What is the issue with using powers for momentum equations?

[taylordnz]

the equation mass X velocity = momentum works but my one of my classes we debated about the photon's momentum.

photons have 0 mass so should have 0 momentum but its momentum has been shown with new solar sail technology.

so to counteract this i have come up with a new theory

volume to the power of density X velocity = momentum

then i found out in practical application it dosent work so we only applied it to quantum physics. but because density is so small and its margin of error would over take it so its

volume X velocity = momentum

still the sum total will still be minute but because photons are so numerous they have over a large area enough momentum to push a solar sail.

do you agree with me?

 

[ahrkron]

So, you are saying that:

1. In a problem that involves classical mechanics, we should use momentum = mass x velocity, and
2. If quantum physics is involved, replace mass with volume

Is this the idea?

A first problem I see is that, when replacing mass with volume, the units will be different, so that all equations that use momentum will need to be changed also,

Second problem: How does nature know when to use each? In some cases, it is not going to be clear cut if the problem is involves quantum effects or not. It would be better to have one equation that applies to both cases, don't you think?

Third, does this really solve your problem? what is the volume of each photon? they may be many, but if each one has a zero volume, you are back in square one.

 

[russ_watters]

Fourth, we already have perfectly good equations describing light's momentum and explanations for them. No need to invent another. Try:

http://math.ucr.edu/home/baez/physics/Relativity/SR/light_mass.html

or

http://plus.maths.org/issue5/qm2/

 

[taylordnz]

okay, i got a bit muddled up over the weekend and i have refined it a bit.

momentum = mass X velocity

use it for classical physics and quantum physics but with the exception of massles particles (e.g photons) and then use

momentum = volume X velocity

 

[Integral]

Momentum must have units of

kg \frac m s

your quantity has units

\frac {m^4} s

clearly it is not, and cannot be used as, momentum.

As stated above there already exists an expression for the momentum of a photon which does not directly involve mass, it does have the correct units so is indeed momentum.

 

[quartodeciman]

The magnitude of a momentum can be got from this SR relation:

p = (E² - E₀²)^½/c

. This is just a rearrangement of a famous SR relation:

E² = E₀ + p²c²

.

{E is total energy at maximum speed, E₀ is total energy at rest (when that is possible, otherwise value zero), c is lightspeed, p is magnitude of momentum}


If E₀ is assumed to be zero, then we are left with

p = E/c

.

Nothing here tells us what the direction of a momentum vector should be.

Maybe we take it from

p = nh

for the case of deBroglie electron waves,
{n is wave vector of an electron with perfect momentum magnitude, and this vector has magnitude equal to reciprocal of deBroglie wavelength, h is Planck constant.}

Or maybe we must go to 4-vector momenta or energy-momentum tensors.

 

[Janitor]

As Integral points out, there are problems with your units.

In fact, as soon as you said, "volume to the power of density X velocity," I knew you were on the wrong track. Look at any equation that comes up in physics that has a base to a power, and you will find that the power itself is dimensionless (though the base may have physical units). For instance, if the power is the product of several variables, the units will cancel out in the product. The reason for this is that a power means multiplying the base number by itself a certain number of times. (Okay, this definition needs to be finessed since the power can be non-integer, but you get the drift.)

Sometimes you may see a power that looks at a glance like it is not dimensionless, but in that case you will find that there are "hidden" factors of 1 in it, meaning that units have been chosen so that certain constants such as c or h-bar are equal to one, and the constants are not explicitly shown.