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Edit: I have already tried to delete this several times. I have already posted that I would delete this if I could and it was deleted. I have three very legitimate masters degrees in physics and more than 15 years in the field. This was a silly what if question and not meant to be argued as seriously as it was. That’s why I said it was silly in the title.

Edit 2: half written document because someone asked me about this post today. His questions had very little to do with it so my replies had very little to do with it. Here is the link. Tangentially relevant document I am not willing to completely erase the history of this post here. That’s a troll-trap.

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What if, in your own reference frame, you see your acceleration as increasing over time? That should increase the velocity at which you perceive others moving relative to you. Since at least in principle, your own mass determines how your velocity plateaus as it approaches c from the perspective of the external observer, does that mean that the mass of the external world determines how their velocity plateaus from your perspective, and as a result, the velocity with which you assume you are moving? So— velocities may not be perceived reciprocally in the presence of acceleration??? Could that explain something like dark energy or dark matter????!!!!

Sort of a silly thought experiment, but wondering if anything I said made sense. I want you to know I’m very sleep deprived, haven’t run this by anyone else, and haven’t done any calculations. If true this would be amazing. 95% chance I am having exhausted ridiculous thoughts and I would appreciate my error being pointed out— though my question being heard with an open mind.

If observer 1 sees their velocity change from v1 to v1’ because of their own mass m1, they are really perceiving v2’s velocity and inferring their own. Likewise v2 goes from v2 to v2 ‘ as perceived by observer 2 due to its acceleration and mass m2. But since it’s perceived v2 really reflects how it sees the velocity of observer 1 changing and from that observer two infers their own… then does that mean changes in v1 are really due to m2 in a sense and changes in v2 are really due to m1 in a sense. In other words does the underlying reality reflect that the velocity of each observer as they infer about themselves depends on the mass of the other. What then should we infer about our own mass/velocity in comparison to the universe?

Hope I said this clearly.

Steven

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  • $\begingroup$ Outer reference frames are not accelerating. You have pseudo forces acting on you, outer frames - do not. $\endgroup$ Commented Apr 14 at 7:40
  • $\begingroup$ I don't care. Physics Stack Exchange has neither been kind nor helpful. I would delete if I could. $\endgroup$ Commented Apr 16 at 22:10
  • $\begingroup$ PSE is very helpful as it is seen from the high stream of questions registered here. Just try to relax, calm down. Do not push anything from your brains "by force". When you will be ready,- reformulate your question once more. $\endgroup$ Commented Apr 17 at 7:46
  • $\begingroup$ Thank you. This is scary. $\endgroup$ Commented Apr 18 at 14:56

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I think that the main problem with your argument is that you are treating an accelerated frame and an inertial frame as equivalent, while they are not.

Paraphrasing, you state that if a force is applied to an object, its acceleration (and therefore its velocity at any given time) depends on the mass of the object. This is true also in Newtonian physics ($F=ma$)

In the frame of the accelerated observer, they see the rest of the world whizzing past them. And then you wish to reverse the statement and infer that the velocity of the rest of the world depends on the mass of it. In some way this sounds like applying the principle that all inertial frames are physically equivalent.

Except that the accelerated frame is not inertial. In non inertial frames $F=ma$ is not valid anymore and there are fictitious forces, like the one that seems to be accelerating the rest of the world.

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  • $\begingroup$ I would like to clarify a bit though. The velocities as perceived by the two frames at any given time are symmetric but the v(t) and v(t’) are not interchangeable by swapping t and t’ because t is not equal to t’. So that means a(t) and a(t’) are also not equal, even at a specific time, because dt is not equal to dt’. $\endgroup$ Commented Apr 14 at 16:47

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