How Does Force Work in Special Relativity?

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Discussion Overview

The discussion revolves around the concept of force in the context of special relativity, particularly addressing the validity of applying Newtonian mechanics and the implications of inertial versus non-inertial reference frames. Participants explore the relationship between force, acceleration, and momentum within special relativity and the potential need for general relativity in certain scenarios.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the applicability of force in non-inertial frames, suggesting that the principle of relativity may not hold, and thus the formula F=ma may not be valid.
  • Another participant clarifies that while some methods of special relativity are valid only in inertial frames, special relativity can still address accelerated frames using different mathematical approaches.
  • It is noted that in an inertial frame, one can measure velocities and discuss conservation of momentum, which allows for the discussion of forces as rates of change of momentum.
  • A participant presents the correct formulation of force in special relativity as F=m(d(γv)/dt), emphasizing the role of relativistic momentum.
  • One participant seeks clarification on whether small changes in momentum can be discussed in special relativity, while larger changes would necessitate general relativity.
  • Another participant refutes the idea that special relativity cannot handle accelerating objects, asserting that it can treat them without invoking general relativity.
  • A technical note is provided regarding the proper use of LaTeX formatting for mathematical expressions in the discussion.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of force in non-inertial frames and the necessity of general relativity for certain scenarios. The discussion remains unresolved regarding the extent to which special relativity can handle accelerating frames and objects.

Contextual Notes

Some participants highlight misconceptions about the limitations of special relativity concerning accelerating frames, indicating that further clarification on the mathematical treatment of these scenarios may be necessary.

Angelos
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Hi...

Please can anyone explain me why we can talk about force in special relativity? I thought that special relativity is valid only in inertial reference frames and the force has to product acceleration. And so the reference frame is not inertial and the principle of relativity doesn't have to be valid. And so the formula F=ma doesn't have to be true in this refernce frame. What's wrong with this thinking? My understanding of inertial frames? Thank you very much for explanation.
 
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Note that we don't have problems talking about force in Newtonian kinematics, which also has a notion of a mechanical principle of relativity and a notion of inertial frames.

In the modern presentations, special relativity is physics in a non-curved ("flat") spacetime on R^4. In this view, "special relativity is valid only in inertial reference frames" is not strictly correct. Rather, "some methods of special relativity are valid only in inertial reference frames"... for example, one can only perform a global Lorentz boost to transform from one inertial frame to another inertial frame.

Accelerated frames of reference are akin to using curvilinear coordinates (e.g. polar coordinates) in Euclidean space. Its use may require calculus, in addition to algebra and trigonometry which are often sufficient for many discussions with inertial frames. (Calculation of curvature in these non-rectangular coordinates will still yield zero curvature.) Admittedly, some of these calculus-based methods were developed with General Relativity (where spacetime is not a flat R^4)... but were later seen to be useful in Special Relativity.
 
In an inertial frame you can still measure velocities, observe collisions, and talk about conservation of momentum. From there, it isn't difficult to discuss forces (rates of change of momentum). If two bodies interact with some force, the usual idea would be to describe this from a third reference frame (which isn't interacting or accelerating) hence, what difficulty is there?
 
Last edited:
Angelos said:
Hi...

And so the formula F=ma doesn't have to be true in this refernce frame.

The correct formula in BOTH Newtonian and SR dynamics is F=\frac {dp}{dt}In SR , the above becomes:

F=m \frac {d (\gamma v)}{dt}

since
p=m \gamma v

and

\gamma=\frac {1}{\sqrt(1-v^2/c^2)}
 
So did I understand correctly that we can talk in special relativity about small change in momentum as well as we can talk about classic mechanics only in small velocities? In larger change in momentum we would have to use general relativity right?
 
No. You might try re-reading robphy's response, or perhaps try

http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

FAQ said:
It is a common misconception that Special Relativity cannot handle accelerating objects or accelerating reference frames. It is claimed that general relativity is required because special relativity only applies to inertial frames. This is not true. Special relativity treats accelerating frames differently from inertial frames but can still deal with them. Accelerating objects can be dealt with without even calling upon accelerating frames.

(emphasis mine)

One can deal with accelerating objects without using accelerating frames. This is as true in classical mechanics as it is with relativity.

There are other points to be made, but in the interest of keeping things simple I'll only make this one - that one can deal with accelerating objects without using accelerating frames.
 
nakurusil said:
\gamma=\frac {1}{\sqrt(1-v^2/c^2)}

nakurusil, just a note about LaTeX (since this appears to be a repeating pattern with your posts here), you need to group stuff together with { .. } for an operator like \sqrt to apply to the entire contents.

such as
\gamma=\frac {1}{ \sqrt{1-v^2/c^2} }
 

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