Insights An Introduction to the Generation of Mass from Energy

[lightarrow]

Thank you, vanheese71.
Unfortunately I am able to understand just a part of your very interesting post. I think the "centre of energy" is the key concept here, anyway I can't understand exactly where is the mistake in the computation of the system's (invariant) mass in my two examples.
Regards

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[vanhees71]

There is no error. What you did was using the usual "kinematics" of collisions/decays, i.e., the energy-momentum conservation for the transition from the asymptotic free initial to the asymptotic free final state. What you then called "mass" is in fact the total energy of the system as measured in the center-of-momentum frame. For collisions that's "Mandelstam s", defined by $s=(p_1+p_2)^2$ (where $p_1$ and $p_2$ are then four-momenta of the two incoming particles). For a decay you have only the one unstable particle in the initial state. So there indeed $s=M^2 c^2$, where $M$ is the invariant mass of the unstable particle.

What I wanted to stress is that there's no additional conservation law for invariant mass as in Newtonian physics, and how this can be understood from very fundamental symmetry considerations.

 

[neilparker62]

Sorry, but can't understand what you mean with "generation of mass from energy". I thought mass is conserved in SR (yes I'm serious and I'm speaking of invariant mass).

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The specific example you bought up in your post has been discussed in some detail by others much better qualified than myself so I won't go into that. But I'll just try and explain what I had in mind.

Very simplistically we had the following equation in the article:

$${\left(\frac{m_f}{m_i}\right)}^2 = 1 + \frac{2(\gamma-1)m_1m_2c^2}{(m_1+m_2)}\times\frac{1}{(m_1+m_2)c^2}=1+\frac{2(\gamma-1) \mu c^2}{(m_1+m_2)c^2}=1+\frac{2T_{\mu}}{m_ic^2}$$ Hence:
$$\frac{m_f}{m_i} = \sqrt{1+\frac{2T_{\mu}}{m_ic^2}}$$ For a very rough approximation we can take a one tem binomial expansion of the square root expression: $$\frac{m_f}{m_i} \approx 1+\frac{T_{\mu}}{m_ic^2}\implies m_f \approx m_i+\frac{T_{\mu}}{c^2}$$ The term $\frac{T_{\mu}}{c^2}$ is essentially the 'system's' kinetic energy expressed in units of mass. In a typical 'collide and coalesce' collision in the macro world, this energy is dissipated as heat/sound/deformation etc but here it is not. It results (for example) in the production of kaons (ie new 'massive' particles) as per worked example.

 

[lightarrow]

The specific example you bought up in your post has been discussed in some detail by others much better qualified than myself so I won't go into that. But I'll just try and explain what I had in mind.

Very simplistically we had the following equation in the article:

$${\left(\frac{m_f}{m_i}\right)}^2 = 1 + \frac{2(\gamma-1)m_1m_2c^2}{(m_1+m_2)}\times\frac{1}{(m_1+m_2)c^2}=1+\frac{2(\gamma-1) \mu c^2}{(m_1+m_2)c^2}=1+\frac{2T_{\mu}}{m_ic^2}$$ Hence:
$$\frac{m_f}{m_i} = \sqrt{1+\frac{2T_{\mu}}{m_ic^2}}$$ For a very rough approximation we can take a one tem binomial expansion of the square root expression: $$\frac{m_f}{m_i} \approx 1+\frac{T_{\mu}}{m_ic^2}\implies m_f \approx m_i+\frac{T_{\mu}}{c^2}$$ The term $\frac{T_{\mu}}{c^2}$ is essentially the 'system's' kinetic energy expressed in units of mass. In a typical 'collide and coalesce' collision in the macro world, this energy is dissipated as heat/sound/deformation etc but here it is not. It results (for example) in the production of kaons (ie new 'massive' particles) as per worked example.

In your article (and the previous one) you begin with a particle 1 moving at $v_1$ against a particle 2 initially at rest and all the computations are (implicitly) made in the rest frame of particle 2. Later you write:
"Denoting $m_3$ as $m_f$ – the final post collision mass – and the sum $(m_1+m_2)$ as the pre-collision mass [which you then call it "$m_i$"], we obtain"...
and you write an equation involving the relative velocity v of the two particles.
1) Which is the reference frame here? Is the same as the previous article, that is the particle 2 rest frame? The phrase "relative velocity" could suggest it's not.
2) You call $(m_1+m_2)$ "pre-collision mass" but I can't see which is the physical system which have this mass.

