# Do photons traveling from A to B reach B in the same time if A and B are “moving” away from each other with v>c due to space expansion?

Consider two points A and B at rest with respect to each other. When you fire a photon from A to B, the photon obviously can reach B.

Now consider the same two points which, after the photon is fired, "move" away from each other with a velocity greater than the speed of light due to space expansion. If the expansion is constant, the two point move away from each other at an increasing "velocity".

Because the photon is "taken along" by the expansion it's always moving at c with respect to the expansion.

Is the trip for a photon from A to B, where A and B are at rest with respect to each other (no expansion), exactly the same as the trip from A to B when B recedes from A at v>c, no matter how great v? In other words, does the photon reach B at the same time as in the case of A and B are not moving away from each other due to the space expansion between A and B? Or to put it differently again, is the trip in expanding space just a scaled up version of the trip the photon makes from A to B when space isn't expanding?

• No. Space will expand faster than light can overcome, for speeds at the event horizon. See a different but similar case detailed in physics.stackexchange.com/questions/82810/… – Bob Bee Feb 10 '17 at 1:57
• @Bob Bee-But what about inflation? Wasn't the velocity of light increased by an enormous factor, thereby connecting patches of space that were receding from each other at a speed much greater than light? After inflation, the Universe had a size much greater than the currently observable Universe and event horizons developed around every point in space, revealing (by looking at the CMBR) that the Universe is isotropic (on the average). – Deschele Schilder Feb 10 '17 at 11:18

Now consider the same two points which, after the photon is fired, "move" away from each other with a velocity greater than the speed of light due to space expansion.

This is not really something that can be answered as simply as you might imagine because the apparent velocity between A and B due to metric expansion will depend on their distance from each other. And, obviously, as time increases so will distance and therefore apparent velocity. But I will try to answer your question.

The time it takes for the photon to traverse the distance between A and B when space isn't expanding is the easy part. The photon moves at velocity $c$, so the time it takes is:

$$\Delta t = \frac{\Delta s}{c}$$

where $\Delta s$ is the distance between A and B.

Now comes the hard part. You can model 1-dimensional expansion with the following metric:

$$ds^2 = -c^2 dt^2 + a(t)^2 dx^2$$

where the function $a(t)$ is called the "scale factor." It is a function that varies depending on time and determines the distance between points. To model constant expansion over time, you can simply choose a scale factor whose rate of change is constant over time: $a(t) = kt$ for some constant $k$. This gives us the metric:

$$ds^2 = -c^2 dt^2 + k^2 t^2 dx^2$$

Light travels along null geodesics, i.e. where $ds^2=0$, so the differential equation governing the movement of light in this spacetime is:

$$\frac{dx}{dt} = \frac{c}{kt}$$

This equation is nice and simple, and separable:

$$\frac{k}{c} \int_{x_A}^{x_B} dx = \int_{t_A}^{t_B} \frac{dt}{t}$$

$$\frac{k}{c} (x_B - x_A) = \mathrm{ln} \left ( \frac{t_B}{t_A} \right )$$

which can be re-arranged into:

$$\Delta t = t_A \left [ \mathrm{exp} \left ( \frac{k \Delta x}{c} \right ) -1 \right ]$$

where $\Delta t = t_B - t_A$ is the time it takes the photon to reach point B after leaving A, and $\Delta x = x_B - x_A$ is the coordinate separation between A and B. The initial physical distance between A and B is found from the metric, and can be expressed as: $\Delta s = kt_A \Delta x$. Therefore the time it takes for a photon to traverse the distance $\Delta s$ between A and B, starting at time $t_A$, is given by:

$$\Delta t = t_A \left [ \mathrm{exp} \left ( \frac{\Delta s}{ct_A} \right ) -1 \right ]$$

As you can tell, this is a bit more complicated than the equation in the first scenario. It depends not only on the initial distance between A and B, but also on the time when the photon is released. This should make some sense because the size of this universe depends on time. The moment $t=0$ represents the "big bang" of this universe where the physical distance between all points is zero. So the time it takes for the photon to make the journey depends on how long after the big bang you conduct the experiment. You will find, however, that if you try to solve for a time $t_A$ to start the experiment so that the result is the same as the non-expanding trip time, there is no solution. This means that the trip time will never be the same in the expanding universe as it is in the non-expanding universe, no matter when you try to run the experiment.

(As a side note, it's also interesting to notice that $\Delta t$ in the expanding universe does not depend on the rate of expansion, i.e. $k$.)

The apparent velocity between A and B is given by:

$$v_r = \left ( \frac{\dot{a}}{a} \right ) \Delta s = \frac{\Delta s}{t}$$

Plugging this in gives:

$$\Delta t = t_A \left [ \mathrm{exp} \left ( \frac{v_r}{c} \right ) -1 \right ]$$

The trip time therefore exponentially increases with $v_r$. Hopefully this answers your question!

• @jld-Great! But suppose I'm driving to the beach on a road that's 100(km) long at a speed of 50(km/h). Obviously, it takes two ours to get there. Now I make the same trip again but on a road that doubles in length every hour. So after one hour, the distance from where I started is 100 (km). Still, 100(km) to go. My speed relative to my starting point is now 100(km/h). So now I have a 100 (km) road in front of me while driving 100(km/h). This piece is stretched to 200(km) in one hour, and just like the first part, also the second part takes one hour. I'm at the beach in 2 hours too! Or not...? – Deschele Schilder Feb 10 '17 at 9:07
• @jld-It's like looking at a car driving on a road from a plane high in the sky. When you altitude drops at a constant rate, the road seems to lengthen at a constant rate. But the car started at A reaches B at the same time. – Deschele Schilder Feb 10 '17 at 9:13
• @descheleschilder For your first comment: this is not constant expansion, it's exponential expansion. The length of the road is modeled as $L(t)=L_0 2^{kt}$, and the position of the car is $x(t)=v_0 t$, so what you're looking for here is the solution to the equation $x(t)=L(t)$. The solution is non-trivial: $$t=-\frac{W(L_0 \, k \, \mathrm{ln} 2 \, / v_0)}{k \, \mathrm{ln} 2}$$ where $W(x)$ is the product log function. It is certainly NOT as simple as $t= L_0 / v_0$. – Jold Feb 10 '17 at 17:03
• @descheleschilder For your second comment: one obvious difference between you getting closer to the road and the road itself expanding would be that in the latter case the car doesn't also expand! – Jold Feb 10 '17 at 17:07
• @jld-The road, as seen from an aeroplane at a constant descent speed, seems to expand at a constant rate (maybe that's not totally correct but let's assume it's true), and the car also seems to increase in length because we're getting closer. – Deschele Schilder Feb 10 '17 at 17:38