Why do we need to consider the effective potential in central force problems?

[fayled]

When dealing with central forces, we can find the energy of a particle under the influence of such a force to be
E=0.5m(dr/dt)²+(L²/2mr²)+U(r).
We define the effective potential as the sum of the last two terms.

In my previous experience of dealing with energy considerations to predict motion, analysis of the potential function U(r) has been sufficient. But with central forces it seems necessary to consider Ueff(r). Why is this - why can't I just consider U(r) alone? I can't really find anywhere that properly explains why we need to use it. Thanks.

 

[BruceW]

you can just consider U(r) alone if you want. But then, what is the kinetic energy?

edit: I'm guessing you know this. what I'm trying to say is that it's completely fine to just say the potential is U(r). But because of the special form of the term L²/2mr² it can be very convenient to imagine it as being a term in the 'effective potential'. So, a more interesting question is: what is it about L²/2mr² that makes it suitable for being imagined as an 'effective potential' ?

 

[fayled]

you can just consider U(r) alone if you want. But then, what is the kinetic energy?

edit: I'm guessing you know this. what I'm trying to say is that it's completely fine to just say the potential is U(r). But because of the special form of the term L²/2mr² it can be very convenient to imagine it as being a term in the 'effective potential'. So, a more interesting question is: what is it about L²/2mr² that makes it suitable for being imagined as an 'effective potential' ?

The kinetic energy would just be the other two terms?

I'm not too sure, maybe because the asymptotes for the 1/r^2 make analysis a lot simpler?

 

[BruceW]

yes, the kinetic energy are just the other two terms. so the kinetic energy is 0.5m(dr/dt)²+(L²/2mr²)
Now, generally 'L' will be a function of velocity and position. But luckily, in many cases angular momentum is conserved, so 'L' is just a constant.
So, anyway, your two kinetic energy terms are 0.5m(dr/dt)² and L²/2mr² What is the main difference between these two terms? what are they functions of? And this should give you a hint as to why L²/2mr² is often thought of as a term in an effective potential. (also, think of what the actual potential is a function of).

edit: I guess it's true that the asymptotes for the 1/r^2 make analysis at $r \rightarrow \infty$ a lot simpler. But there is a more general reason for putting the L²/2mr² term as part of an effective potential.

 

[fayled]

yes, the kinetic energy are just the other two terms. so the kinetic energy is 0.5m(dr/dt)²+(L²/2mr²)
Now, generally 'L' will be a function of velocity and position. But luckily, in many cases angular momentum is conserved, so 'L' is just a constant.
So, anyway, your two kinetic energy terms are 0.5m(dr/dt)² and L²/2mr² What is the main difference between these two terms? what are they functions of? And this should give you a hint as to why L²/2mr² is often thought of as a term in an effective potential. (also, think of what the actual potential is a function of).

edit: I guess it's true that the asymptotes for the 1/r^2 make analysis at $r \rightarrow \infty$ a lot simpler. But there is a more general reason for putting the L²/2mr² term as part of an effective potential.

Ok so L²/2mr² is a function of position hence it's use as part of the effective potential, whereas the other KE term is a function of the component of speed pointing radially outwards. That makes sense.

So if we consider say a particle in 1D under the influence of a conservative force, we can simply consider it's potential function to understand it's motion - the useful thing about the potential function is that it is a function of position so the energy considerations tell us where the particle can go. This extends to central foces in 2D, but we add in the extra term which is a function of position to give us a full picture of where the particle can go. Is that the right idea?

The next problem then, is that you said I could just consider the potential alone earlier - but if I did this I would miss out on some important things (as expected otherwise the extra term would be of no use), i.e for gravity I wouldn't even see the possibility of any kind of closed orbit just by looking at the potential function alone, so would there be any way to get around that?

 

[BruceW]

you got it :) yeah, in 2d, that term needs to be there, whether you think of it as part of the kinetic energy, or as part of the effective potential energy. So when I said you can think of the potential alone as U(r), if you did this, then you would have to think of the kinetic energy as 0.5m(dr/dt)²+(L²/2mr²) Which is perfectly fine. It is just nicer to think of the L²/2mr² term as part of an effective potential, since it is a function of position, as you said.

 

[fayled]

you got it :) yeah, in 2d, that term needs to be there, whether you think of it as part of the kinetic energy, or as part of the effective potential energy. So when I said you can think of the potential alone as U(r), if you did this, then you would have to think of the kinetic energy as 0.5m(dr/dt)²+(L²/2mr²) Which is perfectly fine. It is just nicer to think of the L²/2mr² term as part of an effective potential, since it is a function of position, as you said.

Great, thanks for your help :)

 

[BruceW]

no worries!