Not all reflections in 2D are 3D rotations?

Some reflections in the plane can be represented by a rotation in three dimensions, and some cannot: e.g., reflections across the x or y axes can. but a 2D reflection across the line x=y cannot. Thus the question in the summary.

I probably don't understand the question, but whatever reflection you take, you can represent it in 3D as a rotation about the line of reflection at 180 degrees.

nomadreid said:

TL;DR Summary: Is there a criteria to find out whether a given reflection in a real Euclidean plane (in matrix form) can be represented by a 3D rotation?

Some reflections in the plane can be represented by a rotation in three dimensions, and some cannot: e.g., reflections across the x or y axes can. but a 2D reflection across the line x=y cannot. Thus the question in the summary.

I'm not going to try to answer your question, but ask you one instead, so that the question can be clarified a bit:

Why wouldn't you be able to perform a rotation in 3D to represent a 2D reflection over the line y = x? If you choose your axis of rotation to be the line y = x, the rotation works just fine.

Could you please be more explicit about what rotations/reflections you are referring to?

-Dan

Thanks, martinbn and topsquark. Sorry that I wasn't too clear. The idea is that a matrix R for reflection in three dimensions would be

for some k. To be a rotation matrix I would need R^TR = Id and the determinant to be 1, so k=-1.
Using the method outlined to find the axis around which one could rotate for non-symmetric matrices in https://en.wikipedia.org/wiki/Rotation_matrix

I get that u is the 0 vector, which doesn't seem too promising. So I concluded that in order to make the necessary 180 degree rotation around the x=y, I would need to transform the axes and so forth, which does not boil down to a pure rotation. From your replies, the above reasoning is flawed, and I would enjoy finding out where. Thanks.
PS PhysicsForums used to give a chance to view one's post before posting it, but that options seems to have disappeared?

nomadreid said:

Thanks, martinbn and topsquark. Sorry that I wasn't too clear. The idea is that a matrix R for reflection in three dimensions would be
View attachment 332376
for some k. To be a rotation matrix I would need R^TR = Id and the determinant to be 1, so k=-1.
Using the method outlined to find the axis around which one could rotate for non-symmetric matrices in https://en.wikipedia.org/wiki/Rotation_matrix
View attachment 332377
I get that u is the 0 vector, which doesn't seem too promising. So I concluded that in order to make the necessary 180 degree rotation around the x=y, I would need to transform the axes and so forth, which does not boil down to a pure rotation. From your replies, the above reasoning is flawed, and I would enjoy finding out where. Thanks.
PS PhysicsForums used to give a chance to view one's post before posting it, but that options seems to have disappeared?

I would argue that this is still a pure rotation, but no problem.

Okay, so if I understand what you are saying (please verify) what you are asking is:

How do I know if we may represent a reflection in the xy plane with a rotation in 3D about either the x or y axis as the axis of rotation?

The answer is that only a reflection over the x or y axis can be represented as such a rotation.

So the reflection of u = (a, b, 0) over the yz plane is:
##S_{yz} u = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} a \\ b \\ 0 \end{bmatrix} = \begin{bmatrix} -a \\ b \\ 0 \end{bmatrix}##

A rotation of u = (a, b, 0) over the y axis by ##\pi## rad is:
##R_y u = \begin{bmatrix} cos(\pi) & 0 & sin(\pi) \\ 0 & 1 & 0\\ -sin(\pi) & 0 & cos(\pi) \end{bmatrix} \cdot \begin{bmatrix} a \\ b \\ 0 \end{bmatrix} = \begin{bmatrix} -a \\ b \\ 0 \end{bmatrix}##

so they are the same. You can (and probably already have) verify the reflection over the xz plane and rotation over the x axis correlation.

You can use the same argument to show that you can't do this with any other reflection line, if you are only considering rotating over the x or y axes. (But, of course, if you are willing to use the general rotation matrix to rotate about a line that is not the x nor y axis, then you can show that there is still a correlation.)

Of course, none of this works if u has a non-zero z-component...

