# Velocity of light in Galilean transformation

What is the velocity of light in Galilean transformation? Is it infinity?

The speed of light as any speed under Galilean transformation follows the usual vector sum rule. i.e., let it be $c$ the speed of light in a reference frame $R$ and $c’$ the speed of light in another frame $R’$ moving relatively to $R$ with the speed $v$. According to Galilee’s transformation, the speed of light $c’$ should be

$c’=v+c \tag{1}$

As we can see the speed of light can have any value since $v$ and $c$ are unbound.

There is no a priori problem with this reasoning. However, there are theoretical and experimental evidence showing that this is incorrect. A small list of problems would be:

1. If equation (1) is true, then Maxwell’s equations for electrodynamics are not. See this question to confirm that Maxwell’s equations demands for a constant speed of light.

2. Michelson-Morley experiment confirms that the speed of light is constant.

3. Measurement of the speed of particles in particle accelerators shows that the speeds are constrained between $0$ to $c≈3*10^8 \, m/s$.

The simplest transformation that solves all this problems is the so called Lorentz transformations. Another important thing is that Lorentz transformation do become equivalent to Galilee’s transformation for $v<<c$.

• if light takes some relative velocity ,then how does time for an event is same in all inertial frames of reference? – robin Mar 22 '18 at 17:01
• @robin This is the experimental result that tells us that Galilean relativity doesn't hold, because it cannot reproduce the lightspeed invariance observation. So we postulate the simplest relativity that does explain it, as stated in the last paragraph – WetSavannaAnimal Mar 23 '18 at 5:13
• @robin In Newtonian Mechanics the absoluteness of time and space was taken for granted. Using a simple geometrical description of the above primed reference systems and the absoluteness of space it is easy to show that the position of an event in $R’$ is $x’=x_{R’}+x$ ($x_{R’}$ is the instant position of $R’$ origin). Now, admitting that time is the same everywhere ($t’=t$) derivate this equation in respect to it and you’ll get equation (1): $v’=v_{R’}+v$. – J. Manuel Mar 23 '18 at 9:26

Galilean transformations are a low speed approximation and therefore they are not a valid description of the physics when speeds approach the speed of light. Since light travels at the speed of light (obviously :-) that means it cannot be described by Galilean transformations. This is true as long as the speed of light is a universal constant, and it doesn't depend on the numerical value of that speed.

It is often said that special relativity approaches Newtonian mechanics if we take the speed of light to infinity, but all this means is that the range of speeds for which Galilean transformations can be used increases without limit as we increase $c$ towards $\infty$. We can't simply set $c = \infty$ as we can't do arithmetic with $\infty$.

• if light takes some relative velocity ,then how does time for an event is same in all inertial frames of reference? – robin Mar 22 '18 at 17:02

Let's say a light beam travels at a speed of $c$ in one reference frame $R_0$. Now, consider another reference frame $R_1$ where it moves in parallel to the light beam with velocity $v$, then the speed of the light beam will be $c-v$ in the frame $R_1$.

So under Galilean transformation, we see that the speed of light is not invariant.

You can naively set $c=\infty$, so that $\infty=\infty-v$. But this is not exactly correct, because the concept of an infinite speed is not well-defined.

• if light takes some relative velocity ,then how does time for an event is same in all inertial frames of reference? – robin Mar 22 '18 at 17:02
• Time is the same for all inertial frames under Galilean transformation by definition. This is not true for Lorentz transformations. – PeaBrane Mar 23 '18 at 5:07