Re: How do we interpret the following Redshift chart?

On Jul 4, 10:14 pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:
On Jul 4, 5:16 pm, "g...@xxxxxxxxxxx" <g...@xxxxxxxxxxx> wrote:





On Jul 4, 8:54 pm, "g...@xxxxxxxxxxx" <g...@xxxxxxxxxxx> wrote:

On Jul 4, 8:07 am, EricGisse<jowr...@xxxxxxxxx> wrote:

On Jul 4, 2:41 am, "g...@xxxxxxxxxxx" <g...@xxxxxxxxxxx> wrote:

It is confusing and error prone therefore knowing that a linear chart
of redshfit observation vs distance(hubble) represents a fixed
expansion rate of space.

And knowing that instead if both curved ends (past & present ends) are
below the linear then it represents an accelerating expansion rate.

(linear means a chart with a linear straight slope increasing
gradually as it is plotted from present to past distances)

As above but instead:

1. What would a decelerating chart of the expansion rate look like?

2. What if the past distance is curved above the linear and the
present distance is curved below the linear?

3. What if both past and present are curved above the linear?

4. What if past is curved below and present curved above?

How much physics is there left for you to deliberately not understand?- Hide quoted text -

- Show quoted text -

I guess you can only throw insults since the only time you opend your
beak was about tempurature and tensors which me, others and Wikipedia
told you that you were incorrect.- Hide quoted text -

- Show quoted text -

By the way, you promised you would justify your incorrect belief on
temperature in July.

The argument is the same in July as it was in April.

http://groups.google.com/group/sci.physics.relativity/msg/8ec41d88e75...

snip

Let me conclude the reply to all the blind answers you will concure,
THERE IS BOTH AN INVARIANT AND RELATIVISTIC TEMPERATURE YOUR ERROR IS
YOU ARE TRYING TO INPUT YOUR DEDUCTION ON THE WRONG ONE OF THE TWO.

Temperature is not an invariant, as the reference I gave you in April
explicitly explains and whose proof is derived from first principles.


Your words at the link you gave:
"Equating dS and dS' gives T' = sqrt(1 - v^2/c^2) T. A tensor is
invariant under a coordinate transformation - T does not qualify."

Likewise:

"Equating E and E' gives M' = sqrt(1 - v^2/c^2) M. Does this mean M
is not an invariant scalar under coordinate transformation? You're
taking drugs.

You will never admit to being wrong, you already lied that you never
said that Wikipedia was wrong. You're not arguing just with me, you
are arguing with Wikipedia which says Temperature (and Mass) is a rank
0 tensor.

THE VERY SUBJECT ON TENSORS (THEREFORE NOT JUST AN OFF TOPIC COMMENT)
SAYS:

http://en.wikipedia.org/wiki/Tensor
Copied and Pasted Quote: "For example, mass, temperature, and other
scalar quantities are tensors of rank 0"





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