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The Coulomb force constant $k_c$ has si-units of $Nm^2C^{-2}$. The gravitational force constant $G$ has si-units of $Nm^2kg^{-2}$. What does it mean that at very high energies these forces become equal? How can we compare different forces?

Qmechanic
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Raul Laasner
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1 Answers1

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It means that at lower energies the electromagnetic force dominates, and the gravitational effects are barely noticeable if not negligible. If you want to actually test gravity on a microscopic scale in a system that is also subject to electromagnetism (eg charged particles) you have to reach high enough energies for the two forces to be roughly equal in magnitude.

The weak force at high energies becomes equal to the EM force. Physically, why? Because at high energies ($mc^2$ where $m$ the mass of the weak bosons) you can actually create $W$ and $Z$ bosons to carry the interaction.

SuperCiocia
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  • Thanks for the reply. You say "equal in magnitude", but this is exactly what's confusing to me. Does it depend on the types of particles? E.g., I'd imagine the two forces to become equal between a proton and a neutron at lower energies than between a proton and an electron. The sentence that at lower energies EM dominates while at higher energies both are important is understood, but it's not very quantitative. – Raul Laasner Aug 15 '15 at 11:32
  • Let's take the example of the weak force. Quantitatively, you should include a factor proportional to $\frac{1}{q^2-m^2}$ where $q$ is the 4-momentum and $m$ the mass of the weak boson. I'm taking $c=1$. If the momentum-energy regime you are dealing with ($q$) is much lower than the mass of the boson, the factor is small and the interaction is small. As you increase $q$, the denominator decreases so the force becomes bigger. – SuperCiocia Aug 15 '15 at 18:17
  • Pardon my limited knowledge of quantum field theory, but I don't quite see the necessary connections. I understand how the force strengthens with particle energy, but not how it can in some sense become equal to some other force. My current understanding is that at high energies the forces attain similar magnitudes, i.e., they are all equally important in describing physical phenomena, but it doesn't make sense to compare them quantitatively. – Raul Laasner Aug 16 '15 at 09:31
  • Let me put this in very simple terms. Say we have two charged particles of arbitray types with arbitray distance between them. Are the gravitational and EM forces between them always equal at high enough energies? – Raul Laasner Aug 16 '15 at 09:33
  • I think you are thinking of the classical equations for EM and gravity, given by Coulomb and Newton. They are not valid in the quantum(-field theory) regime. A force is not necessarily associated with a particular equation, rather with a gauge field that mediates an interaction. In QFT terms, the EM "force" is the EM field $A_{\mu}$ that can either exist on its own (free photons) or couple to the fermions's wavefunctions (Quantum Electro-Dynamics, charged particles interactions). --> cont – SuperCiocia Aug 16 '15 at 14:21
  • This field term appears in the Lagrangian, and to obtain the equations of motion you apply perturbation theory. You get terms called 'propagators' for force carriers, which are essentially just the first level of approximation to the equation of motion. These propagators, like the term that I have shown in the previous comment, depend on the energy regime of the system. – SuperCiocia Aug 16 '15 at 14:23
  • To give you a physical understanding of what's going on: as you increase the energy, you can create particles out of nowhere such as electrons, positrons or even weak bosons. These participate to the various interactions (EM, weak etc.) and therefore affect its 'magntitude'.

    By the way, gravitation has not yet been successfully written as a QFT so everything that you might have read is likely to be hypothetical.

     – SuperCiocia Aug 16 '15 at 14:25
  • Indeed I was perhaps thinking in too classical terms. My understanding of the situation has now improved, but the "equality" part still bothers me. Would you be able to put the statement that, say, the EM and the weak force become equal at high energies into a mathematical equation? Or perhaps suggest an online resource or a book that deals with this? – Raul Laasner Aug 16 '15 at 19:01
  • The photon propagator is $\propto \frac{1}{q^2}$, while the weak boson propagator is $\propto \frac{1}{q^2-m^2}$. When $q>>m$, $q^2-m^2 \approx q^2$ so the two propagators are kinda equal. – SuperCiocia Aug 16 '15 at 20:57
  • I'm just trying to give you a very intuitive feel for this, I don't want to get lost in the maths of QFT – SuperCiocia Aug 16 '15 at 20:58
  • Ok, now it makes more or less sense to me. Let's call it a day. Thanks for the discussion. – Raul Laasner Aug 16 '15 at 21:50