1. Jan 20, 2005

### QuantumDefect

Hello everybody,

First of all, I want to state that im an undergrad physics major who has yet to take a course in Special Relativity. What I have read in books state that Spacetime is one entity, my question is this: If Spacetime is one entity, is it correct to think that there are 3 separate spacial dimensions and one of time? Or are they really just meshed together and there is only one "true" dimension? And that it just helps us visualize spacetime by separating the qualities of it? Thanks for your time.

2. Jan 20, 2005

### chroot

Staff Emeritus
Spacetime is a four-dimensional manifold, but the space and time dimensions are most certainly different. The metric, in fact, demonstrates this difference. In special relativity, for example, the metric is

$$g = \eta = \left( \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array} \right)$$

Note that the sign of the time component of the metric is different than the sign of the space components.

Though all four dimensions can be treated together as one mathematical entity, space and time are still quite different physically.

- Warren

3. Jan 20, 2005

### dextercioby

Nope,space-time can be thought of a weird (actually curved,but in SR not) form of R^{4}.This should be giving you an idea about space-time in SR seen mathematically,like a geometric concept.

However,the fact that physical quantities are described in this space affects them by the fact that the "0-th dimension" is time-like...This 4D-space is constructed by the requirement that space and time have the same role...It doesn't have "natural" geometrical objects (like vectors) with it.These vectors (called 4-vectors) are built/"assembeld" from ordinary vectors (and scalars) which were found in the Euclidean space of classical dynamics.That's how and where the relativistic theories derive their concepts.Simply taking scalars and vectors from the classical dynamics and "assembling" them into objects (vectors,tensors) in this space...The easiest one was with the coordinates:they took the time (scalar) and three space coordinates (components of a vector,coordinate vector) and built a 4-vector,the coordinate 4-vector,whose components are simply time and 3 space-coordinates...

Daniel.
Daniel.

4. Jan 20, 2005

### Garth

chroot's point about the time dimension being treated differently in the metric, because its component in the metric is negative with respect to the others, has a deeper and interesting significance. The metric actually deals with the squares of displacements in the dimensions:

d$$\tau$$2 = dt2 - [dx2 + dy2 + dz2]/c2

To get to the actual displacements themselves you have to take the square root, hence in relativity theory although time is a dimension as the other space dimensions, it is not exactly the same, it is different in that it bears the same mathematical relationship to them as the imaginary numbers to the real.

That time is not exactly the same as the other dimensions is, to me at least, intuitively obvious.

Garth

Last edited: Jan 20, 2005
5. Jan 20, 2005

### Chronos

It was always easier for me to visualize the time dimension as a displacement of the origin of a 3 dimensional coordinate system.