- #1
jcap
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Let us assume that an observer is stationary at the origin in expanding space. We assume the FRW metric near the origin is given by:
$$ds^2=-dt^2+a(t)^2dr^2$$
Let us assume that the observer measures time by bouncing a light beam at a mirror that is at a constant proper unit distance away from him.
By substituting ##dt=0## in the metric we find that an interval of proper distance ##ds## is given by:
$$ds=a(t)\ dr$$
Integrating we find the proper distance ##S## to an object at co-moving co-ordinate ##r## is given by:
$$S=a(t)\ r$$
Thus if the mirror has constant proper distance ##S=1## then it follows a path in co-moving co-ordinates given by:
$$r=\frac{1}{a(t)}$$
A light beam is described by substituting ##ds=0## in the metric to get the null geodesic:
$$dr=\frac{dt}{a(t)}$$
Integrating we get the path of a light beam:
$$r = \int \frac{dt}{a(t)}$$
The spacetime diagram below shows the observer attempting to measure cosmological time ##t## using the light clock.
One can see that as the mirror gets closer in co-moving co-ordinate ##r## the period of the light clock gets smaller and smaller. Thus the light clock is getting faster relative to cosmological time ##t##.
Now let us make a transformation to conformal time ##\tau## given by the relationship:
$$d\tau=\frac{dt}{a(t)}$$
The transformed radial co-ordinate ##\rho## is given by:
$$\rho\ d\tau = r\ dt$$
Therefore the path of the mirror in conformal co-ordinates is given by:
$$\rho = r\ \frac{dt}{d\tau}$$
$$\rho = \frac{1}{a(t)} \cdot a(t) = 1$$
The path of a light beam in co-moving co-ordinates is given by:
$$dr=\frac{dt}{a(t)}$$
Therefore in conformal co-ordinates we have:
$$d\rho=d\tau$$
On integrating:
$$\rho = \tau$$
Thus in conformal co-ordinates the light clock is described by the diagram below:
One can now see that the light clock period is constant - the clock is working properly.
Thus a light clock measures conformal time ##\tau## rather than cosmological time ##t##.
Is this argument correct?
$$ds^2=-dt^2+a(t)^2dr^2$$
Let us assume that the observer measures time by bouncing a light beam at a mirror that is at a constant proper unit distance away from him.
By substituting ##dt=0## in the metric we find that an interval of proper distance ##ds## is given by:
$$ds=a(t)\ dr$$
Integrating we find the proper distance ##S## to an object at co-moving co-ordinate ##r## is given by:
$$S=a(t)\ r$$
Thus if the mirror has constant proper distance ##S=1## then it follows a path in co-moving co-ordinates given by:
$$r=\frac{1}{a(t)}$$
A light beam is described by substituting ##ds=0## in the metric to get the null geodesic:
$$dr=\frac{dt}{a(t)}$$
Integrating we get the path of a light beam:
$$r = \int \frac{dt}{a(t)}$$
The spacetime diagram below shows the observer attempting to measure cosmological time ##t## using the light clock.
One can see that as the mirror gets closer in co-moving co-ordinate ##r## the period of the light clock gets smaller and smaller. Thus the light clock is getting faster relative to cosmological time ##t##.
Now let us make a transformation to conformal time ##\tau## given by the relationship:
$$d\tau=\frac{dt}{a(t)}$$
The transformed radial co-ordinate ##\rho## is given by:
$$\rho\ d\tau = r\ dt$$
Therefore the path of the mirror in conformal co-ordinates is given by:
$$\rho = r\ \frac{dt}{d\tau}$$
$$\rho = \frac{1}{a(t)} \cdot a(t) = 1$$
The path of a light beam in co-moving co-ordinates is given by:
$$dr=\frac{dt}{a(t)}$$
Therefore in conformal co-ordinates we have:
$$d\rho=d\tau$$
On integrating:
$$\rho = \tau$$
Thus in conformal co-ordinates the light clock is described by the diagram below:
One can now see that the light clock period is constant - the clock is working properly.
Thus a light clock measures conformal time ##\tau## rather than cosmological time ##t##.
Is this argument correct?