Suppose we have a massive non-rotating planet, so that time is dilated on its surface.

There are two cities on the planet, city A and city B, and it takes light 1 ms to travel between them.

But what time will it take from the point of view of an observer in space?

Suppose the city A when sending signal to city B also sends signal to the observer in space. The city B also sends the observer in space a notification when it recieves the signal.

Due to time dilation on the planet's surface, the remote observer should notice that the signal travelled more than 1 ms.

But the speed of light should be the same. Does it mean that the distance between the cities A and B is greater from the point of view of a remote observer? So, the area of the planet is greater as well?

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    $\begingroup$ The light signal will go in a circular orbit only if the radius of the planet is precisely $\dfrac{3GM}{c^2}$. I haven't fully understood your question but seems like sending a photon directly from one place on the sphere to another is a crucial part of your set-up. This will work only if the radius is precisely $\dfrac{3GM}{c^2}$. Apart from that, can you clarify light takes $1$ $ms$ to travel from $A$ to $B$ as observed by whom? The proper time of light itself will certainly be zero. $\endgroup$ – Dvij Mankad Jun 16 '17 at 18:18
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    $\begingroup$ The speed of light measured by a far observer will not be c speed-light.info/speed_of_light_variable.htm $\endgroup$ – user126422 Jun 16 '17 at 18:33
  • $\begingroup$ I think the question is rather more mundane than one needed the photon sphere, he's just trying to use the travel time as clock. But a milisecond is 300 kilometers, so it is perhaps too far for easily rigging up the experiment. Make it 100 microseconds (so the distance is 30 kilometers) and then put the lights in towers or the towns on the walls of a valley so that there is a direct line of sight. $\endgroup$ – dmckee Jun 16 '17 at 23:00

I will have to first transform your question in a different but equivalent one.

Assume there is a direct line of sight between A and B.

Suppose you send a signal from A, that gets reflected from B and is received by A after 2 ms in frame of reference of A. As A and B are stationary wrt to each other, there frame of reference is same, assuming same escape velocity at A and B.

Suppose/assume speed of light is same both ways, from A to B and B to A.

Suppose the observer in space (C) is equidistant from A and from B and is stationary wrt A and B. Meaning same frame of reference (apart from gravitational difference) and suppose light takes same amount of time from A to C and from B to C.

Now your question can be reduced to -

A sends two signals, 1 ms (in A's frame) apart to C. Will C receive the two signals 1 ms apart in its frame of reference?

The answer should be no. It would be more than 1 ms as C's clock running faster than that of A due to gravitational difference.

May be someone can demonstrate it mathematically.

If you assume C in far space (free from the planet's gravity), then there will be no time dilation at C and the time dilation at A would be equivalent of that due to a speed in far space which is equal to the escape velocity at A. This can quantify the difference in time as a special scenario.


I believe you are correct in thinking that since the speed of light is the same to the observer in space and it takes more time for light to travel between A and B according to him, he will measure the distance AB greater. As to the surface area, keep in mind that the Euclidean formula $ S = 4 \pi r^2$ for the surface area of a sphere might not hold, because we are dealing with non-Euclidean geometry caused by gravity and curvature here.


A light-millisecond is a fairly sizable distance on the scale of a planet, so I'll simplify the question by assuming that points A and B on the planet's surface are actually close enough together that vertical lines from A and B are negligibly different from being parallel, and acceleration effects are negligible in the time it takes to send a light pulse from A to B. Under those assumptions, it works to treat A and B as sharing an inertial frame of reference (the "surface frame") during the process of the light pulse going between A and B.

I'll answer your second question first. Using the typical assumptions in a gravitational time dilation problem, the horizontal distance between A and B is the same between the distant observer's inertial frame of reference either before or after having been parallel transported to the surface (hereinafter just called the "distant frame") and the surface frame, $$\Delta x_d=\Delta x_s\ .$$ This fact can be easily deduced from the fact that parallel transport is defined such that it never changes the inner product $g_{\mu\nu} u^{\mu}v^{\nu}$ between any two vectors. So a spatial four-vector pointing in the direction from A to B on the surface must have been parallel transported from a four-vector of the same magnitude at the distant observer (because the inner product of that vector with itself remains constant), and must also be a purely spatial four-vector at the distant observer due to the symmetry of the problem (assuming the distant observer remains at a fixed location above the planet).

In addition to the two frames agreeing on the horizontal distance between A and B, the two frames also agree that the speed of light is the same constant $c$.

The reason these two facts can both hold and still be consistent with the gravitational time dilation is because the two frames are moving relative to each other vertically. The surface frame is stationary relative to the planet's surface, but at the surface the distant frame is moving vertically at a speed equal to the escape velocity at the surface (assuming that the far away observer is too far away to be affected by the planet's gravity, and assuming that the far away observer is at a fixed $r$ Schwarzschild coordinate instead of moving away from or toward the planet).

