Could relativity be consistent if there are multiple light-like fields with different invariant speeds?

Question

My understanding of real physical theory of electromagnetism goes like this:

The Maxwell equations can be used to derive the speed of light; $$\nabla\cdot\textbf{E}=0$$ $$\nabla\cdot\textbf{B}=0$$ $$\nabla\times\textbf{E}=-\frac{\partial\textbf{B}}{\partial t}$$ $$\nabla\times\textbf{B}=\mu_0\epsilon_0\frac{\partial\textbf{E}}{\partial t}$$ gives $$c=\frac{1}{\sqrt{\mu_0\epsilon_0}},$$ and Special Relativity derives from the fact that this $c$ is constant in all reference frames.

Suppose there was another light-like field $$\nabla\cdot\hat{\textbf{E}}=0$$ $$\nabla\cdot\hat{\textbf{B}}=0$$ $$\nabla\times\hat{\textbf{E}}=-\frac{\partial\hat{\textbf{B}}}{\partial t}$$ $$\nabla\times\hat{\textbf{B}}=\mu_0\epsilon_0\frac{\partial\hat{\textbf{E}}}{\partial t}$$ $$\hat{c}=\frac{1}{\sqrt{\hat{\mu_0}\hat{\epsilon_0}}},$$ with $\hat{c}\neq c$

Suppose we require $\hat{c}$ to be constant in all reference frames. Is it still possible to construct Special Relativity consistently with these two different speeds of light-like fields?

EDIT -- to clarify, based on the discussion under Eric Smith's answer: the entire mathematical structure of a hypothetical universe is up for grabs here. Make no assumptions about reality beyond what is necessary to support the premise. (For example, if this yields a universe in which the definition of simultaneity depends not only on your reference frame but on which type of particle you are, that doesn't have to "make sense", the only issue is whether the mathematics can be consistent in any universe.)

Answer 1

No, there cannot be two different invariant speeds, because an invariant speed is a feature of mechanics, not just of the electromagnetic (or other) field. For example, given an invariant speed $c$ and the principle of relativity one can derive the velocity composition formula: $$ u' = \frac{u+v}{1+\frac{uv}{c^2}} $$ If you then do the same with another invariant speed $\hat{c}$ you would obtain: $$ u' = \frac{u+v}{1+\frac{uv}{\hat{c}^2}} $$

Then if $u \ne 0$ and $v \ne 0$ some straightforward algebra shows that $c = \hat{c}$.

Answer 2

No.

The constancy of c is more fundamental than the speed of electromagnetic radiation. It is a basic property of the spacetime in which we live.

As you probably know, if you accelerate a massive particle towards c, it will require more and more energy to do, and you will never reach c.

If, on the other hand, you start with a particle with zero rest mass, like the photon, it will have zero energy until it reaches c. A particle with zero energy has no effect on anything else and might as well be said to not exist.

So, all mass-less particles will move at the same speed c.

Particles with mass will move at a speed that depends on their energy, in other words, not a constant.

Answer 3

A nicely behaved theory should be invariant under a Lorentz Transformation. Let's write Maxwell's equation in a form that transforms nicely under Lorentz Transformations. Define \begin{align} F^{\mu\nu}= \begin{pmatrix} 0&-E_x/c0&-E_y/c0&-E_z/c\\ E_x/c&0&-B_z&B_y\\ E_y/c&B_z&0&-B_x\\ E_z/c&-B_y&B_x&0\\ \end{pmatrix} \end{align} as the electromagnetic field tensor. Maxwell's equation are then written as \begin{align} \sum^3_{\mu=0}\frac{\partial}{\partial x^\mu}F^{\nu\mu}=-\mu_0J^ \nu \end{align} where $x^{\mu}=(ct,x,y,z)$, $J^\nu=(c\rho,\mathbf J)$ is the four-current, $\rho$ is the charge density and $\mathbf J$ is the current. That was a lot of definitions, but now if we transform the electric- and magnetic field in the right way, we see that the field tensor transforms as a proper matrix! \begin{align} \tilde F^{\mu\nu}&={\Lambda^\mu}_{\alpha}F^{\alpha\beta}{\Lambda_\beta}^{\nu}\\ \tilde F&=\Lambda(v/c)\cdot F\cdot \Lambda^T(v/c) \end{align} where $\Lambda(v/c)$ is the Lorentz Transformation matrix, as a function of the velocity over $c$. Because $J$ gets a $\Lambda$ as well, $\mathbf {\tilde J}=\Lambda\mathbf J$, the overall equations of motion are invariant. In classical mechanics this works the same. For example, under a rotation force and acceleration both get a rotation matrix, i.e. $\mathbf{\tilde F}=R\mathbf F$ and $\mathbf{\tilde a}=R\mathbf a$, which makes $\mathbf F=m\mathbf a$ invariant under rotations.

So why can't we have two types of electromagnetism with different speeds of light? A Lorentz Transformation can only have one speed of light (speed of information). If you would transform the field tensor $\hat F$, then either the equations of motion would not be invariant under Lorentz Transformations, or $\hat c$ would depend on the reference frame. Both would be pretty bad for your theory, especially since we have experimentally verified that light only moves at $c$.

Answer 4

Existing answers are sort of correct but the best answer is in one of the comments, namely that this is really about spacetime before we introduce any particular field theory.

The concepts behind relativity theory begin with thinking about how spatial and temporal measurements are defined and what is meant by simultaneity and things like that. By such thinking one can arrive at two basic ways to construct the theory. Method (1) is to assert relativity principle and a maximum speed for signals, and obtain the Lorentz transformation. Method (2) is to assert that spacetime can be treated mathematically as a manifold with a metric, and to assert the Minkowski metric.

On either approach there is one maximum speed for signals, called $c$.

We then come to field theories such as electromagnetism. The Maxwell equations satisfy the requirements of Lorentz transformations if and only if the parameter $c_{wave} = 1/\sqrt{\epsilon_0 \mu_0}$ has a value equal to $c$. The second set of equations proposed in the question do not have this property, so they do not satisfy the requirements of Lorentz transformations and hence they are not consistent with relativity theory.

If you now ask whether there is some other way of constructing either a metric or a coordinate transformation which would allow the notion of two invariant speeds, then I am pretty sure the answer must be 'no', at least in a spacetime of only one timelike dimension. (And physics with more than one timelike dimension is sort of crazy, so crazy that it is debatable whether it should be called physics).

Answer 5

Your universe would experience problems with people at rest in the same place being unable to agree on whether an event was in the past or the future.

Take as a thought experiment c1 as the normal speed of light (and only light) and c2 as the speed of dark light (and the speed limit for matter). And c2 >> c1; c2 in fact being arbitrarily large for simplicity.

Ava is on planet A and Bob is on planet B, they are 1 light year apart. Ava emits a very bright flash then travels to Planet B in 10 days (this is fine because it's below Ava's invariant speed c2). Because Ava's speed is so far below the speed of dark light she experiences almost no relativistic effects. During her journey she observes light travelling at c1 away from her. She observes the flash reflecting off planet B just before she arrives.

On arriving at planet B Ava asks Bob what he thought of the flash but he says he hasn't seen it. Ava says that makes no sense she saw its relection off planet B during the journey.

Bob does some maths and says that the flash should be visible in just under a year, they both agree to meet back then.

In a year they both look towards planet A. Bob sees the bright flash, Ava sees nothing.

So your universe is inconsistent and you'd need to make huge changes to other laws of physics to make it work (probably abandoning the idea that two entities in the same location at rest with each other experience the same events)