Another classic out the purple, fiery, math dimension you inhabit!
@RunningRay9
4 жыл бұрын
Wow, this is a great explanation. This video pretty much clear up all the thing that i don't really feel comfortable following in the original video. I think all the missing/poorly explained detail has been filled now, nice work man
@samhollins2076
4 жыл бұрын
0:09 Only kyle's subs got the "alignment" joke 😂
@JustAlex96
4 жыл бұрын
Soo... driving really fast makes your length contract? That actually explains a lot...
@juliavixen176
4 ай бұрын
Just a technical note, that's kinda important, the ship is _not_ "at rest" with respect to itself precisely because it _is_ accelerating with respect to itself. It would be "at rest" with respect to itself if it was not accelerating (i.e. inertial). The most noticeable consequence of this is that the rigid physical material of the ship feels a force (change in momentum) in the direction it is being accelerated. If it was not accelerating it would not feel a compressive force (and everything inside is weightless). Essentially, pushing on one end of a rigid object, propagates a wave of pressure through the material at the speed of sound, and continuously accelerates each bit of material next to it as the elastic forces within the material restores it's shape... eventually the far distant end of the object gets accelerated forward if there's nothing there to push back... In relativity... the two ends of the object have a relative velocity to each other. The end being pushed is moving faster than the far distant end... which is why the material is compressed... and the two ends of the object are in _different inertial reference frames_ from each other. That's why proper acceleration is sometimes called "absolute motion". It's _moving_ relative to _itself_. (Not "at rest" with respect to itself.)
@massimilianodellaguzzo8571
16 күн бұрын
Wow juliavixen176, you're really good (and I know because I've read your other comments on other videos) 🙃 In my opinion this happens: a) if we consider the motion of the two spaceships in the frame of the "stationary observer" then the distance between the two spaceships does not contract (the length of the rope does not contract and the rope does not break) b) if we consider the motion of the "stationary observer" in the frame of the two spaceships then the distance between the two spaceships contracts (the rope also contracts and does not break) The two motions a) and b) both occur, although motion a) is hidden. If we consider the motion a) then the frame of the "stationary observer" contracts and if we consider the motion b) then the frame of the two spaceships contracts.
@thephysicistcuber175
4 жыл бұрын
Epic physics time needs to be a thing. Change my mind.
@DrDeuteron
2 жыл бұрын
do it.
@SakisStrigas
4 жыл бұрын
Well done my bro! Amazing video as always! 🤘
@MatesMike
4 жыл бұрын
Your content is amazing! I always love the animations
@HAL-oj4jb
4 жыл бұрын
I heard that paradox sometime in my teenage years when i started to get interested in that sort of stuff, but I never understood it lol. Well now I do, awesome video!
@chemistro9440
4 жыл бұрын
man.. tool/reflection goes unreasonably well with deep math or physics stuff
@spencerbrinton1397
3 жыл бұрын
I agree. I also immediately knew the thing at the very start of the video is tool. Win
@Assault_Butter_Knife
4 жыл бұрын
Wow... Very well explained. I didn't exactly catch Kyle's explanations, but here I only had to watch the video twice to get it
@jonashallgren4446
4 жыл бұрын
You're probably the most underrated math youtuber, like look at those sexy animations and that sexy face! edit: underrated with regards to subs
@nathangrant1824
4 жыл бұрын
more content would help with that.
@EpicMathTime
4 жыл бұрын
I must say that both of you are absolutely correct.
@nathangrant1824
4 жыл бұрын
@@EpicMathTime i really do enjoy your explanations, dude. your video on gabriel's horn is one of my all time faves. :D
@austinmitchell2652
4 жыл бұрын
Great to see a new upload from you! This got me thinking about my modern physics class again, and I appreciate the explanation! Also, do I hear a Mario kart race start sound effect in there?
@Smitology
5 ай бұрын
Special relativity is an impressively self-consistent theory, the majority of "paradoxes" people find are simply because they apply classical intuition, the mixing of special relativistic and classical conclusions is what leads to the paradoxes
@mathOgenius
4 жыл бұрын
Wow ! like the quality is offf the charts , like you have improved a loooot since you have started wow ! write kyle hill with @ symbol on the title this will send him that someone mentioned him , I hope he sees this. its at the level of his vid 😁😁😁
@EpicMathTime
4 жыл бұрын
Thanks bud!
