Findin the four-velocity of a given world line

[Raziel2701]

Homework Statement

I want to find the four-velocity of an observer moving on a curve x=2t for t>1 in the line element:

$$ds^2=-x^2dt^2 +dx^2$$

Homework Equations

$$d\tau^2=-ds^2$$

$$u^t=\frac{dt}{d\tau}$$

$$u^x=\frac{dx}{d\tau}$$

$$u\cdot u=-1$$

The Attempt at a Solution

I get the following things:

dx=2dt

$$d\tau^2=4t^2dt^2 -4dt^2$$

I think that usually we'd like to express t in terms of tau to have everything in terms of tau and be able to compute the components of the four-velocity. I have been looking to learn this by example but so far it seems that once you get dtau equal to some ugly-looking stuff in my case, you integrate to get tau and t, then solve for t, plug that value into everything you see a t in and then follow those two formulas for the t and x components of the velocity.

Unless I'm plugging things wrong, I get the following nasty integral:

$$\int d\tau = 2\int dt\sqrt{t^2 -1}$$

Suffice it to say that I can't solve for t in terms of tau because the solution to that integral is a tangled mess.

So I did the following, I calculated the t and x components of the four velocity by just plugging in what dtau, dx and dt were.

I got:

u-x=1/sqrt(t^2 -1)

u-t=1/2sqrt(t^2 -1)

In any case these are wrong because their dot product is not -1. After this I did a lot of things in desperation such as parametrizing the world line as well some rather shady things by setting t=0 in some convenient places in the expression for dtau but no luck.

So obviously I'm missing something, and I've run out of ideas for now as I've done and redone the book's example and it seems to boil down to the fact that my world line should perhaps be broken down into something else, a different parametrization perhaps?

 

[fzero]

u-x=1/sqrt(t^2 -1)

u-t=1/2sqrt(t^2 -1)

In any case these are wrong because their dot product is not -1. After this I did a lot of things in desperation such as parametrizing the world line as well some rather shady things by setting t=0 in some convenient places in the expression for dtau but no luck.

Assuming you mean $$u^x,u^t$$ above, you can certainly show that $$u\cdot u=-1$$ in the given metric.

I don't think that you're meant to invert the solution for $$\tau$$. It's probably enough to express $$x=2t$$ in terms of $$x^\mu(\lambda)$$, where $$\lambda$$ is an affine parameter. You can then express $$u^\mu$$ in terms of this parameter or in terms of $$t$$ as you have above.

 

[Raziel2701]

So when the metric changes, so would the dot product calculation. I did not know that until today. Thank you very much. :)