# Pseudo-Superluminal Motion and the Synchronization of Clocks

Let's consider two points A an B separated by a finite distance in curved space time. A light ray flashes across an infinitesimally small spatial interval at B.

Metric: $$ds^2=g_{00}dt^2-g_{11}dx^2-g_{22}dy^2-g_{33}dz^2$$ ----------- (1)

We may write,

$$ds^2=dT^2-dL^2$$ Where,

$dT= \sqrt{g_{00}}dt$ and $dL=\sqrt{ g_{11}dx^2+g_{22}dy^2+g_{33}dz^2}$

dL is the physical infinitesimal(spatial) at B. Observer at A records the same value of dL But for dT the observers at A and B record different values: $$dT_{A}=\sqrt{g_{A(00)}}dt$$ $$dT_{B}=\sqrt{g_{B(00)}}dt$$

Local Speed of light[observation from B]$$=c=\frac{dL}{dT_B}=\frac{dL}{\sqrt{g_{B(00)}dt}}$$

Non-Local Speed of Light:[speed measured from A]

$$C_{non-local}=\frac{dL}{dT_A}=\frac{dL}{\sqrt{g_{A(00)}}dt}$$

Therefore : $$c_{non-local}=c \times \sqrt{\frac{g_{B(00)}}{g_{A(00)}}}$$ --------- (2)

For a particle moving across an infinitesimally small spatial interval at B wehave in a similar fashion: $$V_{non-local}=V_{local} \times \sqrt{\frac{g_{B(00)}}{g_{A(00)}}}$$ -----(3)

Since $V_{local}<c$ $$V_{non-local}<C_{non-local}$$ ----------- (4)

From (2) we find that the non local speed of light may exceed the local barrier. The non_local speed of a particle may also exceed the local speed barrier.

But the light ray is always ahead of the particle [in purely local or non-ocal observations]according to relation (4). There is no violation of relativity as such.

Average speed of light for a specified path:

Let's consider a light ray is passing from P to Q along some path.Observation is being made from the point P.Time interval observed from P is given by:

$$dT=\frac{dL}{C_{non-local}}$$ ------(5)

$$=\frac{dL}{c\times \sqrt{\frac{g_{00}}{g_{P(00)}}}}$$

$g_{00}$ is the value of the metric coefficient at some arbitrary point of the path and $g_{P(00)}$ is the value of the metric coeff. at P. The average speed of light for the path between P and Q obtained by integrating relation (5),to obtain the time of travelof the light ray, may be different from the constant standard value of “c”

Query: Is it important to take care of such differences[differences in the local and non-local speed of light] in the synchronization of clocks in sensitive experiments?

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One immediate comment: your math is slightly wrong, it should be $dL=\sqrt{g_{11}dx^2+g_{22}dy^2+g_{33}dz^2}$ – David Z May 7 '12 at 5:20
I tried copy-pasting from relation(1) forgetting to change the signs. The basic facts/observations remain unchanged.I have done the editing now – Anamitra Palit May 7 '12 at 5:23

This is more or less correct, the intention is clear, but you are saying it in a little bit of a clumsy way.

Your first statement is that if you have light crossing a certain distance, the distance is the same from the point of view of point A and point B, but the time taken to traverse the distance is different. This is correct in the way you set things up, but it isn't best to think of this as the local speed of light being different at A and at B, because the local speed of light is not defined by how far light goes by how much time it takes as measured from A. It is defined by the metric itself--- if the interval-length of the tangent vector to the path of the light ray is zero, then the light ray is by definition moving at the speed of light locally.

Your metric is purely diagonal. This means that little vectors in the direction defined by the t,x,y,z coordinates are always perpendicular to each other. The surface t=constant is therefore a kind of global notion of time, and when you say "a light beam crossed from A to B in a time $\Delta t_A$ as measured at A, you mean that the light crossed from point A at global time slice t, to point B at global time-slice $t+\Delta t$. The time this takes mismatches at A and B, but this is not an indication that the speed of light is different, because both A and B agree that the tangent vector to the light path is still null.

But A and B do not agree on the rate at which global slices separate from each other in time. This effect is extremely important, it is the main effect of general relativity. The first approximation to the effect of a gravitational field, is that there is a local variation in the clock rate. This is a metric of the form

$$ds^2 = -(c^2 - 2\phi(x,y,z))dt^2 + dx^2 + dy^2 + dz^2$$

With explicit c's (usually you set these to 1), where $\phi(x)$ is the newtonian gravitational potential (divided by c^2). This approximation for the metric is not great--- there are spatial metric terms I ignored which are of the same order as the time metric term for the case of light (this is why Einstein's deflection of light is double Newton's--- this approximation reproduces Newton's value), but ignore these. The effect of the Newtonian potential is to make the clock rate incompatible at different positions, in exactly the way you describe.

