Gold Member
- 4,662
- 372
The question is as follows:
Prove that any timelike vector for which ##U^0>0## and ##\vec{U}\cdot \vec{U} = -1## is the 4-velocity of some world line.
I need to show that there exists a 4-vector ##\vec{x}##, s.t ##\frac{d\vec{x}}{d\tau} = \vec{U}##, where ##\vec{x}## is a world line of some particle.
So far what I have done is: since ##\vec{U}\cdot \frac{d\vec{U}}{d\tau} = 0##, so we have:
$$dU^0/d\tau = U^1/U^0 dU^1/d\tau + U^2/U^0dU^2/d\tau + U^3/U^0 dU^3/d\tau$$
I used the fact that ##U^0>0##, so I can divide by it above.
We also have: ##dx^\mu/d\tau = U^\mu ##. How to continue? Any ideas?
This is not for homework, I am self studying GR and QFT.
Prove that any timelike vector for which ##U^0>0## and ##\vec{U}\cdot \vec{U} = -1## is the 4-velocity of some world line.
I need to show that there exists a 4-vector ##\vec{x}##, s.t ##\frac{d\vec{x}}{d\tau} = \vec{U}##, where ##\vec{x}## is a world line of some particle.
So far what I have done is: since ##\vec{U}\cdot \frac{d\vec{U}}{d\tau} = 0##, so we have:
$$dU^0/d\tau = U^1/U^0 dU^1/d\tau + U^2/U^0dU^2/d\tau + U^3/U^0 dU^3/d\tau$$
I used the fact that ##U^0>0##, so I can divide by it above.
We also have: ##dx^\mu/d\tau = U^\mu ##. How to continue? Any ideas?
This is not for homework, I am self studying GR and QFT.