45

The speed of light is the same in all inertial frames.

Does it change from a non-inertial frame to another? Can it be zero?

If it is not constant in non-inertial frames, is it still bounded from above?

Qmechanic
  • 201,751
Revo
  • 16,956

2 Answers2

23

To elaborate on Mark M's answer:

If you consider an accelerating reference frame with respect to Rindler coordinates (where time is measured by idealized point-particle accelerating clocks, and objects at different locations accelerate at different rates in order to preserve proper lengths in the momentarily comoving reference frames), then light may not move at c, and can in fact even stop.

Specifically, for motion in one dimension, consider the transformations in natural units ($c=1$) between cartesian co-ordinates $(t,x)$ to Rindler co-ordinates $(t_R, x_R)$, for an observer accelerating at a rate of $g$ from an initial position $x_I = 1$, in order to maintain a fixed interval from the origin: $$\begin{align*} t_R &= \tfrac{1}{g}\mathop{\mathrm{arctanh}}\left(\frac{t}{x}\right) \;, & x_R &= \sqrt{x^2 - t^2\,}\;; \tag{C $\to$ R} \\[2ex] t &= x_R \sinh(gt_R) \;, & x &= x_R \cosh(gt_R) \;. \tag{R $\to$ C} \end{align*}$$ A light signal emitted from some initial position $x_\varphi$ along the X-axis follows the trajectory $x = x_\varphi + vt$, where $v = \pm 1$ just gives the direction. Consider the trajectory that it follows in Rindler co-ordinates: $$\begin{align*} x_R^2 = x^2 - t^2 &= (x_\varphi + vt)^2 - t^2 \\ &= x_\varphi^2 + 2x_\varphi vt \tag{as $v^2t^2 - t^2 = 0$} \\ &= x_\varphi^2 + 2x_\varphi vx_r \sinh(gt_R)\;; \end{align*}$$ using the quadratic formula, we obtain $$\begin{align*} x_R &= x_\varphi\Bigl[v\,\sinh(gt_R) + \cosh(gt_R)\Bigr] \;=\; x_\varphi\exp(\pm gt_R) \;, \quad\text{for $v = \pm 1$}. \end{align*}$$ Yes, that's an exponential function on the right. It follows that the speed of a light signal is dependent on position in Rindler co-ordinates: the speed of the light signal emitted at $t = 0$ at $x_\varphi$ is $$ \frac{\mathrm dx_R}{\mathrm dt_R} = \pm gx_\varphi \exp(\pm g \cdot 0) = \pm g x_\varphi \;.$$ We can show that the speed of light is a function only of position as follows. A light emission (to either the left or right) from $x_1 > 0$ at $t_1 = 0$ reaches a position $x_2 = x_1 \exp(\pm g t_2)$ after an elapsed time of $t_2$; its speed at that time is $v_{1\to2} = g x_1 \exp(\pm g t_2)$, which is equal to the instantaneous speed of a light signal sent from the position $x_2$ at time $t_1 = 0$. So in natural units, the speed of light in Rindler co-ordinates is $$ c(x) = gx \quad\Bigl[\text{in non-natural units,}\;\; gx/c\Bigr],$$ where $x$ is the location of the light signal. In particular, any light signal appears to travel at the inertial constant speed $c$ just as it passes them.

This has a few consequences. Light signals sent from positions $0 < x_\varphi < x_I = 1$ will move more slowly in the proper time of the Rindler observer, with light signals moving to the right taking longer than usual to catch up to the accelerating observer, up until it reaches them, at which point it seems to travel at $c$. As we take $x_\varphi \to 0$, light signals in any direction appear to slow to a stop. Such beams of light define the Rindler horizon of the reference frame, cutting away a region of space-time from which the observer cannot obtain any information because they see the objects in it accelerating away too quickly, as with the event horizon of a black hole. Conversely, light signals at positions $x_\varphi > x_I = 1$ may appear to be travelling faster than c.

Niel de Beaudrap
  • 4,481
  • I would replace "not entirely unlike" with "exactly the same" (because the black hole metric is locally Rindler in near horizon form), but +1, nice answer. Also, it helps to say that the Rindler coordinates are polar coordinates for Minkowki space, so that the properties are immediately apparent. – Ron Maimon Aug 09 '12 at 17:04
  • @RonMaimon: I've strengthened the language involved accordingly; but I'm going to keep to a somewhat milder wording in order to prevent confusion of the sort that often happens at the boundary of people's understanding of black holes when they are relying on qualitative rather than quantitative descriptions. – Niel de Beaudrap Aug 09 '12 at 17:08
  • Ok, you have a point. I hope that qualitative picture will turn into Rindler space from rubber sheet, because at least the former gets things qualitatively right. – Ron Maimon Aug 09 '12 at 18:26
  • @NieldeBeaudrap: "... objects at different locations accelerate at different rates in order to preserve proper lengths in the momentarily comoving reference frames". There is no such thing as "momentarily comoving reference frames". This phrase refers to $\Delta t =0$. So frames are either moving at their own rates of acceleration, or are frozen in time, i.e. not moving at all. – bright magus Jan 26 '16 at 12:44
  • This language seems standard enough to me, although I do appreciate your point. On such language, see e.g. Tsamparlis 2010 $\S7.2$, from "But how can the statement “I am accelerating in my own proper frame” be conceived kinematically? The answer is as follows."... – Colin MacLaurin Dec 11 '18 at 09:41
12

The speed of light has a velocity of c in an accelerating frame of reference if you constrain yourself to making local measurements. So, the simple answer is that yes, the speed of light remains constant. However, if you don't take purely local measurements, you can get a different speed depending on your coordinate system. If you use a coordinate system where you, an accelerating observer, are at rest (like Rindler coordinates, where time is measured by accelerating clocks and distance is measured by rulers undergoing Born rigid acceleration) then light may not move at c.

Mark M
  • 1,465
  • 4
    What is the difference between "an accelerating frame of reference" and "a coordinate system where you, an accelerating observer, are at rest"? – Niel de Beaudrap Aug 09 '12 at 15:34
  • @Neil There is more than one coordinate system that an accelerating observer can use. If the accelerating observer measures the speed with an inertial ruler and clock that happen to be instantaneously at rest relative to himself, he'll get c as the speed. However, like I said, he can use Rindler coordinates where his measuring apparatus will be affected by Born rigid acceleration. In that case, he won't get c. – Mark M Aug 09 '12 at 15:38
  • 1
    I see --- but that is not an accelerating frame of reference. What you are saying is that the speed is c in each inertial frame of reference corresponding to the instantaneous velocity he has at any point; but this is not surprising, because by definition they are inertial, and in particular none of them are actually "an accelerating frame of reference", and the collection of them aren't even a single frame of reference. – Niel de Beaudrap Aug 09 '12 at 15:47
  • Right, but the reason I pointed that out was to demonstrate that the fact an accelerating observer will measure a different speed of light is dependent on the coordinate system. For example, the equivalence principle requires local measurements of space and time. Similarly, an accelerating observer can find the speed of light to be c if restricts himself to such measurements. Nevertheless, you are correct. – Mark M Aug 09 '12 at 15:58
  • @MarkM: "If the accelerating observer measures the speed with an inertial ruler and clock that happen to be instantaneously at rest." It is impossible to measure any speed by an accelerating observer with "an inertial ruler and clock instantaneously at rest", because there is no movement at an instant. Movement requires a change of time, by definition. – bright magus Jan 26 '16 at 12:33