Recent content by michael879

Thank you PeroK, I came to a similar conclusion the other night while I was falling asleep, and forgot about it until now. The answer I had arrived at was: in the first case, the Lagrangian is a functional L[x], as I had written. However, once you plug in F=m\ddot{x} x can no longer be varied...

Vanadium, unless I'm mistaking what you mean, I already know how to do it "right". You simply ignore my second case and use the first one to derive the EOM. If I'm wrong, there must be a logical flaw in those 4 lines I placed above. I'm simply asking you to show me what that is, so I can...

If somebody asked me why dividing by zero was an invalid operation I would be able to give more of a detailed answer than just "because it is". That's all I'm asking for here. And as I've mentioned 3 times now, this "invalid" operation is nearly identical to how you integrate out fields...

I don't think you're understanding the point I'm trying to make. Of course changing the functional form of the Lagrangian changes the EOM. Let me abstract this a litte: - Imagine you have some theory T described by Lagrangian L - Then take some theory T' which also includes L but adds some...

I know there is no paradox or problem with the Lagrangian formalism, that's why I put quotes around it. I am trying to understand why this is an invalid operation. How is this any different fundamentally than integrating a heavy field out of a field theory? You're removing a degree of freedom...

This is a really common ambiguity in how people use the word 'particle'. In addition to the fundamental particles of the Standard Model, quantum field theories are also used to describe quasiparticles such as these and phonons. Since it's unclear whether or not the "fundamental particles" are...

This is the 1-dimensional time-independent Schrodinger equation for a free particle. So by using this equation, as opposed to the general one, you're making some assumptions: 1) 1-dimensional: this particle is confined to 1 spatial dimension 2) time-independent: this particle has a fixed energy...

^bump, anyone else? Also, I'd just like to point out there is no logical flaw here, since I added the EOM as an additional assumption in the second example. The "paradox" is that adding the EOM as an assumption results in an inconsistent "EOM" with additional constraints (F=0)

The wave function is a mathematical construct that has no (or many) clear physical meaning. The Schrodinger equation is one of a handful of equations (Dirac and Klein-Gordon are the main alternative single particle equations) governing the evolution of this wave in space and time. The math...

Well in this example I agree it seems pretty dumb, but if you had a Lagrangian with multiple DOF you could conceivably want to remove one. For example, integrating out a field in a QFT, which basically follows this same procedure if I remember it correctly. You use the EOM to replace one field...

Let me begin by saying I know I'm doing something wrong here, but I'm having trouble seeing what it is. This is a reformulation of https://www.physicsforums.com/threads/plugging-eom-into-lagrangian.905099/, where I've reduced the issue to a much simpler problem. Moderators, feel free to close...

I know that in general plugging the EOM into the Lagrangian is tricky, but it should be perfectly valid if done correctly. Can someone help me see what I'm doing wrong here? I know I'm doing something dumb but I've been staring at it for too long Start with the E&M Lagrangian: L =...

So you're saying there's no difference? You can just square any piece of the Lagrangian and it won't change anything?? I understand that you could square the entire Lagrangian, but there are other parts here (e.g. the Einstein-Hilbert Lagrangian, the potential term, and terms for whatever else...

Can someone help me understand how the following two actions are related? S_1 = \int \left(-\dfrac{1}{2}mg_{\mu\nu}\dot{x}^\mu\dot{x}^\nu - U\right) d\tau S_2 = \int \left(-m\sqrt{g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu} - U\right) d\tau Both of them lead to the correct geodesic equation as the...

If anyone could shed some further light on this that would be great, Stevie?