# I am confused about the idea that the speed of light is independent of the speed of the source of light

Imagine that a mast of a sailboat is $$186,000$$ miles long, and the sailboat is moving to the right at a constant $$10$$ miles per second. If you drop a ball from the top of the mast, the ball will land exactly at the base of the mast. The ball has two components to its velocity.

If the experiment is repeated with a pulse of light instead of a ball, then the pulse of light only has one component of velocity. The pulse of light is going directly down at $$186,000$$ miles per second, but the pulse of light is not going to the right at $$10$$ miles per second since the speed of light is independent of the speed of the source of light. Therefore, I would assume the pulse of light would miss the base of the mast by $$10$$ miles.

Would the pulse of light actually hit the base of the mast?

Physics Asked on December 3, 2021

In the reference frame of the sail boat, no one is moving. The laser pointer at the top of the mast is pointed straight down, and to no one's surprise, the laser takes 1 second to traverse the mask and the spot appears at the base of the mast.

In a reference frame in which the boat is moving perpendicular to the laser beam, it looks a little different. While the laser pointer appears to be perpendicular to the direction of motion, something happens when it is turned on: there is a phase ramp across the emitting aperture.

A wave's phase:

$$phi(x, t) = vec k cdot vec x - omega t = k^{mu}x_{mu}$$

is a Lorentz invariant.

In the sail boat frame, at fixed $$t$$, with $$vec k$$ orthogonal to the aperture, $$phi$$ is constant across the aperture. In the moving frame, at fixed $$t'$$, a phase ramp appears (because $$t'=gamma(t-xv/c^2)$$), causing the laser to be emitted and an angle, which happens to follow the moving mast.

From here, you can Lorentz transform the pulse emission and detection events:

$$T_x = (0,0,0,0)$$

$$R_x = (c,0,0,c)$$

and verify that in both frames:

$$Delta s^2 = cDelta t^2 - Delta x^2 = c^2 - c^2 = 0$$

are light-like separated.

Answered by JEB on December 3, 2021

You are not alone if you have a problem with the "speed of light is the same independent of the motion of its source". This SR postulate appears to be completely ignored in the moving light clock experiment. The light clock "thought experiment" only poses a time-dilation problem if we assume that light does behave differently when its source is moving. That is to say, it moves with its source. So, if the light clock works, it undermines the Postulate that the speed of light is independent of the motion of its source.

Answered by James West on December 3, 2021

If the pulse of light is going directly down, it will miss the base of the mast. There ought to be nothing confusing at all about the speed of light being independent of the speed of the source. Light has an E=hf wave nature. Sound has a wave nature too, as do ocean waves and seismic waves. The speed of the waves depend on the properties of the medium. In mechanics a shear wave travels at a speed v = √(G/ρ) where G is the shear modulus of elasticity and ρ is density. In electromagnetism we have a similar expression c = √(1/ε0μ0) where ε0 is vacuum permittivity and μ0 is vacuum permeability.

But is it going straight down? Hindsight's comment probably refers to Lorentz transformation, wherein distances, directions, and other things appear to change when you move. Ah, I see David Hammen has given an answer referring to this. The direction "straight down" depends on the observer. Instead of a pulse of light, imagine you're dealing with a laser beam. An observer in the ship will claim the laser beam is pointing straight down vertically, an observer standing on the gedanken planet won't. There's nothing mysterious about this, have a read of the Wikipeda Aberration of light article for a simple explanation of how directions appear to change when you move. Note that there's a classical explanation and a relativistic explanation that refers to Lorentz transformation and the velocity addition formula.

Answered by John Duffield on December 3, 2021

What you have implicitly done is to create a single preferred frame of reference, the one in which the sailboat's velocity and the velocity of the beam of light are referenced. That is very contrary to relativity theory.

The ball is just a distraction here, as is gravitation. Instead of that sailboat with an impossibly tall mast, imagine a spaceship with that same impossibly long mast. Our spaceship is far removed from any gravitational sources.

Suppose a member on the ship's crew goes to the end of that mast, stops with respect to the mast, and aims a pulse of laser light at the base. The laser will of course hit the base of the mast. From this crew member's perspective, it will take one second for the light to travel from the end of the mast to the base.

Suppose some other crew member is flying around outside the spaceship, such that the spaceship appear to be moving at a velocity $boldsymbol u$ orthogonal to the mast that is orthogonal to the mast. This is analogous to your situation where the sailboat is moving.

This other crew member will also see the laser pulse hit the base of the mast. The magnitude of the pulse of laser light is c from the perspective of this other crew member; the speed of light is a universal constant. However, from the perspective of this other crew member, the direction of that laser pulse is not directly toward the base. The velocity vector of that laser pulse instead has a component orthogonal to the mast, equal to $boldsymbol u$. The component along the mast is a bit less than $c$.

In relativity, velocities don't add linearly as they do in Newtonian mechanics. You need to use the relativistic velocity addition formula to compute the composition of two velocity vectors.

If you use this formula to calculate the velocity of a pulse of light emitted by a source moving at some velocity $boldsymbol u$ (where $u = ||boldsymbol u||$ is less than $c$), you will find that the magnitude of the light pulse is always $c$, regardless of the source's motion, and regardless of the direction in which the pulse was emitted.

Answered by David Hammen on December 3, 2021

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