Accelerating frame transformation

Click For Summary
SUMMARY

The forum discussion focuses on deriving a coordinate transformation between a lab frame (S) and an accelerating frame (S') in Minkowski spacetime, specifically for a rocket with constant proper acceleration. The relationships established are \(t = \frac{1}{a}sinh(a't')\) and \(x = \frac{1}{a'}[cosh(a't') - 1]\). The discussion emphasizes proving that under certain conditions, the constants A, B, and C can be uniquely determined, leading to the transformations \(t = (\frac{1}{a} + x') sinh(a't'/c)\) and \(x = (\frac{1}{a} + x') cosh(a't'/c) - \frac{1}{a}\).

PREREQUISITES
  • Understanding of Minkowski spacetime and its properties
  • Familiarity with hyperbolic functions (sinh, cosh)
  • Knowledge of proper acceleration in relativistic physics
  • Ability to manipulate and solve equations involving multiple variables
NEXT STEPS
  • Study the implications of constant proper acceleration in general relativity
  • Learn about the Lorentz transformations and their applications
  • Explore the concept of simultaneity in different inertial frames
  • Investigate the role of boundary conditions in solving differential equations
USEFUL FOR

This discussion is beneficial for students and researchers in physics, particularly those studying relativity, as well as educators looking to deepen their understanding of frame transformations in accelerating systems.

Fek
Messages
13
Reaction score
0

Homework Statement


[/B]
In Minkowski spacetime we are considering a (series of) frame(s), S', attached to a rocket with constant proper acceleration. The rocket's speed in S is v.

We find with boundary conditions x = 0 at t = t' = 0 the relationships between S and S' (for x' = 0, i.e. at the rocket):

$$t = \frac{1}{a}sinh(a't') $$
$$ x = \frac{1}{a'}[cosh(a't') - 1] $$

Let's use these results to construct a full coordinate transformation from the lab frame x,t, to the accelerating frame x',t'. Try

$$t = Asinh(a't') + B$$
$$ x= Acosh(a't') + C $$

Prove that if
i) surface of constant t' are surfaces of constant time in a frame moving instantaneously at v
ii) t matches with t' at early times and small x', while x agrees with x' at early times

then A, B and C are uniquely determined and that

$$t = (\frac{1}{a} + x') sinh(a't'/c)$$
$$ x = (\frac{1}{a} + x') cosh(a't'/c) - \frac{1}{a} $$

Homework Equations

The Attempt at a Solution


For x to agree with x' at early times (cosh(a't')=1) we know:

$$A = (k + x') $$
$$C = -k$$

Where k is a constant.

Then t and t' to agree at early t' and small x' we can see k= 1/a.

However I don't know what limit condition 1 implies and cannot see a way to make the proof "watertight".
 
Physics news on Phys.org
Condition 1 basically says you can set up a third frame S'' at the same speed as S' relative to S, and that t'' depends only on t' (not x').
Sorry for the brief reply, I'll elaborate when I can.
 

Similar threads

Replies
10
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
2K