Formula for the acceleration felt by an accelerating object?

[granpa]

what is the formula for the acceleration felt by an accelerating object? I've spent days googling and can't get a straight answer. some say its the derivative of proper velocity with respect to proper time but others say its the derivative of proper velocity with respect to coordinate time.

its obviously the instantaneous acceleration of an object as perceived by an second object traveling at contant velocity equal to the velocity of the first object at that instantaneous moment. I've tried drawing a spacetime diagram and solving equations of lines and doing coordinate transformations but my formulas are a mess of 1/v's and vdeltav's and garbage that doesn't simplify to alpha=a*gamma^3 that I read about.

 

[granpa]

http://en.wikipedia.org/wiki/Four-acceleration
In special relativity, four-acceleration is a four-vector and is defined as the change in four-velocity over the particle's proper time:
http://upload.wikimedia.org/math/b/6/d/b6de33bcaad9039d28577b89b0a60ae3.png
where
http://upload.wikimedia.org/math/f/d/d/fdd106b836057e1ca3479a6ffb79596d.png

and γ_u is the Lorentz factor for the speed u. It should be noted that a dot above a variable indicates a derivative with respect to the time in a given reference frame, not the proper time τ.
In an instantaneously co-moving inertial reference frame u = 0, γ_u = 1 and \dot\γ_u = 0, i.e. in such a reference frame
http://upload.wikimedia.org/math/7/5/1/7515c263351fe55c99643cb76a9e3acd.png
Therefore, the four-acceleration is equal to the proper acceleration that a moving particle "feels" moving along a world line.http://en.wikipedia.org/wiki/Four-vector
the four-velocity of an \mathbf{x}(\tau) world line is defined by:
http://upload.wikimedia.org/math/b/6/b/b6b9bfa43887f036a42076eaa3355fdd.png
http://upload.wikimedia.org/math/0/5/e/05e21017a2762709240b89ee4314505f.png

 

[Fredrik]

what is the formula for the acceleration felt by an accelerating object? I've spent days googling and can't get a straight answer. some say its the derivative of proper velocity with respect to proper time but others say its the derivative of proper velocity with respect to coordinate time.

its obviously the instantaneous acceleration of an object as perceived by an second object traveling at contant velocity equal to the velocity of the first object at that instantaneous moment. I've tried drawing a spacetime diagram and solving equations of lines and doing coordinate transformations but my formulas are a mess of 1/v's and vdeltav's and garbage that doesn't simplify to alpha=a*gamma^3 that I read about.

You're right about the definition. The formulas get easier if you use the "rapidity" $\phi$, defined by $v=\tanh \phi$. Example:

$$a=\frac{dv}{dt}=\frac{d\tau}{dt}\frac{d}{d\tau}\tanh\phi=\frac 1{\gamma^3}\frac{d\phi}{d\tau}=\frac\alpha{\gamma^3}$$

 

[granpa]

thanks.

ok. I did a calculation. not for acceleration but for relative velocity. if a stationary observer measures the coordinate velocity of object A to be v and the coordinate velocity of object B to be v+dv then A will measure the velocity of B to be dv*gamma^2.

so the derivative of that with respect to proper time should be a*gamma^3. which is what a read earlier. now I have to look up where I read that.

 

[granpa]

ok. here's what I did:

coordinate velocity=v=v(t)=t
coordinate acceleration=a=dv/dt=1
gamma=g=(1-v^2)^-1/2
dg/dt=tg^3
proper velocity=b=v*g

db/dt=d(vg)/dt=vdg/dt+gdv/dt=vtg^3+g

proper acceleration=a*gamma^3=gamma^3

a*g^3=vtg^3+g
g^3=g(vtg^2+1)
g^2=vtg^2+1
vt=(g^2-1)/g^2

this appears to be correct. whew.using db/dtau one gets
vt=(g-1)/g^2
which as obviously wrong.

 

[granpa]

if that's the case then I don't understand this quote:
' The proper acceleration 3-vector, combined with a null time-component, yields the object's four-acceleration ' (http://en.wikipedia.org/wiki/Proper_acceleration). below that proper acceleration is clearly and unambiguously defined as rate of change of proper velocity with respect to coordinate time.

yet, In special relativity, four-acceleration is defined as the change in four-velocity over the particle's proper time.

the three spacelike components of 4-velocity define a traveling object's proper velocity \gamma \vec{u} = d\vec{x}/d\tau i.e. the rate at which distance is covered in the reference map-frame per unit proper time elapsed on clocks traveling with the object.

nor do I understand this:
A=(0,a)
Therefore, the four-acceleration is equal to the proper acceleration that a moving particle "feels" moving along a world line.
(http://en.wikipedia.org/wiki/Four-acceleration)

 

[granpa]

heres a simpler solution.

go to http://calc101.com/webMathematica/derivatives.jsp#topdoit
enter 'v[t]/sqrt[1-v[t]^2]' into the 'take the derivative of' box.
enter 't' into the 'with respect to' box
enter 'y' into the 'and again with respect to' box. (doesnt do anything but you must enter something)
push the 'do it' button

the result is a*gamma^3