3

Does special relativity predict the possibility of particle creation? Is it due to the relativistic energy-momentum dispersion $$E^{2}=m^{2}c^{4}+p^{2}c^{2}$$ which shows that mass is simply a form of energy, and as such other forms of energy can be converted into mass energy, or equivalently, mass energy can be converted into other forms of energy, implying the possibilities of matter creation and annihilation respectively?

Given this, would it be correct to say that Newtonian mechanics cannot account for matter creation/annihilation since there is a clear distinction between mass and energy in this theory, and hence there is no way to describe such a conversion that would be required to create/annihilate matter?

Feyn_example
  • 229
  • 1
    Well the prediction of antimatter is linked directly to the discovery of the Dirac equation which was an attempt to find a relativistic version of the Schrodinger equation (by taking the `square root' of the Klein Gordon equation), so in that way antimatter is a prediction of relativistic quantum mechanics. – gautampk May 16 '17 at 21:55
  • @gautampk But, before taking quantum mechanics into consideration, if one treats particles classically is special relativity enough to predict (or correctly describe) particle creation? In several introductory notes that I've read on SR the author discusses relativistic kinematics and dynamics and the notion of threshold energies for particle creation without even mentioning quantum mechanics. – Feyn_example May 16 '17 at 22:00
  • 1
    It is my understanding that you can't have any notion of 'where particles come from' without quantum mechanics. Of course, you can posit that there exist particles and then do kinematic calculations, which is what you do in classical mechanics, but there's no concept of the dynamics of the particles -- where they come from, why they have the properties they do etc. Threshold energies and particle creation are definitely quantum mechanical concepts. That doesn't mean there weren't older non-quantum attempts at explanation, but they aren't really of use anymore. – gautampk May 16 '17 at 22:12
  • @gautampk Fair enough. Is it not possible to draw any conclusions on the subject from the relativistic energy-dispersion relation? I mean, doesn't it imply that mass is not necessarily conserved - it can be converted into other forms of energy (what is conserved is the total energy). This is clearly different from Newtonian mechanics where mass is always conserved (it can neither be created or destroyed in this sense). – Feyn_example May 16 '17 at 22:26
  • Relativistic dispersion relation isn't really a statement about energy conservation so much as it is a statement about frame invariance. $E^2 - p^2 = m^2$ (natural units so $c=1$), is really saying that $m$ is the magnitude of the energy-momentum four vector. As a magnitude it is a scalar, and therefore must be invariant under a change of basis. Conservation is a property of a specific theory (or its Lagrangian), but the dispersion relation is a statement about the entire framework of relativity. Of course, if a theory is Poincare invariant then both $E$ and $p$ are conserved so $m$ is also. – gautampk May 16 '17 at 22:53
  • @gautampk Can one not draw any conclusions from the relativistic dispersion relation then? It seems to at least imply that mass is simply a particular form of energy, and we know that energy can be converted from one form to another, hence kinetic energy could be converted into mass and vice versa (for example). Just to clarify, I'm thinking of system of two (or more) particles here, so if I've understood correctly, the total energy and momentum of the system is conserved, as is the total mass, but that doesn't mean that the mass of an individual particle has to be conserved. – Feyn_example May 17 '17 at 08:25

1 Answers1

3

It is the difference between necessary and sufficient. Special Relativity is necessary in order to have the generation of particles from energy, the four vector algebra defines invariant masses for systems of particles which is impossible in Newtonian mechanics, where there is conservation of mass.

It is not a sufficient theory for describing nuclear and particle interactions. Quantum mechanics and quantum number conservation are necessary for building any coherent theoretical model of particle generation and interactions. At the moment the quantum mechanical standard model of particle physics which incorporates special relativity also, is sufficient to describe and predict the overwhelming majority of particle interactions. Nuclear physics has shell models and liquid drop models to phenomenologically fit the data and have a predictive ability.

