I How Do Dual Vectors Differ From Regular Vectors in Physics?

[pixel]

I've been looking at various online sources for relativity and have some confusion about "dual vectors." I'm hoping for some very basic information/examples from physics, not abstract mathematical concepts from the field of vector spaces.

1. In addition to the vector/dual vector distinction, there is also the distinction between contravariant/covariant forms of vectors. I believe covariant vectors and dual vectors transform the same way. Are they essentially the same thing?

2. Someone gave the electric field as an example of a dual vector because it is found as the gradient of the potential field and transforms as a covariant vector. One colloquial defining property of dual vectors is that they "eat vectors and produce a scalar." An example is -E_μdx^μ = work done. But one can take any two regular vectors and form the sum F^μG^μ which would also be a scalar. So how are dual vectors different in this regard?

 

[robphy]

Possibly useful:
https://www.physicsforums.com/threa...ion-and-operations-of-gr.968301/#post-6149583

To tie in some physics, it is helpful to think of the electric field 1-form [dual vector, covector]
as a map from displacements to voltages. The units of electric field are "Volts/meter".
While an electrostatic field---as the gradient of a scalar potential---is a good example,
the electric field need not be electrostatic and still retain its identity as a 1-form.
(In Gauss's law, the electric field is best thought of as a two-form.)

Similarly, a [conservative] force maps displacements to work-done and thus has units of "Joules/meter".
Momentum can be thought of as something that maps displacements to "actions" and this has units \rm J\cdot s/m
Up to unit conventions, the magnetic field in Ampere's law can be thought of as something that maps displacements (along a closed curve) to currents (enclosed).
The wave-vector \vec k is better thought of as the gradient of a phase, which maps displacements to phase-changes and thus has units of \rm rad/m.

 

[pervect]

Abstractly, vectors are "vector spaces", and represent anything that can be multiplied by numbers (scalars), and added together. The additive operator must commute, so $\vec{a}+\vec{b}=\vec{b}+\vec{a}$, the scalar multiplication must be distributive, so that $a(\vec{b}+\vec{c}) = a\vec{b} + a\vec{c}$, and the whole space has to be closed.

The duality principle says that a linear map from a vector (space) to a scalar is another vector space of the same dimension as the original vector space. It's called the "dual space". The duality principle also says that the dual of a dual space is the same as the original space. While the dual space has the same dimension as the original space, but it's not the "same space".

Proofs of this can be found in most linear algebra textbooks, but it's all rather abstract and I've long since forgotten the details.

Covariant and contravariant vectors are duals of one another. The concept also comes up in pre-tensor analysis as "row vectors" and "column vectors".

Because dual vectors are maps from vectors to scalars, the pairing of a dual vector (a map from a vector to a scalar) and a vector gives a scalar. This is called the "natural pariing".

An application to the previously mentioned row and column vectors. If you have a row vector and a column vector, the matrix product of the row vector with the column vector gives you a scalar. This is all inear, so this makes a row vector a linear map from a column vector to a scalar, and hence the row vector is the dual of the column vector. The reverse is also true.

Graphically, my favorite technique is to represent vectors by little lines with arrows, and dual vectors as "stacks of plates". The number of plates that the arrow "pierces" is the scalar assocaited with the natural pairing of the dual vector and the vector. This is used in textbooks such as MTW's "Gravitation", which is a GR textbook.I

 

[geordief]

Can I ask a question in this thread as I have been looking into this also over the past few weeks,..?

Does the distinction between contravariant and covariant vectors only apply when the base vectors are not orthogonal (ie are they identical otherwise?)

If the bases are not orthogonal what happens when a covariant vector is added to a contravariant vector?

Is the result the same irrespective of the order of the operation?

 

[Orodruin]

Does the distinction between contravariant and covariant vectors only apply when the base vectors are not orthogonal (ie are they identical otherwise?

Covariant and contravariant vectors generally belong do different vector spaces. Only when you have a non-degenerate 2-form is there a connection between them.

If the bases are not orthogonal what happens when a covariant vector is added to a contravariant vector?

You cannot add a covariant and a contravariant vector. They belong to different vector spaces.

 

[geordief]

Covariant and contravariant vectors generally belong do different vector spaces. Only when you have a non-degenerate 2-form is there a connection between them.You cannot add a covariant and a contravariant vector. They belong to different vector spaces.

Thanks,you are helpful. Back to the drawing board

 

[stevendaryl]

I've been looking at various online sources for relativity and have some confusion about "dual vectors." I'm hoping for some very basic information/examples from physics, not abstract mathematical concepts from the field of vector spaces.

1. In addition to the vector/dual vector distinction, there is also the distinction between contravariant/covariant forms of vectors. I believe covariant vectors and dual vectors transform the same way. Are they essentially the same thing?

