Time measured by free-fall observer near object?

[johne1618]

What is the time measured by an observer who is in free-fall near an object?

The Schwarzschild metric is given by:

$\large c^2 d\tau^2 = (1 - \frac{r_s}{r}) c^2 dt^2 - ( 1 - \frac{r_s}{r})^{-1} dr^2 - r^2(d\theta^2 + \sin^2 \theta d\phi^2)$

Now for an observer at a fixed position one uses:

$dr = d\theta = d\phi = 0$

To give an element of proper time as:

$\large d\tau = \sqrt{1 - \frac{r_s}{r}} dt$

This is a time interval experienced by an observer who is stationary in the object's gravitational field.

But I want the time measured by a free-fall observer so I can't assume that his position is fixed.

Perhaps this is how to do the calculation.

I work out the world line for a radial light beam using:

$d\tau = d\theta = d\phi = 0$

Substituting into the above Schwarzschild metric I find the worldline of a lightbeam given by

$\large (1 - \frac{r_s}{r}) \ c \ dt = dr$

Now a free-fall observer experiences a flat spacetime locally. He must use a time element $d\tau$ such that a light beam's worldline is diagonal so that

$\large c \ d\tau = dr$

Therefore his time element must be:

$\large d\tau = (1 - \frac{r_s}{r}) \ dt$

Does this make sense?

 

[Mentz114]

The four velocity of the LeMaitre free-faller is
$$U=\left(\frac{r}{r-2\,m}, -\frac{\sqrt{2}\,\sqrt{m}}{\sqrt{r}} , 0 , 0\right)$$
which agrees with your result.

 

[johne1618]

The four velocity of the LeMaitre free-faller is
$$U=\left(\frac{r}{r-2\,m}, -\frac{\sqrt{2}\,\sqrt{m}}{\sqrt{r}} , 0 , 0\right)$$
which agrees with your result.

Great - thanks!

I'm also interested in calculating the correct time interval we should use with the Friedmann Walker metric

$\large ds^2 = -dt^2 + a^2(t) [ \frac{dr^2}{1-kr^2} + r^2(d\theta^2+\sin^2\theta d\phi^2)]$

As co-moving observers we are also in free-fall.

Thus using $ds = d\theta = d\phi = 0$

The worldline of a radial light beam is given by

$\large \frac{dt}{a(t)} = \frac{dr}{1-kr^2}$

Thus the interval of proper time $d\tau$ for a co-moving observer with a locally flat spacetime (assuming k is small) is given by

$\large d\tau = \frac{dt}{a(t)}$

which I believe is called an interval of conformal time.

I think many people believe we should use cosmological time $t$ as our proper time but according to the above reasoning that is not correct because we are in free-fall (in a local inertial frame) and therefore our proper time is conformal time.

 

[Mentz114]

I don't think that's right. If you divide the line interval by dτ² and set the spatial velocities to zero you get dt/dτ = 1 for this co-moving frame.

 

[johne1618]

I don't think that's right. If you divide the line interval by dτ² and set the spatial velocities to zero you get dt/dτ = 1 for this co-moving frame.

Could you elaborate a little more?

 

[Mentz114]

I can't eleborate without doing the simple algebra and tex'ing it up which I don't have time for. Get a pencil, paper and eraser and do as I suggested. If you have a problem then you could ask about specifics. Remember that ds=dτ.

 

[johne1618]

Ok - this is what I think you did.

For simplicity starting with the FRW metric with k=0 and only radial spatial component:

$\large d\tau^2 = dt^2 - a(t)^2 dr^2$

divide through by $d\tau^2$

$\large 1 = \frac{dt^2}{d\tau^2} - a(t)^2 \frac{dr^2}{d\tau^2}$

For a co-moving frame spatial velocity is zero so

$\large \frac{dr}{d\tau} = 0$

Therefore

$\large \frac{dt}{d\tau} = 1$

Is this right?

 

[Mentz114]

Your calculation is correct.