Understanding Momentum: Definition, Intuition, and Properties

[Studiot]

Can momentum have any useful applications in a system where objects/particles move but are not liable to collide/interact at all? Need it be defined in the first place in such a system?

Yes of course, the interaction of a particle (or stream of them) and a field changes the momentum of the particle(s).

There is a simple secondary school experiment which discusses this in terms of the interaction of alpha, beta and gamma rays with a magnetic field.

Do you know this?

 

[sophiecentaur]

Are you suggesting that momentum is not exchanged or changed then under those circumstances? Is that not where the idea of photon interaction comes in? Alpha particle moves to the right - magnet moves to the left. If not then we have reactionless propulsion. (And you could carry your own magnetic field with you yet still get a force)

 

[Studiot]

Are you suggesting that momentum is not exchanged or changed then under those circumstances?

Again I said no such thing.

Indeed quite the opposite.

a field changes the momentum of the particle(s).

However Hassan specifically asked about the situation where the particles

objects/particles move but are not liable to collide/interact at all?

A single or stream of positive or negative particles passing through a magnetic or electric field do not collide or interact with other particles.

They interact with the field (which can have no momentum of its own) however.

Which is what I said.

 

[sophiecentaur]

So where does the momentum come from? If the particle's motion is changed then there has to be some. The concept of a Field is an artifice, because all fields are there because of the presence of some particle(/s) somewhere -just like the Earth is there for a bouncing ball but we 'ignore it'. In all cases there is a change in momentum which may be regarded as being magicked away (not very satisfactory). But if you regard the change in direction as being due to a photon interaction you're back in business. The deviated particle will have launched its own photon i.e. perturbed the Field.

 

[Delta Kilo]

What does "boost" mean, please?

Boost is a Special Relativity term. It just means changing observer's point of view from one coordinate system to another, moving at a constant speed relative to the first one. It is a fancy kind of rotation of 4D space-time coordinate system in X-T plane. During the boost, spatial and time coordinates get mixed up a little, just like X and Y get mixed up when rotating in XY plane.

I got into this thread in response to a bald statement that the two are "the same".

They are indeed the same but in different dimensions :)
Four-momentum 4-vector can be loosely interpreted as a "rate of flow of mass through 4D space-time" with respect to its own "proper time". Thus momentum (p¹,p²,p³) becomes a flow of mass through space and the energy term p⁰ is a "flow of mass through time". Kinetic energy turns out to be Lorentz factor correction to the total relativistic energy in the limit v<<c.

PS: Relativistic Caveat: This is relativistic treatment of momentum. There is no such obvious connection between momentum and KE in Newtonian physics.

 

[sophiecentaur]

I'm fine with that but that Relativistic Caveat should be writ large and clear for the benefit of all snooker players and others.

 

[Islam Hassan]

Can you think of an example of such a system and how it could be of interest in Physics?

Not really no, I admit it was more an epistemological question than a purely physical one. There may be limited, closed systems of such a nature of which I do not know however.

IH

 

[Studiot]

But if you regard the change in direction as being due to a photon interaction you're back in business. The deviated particle will have launched its own photon i.e. perturbed the Field.

And if you don't?


I have been consistently dealing classically with Hassan's query.

Classically momentum is a vector so yes when a beta particle undergoes a change of direction (but not magnitude) due to its interaction with a magnetic field edit: I am considering this in terms of force (the Lorenz force) = rate of change of momentum, not particle exchange.
Yes in modern terms (and strangely in ancient science and religion too) anything any particle does anywhere in the universe affects all other particles in the universe to some extent or other.

But in classical terms we isolate a section of the universe and consider what happens within that microcosm.

That is what I understand Hassan's question to mean viz

Can momentum in a stream of beta particles be changed without the beta particles bumping into themselves or any other material object.

To which the answer is an unequivocal classical yes.

If I have misunderstood that question then I need to think again.

