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Imagine I am floating in space some large distance X above a neutron star or high mass object and I am using rocket boosters to stay stationary relative to the object. Assume no other forces acting on me or the object and no weird things with the neutron star like magnetic fields or extreme temperatures, it’s just an object of very high mass. Using the laws of motion but excluding special and general rel I calculate that by using my rocket boosters and gravity I can accelerate past light speed before I will reach the neutron star. Obviously this is impossible. Now let’s say I accelerate towards the object and turn my rocket boosters on full blast to accelerate me more. Assume the most powerful rocket boosters imaginable. I know that I can never break light speed before I hit the neutron star but what will my reasoning for this be. What will I actually experience? What will my excuse be as to why I did not reach light speed before impact if you hypothetically asked me after my death? As I approach light speed in my reference frame will I see the distance to the neutron star length contract so that my distance to it shrinks and I don't have enough distance to accelerate past light speed? Or does length contraction not happen in an accelerating reference frame?

garyp
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murram20
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  • According to the principle of equivalence: Locally, the effects of a gravitational field on a mechanics experiment are identical to the effects of an acceleration of the observer's frame of reference. Locally, the effects of a gravitational field on an experiment in mechanics and electromagnetism are identical to the effects of an acceleration of the reference frame of the observer. So, we can consider that the acceleration is due locally to a an additional mass of the star and apply the "artiery" of general relativity. – The Tiler Feb 04 '23 at 12:45
  • I’ve never studied general relativity so am confused. I’ve studied special relativity though so I’m wondering if any length contraction would happen in this accelerating reference frame? Like if i travelled towards any object at 99.9% light speed at constant velocity my measured distance to that object would be length contracted significantly but if i reach 99.9% while undergoing acceleration would there be the same or similar length contraction in my reference frame? – murram20 Feb 04 '23 at 12:53
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    I think that you can do the same experiment without the neutron star in the picture. If you're far enough, it's just a force added to your thrusters. If you aren't far enough to neglect GR, you have no option but use GR. The standard treatment of the uniformly accelerating observer in SR (i.e. flat spacetime, no masses bending it—and since time is much “squishier” than space, not just extremely gravitating masses, any masses whatsoever) is done in hyperbolic Rindler coordinates, which become the Rindler metric when generalized. Also, you'll survive, that's a plus! :) – kkm -still wary of SE promises Feb 04 '23 at 13:06
  • Ok let’s look at it like this then. You are in a universe with two objects only, yourself in a rocket and a neutron star. There is nothing else in the universe. Imagine you start at x = infinity away from the star. GR is now irrelevant. You accelerate towards it to 99.999% light speed and want to break the light speed limit. You then turn off your thrusters and stay at constant speed. You were at x = close to infinity away and GR was negligible but now your distance to the star has length contracted. Now you dont have enough time or distance to break light speed before crashing? – murram20 Feb 04 '23 at 13:16
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    No, your velocity cannot exceed $c$, neither in SR nor in GR. For SR, it's almost follows from its foundational postulate: light in a vacuum as observed is propagating at the constant velocity $c$ for any inertial observer ($c$ is a constant of nature, not “speed” of anything!). GR is SR locally for any observer, and globally absent stress-energy, in the weak-field limit. There is no “magic,” it follows from the math of these theories. If your velocity were to exceed $c$ relative to me, I'd measure your velocity as a complex number, $\sqrt(something_negative)$, and it's non-physical. – kkm -still wary of SE promises Feb 04 '23 at 13:26
  • BTW, welcome to Physics.SO, and it's a pleasure to see an inquisitive mind here, really! Let me explain something: when replying to a comment, type @, and it will drop down a list of usernames in the discussion; pick the one that you're responding to—I saw your reply by pure chance! The topic-starter (whoever's answer or comment you're replying to) will be notified without that, but other commenters won't. And as for your question, it's a stubborn math, the language of physics, that doesn't let your velocity exceed $c$. It captures observations, but we speak in these symbols after all. – kkm -still wary of SE promises Feb 04 '23 at 13:33
  • I agree with kkm that the neutron star adds unnecessary complication. You first need to understand what happens at relativistic speed in flat spacetime before you go to GR. Special Relativity can handle acceleration, if you're careful, but it doesn't handle gravity. – PM 2Ring Feb 04 '23 at 13:49
  • @kkm But when you add acceleration in you get paradoxes. I am specifically talking about black hole information paradox here. An observer will see an object never cross an event horizon but the person falling in will say that they did cross it. I have a theory where length contraction here would play a role in solving the paradox of whether they crossed the horizon or not. Characters are so limited on this that there isnt enough space to explain anything properly. – murram20 Feb 04 '23 at 13:51
  • @pm2Ring ok weird i didnt know special relativity could handle acceleration. But if einstein showed that a a reference frame in a gravitational field is identical to an accelerating frame then why cant special rel be compatible with acceleration towards a massive body? – murram20 Feb 04 '23 at 13:57
  • Well, constant acceleration is equivalent to a uniform gravity field. But the gravity around a star, planet, etc, isn't uniform: it gets stronger as you get closer. So gravity is only locally equivalent to acceleration, in a "chunk" of spacetime that's small enough so that the spacetime curvature is negligible. – PM 2Ring Feb 04 '23 at 14:10
  • @pm2Ring yeah true, i’ll have to think about this more. – murram20 Feb 04 '23 at 14:38
  • I think I'll write an answer. It won't be the highest quality one on the site; I won't do any math, but rather sort out your misconceptions—it may help others. There are too many to help you in the comments. Conceptually, both SR and GR are simple enough to understand the physics. Strict math is complex to do in GR, so if you can ignore its corrections, always go with SR. Similarly, if you can ignore it, Newton gravity is the way to go. It's enough for NASA to calculate trajectories in the Solar System, even for solar probes, where fixed correction work like a charm. – kkm -still wary of SE promises Feb 05 '23 at 04:53
  • In a day or two. I have work to do, and want to free up my mind for writing the answer. I will remove this and above messages when I'll get to it. Both your striving to understand and persistence in this are commendable indeed: many students remember the formulas, can solve differential equations, but do not understand main ideas of relativity at all. You can build up the math to express your intuition: it's the only language to speak about physics. But, IMO, it's useless to go into math when you don't have intuition. Like, someone learns a foreign language but speaks nonsense; no point. – kkm -still wary of SE promises Feb 05 '23 at 05:09
  • BTW, I dunno if you know the difference between theory, model, hypothesis and intuition. I'll be blunt, but I wish you only good. Remember once and for all: the most obscene words you can utter in a chat that shows that you're missing basics is “I have a theory fixing paradoxes in Einstein's relativity.” You don't. But this is a cue for all who know physics to get up and leave you in a company of cranks, each with their own pet theory correcting “Einstein's misunderstanding.” Such a blunder may push you into a runaway self-learning disaster, crash-landing you in crank pseudoscience. – kkm -still wary of SE promises Feb 05 '23 at 22:01
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    @kkm yeah you are right. It’s a hypotheses. – murram20 Feb 06 '23 at 17:40

