Deviations from classical action and D’alembert's principle, why is D’alembert's principle not applicable to QM?

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We can derive action principle from dAlemberts principle. Does deviations from classical paths violate the d’alemberts principle? Like the inertial force not balancing external force in a small element of interest? What would this mean in terms of general relativity?

So if an inertial force doesnt balance external force, in a non stationary path is there anything at all we could take as geodesics?

My derivation was kinda wrong but it still comes to same thing, dalemebert principle should have been independent of physical theory gr qm etc. some say its not applicable but I don’t see why
 
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planck999 said:
We can derive action principle from dAlemberts principle. Does deviations from classical paths violate the d’alemberts principle? Like the inertial force not balancing external force in a small element of interest? What would this mean in terms of general relativity?

So if an inertial force doesnt balance external force, in a non stationary path is there anything at all we could take as geodesics?

My derivation was kinda wrong but it still comes to same thing, dalemebert principle should have been independent of physical theory gr qm etc. some say its not applicable but I don’t see why
I would guess since it's a classical mechanics principle, it adheres to the three rules of Newton.
But Q and R theories don't adhere to this third law.
 
planck999 said:
What would this mean in terms of general relativity?
This is the QM forum, not the relativity forum. If you are asking, what would it mean in the context of quantum field theory in curved spacetime, then we can discuss that in this forum. But I think you need to clarify your question first.
 
loop quantum gravity said:
I would guess since it's a classical mechanics principle, it adheres to the three rules of Newton.
But Q and R theories don't adhere to this third law.
Because action principle does apply to r and q. And it gives the same results with action principle. Also, its derivation doesnt compeletely depend on newtonian mechanics, it should be valid for arbitraty coordinates such as fields.
PeterDonis said:
This is the QM forum, not the relativity forum. If you are asking, what would it mean in the context of quantum field theory in curved spacetime, then we can discuss that in this forum. But I think you need to clarify your question first.
I asked because in qm there are non stationary paths. So, since dalembert is connected to gr I thought maybe there would be a connection. Because whole gr is based on acceleration=gravity and geodesics=frames without acceleration.
 
planck999 said:
I asked because in qm there are non stationary paths.
Yes, that's because QM is not a classical theory. GR is a classical theory.

planck999 said:
since dalembert is connected to gr I thought maybe there would be a connection.
Since, as you note, QM includes non-stationary paths, d'Alembert's principle can't possibly hold for QM. If your question is why that is, the only answer is that, as above, QM is not a classical theory, and it has to account for experimental data that cannot be accounted for by a classical theory. That's why it has to include non-stationary paths in the first place.
 
PeterDonis said:
Yes, that's because QM is not a classical theory. GR is a classical theory.


Since, as you note, QM includes non-stationary paths, d'Alembert's principle can't possibly hold for QM. If your question is why that is, the only answer is that, as above, QM is not a classical theory, and it has to account for experimental data that cannot be accounted for by a classical theory. That's why it has to include non-stationary paths in the first place.
And, we interpret non stationary phases physically in an intuitive manner. What would be the meaning of non dlamabertian paths? And in a non stationary path how would light bending in an inertial elevator thought experiment work?
 
planck999 said:
What would be the meaning of non dlamabertian paths?
Are you familiar with the path integral formulation of QM? That is the formulation that uses these paths and explains why.
 
planck999 said:
in a non stationary path how would light bending in an inertial elevator thought experiment work?
You're misunderstanding the role the paths play in QM. In QM, all of the paths are included in the machinery that determines the probability amplitudes. You can't pick out just one and ask what its effect is; they're all present in any experiment.

The light bending in an inertial elevator thought experiment does not involve any quantum effects, so the classical description is sufficient; the QM description would simply reduce to the classical one.
 
I misunderstood one thing in path integral formulation I guess. Meaning of different phases cancel out at macroscopic level isn’t all paths are present but it collapses into one of them thing, its more like all paths are present and their superposition is the only observable we get out of them and macroscopically it is classical. Am I right, can you observe a particular path or is the only observable their superposition? Like wavefunction can collapse into a particular spin but i guess same cant be said for paths?
 
planck999 said:
can you observe a particular path
Not in QM, no.

planck999 said:
or is the only observable their superposition?
Even that's not quite a direct observable; it gives you the probability amplitude, but all you can actually observe is the squared modulus of that, the probability.
 
planck999 said:
wavefunction can collapse into a particular spin
If you're using an interpretation of QM that includes collapse, yes.

planck999 said:
i guess same cant be said for paths?
Paths aren't wave functions, and even in collapse interpretations of QM, they don't collapse, no. Collapse isn't a concept that has any meaning for the individual paths in the path integral.
 
Thanks for your explanation. Then, a non stationary path cant create weird phenomena in gr, like a non inertial geodesic. Anything beyond this goes to attempts trying to unify qm and gr and there is still no validated answer for that.