Negative time Minkowski metric

[tickle_monste]

What does it mean physically that the sign on the time component is opposite those of the spatial components of the Minkowski metric?

 

[atyy]

It means that the speed of light is the same for all inertial observers.
For the x frame: x²=c²t²
For the x' frame: x'²=c²t'²
So: x²-c²t²=x'²-c²t'² , which is the scalar product, which is computed using the metric.

I cheated, but it's a good cheat.

 

[raul_l]

In other words, in order for the space-time interval to be invariant under Lorentz transformations the time component simply has to have an opposite sign compared to the spatial components. And yes, it also preserves the constancy of the speed of light.

 

[A.T.]

What does it mean physically that the sign on the time component is opposite those of the spatial components of the Minkowski metric?

If you don't like minus signs, you can rearrange the equation so it contains none. This doesn't change the meaning of the formula, but maybe the geometrical interpretation.

Different signs of the time and space components mean that coordinate time is not a dimension just like the space dimensions but a bit different. If you don't like "special dimensions" you can regard proper time \tau as the temporal dimension, and the coordinate time t as the space-time interval. Then you have no minus signs:

dt^2=d\tau^2+dx^2+dy^2+dz^2

But this is not Minkowski space time anymore.

 

[DrGreg]

If you don't like minus signs, you can rearrange the equation so it contains none. This doesn't change the meaning of the formula, but maybe the geometrical interpretation.

Different signs of the time and space components mean that coordinate time is not a dimension just like the space dimensions but a bit different. If you don't like "special dimensions" you can regard proper time \tau as the temporal dimension, and the coordinate time t as the space-time interval. Then you have no minus signs:

dt^2=d\tau^2+dx^2+dy^2+dz^2

But this is not Minkowski space time anymore.

The problem is, that differential equation doesn't actually define a coordinate τ.

These equations would:

t=\sqrt{T^2 + X^2}
x = X​

where (t,x) are standard Minkowski coordinates (let's ignore 2 dimensions) and (T,X) are new coordinates we want to define.

From which

dt = \frac{TdT + XdX}{\sqrt{T^2 + X^2}}
dx = dX
d\tau^2 = dt^2 - dx^2 = \frac{T^2(dT^2 - dX^2) + 2TXdTdX}{T^2 + X^2}​

In this coordinate system, proper time is defined by the last equation above, not by τ = T.

The point of all this is, this is a valid coordinate system to use (although valid only in the region |x| ≤ |t|), but it's technically incorrect to describe T as "proper time". It coincides with proper time only along straight worldlines through the origin.

 

[javierR]

You can also say that it means that space-time is a hyperbolic rather than euclidean metric space. The statement that different observers have different notions of space and time measurements, with an invariant speed of light, can be described by saying that paths of observers are in a hyperbolic space. This also means that there is a natural partition for paths and vectors into three categories: space-, light- and time-like.

 

[A.T.]

t=\sqrt{T^2 + X^2}
x = X​

I think you demand a 1:1 mapping between (x,t) and (X,T) here. But there is no such corespndence between (x,t) and (x,\tau). An event in Minkowski space-time, doesn't have a corresponding single point in space-propertime. In Minkowski space-time you see that two objects meet at (x,t) when their worldlines cross there. In space-propertime(x,\tau) you see they meet if they are at the same (x) (but eventually different (\tau)) after traversing worldlines of the same length:
dt=\int \sqrt{dx^2 + d\tau^2}

 

[DrGreg]

I think you demand a 1:1 mapping between (x,t) and (X,T) here. But there is no such correspondence between (x,t) and (x,\tau). An event in Minkowski space-time, doesn't have a corresponding single point in space-propertime. In Minkowski space-time you see that two objects meet at (x,t) when their worldlines cross there. In space-propertime(x,\tau) you see they meet if they are at the same (x) (but eventually different (\tau)) after traversing worldlines of the same length:
dt=\int \sqrt{dx^2 + d\tau^2}

Fair enough, but that does mean that "space-propertime" is a very limited concept and pretty difficult to grasp. You can plot a single worldline in space-propertime, but an isolated "point", not lying on any worldline, has no meaning, and if you plot more than one worldline on the same graph it's going to get pretty confusing, as the intersection of two lines has no physical significance, and a single event in spacetime could be mapped to multiple distinct points in space-propertime on different worldlines.

 

[Phrak]

...
dt^2=d\tau^2+dx^2+dy^2+dz^2
But this is not Minkowski space time anymore.

OK, but does this hold true?

d\tau^2+dx^2+dy^2+dz^2 = d\tau'^2+dx'^2+dy'^2+dz'^2

 

[A.T.]

Fair enough, but that does mean that "space-propertime" is a very limited concept and pretty difficult to grasp.

It is actually just like normal Euclidian space with dimensions of the same kind, where everything 'moves' at the same rate. For layman it is not more difficult to grasp, than the pseudo-Euclidian Minkowski space-time. I see it as complimentary tool.

You can plot a single worldline in space-propertime, but an isolated "point", not lying on any worldline, has no meaning, and if you plot more than one worldline on the same graph it's going to get pretty confusing,

It depends what you want to show. You can visualize different propertimes for two worldlines very well directly in a space-propertime graph. Like the usual twins for example. And you still see the coordiante time in the diagram, as the length of the world lines.

as the intersection of two lines has no physical significance,

Just like in a purely spatial graph, where the intersection of two paths doesn't imply a meeting point. That was exactly the point, that propertime is just like a 4-th space dimension.

OK, but does this hold true?
d\tau^2+dx^2+dy^2+dz^2 = d\tau'^2+dx'^2+dy'^2+dz'^2

If you mean two different frames of reference observing the same object: no
If you mean two different objects observed in the same frame of reference: yes