I am reading the Landau & Lifshitz on mechanics to understand how we find the free particle Lagrangian, and there are some things that I don't understand.

First, he defines an inertial frame as the following. He says that an inertial frame is a frame in which space is homegenous and isotropic, and time is homogenous.

What does that exactly mean?

I will only take the example of homogenous space for the next part of my question.

I have read on various topic on this website that homogenity of space means if I change $q \rightarrow q+q_0$ in the Lagrangian, then it will be unchanged.

Im ok with this definition but at this moment the book didn't talk about this at all.

So I guess that they want to say that the equation of motion will be unchanged if the position of the particle is changed from $q$ to $q+a$ in the frame (I talk about the differential equation of motion, and not the solution $q(t)$ that will change because of different initial condition).

First question : Am I right with this definition?

Ok, then imagine that I have a particle at rest (so, it doesn't feel any force). I take an accelerated frame in the $x$ direction at constant acceleration.

The equation of motion in this frame of the particle will be : $\ddot{x}=a$

If I use the definition of an inertial frame given by Landau, I would find that this frame is an inertial frame : if I change $x \rightarrow x+x_0$, the equation of motion will be unchanged. And I can do the same for time : $ t \rightarrow t+t_0$, and for isotropy : I have the homogeneity everywhere so it is necesseraly isotropic.

Where is my mistake? Because of course an accelerated frame is not inertial. What did I misunderstood in Landau's Book?

[edit]: I just read from the first answer from here Mechanics Landau Galilean Principle what is the definition of homogenous space, time and isotropic space.

But even with this definition I don't see why an accelerated frame will not be inertial according to Landau definition of an inertial frame.

  • $\begingroup$ If $q$ is a position, then $a$ is a relative distance in the formula $q+a$. It is synonym of a translational symetry. You changed your position regarding the particle. If you write $\ddot{x}=a$ then $a$ is an acceleration! You can't write a transformation $x \rightarrow x+a$. $\endgroup$ – G.Clavier Jan 19 '17 at 17:23
  • $\begingroup$ I just choosed bad notation. Of course the $a$ of $x+a$ is not the same than the one of my acceleration. I edited my question $\endgroup$ – StarBucK Jan 19 '17 at 17:29
  • $\begingroup$ Then in your new referential, the transformation is no more $x \rightarrow x+a$. You have to integrate the equation of motion of your referential. You will find that the coordinates transformation is such as $x \rightarrow x(t) = \frac{1}{2}at^2+a(0)t+x(0)$. Well this is indeed not a simple translation. Welcome to the marvelous world of relativity. You can only pass from one inertial referential to another if one has a rectilign and uniform motion regarding the other (i.e $\dot{x}=a$). I let you do the maths to convince yourself. ;) $\endgroup$ – G.Clavier Jan 19 '17 at 17:35
  • $\begingroup$ Hmmm I'm not sure I understood well. I call $x_{ref}$ the position in the accelerated frame. If I change $x_{ref}(t) \rightarrow \tilde{x}_{ref}(t)+x_0$, then I will still have $ \tilde{\ddot{x}}_{ref}(t)=a $. The equation of motion is then exactly the same. For me you just wrote the solution of the equation of motion but I don't understand where you did the transformation. $\endgroup$ – StarBucK Jan 19 '17 at 17:45
  • $\begingroup$ From one referential to the other, the equation of motion of your particle has actually changed and Newton's law are not applicable anymore. Take a particle at position $x=0$ without any speed and no force acting on it. In the case of constant speed referential, $x'=at$, $\dot{x}'=a$ the apparent movement is linear and uniform. In your case, $\dot{x'}=at+a(0)$, $\ddot{x'}=a$. Your particle is gaining kinetic energy in your new referential so you have to introduce a force. But this force has no physical meaning except to illustrate the acceleration of your referential. $\endgroup$ – G.Clavier Jan 19 '17 at 17:50

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