Register to reply

Unique volume element in a vector space

by yifli
Tags: element, space, unique, vector, volume
Share this thread:
yifli
#1
Jun16-11, 05:14 PM
P: 70
Given two orthonormal bases [itex]v_1,v_2,\cdots,v_n[/itex] and [itex]u_1,u_2,\cdots,u_n[/itex] for a vector space [itex]V[/itex], we know the following formula holds for an alternating tensor [itex]f[/itex]:
[tex]f(u_1,u_2,\cdots,u_n)=\det(A)f(v_1,v_2,\cdots,v_n)[/tex]
where A is the orthogonal matrix that changes one orthonormal basis to another orthonormal basis.

The volume element is the unique [itex]f[/itex] such that [itex]f(v_1,v_2,\cdots,v_n)=1[/itex] for any orthonormal basis [itex]v_1,v_2,\cdots,v_n[/itex] given an orientation for [itex]V[/itex].

Here is my question:
If [itex]\det(A)=-1[/itex], then [itex]f(u_1,u_2,\cdots,u_n)=-1[/itex], which makes [itex]f[/itex] not unique
Phys.Org News Partner Science news on Phys.org
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds
micromass
#2
Jun16-11, 05:21 PM
Mentor
micromass's Avatar
P: 18,040
Hi yifli!
I don't quite see why your reasoning implies that f is not unique. What's the other tensor that makes [itex]f(v_1,...,v_n)=1[/itex]?
yifli
#3
Jun16-11, 09:37 PM
P: 70
Quote Quote by yifli View Post
Here is my question:
If [itex]\det(A)=-1[/itex], then [itex]f(u_1,u_2,\cdots,u_n)=-1[/itex], which makes [itex]f[/itex] not unique
Quote Quote by micromass View Post
Hi yifli!
I don't quite see why your reasoning implies that f is not unique. What's the other tensor that makes [itex]f(v_1,...,v_n)=1[/itex]?
Sorry for the confusion. Actually I mean [itex]f(u_1,u_2,\cdots,u_n)[/itex] is not necessarily equal to 1, it may be equal to -1 also, even though [itex]f(v_1,v_2,\cdots,v_n)[/itex] is made to be 1

micromass
#4
Jun16-11, 09:39 PM
Mentor
micromass's Avatar
P: 18,040
Unique volume element in a vector space

Yes, but the point is that f of every basis with the same orientation will give you 1. If you take the opposite orientation, then you will get -1 of course.
yifli
#5
Jun16-11, 09:47 PM
P: 70
Quote Quote by micromass View Post
Yes, but the point is that f of every basis with the same orientation will give you 1. If you take the opposite orientation, then you will get -1 of course.
So given a fixed orientation, it's impossible for [itex]\det(A)[/itex] to be -1?

What does it mean for [itex]\det(A)[/itex] to be -1?
micromass
#6
Jun16-11, 09:51 PM
Mentor
micromass's Avatar
P: 18,040
Quote Quote by yifli View Post
So given a fixed orientation, it's impossible for [itex]\det(A)[/itex] to be -1?

What does it mean for [itex]\det(A)[/itex] to be -1?
The transition matrix between two orthonormal bases has det(A)=1 if and only if the bases have the same orientation and has det(A)=-1 if they have opposite orientation.
I like Serena
#7
Jun17-11, 01:44 AM
HW Helper
I like Serena's Avatar
P: 6,187
A orthonormal transformation matrix with det(A)=1 is a rotation.
A orthonormal transformation matrix with det(A)=-1 is a rotation combined with a reflection.

So the volume sign would be invariant.


Register to reply

Related Discussions
Volume of element in k-space Atomic, Solid State, Comp. Physics 3
Prove that the additive inverse -v of an element v in a vector space is unique. Precalculus Mathematics Homework 7
Prove that the additive identity in a vector space is unique Precalculus Mathematics Homework 5
Checking for a unique element Programming & Computer Science 2
Volume element in phase space Advanced Physics Homework 0