0
$\begingroup$

Given a metric, we find out the null and timelike geodesic which helps us conclude that how the trajectory of various particles will become in a particular curvature of spacetime.

But I don't understand that with respect to whom these geodesic are calculated. Are these geodesic observed by an observer who is sitting at infinity or are these geodesic according to a person who is actually travelling along these geodesics? Will the null geodesic be same for all the observers regardless of which frame they are in, because light should travel at speed c in every frame so is it true that light take the same path according to every observer?

How do we differentiate between the two observers by just looking at a metric and calculating the geodesic equation?

The situation should change when we consider a local inertial observer because in his frame of reference the spacetime is flat, so for him laws of special relativity would be valid. What would he see when he is falling into a black hole from a finite distance away?

Improve this question
$\endgroup$
1
  • $\begingroup$ Note that a geodesic is an observer-independent notion. $\endgroup$
    – Qmechanic ♦
    Sep 10, 2017 at 9:41

1 Answer 1

Reset to default
0
$\begingroup$

First it is important to note that all geodesics are observer independent, thus it makes no difference wrt whom the geodesics are calculated. A null geodesic is a null geodesic, and a timelike geodesic is a timelike geodesic no matter the observer. It makes most sense to calculate in the frame of reference of the coordinate system, so this is what is done.

Your assertion that the speed of light is the same in every frame is not technically correct. The speed of a ray of light in a local frame is always $c$, but the speed the same ray as measured by some distant observer may not be.

Your observer falling into a black hole cannot be assumed to be in minkowski space for his journey. Along this path he will travel through much curvature.

Improve this answer
$\endgroup$
10
  • $\begingroup$ Can you tell me how the geodesics are observer independent? $\endgroup$
    – Khushal
    Sep 11, 2017 at 19:31
  • 1
    $\begingroup$ They are defined as the paths of extremal distance (intuitively shortest or longest path) between two points in space time. Thus they rely on properties of the spacetime itself, the metric, not any observer within it. $\endgroup$
    – Eddy
    Sep 11, 2017 at 19:35
  • $\begingroup$ Can I say that given a metric, there are unique geodesics, and all the observers will observe objects moving on these geodesics in the same fashion regardless of their position? Also when I change the coordinate system the metric changes and so does the geodesic. So geodesic depends on metric (or the coordinate system in which the metric is written). Each coordinate system will give its own geodesics. Are these statements correct? $\endgroup$
    – Khushal
    Sep 11, 2017 at 20:33
  • 1
    $\begingroup$ The geodesics are part of the structure of space time itself. Consider another, simpler, manifold, say a sphere. The geodesics (shortest path between two points) are the great arcs. This is independent of any coordinate system or entity on the sphere. You can write the metric for the sphere in any coordinate system you like (cartesian, cylindrical, spherical etc) and calculate the geodesics in that coordinate system, the equations will look different but represent the same curves. It works the same in GR. $\endgroup$
    – Eddy
    Sep 11, 2017 at 20:45
  • $\begingroup$ By looking at a metric can we conclude that the particular metric represents an infalling observer or an asymptotic observer $\endgroup$
    – Khushal
    Sep 14, 2017 at 7:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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