# Landau's 4-Accelaration Problem

I was going through Lev and Landau's 2nd volume on classical field theory. I came across a doubt. It's stated in the solution that we go to the frame having particle's velocity zero, which basically means frame is attached to the particle. If v is set 0 then how is the equation derived by differentiating the 4-velocity? We sure can directly assume that the 4-accelaration is what it is given in the solution but then what's the need of providing that v=0?

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– user191954
Oct 5, 2018 at 18:55
• Okay sure. Next time I'll take care. Can you provide the answer? Oct 5, 2018 at 19:17

What you are looking for is the concept of an inertial reference frame in which the particle is instantaneously at rest, usually called the frame instantaneously at rest (duh). It is an inertial reference frame which measures the 4-velocity of the particle to have only the time component. When the particle is not being acted upon by a force, the frame instantaneously at rest does not change, since the rest frame of the particle is itself an inertial frame; but when the particle is accelerated, at each instant we will have a different inertial reference frame that will fulfill this condition. This is why the description of the problem in Landau adds the parentheses "(at each instant of time)"; because there is no unique inertial reference frame in which the 4-acceleration has only spatial components at all times.

So the 4-acceleration has the component form given only instantaneously, in any given inertial reference frame. What Landau did there was to write the requirement that the motion be uniformly accelerated in an invariant form -- and this is done by requiring that $$w_iw^i = -\dfrac{w^2}{c^4} =$$ constant. So the need of providing $$v=0$$ is justified by the fact that the components which the 4-acceleration has change according to the frame; the only inertial frame (or rather a class of frames) in which the components are purely spatial is the one instantaneously at rest.

• No problem, I'm glad I could help :) Oct 5, 2018 at 22:48

The way they write $$w^i = (0, w/c^2,0,0)$$ just like that is a little deceptive, it's a result only valid for $$v=0$$.

You can calculate the four-acceleration explicitly by differentiating (forgive me setting c=1):

$$w^0 = \frac{du^0}{ds} = \gamma\frac{d}{dt}[\gamma] = \gamma\frac{\partial}{\partial v^i}\left[\frac{1}{\sqrt{1-v^2}}\right] a^i = \gamma^4 v_i a^i$$

$$w^1 = \frac{du^1}{ds} = \gamma\frac{d}{dt}[\gamma v^1] = \gamma\frac{\partial}{\partial v^i}\left[\frac{v^1}{\sqrt{1-v^2}}\right] a^i = \gamma a^i \left( \gamma \delta^1_i + \gamma^3 v^1 v_i \right) = \gamma^2 a^1 + \gamma^4 v^1 a^i v_i$$

etc.

If the velocity is 0 this reduces to what's stated in the solution, but you're right in that its not trivial, and generally you can't assume the spatial components of the four-acceleration are equal to the normal acceleration.

• Thanks! Your maths and the other answer's theoretical explanation helped! :) Oct 5, 2018 at 21:14