'one form' vector?

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'one form' vector??

Hi
I am reading a book 'Relativity Demystified'
Please help with the following, I will appreciate any little help.

What is 'one form' vector??
In the book the author says that --> A vector V can be represented with covariant Va. This type of vector is called one form.

I also didnt understand anything about the word 'covariant'.....

Please explain and tell me if any other free resource is available on web to understand this in better way.

Abhishek Jain
 

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  • #2
pervect
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I'll try and give a brief explanation. If you want more info, the topic is covered briefly in linear algebra textbooks, and in more depth in tensor analysis.

Basically, a one-form is a map from a vector to a scalar. (A scalar is just a number, in this context a real number).

If a vector space has a dimensionality of 'n', the dual vector space (since a dual of a vector is a map of a vector to a scalar, a dual vector space is the abstract space defined by such maps) is also a vector space and has the same dimensionality of n as the original vector space. The interesting thing is that the dual of a dual of a vector space is equivalent to the original vector space. This much is mentioned in most linear algebra textbooks.

The existence of a metric, which gives vectors a length, ultimately defines the mapping of vectors to their duals via the relationship

[tex]u_a = \sum_{b=1..n} g_{ab} u^b[/tex]

where [itex]u_a[/itex] is a dual vector (also called a covariant vector, unless I've gotten it backwards, or a cotangent vector and [itex]u^a[/itex] is a vector (also called a contravariant vector and sometimes a tangent vector).

Pictorially, vectors are usually represented as little arrows with heads that indicate their directions. A graphical representation of a one-form used by some textbooks (most notably, MTW's textbook "Gravitation" ) is a stack of plates. The number of plates "pierced" by the arrow is the scalar defined by the composition of the one form and the vector. Remember that the one-form is a map from a vector to a scalar, thus if you compose a one-form and a vector, you get a scalar.
 
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  • #3
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A covariant tensor of rank 1, or a covariant vector, is often called a one-form. Other names are "bra" or "column vector". A "ket" or "row vector" is a contravariant vector, said to be "dual" to the one-form. The product of covariant and contravariant vectors is a scalar (see pervect, above).
 
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  • #4
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Just in case you are interested as to why they are called one forms: (if i remember correctly), an n-form is a fully antisymmetric covariant tensor of rank n.
 
  • #5
quasar987
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Hi
I am reading a book 'Relativity Demystified'
Please help with the following, I will appreciate any little help.

What is 'one form' vector??
In the book the author says that --> A vector V can be represented with covariant Va. This type of vector is called one form.

I also didnt understand anything about the word 'covariant'.....

Please explain and tell me if any other free resource is available on web to understand this in better way.

Abhishek Jain
Just maybe to give it to you as simple as possible: if you don't plan on going deep into the subject, maybe just knowing that the word "covariant vector" only means that we are talking about the vector whose components are written with the index on the bottom, like so: [itex]V_a[/itex], while "contravariant vectors" are written with the index on top, like so: [itex]V^a[/itex], and that the two are linked via

[tex]V_a = \sum_{b=1..n} g_{ab} V^b[/tex]

And in the case of special relativity, g_{ab} is a very specific kind of metric which makes it so that contravariant and covariant vectors in the context of special relativity are always related by

[tex]V_0=-V^0, \ \ \ V_1=V^1, \ \ \ V_2=V^2, \ \ \ V_3=V^3[/tex]
 
  • #6
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...maybe just knowing that the word "covariant vector" only means that we are talking about the vector whose components are written with the index on the bottom, like so: [itex]V_a[/itex], while "contravariant vectors" are written with the index on top, like so: [itex]V^a[/itex]...
It may be interesting to note that, historically, there has been an exception made for the space-time coordinates. The notation [itex]dx_i[/itex], with the index on the bottom, has often entered the equations as a contravariant tensor, just because that is the way the indices were traditionally written. Einstein wrote it this way in his original paper and apparently assumed that there would be no confusion. But it's best to write the contravariant coordinate tensor as [itex]dx^i[/itex].
 
  • #7
An example

I am also beginning to learn some of these things. Let me share a simple example that really helped me.

Imagine you are driving a car across a field, and your car is equipped with a thermometer. Your car is moving at a rate of 10 meters per second, and the temperature rises as you move at a rate of 0.1 degrees per meter (this is the gradient "vector" of the temperature). So, if you want to know how the measured temperature changes with respect to time (not distance) you can see that it is 1 degree per second.

Now let's say that you want to use kilometers instead of meters (this is a change of coordinates).

Your velocity becomes 0.01 kilometers per second and the gradient of temperature is then 100 degrees per kilometer. Then, you ask, how does the rate of change of temperature with respect to time change under the transformation from meters to kilometers? 0.01 kilometers/second * 100 degrees/kilometer = 1 degree per second! The quantity is invariant under the change.

