Is time in Lorenz transformation an imaginary dimension?

Question

Virtual time?

if not why put an imaginary number(i) in the equation of transformation

Answer 1

Hi. No, it is an effect. It is real, not imaginary.

Answer 2

Lorentz Transformations leave the line element

ds² == dx² + dy² + dz² - c²dt²

invariant.

Note that the special theory of relativity is about

1) Relativity: Physical relations are covariant under coordinate transforms between systems that move at constant velocity with respect to each other.

2) Constancy of velocity of light: In each of such coordinate systems, light travels with the same speed c.

From this follows:

==> c = |dr/dt| = √(dx² + dy² + dz²)/dt²)


==> c²dt² = dx² + dy² + dz²


<=> 0 = dx² + dy² + dz² - c²dt² =: ds² (line element)


Just one defintion of a 4-vector:

(x,y,z,ct) =: u


[+ 0 0 0]
[0 + 0 0] =: η (matrix representation of the Minkowski Metric)
[0 0 + 0]
[0 0 0 - ]

==> ds² = du.η.du


or with the complex imaginary number in the 4th slot:

(x,y,z,ict) =: w

==> ds² = dw.dw

To generate this strange Minkowski Metric (+++- or ---+), you either choose the tensor notation and bother with signs or you choose the complex notation and use the fact that i = √-1.


btw, dw.dw isn't an inner product.