In theoretical physics, is the Graviton related to the fundamentals behind the $F = ma$ relationship? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-23T18:29:16Z https://physics.stackexchange.com/feeds/question/431198 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/431198 -1 In theoretical physics, is the Graviton related to the fundamentals behind the $F = ma$ relationship? M Ferguson https://physics.stackexchange.com/users/192309 2018-09-27T22:08:20Z 2018-09-28T05:22:05Z <p>If the Graviton transmits the gravitational force to nuclei within atoms, and an object with mass gains energy by being accelerated (change velocity / change time), does that imply that <span class="math-container">$F = ma$</span> is caused by the interaction between Gravitons and nuclei of the object?</p> <p>I've for a long time been in disdain over the concept of potential energy, and it would seem like the answer to this question would be yes if you only considered the <span class="math-container">$y$</span>-axis considerations. If you lifted an object 10 meters, we would say it had gained some amount of potential energy. It seems like the reason behind that potential energy is the underlying fundamentals of what causes 'F = ma', where the gravitational field simply decelerates the object as it is lifted.</p> <p>One problem with this however, is that 'F = ma' also applies to horizontal movement not affected by potential energy. So Gravitons would need to also affect nuclei even if they are accelerated across an unvarying gravitational field. I'm not sure how to justify this inconsistency which is why I'm curious. Any answers are appreciated.</p> https://physics.stackexchange.com/questions/431198/-/431222#431222 1 Answer by Avantgarde for In theoretical physics, is the Graviton related to the fundamentals behind the $F = ma$ relationship? Avantgarde https://physics.stackexchange.com/users/133418 2018-09-28T01:15:18Z 2018-09-28T01:15:18Z <p>No, but you can relate Newton's second law of motion <span class="math-container">$(1)$</span> to Newton's law of gravity <span class="math-container">$(2)$</span>. Instead of gravitons, just think of a gravitational field <span class="math-container">$\phi$</span>.</p> <p><span class="math-container">$(1)$</span> <span class="math-container">$F = m_i a$</span> tells us an object's acceleration when a force is applied. <span class="math-container">$m_i$</span> is the inertial mass. It is the characteristic of a system to resist acceleration.</p> <p><span class="math-container">$(2)$</span> <span class="math-container">$F = - m_g\nabla \phi$</span> tells us the motion of the object under a gravitational potential <span class="math-container">$\phi$</span>. <span class="math-container">$m_g$</span> is the gravitational mass. It is the characteristic of the system that tells it how to react in a gravitational field.</p> <p>A priori, <span class="math-container">$m_i$</span> and <span class="math-container">$m_g$</span> signify different characteristics. But as was noticed by Newton, Galileo and others, objects of different masses/compositions fall at the same rate of acceleration under a gravitational field. It was therefore postulated that <span class="math-container">$m_i = m_g$</span>. This has been experimentally tested to very high accuracy.</p> <p>Assuming the equality of <span class="math-container">$m_i = m_g = m$</span>, <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> can be combined such that we get <span class="math-container">$m(a + \nabla \phi) = 0$</span>. In other words, in the freely falling frame, an object does not experience any 'force'.</p> <p>This motivated the formulation of the weak equivalence principle, and further generalizations of the <a href="https://en.wikipedia.org/wiki/Equivalence_principle" rel="nofollow noreferrer">equivalence principle</a>, and lead to, of course, general relativity.</p>