So where do vectors and covectors/one forms come into this?

Your velocity is a contravariant vector (or just a vector) because it changes in a certain way under the transformation.

The gradient, on the other hand, is covariant, because it changes exactly by the inverse of the change in velocity.

And the rate of change of temperature with respect to time, which is a scalar, is not changed by the transformation.

In general, tensors can have both upper (contravariant) and lower (covariant) indicies. Vectors are tensors with one upper index, One Forms/Covectors are tensors with one lower index. Scalars have no indices at all.

The only problem with this example I think is that time is a very bad parameter when you're talking about spacetime.
 
  • #8
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Reading the Same Book -- Errata Help?

Hi,

I'm reading the same book. There are a lot of things that the author
just covers and I don't find sufficent information. Also, so far I've found
a number of mistakes. So now when I come to something I confused on
I don't know if its a mistake or I'm not getting something or both.

Do you happen to have any errata sheet? and/or any tips.

Thanks,
Josh
 
  • #9
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Hi Jainabhs and all respondents to this thread. I have nothing to add but my confusion and was not sure whether starting another thread was appropriate for my problem.

My problem is related to this thread. I understand the difference between vectors and one forms and am familiar with Pervects explanation of one forms and the fact that one forms act on vectors ( and vice versa ) to give a scalar. All this I have read in many textbooks and web sites.

A vector is an object which we can describe by using co-variant or contravariant components depending on which set of non orthogonal axes we use to read co-ordinates from and that in normal cartesian coordinates there is no difference between the two. I understand how to generate the reciprocal axes or basis vectors required. However the vector remains a vector no matter how we describe it. I have no problem with one forms and their visualization and with Pervect's description of them. I just cannot relate a certain description of a vector ( co or contra ) with a one form. I know it is I who have the problem and this is a genuine request for enlightenment. No matter how many times I read the many texts on the subject the problem will not go away.

My mind is stuck on the fact that a vector is a vector and a one form is a one form and if it is the dual of a vector why a one form too cannot be co or contra variant. I need to understand as i know the difference is important in many descriptions of the physical world.

Help !!! Mateinste.
 
  • #10
cristo
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I just cannot relate a certain description of a vector ( co or contra ) with a one form.
A covariant vector is a one form. They are two different names for the same thing.
 
  • #11
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Thank you Cristo. But this is exactly my problem. I cannot grasp how a different representation of a vector makes it a different object. Surely a one form and a vector behave differently.

Matheinst.
 
  • #12
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A covariant tensor of rank 1, or a covariant vector, is often called a one-form. Other names are "bra" or "column vector". A "ket" or "row vector" is a contravariant vector, said to be "dual" to the one-form. The product of covariant and contravariant vectors is a scalar (see pervect, above).
When one uses the term "bra" then one is (usually?) working in quantum mechanics. I've never seen the term "bra" used in relativity.

Pete
 
  • #13
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Good Book on Tensor Math

Can someone recommend a good book on Tensors
and Tensor Math?

Thanks,
Josh
 
  • #14
Hurkyl
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A vector is an object which we can describe by using co-variant or contravariant components
You cannot pick and choose! If you're dealing with a tangent vector, you have to use contravariant components. If you're dealing with a one-form, you have to use covariant components. And if you're working with a higher tensor, you have to use the "type" of components appropriate for its rank.

Although the collection of all tensors of a particular rank form a vector space, and thus it would be appropriate to call any tensor a "vector", the language of differential geometry is specialized to better suit its needs -- AFAIK the word "vector" is usually reserved specifically for tangent vectors nowadays.
 
  • #15
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Thank you Hurky. Point taken. I think you have helped. I think maybe another mistake I was making was not realising that a covariant vector is only one example of a one form which by definition is anything that acts on a vector and returns a scalar.

Thanks again Matheinste.
 
  • #16
robphy
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Thank you Hurky. Point taken. I think you have helped. I think maybe another mistake I was making was not realising that a covariant vector is only one example of a one form which by definition is anything that acts on a vector and returns a scalar.

Thanks again Matheinste.
(emphasis mine)

A one-form acts LINEARLY on a vector and returns a scalar.
 
  • #17
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Hello Robphy. Missing out linearity was a slip on my part.

Thanks.

Matheinste.
 
  • #18
robphy
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Thank you Cristo. But this is exactly my problem. I cannot grasp how a different representation of a vector makes it a different object. Surely a one form and a vector behave differently.

