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.