Coordinate transformation

1. Jan 29, 2007

Logarythmic

How can I identify the coordinate transformation that turns

$$ds^2 = \left(1+\frac{\epsilon}{1+c^2t^2}\right)^2c^2dt^2 - \left(\frac{\epsilon}{1+x^2}\right)^2x^2 - \left(\frac{\epsilon}{1+y^2}\right)^2y^2 - \left(\frac{\epsilon}{1+z^2}\right)^2z^2$$

into the Minkowski metric

$$ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2$$?

2. Jan 29, 2007

George Jones

Staff Emeritus
You forgot a bunch of d's in the first equation, and you should use different symbols for the coordinates in the two equations - maybe primes in the second.

What happens if you identify corresponding terms in the two metrics?

3. Jan 30, 2007

Logarythmic

My equations looks just like that, but I guess they are wrong. (P. Coles, Cosmology)
What dou you mean by identify?

4. Jan 30, 2007

whatta

Maybe
$$\left(1+\frac{\epsilon}{1+c^2t^2}\right)^2c^2dt^2 = c^2dT^2$$
$$\left(\frac{\epsilon}{1+x^2}\right)^2dx^2 = dX^2$$
etc?

5. Jan 30, 2007

Logarythmic

Yes that I can see, but that's not a transformation for the whole metric?

6. Jan 30, 2007

George Jones

Staff Emeritus
I'm not sure what you mean by this.

You're looking for coordinate transformations, i.e., $X = X \left( t, x, y, z)$, etc. Then, e.g.,

$$dX = \frac{\partial X}{\partial{t}} dt + \frac{\partial X}{\partial{x}} dx+ \frac{\partial X}{\partial{y}} dy + \frac{\partial X}{\partial{z}} dz.$$

A new coordinate does not have to depend explicitly on all of the old coordinates, i.e., some of the terms in the above expansion can be zero.

7. Jan 30, 2007

Logarythmic

So I can just give the answer to the problem as whatta did above?

8. Jan 30, 2007

whatta

no you probably are supposed to solve that in the form of x(X), y(Y)... t(T) or viceversa

9. Jan 30, 2007

Logarythmic

Like

$$t(T)=T+\frac{\epsilon}{c}\arctan{cT}$$
$$x(X)=\epsilon\arctan{X}$$
$$y(Y)=\epsilon\arctan{Y}$$
$$z(Z)=\epsilon\arctan{Z}$$

so that

$$dt = \left(1+\frac{\epsilon}{1+c^2T^2}\right)dT$$
$$dx = \left(\frac{\epsilon}{1+X^2}\right)dX$$
$$dy = \left(\frac{\epsilon}{1+Y^2}\right)dY$$
$$dz = \left(\frac{\epsilon}{1+Z^2}\right)dZ$$

and then

$$ds^2=c^2dt^2-dx^2-dy^2-dz^2 = \left(1+\frac{\epsilon}{1+c^2T^2}\right)^2c^2dT^2 - \left(\frac{\epsilon}{1+X^2}\right)^2dX^2 - \left(\frac{\epsilon}{1+Y^2}\right)^2dY^2 - \left(\frac{\epsilon}{1+Z^2}\right)^2dZ^2$$.

Is this correct?

10. Jan 30, 2007

dextercioby

It certainly looks so.

11. Jan 30, 2007

whatta

except that you have lost -1 somewhere

12. Jan 30, 2007

Logarythmic

I don't feel that positive..?