 

[Dadface]

They are indeed the same but in different dimensions :)
Four-momentum 4-vector can be loosely interpreted as a "rate of flow of mass through 4D space-time" with respect to its own "proper time". Thus momentum (p¹,p²,p³) becomes a flow of mass through space and the energy term p⁰ is a "flow of mass through time". Kinetic energy turns out to be Lorentz factor correction to the total relativistic energy in the limit v<<c.

PS: Relativistic Caveat: This is relativistic treatment of momentum. There is no such obvious connection between momentum and KE in Newtonian physics.

Where in your references does it state or imply that they(energy and momentum)"are indeed the same"? Early on in the Wiki article it is given that Po=E/c where P and E are different and with different units.

 

[sophiecentaur]

And if you don't?


I have been consistently dealing classically with Hassan's query.

Classically momentum is a vector so yes when a beta particle undergoes a change of direction (but not magnitude) due to its interaction with a magnetic field edit: I am considering this in terms of force (the Lorenz force) = rate of change of momentum, not particle exchange.
Yes in modern terms (and strangely in ancient science and religion too) anything any particle does anywhere in the universe affects all other particles in the universe to some extent or other.

But in classical terms we isolate a section of the universe and consider what happens within that microcosm.

That is what I understand Hassan's question to mean viz

Can momentum in a stream of beta particles be changed without the beta particles bumping into themselves or any other material object.

To which the answer is an unequivocal classical yes.

"Material Object"? When does anything hit one of those? It's all fields or photons, depending on what's of interest at the time. When a charge accelerates, it radiates EM as a photon / photons, doesn't it? That photon will, eventually, effect the system producing the original field. Where is the difference between that and what you refer to as a particle particle interaction? Afaics, it's only a matter of relative distance.

 

[jetwaterluffy]

Yeah, momentum is how hard something hits you ... that's not a bad way to think of it.

If it's moving faster, it hits you harder. If it's heavier, it hits you harder. If it's both heavier and faster, you really should get out of the way.

No, I wouldn't call that a good analogy. A bullet "hits you harder" than an ocean liner, but an ocean liner has vastly more momentum. A better analogy would be "if you are trapped in front of a wall, momentum is how much will an object will crush you.".

Is it correct to say that momentum is of no use unless we are talking about collisions of a sort?

Can momentum have any useful applications in a system where objects/particles move but are not liable to collide/interact at all? Need it be defined in the first place in such a system?

IH

You could say the same thing about velocity.

 

[sophiecentaur]

No variable is much use except to describe some sort of relationship between two things.

Momentum, like Electrical Resistance, is one of those quantities that involve two other, more familiar, quantities. People wast a lot of time trying to put them into 'more simple' terms: eg "Resistance is how hard you have to try to push a current blah blah".

If you just stick with the mass times Velocity definition (or h/λ) and use it enough times in real situations then you can stop trying to find a chatty descriptions for momentum.

 

[Delta Kilo]

Where in your references does it state or imply that they(energy and momentum)"are indeed the same"? Early on in the Wiki article it is given that Po=E/c where P and E are different and with different units.

Different units for time and space are there for our convenience and historical reason. c is just a conversion factor between units, you'll get the same mess if you measure X in feet and Y in meters. But the dead giveaway is E and p get mixed up during boosts, for example in one coordinate system you see an object with energy E and momentum p=(px,py,pz), but when you look at the same object from another coordinate system, moving with velocity v along x direction, you see linear combinations of those:
E'/c = γE/c - βγp_x, p_x' = γp_x - βγE/c, p_y'= p_y, p_z'= p_z, where β=v/c, γ=1/√(1-β²) are Lorentz factors.
This can also be written as:
E'/c = E/c cosh θ - p_x sinh θ, p_x' = p_x cosh θ - E/c sinh θ, p_y'= p_y, p_z'= p_z
Compare this with spatial rotation in XY plane:
E' = E, p_x' = p_x cos α - p_y sin α, p_y' = p_y cos α + p_x sin α, p_z' = p_z.
This just goes to show that E and px,py,px are not independent quantities but components of the same thing.

