espen180 said:
Okay, if I want to find an expression for the acceleration of a particle without assuming dr/ds=0 initially, I have to solve
\frac{\text{d}^2t}{\text{d}\tau^2}+\left(1-\frac{r_s}{r}\right)^{-1}\frac{r_s}{r^2}\frac{\text{d}t}{\text{d}\tau}\frac{\text{d}r}{\text{d}\tau}=0
\frac{\text{d}^2r}{\text{d}\tau^2}+c^2\left(\frac{r_s}{2r^2}-\frac{r_s^2}{2r^3}\right)\left(\frac{\text{d}t}{\text{d}\tau}\right)^2-\left(\frac{r_s}{2\left(1-\frac{r_s}{r}\right)r^2}\right)\left(\frac{\text{d}r}{\text{d}\tau}\right)^2=0
So I figure the first step is to substitute the dt/dτ. The problem is that
\text{d}t=\text{d}\tau \sqrt{\left(1-\frac{r_s}{r}\right)\left(1-\left(\frac{\frac{\text{d}r}{\text{d}t}}{c^2}\right)^2\right)}
which, if correct, would mean that I get a mix of derivatives of r wrt. t and τ. So I have no idea where to start, or if the equations are analytically solvable.
Hi espen,
The equations are analytically solvable and all the information you need is already contained in the Schwarzschild metric for radial motion only:
c^2 d\tau^2 = (1-r_s/r) c^2 dt^2 - \frac{1}{(1-r_s/r)} dr^2
Divide through by (1-r_s/r) d\tau^2 and rearrange:
c^2\left(\frac{dt}{d\tau}\right)^2 = \frac{c^2}{(1-r_s/r)} + \frac{1}{(1-r_s/r)^2} \left(\frac{dr}{d\tau} \right)^2
Substitute this expression for (c dt/d\tau)^2 into your second equation and you obtain:
\frac{\text{d}^2r}{\text{d}\tau^2}+\left(\frac{r_s}{2r^2}-\frac{r_s^2}{2r^3}\right)\left( \frac{c^2}{(1-r_s/r)} + \frac{1}{(1-r_s/r)^2} \left(\frac{dr}{d\tau} \right)^2 \right)-\left(\frac{r_s}{2\left(1-\frac{r_s}{r}\right)r^2}\right)\left(\frac{\text{d}r}{\text{d}\tau}\right)^2=0
Simplify the above:
\frac{\text{d}^2r}{\text{d}\tau^2} = -\frac{c^2 r_s}{2r^2} = -\frac{GM}{r^2}
This result means that the acceleration using these coordinates is independent of the falling velocity dr/ds because we have made no assumption of dr/ds=0. It also means that the acceleration is non-zero for any value dr/ds and so it is proved that the claim by Starthaus that the acceleration is zero when dr/ds=0, is false.
Now you may have heard that acceleration IS dependent on velocity and also that the time dilation factor is the product of time dilation due to gravitational potential and time dilation due to velocity. The above calculations seem to contradict this, but it all makes sense when you are careful about who makes the measurements. The following is an attempt to clarify various coordinate measurements and define a notation convention:
Schwarzschild coordinate measurement:
These are measurements made by a stationary observer at infinity and use the notation r, t, dr and dt.
Local measurements:
These are measurements made by a stationary observer at r of a free falling particle passing r and are indicated by the use of primed variables, e.g. r', t', dr' and dt'.
Co-moving measurements:
These are measurements made by an free-falling observer that is co-moving with respect to the free falling particle and local to the particle. In these coordinates the, free falling particle's velocity is always zero and the acceleration of the particle is always zero.
These measurements are indicated by the zero substript as r_o, t_o, dr_o and dt_o.
In the Schwarzschild metric dt_o means the same as the infinitesimal proper time interval d\tau or ds/c.
----------------------------------------------
Using these definitions and considering a free falling particle with radial motion only, the Schwarzschild coordinate acceleration according to an observer at infinity is:
\frac{d^2 r}{dt^2} = -\frac{GM}{r^2}\left((1-r_s/r)-\frac{3(dr/dt)^2}{(1-r_s/r)}\right)
(Note that this measurement of acceleration is velocity dependent.)
This equation can be obtained from the general equation I derived earlier in post #48:
kev said:
... the general coordinate radial acceleration of a freefalling particle in the metric:
a = \frac{d^2r}{dt^2}= \frac{\alpha^2 H_c^2}{r^4}(r-5M) +\frac{M}{r^2}(2\alpha-3K_c^2\alpha^2)
by setting the angular velocity constant H_c to zero and substituting the full form of K_c back in.
The acceleration of the free falling particle according to a local observer that is stationary at r is:
\frac{d^2r '}{dt' ^2} = - \frac{GM}{r^2}\left(\frac{1-(dr '/dt ')^2}{\sqrt{1-2GM/r}}\right)
The derivation of the above equation can found in post #
https://www.physicsforums.com/showpost.php?p=2747788&postcount=345".
The acceleration of the free falling particle, using a mixture of distance measured by the Schwarzschild observer at infinity and time as measured by a co-free-falling observer is:
\frac{d^2r}{dt_o^2} = -\frac{GM}{r^2}
This is the expression I derived at the start of this post from your acceleration equation which is in turn obtained from the derivation in your document
http://sites.google.com/site/espen180files/Schwartzschild.pdf?attredirects=0&d=1 (using equation (9))
The acceleration of the free falling particle, using distance and time measured by a co-free-falling observer is:
\frac{d^2r_o}{dt_o^2} = 0
This I suspect is something like the acceleration derived by Starhaus, but he is using co-moving in an inconsistent way (the observer is not co-rotating in his orbital calculations) and he using the symbol dr to mean distance measured by a free falling observer when everyone else is using dr to mean distance measured by the Schewarzschild observer. This is why Starthaus said the velocity dr/dt of a stationary particle at r is not zero, because by his definition of dr/dt being measured by a free falling observer, the stationary particle does not appear to be stationary. Starthaus never defines his variables in physical terms, never checks the physical implications or conclusions of his equations and never checks if the symbols he is using have a different physical meaning to the symbols being used by everyone else. He just blindly applies calculus to symbols (defined differently to everyone else) and when his results do not look the same as everyone elses (not surprising really) he declares everyone else to be wrong. (For someone who is supposed to be good at calculus, his basic algebra is surprisingly shaky too.)