That's exactly the way to think about dual vectors: They are functions that "eat vectors and produce a scalar". If you have a "dot" product, then you can convert any vector into a dual vector:

If $u$ is a vector, then you define a corresponding function $f_u$ that acts on vectors this way:

$f_u(v) = u \cdot v$

So the covariant form of a contravariant vector always implicitly involves the dot product, which is more properly called the "metric tensor". If $u$ and $v$ are vectors, and $g$ is the metric tensor, then $u \cdot v = g(u,v)$. In terms of components, $g(u,v) = \sum_{i j} g_{ij} u^i v^j$

2. Someone gave the electric field as an example of a dual vector because it is found as the gradient of the potential field and transforms as a covariant vector. One colloquial defining property of dual vectors is that they "eat vectors and produce a scalar." An example is -E_μdx^μ = work done. But one can take any two regular vectors and form the sum F^μG^μ which would also be a scalar. So how are dual vectors different in this regard?

No, you cannot take two vectors $F^\mu$ and $G^\mu$ (by convention, raised indices indicate vectors and lowered indices indicate dual vectors) and form the sum $\sum_\mu F^\mu G^\mu$. The latter sum is not a meaningful quantity. What is meaningful, as I said, is $\sum_{\mu \nu} g_{\mu \nu} F^\mu G^\nu$. You need the metric tensor $g$ to convert a vector into a dual vector.

 

[Orodruin]

If you have a "dot" product, then you can convert any vector into a dual vector

It is not necessary that this is an inner product. All that is required is a non-degenerate 2-form. For example, in symplectic geometry you use a symplectic form rather than a metric. This is used in Hamiltonian mechanics to relate Hamiltonian vector fields to the exterior derivative of the Hamiltonian function.

 

[geordief]

If you have a "dot" product, then you can convert any vector into a dual vector

So one can convert any vector into a covector (and vice versa?) but the two vectors (in their different spaces) cannot act on each other directly ?(indirectly?)

 

[Orodruin]

So one can convert any vector into a covector (and vice versa?) but the two vectors (in their different spaces) cannot act on each other directly ?(indirectly?)

What do you mean by "act on"? By definition a dual vector acts on a tangent vector to produce a scalar. What you cannot do is to add a tangent vector and a dual vector.

 

[geordief]

What do you mean by "act on"?

By "act on" I had in mind "add to" .

Still ,if one converts a covector to a vector as stevendaryl suggested ,is it then possible to add this vector to one of the vectors in the covectors' dual space (the original vector space) ?

Can you "switch" between the vector space and its dual space in this way?

 

[stevendaryl]

So one can convert any vector into a covector (and vice versa?) but the two vectors (in their different spaces) cannot act on each other directly ?(indirectly?)

Right, two vectors (or two dual vectors) can only act on each other if you have a metric.

I think it's worthwhile to think about what it's like to have a space without a metric. That to me illustrates the importance of metrics and the importance of distinguishing vectors from dual vectors.

Here's a simple example of such a space, from thermodynamics. If you have a quantity of a gas, then the state of that gas can be characterized by two thermodynamic variables: $V$ (volume) and $T$ (temperature) (or we could have chosen volume and pressure, or some other combination of quantities). The state is then a point on a two-dimensional space. In this two-dimensional space, we can define two different notions of vector-like quantities.

Tangent vectors:

If the state $S$ of the system is changing as a function of time, then you can represent the rate of change by a vector: $\overrightarrow{v} = \frac{dS}{dt}$. This vector has two components: $v^V = \frac{dV}{dt}$ and $v^T = \frac{dT}{dt}$. Note that the units of the two components are different. $v^V$ has units of "volume per unit time" and $v^T$ has units of "temperature per unit time". It would be completely meaningless to take the "dot" product of $\overrightarrow{v}$ with itself: $\overrightarrow{v} \cdot \overrightarrow{v} =$ some combination of terms of the form temperature-squared per time-squared and volume-squared per time-squared. That quantity doesn't have consistent units.

Dual vectors:

We can construct a different kind of vector, as well. If we have a function such as $P$ (pressure) that depends on both $V$ and $T$, then we can construct a different kind of vector that characterizes how $P$ changes with "location" in the state space: $\overrightarrow{\omega} = \nabla P$, with components $\omega_V = \frac{\partial P}{\partial V}$ and $\omega_T = \frac{\partial P}{\partial T}$. Note again that the components do not have the same units, so it makes no sense to take the dot product of such a vector with itself.

However, it does make sense to take the dot-product of a vector and a dual-vector:

$\nabla P \cdot \frac{dS}{dt} = \frac{\partial P}{\partial V} \frac{dV}{dt} + \frac{\partial P}{\partial T} \frac{dT}{dt}$

This combination does have consistent units: pressure per unit time. It gives the time rate of change of $P$ as the state changes.

