B Acceleration at Near-Light Speed: Answers

derek10
Messages
54
Reaction score
0
Hi
I understand that accelerating you get closer and closer to the speed of light wrt anything in an asymptotic way, but would acceleration still act the same way as non relativistic speeds (inertia, gyroscope, etc?) even if the speed won't increase almost at all (for example at 0.9999 c)
Thank you
 
Physics news on Phys.org
What do you mean by "act the same way"? If you have a rocket that moves at 0.9999 c relative to Earth, you would not notice anything special in the rocket, and you can use the rocket to accelerate as usual*. That is one of the fundamental principles of physics: the laws of physics are the same in every reference frame. Why should you have to care about your speed relative to Earth? There is nothing special about Earth.

*Observers on Earth will measure a different acceleration than you in your rocket.
 
  • Like
Likes derek10
Someone moving close to the speed of light relative to you can consider themselves at rest and you as moving close to the speed of light. You can both ignite rockets and feel a 1g acceleration. Both of you will report that you are accelerating at 1g and the other as accelerating much less. Both of you can spin up gyroscopes and find they look normal, but the other guy's will look odd - see the illustration half way down http://asia.iop.org/cws/article/news/50366
 
  • Like
Likes derek10
Thank you. I mean accelerating constantly from the Earth toward c , so it will still accelerate but the speed increase of this acceleration will slow down with respect to the earth, but not with other objects, right?
 
Yes. If you are accelerating in your direction of motion with respect to some object, if you feel an acceleration ##a## then an observer in an inertial frame will observe an acceleration ##\gamma^3a##, where ##\gamma=(1-v^2/c^2)^{-1/2}## is the Lorentz gamma factor and ##v## is your speed with respect to that object. So as you approach light speed (with respect to me) I will see your acceleration drop while you will feel it remaining constant.
 
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...
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...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
Back
Top