Relativistic Force Transformation: A Comprehensive Guide to Special Relativity

In summary, the conversation discusses the relativistic transformations of force, specifically in the context of special relativity. The participants mention using second Newton's law in two reference frames, S and S', and using Lorentz transformations to find the transformed force components. The correct approach is to use the Minkowski force, which is a four vector, and transform the components using the Lorentz transformations. The conversation also suggests using a textbook, such as "Special Relativity" by A.P French, for further understanding.
  • #1
N-Gin
56
0
I'm trying to find relativistic transformations of force which has two components [tex]F_{x}[/tex] and [tex]F_{y}[/tex] and velocities [tex]v_{x}[/tex] and [tex]v_{y}[/tex]. I'm not sure if I have right idea so I would be grateful if someone could check it out.

First of all, we have second Newton's law in special relativity (in this case for reference frame S)

[tex]
F_{x}=\frac{d}{dt}\left(\frac{m_{0}v_{x}}{\sqrt{1-\frac{v_{x}^2+v_{y}^2}{c^2}}}\right)
[/tex]

[tex]
F_{y}=\frac{d}{dt}\left(\frac{m_{0}v_{y}}{\sqrt{1-\frac{v_{x}^2+v_{y}^2}{c^2}}}\right)
[/tex]

When I derive everything, I get equations including velocities [tex]v_{x}[/tex], [tex]v_{y}[/tex] and accelerations [tex]a_{x}[/tex] and [tex]a_{y}[/tex].

Then I use second Newton's law in other reference frame, S' moving with velocity [tex]V[/tex] relative to frame S.

[tex]
F_{x}'=\frac{d}{dt}\left(\frac{m_{0}v_{x}'}{\sqrt{1-\frac{v_{x}'^2+v_{y}'^2}{c^2}}}\right)
[/tex]

[tex]
F_{y}'=\frac{d}{dt}\left(\frac{m_{0}v_{y}'}{\sqrt{1-\frac{v_{x}'^2+v_{y}'^2}{c^2}}}\right)
[/tex]

Then I use Lorentz transformations to find [tex]a_{x}\rightarrow a_{x}'[/tex], [tex]a_{y}\rightarrow a_{y}'[/tex] and [tex]v_{x}\rightarrow v_{x}'[/tex], [tex]v_{y}\rightarrow v_{y}'[/tex].

When I derive formulas for [tex]F_{x}'[/tex] and [tex]F_{y}'[/tex] and plug [tex]a_{x}'[/tex],[tex]a_{y}'[/tex], [tex]v_{x}'[/tex] and [tex]v_{y}'[/tex] I should get [tex]F_{x}\rightarrow F_{x}'[/tex] and [tex]F_{y}\rightarrow F_{y}'[/tex].

I'm not sure if this is the right way to do it because I don't have any literature concerning this. I would like to hear your opinion.

Sorry for eventually bad English and thanks in advance!
 
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  • #2
Yes, this is right. You should end up with Fx = Fx' and Fy = gamma Fy'.
 
  • #3
If you try your way you have to also transform d/dt.
That will lead to a mess.
The right answer is not as simple as your answer or Jason's.
F is not a four vector. To transform F use the 'Minkowski force', which is a four vector.
The Minkowski force is [tex]{\cal F}=[\gamma {\bf v\cdot F};\gamma{\bf F}][/tex].
The result is not simple unless F is in the rest system (v=0).
 
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  • #4
There's a (small) mistake in my first post.

[tex]
F_{x}'=\frac{d}{dt'}\left(\frac{m_{0}v_{x}'}{\sqrt{ 1-\frac{v_{x}'^2+v_{y}'^2}{c^2}}}\right)
[/tex]

[tex]
F_{y}'=\frac{d}{dt'}\left(\frac{m_{0}v_{y}'}{\sqrt{ 1-\frac{v_{x}'^2+v_{y}'^2}{c^2}}}\right)
[/tex]

I'm not so familiar with Minkowski force. I know all of this is complicated but it really doesn't matter. I just want to know if my thinking is right.

I will try to solve this my way and see what it looks like. I really don't see anything wrong.
 
  • #5
I tried to say it politely, but your thinking is completely wrong.
You can't transform the force by just transforming the coordinates.
You have to put the components of a 4-vector into the Lorentz transformation.
Don't even try until you are so familiar with the Minkowski force.
You need a textbook. What is your level?
 
  • #6
clem said:
What is your level?

I'm a first year college. We are studying special relativity in general physics. Now, I know how to transform a force acting only in x or y direction. Also, reference frame S' is moving at constant velocity pointing towards positive direction of x (relative to S). Since everything we learned about special relativity are Lorentz transformations, mass and energy, I hoped that this would be enough to find force transformations. I guess I took this thing too serious...
 
  • #7
N-Gin said:
I'm a first year college. We are studying special relativity in general physics. Now, I know how to transform a force acting only in x or y direction. Also, reference frame S' is moving at constant velocity pointing towards positive direction of x (relative to S). Since everything we learned about special relativity are Lorentz transformations, mass and energy, I hoped that this would be enough to find force transformations. I guess I took this thing too serious...

Your approach is sound, although there are more elegant ways of doing it via the transformation properties of the four force from the four momentum which you will come across later on.

Using your approach: Rather than trying to work out how force transforms between S and S' generally, it's easier to work out how it transforms from the proper frame of the particle experiencing the force, S' say, to any other frame S. You'll end up with Fx' = Fx, Fy' = gamma Fy.

If you get stuck, page 211-214 in Special Relativity by A.P French gives an identical working.

A Google book search on "force in relativistic mechanics" which will give a link to the above reference.
 

1. What is the purpose of studying relativistic force transformation?

The purpose of studying relativistic force transformation is to understand how forces behave in the context of special relativity, which is a fundamental theory of physics that describes the behavior of objects moving at high speeds. By understanding how forces transform in different frames of reference, we can gain a deeper understanding of the principles of relativity and their applications in various fields of science and technology.

2. What is the difference between relativistic and classical force transformation?

The main difference between relativistic and classical force transformation lies in their underlying principles. Classical force transformation is based on Newton's laws of motion, which assume that time and space are absolute and do not change. Relativistic force transformation, on the other hand, takes into account the effects of time dilation and length contraction, which are essential components of the theory of special relativity.

3. How does relativistic force transformation affect the concept of mass?

In special relativity, mass is no longer considered a constant quantity but is instead a dynamic quantity that depends on the speed of an object. This means that as an object approaches the speed of light, its mass increases, and it becomes more difficult to accelerate. Relativistic force transformation takes this into account and allows us to calculate the force needed to accelerate an object at relativistic speeds.

4. Can relativistic force transformation be applied to both macroscopic and microscopic systems?

Yes, relativistic force transformation can be applied to both macroscopic and microscopic systems. While the effects of special relativity may not be noticeable in everyday objects, they become significant at high speeds. At the microscopic level, relativistic force transformation is essential in understanding the behavior of particles in particle accelerators, such as the Large Hadron Collider.

5. Are there any practical applications of relativistic force transformation?

Yes, there are many practical applications of relativistic force transformation in various fields such as astrophysics, particle physics, and engineering. For example, it is crucial in understanding the behavior of objects in high-speed collisions, designing spacecraft and satellites, and calculating the effects of time dilation in GPS systems. Relativistic force transformation is also essential in the development of technologies that utilize particle accelerators, such as medical imaging and cancer treatment.

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