Symmetry in Objects: Is Every Object Symmetric & How to Prove It?

  • Thread starter Thread starter khotsofalang
  • Start date Start date
khotsofalang
Messages
21
Reaction score
0
is there a formal definition of symetry? suppose i was to make an assertion that every object in space is symetric at least about one certain axis in space, is this assertion true? why or why not...how would you go about the proof?
 
Physics news on Phys.org
khotsofalang said:
is there a formal definition of symetry? suppose i was to make an assertion that every object in space is symetric at least about one certain axis in space, is this assertion true? why or why not...how would you go about the proof?

one definition of symmetry is an isometry. I believe there are manifolds with trivial isometry groups i.e. the only isometry is the identity. Since any manifold can be embedded isometrically in space, you conjecture is false.

In any dimension, it should be easy to construct measurable sets in space that have no symmetries under reflection - but I am not sure
 
Your hand is a 3D object that has no symmetries.
 
Even functions are symmetric with respect to the y-axis. This is expressed in the relation
f(x) = f(-x). For a 3-D object, a similar expression of symmetry might be
f(x,y,z) = f(-x,-y,z)
 
but how do we formally define symetry? is it just in terms of isometry? if you were to place you hand infinite distances away, then your hand gets very tiny and converges to a point in space which is definitely isometric about some point...eventually becoming symetric!
 
Before you get carried away with mindless sophistry, take a little time and peruse the following article:

http://en.wikipedia.org/wiki/Symmetry

I think you can glean from these scribblings how mathematics treats the definition of symmetry.
 
Hello! There is a simple line in the textbook. If ##S## is a manifold, an injectively immersed submanifold ##M## of ##S## is embedded if and only if ##M## is locally closed in ##S##. Recall the definition. M is locally closed if for each point ##x\in M## there open ##U\subset S## such that ##M\cap U## is closed in ##U##. Embedding to injective immesion is simple. The opposite direction is hard. Suppose I have ##N## as source manifold and ##f:N\rightarrow S## is the injective...
Back
Top