Why Are the Space-Like Components of the 4-Acceleration Non-Zero in the IRF?

[Lorna]

Homework Statement

I read in my textbook that "in the instantanuous reference frame (IRF) $$[u^{'\gamma}]=(c,\overline{0})$$ imply that the components of the 4-acceleration in the IRF are $$[a^{'\gamma}]=(0,\overline{a'})$$"

Homework Equations

The Attempt at a Solution

I don't know why they got a' for the space-like components of the 4-acceleration. Isn't the accelaration = du/d$$\tau$$ if so then using $$[u^{'\gamma}=(c,\overline{0})$$ we get zero for both the time-like and space-like components of the 4-acceleration.

 

[George Jones]

Think back to first-year calculus, and suppose $f\left(x\right) = x^3$. Then, $f\left(2\right) = 8$, and $8$ is a constant, so

$$\frac{d}{dx}8 = 0,$$

but

$$\frac{df}{dx}\left(2) = 12 \ne 0.$$

Just because the 4-velocity has a particular value (which could have zero spatial part) in a particular frame at a particular instant doesn't mean that the 4-velocity is constant. If the 4-velocity is not constant, then the 4-acceleration doesn't have to be zero.

Throw a ball in the air. When the ball reaches its greatest height, its (spatial) velocity is zero, but its (spatial) acceleration still has magnitude $g$

 

[robphy]

In addition, note that the 4-acceleration of a particle is orthogonal to its 4-velocity. (Why?)

 

[Lorna]

That example was very helpful. thanks a lot.

Thanks robphy too.