 

[Delta Kilo]

No variable is much use except to describe some sort of relationship between two things.

Momentum, like Electrical Resistance, is one of those quantities that involve two other, more familiar, quantities. People wast a lot of time trying to put them into 'more simple' terms: eg "Resistance is how hard you have to try to push a current blah blah".

Not really. Its most important feature, namely momentum conservation law, does not follow from p=mv. Momentum survives in places where the both mass and velocity go south, like momentum of a photon. In fact, Newton mechanics is derived from conservation laws and not the other way around.

For example, you wouldn't say that force is nothing more than just a mass times acceleration, as in F=ma. Force is a useful concept on its own, used in many places where neither m nor a is well-defined, like force of a compressed spring for example. But... F=ma is just a time derivative of p=mv, so it must have the same fundamental significance.

 

[James_Harford]

One way to intuitively disentangle momentum from energy of motion is to consider a moving mass (e.g. vehicle) brought to a stop by a constant force (e.g. brakes). Then time to stop "measures" momentum, and distance to stop "measures" energy.

More specifically, for a given force, different masses with the same momentum take the same time to stop. Different masses with the same energy take the same distance to stop. The concepts are different, but have a pretty symmetry when thus described.

 

[James_Harford]

The concepts are different, but have a pretty symmetry when thus described.

I should mention that the above intuitive description is correct even for relativistic masses.

 

[sophiecentaur]

Newton mechanics is derived from conservation laws and not the other way around.

.

Thats absolutely fine but people ask many questions which are totally in the context of Newtonian Laws. There are perfectly reasonable and helpful answers to those questions which are also in terms of Newton.
I wish people would actually read the questions and interpret them in the context of elementary Science first. Many of the questions are posted by School and College students and they are based on the Newtonian stuff they learn on their courses. Can it really help them if thy are hit with all that advanced stuff, involving Vectors and Relativistic niceties?

How many of the people who supply these confusing answers could, hand on heart, say that their answer would have been any use to them when they were 17 years old and had missed some point that their teacher had made earlier on that day?

When youv'e been told about dimensional analysis as a way of checking that you got an A level Physics formula right you will only be confused when someone glibly tells you that, in another frame, the dimensions are all wrong.

Perhaps we should have a more strict system for the placing of posts so that people who want 'the easy answer' are protected from the 'clever answers' - at least at the start of their thread.

 

[sophiecentaur]

One way to intuitively disentangle momentum from energy of motion is to consider a moving mass (e.g. vehicle) brought to a stop by a constant force (e.g. brakes). Then time to stop "measures" momentum, and distance to stop "measures" energy.

More specifically, for a given force, different masses with the same momentum take the same time to stop. Different masses with the same energy take the same distance to stop. The concepts are different, but have a pretty symmetry when thus described.

That is a good, down-to-earth answer and I'm sure it's the sort of thing the original questioner was after. I hope he's still with us and can read it.

 

[Islam Hassan]

That is a good, down-to-earth answer and I'm sure it's the sort of thing the original questioner was after. I hope he's still with us and can read it.

Agreed 100%, very elegant way to put it, thanks.

IH

 

[Studiot]

Spot on, James.

This is easy to prove with high school Physics.

For energy, E

\begin{array}{l}<br /> E = \frac{{m{v^2}}}{2} = \int_0^x {Fdx} = F\int_0^x {dx} \\ <br /> E = Fx \\ <br /> \end{array}

Where the kinetic energy is lost as work against the Force F.

So if F is the same constant force for Energy E₁ and E₂ then if E₁ = E₂ ; x₁ =x₂

Alternatively the formula also shows that if F is not constant, but has the same variation with distance for two cases then the work done is still the same so the distances are still the same. However we have to find an expression for this variation to perform the integration for actual numbers.

