Is the Norm of Four-Acceleration Always Equal to Proper Frame Acceleration?

[johnahn]

I saw that the norm of four acceleration is equal to the magnitude of proper frame's acceleration.

So, if the observer moves in x direction, following equation about norm of it's 4 acceleration is like that

-(d^2 t / dτ^2) + (d^2 x / dτ^2) = d^2 x / dt^2

In comoving frame(proper frame), proper time is equals to t, so the first term in the left side vanishes, and the remain term is equal to the right side.

I tried to proof this relationship generally... but it's very hard for me.

Now I confused whether the general proof exists, or not...

 

[Bill_K]

johnahn, Here's the general relationship. It gets messy! If the 4-velocity is v = (γv, γc) so v·v = c², then the 4-acceleration is

a ≡ dv/dτ = (γ²a + γ⁴(v·a)v/c², γ⁴(v·a)/c) where a is the 3-acceleration, a = dv/dt.

The norm of this is a·a = -γ⁶a·a + γ⁶/c²[(v·a)² - (v·v)(a·a)].

In the special case that v and a are parallel, it reduces to |a| = γ³|a|.

 

[Dale]

I saw that the norm of four acceleration is equal to the magnitude of proper frame's acceleration.
...
I tried to proof this relationship generally... but it's very hard for me.

Now I confused whether the general proof exists, or not...

The norm of the four-acceleration is invariant. So if you prove it in one frame then you have proved it in all frames.