How Does Relativistic Force Impact Acceleration Direction?

[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.

 

[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.

 

[M98Ranger]

Your approach is correct. The x-component of the force, eEo, is only present on the left side of the equation, so it must also be present on the right side. This means that the coefficient of the x-component on the right side must be zero in order for the equation to balance. This results in the equation you derived for ax, which does indeed show that there is a non-zero x-component of acceleration even though the x-component of force is zero.

As for the sign of ax, your reasoning is also correct. The x-component of velocity, ux, decreases as the y-component, uy, increases, so the x-component of acceleration must be negative in order to maintain the overall direction of motion. And as c goes to infinity, the relativistic effects become negligible and the equation reduces to the classical case where ax goes to zero.

Overall, it's important to understand that in special relativity, force and acceleration are not necessarily in the same direction, and this equation shows how they are related at relativistic speeds. Keep in mind that this is just one example and the equation can take on different forms depending on the specific problem at hand.