# Tag Info

1

What's true is that it's law of conservation of energy. It particularly states: Q= Del(U) + W, U: internal energy. That is, the total energy given to a system does two things, first it makes the system do the desired useful work and second it changes the internal energy of the system. The work can be positive or negative according to W=q(dv). Example: ...

-1

The previous answer, in my opinion is half-baked. The use of the delta term is quite misleading, as pointed out in the comments. Now, according to almost every leading textbook and my Professor, who is also a leading author, the I Law of Thermodynamics can be stated as: Q = ΔE + W where Q is the heat transfer across the system boundary, W is the work ...

1

Normally, when combining Lagrangians, we often leave the constant multiplying factor to be determined by experiment. For example, if $\mathcal{L}_{k}$ is the kinetic term (for a system of charges and the electromagnetic field), and we choose to describe the electromagnetic coupling by $\mathcal{L}_{int} = A_\mu J^\mu$, then we combine them as ...

0

Here is an explanation of electrical flow that might illuminate this matter: http://amasci.com/amateur/elecdir.html

1

That there are two distinct types of electric charge is a metaphysical fact. But nature is indifferent to what we choose to label these charges; up / down, left / right, positive / negative, black / white, etc. Electrons will still flow to the plate in a CRT regardless of how we choose to label the polarity of the charge on the electron and plate. ...

1

Benjamin Franklin proposed electric fluid theory and considered electric current to be flow of a charged fluid. He meant to use positive to denote a surplus of the fluid, negative as a deficit of it. No one knows how he came up with the choice, but it became the convention and as a result lead also to the labeling of charge. I know of no fact that could ...

0

As Zeldredge said the name is arbitrary and does not matter electron could have been positive and protron negative just the name.

1

What is wrong with my reasoning? Opposite charges attract because one of the charges has a negative sign. The force on the negatively charged particle is thus $$\vec F_- = \frac{kQ(-Q)}{r^2}\hat r = -Q\,\frac{kQ}{r^2}\hat r = -Q\,\vec E_+$$ The force on the negatively charged particle is opposite the direction of the field from the positively ...

0

The electric field lines show the direction of the electric force acting on a unit positive charge at a particular point in space. So, therefore, the force acting on a negative charge, as is in your question, will act in the opposite direction shown by the electric field lines. I believe that the fact that field lines are defined in terms of a unit positive ...

-1

Electron has less mass and therefore it start accelerating first therefore it is considered that dipole moment from negative to positive when placed in electric field

3

I don't think it's very likely, but one other thing I can think of: when the sign before the $\omega$ is a minus then the wave represents a wave travelling to the "right" - positive x direction - and maybe your TA wants only waves travelling to the right. ^^ In case you don't know why the minus sign represents a wave travelling in the positive x direction: ...

2

You may have lost marks for leaving out the imaginary unit $i$. If not, then understand that either $e^{\pm i\,\omega\,t}$ can be used to represent the real signal with positive frequency $\omega$ - it's wholly a question of convention. But once you have made the choice you must stick with it and the choice has implications throughout all the equations of ...

0

I think that you are missing some $i$ or $j$ (complex $\sqrt{-1}$ in your expression, like in $exp[j(\omega t-\phi)]$. consult this page of the book: The Light Fantastic: A Modern Introduction to Classical and Quantum Optics and read the equation 5.10 and the side note: The choice of the complex form ... is made for convenience. The frequency $\omega$ ...

2

Your question apparently stems from a lack of understanding of the different pictures in quantum mechanics, that are Schrödinger picture, Heisenberg picture and Interaction picture. In the Schrödinger picture, states are time-evolving, while observables are time-independent. The density matrix is another (more general) way of writing the state vector; its ...

2

Actually, you can use duality: the normal states of quantum mechanics are objects of the (unique) predual of the von Neumann algebra of quantum observables. Using a concrete example: if the algebra of observables are the bounded operators on a Hilbert space, the predual are the trace class operators. Of them, the normal states are the ones positive, ...

7

$\rho_\psi$, the density matrix, is not an observable/operator evolving in the sense of the Heisenberg equation of motion $$\mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}t} A = - [H,A]$$ since it is defined, as you correctly write, as a projector on states, hence it is time-dependent in the Schrödinger picture (since there the states it projects on are ...

0

I think I figured out my confusion. With these limits defined, in order for cosine to be correct, I have to redefine cos=-adj/hyp. Then everything else works out fine. Weird. It feels wrong to just redefine cosine, but it's true under these defined limits. Is that right, is that something you have to do sometimes? Redefine a trig function? The rest of the ...

0

Over the length of the wire from -infinity to +infinity the angle theta varies from -pi/2 to +pi/2 and cos(theta) ranges from 0 through 1 and back to 0. It's never negative and the distance from any point on the wire to P is always positive. I think you create an unnecessary problem by treating the line from wire segment dl to P as a directional vector ...

0

You can change the sign of your angle, or you can swap a and b, but you can't do both. When you swap a and b the angle gets increased by 180 degrees (which is the same as changing the sign of your angle), if you do this and change the sign of the angle (again) the two cancel each other out.

