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Is momentum valid?

Anonymous Thu 29 Oct 2015 11:46:13 No.7623816 View Reply Report Delete

Why there is momentum when you can use velocity to figure out the amount of motion?

https://www.youtube.com/watch?v=IuJpd4Q54Fw

Anonymous

Anonymous Thu 29 Oct 2015 11:48:57 No.7623819 Report Delete

Quoted By: >>7623837 >>7623842

>>7623816
Momentum is more fundamental then velocity.

It's weird, I know, but think about it.

You know of momentum as p = mv

But light, made up of massless particles, has momentum. And more than that, even though all light moves at the same speed, it can have different momentum.

The correct thing to do is to define the velocity in terms of the momentum.

Anonymous

Anonymous Thu 29 Oct 2015 11:54:38 No.7623825 Report Delete

Quoted By: >>7623847 >>7623868 >>7623888

>>7623816
Momentum has the extremely important property that it is conserved. Also it is not always dependent upon just velocity.

i.e.
Momentum can be found more generally used the Lagrangian definition. $p = \frac{{\partial \mathcal{L}}}{{\partial \dot x}}$

This definition allows momentum to vary depending on your system.

If you take a simple classical mechanical system with $\mathcal{L} = KE - PE = \frac{1}{2}m{{\dot x}^2} - V\left( x \right)$ then you will get a momentum $\frac{{\partial \mathcal{L}}}{{\partial \dot x}} = m\dot x$ .

But if you take a more complicated system like a relativistic particle in a EM field,
$\mathcal{L} = - m{c^2}\sqrt {1 - \frac{{{{\dot x}^\mu }{{\dot x}_\mu }}}{{{c^2}}}} - \frac{e}{c}{A_\mu }{{\dot x}^\mu }$ ,
You get a more complicated momentum like ${p_\mu } = \frac{{\partial \mathcal{L}}}{{\partial {{\dot x}^\mu }}} = \frac{{m{{\dot x}_\mu }}}{{\sqrt {1 - \frac{{{{\dot x}^\mu }{{\dot x}_\mu }}}{{{c^2}}}} }} - \frac{e}{c}{A_\mu }$ .

Anonymous

Anonymous Thu 29 Oct 2015 12:02:06 No.7623837 Report Delete

Quoted By: >>7623855

>>7623819
Different light sources have different "momentums" because it is not only velocity, it is something moving obviously. So it has energy, which is defined by the Planck constant (?) and the wavelength of light. Still don't see why we need momentum.

Anonymous

Anonymous Thu 29 Oct 2015 12:04:59 No.7623842 Report Delete

Quoted By: >>7623855 >>7623890

>>7623819
B-but light wouldn't have momentum if not for the velocity... It has no rest mass.

Anonymous

Anonymous Thu 29 Oct 2015 12:07:47 No.7623847 Report Delete

Quoted By: >>7623851 >>7623855

>>7623825
Isn't the velocity always conserved as well? Sure it can be dissipated in the form of heat and sound but it is always transmitted to something else even if just at the molecular level.

Anonymous

Anonymous Thu 29 Oct 2015 12:14:51 No.7623851 Report Delete

Quoted By:

>>7623847
>Isn't the velocity always conserved as well?
No

Anonymous

Anonymous Thu 29 Oct 2015 12:17:47 No.7623855 Report Delete

Quoted By: >>7623879

>>7623847
Picture a frictionless pool table.

The cue ball has a large mass M and the billiard balls have a smaller mass m.

You hit the cue ball, it is travelling at a velocity V, until it strikes one of the smaller balls.

Is velocity conserved? No the small billiard ball will have a higher velocity than the cue ball because in this situation p = mv is a good definition for momentum.

>>7623842
>>7623837
Lights momentum is based off of it's wavelength, not velocity. It's total energy is E = pc, it's p = hf/c. If Photons all had the same momentum, they would all have the same Energy.

Anonymous

Anonymous Thu 29 Oct 2015 12:28:20 No.7623868 Report Delete

Quoted By:

>>7623825
Also, when doing quantum mechanics momentum gets even weirder in that it becomes an operator.
i.e. ${p_\mu } = - i\hbar {\partial _\mu }$

Anonymous

Anonymous Thu 29 Oct 2015 12:35:04 No.7623879 Report Delete

Quoted By: >>7623888

>>7623855
That is a fair point, but you could also think of it as the mass and velocity separately that when combined gives you the final velocity.

Anonymous

Anonymous Thu 29 Oct 2015 12:44:45 No.7623888 Report Delete

Quoted By:

>>7623879
You could. And sure that would apply in some situations. But the velocity is still not conserved.
>>7623825
Is a better way to think about it, because p = dL/dx* works for all systems, classical, thermodynamic, quantum, field, relativistic, all of them, can have momentum defined in this way, have that momentum be a conserved value, and mean the same thing for all systems. Same thing with using H as the energy. H = Sum(x*i,pi) - L is just as universal, and conserved for the same reasons.

Anonymous

Anonymous Thu 29 Oct 2015 12:48:04 No.7623890 Report Delete

Quoted By: >>7624234

>>7623842
That's because light is not a particle it is a wave. A photon is like a phonon, you wouldn't say that mass is made out of phonons, would you?

Anonymous

Anonymous Thu 29 Oct 2015 15:41:05 No.7624224 Report Delete

Quoted By:

momentum applies this 2 dimensional representation of the properties of a motion to any relatable 3rd dimension.

Anonymous

Anonymous Thu 29 Oct 2015 15:47:50 No.7624234 Report Delete

Quoted By:

>>7623890
this.
lights momentum is applied to the 4th dimension, where the particle momentum is applied to the light as it travels through 3d space. the thing is that time us not perceivably "travelling", we are, and that is inferred by change in the observable environment described by the 3d measures. So then here momentum would be describing the tendency and degree of the rate of change that this particular light wave applies in its presence within the environment. From there the argument becomes a more philosophical one due to factors of will and measures of deviancy to a will that is dependent on many other factors outside of its domain.

Anonymous

Anonymous Thu 29 Oct 2015 21:51:29 No.7625095 Report Delete

Quoted By:

You took basic physics, so you recognize this, right?

$F = m a$

Right? Well Newton originally thought it as

$F = \frac{\Delta P}{t}$

Which makes sense. A force is just equal to the rate at which momentum changes.

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Is momentum valid?