@Primitarian
8 ай бұрын
So could we explain the expansion of the universe in the same way? From Reference S, the universe appears to be moving away from us, but from Reference I, the rest of the universe is moving at the same speed as us, it's just that we observe the rest of the universe as undergoing length contraction.
@matron9936
4 жыл бұрын
Great work
@dialectphilosophy
4 жыл бұрын
Great video! Definitely addresses some of the main confusions in a succinct way. I’ve always found the rules of spatial contraction to be somewhat ambiguously explicated within the context of SR. A cursory glance suggests that space contracts about the center of mass of the moving frame of reference. Indeed a boost translation leads to the conservation of a center of mass value just as a spatial translation leads to the conservation of momentum. This seems to imply that the distribution of mass determines the “pivot” point about which space begins contracting. But as the Bell Paradox points out, if the motion is not communicated about the center of mass, but rather stems from unequal distributions within the mass itself, the object under motion will experience some type of deformation. So in addition to the contraction of actual space itself, you have a second type of contraction due to internal tensions of the matter itself. Then it gets confusing how to distinguish between interior forces of contraction and the contraction of space itself since for any given object its motion will not be communicated across a single point and interior forces of deformation will be always be at work.
@DrDeuteron
2 жыл бұрын
There are no forces of contraction. A moving object remains its proper-length in its rest-frame (which is moving relative to Earth). In the Earth's frame, the object is shorter. The question "about which point does this contraction occur" does not have physical meaning; rather, it depends on a choice of coordinates. The center-of-mass is a fine choice, but not required: nature does no care about our choice of coordinates.
@massimilianodellaguzzo8571
3 жыл бұрын
If two equal distances are in relative motion to each other at speed v, the distances overlap. (the distance between the two spaceships tied by a string and the distance between two points on the frame of the Earth) In my opinion, if we denote with t the elapsed time in the frame of the Earth and if we denote with t_1 the elapsed time in the frame of the two spaceships (tied by a string), it is t = t_1. (and two equal distances overlap, in this case no problem and the string does not break) The length of the string contracts in the Earth's frame and the length of the Earth contracts in the frame of the two spaceships, the two lengths are equal! (suppose they are equal to d, even if in two different frames) d = gamma * v * t and d = gamma * v * t_1. When two distances overlap it is t = t_1 = d / (gamma * v), for each value of d. If a spaceship is moving at speed v not tied to a second spaceship, then the situation is different. (there are no two distances in relative motion between them)
@HDitzzDH
4 жыл бұрын
@Epic Math Time Quick question, would you say that evaluating problems like 8/2*4 are ambiguous? Most people have heard of the "left to right" rule if two operations has the same priority, what would be the best answer here? Clearly we're lacking parentheses and the division sign is pretty outdated, thus is why so many claim it's a poorly written problem.
@EpicMathTime
4 жыл бұрын
I would say that it is not ambiguous if the order of operations is strictly followed like a robot (8/2*4 = 4*4 = 16) but can be ambiguous in literature.
@HDitzzDH
4 жыл бұрын
@@EpicMathTime Well sure but why perform the division first? Isn't it true that if you have operations of the same priority it shouldn't matter which order you do it? I've just been watching people like the math professor Keith Devlin who explains why it indeed is ambiguous, since we're lacking parentheses and the "left to right" rule isn't a thing in most places.
@EpicMathTime
4 жыл бұрын
The order does not matter at all, but you certainly have to preserve which numbers are associated to which operations. That is, 8/2*4 can also be evaluated as 8*4/2, or 4/2*8, it doesn't matter the order. As long as you actually divide by 2 and actually multiply by 4. This is not conceptually different than 5-3+2. I can do it in any order, as long as I actually subtract 3, and actually add 2. What I cannot do is subtract both 3 and 2, which is what one does if they say: 5-2+3 = 5-5 = 0.
@HDitzzDH
4 жыл бұрын
@@EpicMathTime I think the problem occurs when you divide the 2, some think that you might need to divide the entire expression (2*4) which would give the answer 1, this depends on how we put the parentheses, when it comes to sums like your example it doesn't matter where we put parentheses, it always gives the same answer.
