From my understanding, we can do math in an accelerating frame of reference as long as "fictitious"

Sam Hardin 2022-07-14 Answered
From my understanding, we can do math in an accelerating frame of reference as long as "fictitious" force terms are correctly added. From this point of view, is there anything wrong with viewing the Earth as stationary, and the rest of the universe rotating around it, at least kinematically? And, if so, wouldn't several cosmological objects move faster than light in this frame of reference? How can this be?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

lywiau63
Answered 2022-07-15 Author has 13 answers
You're right - but so too are the photons. They're moving "faster than light", too, in this frame!
You see, talking of "light speed limits" is really just pseudo-Newtonian mechanical heuristics on top of the actual gist of relativity theory, which is that information can transmit between some places in space-time and not others. These relations, or predicates on events, are what must not be violated.
However the predicates that tell you "yes or no" to "Can you send a message from A to B?" take different forms depending on the coordinate system. For "flat" spacetime, in a certain special class of coordinate systems, they show up as a "speed limit", but are more subtle in others. This is why the theory here is called "special" relativity (SR).
All the "weird" effects - time dilation, length contraction, and so forth can be thought of as what a Universe with minimum imposed communication latencies looks like from the viewpoint of someone inside it subject to those constraints.
Not exactly what you’re looking for?
Ask My Question
Willow Pratt
Answered 2022-07-16 Author has 5 answers
Take a simple example: You are standing on a conveyor and a friend is running behind you to try to reach you, 10m away. He is walking 1m/s and it will take him 10 seconds to reach you.
The conveyor may be moving at 2m/s and that would not change the situation. but you could work out what happens using the ground as a reference. Your speed would be 2m/s your friends' speed 3m/s and the outcome would be the same.
So no, there is nothing wrong with a geocentric reference frame if you calculate these kinds of problems. in fact we do it all the time. Your car doesn't care about the angular speed of the milky way spirals, it just shows you km/s with respect to the ground.
But that is cutting short the actual problem with a geocentric world model, where gravity would be hard to explain.
And no, things would not move at above light speed, when you use a geocentric frame of reference, it's just that you would observe things in a different time scale than from other reference frames, because speed is dilating time. What is important to keep in mind concerning reference frames is that at non-relativistic speeds, we can use them interchangeably, but actually, this is "wrong" in the sense that any relative motion is relativistic.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-05-07
"If the frame of reference is translated or rotated, the vector doesn't change."
Although the length of the vector won't change, the angle that this vector makes with the positive direction of the newly defined x-axis changes, no? Hence, how is it possible to state that the vector doesn't change? Have I misunderstood the basic definition of a vector?
asked 2022-07-14
A puck is placed on a friction less disk that is rotating with angular velocity ω . What is the equation of motion with respect to the rotating frame of the puck?
The contradiction that I can not resolve is that it seems really obvious that the puck will rotate with angular velocity ω in the frame of the disk. But if look at the puck from the frame of reference of the disk, we can see that there is a pseudo centrifugal force acting on the puck. This is the only force that acts on the puck in this frame.
Hence, the puck should accelerate radially outward in this frame...
asked 2022-07-16
Let’s assume we have 2 different observers. Observer 1 sits in space and observer 2 sits in a space lab which is in a free fall state toward the Earth. We further assume that observer 2 in the space lab does not have any information regarding its surrounding.
1. Is there anyway that observer 2 can figure out whether any frame attached to the space lab can be regarded as an inertial or non-inertial frame?
2. Lets assume there is a 1kg ball inside the space lab which is also free falling together with the space lab? Observer 1 (with its true inertial frame) can measure the acceleration of the ball and deduce that 9.8Newton force is acting on the ball. On the other hand, observer 2 measures the acceleration of the ball as 0, and hence, deduces that the net force acting on the ball is 0. As the observer 2 has no way of knowing that space lab is not an inertial frame, he will not have any doubt about his force measurement. Does this mean that definition of force is dependent on the frame of reference?
asked 2022-07-22
Orbital velocity of a circular planet is a R , where a is the centripetal acceleration, and R is radius of the planet. With v 1 as the tangential velocity of the rotating planet at the equator.
On the non rotating body, suppose that the orbital velocity is v 0 , and, for an object launched on the rotating body's "equator", that the orbital velocity will be in the form of v 1 + v 2 (the body and the object both going counterclockwise). Now, I half-hypothesized v 0 = v 1 + v 2 = a R and v 2 = a R where a is the "true" rotating body's acceleration, able to be calculated from the rotating frame of reference as
a = a v 1 2 R
The logic was that from rotating body's reference frame, the object would be traveling at v 2 , less than v 0 because of the centrifugal force, so v 2 has to be the orbital velocity if the gravity was "weakened" by centrifugal force.
Tried to solve for a and comparing it to the value, got from rotating frame of reference, ending up with
a = a ( v 1 ( v 0 + v 2 ) R )
Something's not right, and if I had to choose, I would guess the v 0 = v 1 + v 2 , that acceleration is not the same on those two planets, but I don't know how it would change, or why.
asked 2022-05-10
An inertial frame of reference is a frame of reference which is not accelerating. All laws of physics are the same measured from an inertial frame of reference. A rest frame is a frame of reference where a particle is at rest.
Does this mean that a rest frame could possibly be non-inertial (that is, accelerating), but the particle with respect to his rest frame would have a velocity of 0? What kind of velocity? And what exactly would it mean to be at rest with respect to a possibly accelerating frame of reference?
What are the differences and relations between rest frame and inertial reference frame?
asked 2022-05-09
Formula for the Bekenstein bound
S 2 π k R E c
where E is the total mass-energy. That seems to imply that the presence of a black hole in the region is dependent on an observer's frame of reference. Yet, my understanding is that the Bekenstein bound is the maximum entropy that any area can withstand before collapsing into a black hole.
Does this mean that the existence of black holes is observer dependent? Or that even if an observer does not report a black hole in their frame, one is guaranteed to form there in the future?
asked 2022-04-26
Can you identify the direction of your movement? Let's say ship moving nose forward, or tail forward? Is there a sense of "direction of movement" in such an inertial frame of reference?

New questions