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Relativistic centripetal force 
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#181
May1210, 08:54 AM

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Those metrics are incomplete, please provide the remaining terms. Especially all of the terms involving [itex]\omega[/itex]. I don't know why you are being so evasive about this.



#182
May1210, 08:59 AM

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#183
May1210, 09:06 AM

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#184
May1210, 11:05 AM

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I find your evasiveness very disturbing. It is not as though it is unreasonable to ask for the metric. 


#185
May1210, 11:17 AM

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You only need the first term (the coefficient for [tex]dt^2[/tex]) of the "standard" metric, you get the potential through a simple identification. [tex]ds^2=\left(1+\frac{2\Phi}{c^2 }\right)^{1}dr^2+r^2(d \theta^2 +\sin^2 \theta d \phi^2)c^2 \left(1+\frac{2\Phi}{c^2}\right)dt^2[/tex] 


#186
May1210, 01:10 PM

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Thanks for posting the "standard" metric.



#187
May1210, 01:43 PM

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If you use the weak field approximation, you get the result I showed you. If you use the strong field approximation: [tex]ds^2=e^{2\Phi/c^2} dt^2....[/tex] you get : [tex]\Phi/c^2=\frac{1}{2}ln(1\frac{r^2\omega^2}{c^2})[/tex] [tex]\vec{F}=grad(\Phi)=\frac{r\omega^2}{1r^2\omega^2}[/tex] Now: [tex]\omega=\frac{d\theta}{dt}=\frac{d\theta}{d\tau}\frac{d\tau}{dt}=\omega_ {proper}\sqrt{1r^2\omega^2}[/tex] So: [tex]\vec{F}=r\omega_{proper}^2[/tex] Same exact result as in chapter 92, expression (97) page 247 in Moller ("The General Theory of Relativity") 


#188
May1210, 05:09 PM

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Is this the correct full expression for the strongfield approximation metric: [tex]ds^2=\left(e^{\frac{2\Phi}{c^2 }}\right)^{1}dr^2+r^2(d \theta^2 +\sin^2 \theta d \phi^2)c^2 \left(e^{\frac{2\Phi}{c^2}}\right)dt^2[/tex] 


#189
May1210, 06:08 PM

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#190
May1310, 02:46 PM

P: 3,967

[tex]a_0=r\omega_{proper}^2[/tex] using your definition: [tex]\omega=\frac{d\theta}{dt}=\omega_{proper}\sqrt{1r^2\omega^2}[/tex] This means in your coordinates, the coordinate acceleration is: [tex]a=r\omega_{proper}^2(1r^2\omega^2) = r\omega^2 = a_0 \gamma^{2} [/tex] and the fundemental relationship [tex]a_0 = a\gamma^2[/tex] between proper and coordinate centripetal acceleration, given by myself in #1 and later by Dalespam and others is correct. 


#191
May1310, 03:57 PM

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Besides, post #3 shows that the naive transformation of force you attempted is wrong. 


#192
May1410, 02:59 PM

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The transformation of force that you object to in #3 is the perfectly standard Lorentz transformation of force and unless you are claiming the Lorentz transformations are wrong, there is no need for me to derive them. You seem to think that angular acceleration might somehow make the transformation of centripetal force different from the transformation of linear transverse force, but if you understood the ramifications of the clock postulate you would know that acceleration has no effect on time dilation or length contraction and we only need to consider the instantaneous tangential velocity of a particle to work out the transformations of an accelerating particle. So all I did was apply the Lorentz transformation (which does not need deriving because it is an accepted standard result) and the clock postulate (which does not need deriving because it is a postulate supported by experimental evidence). Your main objection seems to be that, even though I get the correct result, the method I used is not complicated enough. 


#193
May1410, 04:59 PM

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#194
May1710, 04:19 AM

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#195
May1710, 08:36 AM

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The type of hacks you keep attempting don't count as correct derivations , even if they produce the correct results by accident. LOL. You "forgot" the files that I wrote about accelerated motion in SR. You "forgot" the relativistic transforms for rotation that I tried to teach you. I simply tried to teach you how to use the correct SR equations as they apply to rotation. 


#196
May1710, 09:03 AM

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