# Minkowski metric

1. Nov 9, 2012

### jaljon

hello

Whic one of these to metric are Minkowski metric
ds^2 =-(cdt)^2+(dX)^2

ds^2 =(cdt)^2-(dX)^2

and what about timelike (ds^2<0) and spacelike (ds^2>0) for each metric?

With my appreciation to those who answer

2. Nov 9, 2012

### bcrowell

Staff Emeritus
Welcome to PF!

These are both ways of writing the Minkowski metric. People refer to this as (-+++) and (+---) signatures. It doesn't matter which one you use as long as you're consistent. It can be confusing reading the literature, because different people use different signatures.

3. Nov 9, 2012

### Staff: Mentor

Both are considered the same metric, just with a different sign convention. My personal preference is the first one, but both are well accepted.

When I want to use a metric with positive timelike intervals squared I tend to use $c^2d\tau^2=c^2dt^2-dX^2$. It is just a convention, but that is my preference.

Last edited: Nov 9, 2012
4. Nov 9, 2012

### Fredrik

Staff Emeritus
The first convention seems to be more popular in texts about classical SR and GR. The second convention seems to be more popular in books on quantum field theory.

5. Nov 9, 2012

### robphy

6. Nov 10, 2012

### jaljon

bcrowell, DalesPam, robphy Thank you very much

my queastion is

if I use first one

ds^2 =-(cdt)^2+(dX)^2

spacelike: (ds^2>0) then -(cdt)^2+(dX)^2>0 then (dX/dt)>c

if I use second one

ds^2 =(cdt)^2-(dX)^2

spacelike: (ds^2>0) then (cdt)^2-(dX)^2>0 then (dX/dt)<c

we know in light cone spacelike out of cone that mean (dX/dt)>c but why second one (dX/dt)<c

7. Nov 10, 2012

### George Jones

Staff Emeritus
No, for this convention for the metric, spacelike means ds^2 < 0.

8. Nov 10, 2012

### robphy

spacelike: outside the lightcone... so "faster than light"
(dx/dt)^2 > c^2 (i.e. either (dx/dt) > c or (dx/dt) < -c).

Thus, spacelike means dx^2 > c^2 dt^2
("larger square-of-the-magnitude of the spatial-part than that of c-times-the-temporal-part")

So, spacelike is dx^2 - c^2 dt^2 > 0.

Now on to the conventions...
If ds^2 = -c^2 dt^2 + dx^2 (-+++), then spacelike is ds^2 > 0 (in -+++... that is, "+ for space").

If ds^2 = c^2 dt^2 - dx^2 (+---), then spacelike is ds^2 < 0 (in +---... that is "- for space").

9. Nov 10, 2012

thanks