Is there no energy conservation?

[nevo]

When we look at a particle's rest frame
its energy is Mc^2.
But the particle decay has some width, so the products of the particle,generally, will have total energy different than Mc^2.
How can it be, is there no energy conservation?

 

[ZapperZ]

When we look at a particle's rest frame
its energy is Mc^2.
But the particle decay has some width, so the products of the particle,generally, will have total energy different than Mc^2.
How can it be, is there no energy conservation?

Did you ever look at the kinetic energy of the decay products?

Zz.

 

[Hootenanny]

There certainly is conservation of energy, the two decay particles may have some non-zero velocity in the original particle's rest frame. Energy and momentum are always conserved (in a closed system).

Edit: Dammit Zz.

 

[nevo]

At the beginning the two decay product did not exist.
For instance when a rho particle decay into to pions,
its energy in its rest frame is Mc^2.
But the decay products, the two pions,generally will have different energy than the original rho.
This is, as I understand, the meaning that the decay has a width.

 

[Hootenanny]

The decay width is another name for the decay rate of a species. However, to answer what I believe your question to be; due to the HUP any particle with a finite lifetime has a non-zero mass distribution (has some uncertainty in the mass), this will result in a non-zero mass distribution for the products.

 

[nevo]

What do you mean by some uncertainty in the mass, is this like uncertainty
of the momentum?
Is the wave function of the particle is a superposition of eigen vectores with different mass eigen value?
Isn't the mass of the particle is the exact value that appear in the hamiltonian?

 

[ZapperZ]

What do you mean by some uncertainty in the mass, is this like uncertainty
of the momentum?
Is the wave function of the particle is a superposition of eigen vectores with different mass eigen value?
Isn't the mass of the particle is the exact value that appear in the hamiltonian?

You do know that in these experiments, they don't measure just one decay, don't you? They measure a gazillion to get the statistics. So what do you think the "mass width" here means?

Zz.

 

[nevo]

First of all, I asked about the "mass width" because I don't understand this
expression.
Second, the fact that "They measure a gazillion to get the statistics" is gust a technical mean to learn about the physics, it is still meaningful to ask about one particle.
When I check about the muon mass in wikipedia or in any other place I always find gust one number, I never saw the width of its mass only of its decay, but even if there is such a thing as "mass width" it must influence the hamiltonian and also the one particle wave function.