Relativistic contraction factor

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Is it correct to state that the relativistic contraction factor is the ratio of the geometric mean to the arithmetic mean of the terms C and v?
Is it correct to state that the relativistic contraction factor is the ratio of the geometric mean to the arithmetic mean of the terms C and v?
 
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south said:
TL;DR Summary: Is it correct to state that the relativistic contraction factor is the ratio of the geometric mean to the arithmetic mean of the terms C and v?

Is it correct to state that the relativistic contraction factor is the ratio of the geometric mean to the arithmetic mean of the terms C and v?
No. The relativistic contraction factor is ##\sqrt{1-v^2/c^2}##. It is for example ##1## for ##v=0##.
 
Sagittarius A-Star said:
No. The relativistic contraction factor is ##\sqrt{1-v^2/c^2}##. It is for example ##1## for ##v=0##.
Thank you very much. Best regards.
 
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Why
Sagittarius A-Star said:
No.
?

For ##c+v## and ##c-v##, the arithmetic mean is
$$\frac{(c+v)+(c−v)}{2}$$
The geometric mean is
$$\sqrt{(v+c)\cdot (v−c)}$$
so I can write the two as a ratio:
$$\frac{\sqrt{(c+v)⋅(c-v)}}{\frac{(c+v)+(c−v)}{2}}$$
which simplifies to
$$\frac{\sqrt{c^2-v^2}}{c}$$
I bring ##c## under the root by squaring it
$$\sqrt{\frac{c^2-v^2)}{c^2}}$$
Then I write the difference with two fractions
$$\sqrt{\frac{c^2}{c^2}-\frac{v^2}{c^2}}$$
the ##c^2## cancel out so i get
$$\sqrt{1−\frac{v^2}{c^2}}=\gamma^{-1}$$
 
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Right. Didn't pay attention
 
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south said:
OP immediately wanted to delete the post when he noticed the error and couldn't find the delete option.
I don't think there is one, but you can reply saying "posted in error, please don't waste your time replying" and then report your original post and ask for it to be deleted.
 
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Ibix said:
I don't think there is one, but you can reply saying "posted in error, please don't waste your time replying" and then report your original post and ask for it to be deleted.
Thanks