Regarding Contravarient Vector Transformations

  • #1
Hello,

I have a question regarding the contravarient transformation of vectors.

So the formula:

V'n = dx'n / dxm Vm

So in words, the nth basis vector in the ' frame of reference over the mth (where m is the summation term) basis vector in the original frame of reference times the mth coordinate of the V vector in the original frame of reference.

My question is, how do you determine the basis vectors in the original frame of reference? Arent they just (1 0 0), (010) and (001)?
 

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  • #2
PeroK
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The ##x^m## and ##x'^n## are coordinates, not basis vectors.
 
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  • #3
The ##x^m## and ##x'^n## are coordinates, not basis vectors.
Ah ok. But if I'm trying to convert a Vector from one reference frame to another, how can I know the component of the vector in the new reference frame. Isnt that what I'm calculating? And what is the difference between V^m and x^m ?
 
  • #4
PeroK
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Ah ok. But if I'm trying to convert a Vector from one reference frame to another, how can I know the component of the vector in the new reference frame. Isnt that what I'm calculating? And what is the difference between V^m and x^m ?
The formula in your original post gives you the components of a vector ##V'^n## in one coordinate system (using the ##x'## coordinates) in terms of the components of the vector in the other coordinate system (using the ##x## coordinates).

Note that talking about the components of the vector in a coordinate system implies that you have a mechanism for defining a set of basis vectors (at each point) based on that coordinate system. This is called a coordinate basis.
 
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  • #5
The formula in your original post gives you the components of a vector ##V'^n## in one coordinate system (using the ##x'## coordinates) in terms of the components of the vector in the other coordinate system (using the ##x## coordinates).

Note that talking about the components of the vector in a coordinate system implies that you have a mechanism for defining a set of basis vectors (at each point) based on that coordinate system. This is called a coordinate basis.
Okay. Sorry, but this is all very new to me.

So lets assume that I have a vector ##V## in the ## x^m ## frame of reference and it's components are:

## \left( 1, 2, 3 \right) ##

With the basis vector ## \hat{i} , \hat{j} ,\hat{k} ## one can write ## V = 1 * \hat{i}, 2 * \hat{j}, 3 * \hat{k} ##

And lets say that I have a second coordinate system with coordinates ## \bar{x^n} ## and that the new basis vectors are

## \hat{i}' = 2 * \hat{i}, \hat{j}' = 2 * \hat{j} , \hat{k}' = 2 * \hat{k} ##

Since this is a contravarient transformation and the Vector itself (direction and magnitude) will stay the same, only the components will change. So what was ## V = \left(1, 2, 3\right) ## will change.

This much is clear to me (correct me if I was wrong).

Now, the formula

## \bar {V^n} = \frac{\partial \bar{x^n}}{\partial x^m} V^m ##

Now I know that I would insert 1, 2 and 3 in for ## V^n ## (1 when n = 1, 2 when n = 2 and 3 when n = 3) but what isn't clear is what one would insert for ## \partial \bar{x^n} ## and ## \partial x^m ## . You said that ## \partial \bar{x^n} ## and ## \partial x^m ## is a coordinate but I dont understand what of. Can you complete the example with the values above?

Thank you very much :D
 
  • #6
PeroK
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Doesn't your textbook cover this?

In the case of a simple linear transformation between Cartesian coordinate systems, the coefficients represented by the partial derivatives are constants.

A good example is to use polar coordinates for the second system. Then the basis vectors depend on position and the components of ##V'## depend on the point at which the vector is located.

You should study that as a good non trivial example of this transformation formula.
 
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  • #7
Doesn't your textbook cover this?

In the case of a simple linear transformation between Cartesian coordinate systems, the coefficients represented by the partial derivatives are constants.

A good example is to use polar coordinates for the second system. Then the basis vectors depend on position and the components of ##V'## depend on the point at which the vector is located.

You should study that as a good non trivial example of this transformation formula.
I'm trying to learn this using the internet, so sadly no textbook. But thanks for your explanations, I think I have a better understanding now.
 
  • #8
@PeroK ok, so after scratching my head for a couple of hours I think I get it now. Is this correct?

Suppose we have a coordinate system ## x^n = (x^1, x^2, x^3) ## and another ## \bar{x^m} = (x^1 * sin(x^2) * cos(x^3), x^1 * sin(x^2) * sin(x^3), x^1 * cos(x^3)) ## .

