# Problem on speed of light

As I was reading special relativity, my book says the speed of light is $c$ with respect to any other thing. Does that mean the speed of an individual photon is $c$ even with respect to another photon? I mean, shouldn't the relative velocity be zero?

• Related: physics.stackexchange.com/q/16018. That comes awfully closing to being a duplicate because the answer is "the question doesn't make sense because it presupposes a rest frame for one of the photons." Also physics.stackexchange.com/q/68600 . – dmckee Jul 12 '14 at 16:26
• I've answered his question in detail. While it's true that the answer to the question you posted must be mentioned, it doesn't strike at the heart of what he's asking. – ticster Jul 12 '14 at 16:31
• Which book? Which page? – Qmechanic Jul 12 '14 at 17:11

Does that mean the speed of an individual photon is c even with respect to another photon? I mean, shouldn't the relative velocity be zero?

When we write

"the speed with respect to X"


we mean precisely

"the speed as observed from the inertial frame of reference in which X is at rest"


Thus, if it is true that the speed of an individual photon is $c$ as observed from any inertial frame of reference, it logically follows that there is no inertial frame of reference in which a photon is at rest.

In other words, if it is true that the speed of an individual photon is $c$ as observed from any inertial frame of reference, then the phrase with respect to another photon presumes something that isn't true.

The whole point of special relativity is that imposing this postulate (on the constancy of the speed of light) forces a reunderstanding on the very geometry of space and time. This re-understanding can be summarized by the Lorentz transformations. These describe the coordinate transformations required when transferring measurement from one observer in reference frame $S$ to another, $S^\prime$, that is moving relative to the first one. If the motion is along the x-axis, these are :

\begin{eqnarray} t^\prime &=& \gamma (t - vx/c^2) \\ x^\prime &=& \gamma (x - vt) \\ y^\prime &=& y \\ z^\prime &=& z \end{eqnarray}

Where $\gamma = 1 / \sqrt{1 - v^2/c^2}$. These relations are quite different than the more intuitive Galilean relations, and herein lies the answer to your question. Notice that we hit a singularity for $v = c$, so we can't answer your exact question. In other words, it is impossible to see things from the point of view of an observer moving at the speed of light. However, let us change your question a bit in a way that makes it possible to answer your conundrum. You are observer $O$. Relative to you, observer $O^\prime$ is moving at $0.9c$. You observe a photon moving toward $O^\prime$ at speed $c$. Now, let us transform the motion of that photon as seen by $O$ to the reference frame of $O^\prime$. Of course, if you were to use the classical Galilean transformations, you would find $1.9c$. However, if you were to use the relations given above, you would again find that the speed of the photon is $c$. Those relations were specifically designed to obtain that answer.

The idea behind special relativity is that the laws of physics do not change from one reference frame to another (this is key). So, if you're moving at velocity v, a photon traveling in that reference frame will travel at c (not c - v, which is the classical Galilean transformation).

I believe this is what your book means when it says "with respect to any other thing."

Now, about your question regarding two photos - I don't believe that you can travel at exactly c without some consequences. For instance, the look at the equation for time dilation - it blows up for v = c (the Lorentz factor has a singularity).