# Can nothing other than light have speed independent of source?

Is it impossible for there to be some phenomenon that travels at a different speed than light to have speed independent of the source? Because if there were such a phenomenon there would be competing formulas for time dilation, correct?

To be clear, this is not a faster-than-light question; the phenomenon could have constant speed lower than light and still be problematic if I'm not mistaken.

• Don't gravitational waves also move at the speed of light? Aug 8, 2021 at 16:54
• Related/possible duplicate: physics.stackexchange.com/q/126694/50583 and its linked questions Aug 9, 2021 at 23:36

Fundamentally, the speed of light has less to do with light, and more to do with a universal speed limit, which is given by the value of $$c$$.

Provided an object is massless, it would move at the speed given by $$c$$, and would meet the criteria that its speed is independent of the motion of source or observer.

Anything with mass would not move at $$c$$, and nothing can move faster than $$c$$, and so we would conclude that $$c$$ would be the only speed something would move, such that this speed is a constant, independent of the motion of the source and observer.

• Why do all massless objects have to move at c exactly? And is there a reason an object with mass cannot have speed independent of source? Aug 8, 2021 at 9:59
• @TeoGelles The argument I'm familiar with: Any object moving slower than $c$ has a momentary inertial rest frame (you just take any reference frame where the object has a velocity and boost by the velocity to cancel it). But in that frame the object has no kinetic energy (because you chose this frame to annihilate it), so the only energy it has in this frame is the rest energy $E=mc^2.$ But if $m=0,$ in this frame $E=0$ and the object has no capacity to "do" anything at all. So massless subluminal entities have no physical relevance.
– HTNW
Aug 8, 2021 at 20:14

All waves in a medium (e.g. sound waves, surface waves in water) travel with a speed that is independent of that of the source (that is ignoring dispersion, which would create a velocity dependence via the Doppler effect)

In a vacuum, it is only light (i.e. electromagnetic waves), gravitational waves, or other massless objects that have a speed independent of that of the source (namely the speed of light c)

• This is the correct answer. It should not have been downvoted. Aug 8, 2021 at 13:32
• Sound waves can readily be experienced, e.g. the sound from airplanes going by. The pitch of the sound changes (Doppler effect), but not the speed of the sound. Aug 8, 2021 at 16:53
• What about gravity waves? Aug 8, 2021 at 18:28
• @user45664 You mean gravitational waves? Gravity waves would be something else (see en.wikipedia.org/wiki/Gravity_wave ). I have edited my answer to account for the former Aug 8, 2021 at 19:51
• @jamesqf in presence of dispersion, Doppler effect will make speed of wave dependent on the speed of the source. Aug 8, 2021 at 21:51

# Overview

At first glance, this seems a simple question of writing down the Lorentz transforms for velocities. However, then raises two further questions as to why Lorentz transforms and why can we not have two or more Lorentz groups on spacetime. Additionally, I saw you asking in a comment why massless objects have to travel at the speed of light which I will also include. First the easier part:

# Lorentz Transformations on Velocity

If we assume that transformations between frames of reference are Lorentz transforms:

$$\begin{cases}ct'=\gamma\left(ct-\beta x\right)\\x'=\gamma\left(x-\beta ct\right)\\y'=y\\z'=z\end{cases}$$

You are correct in considering time dilation, but we also need to consider length contraction:

\begin{align}u'_x=\frac{x'}{t'}&=\frac{\gamma\left(x-\beta ct\right)}{\gamma\left(t-\frac{1}{c}\beta x\right)}\\&=\frac{u_x-v}{1-\dfrac{u_xv}{c^2}}\tag*{As u_x=\frac{x}{t}}\\u'_y=\frac{y'}{t'}&=\frac{y}{\gamma\left(t-\frac{1}{c}\beta x\right)}\\&=\frac{u_y}{\gamma\left(1-\dfrac{u_xv}{c^2}\right)}\tag*{As u_y=\frac{y}{t}}\\\text{Similarly }u'_z=\frac{z'}{t'}&=\frac{z}{\gamma\left(t-\frac{1}{c}\beta x\right)}\\&=\frac{u_z}{\gamma\left(1-\dfrac{u_xv}{c^2}\right)}\tag*{As u_z=\frac{z}{t}}\end{align}

Then if we consider an object travelling at a velocity $$\vec u$$ in a frame of reference $$S$$ then using both time dilation and length contraction we find that the velocity of an object in a frame of reference moving at a velocity $$v$$ in the $$x$$-direction relative $$S$$ is given by:

$$\vec u'=\begin{pmatrix}\dfrac{u_x-v}{1-\dfrac{u_xv}{c^2}},&\dfrac{u_y}{\gamma\left(1-\dfrac{u_xv}{c^2}\right)}&\dfrac{u_z}{\gamma\left(1-\dfrac{u_xv}{c^2}\right)}\end{pmatrix}$$

where $$\gamma\equiv\left(1-\beta^2\right)^{-\frac{1}{2}}$$ and $$\beta\equiv\dfrac{v}{c}$$