As you can see, ## \bar{x^m} ## is polar coordinates. If I have the vector ## V^n = \left(1, 2, 3\right)## and I want to calculate ## \bar{V^1} ## , then I would do the following:
## \bar{V^1} = \frac{\partial \bar{x^1}}{\partial x^1} V^1 + \frac{\partial \bar{x^1}}{\partial x^2} V^2 + \frac{\partial \bar{x^1}}{\partial x^3} V^3 ##
And inserting what I think are the correct vales you would get:

## \bar{V^1} = \frac{\partial x^1 * sin(x^2) * cos(x^3)}{\partial x^1} *1 + \frac{\partial x^1 * sin(x^2) * sin(x^3)}{\partial x^2} *2 + \frac{\partial x^1 * cos(x^3)}{\partial x^3} *3 ##

## \bar{V^1} = sin(x^2)cos(x^3) + 2(x^1sin(x^3)cos(x^2)) - 3(x^1sin(x^3)) ##

Is this the correct understanding? This is what I think you meant by #x^n and \bar{x^m} ## being coordinates.

Thank you for your help :D
 
  • #9
PeroK
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@PeroK ok, so after scratching my head for a couple of hours I think I get it now. Is this correct?
You've got the idea, but I think you need to find a textbook. There are free ones online these days if you look around.

If you are studying this for GR, then Carroll's GR notes have all this mathematics and are here:

https://arxiv.org/pdf/gr-qc/9712019.pdf
 
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  • #10
You've got the idea, but I think you need to find a textbook. There are free ones online these days if you look around.

If you are studying this for GR, then Carroll's GR notes have all this mathematics and are here:

https://arxiv.org/pdf/gr-qc/9712019.pdf
Okay, thanks a lot for your help on this. I bought a textbook on QFT and this is one of the things I need to know. I'm 16 so buying the textbook was a bit ambitious looking back but I'm still going to give it a shot. Thanks for the link btw. Have a nice day :D
 
  • #11
PeroK
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Okay, thanks a lot for your help on this. I bought a textbook on QFT and this is one of the things I need to know. I'm 16 so buying the textbook was a bit ambitious looking back but I'm still going to give it a shot. Thanks for the link btw. Have a nice day :D
You need to be careful that QFT is so advanced that trying to learn it could be counter-productive. And, to be honest, unless you already know SR, QM, Classical Electromagnetism and probably GR and a good bit of particle physics as well, there is no possibility of making sense of QFT.
 
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  • #12
@PeroK just out of curiosity, what is this specific formula called? I'd like to learn more and I couldn't find exactly what I was looking for.
 
  • #13
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@PeroK just out of curiosity, what is this specific formula called? I'd like to learn more and I couldn't find exactly what I was looking for.
It's just the transformation rule for vector components. In general, a vector may be defined as something that satisfies that transformation rule.
 
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  • #14
It's just the transformation rule for vector components. In general, a vector may be defined as something that satisfies that transformation rule.
Okay, thanks
 
  • #15
Ibix
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@PeroK just out of curiosity, what is this specific formula called? I'd like to learn more and I couldn't find exactly what I was looking for.
The matrix whose ##\mu\nu## element is ##\partial x'^\mu/\partial x^\nu## (note they should be partial derivatives not total derivatives) is called the Jacobian matrix, which you can search for.
 
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  • #16
Let's say I have coordinate system ## x^n = (x^1, x^2) ## and another ## x'^m = (2x^1, 2x^2) ## . I know this isn't the best example. Let's say I have a vector ## V=(1,2) ## Now according to the formula:
## V'^m = /frac{/partial x'^m}{/partial x^n} V^n ##

Now if I'm looking for the first component of ## V'^m ## and I insert ## 2x^1 ## for ## /partial x'^m ## and ## x^1 ## for ## /partial x^n ## then I get 2, don't I? So I would multiply 1 by 2 to get 2 even though I should get 0,5. What am I doing wrong?
 
  • #17
PeroK
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Let's say I have coordinate system ## x^n = (x^1, x^2) ## and another ## x'^m = (2x^1, 2x^2) ## .
That's not a coordinate transformation. A coordinate transformation would take the form: $$x_1 = 2x'_1, \ x_2 = 2x'_2$$ You are confusing coordinates with components again.
 
