There is a difference between the concept of the speed of light and the velocity of light. are both of them constant ($dc=0$ and $dv_c=0$)? if yes, why?.
1
-
4$\begingroup$ Possible duplicate of How to bend light. $\endgroup$– MostafaCommented Jun 2, 2013 at 20:38
Add a comment
|
5 Answers
$\begingroup$
$\endgroup$
0
Hint: Velocities include directions. Light can travel in different directions. So...
$\begingroup$
$\endgroup$
3
The speed of light is constant. The velocity of light should be, unless light changes direction. Speed is the magnitude of velocity, a scalar quantity (has size, but not direction), whereas velocity is a vector, which has both magnitude and direction. c is defined as speed, which has only magnitude. There is no definition of the velocity of light, but if the light was traveling north when last measured, velocity could be negative, positive, or zero based on the direction traveled.
-
1$\begingroup$ The last sentence doesn't make much sense. Vectors aren't positive or negative. $\endgroup$– user4552Commented Jun 3, 2013 at 3:06
-
$\begingroup$ The vector can be defined as negative, positive, or zero depending on the direction and coordinate system. By themselves, vectors are not positive or negative, only in reference to a coordinate system. $\endgroup$ Commented Jun 5, 2013 at 16:53
-
2$\begingroup$ No, vectors can have positive and negative components. In any coordinate system, there will be vectors whose components are a mixture of positive and negative values. $\endgroup$– user4552Commented Jun 6, 2013 at 19:18
$\begingroup$
$\endgroup$
The speed of light in vacuum is constant, otherwise is could not be well defined. It is called a Universal Constant for this reason. It does not vary even with your reference frame. That is to say, if you are standing next to your friend who is holding a flash light, the light will appear to be moving the 'same speed' to both of you, even if you are running in one direction. This is different from if your friend were to throw a ball, which would then appear to be traveling at different velocities depending on your reference frame: i.e. it may be moving faster or slower in your reference frame depending on whether you are running in the same or opposite direction as the ball.
The speed of light in vacuum is given by
$c = 299,792,458$ meters per second
Velocity, as cuabanana stated, is just the speed defined with a direction. The magnitude of the velocity (i.e. speed) is constant, given by the same number above.
If light is propagating through a transparent medium, such as glass, it will actually travel slower than the speed of light in vacuum, $c$. The speed of light in a in a medium is given by:
$$v_\text{light} = \frac{c}{n}$$
where $n$ is called the index of refraction of the material. The index of refraction for vacuum is 1: hence the speed of light in vacuum is $c$. The index of refraction for glass is about 1.5, and thus the speed of light in glass is about $c/1.5 \approx 200,000,000$ meters/second.
$\begingroup$
$\endgroup$
There are two questions here -- is the velocity of light constant, and is it invariant?
The direction/velocity of light changes whenever it interacts with something. This includes gravitational deflection, since things have to change direction in curved spacetime in one sense or another. The velocity isn't constant.
Is it invariant under Lorentz boosts in perpendiculal directions? No. The speed is invariant, but the velocity isn't. This should be fairly clear, but you can prove it with brute force --
We need to apply a boost to light's four-velocity, but the four-velocity of light is actually infinite -- it's (infinity, infinity, 0, 0), except the infinities satisfy a certain relation in the sense of being related through a limit. So we consider an object traveling at speed $w$ in the $x$-direction, boost $v$ in the $y$-direction and let $w\to c$. The four-velocity transforms under this boost as:
$$\left[ {\begin{array}{*{20}{c}}{\gamma (w)}\\{w\gamma (w)}\\0\\0\end{array}} \right] \to \left[ {\begin{array}{*{20}{c}}{\gamma (v)\gamma (w)}\\{w\gamma (w)}\\{ - v\gamma (v)\gamma (w)}\\0\end{array}} \right]$$
The conventional 3-velocity can be extracted here by considering $dx/dt$, $dy/dt$:
$$\frac{{dx}}{{dt}} = \frac{{dx/d\tau }}{{dt/d\tau }} = \frac{{w\gamma (w)}}{{\gamma (v)\gamma (w)}} = \frac{w}{{\gamma (v)}}$$ $$\frac{{dy}}{{dt}} = \frac{{dy/d\tau }}{{dt/d\tau }} = \frac{{ - v\gamma (v)\gamma (w)}}{{\gamma (v)\gamma (w)}} = - v$$
Taking the limit as $w\to 1$, you get a 3-velocity of $(1/\gamma(v),-v, 0)$ -- one may confirm that this is not equivalent to the original three-velocity that was $(1,0,0)$, but nonetheless has the same magnitude (speed is invariant).
answered Jun 19, 2013 at 4:17
$\begingroup$
$\endgroup$
7
Velocity of light in higher in Casimir vacuum. It is called Scharnhorst effect.
-
$\begingroup$ Higher than what? How high? $\endgroup$ Commented Jun 3, 2013 at 2:52
-
$\begingroup$ @Brandon Enright higher than the constant c. $\endgroup$– AnixxCommented Jun 3, 2013 at 2:53
-
$\begingroup$ This is a hypothetical effect. There are other effects where the speed of light is higher than $c$, but they don't imply a signal going faster than $c$. $\endgroup$– fffredCommented Jun 3, 2013 at 2:58
-
1$\begingroup$ @Anixx, I do not agree: the question is between velocity and speed. The effect you cite has not even been experimentally proven so you can not use it as an example of non-constant $c$. $\endgroup$– fffredCommented Jun 3, 2013 at 3:18
-
1$\begingroup$ That effect is pure fantasy. Shame on wikipedia for even having an article about it. $\endgroup$– user10851Commented Jun 3, 2013 at 3:20