Inertial reference frame and Bondi k-calculus

In summary, the concept of an inertial reference frame is essential in physics, representing a viewpoint where an observer experiences no net force, allowing for the laws of motion to apply uniformly. Bondi k-calculus, developed by Hermann Bondi, introduces a framework for analyzing spacetime and electromagnetic fields within the context of special relativity. This calculus employs a parameter 'k' to describe the relative motion between different reference frames, facilitating the understanding of relativistic effects such as time dilation and length contraction. Together, these concepts enhance the comprehension of dynamics in relativistic contexts.
  • #1
cianfa72
2,471
255
TL;DR Summary
About the definition of standard coordinate time in SR context compared to Bondi k-calculus
Hi, I was reading the Bondi k-calculus as introduced in R. d'Inverno book and Bondi k-calculus.
As far as I understand, in the context of SR, the radar time coordinate from an inertial observer/clock is basically the same as the coordinate time ##t## assigned to any event from an inertial frame in which the coordinate clocks at rest in it have been Einstein's synchronized with that inertial "master" clock.

Is that the case ? Thanks.
 
Physics news on Phys.org
  • #2
Yes, radar coordinates reduce to a standard inertial frame for an inertial radar.
 
  • Like
Likes robphy
  • #3
Dale said:
Yes, radar coordinates reduce to a standard inertial frame for an inertial radar.
Ok, so in 3D space one can "recover" the event's standard spatial coordinates ##x,y,z## (w.r.t. the radar location) from the event's distance ##d = \frac c 2 \left (T_2 - T_1 \right )## and the direction the radar echo comes from (##T_1## and ##T_2## are proper times along the inertial radar worldline).
 
Last edited:
  • #4
A question somewhat related to the topic. By definition of inertial frame (aka inertial chart) the spacetime path of free objects (zero proper acceleration as measured by accelerometers attached to them) are described in such a chart by paths with zero coordinate acceleration.

What about the spacetime path of objects at rest in a such inertial chart ? My reasoning is as follows:

Consider the spacetime timelike path described by fixed spatial coordinates ##(x,y,z)## in that inertial chart -- call it ##\alpha (\tau)##. At any point in spacetime there is unique geodesic in a given spacetime direction (therefore in particular there is a unique timelike geodesic in a given timelike direction). Take the timelike geodesic tangent to timelike path ##\alpha## at event ##P## -- call it ##\beta(\tau)##. It is at rest in the inertial frame at ##P## therefore it must coincide with ##\alpha## everywhere (otherwise ##\beta## would not have zero coordinate acceleration at ##P##).
 
Last edited:
  • #5
cianfa72 said:
What about the spacetime path of objects at rest in a such inertial chart ?
The definition of "inertial chart" already gives you the answer to this question. Your roundabout "reasoning" implicitly makes use of the fact about inertial charts that answers the question, so you actually aren't proving anything, you're just laboriously restating the definition.
 
  • Like
Likes Ibix
  • #6
PeterDonis said:
The definition of "inertial chart" already gives you the answer to this question. Your roundabout "reasoning" implicitly makes use of the fact about inertial charts that answers the question
My definition of inertial frame/chart: a reference frame/chart such that any free body worldline (i.e. with zero proper acceleration) has zero coordinate acceleration in it.

The above definition does not require/imply, at first glance, that a body at rest in it must have zero proper acceleration.
 
  • #7
cianfa72 said:
My definition of inertial frame/chart
Your definition is irrelevant. What matters is the standard definition in the literature.
 
  • Like
Likes Motore
  • #8
The OP question has been answered. Thread closed.
 

Similar threads

Replies
54
Views
2K
Replies
62
Views
5K
Replies
101
Views
5K
Replies
78
Views
6K
Replies
26
Views
3K
Replies
15
Views
2K
Replies
34
Views
3K
Back
Top