I am currently studying Classical Mechanics, 5th edition, by Kibble and Berkshire. Chapter 1.2 Newton's Laws says the following:

However, this does not completely remove the difficulty, for there is still an apparent contradiction between this classical electromagnetic theory and the principle of relativity discussed in §1.1. This arises from the fact that if the speed of light is constant with respect to one inertial frame — as it should be according to electromagnetic theory — then the usual rules for combining velocities would lead to the conclusion that it is not constant with respect to a relatively moving frame, in contradiction with the statement that all inertial frames are equivalent. This paradox can only be resolved by the introduction of Einstein’s theory of relativity (i.e., ‘special’ relativity). Classical electromagnetic theory and classical mechanics can be incorporated into a single self-consistent theory, but only by ignoring the relativity principle and sticking to one ‘preferred’ inertial frame.

The aforementioned "principle of relativity" is described as follows:

Given two bodies moving with constant relative velocity, it is impossible in principle to decide which of them is at rest, and which moving. This statement, which is of fundamental importance, is the principle of relativity.

I don't understand the authors' description of the problem. I would greatly appreciate it if people would please take the time to explain this to me.

This initially seems like it might be related, but upon reading, it doesn't seem like it addresses my question.


2 Answers 2


I'm not sure if I've understood the question exactly, but it seems to be asking what exactly it is about the (Galilean) principle of relativity that makes (Classical) Electromagnetic theory incompatible with it.

The Principle of Relativity, as you say, basically states that there's no physical experiment that you can conduct in two different inertial frames moving with respect to each other to tell "which" of them is "actually" moving: the very question makes no sense. This is a consequence of the way that space and time transform when you move from one inertial frame to another.


Let's try to put this into a mathematical formalism using our "common-sense". Suppose you had two people, one person ($S^\prime$) in a train, moving rightwards with a velocity $v$, and another on a platform ($S$), and if the person on the platform would like to know what the coordinates of an event $(x,t)$ would look like to someone on the train $(x^\prime, t^\prime)$. For simplicity, we can choose $x=0=x^\prime$ and $t=0=t^\prime$. Using "common sense", it's a simple enough exercise to show from the above figure that:

\begin{equation*} \begin{aligned} x^\prime &= x - vt\\ t^\prime &= t \end{aligned} \end{equation*}

These are called the Galilean Transformations, and they relate the way that the person in $S$ observes the world, to the way that the person in $S^\prime$ observes the world. A natural consequence of this is that the two observers will never agree upon the velocities of a third object. For example, suppose there was a dog running (rightwards) on the train, and both observers were measuring its velocity. The velocity measured by someone on the platform would be

$$u = \frac{\Delta x}{\Delta t},$$

while that observed by someone on the train would be $$u^\prime = \frac{\Delta x^\prime}{\Delta t^\prime}.$$

Using the above Galilean Transformations, and the fact that $v$ is a constant, it's quite trivial to show that a natural consequence is that $$u^\prime = u - v.$$

In other words, the velocity of the dog as observed from the train would be smaller than the velocity of the dog as observed from the platform by exactly the relative velocity of the train and the platform. But then, this is just common sense, right? Furthermore, we can do the same trick above to show that both observers would always agree upon the acceleration of the dog, since:

$$a^\prime = \frac{\Delta u^\prime}{\Delta t^\prime} = \frac{\Delta u}{\Delta t} = a,$$

as should be expected since the frames are inertial, but which also means that the physical laws in both frames will have the same form, since Newton's Laws only deal with accelerations.

An important consequence, however, of this "velocity addition" formula is that there is no (finite) velocity that observers in $S$ and $S^\prime$ will agree upon, since if this were the case, you would need to have

\begin{equation*} \begin{aligned} u^\prime &= u - v \text{ (which is always true according to Galilean Relativity)}\\ u^\prime &= k = u \text{ (for some frame-independent speed $k$)} \end{aligned} \end{equation*}

The only value of $k$ that solves the above equations is $k=\infty$. The only frame-independent speed (i.e.\ the only speed that all observers in all inertial frames can agree upon) is infinity.

Since this is a direct consequence of the Galilean Transformations, if we had another theory that required a frame-independent speed that was finite, then we would have to say that that theory was not compatible with the Galilean Principle of Relativity, and this is precisely what Classical Electromagnetism required. According to it, there is a speed, $c$, which -- despite being very large -- is not infinite, and which everyone in all inertial frames could agree upon. Thus, if Electromagnetism was to be considered "right", we require either that our "common-sense" Galilean Relativity is wrong, or that Classical Electromagnetism is not compatible with Principle of Relativity.

