Distribution of Energy: Deriving T from m, V, v0, & θ

[abccdef125]

Hi,
i´m reading Landau´s mechanics and in the 4th chapter in the part disintegration of particles (16) i can´t figure out the following thing. We have a function T=(1/2)m(V²)+(1/2)m(v₀²) + mVv₀cos(θ) , where V and v₀ are constant in this problem. We also know the distribution of θ which is (1/2)sinθdθ. The goal is to get a distribution of T. Could you please explain to me how to get it? The book derives it in a way i can´t understand. Thanks in advance.

 

[Simon Bridge]

Welcome to PF;
The trick is to understand what the distribution means - i.e. what would the "distribution of T" tell you, once you have it? What is the "distribution of $\small \theta$" telling you?

 

[abccdef125]

Thanks, i went through it once more with your advice in mind and i finally understand it.

 

[Simon Bridge]

Well done :)

The next step is to find someone who is struggling and see if you can get them to understand.

 

[PJC01]

Hello there,

The equation you provided for T is the total kinetic energy of a particle, which is the sum of its translational kinetic energy (1/2)m(V^2), its initial kinetic energy (1/2)m(v0^2), and the work done on the particle by a force F (mVv0cosθ). This equation is derived from the work-energy theorem, which states that the work done on a particle by a force is equal to the change in its kinetic energy.

To derive the distribution of T, we first need to understand the distribution of θ. In this case, it is given by (1/2)sinθdθ, which represents the probability of a particle having an angle θ within a small range of θ and is normalized to 1. This distribution is often referred to as the "solid angle" because it represents the area subtended by a cone with an angle θ.

To get the distribution of T, we need to integrate the total kinetic energy equation over all possible values of θ. This is because the distribution of θ represents all possible directions in which the particle can move, and we need to account for all of them.

Integrating (1/2)sinθdθ over all possible values of θ, we get 1, which means the distribution is normalized to 1. This makes sense because the probability of a particle having some angle θ must be equal to 1.

Next, we integrate the total kinetic energy equation over all possible values of θ. This gives us the distribution of T, which is a function of m, V, and v0. The resulting distribution is a probability distribution, which represents the probability of a particle having a certain kinetic energy T.

I hope this explanation helps you understand the derivation of the distribution of T in a more clear manner. If you have any further questions, please feel free to ask. Good luck with your studies!