For momentum, p

Force is the rate of change of momentum.

\begin{array}{l}<br /> F = \frac{{dp}}{{dt}} = {\rm{constant}} \\ <br /> {\rm{dp = Fdt}} \\ <br /> \int_0^p {dp} = \int_0^t {Fdt} = F\int_0^t {dt} \\ <br /> p = Ft \\ <br /> \end{array}

We have a similar calculation to the energy one and a similar line of reasoning shows that for a constant force

If p₁=p₂ then t₁=t₂

and also that for a variable force the stopping times are still the same, this in this case so long as the variation with time is the same, the stopping times are still equal. Again we need an expression to perform the numerical calculation.

 

[Dadface]

This just goes to show that E and px,py,px are not independent quantities but components of the same thing.

In some contexts "not independent but components of the same thing" perhaps but they are not the same.Momentum is momentum and energy is energy.
Please read and take into account sophiecentaurs post 47 above.

 

[_PJ_]

Momentum is the massive component of the kinetic energy of a body when relative motion is considered.

 

[Studiot]

Momentum is the massive component of the kinetic energy of a body when relative motion is considered

Would you like to elaborate on this?
I sort of guess that you are identifying the kinetic energy as the dot product of two vectors viz the momentum vector and the velocity vector. But where does the factor of one half come in?

 

[_PJ_]

The 'other half' (I'm, guessing you mean as in KE=0.5*m*v*v) is for the remaining velocity contribution. It's from the integration of v

 

[Studiot]

Perhaps a derivation?

 

[Delta Kilo]

In some contexts "not independent but components of the same thing" perhaps but they are not the same.Momentum is momentum and energy is energy.

I was answering specific question. And I clarified it as "same but in different dimensions". And I did add Relativistic Caveat as requested, didn't I?

Please read and take into account sophiecentaurs post 47 above.

I understand the need to keep things simple, but I'm against dumbing them down. For, example this "down-to-earth" explanation looks sort-of-alright. But it only works in a very specific case where the force is constant and is known in advance. This case is in fact very rare as we don't usually know the forces and they tend not to stay constant. The example with vehicle is in fact incorrect because braking force depends on the weight of the car, so if you try measuring momentum of the car with the same initial speed but different loads you will not get expected results.

And I can't think of any typical school or college-type problem where this kind of "intuition" would be helpful. In fact, most of the time when you deal with momentum, the forces are unknown, like the proverbial billiard balls or rocket equation, and you just use conservation law directly.

On the other hand, it would be very nice to grasp the connection between momentum conservation and Newton's laws of motion as soon as possible. It doesn't require anything beyond school programme to see the that p=mv and F=ma are basically telling us the same thing, but it makes everything so much neater.

It is also worthwhile to mention that momentum conservation comes directly from translational symmetry, i.e. the fact that laws of physics do not depend on position in space (without going into details of Noether theorem of course). It just wraps it up nicely.

 

[sophiecentaur]

I agree with most of what you've written except the phrase "dumbing them down". It isn't dumbing down to establish a good understanding of things at a very worldly level. Believe it or not, there are many people for whom SR is and will always be MAGIC. They may be able to reiterate the consequences of traveling at near c but the associated sums and beyond are just too much. Nontheless, many of these same people can do calculations involving colliding cars and snooker balls.
The A level Mechanics level is a kind of 'base camp' from which a few hardy souls can scale the mountain.
Talking of dumbing down- I used to kick my Sons' arses when they would wind me up with "Momentum is speed times weight". They still do, at the age of 40.

 

[James_Harford]

I understand the need to keep things simple, but I'm against dumbing them down. For, example this "down-to-earth" explanation looks sort-of-alright. But it only works in a very specific case where the force is constant and is known in advance. This case is in fact very rare...

It is not really "dumbed down" if it is correct in its domain. Nor is it just a matter of keeping things simple, but of starting out simple if it can be properly related to everyday experience, and then elaborating step by step upon that insight.

For example, since the intuitive description happens to occur along a line, it can be formulated as (1) Δp=FΔt and (2) ΔE = FΔx, where F is constant. Since Δt is any nonzero value. This includes as a special case,

(1) dp = Fdt
(2) dE = Fdx.

which answers your first objection, and so on...