by by ZFC in 1908-1922.
Keep in mind that all of these developments really occurred over the period of many years, not on specific dates, so what we really have is an overlapping set of time periods, with people working independently of one another and not necessarily producing consistent systems.
The basic notions that we really need in order to do linear algebra are algebraic, i.e., essentially syntactical facts, e.g., the property $(atextbfu)cdottextbfv=a(textbfucdottextbfv)$. When you look at an axiom like this one and then apply it, it makes no difference whether or not the objects $a$, $textbfu$, and $textbfv$ belong to certain sets. The notion of a set came later than the notion of a vector space. The fact that people today usually use ZFC as a foundation and define things like vector spaces in terms of operations on some set does not mean that it has to be done that way, was historically done that way, or is always convenient to do that way.
So for example, suppose an object starts at rest and is accelerated by a constant force $textbfF$, so that $textbfF=(2m/t^2)textbfx$. The identities that define the properties of the inner product have the same syntactical form, and therefore lead to the same results, regardless of the fact that it wouldn't make sense to talk about $textbfF+textbfx$. You can find the work done by this force, and while you're doing that, you can freely use the identity $(atextbfu)cdottextbfv=a(textbfucdottextbfv)$.
If you like, you can describe forces and displacements as different vector spaces with some machinery to connect them so that you can do dot products, or if you like, you can think of them as both belonging to some sort of space that has some forbidden operations, such as not being able to some additions. It doesn't matter which of these you do, and in practice nobody does anything like this formally.
Keep in mind also that the physicist's definition of a vector is more restrictive than a mathematician's, e.g., if you form an ordered pair consisting of $(S,q)$, where $S$ is the current value of the S&P 500 stock market index, and $q$ is an electric charge, then to a mathematician, this is a vector that lives in some vector space, but to a physicist this is not a vector at all, because it doesn't transform as a vector.
answered 18 mins ago
Ben Crowell
45.2k3147275
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I assume you're referring to the fact that they belong to different vector spaces because they have different units. The difficulty you're running into is just an indication of an inconvenience in the set of definitions and foundations for mathematics that currently happens to be popular. To understand this, it may be helpful to look at the history.
Something like what we today define as a vector space was first defined by Peano in 1888.
Vectors, and including the words "vector" and "scalar," were defined by Gibbs around 1888, as a way of simplifying the quaternion system for his students at Yale.
Physicists' modern definition of a vector, which involves its transformation properties, was standardized ca. 1930-1950.
Set theory was cast into something like its present form by by ZFC in 1908-1922.
Keep in mind that all of these developments really occurred over the period of many years, not on specific dates, so what we really have is an overlapping set of time periods, with people working independently of one another and not necessarily producing consistent systems.
The basic notions that we really need in order to do linear algebra are algebraic, i.e., essentially syntactical facts, e.g., the property $(atextbfu)cdottextbfv=a(textbfucdottextbfv)$. When you look at an axiom like this one and then apply it, it makes no difference whether or not the objects $a$, $textbfu$, and $textbfv$ belong to certain sets. The notion of a set came later than the notion of a vector space. The fact that people today usually use ZFC as a foundation and define things like vector spaces in terms of operations on some set does not mean that it has to be done that way, was historically done that way, or is always convenient to do that way.
So for example, suppose an object starts at rest and is accelerated by a constant force $textbfF$, so that $textbfF=(2m/t^2)textbfx$. The identities that define the properties of the inner product have the same syntactical form, and therefore lead to the same results, regardless of the fact that it wouldn't make sense to talk about $textbfF+textbfx$. You can find the work done by this force, and while you're doing that, you can freely use the identity $(atextbfu)cdottextbfv=a(textbfucdottextbfv)$.
If you like, you can describe forces and displacements as different vector spaces with some machinery to connect them so that you can do dot products, or if you like, you can think of them as both belonging to some sort of space that has some forbidden operations, such as not being able to some additions. It doesn't matter which of these you do, and in practice nobody does anything like this formally.
Keep in mind also that the physicist's definition of a vector is more restrictive than a mathematician's, e.g., if you form an ordered pair consisting of $(S,q)$, where $S$ is the current value of the S&P 500 stock market index, and $q$ is an electric charge, then to a mathematician, this is a vector that lives in some vector space, but to a physicist this is not a vector at all, because it doesn't transform as a vector.
answered 18 mins ago
Ben Crowell
45.2k3147275
I assume you're referring to the fact that they belong to different vector spaces because they have different units. The difficulty you're running into is just an indication of an inconvenience in the set of definitions and foundations for mathematics that currently happens to be popular. To understand this, it may be helpful to look at the history.
Something like what we today define as a vector space was first defined by Peano in 1888.
Vectors, and including the words "vector" and "scalar," were defined by Gibbs around 1888, as a way of simplifying the quaternion system for his students at Yale.
Physicists' modern definition of a vector, which involves its transformation properties, was standardized ca. 1930-1950.
Set theory was cast into something like its present form by by ZFC in 1908-1922.
Keep in mind that all of these developments really occurred over the period of many years, not on specific dates, so what we really have is an overlapping set of time periods, with people working independently of one another and not necessarily producing consistent systems.
The basic notions that we really need in order to do linear algebra are algebraic, i.e., essentially syntactical facts, e.g., the property $(atextbfu)cdottextbfv=a(textbfucdottextbfv)$. When you look at an axiom like this one and then apply it, it makes no difference whether or not the objects $a$, $textbfu$, and $textbfv$ belong to certain sets. The notion of a set came later than the notion of a vector space. The fact that people today usually use ZFC as a foundation and define things like vector spaces in terms of operations on some set does not mean that it has to be done that way, was historically done that way, or is always convenient to do that way.
So for example, suppose an object starts at rest and is accelerated by a constant force $textbfF$, so that $textbfF=(2m/t^2)textbfx$. The identities that define the properties of the inner product have the same syntactical form, and therefore lead to the same results, regardless of the fact that it wouldn't make sense to talk about $textbfF+textbfx$. You can find the work done by this force, and while you're doing that, you can freely use the identity $(atextbfu)cdottextbfv=a(textbfucdottextbfv)$.
If you like, you can describe forces and displacements as different vector spaces with some machinery to connect them so that you can do dot products, or if you like, you can think of them as both belonging to some sort of space that has some forbidden operations, such as not being able to some additions. It doesn't matter which of these you do, and in practice nobody does anything like this formally.
Keep in mind also that the physicist's definition of a vector is more restrictive than a mathematician's, e.g., if you form an ordered pair consisting of $(S,q)$, where $S$ is the current value of the S&P 500 stock market index, and $q$ is an electric charge, then to a mathematician, this is a vector that lives in some vector space, but to a physicist this is not a vector at all, because it doesn't transform as a vector.
answered 18 mins ago
Ben Crowell
45.2k3147275
edited 12 mins ago
answered 18 mins ago
Ben Crowell
45.2k3147275
answered 18 mins ago
Ben Crowell
45.2k3147275
answered 18 mins ago
Ben Crowell
45.2k3147275
45.2k3147275
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1
Don't both vectors live in $mathbbR^3$?
– ZeroTheHero
1 hour ago