Recent content by Ryder Rude

Okay. But how should I interpret ##\psi (x)|0\rangle## and ##A^{\mu} (x) |0\rangle##? Both of these objects have have extra index ##\mu## or ##\alpha##, which really shouldn't be there in the state-vector describing a single photon or an electron/positron.

So I should think of ##\psi (x) |0\rangle## as just an absract object with spinor indices rather than a particle at a position ##x##, right? And the same logic applies to ##A^{\mu} (x)|0\rangle##? Also, can you please explain your interpretation of this as a correlation function? Why is it...

I can understand how ##\phi (x)|0\rangle## represents the wavefunction of a single boson localised near ##x##.I don't understand how the same logic appies to ##A^{\mu}(x)|0\rangle## and ##\psi |0\rangle##. Both of these operators return a four component wavefunction when operated on the vaccuum...

Sorry I was viewing it on mobile. I had to slide it to see it full

You accidentally wrote just ##R##

It should be ##R + \frac{L}{2}## for the CoM, right?

We could still talk about a point of application in 1D, line of application in 2D, and plane of application in 3D. I think the fact that it's a property of the forces acting instead of a property of the system makes it way less fundamental than CoM. This point is a good thing to talk about only...

Yeah, I got that from your previous reply. But what additional conditions are missing? The formula was like taking the weighted average of the distances by using forces at the points as weights.

Not exactly the same effect (as in, the particles won't move in the exact same way) But maybe the same net effect? (as in, if you add up the add up the angular and linear momenta of the prticles, you get the same result in both cases) Not sure what you're saying. I'm not arguing against...

Did I miss anything? There are not many links about 'Point of application'. It's not on even wikipedia

From what I found, I think we're trying to find the point on which applying the net force would produce the same linear and angular acceleration of the system, as the individual forces together produce. So in this sense, the individual forces can be thought of as acting together on this point.

It's worth knowing as a concept though. We don't use Cramer's rule practically, but we still know it. 'Center of force' should be taught in general instead of just 'center of gravity'. You didn't say if my analogy was correct in the post you replied to

Why wouldn't it be correct? The center of mass point literally follows the path that a particle of mass ##M## would've followed if the forces on the system were thought of as directly acting on that point. Also, how do we actually summarise the external forces? What's the formula? I can't seem...

And if we do define it, its purpose will be related to measure the angular acceleration of the system? Here I'm making an analogy with center of mass: If we track the location of the center of mass over time, and find it to be accelerating with acceleration ##\vec{a}##, then we conclude that...

But that is also the whole idea of center of mass. It is the point on which the external forces can be thought of as acting directly How can center of gravity be describing the same idea and also be different from center of mass? What if we have a situation where, say, some countable number...