Coordinate Rotation: Find Transformation Matrix for 120° Rotation

[AndresPB]

Homework Statement

[/B]Hello, I am seeking help solving the following problem: find the transformation matrix that rotates a rectangular coordinate system through an angle of 120° about an axis making equal angles with the original three coordinate axes.

Homework Equations

none, we need to find a Matrix

The Attempt at a Solution

I've thought about it but my problem is trying to figure out how a 120° rotation would be, wouldn't the answer be different if I choose a different axis (one rotation around x₁, another matrix if I choose x₂, a different one if I choose x₃) each time? Thanks.

 

[stevendaryl]

[deleted]
This is all wrong. Sorry.

 

[AndresPB]

So the answer would be (0,0,1;1,0,0;0,1,0) and in another case (0,1,0;0,0,1;1,0,0)? Each ; is another row in a matrix

 

[stevendaryl]

So the answer would be (0,0,1;1,0,0;0,1,0) and in another case (0,1,0;0,0,1;1,0,0)? Each ; is another row in a matrix

I haven't worked out the answer, but you're looking for a single square matrix that works for all vectors. So I'm not sure what you mean by "another case". What cases are you talking about?

Just write down the three equations for transforming \vec{A} = (A_x, A_y, A_z) into \vec{A'} = (A'_x, A'_y, A'_z):

A'_x = ...
A'_y = ...
A'_z = ...

where the "..." is some expression involving A_x, A_y, and A_z. Then you can convert those three equations into a single matrix equation.

 

[AndresPB]

I haven't worked out the answer, but you're looking for a single square matrix that works for all vectors. So I'm not sure what you mean by "another case". What cases are you talking about?

Just write down the three equations for transforming \vec{A} = (A_x, A_y, A_z) into \vec{A'} = (A'_x, A'_y, A'_z):

A'_x = ...
A'_y = ...
A'_z = ...

where the "..." is some expression involving A_x, A_y, and A_z. Then you can convert those three equations into a single matrix equation.

I mean that depending on the axis I'm rotating the solution will be diferent, I don't need a generalization, as the problem says we need to rotate 120 degrees through and axis and there is an specific solution so that the angles are equals from the original. So if i choose to rotate x2 the matrix will be different if I choose to rotate around x3? I am getting a bit confussed, the problem is taken from cap 1 exercise 9 from Marion Thorton - Classical Mechanics, thanks a lot

 

[haruspex]

So the answer would be (0,0,1;1,0,0;0,1,0) and in another case (0,1,0;0,0,1;1,0,0)? Each ; is another row in a matrix

Yes, if you assume the 'equal angles' constraint refers to the angle to the positive half axis in each instance (x, y, z).

 

[AndresPB]

Yes, if you assume the 'equal angles' constraint refers to the angle to the positive half axis in each instance (x, y, z).

Thanks a lot, can you assume the equal angeles constraints refers to the angle to the negative half axis? or can you achieve another solution?

 

[haruspex]

Thanks a lot, can you assume the equal angeles constraints refers to the angle to the negative half axis? or can you achieve another solution?

If it is consistently the positive half or consistently the negative half then it will be the result you found. Otherwise there will be three other possibilities.

 

[Charles Link]

I think I have a solution for you. Normally, a 3-D rotation about an arbitrary axis can be somewhat complicated, but this axis makes equal angles with the three coordinate axes. (The axis of rotation is along (1,1,1)). There is a 3-fold axis of symmetry for this diagonal of the cube so that a 120 degree rotation about this axis will convert (1,0,0) to (0,1,0). It will also convert (0,1,0) to (0,0,1) and it will convert (0,0,1) to (1,0,0). (Each of these are column vectors.) (Notice each vector itself only rotates 90 degrees for the 120 degree rotation, but that is because they are not perpendicular to the axis of rotation.) We also must have (1,1,1) rotates into (1,1,1). (It stays the same.) I think you will find the rotation matrix you need is 0 0 1 in the first row, 1 0 0 in the second row, and 0 1 0 in the third row. (I don't know how to do matrices in Latex). This will convert any arbitrary x y z to an x' y' z' for the 120 degree rotation about the (1,1,1) axis.

 

[haruspex]

I think I have a solution for you. Normally, a 3-D rotation about an arbitrary axis can be somewhat complicated, but this axis makes equal angles with the three coordinate axes. (The axis of rotation is along (1,1,1)). There is a 3-fold axis of symmetry for this diagonal of the cube so that a 120 degree rotation about this axis will convert (1,0,0) to (0,1,0). It will also convert (0,1,0) to (0,0,1) and it will convert (0,0,1) to (1,0,0). (Each of these are column vectors.) (Notice each vector itself only rotates 90 degrees for the 120 degree rotation, but that is because they are not perpendicular to the axis of rotation.) We also must have (1,1,1) rotates into (1,1,1). (It stays the same.) I think you will find the rotation matrix you need is 0 0 1 in the first row, 1 0 0 in the second row, and 0 1 0 in the third row. (I don't know how to do matrices in Latex). This will convert any arbitrary x y z to an x' y' z' for the 120 degree rotation about the (1,1,1) axis.

