Nuclear reactions and Einstein's famous formula


by JanEnClaesen
Tags: einstein, famous, formula, nuclear, reactions
JanEnClaesen
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#1
Nov23-13, 08:17 AM
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In high school, we calculated the reaction-energy by equating the mass difference and the energy difference with proportionality factor cē.
How does Einstein's theory of relativity suddenly enter nuclear physics? It startles me because both fields developped historically completely divergent. Aren't there nuclear physical ways to calculate this energy? In retrospect, the calculations seem too simple to be true.
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dauto
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#2
Nov23-13, 08:52 AM
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The theory of relativity is a universal theory. It applies to all of physics, including nuclear physics.

Sometimes calculations are both simple and true.
JanEnClaesen
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#3
Nov23-13, 09:21 AM
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So is quantum mechanics...

dauto
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#4
Nov23-13, 09:37 AM
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Nuclear reactions and Einstein's famous formula


Quantum mechanics is... (Half an answer to half a question)
JanEnClaesen
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#5
Nov23-13, 09:44 AM
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My knowledge is too incomplete to ask full questions. But as far as I know, quantum mechanics relates the same variables but is not consistent with general relativity and hence might provide a different answer.
dauto
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#6
Nov23-13, 09:49 AM
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General relativity is a theory of gravity which plays no role in nuclear physics. Relativity on the other hand is a universal meta-theory that applies to all physics including nuclear physics.
JanEnClaesen
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#7
Nov23-13, 10:16 AM
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I see, special relativity isn't a special case of general relativity.
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#8
Nov23-13, 10:33 AM
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Quote Quote by JanEnClaesen View Post
I see, special relativity isn't a special case of general relativity.
No, it's more that:

1) Special relativity applies in the special case of flat space-time, whereas general relativity applies in the general case of flat or curved spacetime.
2) Although special relativity is therefore a special case of general relativity, there are a large number of problems, including just about everything at and below atomic scale, for which we don't use general relativity. That's not because it's not applicable (it is the more general theory, after all), it's because the general relativistic effects are so insignificant that we can safely ignore them.
3) SR calculations are so much simpler than GR calculations that no one ever uses GR except when it's needed.
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#9
Nov23-13, 12:56 PM
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Quote Quote by JanEnClaesen View Post
I see, special relativity isn't a special case of general relativity.
Special relativity is a special case of general relativity since we can obtain the former by applying the latter to a flat space case. Both of those theories are instances of relativistic theories. There are other conceivable ways to apply the concepts of relativity to gravity and obtain alternate relativistic theories of gravity other than general relativity. In that sense General relativity is a special implementation of relativity to gravity. None of that matters here though because 1) gravity plays no role in nuclear physics. 2) there really isn't any obvious inconsistencies between GR and quantum mechanics. What we do have is incomplete understanding of how to apply quantum mechanics to general relativity, which is a different thing. The only situation where there appears to be a real contradiction between GR and QM is the B-hole information paradox, but that paradox - like most physical paradoxes - Is like just an apparent paradox, not a real paradox.
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Nov23-13, 01:29 PM
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Quote Quote by JanEnClaesen View Post
In high school, we calculated the reaction-energy by equating the mass difference and the energy difference with proportionality factor cē.
How does Einstein's theory of relativity suddenly enter nuclear physics? It startles me because both fields developped historically completely divergent. Aren't there nuclear physical ways to calculate this energy? In retrospect, the calculations seem too simple to be true.
There are nuclear physical ways to calculate the energy difference. For instance, in nuclear fission, the energy difference is mainly (other nuclear effects enter, but are smaller) due to the difference in the Coulomb energies of the parent nucleus and its daughters.
E=mc^2 holds for any process, it is just large enough in nuclear processes to be seen.
When hydrogen burns, the mass of H_2O is less than the mass of H_2 and O, but this difference is too small to be seen.
JanEnClaesen
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#11
Nov23-13, 05:59 PM
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Is the binding energy a measure of the internal motion in the nucleus?
Drakkith
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Nov23-13, 08:25 PM
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Quote Quote by JanEnClaesen View Post
Is the binding energy a measure of the internal motion in the nucleus?
It's a result of the interaction of the strong force and the EM force on the particles in the nucleus. If you take Hydrogen and start adding the right mix of nucleons to it you will increase the binding energy per nucleon thanks to the strong force. However, once you hit iron and nickel, adding further nucleons results in a higher binding energy per nucleon because the size of the nucleus starts to become greater than the range of the strong force. You have protons that are repelling each other through their positive charges, yet aren't attracting each other through the strong force anymore because they are too far away from each other. This is why very heavy elements are unstable. The protons start to feel massive repulsion from the combined force of every other proton in the nucleus, but they only feel the strong force from a few of the nucleons that are nearby.


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