As promised, the continuation.

Well, this is wrong.

Light will travel at C from its point of origin.

If the rocket travels at C, the light in the rocket will actually standing still.

No, it will not. Light is made of electromagnetic waves. Light "standing still" will result in a static electromagnetic wave in a vacuum, which in Maxwell's equations is an absurd result, and cannot be derived using Maxwell's equations. This is because magnetic fields in a vacuum are generated by

*changing* electric fields, and vice versa. The way light propagages means that, if it were possible to move with the light wave, then this is exactly what we'd see — a

*static, unchanging* magnetic and electric wave with no source in a vacuum. And again, there's no allowance in Maxwell's equations for the speed of the observer; carrying through with the equations yields a result that demands an electromagnetic wave propagating at c, even when moving at c yourself.

*Physicists in the nineteenth century knew this.* Yes, it was mind-blowing, but it is also what Maxwell's equations were telling them, and those equations were proving their worth otherwise. They knew that a static electromagnetic wave is

*not* permitted by Maxwell's equations;

**if it were, Einstein would never needed to come up with special (and general) relativity in the first place!** Physicists were coming up with ways around this precisely because they knew what you described above is not possible.

Nobody has

*ever* observed a light wave "standing still." Nobody has

*ever* made any observation that

*supports* that proposition, either. Again, if it did, the Michelson-Morley experiment would have yielded a positive result in detecting the aether wind. The aether wind would be effectively light being slowed down a bit by motion. But no, there was no detection even with thorough testing over a wide range of velocities.

You have not given me any reason to believe that you are correct, while on my side I have plenty of reason to believe that you are incorrect. Yours are just assertions. You simply

**assert** that a light wave

*can* be stationary from some observer, even though currently physics has firmly established that it isn't true, and as such only a physical experiment could knock that principle over.

TL;DR — the

*existence* of special relativity is a result of physics knowing, even in the nineteenth century, that a stationary electromagnetic wave is an absurdity. Your proposition wouldn't even survive physics pre-Einstein.

This is the great distortion of the interpretation of Lorenz.

You keep asserting that, but I have yet to see you use it even once. How could you know what I say is "distorted" if you don't even bother finding out what it says?

I don't think you even know what the Lorentz transformation is, or what it implies. Here, let me demonstrate that in all frames of reference, where v < c, the speed of a light beam will always be observed at going c, using the Lorentz transformation that I supposedly don't understand.

The Lorentz transformation is for some velocity v in the x+ direction:

t' = γ(t - xv/c²)

x' = γ(x - vt)

y' = y

z' = z

where γ = 1/sqrt(1-v²/c²)

Let's ignore the y and z axis. Similar algebra will reveal that other direction will yield a similar result.

In frame S, a photon begins at P

_{0} = (x

_{0}, t

_{0}) and after some time, T, arrives at P

_{1} = (x

_{0} + cT, t

_{0} + T). The difference between the two points is ΔP = (Δx, Δt) = (x

_{0} + cT - x

_{0}, t

_{0} + T - t

_{0}) = (cT,T). Pretty straightforward.

The velocity of the photon in S is u = cT/T = c.

In frame S' (moving at v in S), the two points are P'

_{0} = (x'

_{0}, t'

_{0}), and P'

_{1} = (x'

_{1}, t'

_{1}), and ΔP' = (x'

_{1} - x'

_{1}, t'

_{1} - t'

_{0}). Using the Lorentz transformation on these two points and a little algebra reveals that ΔP' = (γ(Δx-vΔt), γ(Δt-vΔx/c²).

The velocity of the photon in S' is u' = γ(Δx-vΔt)/γ(Δt-vΔx/c²) = (cT-vT)/(T-vcT/c²) = (c-v)/(1-v/c) = c·(1-v/c)/(1-v/c) = c.

Oh, look at that! As I asserted, a particle that has velocity c in one frame of reference will have velocity c in another frame of reference. Of course, if v = c, the Lorentz transformation blows up on you (it divides by zero), so there can be no Lorentz transformation into a frame going at c in any other frame.

So, you are catagorically NOT using a Lorentz transformation. Vectors traveling at c are eigenvalues of this transformation — they remain unchanged up to being stretched or squished; ergo, if any particle traveling at c in one frame do not travel at c in another frame, you have not used a Lorentz transformation.

Which is basically what's been happening.

