Re: Any theory that yields a twin paradox must be dubbed bull***

On May 18, 11:23 am, Albertito <albertito1...@xxxxxxxxx> wrote:
On May 18, 5:01 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:



On May 18, 10:25 am, Albertito <albertito1...@xxxxxxxxx> wrote:

On May 18, 3:59 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:

On May 18, 9:11 am, Strich-Reply-To-Idiots <strich.9...@xxxxxxxxx>
wrote:

On 14 Maj, 12:35, PD <TheDraperFam...@xxxxxxxxx> wrote:

On May 14, 8:27 am, "kens...@xxxxxxxxxx" <kens...@xxxxxxxxxx> wrote:

On May 13, 4:21 pm, "Paul B. Andersen" <paul.b.ander...@xxxxxxxxxxxx>
wrote:

Albertito wrote:
On May 13, 2:21 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:
On May 13, 8:11 am, Albertito <albertito1...@xxxxxxxxx> wrote:

On May 13, 1:40 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:
On May 12, 5:59 am, Albertito <albertito1...@xxxxxxxxx> wrote:
In SR, the twin paradox is not referred to the travelling
twin that aged slower than the stay-at-home twin, but to
the stay-at-home twin that, in the rest frame of the travelling
twin, aged slower. So, under SR, it is impossible to state
which twin was the travelling twin and which one was the
stay-at-home twin.
That's incorrect, Albertito, and that is the *whole* point of the twin
puzzle.
There is no symmetry between the twins, and in fact there is a very
important asymmetry. If you take the position that you SHOULD be able
to consider them symmetrically, then you're just not accepting the
laws of physics.
PD
The dogmatic laws of Einstein's relativity are not
the laws of physics unless you lived at Shangri-Lahttp://en.wikipedia.org/wiki/Shangri-La
The laws of physics are what nature says they are. We discover them by
inference and by experimental test. There is no dogma afforded by this
scheme.

If you insist that the laws of physics are what YOU think they should
be, regardless of experimental evidence, then it is YOU who are
manufacturing a Shangri-La.

PD

Did you discover by inference and by experimental test
the dogmatic law that states:

"Light in vacuum propagates with the speed c (a fixed constant)
in terms of any system of inertial coordinates, regardless of
the state of motion of the light source."

????????????????????

And you don't you know what a postulate is?

You know, Albertito, the theory that follows from the two postulates
of SR is a consistent theory, with no contradictions.

Sure there is contradiction. For example: In Einstein's train gedanken
the SR postulate claims that the speed of light in the train is
isotropic and at the same time the SR concept of Relativity of
Simultaneity claims that the speed of light in the train is
anisotropic

No, it does NOT say that. You have misread. The velocity c+v is NOT
the speed of light in the train. It is a *closing speed*, NOT the
speed of light.

and that's why the train observer does not sees the light
fronts from the strikes arrive at him simultaneously.

Ken Seto

And what's more, its a thoroughly tested theory, well confirmed
and never falsified.

Here is an example of a set of postulates that leads to an
inconsistent theory:http://home.c2i.net/pb_andersen/pdf/Albertito.pdf

--
Paul

http://home.c2i.net/pb_andersen/-Hidequotedtext-

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Please define for us what "closing speed" is. I cannot find it in the
reputable textbooks.

Which reputable books are you looking in? Frankly, I'm surprised.

Tell us how that is different from the standard
"speed".

Closing speed is the mathematical subtraction of two speeds measured
in a particular reference frame. It is not a directly measured (or
measurable) quantity.

PD

Let's see:
If v_ba is the velocity of body B in the rest frame of body A
and v_cb is the velocity of body C in the rest frame of body B,
then we have the vectorial sum

v_ca = v_ba + v_cb

as the velocity of body C in rest the frame of body A.

No, certainly not. This doesn't match experiment.

The velocity of c in the rest frame of body A is
v_ca = (v_ba + v_cb)/(1 + v_ba*v_cb/c^2)

If we substract v_cb on both sides, we get

v_ca - v_cb = v_ba.

therefore, is v_ba a "closing velocity"? Is it measurable?

The *measured* velocity of B in the frame of A will not correspond to
the number arrived in the way you just calculated it.

We see that v_ba is a real velocity, and it is measurable.
You only have to measure the velocity of body C in the rest
frame of A and subtract the velocity of body C in the rest
frame of body B.

Numerically, it yields the wrong number.

Easy, isn't it? Why do you need nonsensical
definitions like "closing speed"? In a nonsensical theory,
you need nonsensical definitions in order to stomach it,
But, sooner or later it will produce an indigestion.

:-)

Einstein addition of velocities is not a law of Nature.

No, but it follows as a consequence from the laws of nature. And it
agrees with observation. This is usually sufficient in science.

in v_ca = (v_ba + v_cb)/(1 + v_ba*v_cb/c^2), you must
take the limit as c -> oo,

Why "must" you do that? If you take c to be a value that it does not
have, then you will get results that do not match experiment.

and then it yields

v_ca = v_ba + v_cb,

that is the natural addition of velocities.

I'm sorry, but what is more "natural" about it? If measurements from
*nature* tell you that v_ca = (v_ba + v_cb)/(1 + v_ba*v_cb/c^2) gives
you the right answer and v_ca = v_ba + v_cb gives you the wrong
answer, then it is the former that is more natural, not the latter.

In Einstein's
relativity you identify real positions of bodies with
aparent position, and real velocites with apparent
velocities, because you apply distorted and nonsensical
rules.

What is *measured* is real in science. If you choose to adopt a
"reality" that does not match what is measured, for the sake of
appealing to things you find more intuitively appealing, then you are
no longer doing science. You are doing religious mythology.

PD


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