-Dan

Addendum: In the upper right corner you will see an icon that looks like a small piece of paper with a magnifying glass over it. That's the "preview" button. If you are still having troubles with this, it's probably because you have to refresh the page in order to see the first LaTeX code in the thread. Now that some has been posted, that problem should go away.

Thanks for the quick reply, Dan.
"Okay, so if I understand what you are saying (please verify) what you are asking is:

How do I know if we may represent a reflection in the xy plane with a rotation in 3D about either the x or y axis as the axis of rotation?

No, that is not what I am asking. The procedure for finding the axis and angle of a given 3x3 rotation matrix ---- over an axis other than the x, y, or z axes ---- is given in the Wikipedia page cited; for example, this gives clear answers for the result of any pair of two successive different 90 degree airplane rotations (out of pitch, roll and yaw) (e.g., 120 degree rotation over (1,1,1), etc.). But for the matrix given for reflection over x=y, I get no answer (even though intuitively it should be a 180 rotation over (1,1,0).

nomadreid said:

Thanks for the quick reply, Dan.
"Okay, so if I understand what you are saying (please verify) what you are asking is:

No, that is not what I am asking. The procedure for finding the axis and angle of a given 3x3 rotation matrix ---- over an axis other than the x, y, or z axes ---- is given in the Wikipedia page cited; for example, this gives clear answers for the result of any pair of two successive different 90 degree airplane rotations (out of pitch, roll and yaw) (e.g., 120 degree rotation over (1,1,1), etc.). But for the matrix given for reflection over x=y, I get no answer (even though intuitively it should be a 180 rotation over (1,1,0).

Sorry, I'm probably still not answering your question, but I'm also not certain which method on the page you are using.

Note, however, that a rotation about any axis in the xy plane can be represented as ##R_z( \theta ) R_x( \pi ) R_z( -\theta )##, where ##\theta## is the angle between the axis of rotation and the +x axis.

Hopefully that helps!

-Dan

Thanks, Dan.

topsquark said:

but I'm also not certain which method on the page you are using.

In the Wikipedia page cited, scroll down to the section

Conversion from rotation matrix to axis–angle​

​

topsquark said:

a rotation about any axis in the xy plane can be represented as , where is the angle between the axis of rotation and the +x axis.

Yes, I got those matrices (where theta is pi/4), but I was looking for a single rotation around a single axis (obviously other than x,y, or z). As far as I can see, it doesn't exist. For which matrices a single rotation does not exist, and which matrices can be represented by a single rotation matrix , is something that I will have to keep digging to find out. So I will leave it at that in this forum, with my thanks for the responses. At least I found out where the review button was

nomadreid said:

Thanks, Dan.

In the Wikipedia page cited, scroll down to the section

Conversion from rotation matrix to axis–angle​

​

Yes, I got those matrices (where theta is pi/4), but I was looking for a single rotation around a single axis (obviously other than x,y, or z). As far as I can see, it doesn't exist. For which matrices a single rotation does not exist, and which matrices can be represented by a single rotation matrix , is something that I will have to keep digging to find out. So I will leave it at that in this forum, with my thanks for the responses. At least I found out where the review button was

But there is one:
##R_{y=x}(\pi) = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} = R_z(\pi/4) R_x(\pi) R_x(-\pi/4)##

It's a rotation matrix, it just isn't ##R_x##, ##R_y##, or ##R_z##.

-Dan

Thanks, Dan. (The delay in my reply is due to differing time zones.)

Yes, that is the same matrix as I noted above. However, I have found my problem. . I had lamented that the method for finding the axis given in the cited Wikipedia article did not give me the results despite the fact that the matrix was not symmetrical. Alas, I didn’t notice another condition for the method to work was that R^T-R had to be non-zero, which is not the case with this matrix. So, my assumption that this method should work in this case was the killer.

Going to an alternate (albeit slightly more tedious but applicable) method in the same section of that article: that any 3D rotation matrix around an axis (u_x,u_y,u_z) at an angle θ should look like this:

which gives the matrix you listed when θ=π and (u_x,u_y,u_z)=(1/√2 , 1/√2, 0), corresponding to the original intuition.

Hence all is well, and on one side I red-facedly regret the bother which I have caused, and on the other side I am grateful for the help in getting me to find my error.