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

  • #1
tellmesomething
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Homework Statement
Q1) Why did he make the cylinder's height infinite and the thick slab's plane infinite...?

Q2)I dont understand what he meant by them having dimensional symmetry...symmetry of what? electric fields? and how did he get that it will 1, 2 or 3..Can someone explain this...?

Context below..
Relevant Equations
None
I derived an expression for the electric field due to solid uniformly charged non conducting spherical volume to be
$$ \frac{ Qz} {4π\epsilon R^3 } $$ where z is the distance of the point from the center and it is less than the radius R I.e the point lies inside the sphere...

This in terms of volume charge density can also be written as $$ \frac{\rho z } { 3 \epsilon } $$

A teacher whose video lessons i follow then gave us the expressions for the field at a point inside a cylinder whose height is infinite but cross sectional area is finite

$$ \frac{ \rho z } {2 \epsilon} $$
Where z is the distance from the line of symmetry ..


and the field at a point inside a thick slab whose length and thickness is infinite but breadth is finite...

$$ \frac{ \rho z } { \epsilon} $$
Where z is the distance from a plane of symmetry..

He then went on to explain that the sphere has 3d symmetry the cylinder has 2d symmetry and the slabs have 1d symmetry..and thats a way to remember these results...

Q1) Why did he make the cylinder's height infinite and the thick slab's plane infinite...?
Q2)I dont understand what he meant by them having dimensional symmetry...symmetry of what? electric fields? and how did he get that it will 1, 2 or 3..Can someone explain this...?
IMG_20240521_194029.jpg
 
Last edited:
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  • #2
tellmesomething said:
Q1) Why did he make the cylinder's height infinite and the thick slab's plane infinite...?
So that there is only parameter in each case.
tellmesomething said:
Q2)I dont understand what he meant by them having dimensional symmetry...symmetry of what? electric fields? and how did he get that it will 1, 2 or 3.
He means geometric symmetry of the object, but I cannot see a basis for that categorisation.
The sphere does have more symmetry than the other two since you can swap all the axes around, but I would not have thought the infinite cylinder had any more symmetry than an infinite slab. In both cases you can swap two axes.
It may be a reference to dimensionality of Lie groups, in which case I defer to anyone with some familiarity with those.
 
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  • #3
tellmesomething said:
Q1) Why did he make the cylinder's height infinite and the thick slab's plane infinite...?
Suppose the slab is very large but finite. For a given value of z, the field’s magnitude and direction will also depends on where you are (x,y) in the slab.

For example, the field near an edge or corner will be very different to the field near the centre, even for the same z.

That’s because the field at any point is the (vector) sum of the fields from all surrounding charges. The distribution of surrounding charges is very different for a point near the centre compared to a point near an edge or corner.

But if the slab is infinite, there are no corners or edges. You can travel in any (xy) direction and the distribution of charges around you never changes. So the field depends only on z.

As a result, the field can have no components in the x and y directions. The field acts only in the z direction and this makes its calculation simple (e.g. using Gauss’s law).

You should be able to apply a similar argument to the cylinder for yourself.

tellmesomething said:
Q2)I dont understand what he meant by them having dimensional symmetry...symmetry of what? electric fields? and how did he get that it will 1, 2 or 3..Can someone explain this...?
Don't like the terminology. Better to use the terms spherical symmetry, cylindrical symmetry and planar symmetry; these terms describe the charge distributions in your 3 cases.

EDIT: Aha. I see @haruspex had already replied.
 
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  • #4
First of all, none of these are fields without a specified direction.

tellmesomething said:
A teacher whose video lessons i follow
This is an unhelpful reference. It does not allow us to look at what you have seen, which is crucial to see what it actually says rather than your recollection of it.
 
  • #5
Steve4Physics said:
Suppose the slab is very large but finite. For a given value of z, the field’s magnitude and direction will also depends on where you are (x,y) in the slab.

For example, the field near an edge or corner will be very different to the field near the centre, even for the same z.

That’s because the field at any point is the (vector) sum of the fields from all surrounding charges. The distribution of surrounding charges is very different for a point near the centre compared to a point near an edge or corner.

But if the slab is infinite, there are no corners or edges. You can travel in any (xy) direction and the distribution of charges around you never changes. So the field depends only on z.

As a result, the field can have no components in the x and y directions. The field acts only in the z direction and this makes its calculation simple (e.g. using Gauss’s law).