Now we are interested in speed not velocity so:

\begin{align}u'&=\sqrt{u_x'^2+u_y'^2+u_z'^2}\\&=\frac{1}{1-\dfrac{u_xv}{c^2}}\sqrt{u_x^2-2u_xv+v^2+\gamma^{-2}\left(u_y^2+u_z^2\right)}\\&=\frac{1}{1-\dfrac{u_xv}{c^2}}\sqrt{u_x^2-2u_xv+v^2+\left(1-\beta^2\right)\left(u_y^2+u_z^2\right)}\\&=\frac{1}{1-\dfrac{u_xv}{c^2}}\sqrt{u^2-2u_xv+v^2-\beta^2\left(u^2-u_x^2\right)}\\&=\frac{1}{1-\dfrac{u_xv}{c^2}}\sqrt{\gamma^{-2}u^2+\beta^2u_x^2-2u_xv+v^2}\\&=\frac{u}{1-\dfrac{u_xv}{c^2}}\sqrt{\gamma^{-2}+\beta^2\frac{u_x^2}{u^2}-2\frac{u_xv}{u^2}+\frac{v^2}{u^2}}\end{align}

If we take $$u\le c$$ then $$\beta\le\frac{v}{u}$$ with equivalence only when $$u=c$$ and so:

\begin{align}\frac{u'}{u}&\le\frac{1}{1-\dfrac{u_xv}{c^2}}\sqrt{1-\beta^2+\frac{u_x^2v^2}{u^4}-2\frac{u_xv}{u^2}+\frac{v^2}{u^2}}\\&=\frac{1}{1-\dfrac{u_xv}{c^2}}\sqrt{\left(1-\frac{u_xv}{u^2}\right)^2-\beta^2+\frac{v^2}{u^2}}\quad\text{with equivalence only when u=c}\\\therefore\frac{u'}{u}&\le1\quad\text{found by substituting in u=c}\end{align}

Thus, we only get the constant velocity if $$u=c$$.

As you can see from the above inequality the speed of light is a speed limit. Note also that we never mentioned the class of object we were calculating the velocity of, in fact, it need not even be a physical object but could be another frame of reference. This is important because this means that the speed limit is a property of spacetime not of the objects within spacetime.

# Why Lorentz Transforms

First of all, because it is what we observe experimentally. But what other options are there? Well In General Relativity we generalise spacetime allowing it to take on the form of a (pseudo-)Riemannian manifold and so the invariant distance between points on the manifold is given by $$\text{d}s^2=g_{\mu\nu}\text{d}x^{\mu}\text{d}x^{\nu}$$ where $$g_{\mu\nu}$$ is the metric tensor. Now we can always choose the coordinates so that at a given point in spacetime the metric is diagonal with either $$+1$$ or $$-1$$ for each of the diagonal elements. As we observe space to be isotropic then we must give all the space coordinates the same sign and so without loss of generality we can make this sign negative. Thus, the metric must the form $$\operatorname{diagonal}\left(\pm1,-1,-1,-1\right)$$. In both cases, translations are allowed as we are using a differential distance. But if we take the negative sign we have locally Euclidean space and the only possible additional transforms are rotations and reflections - see the orthogonal group for more detail. But If we take the positive sign then we have locally Minkowski space and the only possible additional transformations are Lorentz transforms and reflections - see the Lorentz group for more details. Now we observe Lorentz transforms in experiments so we know which of the two possibilities we live in. It now should be clear that Lorentz transformations apply to the spacetime, not the objects within.

# Why not multiple values of $$c$$

Now we know why we have Lorentz Transforms and we know that the group of Lorentz transforms for a given value of $$c$$ impose a speed limit of $$c$$. Now imagine if we allowed $$c$$ to take on more than one value. What would this imply? Let's say you are stood on a train platform and watch a train pass. On the train is a person walking towards the front of the train at a velocity $$u$$ relative to the train. Now, what speed do we see the person walking? Well we have established that the Lorentz transforms are property of spacetime and not the object so if we had multiple values of $$c$$ then each value of $$c$$ would give a different velocity you would observe for the person walking and as you can only observe a single velocity then there must only be a single value for $$c$$.

Now maybe you are not convinced that the Lorentz transformation has to be a property of spacetime so let's for a second consider what would happen if $$c$$ did depend on the object. Consider an object constructed of particles some of which were affected by one value of $$c$$ and some of which were affected by some other value of $$c$$. In some frame of reference, the object may be moving relative to the observer but it would appear to be unchanging relative to its comoving coordinates. Now a different observer would then observe the object tearing apart in their reference frame as the particles of different $$c$$ in this frame must have a relative velocity to each other! Now we don't observe this, but would it even be a consistent theory? Well, consider two observers separated in space moving relative to the object such that it appears not to be tearing apart. Now if one of the observers speeds up to move to the other observer the object would appear to them to tear apart, then when they reach the other observer and slow down the object will stop tearing apart further but will still be torn apart from their perspective. But the observer already there would argue that they see the object still in tact!