You can see that it makes intuitive sense for the distant frame to be moving at the escape velocity $v_e$ at the planet's surface, if you consider a projectile fired from the surface toward the distant observer at a speed $v_e-\epsilon$ that's a smidgen less than the escape velocity. A local inertial frame of reference centered on the projectile is a good way of picturing what parallel transport of a frame of reference is. At some point along its orbit, the projectile will be near the distant observer, and moving only at speed $\epsilon$ relative to the distant observer, so the projectile and the distant observer can at that point be considered to be using nearly the same inertial frame of reference. At a later point in its orbit, the projectile will again be at the planet's surface, again moving at speed $v_e-\epsilon$. Whether the frame shared by the projectile and the distant observer when they are near each other is parallel transported backward or forward in time to the surface, either way the frame is moving at a speed near $v_e$ at the surface. (The magnitude of the frame's motion is important, but the direction is unimportant in what follows.)

According to the surface frame, the time it takes the pulse to go from A to B is just $$\Delta t_s=\frac{\Delta x_s}{c}\ .$$ But according to the distant frame, it takes longer that that, because the pulse travels not only horizontally by a distance $\Delta x_d=\Delta x_s$, but also travels vertically by a distance $v\Delta t_d$. In other words, $$\Delta t_d=\frac{\sqrt{(\Delta x_d)^2+(v_e \Delta t_d)^2}}{c}\ .$$ Solving for $\Delta t_d$ gives $$\Delta t_d = \gamma \Delta t_s\ ,$$ where $$\gamma =\frac{1}{\sqrt{1-v_e^2 / c^2}}$$ is the Lorentz factor for the escape velocity at the surface.

At the level of analyzing the pulse between A and B, it's the exact same physics as with relative velocity time dilation. Calling it "gravitational" time dilation is just a way of pointing out that one of the two frames involved was obtained by parallel transporting some far away frame to the surface.


A and B are on earth, but must be closer together than 1 ms (300 km) so they can see each other from the tops of reasonable height towers. An observer on earth using his own meter stick measures a distance $\Delta x$ between A and B. Using his own clock he also measures the time $\Delta t$ for a light pulse to travel between A and B. He confirms that $\frac{\Delta x}{\Delta t}$ yields the standard speed of light $c=3 \ 10^8 \ km/sec$.

Then, yes the observer far out in space and using his own clock will see $\Delta t'>\Delta t$. He will also see (by the angular separation between A and B) that $\Delta x'=\Delta x$ and conclude that the light (which is deep in the earth's gravitational potential compared to the far out observer) traveled between A and B with a speed $c'=\frac{\Delta x'}{\Delta t'}$ where $c'<c$. This is the same effect as seen in the Shapiro Delay where the far out observer sees a radar signal passing near the sun travel at $c'<c$.

What caused the confusion in your question was your statement: “But the speed of light should be the same….”. It isn't. It is true that the coordinate speed of light is the same $c$ in all frames with a Minkowski metric and that Lorentz transformations leave the Minkowski metric unchanged. However, the Scwarzschild field (and Gravity Waves) do strain transformations (which are not Lorentz transformations) to the coordinates and to the metric indices. For these strains and for Lorentz transformations, the proper length $g’_{\mu\nu} \Delta x’^{\mu}\Delta x’^{\nu}=\eta_{\mu\nu} \Delta x^{\mu}\Delta x^{\nu}$ remains unchanged. These strain transformations do not leave the Minkowski metric $\eta_{\mu\nu}$ invariant. Instead the strains turn the Minkowski metric into the familiar Schwarzschild metric. The new $ g’_{\mu\nu} $ Schwarzschild metric tells you that the far away observer will measure $c'<c$ for the new coordinate speed of light passing at a radius r in the Schwarzschild field. The proper length along the path of a photon is 0, $\Delta r=0$ between A and B, and $\Delta x’= r \Delta \theta$ in the Schwarzschild coordinates. $$ 0=(1-2GM/r)(c\Delta t’)^2-(\Delta x’)^2 $$ $$ c’=\frac{\Delta x’}{\Delta t’}=(1-2GM/r)^{1/2}c $$ The summary answer to the question is: The far away observer sees the same distance AB (for any AB direction perpendicular to the radius vector) and the same area of the earth as the observer on earth. The far away observer sees the light take more time $\Delta t'>\Delta t$ to go from A to B, and the light appears to go slower than the usual c such that $\Delta x'=c'\Delta t'=c\Delta t=\Delta x$.


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