Solve 4-velocity Problem with 3-velocity \underline{v}

[Deadstar]

Homework Statement

4-velocity $$\overline{U} = (U^0, U^1, U^2, U^3)$$ is related to 3-velocity $$\underline{v}$$ by $$\overline{U} = \gamma(v) (1,\underline{v})$$

Express $$U^{\alpha}$$ in terms of $$\underline{v}$$, where $$\alpha$$ represents the spatial components and takes values 1,2,3.

Homework Equations

$$\gamma(v) = \frac{1}{\sqrt{1-v^2}}$$.I can't see how this is possible.

$$U^{\alpha} = \frac{dx^{\alpha}}{d \tau}$$

$$v^{\alpha} = \frac{1}{\gamma (v)} \frac{dx^{\alpha}}{d \tau}$$

How can you do it for just v..?!?

 

[gabbagabbahey]

I can't see how this is possible.

$$U^{\alpha} = \frac{dx^{\alpha}}{d \tau}$$

$$v^{\alpha} = \frac{1}{\gamma (v)} \frac{dx^{\alpha}}{d \tau}$$

How can you do it for just v..?!?

Combining the two equations, you should immediately see that $U^{\alpha}=\gamma(v)v^{\alpha}$ (Remember, $\alpha$ varies from one to 3 only, for the spatial components of 4-velocity)

Now you need only express $\gamma(v)$ in terms of the components of $v$...

hint: $|\vec{v}|^2=v_{\alpha}v^{\alpha}$

 

[Deadstar]

Combining the two equations, you should immediately see that $U^{\alpha}=\gamma(v)v^{\alpha}$ (Remember, $\alpha$ varies from one to 3 only, for the spatial components of 4-velocity)

Now you need only express $\gamma(v)$ in terms of the components of $v$...

hint: $|\vec{v}|^2=v_{\alpha}v^{\alpha}$

Thanks for the reply. I had already got the first part. Just a bit odd notation that threw me off at first.