@EpicMathTime
4 жыл бұрын
Well, I can't answer what some may think, all I can say is that the standard convention and all programming languages parse 8/2*4 as 16, and the expressions 5-3+2 and 5/3*2 are in perfect analogy with respect to the standard order of operations. The order of operations doesn't care what / looks like, only that it is symbolizing the operation of division.
@2false637
4 жыл бұрын
This video was amazing! Keep it up!
@somewhatblankpaper1423
4 жыл бұрын
Interesting video! Keep up the good work!
@mastershooter64
4 жыл бұрын
the math boiis challeging physics bois?
@michalchik
3 жыл бұрын
Not to be confused with Bell's inequality
@DrDeuteron
Жыл бұрын
but what if each ship has half of an entangled quantum state?
@imasiontist653
3 жыл бұрын
Is that fear inoculum I hear?!
@evgeniyan2426
2 жыл бұрын
Let two rockets don't move at all. But reference frame moves with acceleration in relation to them. This is absolutely equivalent problem. So, why the string should break ?
@EpicMathTime
2 жыл бұрын
It is clearly not an equivalent problem. In your reference frame moving relative to the ships, the ships will move closer together due to length contraction; that violates the premise. In this problem, the ships remain at a constant distance apart in the observer's frame.
@evgeniyan2426
2 жыл бұрын
This is equivalent formulation of problem. Not equivalent frames. Read carefully ! Really we have two frames - rockets and moving. So we have two possibilities: a. To fix frame and move rockets. b. To fix rockets and move frame. a. and b. absolutely equivalent problems. But b more clear and easier. In this frame distant is constant and no string break. (This what you say "observer" frame). Relativity transformations are continuous.(They are transformations of rocket - unmoving frame up to moving one). So in moving frame the break is impossible. In a. formulation all the same. But in reverse logic - which wasn't done correctly in this video. P.S. the very short disprove of this video: Relativity transformations are continuous ones. So its impossible to have broken and unbroken string in different frames - this situation in video. End of disproof.
@EpicMathTime
2 жыл бұрын
This premise of this problem implies that the ships remain a constant distance apart in the _observer's_ frame of reference. In both of your situations a and b (which are equivalent to each other) the ships remain a constant distance apart in the _ship's_ frame of reference, but not in the observer's frame of reference. Therefore neither of these are equivalent formulations of the problem. You never have a "broken and unbroken string in different frames." In your problem, the string does not break (in any reference frame). In Bell's Spaceship Paradox, the string does break (in all reference frames).
@EpicMathTime
2 жыл бұрын
I suspect there is a language barrier between us, so maybe reading the Wikipedia article in your first language will make it clear why your formulation is not equivalent to the problem that is posed. en.m.wikipedia.org/wiki/Bell%27s_spaceship_paradox
@evgeniyan2426
2 жыл бұрын
@@EpicMathTime read my P.S. And there is no language, this is your and both Wikipedia mental problem.
@Hataldir
4 жыл бұрын
Math + Tool. Subscribed.
@NonTwinBrothers
2 жыл бұрын
Also nice vid
@EpicMathTime
2 жыл бұрын
Thanks for watching!
@h2_
4 жыл бұрын
What if the ships are not actually attached? Suppose the front ship is trailing a refueling hose that functions in vacuum by just squirting fuel a few centimeters in front of some intake port on the trailing ship, or something, so no contact. Will the hose ever break? If the hose can refuel the trailing ship as long as it is within a certain distance d of the second ship, can you find when it will be out of range and incapable of refueling the second ship?
@SpaghettiToaster
4 жыл бұрын
You're just overcomplicating the problem. The rules are the same, no matter the specific way you imagine the connection. The point is that, from the reference frame of a point inbetween the two ships, they're moving apart from each other. All consequences follow naturally and can be calculated just as if the ships were cars travelling apart from each other on earth.
@nujuat
2 жыл бұрын
Lmao this was a problem in my physics assignment
@se7964
3 жыл бұрын
I'm a little confused and looking for clarification: as things travel at faster velocities, they see lengths contract. Wouldn't this make the rear spaceship see the distance to the front spaceship shorten, not get longer?
@EpicMathTime
3 жыл бұрын
The front ship generally has a very low speed relative to the rear ship (and vice versa), so there's no significant relativistic effect.