This is what Andres already found in post #3, and I corroborated in post #6. That's lucky, because otherwise your post would have violated the homework forum rule of not posting complete solutions.

 

[AndresPB]

I think I have a solution for you. Normally, a 3-D rotation about an arbitrary axis can be somewhat complicated, but this axis makes equal angles with the three coordinate axes. (The axis of rotation is along (1,1,1)). There is a 3-fold axis of symmetry for this diagonal of the cube so that a 120 degree rotation about this axis will convert (1,0,0) to (0,1,0). It will also convert (0,1,0) to (0,0,1) and it will convert (0,0,1) to (1,0,0). (Each of these are column vectors.) (Notice each vector itself only rotates 90 degrees for the 120 degree rotation, but that is because they are not perpendicular to the axis of rotation.) We also must have (1,1,1) rotates into (1,1,1). (It stays the same.) I think you will find the rotation matrix you need is 0 0 1 in the first row, 1 0 0 in the second row, and 0 1 0 in the third row. (I don't know how to do matrices in Latex). This will convert any arbitrary x y z to an x' y' z' for the 120 degree rotation about the (1,1,1) axis.

thanks a lot, very good explanation

 

[Charles Link]

This is what Andres already found in post #3, and I corroborated in post #6. That's lucky, because otherwise your post would have violated the homework forum rule of not posting complete solutions.

Thank you. I'm new to this section of the website. I just now read through the rules.

 

[stevendaryl]

Yes, it depends on the axis, of course.

The simplest case of rotation is rotation within the x-y plane through an angle \theta. That means rotation about the z-axis. A vector with components (x,y) is transformed into a vector with components (x', y') where

x' = x cos(\theta) - y sin(\theta)
y' = y cos(\theta) + x sin(\theta)

So the question is, how to generalize that to the case of rotating an arbitrary vector \vec{A} about an arbitrary axis of rotation \hat{B} (\hat{B} should be a unit vector to make it simpler). The way I know how to do it is to use cross-products:

\vec{A'} = \vec{A} cos(\theta) + (\vec{A} \times \hat{B}) sin(\theta)

This is actually WRONG. It only works if \vec{A} is perpendicular to \hat{B}.

 

[stevendaryl]

This is actually WRONG. It only works if \vec{A} is perpendicular to \hat{B}.

The general case is sort of a mess, but this is what I think it is:

\vec{A'} = (\vec{A} \cdot \hat{B}) \hat{B} (1 - cos(\theta)) + \vec{A} cos(\theta) - (\vec{A} \times \hat{B}) sin(\theta)

where \vec{A'} is the result of rotation \vec{A} about axis \hat{B} through an angle \theta.

 

[Charles Link]

Most often, I have seen 3-D rotations as successive rotations about the different (and rotated) coordinate axes. I have also seen rotations about an arbitrary axis in an advanced Quantum Mechanics course and there the solution was something like $R= exp(i*\theta*n*J*2*\pi/h)$ where "n" is a unit vector along the axis of rotation and J is the angular momentum operator (which can be a matrix), and $\theta$ is the angle of rotation. This one turned out to be somewhat simple, but I think the general case is somewhat complex. There are no doubt other algebraic solutions to the general case besides the Q.M. solution that we were shown in the advanced course.

 

[AndresPB]

Thank you all for your help, if it is not to much trouble can someone model a cube rotating 120° degrees throughout (1,1,1)? is there a program for doing this? Also do we need to consider if we rotate clockwise or not clockwise? Is there a convention when someone says "rotate through an axis"?

 

[Charles Link]

An interesting question you are asking about a clockwise rotation of 120 degrees. This should be the same as a counterclockwise rotation of 240 degrees or two 120 degree counterclockwise rotations. Squaring the R matrix for a 120 counterclockwise rotation does give the 120 degree clockwise matrix, and just as a check, three 120 degrees rotations of either clockwise or counterclockwise rotations does indeed give us the identity matrix.

 

[stevendaryl]

An interesting question you are asking about a clockwise rotation of 120 degrees. This should be the same as a counterclockwise rotation of 240 degrees or two 120 degree counterclockwise rotations. Squaring the R matrix for a 120 counterclockwise rotation does give the 120 degree clockwise matrix, and just as a check, three 120 degrees rotations of either clockwise or counterclockwise rotations does indeed give us the identity matrix.

The usual convention is counter-clockwise, or the right-hand rule. The right hand rule says: Open your right hand. Point the thumb of your right hand in the direction of the axis. Now close your right hand. Your fingers are moving in the direction of increasing \theta.