Frame velocity v. object velocity in derivation of 4-velocity

[EricTheWizard]

I've been learning about 4-velocity and all the "proper" 4-vectors recently, and if I understand correctly, proper velocity η (the 3-vector) is related to ordinary velocity by the relation \vec\eta = \frac{\vec u}{\sqrt{1-\frac{u^2}{c^2}}}, where u is ordinary velocity of an object within a certain frame, but that it is derived from the lorentz transformations for ordinary and proper time, which give the relation \frac{d\vec x}{d\tau} = \frac{d\vec x}{dt}\frac{dt}{d\tau}=\frac{\vec u}{\sqrt{1-\frac{v^2}{c^2}}} where v is the velocity of the reference frame. What I don't understand is how v made this leap to u, or more clearly, how the frame velocity became ordinary velocity.

 

[homology]

Not entirely sure what you're asking but I'll take a stab at it by thinking out loud..

Suppose we see some object moving at constant speed u along the x direction (just to make things simpler). To find out how fast its moving I take a derivative with respect to proper time, dx^{\beta}/d\tau and it will have components (u^0, u^1,0,0)=(\gamma, \gamma u,0,0) where u is how fast I measure (using my clocks and rods) the object moving.

So what's in \gamma? The speed u, so \gamma=(1-u^2)^{-1/2} because that is what I observe. Also, you can check by working out the boost to the object's rest frame:

<br /> \bar{u}^{\alpha}=\Lambda^{\alpha}_{\beta}u^{\beta}<br />

which results in \bar{u}^{\alpha}=(1,0,0,0) as it should.

So where are you getting your v? Is there another frame you're not including in your question?

 

[EricTheWizard]

Ahh I understand now, I had this picture in my head of some extraneous frame I guess I didn't need... Thanks for clarifying.