Finding the 4-velocity of a world line

[trezm]

Hello! This is my first post, and I was wondering if anyone could help my uncertainty about this problem.

Homework Statement

given an observer's world line, X = 2T and the metric ds² = -X²dT² + dX², find the components of the observer's 4-velocity

Homework Equations

I'm using the equation that ds²/dτ² = u = e₀

The Attempt at a Solution

Taking our equation for ds² and dividing it by dτ², I solve for dX²/dτ² = u², keeping in mind that this also equals -1, to get that dX²/dτ² = X²dT²/dτ² - 1.

Is the correct way to think about it? And from the observers frame, is her 4-velocity u = ( X²dT²/dτ² - 1, 0 , 0, 0 )?

Thanks so much for your help!

 

[mark726]

Hello there! It looks like you are on the right track with your solution. Just to clarify, the equation for the 4-velocity should be u = (X2dT/dτ, dX/dτ, 0, 0). This is because the 4-velocity is a vector with four components, representing the observer's velocity in the four dimensions of space-time. So, the first component represents the observer's velocity in the time dimension, and the second component represents the observer's velocity in the space dimension.

In terms of your approach, you are correct in dividing the metric by dτ2 to get the 4-velocity. Another way to think about it is to use the fact that u = dx/dτ, where x is the observer's position in space-time. So, in this case, x = (X, T), and taking the derivative with respect to τ gives u = (dX/dτ, dT/dτ).

I hope this helps clarify things for you. Keep up the good work and don't hesitate to ask for further clarification if needed. Good luck with your studies!