2

You are using the repulsive force as the force acting to move the charge from B to A(which is not actually moving the charge). We need an external force to move the charge from B to A, which will be taken into consideration(to calculate workdone).

0

There is this famous reality: Addition or subtraction of any constant to potential energy doesn't change the equations of motion. In your case $U_0$ is just a constant that one can add or subtract freely. You can assume that system had an initial constant potential energy (independent of your generalized coordinates of course) just before you started to ...

0

As another answer points out, a constant can be added to the potential energy without affecting the equations of motion. Often, we impose the boundary condition that the potential energy is zero 'at infinity'. For the case of a central gravitational (attractive) force, imposing the "zero at infinity" boundary condition means that the gravitational ...

0

You have this negative quantity because you need to pick a zero point for energy. It's a sort of needing of an arbitrary costant. But another important thing is that the system you're considering is a legacy system. Now I tell you what it is: A legacy system is a particular system where a force operate with a big power, so for separating the two objects of ...

3

Negative energies are totally fine, because you had to pick a zero-point for energy. In your calculation you picked it to be at infinity. You could have chosen the zero-point for potential energy in such a way that your system had zero energy, or whatever. Only changes in energy are meaningful, in general. Consider this: what happens if you add energy to ...

3

A good question, you are right the frequency remains constant (unless you have Doppler effects due to relative movement, but that's not your question). For visible light, refraction properties are quite often in question and as such it make sense to speak in terms of wavelength. As you go even higher in "frequency", physicists start talking in keV and MeV ...

2

Relativists tend to use the proper time, $d\tau$, and the proper distance, $ds$, interchangably. If you're working with proper time you'd expect the equation for it to look like: $$d\tau^2 = dt^2 + \text{other terms}$$ while if you're working with proper distance you expect: $$ds^2 = dx^2 + dy^2 + dz^2 + \text{other terms}$$ The sign problem comes ...

0

Could you provide a simple reason for these two conventions? The reason behind the (-,+,+,+) convention (the "mostly plus metric") is that a positive length in 3 dimensional space (e.g., the distance from my head to my toes) should still be a positive length in 4 dimensional space-time. Why should the distance from my head to my toes all of a sudden ...

1

I think you have the answer for your second question. For your first question let me clarify: $$\langle x\ |\ \hat{x}\ |\ p\rangle \overset{(1)}{=} \int dx'\langle x\ |\ \hat{x}\ |\ x'\rangle\langle x'\ |\ p\rangle \overset{(2)}{=} \int dx'x'\langle x\ |\ x'\rangle\langle x'\ |\ p\rangle \overset{(3)}{=} x\langle x\ |\ p\rangle ... 0 One never pluralizes unit abbreviations. Your link goes to the BIPM, the body responsible for maintaining the definitions of the international system of units, and is authoritative. The folks at NIST agree and address most of your questions. I would say The pipe is 0.75 m long. or The pipe is 75 centimeters long. or even The pipe is ... 3 cms and kgs are wrong. The SI units are abbreviations which are also used in the plural. You will write 2.6 m/s or 1 m/s, but say "2.6 meters per second" or "1 meter per second" respectively. Keep in mind the SI units are also used in tons of other languages that do not form the plural by attaching an -s. The units look the same in those languages. (e.g. ... 0 Second, if I try to do the same thing by using the p-basis representation of \hat{x}, I get into more trouble:$$\langle x\ |\ \hat{x}\ |\ p\rangle=\int dp'\langle x\ |\ \hat{x}\ |\ p'\rangle\langle p'\ |\ p\rangle=\int dp'i\hbar\frac{\partial}{\partial p'}\langle x\ |\ p'\rangle\langle p'\ |\ p\rangle=\int dp' ...

1

This question (v6) [concerning the overall minus sign in OP's calculation] is essentially a Fourier transformed version of e.g. this Phys.SE post, see Emilio Pisanty's answer and my answer. The main point is again that the derivative in the momentum Schrödinger representation $$\hat{x}~=~i\hbar\frac{\partial}{\partial p}, \qquad \hat{p}~=~p,$$ acts on the ...

0

I think this is really about which way you count current and voltage to be positive. For every element in a network you can define a a current and a voltage. If voltage and and current point the same direction, it's called "receptor". If they are in opposite direction its' called "generator". The most common convention is to use "receptor" for resistors, ...

2

I recently answered a similar question here. The ideal capacitor equation $$i_C = C\frac{dv_C}{dt}$$ assumes the passive sign convention which means that the reference direction for $i_C$ is into the positive labelled terminal. When you write $$iR = v_C$$ it is necessarily the case that $$i_C = - i$$ To see this, assume that both positive labelled ...

2

This is a common question. The issue is that the "Q" in $i = dQ/dt$ is not the same as the $Q$ that represents the charge on the capacitor. The variable $Q$ in use here is simply the charge on the capacitor. No problem. When the capacitor discharges the quantity of charge that is introduced into the circuit after a time $\delta t$ has elapsed is ...

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