# Tag Info

## New answers tagged definition

0

The boundaries aren't hard and fast, but here's the basic heirarchy. fluorescence: photons emitted as electrons randomly drop to a lower state superfluorescence: photons emitted with a "little bit" of stimulated emission but the overall emission pulse doesn't show measurable coherence. superradiance: as you wrote, localized bursts of stimulated ...

0

The moment of intertia is defined by the distribution of mass of an extended object with respect to a (rotation) axis and as such defined as $M = \int m_r r^2 dr$. So unless you want to consider an elastic object which deforms in the presence of other nearby bodies due to external forces like gravity or electro-magnetic forces, the moment of inertia of an ...

1

It is necessary because the operational definition is what the quantity really is. As Eddington put it “A physical quantity is defined by the series of operations and calculations of which it is the result.” Eddington A.S., 1923, The Mathematical Theory of Relativity, CUP. Assuming that something else is the physical quantity is metaphysics. There is no ...

1

The laws of physics are mathematical expressions of variables that stand for numbers. In order to actually use a law, rather than just looking at it and admiring it, we need some way of measuring the quantity so as to provide a number to use in the law. That is, in order to use the law it is sufficient to have an operational definition.

1

Why is it only necessary to give operational definition of time (or force, time, length...) and not also what they really are? It is only necessary since, to the extent that we can indeed determine what a physical entity "really is", we must do so after having defined it operationally and having performed experiments with it. It would be pointless ...

1

Tensor is a multi dimensional vector in colloquial language. Where the variations in one direction effects the other. In Newtonian mechanics we assume all forces, velocities etc that are mutually orthogonal $\Rightarrow$ mutually independent. $$F=F_x\vec i + F_y\vec j +F_z\vec k$$ $F_x\vec i . F_y\vec j =0$ since $\vec i . \vec j =0$ Whenever we apply some ...

2

Let me be specific here. Speed is not the magnitude of velocity but instantaneous speed is the magnitude of instantaneous velocity. As pointed out by Brain Stroke Patient, this happens in the limiting case as $lim \Delta t \rightarrow 0$. In that case, $|\Delta\vec{x}|\rightarrow D$. I think your confusion is that distance travelled and displacement aren't ...

1

$\vec{v} = \Delta \vec{x}/\Delta t$ for infinitesimally small displacement and time. In that limit D is equal to $\Delta x$. If the displacement isn't infinitesimal then you have the average velocity and that need not be equal to, in magnitude, to the average speed.

3

Equations of motion (EOM) are typically the equations that determine the time-evolution of the system. E.g. in Newtonian mechanics, Newton's 2nd law is the EOM. (One should avoid referring to the kinematic $suvat$-equations as EOM to avoid confusion.) For Lagrangian systems, the Euler-Lagrange (EL) equations are referred to as EOM, even in case of EL ...

0

The practical (although probably not the most rigorous) definition comes from looking at the derivatives in the equation. Equations of motion describe the evolution of the position of particles, so the independent variable is time (or a parameter like proper time) and the dependent variables are spatial position (or spacetime event locations). You'll see a ...

4

Field equations tell you how fields change in spacetime, whereas equations of motion tell you how arbitrary physical objects move in spacetime. In other words, field equations are equations of motion for a field. Normally the term EOM is used in classical mechanics to denote the motion of a single body or system of bodies (though it certainly is used in ...

8

There is no "precise distinction" between these terms. The "field equations" are just important equations of a field theory, which may or may not be the equations of motion for that theory. And even which equations are "equations of motion" is not unique! In the Lagrangian formulation of electromagnetism coupled to a charged ...

1

There isn't a standard, unique notation for this (or technically anything, really). Some of what you have written looks like what I have seen in the past, although... $V_{ab}$ could instead also be $V_b-V_a$. With what you have $V_{ab}=-\Delta V_{ab}$, which is odd. I have never seen $\Delta V_a$ before, but usually if one is specifying just a single $V$ ...

-2

Well, its about the concentration difference in electricity. It's basically the amount of change in potential energy when a unit charge goes from one position to another.

3

Voltage is the potential difference between two points.

0

The fundamental tensor equation of relativistic mechanics of continous matter is $$K^\mu = \partial_\mu T^{\mu\nu}$$ where $K^\mu$ is the 4-force-density acting on material medium and $T^{\mu\nu}$ is the energy-momentum-stress tensor of the system. Consistently for $\nu=0$ we obtain the equation of continuity and for $\nu=1,2,3$ the 3-vector equation of ...

2

For practical purposes, the Hamiltonian formulation does express conservation of energy and momentum in generalised coordinates. It is possible to construct counter-examples, using time varying coordinates. Time varying coordinates mess with the definitions of energy and momentum, but this is artificial. Coordinates are a human choice, and the sensible ...

0

Hamiltonian of a system need not necessarily be defined as the total energy $T$+$V$ of a system. It is some operator describing the system which can be expressed as a function in terms of the variables of phase space. Speaking physically, it is the Legendre Transformation of the Lagrangian of a System. The Lagrangian of a System is that function, which ...

0

In my understanding a frame of reference is a structure that quantifies the term "at rest". To go into detail we first have to clearify the framework we are in, i.e. the spacetime structure. For Newtonian mechanics this would be the Galileian spacetime. In beginner courses this is often introduced as $M=\mathbb{R}\times\mathbb{R}^d$, a Cartesian ...

0

In first year physics, a chosen coordinate system defines the frame of reference, and as Safesphere points out, it is chosen relative to some object (like the earth).

3

Time to jump into the fray. This equation here $$W=\int\mathbf F\cdot\text d\mathbf x$$ is just the definition of the work $W$ done by a force $\mathbf F$ along some path that you are performing the integral over. It is always applicable, as it is a definition. However this equation $$W=\Delta K$$ is only valid when $W$ is the total work being performed on ...

1

But if you imagine lifting up a rock from the ground at a constant speed, am I not doing work on the rock by converting the chemical energy stored in my muscles into the potential energy of the rock? Yes, you do. There are a few things to be noted here. First, as you lift it up, the rock has to move which serves as it's kinetic energy. As you lift the rock, ...

1

I think maybe you are confused by an example that is often given to emphasize that the physics definition of "work" doesn't always match how it is used in everyday speech. If you move a rock horizontally at constant speed, then you are doing no work on the rock. Its kinetic and potential energy are both the same through the whole process. This is a ...

2

The problem with work is the word we use for it. Since we work everyday, we're used to associate "physical work" and "effort", and that's confusing. Physical work is a well defined quantity, but it needs 3 surnames, and this is usually omitted. You must specify these three parameters: Work done [by some force] [on some system] [along this ...

0

I think perhaps the easiest (perhaps way too didactic) method to understand it is in terms of currencies. For example, think of dollars (for potential energy), euros (for kinetic energy) and Yens (in the case of muscles chemical energy). In the example of lifting up a stone (supposing somehow it was originally in motion), you are exchanging Yens for dollars (...

1

Assume you have two opposite forces of the same magnitude acting on a particle. The total work is zero, and there is no change in kinetic energy. However one of the forces made positive work on the particle and the other negative work. Whatever did positive work lost some form of energy, and the one that did negative work won some energy. The net effect on ...

0

My professor explained it to me by giving me the spring example. Suppose there is a spring attached to a block of mass m at one end and wall at the other.Neglect friction everywhere. If we now pull the mass with constant velocity. Let's us see what changes happens: In the above example there are 3 objects that are me pulling the block , the block , and the ...

1

Scalars are tensors of rank 0. Vectors are tensors of rank 1. Both the objects you describe are therefore tensors. The main difference is notation. $v^i e_i$ might be better written $v^i \mathbf e_i$. This is a vector, in which the basis vector $\mathbf e_i$ is explicit. In index notation, we omit to write the basis vectors. The same vector is written $v^i$ ...

0

There is a lot of abuse of notation in GR I find. I would add something specifically about scalars: $v^i v_i$ is a scalar because it is shorthand for an inner product between two elements of the vector space. An inner product in this scenario is a map which takes two vectors and gives you a number (in $\mathbb{R}$ or $\mathbb{C}$ usually). Let's call the map ...

-2

$v^ie_i$ is an element of the vector space, but it's not a vector in the same sense as $v_i$. This is because we define vectors/tensors based on their transformation properties. $v_iv^i$ is a scalar. Points addressed above When talking about tensors, we usually just mean the components $v_i$ (In this case, the $v_i$s transform as a vector) We usually have a ...

1

Here is a very simple examples of collective modes: the normal modes of a mass-spring chain. This is pretty close to phonon modes in a solid state system. But the same principle applies to any other situation where many parts of a system are coupled and oscillate at the same frequency. https://www.youtube.com/watch?v=WE-HB8DBFU4

Top 50 recent answers are included