A proof for the Stationary Earth, Part 2

[Sparko]

I thought that was kind of obvious by the conclusions he was drawing from them


Jim

i generally just skip john's posts like that because I already know he doesn't have a clue what he is talking about, but I assumed he at least knew what the formulas were supposed to represent.

[DesertBerean]

Even a math dunce like me know you must define the terms you're using so others can evaluate the accuracy of your calculations.

[JohnMartin]

Quote Originally Posted by JohnMartin View Post
The value of v does not change at x,y,z,t and x',y'z',t', so v is invariant with respect to both x,y,z,t and x',y'z',t'.
Again, what are (x,y,z,t) and (x',y',z',t') you leave this completely unexplained. As such basically everything you say is nonsense. Do you mean two different points, at two different times in a euclidian space? Or do you mean the same point in two different inertial reference frames? I assume you mean the latter, because if you mean the former, your argument is incoherent.

The later. The same point in two different inertial reference frames.

v = v at both x,y,z,t and x',y'z',t', so v is invariant apparent from any transform.

v = v is always true, no matter what. Any quantity is always equal to itself. Again, this is slobby notation, as I assume you mean v = v', namely the velocity at the one point, and the velocity at the other. .

See this makes me think that you're merely talking about two different spatial points, each with an associated velocity which just happens to be the same. Who cares about that? If I'm running along at 2 meters per second, and a bicyclist somewhere else, is going in the same direction at the same speed... then so what? That's not what an invariant means. That's an equality... the two velocities would be equal to each other.

Something is invariant, if during a kind of transformation it is unchanged.

v is a fundamental variable of gamma and is not subject to a transform. If "invariance is something you apply to transformations", as you say, then c is invariant and so is v.

c is always invariant under a lorentz transformation (going from one inertial frame to another). If v is invariant, then v = c.

No, v=c is false. An object in SR always has a v<c. Furthermore, v is fundamental to gamma, so v is not transformed. There is no transform of the gamma transform.

You said this - If something is the same after a lorentz transformation, then its an invarient - this is false.

Nope this is true.

²/c²)^1/2

Eq 2) t'=(t-vx/c²)/(1-v²/c²)^1/2

Let us call the v in the numerator as v_n and v in the denominator as v_d. We see in Eq 1), v_nd in the denominator, (1-v²/c²)^1/2 is used to modify v_n. v_d is used along with c in the denominator to modify v_n. As c in the denominator is invariant, then so too, v_d must also be invariant. Why? Because c is assumed to be invariant in all frames and is used as the benchmark value to modify all other values. Such use of c, must then be consistently applied to the use of v_d. v_dd is applied consistently with c in the denominator, then v_d is an invariant velocity, modifying v_n. v_ndinvariantt_invariant

Eq 4) x = (x-v_variantt_variant) . gamma

Eq 3) is the non controversial formula for x, whereby v and t are standard, invariant variables. v is the velocity at x, which does not change by shrinking to another variant velocity. Likewise, t does not dilate and become t'. What then happens to v when multiplied by another variable in gamma? v changes from v_invariant to v_variantd at x, and v_n /(1-v_d²/c²)^1/2_n and a variant v_d

You should have said this - If something is the same after a lorentz transformation, then there is a contradiction. For the same would be the same both before and after the transform, but the transform is a variable other than 1. Hence, the same cannot be the same both before and after a transform. The implied contradiction within your statement is another indication that SR theory is false.

The identity transformation, basically going from the inertial reference frame... to the same reference frame, is one special case of the lorentz transformation. And yes, c is invariant under this transformation. So are all quantities. Its impossible for any number to not be invariant under an identity transformation.

However c, and only c, is invariant under a lorentz transformation of any kind, and any rotation. Its tedious to prove it for the general case from the lorentz transformation alone, however I don't mind showing the proof for one of the simpler cases.

SR wants us to believe only c is invariant, but the SR equations assume v_d

Lets take two inertial frames of reference, moving in the same direction. One at v, the other at v'. In that case we can calculate what any velocity would be like after the reference frame shift (lorentz transformation).

In this case its given by the simple velocity addition formula from the Special Theory of Relativity.

In the special case that u is equal to c, then we have for any difference of velocities (v - v').

So if u = c, then u' = c after a lorentz transformation.

However there is no u, less than c, which will be the same after a lorentz transformation. Only c is invariant under all lorentz transformations.

Same problems as posed above. v is both variant in the numerator and invariant in the denominator.

SR is sophistry.

JM

[JohnMartin]

i generally just skip john's posts like that because I already know he doesn't have a clue what he is talking about, but I assumed he at least knew what the formulas were supposed to represent.

I generally skip your posts, because I know you are an inventor. Example, see your recent ramble about the SR formulas.

JM

[Sparko]

I generally skip your posts, because I know you are an inventor. Example, see your recent ramble about the SR formulas.

JM

which one is that?