Uh-oh. It turns out, I didn't properly calculate ##||\ddot{\mathbf r}||##; it is still 1/2 in cylindrical coordinates. One needs to pay attention that
##\ddot{\mathbf r} = (\ddot{r}-r\dot{\theta}^2,2\dot{r}\dot{\theta}+r\ddot{\theta},\ddot{z} )##
This means that changing the coordinates...
To add this - even in the same coordinate system a curve can be reparametrized from a non-unit-speed to unit-speed. If a curve is already unit-speed in some coordinate system, I am uncertain why it would still be unit-speed in any other coordinate system.
HallsofIvy, I hear you :) "Unit speed" seems to be a term used in math.
It turns out, ##||\ddot{{\mathbf r}}||## depends on the coordinate system; it wasn't obvious to me that ##||\dot{{\mathbf r}}||## doesn't.
Hello,
If for a curve in Cartesian coordinates ##||\dot{{\mathbf r}}||=\mbox{const}## (i.e. the curve is constant speed) will the speed of the curve change in cylindrical and spherical coordinates?
Could someone experienced share how the transition from flat Euclidian space to curved space...
Yes, it makes sense now. Since ##\ddot{\mathbf r}=0##, then the curve is a geodesic.
The conclusion then is:
(1) The expressions of differential geometry are always true for curves irrespective of the type of coordinate system used
(2) The values of mathematical entities are different in...
I have managed to understand this. Then ##\kappa## has different values in Cartesian, spherical and cylindrical coordinates, but the formula is always ##\kappa=||\ddot{{\mathbf r}}||##, no matter what coordinates we are working with.
My apologies, I didn't quite get this...
Hi Ssnow, very interesting point. Let me make sure I understand you correctly.
The curvature of a 3-d unit-speed curve is given by
##\kappa=||\ddot{{\mathbf r}}||##
and the principal unit normal vector is given by
##{\mathbf n}=\frac{\ddot{{\mathbf r}}}{||\ddot{{\mathbf r}}||}##
Are these...
Hi,
##x(s)=\cos\frac{s}{\sqrt{2}}##
##y(s)=\sin\frac{s}{\sqrt{2}}##
##z(s)=\frac{s}{\sqrt{2}}##,
it is a unit-speed helix. Its curvature is ##\kappa=||\ddot{r}||=\frac{1}{2}##. Principal unit normal is ##{\mathbf n}=(\cos\frac{s}{\sqrt{2}},\sin\frac{s}{\sqrt{2}},0)##. So far so good...
But the...
It seemed to me that the "coordinate time" ##t## would appear to be a reasonable parameter, because I understand it to be the reading of a stationary clock; while the "proper time" ##\tau## would be the reading of a clock traveling with the ray, and it would not be convenient to use ##\tau## as...
Hello,
Is this parameterization correct? -
##r(t) = R = \mbox{const}##
##\theta(t) = 0 = \mbox{const}##
##z(t) = ct##
##t = t##
This is supposed to be the null geodesic curve in the case of a light ray, emitted at point {##r=R,\theta=0,z=0,t=0##} parallel to the ##z-##axis in flat spacetime...
There is this one last question - the above discussion was about a metric tensor with all diagonal elements equal to zero, which corresponds to null coordinates. What would be the meaning of a metric tensor where one (or some) of the diagonal elements are zero. This seems to suggest that the...
Thank you to all who posted in this thread. What was said about the diagonal zeroes being due to the choice of coordinates is very useful and I will keep thinking about it and it seems to me to tie well to other posts say this one...
http://en.wikipedia.org/wiki/Skew_coordinates
That is the question indeed... when you think about it, a skew-symmetric matrix has zero diagonal elements because ##a_{ij} = -a_{ji}##
would this mean the metric tensor cannot have zero diagonal elements as ##g## is symmetric?
If there are some nonzero off-diagonal components of ##g_{jk}## then the coordinate system is not orthogonal. Thinking along these lines, probably there is a meaning behind zero diagonal elements