# Force & acceleration equation

1. Aug 1, 2005

### mmwave

While working on a problem in special relativity I ran into the following equation and am trying to understand it before going on. It was accompanied by the vague statement that at relativisitic speeds, force & acceleration aren't necessarily in the same direction.

$$\vec{F} = m_{o}\gamma \vec{a} + m_{o} \vec{u} \gamma^{3}(\vec{a} \cdot \vec{u}) / c^{2}$$

where
$$\gamma \equiv \frac{1}{\sqrt{1 - u^2/c^2}}$$

To understand this, I apply it to the case where
$$\vec{F} = eE_{o} \hat{y}$$ and $$u(t=0) = u_{o} \hat{x}$$ and $$u_{x}(t=0)$$ is nearly equal to c.

Now, looking at the x and y components, since the left side has only y component, so must the right side. Therefore the coefficients of $$\hat{x}$$ must be zero which gives me the following equation:

$$a_{x} + \gamma^{2}u_{x}a_{x}/c^{2} + \gamma^{2}u_{y}a_{y}/c^{2} = 0$$ with solution $$a_{x} = -\gamma^{2}u_{y}a_{y}/c^{2}* \frac {1} { (1+\gamma^{2}u_{x}/c^{2})}$$

question: Is this the right approach? am I right that the x component must vanish? Notice that for the relativistic case, even though the x component F is zero, there is still an x component to the acceleration.

I like the sign of ax because it should be negative (the x component of u must decrease as the y component increases since the speed must be less than c) and as c goes to infinity (classical case), ax goes to zero. To do this, there must be a negative x component of the acceleration.

2. Aug 8, 2005

### mmwave

When I carried on with this method, I got a mess! substituting my expression for ax into the y component gave me a nonlinear differential equation involving both ux and uy and their derivatives. This is way to advanced for the book so I must be using the wrong approach.