Relative force

  • Thread starter Yayness
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  • #1
Yayness
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One reference frame A uses a force F on another reference frame B.
Both A and B counts the time it takes before A stops using the force F on B. They end up with different results.

Will B feel a different value of the force coming from A?
Will F*t be different in the two reference frames? (Where t is the time it takes before A stops using the force.)
Will B use the same force on A as the force A uses on B?
 

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  • #2
tiny-tim
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Hi Yayness! :smile:

Tell us what you think, and then we'll comment. :wink:
 
  • #3
Dale
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One reference frame A uses a force F on another reference frame B.
Reference frames are coordinate systems. They are mathematical constructs, not physical objects. How do you apply a force to a mathematical object like a coordinate system?
 
  • #4
Yayness
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I am quite new to relativity, but I have read about it and tried to derive different formulas. (So far, I have derived formulas for length contradiction, time dilation and velocity addition for instance.)

I read about relativity on the internet and I saw this formula:
[tex]a=\frac{F}{m}(1-\frac{v^2}{c^2})^{\frac{3}{2}}[/tex]
Where a is the acceleration of an object, observed by an inertial frame. F is the force used on the object (by the inertial frame, I believe), m is the rest mass of the object, and v is the velocity relative to the inertial frame.

I tried to derive this formula, and so far, I have managed to derive that
[tex]a=(1-\frac{v^2}{c^2}) a_0[/tex]
where a0 is the acceleration felt by the accelerating object, and a is the acceleration observed by the inertial frame.

Relative to itself, the object is at rest, so the force it feels does only depend on its rest mass and the acceleration it feels. If we let F0 be the force it feels, then
F0=a0m

If I put that in the formula I derived, I get that:
[tex]a=\frac{F_0}{m}(1-\frac{v^2}{c^2})[/tex]
So if the formula I am trying to derive is correct, it means that
[tex]F_0=\sqrt{1-\frac{v^2}{c^2}} F[/tex]
where F is the force the inertial frame uses on the object, and F0 is the force the object feels.
Is this correct, and if so, why?
 
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  • #5
Yayness
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Reference frames are coordinate systems. They are mathematical constructs, not physical objects. How do you apply a force to a mathematical object like a coordinate system?
I mean two objects where one of them is at rest in one reference frame, and the other one is at rest in another reference frame.
If one object uses a force on the other one, will the value of this force be different in the two reference frames?
 
  • #6
jason12345
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One reference frame A uses a force F on another reference frame B.
Both A and B counts the time it takes before A stops using the force F on B. They end up with different results.

Will B feel a different value of the force coming from A?
Will F*t be different in the two reference frames? (Where t is the time it takes before A stops using the force.)
Will B use the same force on A as the force A uses on B?

These are all good questions that highlight the problem of what exactly is "force" in relativistic dynamics.

Have a think about what proper force is.
 

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