- #1
JD96
- 33
- 0
Hello,
in this section of the wiki article on Rindler coordinates it is stated that the proper acceleration for an observer undergoing hyperbolic motion is just "the path curvature of the corresponding world line" and thus a nice analogy between the radii of a family of concentric circles and the proper acceleration of a family of rindler observers can be drawn. My question is how the curvature of a worldline given in the form ## ct(x)## is defined (since I don't know anything about differential geometry the wiki article about path/geodesic curvature doesn't help). Using ##k(x)=\frac{f''(x)}{(1+(f'(x))^2)^{3/2}}## to compute the curvature doesn't work, since in the case of a hyperbolic worldline the result isn't the proper acceleration. But when exchanging the plus sign in the denominator with a minus sign the curvature is apart from a factor of i (using the (-+++ convention) equal to the proper acceleration. So I guess ##k(x)=\frac{f''(x)}{(1-(f'(x))^2)^{3/2}}## is the correct expression for curvature in Minkowski space (for a worldline in 1+1d), but I still don't know how to get there.
The definition of curvature, I think is the most intuitive one, is to define the curvature of a circle to be ##1/r## and then to define the curvature at a point on the graph of a function to be that of the osculating circle to that point. My question is whether the curvature of a worldline (again in the simplified case of only one spatial dimension) can be defined in similar way. To be more concrete by defining the curvature of a hyperbolic worldline to be the proper acceleration and then using osculating hyperbolae to arrive to the expression I wrote above?
Thanks in advance
in this section of the wiki article on Rindler coordinates it is stated that the proper acceleration for an observer undergoing hyperbolic motion is just "the path curvature of the corresponding world line" and thus a nice analogy between the radii of a family of concentric circles and the proper acceleration of a family of rindler observers can be drawn. My question is how the curvature of a worldline given in the form ## ct(x)## is defined (since I don't know anything about differential geometry the wiki article about path/geodesic curvature doesn't help). Using ##k(x)=\frac{f''(x)}{(1+(f'(x))^2)^{3/2}}## to compute the curvature doesn't work, since in the case of a hyperbolic worldline the result isn't the proper acceleration. But when exchanging the plus sign in the denominator with a minus sign the curvature is apart from a factor of i (using the (-+++ convention) equal to the proper acceleration. So I guess ##k(x)=\frac{f''(x)}{(1-(f'(x))^2)^{3/2}}## is the correct expression for curvature in Minkowski space (for a worldline in 1+1d), but I still don't know how to get there.
The definition of curvature, I think is the most intuitive one, is to define the curvature of a circle to be ##1/r## and then to define the curvature at a point on the graph of a function to be that of the osculating circle to that point. My question is whether the curvature of a worldline (again in the simplified case of only one spatial dimension) can be defined in similar way. To be more concrete by defining the curvature of a hyperbolic worldline to be the proper acceleration and then using osculating hyperbolae to arrive to the expression I wrote above?
Thanks in advance
Last edited: