New answers tagged

0

It's an axiom which leads to a formalism that has great agreement with experiment. Asking why an axiom is what it is isn't really useful.


2

Gravitational field is a vector field and is determined by negative gradient of the gravitational potential. $$\vec g=-\vec\nabla \phi$$ Frome equation above, it is obvious that $|\vec g|=|-\vec\nabla \phi|$ (magnitude of $\vec g$ is equal to magnitude of $-\vec \nabla \phi$) and we know that $|-\vec\nabla \phi|$ is a non-negative quantity. You have made a ...


6

For an open orientable surface there are two possible, equivalent normals: $\vec n$ and $-\vec n$. The usual convention is that you choose a direction in which the perimeter of the surface is traversed and define the positive direction as the direction given by the right hand rule, as shown in the following picture. This can also be done for non-simply ...


1

By convention, for a flat lamina or a plane surface, the area vector is a vector whose magnitude is the area of the surface and whose direction points in a direction perpendicular to the surface. If you have a curved surface, then you have to consider elemental areas, i.e: small patches of area denoted by $dA$ whose direction is perpendicular to the small ...


1

They are linked by the "law of the right hand": The preferred direction of dℓ⃗ dℓ→ along the loop is that from the palm to fingertips of your right hand when it surrounds the loop. Then, the associated preferred direction of dA⃗ dA→ is indicated by the thumb.


0

According to the CPT thoerem, the choice of positive parity for particles and negative parity for antiparticles is just as arbitrary as the choice of positive charge for protons and negative charge for electrons.


0

Here are a few examples of simple notation you could adopt. Rotations in $d$ dimensions can be represented using $d\times d$ matrices (call them $R$) such that $R^TR$ is the identity matrix. If your initial vector that describes a 'positive' direction is $v$, then you can express directions relative to $v$ as $R v$, where $R$ is one of these rotation ...


0

I'm not aware of any such convention. You can always show the net force vector acting on a free body as long as it is clearly labeled as such, to avoid confusion with any other applied forces. I personally wouldn't include a net force vector unless there was a good reason to, like to illustrate some accompanying discussion.


6

Whenever you see a function that looks like: $$ y = \tfrac{1}{2}kx^2 $$ there's a good chance it came from integrating the function: $$ \frac{dy}{dx} = kx $$ For example your distance function comes from integrating the velocity $v = at$: $$ y(t) = \int v\,dt = \int at\,dt = \tfrac{1}{2}at^2 $$ The spring energy function comes from integrating the ...


0

This result is consistent with the principle that says "the work is negative whenever the electric potential increases", and the one that states "the electric potential increases whenever a charge moves in the opposite direction of the force applied to it by the electric field of another charge". My question would be whether my answer is wrong and I am ...


1

Physicists do you the rules of calculus but can be sloppy in their notation. The symbols $\delta, \Delta$ and $d$ often seem to be used interchangeably to mean a (small or infinitesimal) change in something or better still a final value minus an initial value. So in your equation $dU$ is the change in internal energy of a system or final internal energy ...


0

$U$ is the electrostatic potential energy of the system of the three protons. To assemble the three protons coming from infinity external work needs to be done so the electrostatic potential energy is positive. So the textbook answer has computed the external work which needs to be done to bring the protons together. Another approach is to compute the work ...


0

I don't think it matters really, it's a matter of convention. In your picture if we start at infinity: $$ W = \int_\infty^r dr' F(r')= C \int_\infty^r \frac{1}{r'^2} = \frac{-C}{r} \\ E_{pot} = -Q\int_\infty^r dr' E(r') = \frac{C}{r} $$ Energy conservation holds:$E_{pot}+W=0.$ Depending upon how you define work i.e. if the the system "carries out the work" ...


3

The operator $aa^\dagger$ is easily seen to be nothing else than $$ aa^\dagger = a^\dagger a + 1$$ which is equivalent to $[a,a^\dagger]=1$, a commutator that easily follows from $a\sim (x+ip)/\sqrt{2}$ with some extra coefficients and $[x,p]=i\hbar$, so $aa^\dagger$ is the number operator plus one, i.e. taking values $1,2,3,4,\dots $ instead of $0,1,2,3,\...


1

Work out how a plane wave propagates. If its intensity grows in the direction of propagation, then you need to switch the sign of the imaginary part! It depends on whether you define $e^{+i\,k\,z}$ or $e^{-i\,k\,z}$ as your diffraction operator in the direction of propagation. If it's $e^{+i\,k\,z}$, then a lossy dielectric always has a positive imaginary ...


0

Given mild differentiability conditions on the metric (I thought this might be $C^\infty$ which is not very mild, but see comments to this answer by 0celo7 below) then, for any point $p$, you can always pick a coordinate system $\left\{x^i\right\}$ which is locally flat, which means that $g_{ij}(p) = \pm\delta_{ij}$ -- tangent vectors along coordinate ...



Top 50 recent answers are included