# Studying special relativity and using the idea of four vectors

 P: 92 We are currently studying special relativity and using the idea of four vectors. The position 4-vector has been defined in class as $$\vec{X} = (ct, \vec{r}) \ \ \ \ \ \ \ where \ \vec{r} = (x, y, z) \ \$$(ie: the usual 3 space position vector) and the velocity 4-vector is then $$\vec{V} = \frac{d\vec{X}}{d\tau} \ \ \ \ \ \ \ \ \ where \ \tau \ \ is \ the \ proper \ time$$ substituting $$\ \ \ \tau = \frac{t}{\gamma} \ \ \$$ we got $$\vec{V} = \gamma (c, \vec{v})$$ The part I don't understand is why we differentiate with respect to the proper time, rather than the time as seen by an observer in the frame with which we want the velocity with respect to? To illustrate my concerns, consider the simplified scenario of an object travelling at constant speed u in the +x direction of a frame S. I would think that if we wanted to find its velocity according to an observer in frame S then we would need to calculate the distance it travels, divided by the time that it took with all quantities being measured by an "observer" in S. Yet it seems from the definition of a 4-velocity I'm supposed to divide by the time it takes according to the frame stationary to the object (the proper time). This to me doesn't seem logical since were trying to find the velocity wrt S. I have tried to come up with an argument to explain why we use the proper time, rather than the time measured by S. Perhaps in the limit that $$dt$$ goes to zero, $$d\vec{X}$$ goes to zero as well, so the time it takes to move a distance $$d\vec{X}$$ approaches the proper time. Is this the correct explanation, or am I way off? Your help is greatly appreciated.
 Emeritus Sci Advisor P: 7,662 The 4-velocity is not the velocity. If you wanted to calculate the velocity, you'd do exactly what you described. But velocities do not transform as 4-vectors. 4-velocities do transform as 4-vectors., because 'tau' is the same for all observers Take a look at the definition again (dt/dtau, dx/dtau, dy/dtau, dz/dtau) transforms exactly like (t,x,y,z) because tau is the same for all observers. This means you can transform 4-velocities with the Lorentz transform just like any other 4-vector. If the rationale of the 4-velocity vector doesn't seem obvious, be patient. Aside from their transformation properties, you'll find out (for instance), that the energy and momentum of an object are given by $$\vec{E} = m \vec{u}$$ where u is the 4-velocity. So it's definitely a useful concept.
 Sci Advisor HW Helper PF Gold P: 4,139 Studying special relativity and using the idea of four vectors Note that the 4-velocity $$\widetilde{V} = \gamma (c, \vec{v})$$ is (1/c) times "a [future] timelike unit 4-vector tangent to the worldline of the particle": \begin{align*} \widetilde{V}\bullet \widetilde{V} &= \gamma^2 (c^2- \vec{v}\cdot\vec{v}) \\ &= c^2\gamma^2 (1- v^2/c^2)\\ &= c^2\ \end{align*} So, $$\widetilde{V}/c$$ is a unit 4-vector. Without that $$\gamma$$, the norm of $$\widetilde{V}$$ would not be a scalar.