# Transformation of Force

1. Feb 11, 2010

### N-Gin

I'm trying to find relativistic transformations of force which has two components $$F_{x}$$ and $$F_{y}$$ and velocities $$v_{x}$$ and $$v_{y}$$. 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)

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

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

When I derive everything, I get equations including velocities $$v_{x}$$, $$v_{y}$$ and accelerations $$a_{x}$$ and $$a_{y}$$.

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

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

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

Then I use Lorentz transformations to find $$a_{x}\rightarrow a_{x}'$$, $$a_{y}\rightarrow a_{y}'$$ and $$v_{x}\rightarrow v_{x}'$$, $$v_{y}\rightarrow v_{y}'$$.

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

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.

2. Feb 11, 2010

### jason12345

Yes, this is right. You should end up with Fx = Fx' and Fy = gamma Fy'.

3. Feb 11, 2010

### clem

If you try your way you have to also transform d/dt.
That will lead to a mess.
F is not a four vector. To transform F use the 'Minkowski force', which is a four vector.
The Minkowski force is $${\cal F}=[\gamma {\bf v\cdot F};\gamma{\bf F}]$$.
The result is not simple unless F is in the rest system (v=0).

Last edited: Feb 11, 2010
4. Feb 12, 2010

### N-Gin

There's a (small) mistake in my first post.

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

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

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. Feb 12, 2010

### clem

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. Feb 12, 2010

### N-Gin

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. Feb 12, 2010

### jason12345

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.