# Why isn't velocity a dimension in SR?

1. Jan 20, 2012

### vinven7

Hello all,

Special relativity tells that space and time should be seen as a single four dimensional space time. Even the metric for SR has four components, x1, x2, x3, and x4 = ict. The Lorentz transform tells us how to convert these coordinates from this to another providing that we are moving at a constant velocity v in an inertial frame. Therefore, to completely express an event in a coordinate transformed spacetime, we need, not just the current space and time coordinates but also our velocity.
Now, dimensions are also defined as the minimum number of quantities that are required to completely identify a "point" in our manifold. By this definition, shouldn't velocity be counted as a dimension as well?
To put in other terms: Suppose I tell you that I am at x,y,z at time t but don't tell you what my v is - if we are in different frames of reference, then you still cannot compute my exact coordinates. It seems to me that the velocity v is as critical a quantity as the other four and hence should be accorded the title of a dimension.
If my question is clear to you, what are your thoughts? Cheers!

2. Jan 20, 2012

### ghwellsjr

We are both in the same Frame of Reference. You should think in terms of one Frame of Reference in which you specify your postion as a function of time and my postition as a function of time which don't have to be constant velocities. If you don't do that, then of course no one can make any sense of what's going on or how to use the Lorentz transform. When you use the Lorentz transform to get to another Frame of Reference moving with some speed with respect to the first one, it doesn't have to be a speed that either of us is traveling at, it can be any speed. You might want to transform to a speed in which we are both traveling in opposite directions, it doesn't matter.

3. Jan 20, 2012

### elfmotat

The Minkowski metric has 16 components:

$$\eta_{\mu \nu }=\begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

x4=ct if you're using the Minkowski metric. If you're using a positive definite metric, then you could use x4=ict.

No, because you can map a unique coordinate to every point in spacetime without knowing anything about the velocities of anything. It's only when you transform into another set of coordinates that you need this information.

4. Jan 20, 2012

### Staff: Mentor

The velocity is a parameter of the Lorentz transform (boost), just like the angle is a parameter of rotations. You don't need to add dimensions just because you have a transform which takes a parameter.