What Coordinate Transformation Converts a Complex Metric to Minkowski Space?

[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

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

 

[George Jones]

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

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

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?

 

[Logarythmic]

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

 

[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?

 

[Logarythmic]

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

 

[George Jones]

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

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.

 

[Logarythmic]

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

 

[whatta]

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

 

[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?

 

[dextercioby]

It certainly looks so.

 

[whatta]

except that you have lost -1 somewhere

 

[Logarythmic]

I don't feel that positive..?