Exercise 2.5 in Misner, Thorne and Wheeler

[TerryW]

If the squared length of a 4-velocity is -1, how can you have a component v⁰ =0?

I've played with equation (2.35) and produced a result of v = γ(-u², -u¹, -u², -u³), which doesn't have a squared length of -1.

Can anyone help out?



TerryW

 

[Bill_K]

In Exercise 2.5, u is the particle's 4-velocity with squared length -1, whereas v is a newly defined spacelike vector, apparently the projection of u onto the observer's space section.

 

[TerryW]

Thanks

Hi Bill_K,


So my result is probably OK. A few words along those lines in the text hinting at the special nature of v might have helped, it does come out of left field a bit.

Thanks for your prompt reply. You obviously don't subscribe to the view I saw in one post that if you are working through MTW, you should need any help!


Regards

TerryW