What are common signs of a faulty wheel hub bearing?

[andert]

My preferred way to think of momentum is as part of the stress-energy-momentum tensor which is a covariance matrix that becomes diagonal (assuming an orthogonal coordinate system) when the observer and the object are at rest with respect to one another. Momentum forms the time-space components (assuming a symmetric tensor) of the tensor. So we can think of momentum of a massive particle as the flux of its internal energy in the observer's frame. This means that momentum depends on the relative frames of the observer and the object, i.e. it depends on your state of motion.

A fascinating fact is that momentum is conserved in every reference frame even though it changes (covaries) from frame to frame, i.e. in the absence of forces \partial_\mu T^{\mu\nu} = 0 where, e.g., T^{\mu\nu}=mu^\mu u^\nu is the stress-energy-momentum tensor of a basic particle. This is quite different from other properties of particles. For example, for electric current, \partial_\mu J^{\mu} = 0. There's no second Lorentz index on the current. So while the conservation of the stress-energy-momentum tensor says that a Lorentz 4-vector, \partial_\mu u^\mu (mu^\nu), is conserved in an arbitrary frame, the conservation of the electric current only says that a scalar, the rest charge, q, is conserved in every frame.

 

[jtbell]

(2) \enspace \frac{m_0}{\sqrt{1-\left ( \frac{v}{c} \right )^2}},

Correction:

(2) \enspace \frac{m_0 v}{\sqrt{1-\left ( \frac{v}{c} \right )^2}}

 

[Rasalhague]

My preferred way to think of momentum is as part of the stress-energy-momentum tensor which is a covariance matrix that becomes diagonal (assuming an orthogonal coordinate system) when the observer and the object are at rest with respect to one another. Momentum forms the time-space components (assuming a symmetric tensor) of the tensor. So we can think of momentum of a massive particle as the flux of its internal energy in the observer's frame. This means that momentum depends on the relative frames of the observer and the object, i.e. it depends on your state of motion.

Is the quantity you call "momentum" that labelled "momentum density" on the diagram at the top right of this page, rather than "energy flux", the name they give to the time-space components? (Not that that affects the numbers if it's symmetric.)

If there was only one particle and it was at rest with respect to the "observer" (i.e. reference frame), would all the components vanish except T_{00}?

 

[danielatha4]

Momentum can be thought of as the rate at which an object comes to rest at its center of mass.

 

[Galap]

I don't want to know what it's like, I want to know what it IS.

On an unrelated note, I read this excerpt from a Hyperphysics article on voltage;

"Like mechanical potential energy, the zero of potential (of voltage) can be chosen at any point, so the difference in voltage is the quantity which is physically meaningful."

What does that mean? It's really bothering me.

Momentum ends up being the quantity that is physically 'real' or 'meaningful', as it is conserved in all frames of reference in both relativity and QM. Both mass and velocity (the components that classically make up momentum) are not conserved and are interchangeable, etc. It is a measure in a way of how much stuff there is in a system.

 

[gato_]

(1) \enspace m\textbf{v}.
(2) \enspace \frac{m_0 \,v}{\sqrt{1-\left ( \frac{v}{c} \right )^2}},
(3) \enspace E^2 = \left ( pc \right )^2 + \left ( mc^2 \right )^2,
(4) \enspace E = pc
(5) \enspace p = \frac{h}{\lambda},

Actualy, all these definitions are related. All of them can be interpreted as infinitesimal generators of translations
\phi(x+\delta x)\sim \phi(x)+\delta x^{\mu}\partial_{\mu}\phi\equiv \phi(x)+\delta x\cdot p[\phi(x)]
(A factor i\hbar is required in the quantum definition for mathematical reasons).
So, first fact, momentum is related to space and time translations (in relativity, energy is the momentum component related to time translations). There is a similar derivation for the energy momentum tensor.
The mathematical fact is the momentum, on whatever the representation you choose to act on, is conserved whenever the underlying system is invariant under translations. And the physical principle is that the undelying space on which matter exists is itself invariant under those translations (or have other symmetries leading to other conserved quantities). All the physical interactions may be thought off as local momentum interchanges.
On a more technical side, it could be said that SO(3,1), the group representing the symmetry of flat space in general relativity has two (Cassimir) invariants: P^{2}=M^{2} and spin. That means any irreducible representation of this group (that is, any free particle) can be labeled by those two. Eq (3) merely states that a free particle has a constant momentum. Of course, when you consider the rest of interactions, Electromagnetic weak and strong, you have to add additional labels.

 

[earn1getmoney]

after reading other answer everybody define only its exact definition of it nobody tell its practical feel as a engineering student what i feel is that it is a impact which act on a body when it hurt or get collide with some other body mean more momentum more impact also we can say a type of inertia acting for a long time
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