# Definition for curvature of worldline in Minkowski space

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1. Aug 24, 2015

### JD96

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 similiar 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?

Last edited: Aug 24, 2015
2. Aug 24, 2015

### Mentz114

The curvature of a worldline ( a member of a congruence) is a rank-2 tensor $\nabla_\nu \dot{x}_\mu$ and proper acceleration is the part in the direction of $\dot{x}$, viz $\ddot{x}_\mu=\nabla_\nu \dot{x}_\mu \dot{x}^\nu$. So you are fairly close.

3. Aug 24, 2015

### Orodruin

Staff Emeritus
I suggest you first think about how you could express the curvature of a parameterised curve in ordinary euclidean space in terms of derivatives with respect to the curve parameter. Once you have done that, you should find it easier to simply carry over that result to minkowski space with minor modifications.

Based on the thread level, I am not sure that this helps the OP.

4. Aug 24, 2015

### JD96

Thanks for your response! I see that you are referring to the definition $|\frac{d\vec u_{T}(t)}{dt}|$, where $\vec u_{T}(t)$ is the unit tangent vector to a curve parameterized by $t$. I think to apply this definition to minkowski space I could use proper time as a natural parameterisation and the four velocity as a tangent vector, but is the four velocity independent of the metric signature considered a unit vector or to put it another way is the square of the length of a timelike unit vector equal to 1 or -1 in Minkowski space? (not that it makes a big difference, I am just curious).

Anyway, using the definition for curvature you indicated it is now clear to me how it follows from the definition that the curvature equals the proper acceleration, since the derivative of the four velocity with respect to proper time yields the four acceleration. (Is this what you @Mentz114 explained to me in your post?)

Last edited: Aug 24, 2015
5. Aug 24, 2015

### bcrowell

Staff Emeritus
If you want the simplest possible definition (but not necessarily the easiest one with which to perform calculations), then the following is probably it. Given a point P on a world-line W, define Minkowski coordinates such that P is at the origin and W is tangent to the t axis at P. In these coordinates, the only natural and physically reasonable way to define the curvature as a single real number is $|d^2 x/dt^2|$. This is the magnitude of the proper acceleration, which is the acceleration measured in the inertial frame instantaneously moving with W at P. This definition is simple to state, but does not immediately tell you how to specify this quantity's direction or transform it to other frames.

6. Aug 24, 2015

### JD96

@bcrowell Thanks for your input!

So to make sure I got it right, are you defining curvature as the coordinate acceleration in the momentarily comoving frame and it is problematic to obtain its direction due to the fact that only its norm is an invariant quantity?

7. Aug 24, 2015

### bcrowell

Staff Emeritus
It's not problematic to obtain its direction, and the definition I gave wasn't stated as the norm of an invariant quantity. I was just giving the most straightforward possible definition, using only freshman calculus.

If you want something a little fancier, then the measure of curvature to use is the acceleration four-vector. That includes direction information, and it has standard transformation properties.