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  • #18
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PS again use polar coordinates as your example. We write: $$x = r\cos \theta, \ y = r \sin \theta$$
 
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  • #20
Ibix
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Let's say I have coordinate system ## x^n = (x^1, x^2) ## and another ## x'^m = (2x^1, 2x^2) ## . I know this isn't the best example.
What you seem to be attempting to define here is two coordinate systems, both simple Cartesian grids but one twice the size of the other. If so, the way you are writing it is completely wrong.

The idea is that any point can be identified by coordinates in either system. What you need to do is think of a point and write down its coordinates in both systems. Then another point and another and another - and then you have to figure out a general rule for how the ##x^1## coordinate of a point is related to the ##x'^1## and ##x'^2## coordinates of the same point in the other system, and how the ##x^2## coordinate of the point relates to its primed coordinates. In your case this is easy: ##x^1=2x'^1## and ##x^2=2x'^2## (or possibly divided by two depending which system you intended to have the larger grid). In the case @PeroK is urging you to consider in #18 the relationships are more complex but you can see that what his maths does is take ##r,\theta## coordinates and expresses the same point in Cartesian coordinates.
Let's say I have a vector ## V=(1,2) ##
You can't have vectors yet. First you need to specify your basis vectors. There is a lot of formalism to this that I'm not going to try to explain, but will be a bit sloppy. What I presume you have done is defined a unit vector ##\mathbf{i}## that points in the direction of increasing ##x^1## coordinate and a unit vector ##\mathbf{j}## that points in the direction of increasing ##x^2## coordinate, and you intend ##V=(1,2)## to mean that there is a vector at some point that can be written as ##\mathbf{i}+2\mathbf{j}##.

Note that this technique of using basis vectors that point along the direction of the coordinate grid is often extremely convenient (it's called a coordinate basis), but it is not the only way of picking basis vectors and other systems are used!
Now according to the formula:
## V'^m = \frac{\partial x'^m}{\partial x^n} V^n ##
You have your slashes the wrong way round in the LaTeX, by the way - you need \, not /. I fixed them in the quote.

We've written down the relationships between the two coordinate systems so you should now be able to write the partial derivatives. Then you can apply your formula to your vector whose components are (2,1) in the unprimed coordinate basis to calculate its components in the primed coordinate basis. You should also be able to tackle PeroK's polar coordinate example.
 
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  • #21
What you seem to be attempting to define here is two coordinate systems, both simple Cartesian grids but one twice the size of the other. If so, the way you are writing it is completely wrong.

The idea is that any point can be identified by coordinates in either system. What you need to do is think of a point and write down its coordinates in both systems. Then another point and another and another - and then you have to figure out a general rule for how the ##x^1## coordinate of a point is related to the ##x'^1## and ##x'^2## coordinates of the same point in the other system, and how the ##x^2## coordinate of the point relates to its primed coordinates. In your case this is easy: ##x^1=2x'^1## and ##x^2=2x'^2## (or possibly divided by two depending which system you intended to have the larger grid). In the case @PeroK is urging you to consider in #18 the relationships are more complex but you can see that what his maths does is take ##r,\theta## coordinates and expresses the same point in Cartesian coordinates.

You can't have vectors yet. First you need to specify your basis vectors. There is a lot of formalism to this that I'm not going to try to explain, but will be a bit sloppy. What I presume you have done is defined a unit vector ##\mathbf{i}## that points in the direction of increasing ##x^1## coordinate and a unit vector ##\mathbf{j}## that points in the direction of increasing ##x^2## coordinate, and you intend ##V=(1,2)## to mean that there is a vector at some point that can be written as ##\mathbf{j}+2\mathbf{j}##.

Note that this technique of using basis vectors that point along the direction of the coordinate grid is often extremely convenient (it's called a coordinate basis), but it is not the only way of picking basis vectors and other systems are used!

You have your slashes the wrong way round in the LaTeX, by the way - you need \, not /. I fixed them in the quote.

We've written down the relationships between the two coordinate systems so you should now be able to write the partial derivatives. Then you can apply your formula to your vector whose components are (2,1) in the unprimed coordinate basis to calculate its components in the primed coordinate basis. You should also be able to tackle PeroK's polar coordinate example.
Thank you so much this really cleared things up.
 
  • #22
Okay, now I think I understand this. I just have one question.

Above I wrote ## \bar{x}^1 = 2x^1, \bar{x}^2 = 2x^2 ##

Now if I use the formula for calculating the components of the vector ## V(1, 2) ## (which uses ## x^1 ## and ## x^2 ## coordinates) in ## \bar{x}^1 ## and ## \bar{x}^2 ## coordinates I get:

## \bar{V}^1 = \frac{\partial \bar{x^1}}{\partial x^1}V^1 + \frac{\partial \bar{x^1}}{\partial x^2}V^2 ##
which equals
## \bar{V}^1 = \frac{\partial 2x^1}{\partial x^1} + \frac{\partial 2x^1}{\partial x^2}*2 ##
which equals
## \bar{V}^1 = 2 + 0 = 2 ##
But common sense says that ## \bar{V}^1 ## should be 0,5.

I understand how it should work in theory, I just don't know why the formula isn't working. I watched a video on the Jacobian and they do the partial derivatives the same way.

UPDATE:

I tried this formula on polar coordinates and I'm also stuck.

## \bar{x}^1 = x^1cos(x^2), \bar{x^2} = x^1sin(x^2) ##

Here ## x^1 ## is r and ## x^2 ## is ## \theta ##

Now to get ## \bar{V}^1 ##

## \bar{V^1} = \frac{\partial x^1cos(x^2)}{\partial x^1} *V^1+ \frac{\partial x^1sin(x^2)}{\partial x^2}*V^2 ##

which equals
## \bar{V^1} = cos(x^2) *V^1+ x^1cos(x^2)*V^2 ##
Now that I have this, I don't know what to insert for x^2 and x^1 since they aren't numbers. Sorry if this is a stupid question and thank you for your help so far.
 
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  • #23
Ibix
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Above I wrote ## \bar{x}^1 = 2x^1, \bar{x}^2 = 2x^2 ##

.....

## \bar{V}^1 = 2 + 0 = 2 ##
But common sense says that ## \bar{V}^1 ## should be 0,5.
What does ##\bar{x}^1=2x^1## tell you about the size of the grid squares in the barred coordinates versus unbarred?
Now that I have this, I don't know what to insert for x^2 and x^1 since they aren't numbers.
Nothing. I'd tend to replace ##x^1## with ##r## and ##x^2## with ##\theta## for readability, but it's not necessary. You've got an answer, at least once you transform the other component.

Sanity check: what does a unit vector pointing radially outwards look like in the ##r,\theta## coordinate basis? What does it look like in the Cartesian coordinate basis? Does your transform turn the first into the second?
 
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  • #24
What does ##\bar{x}^1=2x^1## tell you about the size of the grid squares in the barred coordinates versus unbarred?
The ## \bar{x^1} ## grid squares are twice as big as the ## x^1 ## grid squares so the ## \bar{V^1} ## component should be half as much to compensate for that (contravarient), right?
Nothing. I'd tend to replace ##x^1## with ##r## and ##x^2## with ##\theta## for readability, but it's not necessary. You've got an answer, at least once you transform the other component.
Okay. I found online that you get a transformation matrix as an answer but how do I use that to get the components of the vector if I have no value for r and ## \theta ## ?
Sanity check: what does a unit vector pointing radially outwards look like in the ##r,\theta## coordinate basis? What does it look like in the Cartesian coordinate basis? Does your transform turn the first into the second?
What exactly do you mean by this?

Sorry, I'm a slow learner.
 
  • #25
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The ## \bar{x^1} ## grid squares are twice as big as the ## x^1 ## grid squares
No. When ##x^1=1## then ##\bar{x}^1=2##, so there are two barred grid squares per unbarred square (or consider 10 instead of 2 and think about cm and mm squares). That means a vector expressed in the barred basis needs components twice as large - so your 2 is correct.
Okay. I found online that you get a transformation matrix as an answer but how do I use that to get the components of the vector if I have no value for ##r## and ##\theta##?
You've got the general answer for any ##r## and ##\theta##. You don't need to put numbers in until someone asks you about a vector at a particular point, and then they'll tell you the ##r## and ##\theta## they're interested in.
What exactly do you mean by this?
A sanity check is a simple test you can do to check that your answer isn't totally crazy. So if there's a vector you can write down in both coordinates and bases then you can try your transform to see if it works.

Do you know what are the components of a vector pointing away from the origin in polar coordinates? What about Cartesian coordinates? What happens if you apply your matrix to the polar coordinates version of the vector.
 
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