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[neilparker62]

In your article (and the previous one) you begin with a particle 1 moving at $v_1$ against a particle 2 initially at rest and all the computations are (implicitly) made in the rest frame of particle 2. Later you write:
"Denoting $m_3$ as $m_f$ – the final post collision mass – and the sum $(m_1+m_2)$ as the pre-collision mass [which you then call it "$m_i$"], we obtain"...
and you write an equation involving the relative velocity v of the two particles.
1) Which is the reference frame here? Is the same as the previous article, that is the particle 2 rest frame? The phrase "relative velocity" could suggest it's not.
2) You call $(m_1+m_2)$ "pre-collision mass" but I can't see which is the physical system which have this mass.

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The start point for this article is the following diagram from the given reference

Equation 33 which I have 're-worked' follows from the above so yes the rest frame is where particle 2 is at rest. "Relative" velocity in this case is just the velocity of the other particle. But the technique I have used in my classical articles does not - in principle - require either particle to be stationary since it is based on their velocity relative to each other. Collision impulse is calculated according to the formula $\Delta P=\mu \Delta v$ for completely inelastic collisions and $\Delta P=2\mu \Delta v$ for elastic collisions. $\mu$ is the reduced mass of the colliding objects and $\Delta v$ is their relative velocity.

In this article I sought to find a similar approach for a relativistic collision - hence the formula referred to in my previous post (which brings in the ubiquitous 'reduced mass' parameter). I think there was one example I found somewhere and solved where the two colliding particles were both moving in some reference frame. Perhaps I need to dig it up and include it as part of this article.

For point 2 I would refer you to the worked example where the pre-collision mass is that of the two colliding protons and the post-collision mass is that of the two protons plus the 'generated' kaons.

 

[neilparker62]

Have added another worked example which (I hope) puts the formula through its paces!

 

[lightarrow]

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For point 2 I would refer you to the worked example where the pre-collision mass is that of the two colliding protons and the post-collision mass is that of the two protons plus the 'generated' kaons.

Sorry, I don't have seen your example yet, anyway, concerning the transformation from a ref. frame where both particle 1 and particle 2 have non zero velocity, I find, instead of equation (33) of "Introduction to Relativistic Collisions":

${E_3^0}^2={E_1^0}^2+{E_2^0}^2+2E_1^0 E_2^0+2T_1E_2^0+2T_1T_2+2T_2E_1^0-2c^2\vec{p_1}·\vec{p_2}$ (*)

Then we change frame on K', co-moving with particle 2; $\vec{v_1}'$ is now the relativistic relative velocity between particle 1 and particle 2 and:
$T_2'=\vec{p_2}'=0$
so (*) becomes equal to (33).
I apologize if you have already described this in your last example.

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[lightarrow]

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Only a last comment on the title of this insight, Neil (by the way, from your nickname it seems we have the same age): if energy is always conserved, how can it "generate" something else ("mass")?

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[neilparker62]

Energy is neither created nor destroyed - simply changed from one form to another. I think that would apply to just about any form of energy 'generation'. In a nuclear power plant we generate energy from mass, here we generate mass from energy.

 

[lightarrow]

Energy is neither created nor destroyed - simply changed from one form to another. I think that would apply to just about any form of energy 'generation'. In a nuclear power plant we generate energy from mass, here we generate mass from energy.

Energy cannot be "generated", only transformed or converted. Your examp
Anyway, just my opinion, of course.
Edit: I want to make another example: let's say a particle, in a ref. frame where it's still, could disintegrate completely into photons, in a spherically symmetric pulse of light. Which is the mass of this light pulse? It's not zero, it's the same m as before. But I'm sure a lot of untrained people would say it's zero.
Regards.

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[lightarrow]

Energy cannot be "generated", only transformed or converted. Your examp

Something went wrong here. The phrase should have been:
"Your example shows, in my opinion, that kinetic energy can be converted into rest energy or the reverse".

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[neilparker62]

Sorry, I don't have seen your example yet, anyway, concerning the transformation from a ref. frame where both particle 1 and particle 2 have non zero velocity, I find, instead of equation (33) of "Introduction to Relativistic Collisions":

${E_3^0}^2={E_1^0}^2+{E_2^0}^2+2E_1^0 E_2^0+2T_1E_2^0+2T_1T_2+2T_2E_1^0-2c^2\vec{p_1}·\vec{p_2}$ (*)

Then we change frame on K', co-moving with particle 2; $\vec{v_1}'$ is now the relativistic relative velocity between particle 1 and particle 2 and:
$T_2'=\vec{p_2}'=0$
so (*) becomes equal to (33).
I apologize if you have already described this in your last example.

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lightarrow

Not explicitly no. But I did indicate somewhere in the article that in the Lorentz factor $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ the 'v' was to be taken as a relative velocity in which case $T_2=0$.