You should be able to apply a similar argument to the cylinder for yourself.


Don't like the terminology. Better to use the terms spherical symmetry, cylindrical symmetry and planar symmetry; these terms describe the charge distributions in your 3 cases.

EDIT: Aha. I see @haruspex had already replied.
I dont think I understand..
You're saying that electric field at point p in the the case where the length breadth and width is finite would be dependent on all three of its coordinates.... But when its infinite it would only be dependent on the z axis?...whys that? Because its infinite would the field due to the y and x component cancel out throughout?
Screenshot_2024-05-21-17-20-28-611_com.miui.gallery.jpg
Screenshot_2024-05-21-17-20-19-370_com.miui.gallery.jpg
 
  • #6
haruspex said:
So that there is only parameter in each case.

He means geometric symmetry of the object, but I cannot see a basis for that categorisation.
The sphere does have more symmetry than the other two since you can swap all the axes around, but I would not have thought the infinite cylinder had any more symmetry than an infinite slab. In both cases you can swap two axes.
It may be a reference to dimensionality of Lie groups, in which case I defer to anyone with some familiarity with those.
Okay so symmetry only means how many axes you can swap to get the same end result?
 
  • #7
Orodruin said:
First of all, none of these are fields without a specified direction.


This is an unhelpful reference. It does not allow us to look at what you have seen, which is crucial to see what it actually says rather than your recollection of it.
I quoted him word to word...
 
  • #8
tellmesomething said:
Okay so symmetry only means how many axes you can swap to get the same end result?
It's not a matter of swapping axes. It's a matter of what @Steve4Physics mentioned in post #3, so I will repeat it but differently. Imagine being on a uniformly charged infinite plane. No matter in which direction you walk on this plane and how far from your starting point, you see the same distribution. Of course, if you move along the perpendicular to the plane, you will know it immediately because you can measure the distance between you and the plane and see that it has increased. There is a unique distinguishable direction, the plane perpendicular.

From this it follows that the electric field can only be in a direction perpendicular to the plane. If that were not the case, then there would be a component tangent to the plane and you could draw an arrow in the plane pointing in that direction. But that cannot be the case because the source of any electric field is a charge distribution which, in this case, is uniform and has no in-plane preferred direction.
 
  • #9
kuruman said:
It's not a matter of swapping axes. It's a matter of what @Steve4Physics mentioned in post #3, so I will repeat it but differently. Imagine being on a uniformly charged infinite plane. No matter in which direction you walk on this plane and how far from your starting point, you see the same distribution. Of course, if you move along the perpendicular to the plane, you will know it immediately because you can measure the distance between you and the plane and see that it has increased. There is a unique distinguishable direction, the plane perpendicular.

From this it follows that the electric field can only be in a direction perpendicular to the plane. If that were not the case, then there would be a component tangent to the plane and you could draw an arrow in the plane pointing in that direction. But that cannot be the case because the source of any electric field is a charge distribution which, in this case, is uniform and has no in-plane preferred direction.
So that just means if I integrate the small de fields acting on a point inside the slab (not on the surfaces) throughout the whole volume...id get only the z component as the net field... The x and the y components would be same on either side of this point... Hence they would cancel out?
 
  • #10
tellmesomething said:
So that just means if I integrate the small de fields acting on a point inside the slab (not on the surfaces) throughout the whole volume...id get only the z component as the net field... The x and the y components would be same on either side of this point... Hence they would cancel out?
It means more than that. There are no x and y components. That's what
kuruman said:
From this it follows that the electric field can only be in a direction perpendicular to the plane.
means. The x and y components are zero.
 
  • #11
tellmesomething said:
I quoted him word to word...
Sorry, but that is not good enough. If I had a penny for every instance of a user claiming to quote a source word by word just for it to turn out this was certainly not the case … I would not be rich, but I would have a lot of pennies …

Please do not quote. Give the actual reference.
 
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  • #12
Orodruin said:
Sorry, but that is not good enough. If I had a penny for every instance of a user claiming to quote a source word by word just for it to turn out this was certainly not the case … I would not be rich, but I would have a lot of pennies …

Please do not quote. Give the actual reference.
But its a video in an app Which doesnt allow screenshots or screen recordings...I could use another device to take a snapshot hmm uploading now..
 