@se7964
3 жыл бұрын
@@EpicMathTime Still a little confused, I’ll ask my question differently: say you have a spaceship on earth and a spaceship on Alpha Centauri, and they begin simultaneously accelerating (from earth frame). If they reach relativistic speeds fairly quickly, then the earth-spaceship will see the distance to Alpha Centauri contracted. So if these spaceships have a string between them, does that mean the string will contract too?
@EpicMathTime
3 жыл бұрын
If I am interpretting you correctly, it sounds like you are asking if the string contracts in the reference frame of one of the ships. The ships' speeds relative to earth (or alpha-centauri, or anything else) bears no relevance at all to this. That is, regarding the relativistic effects on the string in a ship's reference frame, the only thing under consideration is how fast the string is moving relative to the ship. If we put you in, say, the rear ship, the front ship and the string are barely moving relative to you (this is what my first comment was saying), so neither of those things would undergo any kind of relativistic effects in your reference frame. Anything moving at relative speeds _relative to you_ will have its length contracted in _your reference frame._ If you are moving relativistic speeds relative to earth, your length contracts in the earth's reference frame; and earth's length contracts in your reference frame. You are never moving at relativistic speeds _relative to the other ship,_ though. Here's an analogy, if it helps. Let's say that the Milky Way is moving towards some other galaxy at 5% light speed. So, I'm sitting in my living room, moving towards that galaxy at 5% lightspeed, and so is my television. Will I observe any relativistic effects on my television? No, of course not, because in my reference frame, the TV is just sitting there - why does our speeds relative to some random galaxy matter? (In this analogy, I am one of the ships, my TV is the other ship, and the random galaxy is alpha-centauri).
@se7964
3 жыл бұрын
@@EpicMathTime Right, length only contracts between relative frames, I definitely understand that. I appreciate your responses, btw, I guess I'm having a lot of difficulty pinpointing what exactly my confusion is. If you stick with me a little longer I might figure it out, it's driving me nuts. Here's what I do understand: an observer in the rest frame not accelerating with respect to the spaceships will see each spaceship undergo kinematical length contraction. The string (presumably also undergoing length contraction) is thus forced to snap because it can no longer span the elongated distance between the ships. Here's where I get lost. According to your video, in the frame of the rear spaceship, he sees the front spaceship traveling faster than himself, creating an extra distance between the ships that is responsible for snapping the string. I guess my first confusion is: why does he see the front spaceship traveling faster? You mention it has something to do with different simultaneities, but I guess I'm not grasping how that's related. Plus, aren't velocities in relativity additive? Meaning if observer A on earth measures the rear rocket ship to have velocity v, while the rear rocket ship measures the front rocket ship to have a different velocity w, then the observer A on earth can't measure the front rocket ship as also having velocity v, rather he has to measure (v+w)/(1+vw/c^2).
@EpicMathTime
3 жыл бұрын
By instructing the ships to remain the same speed in earth's frame, we have equivalently instructed them to move apart from each other in their own frames. Appealing to relativity of simultaneity helps describe the difference between the reference frames, but we don't need it to come to the conclusion that the ships are moving apart. Basically, "why" the ships move apart is ultimately just "because that's how they were instructed to move." All of the relativistic differences between the two frames will reflect that, but it's not how we come to the conclusion. We set the ships to have the same speed in earth's frame; we dont make any claims about how the ships move relative to each other. Let that fall where it may. Since we force the ships to be at the same speed in earth's frame, they remain the same distance in earth's frame, and hence the length of the string is constant in earth's frame. If I gave you a 1 meter long rubber band and said "hey, take this away from me at relativistic speeds, but keep the length of the rubber band constant in _my_ reference frame", you are forced to move the ends of the rubber band away from me in a way that stretches the rubber band. When the rubber band reaches a speed that would give a lorentz factor of 2, the rubber band is at that point, necessarily, stretched to twice its original length in order to remain at its original length in my frame (if you did not move the band in a way that stretched it, the rubber band would be contracted to 1/2 of its proper length in my frame, so you did not do what I told you to do). That's exactly what happens with the ships. The ships were set to move at the same speed at the same time in the earth frame, so they move at the same speed at the same time in the earth frame. What happens in some other frame (like what the ships themselves witness) is "not our problem." In any other frame, whatever needs to happen so that the ships travel at the same speed at all times in the earth frame is exactly what happens.
