If basis change, how metric hcange?

[LuisVela]

Homework Statement

I have a vector space (dim=2) with given basis (x1,x2), and a metric Q=[-1,0;0,1] (Minkowsky metric). Now i perform a transformation on the basis of the form y1=y1(x1,x2), and y2=y2(x1,x2)..the transformation is linear, and the determinat of its matrix is equal to 1.
Does the metric changes?...in that case, how can i find the coeficients of the metric tensor? (coeficients of the matrix)?
Hope i explained myself good enough.

Homework Equations

I don't know if i need to specifie more about the transformation. i think to say that its linear, and detT=1 will suffice.

The Attempt at a Solution

I tryed to go to the definition of the components of the metric tenzor...which will satisfy the equation: [Qij=(yi,yj), where the symbol (.,.) means scalar product.
But of course i don't know how tot ake the scalkar product...i mean..should i use the old metric?
Im really lost here..i don't know what to do...
Thx in advance

 

[CompuChip]

Let's look at it more generally, using linear algebra.

Suppose you have some linear transformation T which corresponds to a matrix A in some specific basis. Do you know how to construct a matrix C that changes basis, and how to use this to represent T in the new basis (i.e. write the matrix belonging to T with respect to the new basis, in terms of A and C)?

 

[LuisVela]

Hi. Yes i do.
Let be [e1,e2] and [f1,f2] basis and C a matrix such that [f1,f2]=C[e1,e2].
Let T be a lin. transf. (represented by the matrix A) of the basis [e1,e2].
...The transformation T (matrix A) expressed in the basis [f1,f2] is equal to C^(-1)*A*C

Go on...

 

[CompuChip]

Well, the metric tensor is just a 2 x 2 matrix.
So in principle, you can construct the change-of-basis and calculate the new metric.

Note that if the basis change is a Lorentz-transformation (i.e it only rotates and boosts the basis vectors) then the metric will be preserved.

 

[LuisVela]

Thank you very much. That will help me.
One more thing though. When you say that the metric is preserved, will that imply that the norm of the vectors will be invariant?
Is just that in my case, i found that the norm of the vectors which expand the new space (the new basis) is cero!...is that possible?

 

[CompuChip]

Yes, if the transformation is a Lorentz transformation then also the norm
||\vec a||^2 = \vec a \cdot \vec a = a^\mu a_\mu
of vectors will be preserved.

If it is not a Lorentz transformation, the norm of a non-zero vector can become zero. For example, consider a (non-invertible!) transformation where you collapse both basis vectors onto the light cone. Any vector will now be projected onto the light cone and have zero norm.

 

[LuisVela]

...(how did you write in latex?)

 

[LuisVela]

Thanks for the reply. It was of great help.

 

[LuisVela]

my transformation is the following:
<br /> u=\frac{1}{\sqrt(2)}(t-x)\\<br /> v=\frac{1}{\sqrt(2)}(t+x)<br />

in the coordinates (t,x) the metric is the minkowsky metric
<br /> \begin{array}{cc}-1&amp;0\\0&amp;1\end{array}<br />

I found that in the new coordinate system the metric is
<br /> \begin{array}{cc}0&amp;-1\\-1&amp;0\end{array}<br />

The basic vetors then (expresed in the new coordinates ) are e_{u}=(1,0) and e_{v}=(0,1)

Now, if you calculate the norm of e_{u} \ e_{u} it will be =0. My question now is: should i calculate their norm using their representation according to the old metric?..i.e.
<br /> e_{u}=\frac{1}{\sqrt(2)}(1.-1)
<br /> e_{v}=\frac{1}{\sqrt(2)}(1.1)
..in which case..its conserved..i.e. its equal to 1.