# Massless Objects

Momentum is a 4-vector and so must transform via Lorentz transforms. This means that the form of the 3-part of the momentum is $$\vec p=\gamma m\vec v$$. Now we simply invert this equation to find $$\vec v$$:

\begin{align}p^2\left(1-\beta^2\right)&=m^2v^2\\v^2&=\frac{p^2c^2}{m^2c^2+p^2}\\\implies\vec v&=\frac{\vec pc}{\sqrt{m^2c^2+p^2}}\end{align}

Now set $$m=0$$ and trivially we see that $$v=c$$.

• I have edited to add an additional example of why Lorentz transforms must be a property of spacetime and not the objects within it. Aug 8, 2021 at 11:46
• That's a nice reductio on $c$ being a property of spacetime rather than objects. Is there a similar logic that determines $c$ must be constant across all of spacetime rather than being variable? Aug 10, 2021 at 1:19
• My initial thoughts are that if $c=c\left(x^\mu\right)$ then the transforms would no longer be linear in $x^\mu$ and so will no longer be memebers of the Lorentz group and thus are not valid on Minkowski spacetime. However, I don't feel confident enought to state this as true and would be intrested to see other peoples answers on this so if you don’t open a question on it I will and will leave a link to it in the comments. Aug 10, 2021 at 7:15
• If you search online for non-linear actions of the Lorentz group it seems as if non-linear actions do exist and so my intuition in the above comment does not hold. Aug 10, 2021 at 7:57

Yes, it is impossible.

I have found the following opening discussion in Landau-Lifshitz$$^1$$ illuminating. Here it is argued that the speed of interactions is maximum, unique and universal.

The interaction of material particles is described in ordinary mechanics by means of a potential energy of interaction, which appears as a function of the coordinates of the interacting particles. It is easy to see that this manner of describing interactions contains the assumption of instantaneous propagation of interactions. For the forces exerted on each of the particles by other particles at a particular instant of time depend, according to this description, only on the positions of the particles at this one instant. A change in the position of any of the interacting particles influences the other particles immediately.

However, experiment show that instantaneous interactions do not exist in nature. Thus a mechanics based on the assumption of instantaneous propagation of interactions contains within itself a certain inaccuracy. In actuality, if any change takes place in one of the interacting bodies, it will influence the other bodies only after the lapse of a certain interval of time. It is only after this time interval that processes by the initial change begin to take place in the second body. Dividing the distance between the two bodies by this time interval, we obtain the velocity of propagation of the interaction.

We note that this velocity should, strictly speaking, be called the maximum velocity of propagation of interaction. It determines only that interval of time after which a change occurring in one body begins to manifest itself in another. It is clear that the existence of a maximum velocity of propagation of interactions implies, at the same time, that motions of bodies with greater velocity than this are in general impossible in nature. For if such a motion could occur, then by means of it one could realize an interaction with a velocity exceeding the maximum possible velocity of propagation of interactions. Interactions propagating from one particle to another are frequently called "signals", sent out from the first particle and "informing" the second particle of changes which the first has experienced. The velocity of propagation of interaction is then referred to as the signal velocity.

From the principle of relativity it follows$$^2$$ that the velocity of propagation of interactions is the same in all inertial systems of reference. Thus the velocity of propagation of interactions is a universal constant velocity (as we shall show later) is also the velocity of light in empty space.

I hope this provides clarity.

Footer

$$^1$$Landau L. D., E. M. Lifshitz; Pg. 1-2; The Classical Theory of Fields, Fourth Revised English Edition, Course of Theoretical Physics Volume 2; Publisher: Butterworth Heinemann

$$^2$$ Let the speed of interactions depend on the speed of an inertial frame wrt. lab. Then by measuring the change in the speed of interactions wrt. lab value, moving inertial observers would be able to tell that they are moving thus violating the principle of relativity.

• I think the absolute question is "why do instantaneous interactions not exist in nature?" Sep 27, 2021 at 0:44
• I think the question was about why there isn't a multitude of $c$ vals...as for why instantaneous interactions don't exist in nature, I don't think there is any fundamental theoretical reason - just that its an experimental fact; during times when it wasn't, instantaneous mechanics worked just fine. Sep 27, 2021 at 11:51
• I meant "I think the OP's question ultimately leads to 'why do instantaneous interactions not exist in nature?' ". But if someone does not want to understand the other side, they wouldn't; or in other words, they would find an excuse not to. Sep 27, 2021 at 12:29

The following are equivalent:

1. In some reference frame, the particle $$X$$ sometimes travels at speed $$c$$.

2. In some reference frame, the particle $$X$$ always travels at speed $$c$$.

3. In every reference frame, the particle $$X$$ sometimes travels at speed $$c$$.

4. In every reference frame, the particle $$X$$ always travels at speed $$c$$.

5. The mass of the particle $$X$$ is zero.

• Does any massless particle have to move at the speed of light, or can it be stationary somehow? Sep 27, 2021 at 0:45
• @Xfce4 : The answer hasn't changed since I posted it. Sep 27, 2021 at 0:58
• Let's take underwater as some reference frame. How does particle photon sometimes travel at speed c in that frame? Sep 27, 2021 at 1:16