Another Schwarzschild question

[DrGreg]

Proper time along a geodesic

I wonder if DrGreg or George can provide us with an equation for the time measured by the clock of an observer falling from r2 to r1 and what she would consider the distance r2-r1 to be? For simplicity, consider only the motion of a observer falling from infinity with an inital velocity of zero.

I don't claim to be an expert in this subject.

According to Woodhouse⁽¹⁾⁽²⁾, the equations of free fall (i.e. geodesics) in the Schwarzschild metric are:

$$E = \left(1 - \frac{r_s}{r} \right) \frac{dt}{d\tau} = \mbox{constant}$$ ...(1)
$$J = r^2 \sin^2 \theta \frac{d\phi}{d\tau} = \mbox{constant}$$ ...(2)​

together with the definition of $d\tau$ via the metric. E is energy and J is angular momentum, and both are determined by the initial conditions of motion.

The above is from a reputable textbook so must be true. What follows is my own working, so might not.

In the case of radial motion, $d\theta/d\tau = d\phi/d\tau = 0$, and the metric simplifies to

$$\left(1 - \frac{r_s}{r} \right) \left(\frac{dt}{d\tau}\right)^2 - \frac{1}{1 - r_s/r}\left(\frac{dr}{d\tau}\right)^2 = 1$$ ...(3)​

Eliminating $dt/d\tau$ between (1) and (3) gives

$$\frac {E^2 - (dr/d\tau)^2}{1 - r_s/r} = 1$$ ...(4)
$$\left(\frac{dr}{d\tau}\right)^2 = E^2 - 1 + \frac{r_s}{r}$$ ...(5)
$$\int_{r_1}^{r_2}\frac{dr}{\sqrt{E^2 - 1 + r_s/r}} = \tau$$ ...(6)​

Feed that into your favourite online integrator. (Good luck, I had problems with one of sites I tried! )

If the particle is released from "rest" ($dr/d\tau = 0$) at radius r₃, then (5) tells you the value of E, and so (6) becomes

$$\int_{r_1}^{r_2}\frac{dr}{\sqrt{r_s/r - r_s/r_3}} = \tau$$ ...(7)​

In particular if $r_3 = \infty$, E = 1 (which makes the integral easy enough to do without computer assistance).

Coordinate time t can be calculated by combining (1) and (5):

$$t = \int \frac{dt}{d\tau} d\tau = \int \frac{dt/d\tau}{dr/d\tau} dr$$ ...(8)​

As for the distance, the simple answer is zero! The particle passes through both events so the distance between events in the particle's frame is zero. But I guess what you really mean is, what is the distance between two hovering buoys at the start and end points? Off the top of my head, I don't know, I'll have to think about that.

References

1. Woodhouse, N.M.J. (2007), General Relativity, Springer, London, ISBN 978-1-84628-486-1, pages 100, 107, 124.

2. Woodhouse, N.M.J. (2003), http://people.maths.ox.ac.uk/~nwoodh/gr/index.html - online lecture notes on which the book (1) was closely based, pages 54-55 (Lecture 12), p59 (L13), p73 (L15).

Online Integration results

Integrate[1/Sqrt[a^2 - 1 + (2*m)/x], x] ==
(Sqrt[-1 + a^2 + (2*m)/x]*x)/(-1 + a^2) - (m*Log[m + (-1 + a^2 + Sqrt[-1 + a^2]* Sqrt[-1 + a^2 + (2*m)/x])*x])/(-1 + a^2)^(3/2)

Integrate[1/Sqrt[(-2*m)/r3 + (2*m)/x], x] ==
-(r3*(2*Sqrt[m*(-r3^(-1) + x^(-1))]*x + Sqrt[m]*Sqrt[r3]*ArcTan[ (Sqrt[r3]*Sqrt[m*(-r3^(-1) + x^(-1))]* (-r3 + 2*x))/(2*Sqrt[m]*(-r3 + x))]))/ (2*Sqrt[2]*m)

 

[stevebd1]

As you know there is usually more than one physical interpretation of a given situation in relativity. The speed of light is always c as measured by a local observer and according to an observer at infinity the speed of light is:

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

The two observations can be combined into one more general equation:

$$c' = \frac {dr}{dt}=\frac{c(1- r_s/r)}{(1 - r_s/R)}$$

where R is the location of the observer and r is the location of the measurement. It is easy to see when R=r the local speed of light c' = c.

