What is the issue with using powers for momentum equations?

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In summary: So, for instance, momentum could technically have a power of 9, but because c is equal to h-bar, it will be shown as having a power of 8 in most cases.Now, if we were to take the power of volume to the power of density, we would be multiplying each number by itself a lot more than 9, meaning that the equation would become very messy and hard to read. Overall, your idea is intriguing, but it would be better to focus on one equation that covers both classical and quantum mechanics.
  • #1
taylordnz
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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?
 
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  • #2
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.
 
  • #3
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/
 
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  • #4
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
 
  • #5
Momentum must have units of

[tex] kg \frac m s [/tex]

your quantity has units

[tex] \frac {m^4} s [/tex]

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.
 
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  • #6
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.
 
  • #7
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.
 

FAQ: What is the issue with using powers for momentum equations?

What is the new equation for momentum?

The new equation for momentum is p = mv, where p represents momentum, m represents mass, and v represents velocity.

How does the new equation for momentum differ from the previous equation?

The previous equation for momentum was p = mass x velocity, which did not include the variable for velocity. The new equation takes into account the object's velocity, making it a more accurate representation of momentum.

What is the significance of the new equation for momentum?

The new equation for momentum allows for a more comprehensive understanding of an object's momentum, as it takes into account both its mass and velocity.

Can the new equation for momentum be applied to all objects?

Yes, the new equation for momentum can be applied to all objects, regardless of their mass or velocity. It is a fundamental equation in physics and is used to describe the motion of objects.

How does the new equation for momentum relate to Newton's laws of motion?

The new equation for momentum is closely related to Newton's laws of motion, specifically the second law which states that the net force acting on an object is equal to its mass multiplied by its acceleration. The new equation for momentum can be derived from this law, making it an important concept in understanding motion and forces.

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