Note: I probably shouldn't use the same symbol, $\cdot$ to mean the action of a dual vector on a vector and to mean the combination of two vectors.

 

[geordief]

So a metric is a mathematical object that allows one to work with two recalcitrant quantities ,like a kind of bridge?

It comes with a bit of extra information that slots into either side of the question and let's them "talk to each other mathematically"?

 

[stevendaryl]

So a metric is a mathematical object that allows one to work with two recalcitrant quantities ,like a kind of bridge?

It comes with a bit of extra information that slots into either side of the question and let's them "talk to each other mathematically"?

Vectors and dual vectors talk to each other just fine. It's two vectors that can't talk without help.

You can think of it like the male and female ends of a garden hose. You can't connect two male ends or two female ends without some kind of adapter.

 

[Orodruin]

Vectors and dual vectors talk to each other just fine. It's two vectors that can't talk without help.

This depends what you mean by "talk to". With two vectors in the same vector space you can always add them (by definition of "vector space"). What you cannot do is map them linearly to scalars, for this you need a vector and a dual vector.

 

[pixel]

I think it's worthwhile to think about what it's like to have a space without a metric. That to me illustrates the importance of metrics and the importance of distinguishing vectors from dual vectors.

What is the metric in your example?

 

[geordief]

Vectors and dual vectors talk to each other just fine. It's two vectors that can't talk without help.

You can think of it like the male and female ends of a garden hose. You can't connect two male ends or two female ends without some kind of adapter.

Thanks . I feel like I have learned quite a bit.

Gooseberries can be useful
(https://dictionary.cambridge.org/dictionary/english/play-gooseberry)
:wink:

 

[Orodruin]

What is the metric in your example?

The entire point was that he was looking at a space without a metric where there is no direct way to relate tangent to dual vectors.

 

[fresh_42]

The entire point was that he was looking at a space without a metric where there is no direct way to relate tangent to dual vectors.

I don't understand this. A dual vector is an element of $V^*$ the vector space of linear functionals of $V$. They automatically relate to vectors via $(v^*,u)\longmapsto v^*(u)$. There is no need for any metric, and they are still directly related.

 

[Nugatory]

I don't understand this. A dual vector is an element of $V^*$ the vector space of linear functionals of $V$. They automatically relate to vectors via $(v^*,u)\longmapsto v^*(u)$. There is no need for any metric, and they are still directly related.

The problem here is the imprecision in that word "relate"... Without a metric there's no way of raising and lowering indices, no dual vector $V_\mu$ that we can say is "related to" the vector $V^\mu$.

 

[Orodruin]

I don't understand this. A dual vector is an element of $V^*$ the vector space of linear functionals of $V$. They automatically relate to vectors via $(v^*,u)\longmapsto v^*(u)$. There is no need for any metric, and they are still directly related.

Relating here does not imply ”mapping a vector and dual vector to a scalar”. It implies a bijection between a vector space and its dual.

 

[George Jones]

Relating here does not imply ”mapping a vector and dual vector to a scalar”. It implies a bijection between a vector space and its dual.

Natural, basis-independent linear bijection.

Since, for finite-dimensional $V$, $V$ and $V^*$ have the same dimension, it is always possible to construct a linear bijection between $V$ and $V^*$, even without a "metric".

 

[pixel]

The entire point was that he was looking at a space without a metric where there is no direct way to relate tangent to dual vectors.

I asked because he gave an example of a vector and a dual vector but I think I understand what you are saying is that the latter was not the dual of the former.

To put things in concrete terms, can you give an example of a vector representing a physical quantity and its corresponding dual vector, where the latter was derived by applying the metric to the former?

 

[fresh_42]

The problem here is the imprecision in that word "relate"... Without a metric there's no way of raising and lowering indices, no dual vector $V_\mu$ that we can say is "related to" the vector $V^\mu$.

Even this makes me wonder.
Let's say we have $\varphi_1\in V^*-\{\,0\,\}$. Then there is a vector $v_1\in V-\{\,0\,\}$ such that $\varphi(v_1)=1$. Now we can split $V=\mathbb{F}\cdot v_1 \oplus \operatorname{ker}\varphi_1$ and continue with $V_2:= \operatorname{ker}\varphi_1$. Where is the problem?

 

[stevendaryl]

What is the metric in your example?

You included the following quote from me:

I think it's worthwhile to think about what it's like to have a space without a metric.

The $V,T$ space is a space without a metric.

 

[stevendaryl]

I asked because he gave an example of a vector and a dual vector but I think I understand what you are saying is that the latter was not the dual of the former.

The two types of vectors ("velocity" vectors with components $\frac{dV}{dt}$ and $\frac{dT}{dt}$ and "gradient" vectors with components $\frac{\partial P}{\partial V}$ and $\frac{\partial P}{\partial T}$) are duals of each other. But there is no metric, so you can't convert the one type of vector into the other type.