1 Answers1

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You will observe that blasting the rocket motors for a short time causes a velocity change of almost all objects in the universe. The amount of velocity change depends on what velocity an object had initially, more initial velocity means less velocity change. When initial velocity approaches c, velocity change approaches zero.

To explain this effect you need to change to an inertial frame and do the explaining in that frame.

Here "velocity" means velocity towards or away from you.

Here is the explanation in an inertial frame: How does SR explain constant light speed where the distance between observer and light source is increasing?

stuffu
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  • But if I apply a constant force with rocket thrust and know my acceleration from gravity alone should always be constant then in my own reference frame i will appear to hit this arbitrary point in time where i stop accelerating. This would break the laws of physics in my frame? Surely i need some fundamental reason/measurement/observation as to why i have stopped accelerating? – murram20 Feb 04 '23 at 12:23
  • @murram20 You're heading towards the neutron star, so the acceleration due to gravity increases (even in Newtonian physics), it's not constant. Why do you think you'd stop accelerating? – PM 2Ring Feb 04 '23 at 12:54
  • If you didnt stop accelerating you’d reach the speed of light which is impossible. But from newtonian calculations you calculated you would exceed light speed. In your reference frame there must be some reasoning as to why you cannot exceed this speed. – murram20 Feb 04 '23 at 13:07
  • @murram20 No, it doesn't work like that. Relativity doesn't limit your acceleration, and you can accelerate indefinitely, with whatever thrust your engines can supply. But that will never let you reach c because of how velocity addition works at relativistic speeds. We have lots of answers about that, here's one of mine: https://physics.stackexchange.com/a/598415/123208 And here's one that derives the velocity addition equation: https://physics.stackexchange.com/a/345492/123208 – PM 2Ring Feb 04 '23 at 13:44
  • @pm2Ring thanks. Unfortunately the maths went way over my head. But just from an experiential point of view im still wondering if my distance to an object appears to shrink or is measured to shrink as i fall towards an object at near light speed? I’m trying to insert a special rel concept into a frame with a large mass which may not be allowed but im trying to visualise what that falling person would experience and if length contraction would happen? – murram20 Feb 04 '23 at 14:01
  • @murram20 Sure, as your speed gets higher (eg, relative to the galaxy), distances get contracted in the direction of travel. – PM 2Ring Feb 04 '23 at 14:14
  • @pm2ring ok nice. That is what i thought. This fits in with a hypothesis i have about black holes and the information paradox. – murram20 Feb 04 '23 at 15:26