The light fired at the tail of a rocket traveling at C, will stay at the tail!

Mere repetition of the same faulty reasoning.

The rocket cannot travel at c. I've been generous up to this point in allowing this sloppy reasoning, but I'm putting my foot down here. The rocket has mass. Only massless particles can travel at c. Ergo, the rocket is traveling strictly less than c, even though it may be traveling arbitrarily close to c. Therefore, in any physically possible scenario the light will eventually reach the nose, even by your ignorant argument.

We have never observed a mass-bearing particle traveling at c. Everything we have observed about particles with mass reveals that they cannot travel at c. Furthemore,

*only* mass-bearing particles can travel slower than c. Massless particles are never observed traveling at any speed other than c. Because rockets occasionally travel slower than c (like standing still on a rocket pad), they must be mass-bearing and therefore speeds of c are not possible for them.

The above is a summary of what is known about c and particles that can and cannot travel at c. The characteristic that makes this difference is mass. The rocket, which has mass, cannot travel at c.

Again, the Lorentz transformation preserves vectors traveling at c. Furthermore, there are no Lorentz transformations taking a frame into one traveling at c.

This is the characteristic of light!

It travels at C which is 300 000 km per hour(about)

Yes, it does, in all reference frames. This includes the rocket, so it must also see light go at c. QED.

Indeed, not only is it the characteristic of light, but it also is the maximum speed that any interaction can occur. It's specialness goes beyond merely being the speed of propagation of the waves of a particular field — it's one of the big damn important constants.

Now, reality has it that the light beam of the rocket, as well as the one from the earth observer will arrive at the same time on the moon.

therefore, the rocket man will see the point of the light beam and the rocket light beam travelling with him, and all 3 arrive at the same instance at the moon.

You talk about reality, yet immediately speak of an unreal scenario.

Again, I refer you to Maxwell's equations which asserts that static electromagnetic waves cannot exist, and therefore a electromagnetic wave seen to be standing stationary is not possible. Even by 19th century standards of physics, your assertion would be false, because if it were true,

*there would be no special relativity for you to argue against.*Therefore, the Rocketman will not be able to measure light at C, because light will slow down in his point of view.

He knows that the light is travelling at C, but from ITS SOURCE!

NOT FROM THE ROCKET'S SPEED ADDED TO C!

The rocketman will observe that the light going at c, even in his frame. It will take the light about one nanosecond by his clock to traverse his wooden 12" ruler. He may

*think* that he is measuring v + c, but any experimenter outside the rocket and seeing the rocket move at v will also observe the light moving at c, by his own rulers and his own clock. I know this is wierd, but it is what special relativity asserts, and what happens in real life by any experiment we care to perform.

What Relativists say is,

<snip>

THE ROCKET MAN WILL ALSO SEE THE LIGHT OUTSIDE HIS ROCKET, THAT WAS FIRED FROM THE EARTH, TRAVELLING AT C RUSHING PAST HIM TO THE MOON, BUT ON EARTH THE LIGHT SEEMS TO TRAVEL WITH THE ROCKET.

More repetition of the same old crap. No, it won't. No "Relativist" will say this. They're called physicists, by the way, and they have forgotten more physics than you will ever know. They are very clear on this, and I have used the Lorentz transformation to demonstrate this: there is no frame of reference that will ever observe light going at any speed other than c.

The rocket cannot travel at c. Ergo, your scenario is impossible. Period. It

*must* travel at less than c, which means that even by your logic, the light beam does

*not* stay at the tail of the rocket because the light beam is still faster, even if only marginally so.

Do you see the incorrect interpretation of Lorenz?

I have seen lots of

*assertions* from you that I have the incorrect interpretation of the Lorentz transformation, but I have yet to see you

*demonstrate* it, just like in your other threads. Above, you will find a demonstration that light traveling at c from one frame will also be traveling at c in another. So tell me again how I have the "incorrect interpretation of Lorentz."

Better yet,

**demonstrate** this, instead of simply spewing about it. Please have this demonstration in your next rounds of posts, or I win.

Your assertions are supported by no experiment or observation, and there is plenty of experiment and observation to the contrary. If a hypothesis doesn't agree with experiment, it's WRONG. Special relativity, and the constancy of the speed of light in all frames, by all means of measurement, is the foundation of all modern physics and is not going to be dismissed by some crank on the internet with a messiah complex.