While gravitational time dilation is expressed as -

$$t' = t \sqrt{1-r_s/r}$$

and gravitational redshift is expressed as -

$$z=\frac{1}{\sqrt{1-r_s/r}}-1$$

Referring to the quote above from post #4, how come the 2 parts of the equation for velocity of light observed from infinity and locally, that relate to Schwarzschild metric, aren't square rooted to match time dilation and redshift (i.e. as below)?

$$c' =c\sqrt{(1-\frac {r_s}{r})}$$

$$c' =\frac{c\sqrt{(1- r_s/r)}}{\sqrt({1 - r_s/R)}}$$

In post #10, this seems to have been taken into account-

The proper time rate for a photon is zero so dtau=0.
For horizontal velocity dr=0.
By setting $d\phi$ to zero we are left with:

$$0=(1-{r_s}/{r}) c^2(dt)^{2}-r^{2} (d\theta)^{2}$$

which rearranges to:

$$r (d\theta)/dt = c\sqrt{1-r_s/r} = c_H$$

which is the coordinate tangential velocity of light according to an observer at infinity.

Is there a difference between coordinate tangential velocity of light and simply velocity of light?

Great thread by the way

 

[yuiop]

Is there a difference between coordinate tangential velocity of light and simply velocity of light C?

Great thread by the way

Hi Steve,

"coordinate tangential velocity of light" is just my long winded way of saying the horizontal speed of light according to an observer at infinity which I said equates to:

$$c_{h} = c_{o}\sqrt{1-r_s/r}$$

where $c_{o}$ is "simply velocity of light" as measured locally in a vacuum when there is no gravitational field present or in this case the speed of light as it measured locally in the gravitational field.

The horizontal speed of light according to the observer at infinity can be calculated from the gravitational time dilation equation as:

$$c_{h} = \frac{dx_{h} }{dt } = \frac{dx_{h} '}{(dt ' / \sqrt{1-r_s/r})} = c_{o}\sqrt{1-r_s/r}$$

I also said the vertical coordinate speed of light is:

$$c_{v} = c_{o} (1- r_s/r)$$

which does not seem to agree with the gravitational time dilation factor. This is because the length contraction has not been taken into account. While local horizontal rulers do not length contract, local vertical rulers do and the equation is:

$$dx_{v} ' = \frac{dx_{v}}{ \sqrt{1-r_s/r}}$$

The vertical speed of light according to the observer at infinity can be calculated from the gravitational time dilation equation in the same way as:

$$c_{v} = \frac{dx_{v} }{dt } = \frac{(dx'_{v} \sqrt{1-r_s/r} )}{(dt' / \sqrt{1-r_s/r})} = c_{o}(1-r_s/r)$$

The local observer always measures the speed of light as 299,792,458 meters/second in a vacuum.

Note that only the component of length that is vertical is subject to gravitational length contraction which has an analogy in SR. In SR, only the component of length parallel to the motion, contracts.

 

[yuiop]

I was just trying to derive this

$$c' = \frac {dr}{dt}=\frac{c(1- r_s/r)}{(1 - r_s/R)}$$

from entry #4 and I realized that I have a more fundamental question: do all observers in this static, spherically-symmetric gravitational field agree on the r coordinate of a particular location, regardless of their own location and state of motion?

Hi Snoopies,

When the observer is at R = infinity the above equation reduces to the more familiar

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

As for the agreement on r by all observers the answer is generally no.

It all depends on what methods you use to make the measurements and how the results are interpreted. It is true that if a hovering local observer measures the velocity and orbital period of a local satellite at his location and calculates the circumference of the orbit and divides by 2pi he will get a value for r that agrees with the coordinate value for r. Because there is no horizontal length contraction he will get a similar result if he physically measures a ring co-located with the orbit using rulers. An observer orbiting at the same radius would get a different value for the circumference and possibly for r depending on how he interprets the results. If the orbiting observer considers himself to be stationary there will be no apparent attraction to the main gravitational body and he would have a very strange view of how physics works because from his point of view gravity appears to be working on the hovering observer who appears to be orbiting but has no effect on himself. He will also discover that orbital velocity is directional. If he launches a test satellite in the opposite direction to the the apparently orbiting hovering observer with the same speed he will see it spiral off into space. If he launches another test satellite in the same direction as the hovering "orbiting" observer but with the twice the speed he will see that it too stays in stable orbit. In other word he has a very asymmetrical view of the gravity and how the universe works if he insists he is stationary but he is perfectly entitled to his point of view according to relativity.

If observers measure r using physical vertical rulers they will get a different measurement from the coordinate radius and if the use light radar signals to measure distances they will get yet another measurement which will vary again depending on where the observer is. However, all observers could agree (if the desire and cooperation was there) to convert all measurements to standard measurements that they could plot on one universal map so that they can meaningfully compare measurements. That map would be called the Schwarzschild metric. That would require the orbiting observer to agree he is in fact orbiting and not stationary, but if he comes around to that point of view he would see that the physics of the universe is lot simpler and more symmetrical than he imagined when he insisted he was stationary.