Let $(M,g)$ be a flat spacetime, $(\gamma:\mathbb{R}\rightarrow M,e)$ an observer (where $e_1(\lambda)$, $e_2(\lambda)$, $e_3(\lambda)$ together with $\dot{\gamma}(\lambda)$ form the basis of the observers frame at $\gamma(\lambda)$) and $\delta:\mathbb{R}\rightarrow M$ a particle.

How does the observer generically observe the speed of $\delta$?

I heard that, in the case where $\gamma(\lambda)=\delta(\lambda)$ for some $\lambda\in\mathbb{R}$, the observer will measure the following velocity vector for the particle:

$$\text{v}_{\delta}:=\sum_{i=1}^3\in^i(\dot{\delta}(\lambda))\cdot e_i$$

where $\in^i$ is the induced dual basis of $T_{\gamma(\lambda)}M$ w.r.t. to the basis $e$.

Say $M=\mathbb{R}^3$ so that we have a visaulizable, 3 dimensional spacetime manifold with two spatial dimensions from the viewpoint of an observer, and

$$g_{ij}=\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{bmatrix}_{ij}$$ so that it is flat. Say we have two worldlines

$$\text{E},\text{P}:\mathbb{R}\rightarrow M,\ \text{E}(\lambda)=(\lambda,0,0),\ \text{P}(\lambda)=(\lambda,1,\lambda\cdot v)$$

for some $v\in\mathbb{R}^+$. Then how will an observer following $\text{E}$, given any frame, perceive the velocity of $\text{P}$?

My first approach went very much wrong. To show effort let me summarize: We can view the two light rays $L_1$ and $L_2$ between $\text{P}$ and $\text{E}$ - at two pairs of $\lambda$-values, $(\lambda_1^P,\lambda_2^P)$, $(\lambda_1^E,\lambda_2^E)$, for $\text{P}$ and $\text{E}$ such that they seperate $\text{E}$ by a time interval of length 1 - as vectors in $\text{E}$s frame $(T_{\text{E}(\lambda_{1/2}^E)}M)$ at this observer's parameter values ($\text{E}$ will 'see' $\text{P}$ at these points in (his/her) time via the world lines of the two lightrays). We can parallely transport $L_1$ to the frame at parameter value $\lambda^E_2$, where $L_2$ lives, and subtract the two. Maybe if we measure $L_2-L'_1$s length with $g$ this will give us something akin to the velocity $\text{E}$ would measure of $\text{P}$? But that turns out not to be the case.

I read online that one can calculate the subjective distance in special relativity (using the tools of the manifold-spacetime) as follows: Measure the travel time observed by $\text{E}$ of a lightray sent out at some $\text{E}(\lambda)$, being reflected by $\text{P}$s worldline and consequently crossing $\text{E}$ again. Multiplying this measured time by the speed of light divided by 2 will give the perceived distance between $\text{E}$ and $\text{P}$. But in our example of $\text{E}$ and $\text{P}$, an ordinarily reflected lightray won't hit $\text{E}$ ever again. Can this method still be applied to measuring distance, further on velocity, in a spacetime manifold?


1 Answer 1


The idea is that in special relativity an observer can in principle (this in principle is very interesting) assign a date to every event in spacetime. A prescription for this was given by Einstein and Poincaré. The idea is that for every point $p\in M$ there exists a unique $\lambda_1\in\mathbb{R}$ such that a light pulse emitted from $\gamma(\lambda_1)$ reaches $p$. This assumes $\gamma$ is time-like. Similarly, there is a unique $\lambda_2\in\mathbb{R}$ such that a light pulse emitted from $p$ reaches $\gamma(\lambda_2)$. Now, up to an additive constant, each curve $\gamma$ induces a proper time function $s:\mathbb{R}\rightarrow\mathbb{R}$ which is independent of parametrization (namely its length). We say that the time of $p$ according to the observer $(\gamma,e)$ (notice that $e$ hasn't played a role so far) is given by $$\frac{s(\lambda_1)+s(\lambda_2)}{2}.$$ This defines a time function $t:M\rightarrow\mathbb{R}$. Notice that this time function has a very nice physical interpretation since loosely speaking the observer assigns the same time interval to the trajectory of the photon getting from it to $p$ and the photon coming back. This is very much related to the constancy of the speed of light. The preimages $t^{-1}(T)$ are the surfaces of simultaneity at time $T$ according to $(\gamma,e)$. A very nice discussion of this is found in Gourgoulhon, Special Relativity in General Frames.

The discussion of measuring spatial distances in this book is then only restricted to nearby points. However, one would hope that the induced metric on $t^{-1}(T)$ would be Riemannian, giving us a notion of distance on this surfaces of simultaneity and thus allowing us to solve your problem. I would believe this allows for a definition of velocity and so on. I don't know why this reference doesn't do this. Instead, their approach is based on the fact that on some neighborhood of the trajectory of the oberver, each point $p\in M$ has a unique $\lambda\in\mathbb{R}$ such that $p\in \gamma(\lambda)+\operatorname{span}\{e^1(\lambda),e^2(\lambda),e^3(\lambda)\}$. Thus, the trajectory of a particle in this neighborhood is described by a map $\vec{x}:\mathbb{R}\rightarrow\operatorname{span}\{e^1(\lambda),e^2(\lambda),e^3(\lambda)\}$. The derivative of this map yields the velocity. In the case of an oberver moving on a straight line as the one you are considering, this neighborhood can be extended to all of spacetime. However, in the presence of curvature this prescription will only work locally since in general the local rest spaces will intersect.

  • 2
    $\begingroup$ "[...] the observer assigns the same time interval to the trajectory of the photon getting from it to $p$ and the photon coming back." Note that although it seems logical, the observer is not forced to make that assignment. We can only measure the round trip speed of light, not its one-way speed. $\endgroup$
    – PM 2Ring
    Jun 28, 2020 at 11:32
  • 2
    $\begingroup$ Einstein's clock synchronization convention adds the assumption that the one way speed equals the round trip speed, but Reichenbach, Grünbaum et al showed that other conventions are possible. John Norton has a nice discussion here. $\endgroup$
    – PM 2Ring
    Jun 28, 2020 at 11:35
  • 1
    $\begingroup$ I am definitely going to read more on this when I get some time! $\endgroup$ Jun 28, 2020 at 17:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.