Is the perception of backwards time possible in relativity during deceleration?

  • Thread starter Thread starter Ookke
  • Start date Start date
  • Tags Tags
    Time
Ookke
Messages
172
Reaction score
0
Hi. A short question regarding relativity of simultaneity, its interpretation and maybe some terminology.

Suppose A and B are distant space stations, at rest with each other, clocks synchronized. A rocket R travels with high speed passing A at time t=0, heading towards B. Not going into details, from R's perspective the clock reading at B could be couple of years ahead.

If the rocket suddenly stops at A, it enters the frame of A and B, where the clocks are synchronized. After stop R will agree with A that the time at B is t=0.

Should we say that from R's perspective time went backwards at B during the deceleration? Of course R could not really watch the event because of distance and finite light speed, but nevertheless R "observed" it (in the sense commonly used in relativity, involving calculations). Or should we think this only a coordinate system change, not an observation at all?
 
Physics news on Phys.org
Ookke said:
If the rocket suddenly stops at A, it enters the frame of A and B, where the clocks are synchronized. After stop R will agree with A that the time at B is t=0.
Setting of clocks is a convention. If, when the rocket stops, it wants its clocks to match those of B, it will need to set them manually. It's like clocks changing in winter (in countries where that happens!) - you simply change your clock to match the "official" time.

The point is that the rocket does not "enter the frame of A and B". It can always be described in any frame - just like anything else. It's motion change means that its clocks now tick at the same rate as A's and B's clocks, but it doesn't mean that they must show the same time.
Ookke said:
Should we say that from R's perspective time went backwards at B during the deceleration?
No. If you want a single coherent perspective for R, it cannot be two inertial frames naively stitched together, precisely because doing that would cause some events to have two coordinates assigned to them and some none. My favourite way of constructing non-inertial coordinates is Dolby and Gull's radar time - see https://arxiv.org/abs/gr-qc/0104077.
 
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...
Abstract The gravitational-wave signal GW250114 was observed by the two LIGO detectors with a network matched-filter signal-to-noise ratio of 80. The signal was emitted by the coalescence of two black holes with near-equal masses ## m_1=33.6_{-0.8}^{+1.2} M_{⊙} ## and ## m_2=32.2_{-1. 3}^{+0.8} M_{⊙}##, and small spins ##\chi_{1,2}\leq 0.26 ## (90% credibility) and negligible eccentricity ##e⁢\leq 0.03.## Postmerger data excluding the peak region are consistent with the dominant quadrupolar...
Back
Top