What does having 3d symmetry 2d symmetry and 1d symmetry mean?

[Orodruin]

See also this: https://www.physicsforums.com/insights/symmetry-arguments-and-the-infinite-wire-with-a-current/

This is about the magnetic field, which is a pseudo vector instead of a vector, but it illustrates how symmetry arguments can be applied.

 

[haruspex]

Yes I got that, the z axis is the cylinder's axis, the y axis is one of the two axes in the cross section of the cylinder... View attachment 345757

The above image is the reference axes I took. In the image below the person has taken the cylinder's axis as y axis and the axis whose field im confused about as z axis.
This is an argument I received for why theres no field along the z axis (in this person's diagram)however I dont understand how its symmetrical thereView attachment 345758
As you say the field is everywhere normal to the cylinder's axis I agree with that but he says otherwise .View attachment 345759

##\vec E$is a function of position. The writer does not make it clear what position is being considered. It seems to be a point on the cylinder’s axis, in which case it is zero. If we consider a point$z\hat z$(in the writer's notation),$z\neq 0$, then the rotation through$\pi$about the x axis maps it to$-z\hat z$and flips the field to$-\vec E(z\hat z)=\vec E(-z\hat z)## (because of the symmetry).
Where's the contradiction?

Edit: I misunderstood the diagram… more below.

 

[Orodruin]

$\vec E$ is a function of position. The writer does not make it clear what position is being considered. It seems to be a point on the cylinder’s axis, in which case it is zero.

It is not on the axis, but $z = 0$. The axes are defined such that the $z$ direction is the tangential direction at the point being considered.

If we consider a point $z\hat z$ (in the writer's notation), $z\neq 0$, then the rotation through $\pi$ about the x axis maps it to $-z\hat z$ and flips the field to $-\vec E(z\hat z)=\vec E(-z\hat z)$ (because of the symmetry).
Where's the contradiction?

The only point being considered is a single point along the x-axis. This is sufficient since it can be mapped to any other point at the same radial distance using translation and rotation.

 

[haruspex]

Yes I got that, the z axis is the cylinder's axis, the y axis is one of the two axes in the cross section of the cylinder... View attachment 345757

The above image is the reference axes I took. In the image below the person has taken the cylinder's axis as y axis and the axis whose field im confused about as z axis.
This is an argument I received for why theres no field along the z axis (in this person's diagram)however I dont understand how its symmetrical thereView attachment 345758
As you say the field is everywhere normal to the cylinder's axis I agree with that but he says otherwise .

@Orodruin has explained to me that the point being considered is on the x axis. Consequently $\vec E_y=\vec E_z=0$ is consistent with the field being normal to the cylinder's axis. What made you think it was not?

 

[tellmesomething]

I mean, it is not very clearly argued. The actual argument is to consider the reflection symmetry in a plane through the central axis and the point you are considering the field at. Since the electric field is a proper vector, the component orthogonal to the reflection plane changes sign under the reflection. However, the charge distribution is the same before and after the reflection and therefore the field must also be the same. The component in the y-direction must therefore equal its own negative, which can only happen if it is zero.

Okay I see, a point on the y axis (in my diagram) would also have field $E_y$ a point the x and y plane would also have fields ($E_x$& $E_y$) but a point on the cylinder's axis would not have any net field.

 

[tellmesomething]

@Orodruin has explained to me that the point being considered is on the x axis. Consequently $\vec E_y=\vec E_z=0$ is consistent with the field being normal to the cylinder's axis. What made you think it was not?

The field is everywhere normal to the cylinder axis

The y axis is also normal to the cylinder's axis...
But I think I get it it also depends on where the reference point is...so

 

[haruspex]

The y axis is also normal to the cylinder's axis...
But I think I get it it also depends on where the reference point is...so

Yes, $\vec E(x\hat x)=f(x)\hat x$, so no $\hat y$ or $\hat z$ term.