anna v
  • 233,453
  • 1
    So would it be correct to say that special relativity allows for particle creation/annihilation to occur since mass, which is simply a form of energy, can be converted into other forms of energy, and so mass is not necessarily conserved. However, special relativity cannot explain the mechanism(s) by which such creations/annihilations can occur, it just predicts that in principle they can occur. Newtonian mechanics, on the other hand, doesn't permit particle creation/annihilation since in Newtonian theory, mass is a distinct concept from energy and can never be created/destroyed. – Feyn_example May 17 '17 at 18:26
  • Yes, it is correct. – anna v May 17 '17 at 18:55
  • Ok great, thanks for your help. Is there a way to show that mass isn't conserved in special relativity or does it boil down to special relativity permitting the non-conservation of mass, but not being able to explain the exact mechanisms? – Feyn_example May 17 '17 at 19:07
  • Also, is it correct to say that whilst the individual masses of particles in a given system may not be conserved, the invariant mass of the total system $m_{total}^{2}=E^{2}{total}-p^{2}{total$ is conserved? (For example, if one considers the interaction $e^{+}e^{-}\rightarrow\gamma\gamma$, then the individual masses of the electron and positron are not conserved, however the total invariant mass of the system is, $m^{2}{before}=E^{2}{before}-p^{2}{before}=E^{2}{after}-p^{2}{after}=m^{2}{after}$) – Feyn_example May 17 '17 at 19:36
  • Yes, the invariant mass replaces the concept of mass. The elementary particles have a unique characteristic invariant mass and the invariant mass of complex particles is built up by the addition of the four vectors. Invariant mass is the "length" of the four vectors. – anna v May 18 '17 at 03:51
  • What I'm slightly confused about is that in the example $e^{+}e^{-}\rightarrow\gamma\gamma$, $m^{2}{before}=4m{e}^{2}$ whereas $m_{after}^{2}=0$, so clearly $m_{before}^{2}\neq m^{2}{after}$. Is the point that $(E{+}+E_{-})^{2}-(m_{e}+m_{e})^{2}\neq 4m_{e}^{2}$ ($E_{\pm}$ are the respective energies of the incoming electron and positron, with $\mathbf{p}{\pm}$ their momenta), however, $(E{+}+E_{-})^{2}-(\mathbf{p}{+}+\mathbf{p}{-})^{2}=(E_{\gamma, ,1}+E_{\gamma, ,2})^{2}-(\mathbf{p}{\gamma, ,1}+\mathbf{p}{\gamma, 2})^{2}$? – Feyn_example May 18 '17 at 09:40
  • .... Thus, in general, the invariant mass of the system as a whole is conserved, i.e. $\left(\sum_{i}E_{i}\right)^{2}-\left(\sum_{i}\mathbf{p}{i}\right)^{2}=\left(\sum{f}E_{f}\right)^{2}-\left(\sum_{f}\mathbf{p}{f}\right)^{2}$, because energy and momentum are individually conserved in collisions, however, the rest masses of the individual particles constituting the system aren't necessarily conserved, since $\left(\sum{i}E_{i}\right)^{2}-\left(\sum_{i}\mathbf{p}{i}\right)^{2}\neq\left(\sum{i}m_{i}\right)^{2}$. – Feyn_example May 18 '17 at 09:58
  • when e+annihilates on e- at rest, as you are assuming in your comment the invariant mass of the gamma pair produced is the "length",sqrt(four dot) of the summed four vector of the two gammas. This will be the same as the invariant mass of the two e+e- before. At LEP e+ and e- have large energies and annihilate to give the Z particle which decays to a lot of others particles. The total invariant mass before and after is the same, it comes from conservation of energy and momentum. The sum is a vector sum.. – anna v May 18 '17 at 10:24
  • Is the invariant mass of a system of particles defined as $\left(\sum_{i}E_{i}\right)^{2}-\left(\sum_{i}\mathbf{p}{i}\right)^{2}$? Is it the correct to say that this is in general not equal to the squared sum of the rest masses of the constituent particles, i.e. it's not equal to $\left(\sum{i}m_{i}\right)^{2}$, and so the individual rest masses of the constituent particles are not conserved in general. The invariant mass of the system is conserved, however, since energy and momentum are individually conserved. – Feyn_example May 18 '17 at 11:04
  • Yes, that is correct . http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html – anna v May 18 '17 at 11:56