  • #13
kuruman said:
It means more than that. There are no x and y components. That's what

means. The x and y components are zero.
But if I take an individual element in the slab and use that to find a small de field at a point x and y component wont be zero right... The total sum of the x component and the total sum of the y component would be zero respectively... I think
 
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  • #14
Orodruin said:
Sorry, but that is not good enough. If I had a penny for every instance of a user claiming to quote a source word by word just for it to turn out this was certainly not the case … I would not be rich, but I would have a lot of pennies …

Please do not quote. Give the actual reference.
Please check I have edited.... The little white mark is because my number was visible...learnt it the hard way that people on physics forums could also be concerned with prank calling numbers 😔
 
  • #15
tellmesomething said:
But if I take an individual element in the slab and use that to find a small de field at a point x and y component wont be zero right... The total sum of the x component and the total sum of the y component would be zero respectively... I think
That would be the case. The idea is to use symmetry to deduce that the integrals and hence the components are zero without actually having to integrate.
 
  • #16
kuruman said:
That would be the case. The idea is to use symmetry to deduce that the integrals and hence the components are zero without actually having to integrate.
This applies to a point inside as well as outside the slab?
 
  • #17
kuruman said:
That would be the case. The idea is to use symmetry to deduce that the integrals and hence the components are zero without actually having to integrate.
Also for the cylinder case... Since the z axis is infinitely long considering the height is along the z axis... The x and y axis are on the cross sectional area of the cylinder... Any point inside the cylinder's net field would only be in the x and y direction?
Screenshot_2024-05-21-20-13-18-097_com.miui.gallery.jpg

Net field could only be calculated in the Same plane then? For example I could not get any field from A on plane B which contains point P...
 
  • #18
tellmesomething said:
Okay so symmetry only means how many axes you can swap to get the same end result?
Very much not. A symmetry is any transformation that leaves the system invariant. This can include any sort of rotations, translations, reflections, or scalings (if we just look at the purely spatial symmetries). How a field transforms under such a symmetry (along with uniqueness of solutions) can severely restrict the possible form of the solutions.
 
  • #19
Orodruin said:
Very much not. A symmetry is any transformation that leaves the system invariant. This can include any sort of rotations, translations, reflections, or scalings (if we just look at the purely spatial symmetries). How a field transforms under such a symmetry (along with uniqueness of solutions) can severely restrict the possible form of the solutions.
I think i get it.....Can you tell me how the cylinder case can also be viewed...how do I rotate the cylinder about the z axis ...?
 
  • #20
Orodruin said:
Very much not. A symmetry is any transformation that leaves the system invariant. This can include any sort of rotations, translations, reflections, or scalings (if we just look at the purely spatial symmetries). How a field transforms under such a symmetry (along with uniqueness of solutions) can severely restrict the possible form of the solutions.
If its symmetrical about an axis then that is Not a solution?
 
  • #21
tellmesomething said:
Okay so symmetry only means how many axes you can swap to get the same end result?
I did not mean to imply that. I was merely looking for a way to explain the teacher's thinking.
If swapping axes leaves the equations of the shape unchanged then that is an example of a symmetry, but there can be others.
 
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  • #22
haruspex said:
I did not mean to imply that. I was merely looking for a way to explain the teacher's thinking.
If swapping axes leaves the equations of the shape unchanged then that is an example of a symmetry, but there can be others.
I see... In the case of cylinders theres this simplification which most standard textbooks use and they assume the charge of the cylinder to lie on an infinitely long rod...how do they get that?
 
  • #23
tellmesomething said:
I see... In the case of cylinders theres this simplification which most standard textbooks use and they assume the charge of the cylinder to lie on an infinitely long rod...how do they get that?
Also for the case of the cylinder equations appear to be symmetrical only about the axis of height...the plane in cross sectional area is where the direction of Electric field Should be ...
 
