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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.

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StarBucK
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  • 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 $x \rightarrow x+a$. – G.Clavier Jan 19 '17 at 17:23
  • 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 – StarBucK Jan 19 '17 at 17:29
  • 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. ;) – G.Clavier Jan 19 '17 at 17:35
  • 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. – StarBucK Jan 19 '17 at 17:45
  • 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. – G.Clavier Jan 19 '17 at 17:50
  • I agree that my particle is not at rest in $R_{ref}$ (my accelerated frame) if it is at rest in $R_0$ and I have to introduce the "virtual" force $F=-m a$.

    But I don't understand the link with all this. Landau says : "An inertial frame is a frame in which space is homegenous and isotropic, and time is homogenous". Here, in my accelerated frame $R'$, if I change $x_{ref} \rightarrow x_{ref}+x_0$, I will have the same equation of motion, so space seems to be homogenous.

    What is precisely wrong in my comment ? Sorry if I misunderstood your answer.

     – StarBucK Jan 19 '17 at 18:02
  • Related: http://physics.stackexchange.com/q/23098/2451 and links therein. – Qmechanic Jan 19 '17 at 19:02
  • Actually as you have to introduce fictious forces that do not depend on the interaction of you particles with another physical element (via mass or charge), the whole space is still homogeneous but not isotropic anymore. – G.Clavier Jan 19 '17 at 20:43

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