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So you're right that (low-energy) gravitational physics is understood because you know how to write down the action. The question is whether the sum over paths

by zachf 5y ago

So you're right that (low-energy) gravitational physics is understood because you know how to write down the action. The question is whether the sum over paths in the path integral should also include a sum over metrics. That sum over metrics is equivalent to saying that gravitons exist (in the same way that the sum over electromagnetic potentials is equivalent to saying that photons exist). Now if you want to include gravitational effects and do it in a consistent way, you have to sum over metrics, meaning you have to have gravitons. That's because trying to treat gravity like it's classical but treating everything else like it's quantum mechanical is inconsistent. For example, classical gravity could tell you which slit an individual electron passed through in a double slit experiment, if you measured the gravitational field accurately enough--you could say definitively that it came through one slit or the other by measuring which way the gravitational field that it generated is pointing. This would destroy the interference pattern and you wouldn't be able to conduct the double-slit experiment at all.

wyager 5y ago
Thank you for explaining! > the path integral should also include a sum over metrics Understood, thanks. > That sum over metrics is equivalent to saying that gravitons exist (in the same way that the sum over electromagnetic potentials is equivalent to saying that photons exist). Could you explain this? I'm not making the connection where "you should sum over this potential" implies "there exists a corresponding particle for the potential". If that implication is true, that feels like a pretty important generalizable principle that someone should have told me in undergrad...
- zachf 5y ago
  I'm going to have to butcher it a bit to make this fit in a comment :) Basically the thing you need to think about is the energy associated with the field itself. For the electromagnetic fields, the energy in the field is like (E^2 + B^2), or in terms of the electromagnetic field tensor F (see [0] if this isn't a familiar concept), it's F^2. You can write F in terms of a potential A, basically F = dA, where d means gradient [1]. So the energy looks like (dA)^2, which is a kinetic energy for the field A, because it tells you that an oscillation in the field costs energy. Integrating over the possible values of the A field is a lot like the integral you would do for the matter fields [2], in particular the math looks basically the same as what you would do for electrons for example [3]. And just like with electrons, the local propagating degrees of freedom are the ones we call particles. So this is how you make dynamics happen in QFT--you integrate over all the possible field values, and the local propagating degrees of freedom are called particles. And for EM and other gauge theories, the math is basically the same as for electrons, and it should be interpreted the same. The same story carries over for gravity, too, where the Einstein-Hilbert Lagrangian (R) is like the EM field strength (F^2), and the metric g enters into it in a similar way as A does for EM (basically, R = (dg)^2). [0] https://en.wikipedia.org/wiki/Electromagnetic_tensor https://en.wikipedia.org/wiki/Electromagnetic_tensor [1] This isn't quite right, you have to make sure everything is gauge invariant, but that's not important for understanding what's going on at a high level. [2] See [1]. [3] If this isn't a story that you understand yet, take some time to study free scalar field theory, and make sure to understand both the path integral method and the canonical method. By seeing how these translate between each other, you can build intuition.

by zachf 5y ago

wyager 5y ago

Thank you for explaining! > the path integral should also include a sum over metrics Understood, thanks. > That sum over metrics is equivalent to saying that gravitons exist (in the same way that the sum over electromagnetic potentials is equivalent to saying that photons exist). Could you explain this? I'm not making the connection where "you should sum over this potential" implies "there exists a corresponding particle for the potential". If that implication is true, that feels like a pretty important generalizable principle that someone should have told me in undergrad...

zachf 5y ago
I'm going to have to butcher it a bit to make this fit in a comment :) Basically the thing you need to think about is the energy associated with the field itself. For the electromagnetic fields, the energy in the field is like (E^2 + B^2), or in terms of the electromagnetic field tensor F (see [0] if this isn't a familiar concept), it's F^2. You can write F in terms of a potential A, basically F = dA, where d means gradient [1]. So the energy looks like (dA)^2, which is a kinetic energy for the field A, because it tells you that an oscillation in the field costs energy. Integrating over the possible values of the A field is a lot like the integral you would do for the matter fields [2], in particular the math looks basically the same as what you would do for electrons for example [3]. And just like with electrons, the local propagating degrees of freedom are the ones we call particles. So this is how you make dynamics happen in QFT--you integrate over all the possible field values, and the local propagating degrees of freedom are called particles. And for EM and other gauge theories, the math is basically the same as for electrons, and it should be interpreted the same. The same story carries over for gravity, too, where the Einstein-Hilbert Lagrangian (R) is like the EM field strength (F^2), and the metric g enters into it in a similar way as A does for EM (basically, R = (dg)^2). [0] https://en.wikipedia.org/wiki/Electromagnetic_tensor https://en.wikipedia.org/wiki/Electromagnetic_tensor [1] This isn't quite right, you have to make sure everything is gauge invariant, but that's not important for understanding what's going on at a high level. [2] See [1]. [3] If this isn't a story that you understand yet, take some time to study free scalar field theory, and make sure to understand both the path integral method and the canonical method. By seeing how these translate between each other, you can build intuition.