@knave5759
4 жыл бұрын
Would have appreciated more of the math behind this. Very fun video tho
@knave5759
3 жыл бұрын
@Robin Hack Thanks for the equations. Since posting this comment, I've gone through a bit of relativity in our "modern physics" course, so it was fun to watch this again with the tools to solve it.
@yogesh_dangi.
4 жыл бұрын
Ok so it's not related to this video but i guess you can help me what is y²> or = o is it +- y or +y when i simplify it and why please help
@EpicMathTime
4 жыл бұрын
The inverse relation of ^2 is +/- sqrt, so you get "+/- y >= 0." That says y >= 0 or -y >= 0. We can then multiply both sides by -1 on the latter to get y = 0 is true for any real number.
@yogesh_dangi.
4 жыл бұрын
@@EpicMathTime thanks a lot you gained a subB-)
@NonTwinBrothers
2 жыл бұрын
Minute physics gang
@new-knowledge8040
4 жыл бұрын
1:50 Imagine that a 4 dimensional environment existed. Let's say that it is composed of 3 dimensions of space, and one dimension of time. Hmmm, that sounds familiar. Anyhow, picture two spaceships that are present within that 4D space-time environment. They are spaced apart from each other in space, but they are always positioned at the same place in the time dimension as both of these spaceships are on the move across that dimension. Suddenly, the two spaceships both explode. Here we have two simultaneous events. This is true since both explosions occurred at the same place somewhere along that dimension known as the dimension of time. However, other observers in their own frames of reference, disagree. Some say that spaceship A exploded first, and others may say that spaceship B exploded first. But there is no way in which any observers conclusions can be proven to be correct. In turn, many people say that simultaneity is relative. But the point is, two events can occur, whether they both occur at the same point in the dimension of time, or not, they do occur. It is what had occurred in the absolute sense within the 4D space-time environment, that will be viewed differently from different frames of reference.
@SpaghettiToaster
4 жыл бұрын
What does that have to do with the video?
@new-knowledge8040
4 жыл бұрын
@@SpaghettiToaster That's a long story.
@SpaghettiToaster
3 жыл бұрын
@Robin Hack And what does that have to do with his comment?
@user-vn7ce5ig1z
4 жыл бұрын
My issue with Kyle's video was that it's essentially no different than a single ship with two thrusters. Since the ships are continually accelerating, the one that starts first (even by a single Planck moment) will constantly get further and further away because of its head-start. And since a single ship with multiple engines would never realistically be able to fire all of them at _exactly_ the same time, the difference would _eventually_ accumulate to a point where they would tear the ship apart.
@EpicMathTime
4 жыл бұрын
The scenario presented by Kyle is different from a single ship, as a single ship does not move as required by the paradox. As for the simultaneity itself, it is certainly possible for the two ships to precisely accelerate simultaneously (and not kind-of simultaneously). Whether or not this is practically acheivable or not for humans shouldn't matter, otherwise why talk about simultaneity at all?
@SpaghettiToaster
4 жыл бұрын
If the thrusters fire at precisely the same time, the ship will still be torn apart. Because from the reference point of the center of the ship, the bow of the ship is moving further ahead of the stern, causing it to break. Either that or the ship is constructed solidly enough to prevent this, in which case the requirements of the scenario are not met and the bow and stern move closer together in their reference frame.
@fittingin7411
4 жыл бұрын
What is your hair loss regiment?
@pendragon7600
4 жыл бұрын
Not convinced. I mean okay, sure, I can follow your explanation and it makes sense, but "Relativity of Simultaneity" sounds like obvious bullshit. And that was kind of the founding assumption of this whole argument.
@EpicMathTime
4 жыл бұрын
The relativity of simultaneity is a very normal part of relativity with zero controversy.
@pendragon7600
4 жыл бұрын
@@EpicMathTime I'm sure, but to somebody like me with a background in math and not much physics (and certainly not relativity), it sounds unintuitive and questionable. Surely we aren't stupid enough to make the mistake again of being unjustifiably arrogant in our physical models? It's somewhat sketchy to pretend we have the knowledge to answer a currently unverifiable thought experiment like this with such confidence. Edit: also is it zero controversy, or "zero" controversy? Basically what I'm trying to say is.. assuming relativity is part of a perfect model of the universe, sure, this is internally consistent within the bounds of that theory. But stating the answer to such a "paradox" so authoritatively is questionable at best.
