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Gravitational force and acceleration in General Relativity. 
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#19
May1010, 05:16 PM

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Actually it wasn't luck. I just refer to conclusions that are rigorously derived by other people that know what they are doing. 


#20
May1010, 05:19 PM

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#21
May1010, 05:28 PM

P: 355

To get the acceleration measured by our observer at r, you need to measure the acceleration vector in the orthonormal basis used by an observer at r. So the observed acceleration is [tex]\frac{d^2 x^\alpha}{d\tau^2} \cdot \hat e^r[/tex] Where [tex]\hat e^r^2=1[/tex] and [tex]\frac{d x^a}{d\tau} \cdot \hat e^r = 0[/tex]. e^r is directed toward larger r value. It is very important to remember that formula for the dot product involves the metric. Because our observer is at constant r, just like our observer at infinity, we can choose [tex]\hat e^r[/tex] to be in the same direction as the observer @ infinity's radial vector, but [tex]\hat e^r[/tex] has a different length (remember, a length of a vector is only defined at a point in spacetime. If we put both of these vectors at the same point, then e^r is bigger by a factor of (12M/R)^(1/2)) 


#22
May1010, 05:53 PM

P: 3,967

So far, in our new "guess what Starthaus is thinking" game, I have guessed in #15 that you are thinking that the relationship between the proper time and the coordinate time of a stationary particle is given by: [tex]\frac{dt}{dS} = \frac{k}{12m/r}[/tex] and you have not denied it. Right there is a mistake in your thinking. 


#23
May1010, 06:00 PM

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#24
May1010, 06:00 PM

P: 355

In particular, you mention dr/ds as being the proper speed, but this is incorrect. The observer at r has to deal with length contraction in addition to the time dilation that you handled. He measures a dr'/ds, not dr/ds. Working with 4vectors instead of the individual coordinates leads to a much cleaner treatment that allows you to transform between observers' measurements easily. Attacking someone and insisting that they are wrong just because they didn't show their work (which he said he would if you asked) usually doesn't get you a nice response... Anyway, the question's been cleared up.. 


#25
May1010, 06:01 PM

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#26
May1010, 06:05 PM

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#27
May1010, 06:11 PM

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#28
May1010, 06:13 PM

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#29
May1010, 06:27 PM

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#30
May1010, 06:30 PM

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#31
May1010, 06:33 PM

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#32
May1010, 06:39 PM

P: 3,967

Your derivation concluded that proper centrifugal acceleration is zero and in another place you claimed proper centrifugal acceleration is defined as [itex]a_0 = a\gamma[/itex] and in another place you got [itex]a_0 = a[/itex]. You never did arrive at the conclusion that everyone else arrived at, that proper centrifugal acceleration is [itex]a_0 = a\gamma^2[/itex]. So can you leave out the condescending tone? 


#33
May1010, 06:40 PM

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#34
May1010, 06:42 PM

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#35
May1010, 06:45 PM

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#36
May1010, 06:49 PM

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