# 4-velocity -> 4-acceleration?

1. Oct 8, 2006

### JabberWalkie

4-velocity --> 4-acceleration?

Hey all, my first post here.

Now, Im trying to derive the 4-acceleration from the 4-velocity by explicitly calculating du/d(tau). So from Hartle we have

u=( gamma, gamma*v)

where gamma=1/Sqrt(1-v^2)

and v is the 3 velocity, taking c=1

So then,

du/d(tau)=( d(gamma)/d(tau), d(gamma*v)/d(tau) )

from the book we have

du/d(tau)=(gamma*(a dot v), gamma*a)

so then we should have

d(gamma)/d(tau)=gamma*(a dot v)

but when i calculate d(gamma)/d(tau) i get....ok im going to show all my steps so people can know what ive done....

d(gamma)/d(tau)=[d(gamma)/dv)]*[dv/dt]*[dt/d(tau)]

=[d(1/(1-v^2)^(1/2))/dv]*[a][gamma]

=[(-1/2)(-2*(v)/(1-v^2)^(3/2)]*[a]*[gamma]

=[ va/(1-v^2)^(3/2)*[1/gamma]

=v*a*gamma^4 != v*a*gamma

does anyone know whats going wrong or what im doing wrong??

Any help would be appreciated, Thanks in advance.

2. Oct 9, 2006

### pervect

Staff Emeritus
You should expect that "proper acceleration", which can be defined as the magnitude of your 4-acceleration, is related to coordinate acceleration by a factor of gamma^3, not by a factor of gamma.

See for instance http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity [Broken])

(which has this result, but doesn't work through the math, unfortunately).

But if we work through your example, we get for the two components of the 4-acceleration

v*a*gamma^4, a*gamma^4

Note that this makes the 4-acceleration vector Minknowski-perpendicular to the 4-velocity, this is another known result.

Now when we take the squared magnitude of the above we get

(1-v^2)*a^2*gamma^8 = a^2*gamma^6

so the square root of this is a * gamma^3, i.e. the magnitude of the 4-acceleration is gamma^3 times the magnitude of the coordinate acceleration, which is what we expect.

Last edited by a moderator: May 2, 2017