@EpicMathTime
4 жыл бұрын
@@pendragon7600 It's a special relativity problem. Without accepting the model of special relativity, at least for the sake of the problem, we can't say much of anything and there is nothing to discuss. Does the string break? In the model of special relativity, yes, it does, and the answer is interesting. If you don't accept special relativity, well, ok, but you can't really talk about the problem at all anymore. The string might not break, it might break, it might turn into an elephant. We have to make some assumptions about the nature of spacetime, and right now relativity is the working model in physics. No physicist would ever say "relativity is true, _definitely"_ but if you ask them a question about spacetime, they aren't going to throw their hands in the air and say "well we can't be sure that our models of the universe are true, so we can't say anything." No, they will explore the question while taking the current models as axiomatic for that exploration, because it's the most useful and productive thing to do. Since your background is in math, this should be very natural for you. When a physicist poses a problem like this, it is a special relativity problem, that is, yes, the model of special relativity is taken as axiom for the sake of discussing a _special relativity_ problem. Whether or not special relativity is "really" an actual truth of our physical universe does not need to be considered. If we are studying Euclidean geometry and prove the Pythagorean theorem, it would be a little weird to say "how do we know that Euclid's model of geometry is a model for _real_ triangles?" We don't, but that's not the point. The point is always "in this model, this is the result." As for the controversy thing, yes, zero controversy. All of relativity can be mathematically derived from the speed of light being constant in all reference frames. So, if one accepts the speed of light being constant, then they necessarily accept the relativity of simultaneity. Is the speed of light _definitely_ constant among all reference frames? No, we don't _know_ that our descriptions are actual rules of the physical universe, we can only say "based on the countless experiments that show this, it sure seems like it." The only logical framework difference between physics and mathematics is that in mathematics, we pick whatever axioms we want because we think they might lead to something cool. In physics, their axioms are chosen based on observation and experimentation. The observation is "the speed of light is constant", special relativity is the deduction from that axiom.
@pendragon7600
4 жыл бұрын
@@EpicMathTime I frequently get myself into discussions like this, where the only "disagreement" is caused by my poor communication. I understand that assuming relativity, the string would break. I'm just not convinced this actually reflects reality.. it's just so counterintuitive that it's more believable (to me) that physicists made a mistake somewhere along the line. This is nothing new I have the same problem with superposition.
@pendragon7600
3 жыл бұрын
@silverrahul Yeah, right, you're missing the point mate. Being counterintuitive is sometimes a symptom of being wrong. I'm not saying it's wrong because it's counterintuitive, rather that I'll need something more than this video to convince me of it.
@stevenutter3614
3 жыл бұрын
Wow minutephysics does such a better job, and in less than 4 minutes.
@user-scienceislove
3 жыл бұрын
relativity is psedu science
@fullfungo
3 жыл бұрын
Prove it
@aashsyed1277
3 жыл бұрын
no
@DrDeuteron
Жыл бұрын
except for the fact that it describes reality, and without it GPS would fail, no particle accelerators would work, communication equipment with exact frequencies and timing would desynchronize as the earth rotates/revolves, police radar would be direction/time dependent....talk about reasonable doubt you're honor (in the US they have to prove you are guilty, you don't need to prove your innocence), heck, you could probably set up trading arbitrage based on the variable propagation speeds, and Maxwell's equation would be inconsistent, which would break quantum electrodynamics...the theory that made the most accurate prediction in the history of Mankind, g-2 for the electron, which is good to 10 (TEN!) digits. Moreover, our interplanetary spacecraft navigation would fail (see post-Newtonian corrections). You got a big hill to climb to prove your point.
@oscarobioha595
4 жыл бұрын
If u have an email, we can have correspondence so that I can shoe u my derivation
@ZombieTubular
4 жыл бұрын
Dude, why the music so loud? Also, why music at all. You want to be heard, right? Cut out the music that competes with you speaking.
@chillagma
3 жыл бұрын
no,it's amazing and it makes it more engaging.
@oscarobioha595
4 жыл бұрын
Well accelerating of a particle as seen by all reference frame. For u to answer this question u might have to use GR. But I think ur wrong
@oscarobioha595
4 жыл бұрын
Bruhhh. Ur WRONG. acceleration is absolute in all reference frame. All ur explanations is based on uniform velocity. It's very easy so derive the absolutism of acceleration in all RF
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