  • #24
tellmesomething said:
If its symmetrical about an axis then that is Not a solution?
I don't understand your question.
If a charge distribution has a certain symmetry then the field must have that symmetry.
The sphere has spherical symmetry, meaning that it can be rotated around its centre in any direction.
The infinite cylinder can be rotated through any angle about its axis, translated to any displacement along its axis ("glide" symmetry) or reflected about any plane normal to its axis or any plane through the axis.
The infinite slab can be rotated to any extent about any axis normal it, translated to any extent in any direction parallel to it, or reflected about any plane normal to it or the plane parallel to it through its centre.
I think that is everything.
https://en.wikipedia.org/wiki/Symmetry
 
  • #25
haruspex said:
The infinite slab can be rotated to any extent about any axis normal it, translated to any extent in any direction parallel to it, or reflected about any plane normal to it or the plane parallel to it through its centre.
... or rotated by ##\pi## around any axis that is within the plane (this is also a result of first reflecting in the plane itself and then in the plane normal to it crossing it in the axis so without this transformation the symmetry group is not closed).

haruspex said:
The sphere has spherical symmetry, meaning that it can be rotated around its centre in any direction.
... or reflected in any plane through its centre. (O(3) rather than SO(3) as the symmetry group ... since you mentioned reflections in the other cases)
haruspex said:
The infinite cylinder can be rotated through any angle about its axis, translated to any displacement along its axis ("glide" symmetry) or reflected about any plane normal to its axis or any plane through the axis.
... or rotated by ##\pi## about any axis through and perpendicular to the cylinder axis. (Same as above, this is also a combination of two reflections.)


... and of course all sorts of combinations not mentioned, such as a combined rotation and translation of the plane.
 
  • #26
haruspex said:
I don't understand your question.
If a charge distribution has a certain symmetry then the field must have that symmetry.
The sphere has spherical symmetry, meaning that it can be rotated around its centre in any direction.
The infinite cylinder can be rotated through any angle about its axis, translated to any displacement along its axis ("glide" symmetry) or reflected about any plane normal to its axis or any plane through the axis.
The infinite slab can be rotated to any extent about any axis normal it, translated to any extent in any direction parallel to it, or reflected about any plane normal to it or the plane parallel to it through its centre.
I think that is everything.
https://en.wikipedia.org/wiki/Symmetry
]
  • I know ive already been very slow in understanding but after reviewing everything i dont understand why Along the y-axis, there is no preferred direction. Any point on the positive y-axis is symmetrically not equivalent to its counterpart on the negative y-axis. Therefore, any electric field component along the y-axis should be considered
I GET the argument along the z axis. But not the y axis for an infinitely long cylinder
 
  • #27
Orodruin said:
... or rotated by ##\pi## around any axis that is within the plane (this is also a result of first reflecting in the plane itself and then in the plane normal to it crossing it in the axis so without this transformation the symmetry group is not closed).


... or reflected in any plane through its centre. (O(3) rather than SO(3) as the symmetry group ... since you mentioned reflections in the other cases)

... or rotated by ##\pi## about any axis through and perpendicular to the cylinder axis. (Same as above, this is also a combination of two reflections.)


... and of course all sorts of combinations not mentioned, such as a combined rotation and translation of the plane.
  • I know ive already been very slow in understanding but after reviewing everything i dont understand why Along the y-axis, there is no preferred direction. Any point on the positive y-axis is symmetrically not equivalent to its counterpart on the negative y-axis. Therefore, any electric field component along the y-axis should be considered
I GET the argument along the z axis. But not the y axis for an infinitely long cylinder
 
  • #28
tellmesomething said:
  • I know ive already been very slow in understanding but after reviewing everything i dont understand why Along the y-axis, there is no preferred direction. Any point on the positive y-axis is symmetrically not equivalent to its counterpart on the negative y-axis. Therefore, any electric field component along the y-axis should be considered
I GET the argument along the z axis. But not the y axis for an infinitely long cylinder
I'm not sure how you are assigning axes.
The field is everywhere normal to the cylinder axis and does not depend on displacement along that axis.
Rotating the charge distribution about that axis does not change it, nor does reflecting it in a plane through the axis.
 
  • #29
haruspex said:
I'm not sure how you are assigning axes.
The field is everywhere normal to the cylinder axis and does not depend on displacement along that axis.
Rotating the charge distribution about that axis does not change it, nor does reflecting it in a plane through the axis.
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...
Screenshot_2024-05-22-22-47-12-617_com.miui.notes.jpg


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 there
IMG_20240523_090232.jpg

As you say the field is everywhere normal to the cylinder's axis I agree with that but he says otherwise .
IMG_20240523_090250.jpg
 
  • #30
tellmesomething said:
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
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.
 
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  • #32
tellmesomething said:
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.
 
Last edited:
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  • #33
haruspex said:
##\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.

haruspex said:
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.
 
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  • #34
tellmesomething said:
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?
 
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  • #35
Orodruin said:
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.
 

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