Re: Relativity contradiction - gurus please explain


"David" <dseppala@xxxxxxxxxxxxx> wrote in message news:bht0o29j3iuk3mhu46vps92oj5440pf5j9@xxxxxxxxxx
On Wed, 13 Dec 2006 18:18:19 GMT, "Dirk Van de moortel"
<dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:


"David" <dseppala@xxxxxxxxxxxxx> wrote in message news:mmg0o29chrjcmfhu3j79qclkv01hnspfkt@xxxxxxxxxx
On Wed, 13 Dec 2006 17:56:30 GMT, "Dirk Van de moortel"
<dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:


"David" <dseppala@xxxxxxxxxxxxx> wrote in message news:oae0o21j3aqn4vdnbf4e8sqafk0fb166qi@xxxxxxxxxx
On Wed, 13 Dec 2006 16:31:58 GMT, "Dirk Van de moortel"
<dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:


"David" <dseppala@xxxxxxxxxxxxx> wrote in message news:6070o2pi2c8mdnbtbspj4hcbqim7j71p71@xxxxxxxxxx
On Wed, 13 Dec 2006 14:07:02 GMT, "Dirk Van de moortel"
<dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:


"David" <dseppala@xxxxxxxxxxxxx> wrote in message news:1a10o2d7egjced3pk4qc49hkbihagls262@xxxxxxxxxx
On Wed, 13 Dec 2006 13:35:13 GMT, "Dirk Van de moortel"
<dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:


"David" <dseppala@xxxxxxxxxxxxx> wrote in message news:1qrvn2tqv31aitmmoff1on2l355a4bjh0v@xxxxxxxxxx

[snip]

X (keeping time t) sees Y (keeping time t') and Z (keeping time t")
move in opposite directions at the same velocity v.
Put g = 1/sqrt(1-v^2).

For two events taking place on clock Y a time dt' apart,
X measures dt = g dt'. Rate = dt'/dt = 1/g.
For two events taking place on clock Z a time dt" apart,
X measures dt = g dt". Rate = dt"/dt = 1/g.
So according to X, the clocks Y and Z "run at the same rate".

For two events taking place on clock X a time dt apart,
Y measures dt' = g dt. Rate = dt/dt' = 1/g
For two events taking place on clock X a time dt apart,
Z measures dt" = g dt. Rate = dt/dt" = 1/g
So for two events taking place on clock X a time dt,
Y and Z measure dt' = dt" = g dt. Rate = dt/dt' = dt/dt" = 1/g
So Y and Z, measure clock X to "run at the same rate".

Y and Z measure each other to move at relative velocity
V = 2v/(1+v^2).
Put G = 1/sqrt(1-V^2)

For two events taking place on clock Y a time dt' apart,
Z measures dt" = G dt'. Rate = dt'/dt" = 1/G
For two events taking place on clock Z a time dt" apart,
Y measures dt' = G dt". Rate = dt"/dt' = 1/G
So Y and Z sea each other's clock "to run at a slower rate".

Which step is not correct or do you fail to understand?

Thank you for the physics/math posting.
The step I fail to understand is that this same logic applies at every
point around the circumference.

Which point of the above is not correct or do you fail to understand?

You say Z sees Y's clock to run at a slower rate than his own.
okay. Let's say Z measures that Y's clock runs at half the rate of
his own.

Why do you always introduce these silly numbers?

Because its easier to talk about things with V=0.866c and times and
numbers related by a factor of 1/2 or 2 instead of writing transforms
at each step (the logic is the same).

No. It is downright silly to use numbers.
You get a result, and you have no idea where it came
from and how it is related to other results, and you have
no way of verifying whether the result makes sense at
all.

What does 1/2 have that 1/G doesn't?
Use variables.

Okay, Z measures Y's clock rate to be 1/G of his clock rate (does that
make you happy) at each point around the circumference of the circle

I am not talking about a circle.

Oh, okay - I don't see conflicts with things when Y and Z are
traveling in straight lines at a constant velocity. The problem I
posted had to do with them moving around in a circle.

No, it did not.
Yes it did. My original post stated," Clock A and Clock B are
actually revolving around a circle centered at (x=0,y=0) in the rest
frame.

Yes. And then you went to the straight line with your words
"we can make L (the radius of the circle) arbitrarily large".

Dirk Vdm


.

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