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?
[Physics] Does the speed of light vary in non-inertial frames
general-relativityinertial-framesreference framesspecial-relativityspeed-of-light
Related Solutions
You have said: If,for instance,the relative motion observed between two frames of reference is that of uniform acceleration, how can we determine which frame is the unaccelerated system? It is obviously not possible. and
Another part of this very question is also: How can we call the occupied frame of reference as being inertial regardless of whether other frames of reference are accelerating with respect to the occupied frame of reference?
Both these questions have been answered below.
Why would it not be possible? If you are in a reference frame which is accelerating at all, then you will experience pseudo-forces(forces whose source is not determined in that frame). That will tell you that your frame is accelerating. Moreover,if the relative motion between two frames is that of uniform acceleration,then both are accelerating! You do not have to determine WHICH is accelerating! The presence of acceleration(uniform or not) for any reference frame, guarantees that you will experience pseudo-force if you are in it. for example, if you throw a ball from a height,it seems to hit the ground after travelling a path perpendicular to ground. but the actual trajectory is not so. as the ball falls it is deflected due to Coriolis force,which is a pseudo-force. so technically the earth is not an inertial frame of reference in any way since we can never point to a source who caused this Coriolis force!
You have said: Resnick states that the frame of reference he occupies is an unaccelerated one. With respect to what? If accelerated motion were to be observed with respect to other frames of reference, how are we to determine that we occupy an inertial frame of reference at all?
According to Resnick he occupies an inertial frame that means, in his frame, Newton's first law holds true. obviously you need a reference object. when we say a car travels at 75m/s then we actualy mean it travels 75m/s with respect to, say,a stationary tree. but it would travel at 50m/s with respect to another car travelling with 25m/s. so you need a reference object.
But then someone talked to me about Principle of Equivalence and not possibly being able to identify what is proper acceleration and what is coordinate acceleration with an accelerometer. Is it true ?
That's not true. By definition, an ideal accelerometer measures proper acceleration.
It appears you (and possibly the acquaintance who talked to you) are mixing and matching concepts from Newtonian mechanics and general relativity. Don't do that! Inertial frames in Newtonian mechanics and general relativity are rather different beasts.
The concept of an inertial frame is extremely important in Newtonian mechanics, not so important in general relativity. Proper and coordinate acceleration are concepts from relativity theory. In Newtonian mechanics, gravitation is a real force, but accelerometers can't measure it.
Newton's first law conceptually provides a way to test whether a frame is inertial: Simply find a test particle on which the net force is zero. Does the object appear to obey or disobey Newton's first law? The only problem: Good luck with this search!
That approach works nicely in general relativity. In fact, Gravity Probe B used exactly this approach. An accelerometer that registers zero acceleration does make for a local inertial frame in GR. Gravity Probe B flew low, so it was subject to drag. It had a free-floating test mass in its core. It used its thrusters to force the main part of the probe to accelerate just so and keep that free-floating mass centered. In doing this, Gravity Probe B was flying inertially from the perspective of general relativity. Because it was flying low, it was subject to some of the finer aspects of general relativity, and scientists could thereby use the observed motion as a test of general relativity.
Best Answer
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.