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
<br /> U=\left(\frac{r}{r-2\,m}, -\frac{\sqrt{2}\,\sqrt{m}}{\sqrt{r}} , 0 , 0\right)<br />
which agrees with your result.

 

[johne1618]

The four velocity of the LeMaitre free-faller is
<br /> U=\left(\frac{r}{r-2\,m}, -\frac{\sqrt{2}\,\sqrt{m}}{\sqrt{r}} , 0 , 0\right)<br />
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.