For the earth's gravitational field $\phi(x,y,z)= gz$, where g is $9.8{m\over s^2}$the potential increases with height as g The difference in the transit time for different positions leads a person at a height h, who labels the time slices according to the faster clock at the higher location, considers the light travelling a shorter distance per time slice than a person who labels the time slices according to the slower clock at a lower position.

The sychronization failure means that clocks at different altitudes must be configured to run at different rates to stay synchronized, and this is the "general relativistic effect" that people talk about with regard to GPS satellites. In addition to their special relativistic time-dilation due to the orbital speed, which correction factor is proportional to $v^2\over 2c^2$ and makes them tick slower, they also have a General Relativistic correction factor proportional to $2\phi\over c^2$, which makes them tick faster than clocks close to the ground.

The virial theorem shows that in a Kepler orbit, the expected value of the Newtonian potential energy is proportional to the expected value of the Newtonian kinetic energy

$$v^2/2 = GM/2r$$ $$\langle {v^2\over 2}\rangle = -{1\over 2}\langle\phi\rangle$$

Where the brackets mean average over one orbit. You can see this directly for a circular orbit from setting the required centripetal equal the Newtonian gravitational force.

$${mv^2\over r} = {GmM\over r^2}$$

But the virial theorem says that it is true on average for elliptical orbits too. So the total slowing-down factor for a clock is

$${1\over c^2}({v^2\over 2} + \phi) = -{1\over 2} {v^2\over 2c^2}$$

And the GR effect is twich as big as the SR effect, and in the opposite direction--- so the orbiting satellite clock is faster than the stationary ground clock by the expected slowing down amount.

The General Relativistic slowing down of clocks near the Earth's surface is also measured by taking atomic clocks on a plane around the Earth, and this is a famous experiment from the 1970s. The effect is equivalent to the redshifting of light as they travel up a gravitational well (the light loses energy as it climbs up the gravitational well, each photon loses frequency). The lost frequency just means that the photon going from A to B keeps its frequency fixed in terms of the clock ticks of A (or in terms of the clock ticks of B) but since the clocks don't match, the frequency as measured locally is different.

These General Relativistic effects are generally known as weak-equivalence principle tests, since they only test that time dilates in the way described by the local metric above. This dilation was derived by Einstein by analyzing a uniformly accelerated frame in special relativity, and deriving the relation between the tick-rate of synchronized clocks in the accelerating frame. He assumed that there is no local difference between the uniform Earth's gravity and constant acceleration, so he was able to find the relation between gravitational potential and the tick-rate for clocks that stay synchronized. The derivation is found in elementary GR books (Schutz is good and especially elementary), and it is also found on the Wikipedia article for gravitational redshift

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Four vectors in the GR perspective: $ds^2=g_{00}dt^2-g_{11}dx^2-g_{22}dy^2-g_{33}dz^2$or,$1=g_{00}(\frac{dt}{ds})^2-‌​g_{11}(\frac{dx}{ds})^2-g_{22}(\frac{dy}{ds})^2-g_{33}(\frac{dt}{ds})^2$.As $ds$ tends to zero the individual components tend to infinitely large values but the norm stays at the value of unity[c=1 in the natural units]. At v=c if the norm is zero we have to define the components forcibly to get zero value for the norm.That would indicate to some discontinuity – Anamitra Palit May 7 '12 at 7:59
On Four Vectors[in flat spacetime]: The 4-vector components are (cdtdτ,dxdτ,dydτ,dzdτ). As v tends to c the four components individually become infinitely large,individually.For v=c the four components become undefined.How do you calculate[rather define] the four components for light if we treat it as a four vector?The three vector concept of light is much simpler. Locally we get the constant un-surpassable value of "c" – – Anamitra Palit May 7 '12 at 8:01
In the previous comments I have referred to the velocity four vector – Anamitra Palit May 7 '12 at 8:17
@AnamitraPalit: Your first comment is correct if you boost the vectors, meaning you try to make a non-null vector null, keeping the length fixed (this is impossible of course)---- you get a diverging spatial component (as the sinh of the relativistic rapidity) and a diverging time compoenent (as the cosh), as you approach the null limit. This is a discontinuity if you insist on taking a null limit by boosting--- the time component blows up. This is physical, it's the blowing up relativistic mass(the energy) of a massive particle as you try to get it to move at the speed of light. – Ron Maimon May 7 '12 at 16:20
@AnamitraPalit: But light always moves at light speed, the velocity vector starts out null, stays null, and doesn't rotate/boost from a non-null value. This does make null things special in a sense--- the null curves have no proper time--- and you parametrize them by affine parameter. An intro to GR book (there are some available online) will make all this clear, and standardize your language (you are saying things in a way that is hard to recognize for someone brought up with traditional terminology). – Ron Maimon May 7 '12 at 16:24