Matheinst.
One only has a "different representation of a vector" [in the context suggested] when there is additional structure around, like a metric tensor. With such a structure, one can take (say) a vector [represented by an arrow] and then obtain its metric-dual covector (a.k.a. one-form) [represented by a family of planes]. Technically, the vector and its metric-dual covector are different objects.... indeed, one cannot add a vector to a covector.. or have a [co]vector act linearly alone on another [co]vector.
 
  • #19
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Hello robphy. you say
---""With such a structure, one can take (say) a vector [represented by an arrow] and then obtain its metric-dual covector (a.k.a. one-form) [represented by a family of planes].""--

That is what I cannot visualise, a vector as an arrow becoming a one form as a sheet of planes. I have no doubt as to the correctness of what you say. All textbooks of course say the same. My mind is unable to grasp it. When I do see it then it will appear obvious.

Matheinste.
 
  • #20
robphy
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Hello robphy. you say
---""With such a structure, one can take (say) a vector [represented by an arrow] and then obtain its metric-dual covector (a.k.a. one-form) [represented by a family of planes].""--

That is what I cannot visualise, a vector as an arrow becoming a one form as a sheet of planes. I have no doubt as to the correctness of what you say. All textbooks of course say the same. My mind is unable to grasp it. When I do see it then it will appear obvious.

Matheinste.
There is a geometric construction for this due to William Burke (Applied Differential Geometry), which I believe is originally due to Jan Schouten (Ricci Calculus, 1924). I am actually going to several physics conferences this summer to present ideas for visualizing tensors. When I have some time later, I'll demonstrate the construction. [It'll be good practice for me to address questions that may arise.]
 
  • #21
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Hello robphy. I look forward to your constructions. Meanwhile I will try to get hold of your recommended books. Thanks a lot.

Matheinste.
 
  • #22
nrqed
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Thank you Cristo. But this is exactly my problem. I cannot grasp how a different representation of a vector makes it a different object. Surely a one form and a vector behave differently.

Matheinst.
I think it's important to realize that a one-form and a vector are completely different things. Using the terminology "covariant and contravariant vectors" confuses things since it gives the impression that one is talking about the smae thing expressed in a different form. They are just different things. But to add to the confusion, a metric can be used to associate to each one-form a vector and vice-versa. But again, it's better to think if the two as being completely distinct entities.
 
  • #23
robphy
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Consider a [for simplicity, 2-dimensional] vector space.
Without a metric tensor, there is no notion of the magnitude of a vector. In particular, there is no way to determine what the unit vectors are [and where their tips are]. In addition, there is notion of orthogonality.

We introduce a [for simplicity] Euclidean metric tensor g, represented by a circle. This curve represents all of the tips of the unit vectors [according to this choice of metric]. That is, g(u,u)=1.

Given a vector Va, we can lower its tensor-index and obtain its metric-dual covector Vb=gab Va, which is represented by a family of parallel planes... as follows.

Through the tip of Va, construct the tangents to the circle. These determine a chord of the circle [generally, a hyperplane]. Through the origin, draw the diameter parallel to this chord. This pair of lines [hyperplanes] represents the metric-dual covector, Vb=gab Va.

With a little geometry, you can show that, in this figure, the spacing between the covector planes is inversely proportional to the length of the vector. In particular, if you draw in the family of planes, the number of plane-piercings by that vector [not counting the one at the origin] is the square of the length of the vector... that is, VbVb=Vb gab Va.

In fact, you can also show that the dot-product operation is captured by this construction. In particular, vectors on the hyperplane through the center are orthogonal to the vector Va.

An important feature of this construction is that it is invertible. Given the covector, you can construct its metric-dual vector.

Some additional steps in the construction are needed to handle some cases that arise. For the non-Euclidean case, the ideas generalize. Details will be in a paper I'm working on.

I'm working on a program to visualize these and other tensorial operations in VPython and Maple. My goals are to visualize tensors and various tensorial-operations in space and in spacetime. That's what I'll be presenting at various conferences this summer.
 

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  • #24
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Interesting. I have one issue:

Through the tip of Va, construct the tangents to the circle.
I initially thought of a vector v such that g(v,v) < 1. Struggling to find any tangents to the circle that passed through the tip, I referred to the diagram to find that you have assumed g(v,v) > 1. Is there a similar construction that works for v of length less than 1, or is there something more obvious that I have missed completely?
 
  • #25
robphy
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Interesting. I have one issue:



I initially thought of a vector v such that g(v,v) < 1. Struggling to find any tangents to the circle that passed through the tip, I referred to the diagram to find that you have assumed g(v,v) > 1. Is there a similar construction that works for v of length less than 1, or is there something more obvious that I have missed completely?
Yes, a vector within the circle requires a little more work.
I don't remember the exact construction right now.
The whole scheme is based on the notion of the "the pole" and "the polar" with respect to a conic section.
 

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