I don't think you can really test a definition. I mean, with v=dx/dt, you could measure the distance your car travels in a certain time and compare that to what your speedometer says. But I think if you really analyze it, you'd find you're not really testing v=dx/dt. That statement is true by...
There are apparently different ways of interpreting Newton's laws (what's a definition, what's an axiom, etc.). But I'm just stating what is, I think, the conventional way of teaching it. Although as I alluded to earlier, I think there's a good deal of confusion about this even among folks...
I mean simply that we conventionally take F=ma to be the definition of force, and insofar as it is a definition, it can't be experimentally tested on it's own. Not anymore than one could experimentally test v=\frac{dx}{dt}
That's not to say we can't test Newton's laws as a whole in the lab...
Perhaps it would help if you explain what your goal is in making this simulation. As others have said, the laws of physics are very different for quantum mechanical objects.
And it's not simply a matter of substituting some different equation for f=ma to find the position. The notions of...
I've actually seen a lot of students get confused on this point, and it doesn't help when a textbook asks you to "derive" ##F=ma##. ##F=ma## is not derived from any other principles or equations. It also does not come from experiment. There is no experiment you can do test ##F=ma##. ##F=ma## is...
That page is more a brief description of what an error bar is. I'm looking more for the history of its development. When did it become common to put error bars around things? Did a bunch of scientists come to some conference one day and discuss how they needed to describe the uncertainties in...
I was reading an 1803 paper by Thomas Young (of double slit fame), "Experiments and Calculations relative to physical Optics". In it, he lists various dimensions of fringes of light and things.
All without any error bars.
It got me thinking, what's the history of error bars in scientific...
Maxwell field commutation relations
I'm looking at Aitchison and Hey's QFT book. I see in Chapter 7, (pp. 191-192), they write down the canonical momentum for the Maxwell field A^\mu(x):
\pi^0=\partial_\mu A^\mu \\
\pi^i=-\dot{A}^i+\partial^i A^0
and then write down the commutation...
In QFT, the commutation relation for the field operator \hat{\phi} and conjugate momentum is
[\phi(x,t),\pi(y,t)] = i\delta(x-y)
Maybe this is obvious, but what would the commutator of \phi or \pi and, say, e^{i k\cdot x} be?
Simple question. Simple answer: We don't really know. This is a philosophical hanging thread in the theory. Measurements are needed to make sense of the theory, but it's hard to define what a measurement is. Look up "measurement problem in quantum mechanics" in Google. So, no, you're not stupid...
Right after I posted that, I figured that's what you meant. And it finally struck me why we don't have to worry about the \vec{k}. Thanks for your help.
Thanks for the reply. I guess I'm missing the last part of that equality F(x) = F(\Lambda^{-1}x). As a test, if I define a function
F(x)=k\cdot x
where k is some arbitrary 4-vector, then F(x) is a Lorentz scalar. If I have some transformation such that x'=\Lambda x, then
F'(x')=k'\cdot...
Homework Statement
Show that
[\hat{\phi}(x_1),\hat{\phi}^\dagger(x_2)] = 0
for (x_1 - x_2)^2 < 0
where \phi is a complex scalar field
Homework Equations
\hat{\phi}=\int\frac{d^3 \mathbf{k}}{(2\pi)^3 \sqrt{2\omega}}[\hat{a}(k)e^{-ik\cdot x} + b^\dagger(k)e^{ik\cdot x}]...