However, it turns out that Classical Electromagnetism is compatible with the theory of relativity, it is just that our "common-sense" Galilean Transformations are not true, and they have to be replaced with the correct "Lorentz Transformations", leading to the Special Theory of Relativity, which is consistent with Electromagnetism. (In fact, it's a fun exercise to how that the magnetic field is an entirely relativistic effect!)

  • $\begingroup$ How exactly do you show that $u^\prime = u - v$? I do not know the standard way of treating $\Delta$ after substitution of $x^\prime = x - vt$. $\endgroup$ May 28, 2020 at 6:56
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    $\begingroup$ I'm not sure I completely understand the question, but let me see if I can answer it: since $x^\prime = x - v t$, you can take the difference between two coordinates and show that $\Delta x^\prime = \Delta x - v \Delta t$, since $v$ is a constant. You would like to relate $\Delta x^\prime/\Delta t^\prime$ to $\Delta x/\Delta t$, and so you can divide the LHS by $\Delta t^\prime$ and the RHS by $\Delta t$, and you'd get $$\frac{\Delta x^\prime}{\Delta t^\prime} = \frac{\Delta x}{\Delta t} - v,$$ and from here you can just use the definition of speed to show $u^\prime = u -v$. $\endgroup$
    – Philip
    May 28, 2020 at 10:11
  • $\begingroup$ Thanks for the response. My point was that, in trying to show that $u^\prime = u - v$, I was thinking that we would have $u^\prime = \dfrac{\Delta(x - vt)}{\Delta t}$, but I wasn't sure if this substation was mathematically valid (given the presence of the $\Delta$). But, if I'm following your comment correctly, $\Delta$ has the distributivity property, and so we have that $u^\prime = \dfrac{\Delta(x - vt)}{\Delta t} = \dfrac{\Delta x - v \Delta t}{\Delta t}$? When you say "take the difference between two coordinates", are you referring to the difference between $(x, y)$ and $(x', y')$? $\endgroup$ May 28, 2020 at 10:35
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    $\begingroup$ Ah, I understand now! The $\Delta x$ is just shorthand notation for some $x_B - x_A$. It turns out that such "differences in coordinates" also satisfy the same transformations (you can show this easily): $$x_B^\prime - x_A^\prime = (x_B - x_A) - v (t_B - t_A), \quad \quad t^\prime_B - t^\prime_A = t_B - t_A,$$ and so if you wanted the speed between two points $A$ and $B$, you could just divide $$\frac{x_B - x_A}{t_B - t_A} = u,$$ and do the same for someone in $S^\prime$: $$\frac{x^\prime_B - x^\prime_A}{t^\prime_B - t^\prime_A} = u^\prime,$$ and use the transformation laws given above. $\endgroup$
    – Philip
    May 28, 2020 at 10:55
  • $\begingroup$ Ahh, I see. And when you say "speed" above, you actually mean velocity, right? $\endgroup$ May 28, 2020 at 11:19

I will give my take on this.

The statement for the relativity principle is that two inertial frames, any two and by extension all of them, are equivalent. Thus, a moving object would be described by the same laws in any reference frame, only changing coordinates accordingly. However, here we have two choices, so to speak. One is to use Galilean transformation laws between two reference frames, Newton's laws would be the same and the kinematics and dynamics the same as well. The problem is that Maxwell's laws and specially the wave equation (for light) you find in them is not invariant under such coordinate transformation $x \rightarrow x'= x - Vt.$ Moreover, they are invariant under Lorentz transformations which is the other choice. The problem is that Newton's laws are not invariant under Lorentz transformations and therefore we need to decide what to do.

In Galilean transformations, to find the speed of any phenomena (i.e. $v$) when changing from a reference frame to a moving one (lets say with speed $V$), we add them as in $v \rightarrow v' = v -V.$ This means that there is always a reference frame for which the phenomena can be at rest. If they are equivalent then light could be stopped and studied at rest in that frame, the wave equation in that frame should have $c=0.$ If one is to accept that light follows that same rule of Galileo, there should be a reference frame for which light is at rest.

For electromagnetic waves that is not the case, one can not find a reference frame where light is at rest, all reference frames, no matter what, measure the same speed for light. Galilean transformations are not correct for Maxwell's theory.

If one accepts that light is always moving then one should accept that there is a preferred reference frame with respect to which light is always moving and that other reference frames have to be studied with respect to that particular one to avoid a photon a rest (lets say to forbid traveling at $c$) and that Maxwell's laws are actually written with respect to that reference frame. Therefore, not all of them are equivalent, one is actually in absolute rest so light have $c$ constant in that frame and is mandatory to use that for light. So, we need to renounce the principle of relativity of equivalence of all reference frames.


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