To put things in concrete terms, can you give an example of a vector representing a physical quantity and its corresponding dual vector, where the latter was derived by applying the metric to the former?

Any time you compute the "length" of a vector, that's what you're doing. You have a velocity vector $\overrightarrow{v}$. When you write $|v| = \sqrt{\overrightarrow{v} \cdot \overrightarrow{v}}$ you're using the metric to convert one of the $\overrightarrow{v}$ into a dual vector.

$|v| = \sqrt{\overrightarrow{v} \cdot \overrightarrow{v}} = \sqrt{g_{\mu \nu} v^\mu v^\nu}$

The quantity $g_{\mu \nu} v^\mu$ is the dual vector $v_\nu$.

You might be asking what's a natural interpretation of $v_\nu$ as a dual vector. Dual vectors are most naturally created as a kind of "gradient" of a scalar function: $\omega_\mu = \frac{\partial \phi}{\partial x^\mu}$. You can certainly reverse engineer a scalar function $\phi$ such that $v_\mu$ are the components of its gradient, but I guess that's not very interesting.

 

[stevendaryl]

Even this makes me wonder.
Let's say we have $\varphi_1\in V^*-\{\,0\,\}$. Then there is a vector $v_1\in V-\{\,0\,\}$ such that $\varphi(v_1)=1$. Now we can split $V=\mathbb{F}\cdot v_1 \oplus \operatorname{ker}\varphi_1$ and continue with $V_2:= \operatorname{ker}\varphi_1$. Where is the problem?

Let's consider my example. We have a 2-D state space for the thermodynamic state of a quantity of a gas, which can be characterized by the coordinate system $V,T$ (volume, temperature). We can use the rate of change of the state as an example tangent vector with components $v^V = \frac{dV}{dt}$ and $v^T = \frac{dT}{dt}$. The point is that there is no meaningful notion of $v \cdot v$. You can certainly make up a way to make sense of that, but it's not likely to be physically meaningful.

 

[fresh_42]

Let's consider my example. We have a 2-D state space for the thermodynamic state of a quantity of a gas, which can be characterized by the coordinate system $V,T$ (volume, temperature). We can use the rate of change of the state as an example tangent vector with components $v^V = \frac{dV}{dt}$ and $v^T = \frac{dT}{dt}$. The point is that there is no meaningful notion of $v \cdot v$. You can certainly make up a way to make sense of that, but it's not likely to be physically meaningful.

Yes, but there is a major difference between physically meaningful and the requirement of a metric or even impossible. The algorithm I mentioned allows a unique identification $\varphi_i \longleftrightarrow v_i$ of basis vectors with the property $\varphi_j(v_i)=\delta_{ij}$. No metric anywhere near, only fundamental properties of the category of vector spaces.

 

[George Jones]

Even this makes me wonder.
Let's say we have $\varphi_1\in V^*-\{\,0\,\}$. Then there is a vector $v_1\in V-\{\,0\,\}$ such that $\varphi(v_1)=1$. Now we can split $V=\mathbb{F}\cdot v_1 \oplus \operatorname{ker}\varphi_1$ and continue with $V_2:= \operatorname{ker}\varphi_1$. Where is the problem?

Requiring that the "relation" between $V$ and $V^*$ be a vector space isomorphism (linear bijection) would mean that
$$\psi_1 \sim v \Rightarrow c \psi_1 \sim c v$$
for $c\neq0$ in $\mathbb{F}$.

Now, let $\phi_1 = c\psi_1$, and run the procedure again. When this is done, $\phi_1 = c \psi_1 \sim \frac{1}{c} v$. Consequently, this procedure does not lead to a natural isomorphism.

 

[fresh_42]

Requiring that the "relation" between $V$ and $V^*$ be a vector space isomorphism (linear bijection) would mean that
$$\psi_1 \sim v \Rightarrow c \psi_1 \sim c v$$
for $c\neq0$ in $\mathbb{F}$.

Now, let $\phi_1 = c\psi_1$, and run the procedure again. When this is done, $\phi_1 = c \psi_1 \sim \frac{1}{c} v$. Consequently, this procedure does not lead to a natural isomorphism.

I do not understand this .

I agree that the isomorphism isn't natural, because certain correspondences are deliberately chosen, but whether I have $\varphi_i(v_j)=\delta_{ij}$ or $\langle \varphi_i , v_j \rangle =\delta_{ij}$ doesn't make a difference. If you stretch both, vector and covector with the same factor, you will get a squared factor anyway: $\begin{bmatrix}2&0\end{bmatrix}\cdot\begin{bmatrix}2\\0\end{bmatrix}=4$, even in the ordinary plane.