Landau Mechanics Chapter 2 Problem 1: Direction of Potential Energy?

[Rick16]

TL;DR

orientation of potential energy

Problem Statement

A particle of mass m moving with velocity v1 leaves a half-space in which its potential energy is a constant U1 and enters another in which its potential energy is a different constant U2.
Determine the change in the direction of motion of the particle.

Beginning of Landau's Solution

The potential energy is independent of the coordinates whose axes are parallel to the plane separating the half-spaces. The component of momentum in that plane is therefore conserved...
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The rest of the solution is straightforward, if only I understood the beginning. Why would the potential energy have this specific direction? The problem statement does not mention anything about it.

 

[Ibix]

It doesn't say that the potential has a direction (which wouldn't make sense for a scalar anyway), it says that if you pick coordinates with one axis parallel to the boundary, the potential's value does not depend on that coordinate. What does that tell you about any force it generates?

 

[Rick16]

It doesn't say that the potential has a direction (which wouldn't make sense for a scalar anyway), it says that if you pick coordinates with one axis parallel to the boundary, the potential's value does not depend on that coordinate. What does that tell you about any force it generates?

I understand the case with a single potential energy field, or perhaps I should rather say force field if I want to speak of the direction of the field. I can orient my coordinate axes such that the force points along one of the axes. The momentum of a particle moving in this field would then only be affected by the field along this coordinate axis and would be conserved in the plane of the other coordinate axes. But in the present case we have two potential energies, which do not necessarily have their gradients pointing in the same direction. Why would the boundary between the half-spaces be perpendicular to both forces?

 

[Ibix]

You don't seem to be reading the problem correctly. It states that the potential is $U_1$ in half of space and $U_2$ in the other half. Can you state where the force due to this potential is non zero?

 

[Matterwave]

It seems your confusion is not reading that the U1 and U2 are constants.

 

[GiorgioPastore]

A possible schematization for this problem is the x-dependent three-dimensional potential function
$$
U(x)=U_1+(U_2-U_1)\frac{\left(1+\tanh\left( \frac{x}{w} \right)\right)}{2}.
$$
Over an interval controlled by the width $w$, $U(x)$ goes from values very close to $U_1$ at negative values of $x$, to values very close to $U_2$ at large values of $x$. In the limit of $w \rightarrow 0$, $U(x)=U_1$ for negative $x$ and $U(x)=U_2$ for positive $x$.

For any positive width $w$, the force has only the $x$ component $F_x(x)$, always the same sign (negative if $U_2>U_1$ and positive if $U_1>U_2$, and it is concentrated in a region of width of a few $w$. Moreover, in the limit $w \rightarrow 0$, $F_x(x)=(U_2-U_1)\delta(x)$, where $\delta(x)$ is the Dirac delta.

I hope that such an example could help to clarify and solve the problem. Of course, there is nothing special about the continuous family of functions I have chosen. Every other family converging towards a step-function equal to $U_1$ in the negative $x$ half-space and to $U_2$ in the positive $x$ half-space will be equivalent.

 

[Rick16]

I don't understand the last post. I think the situation is much simpler. I forgot indeed to consider that the potential energies are constant. Since they are constant, there is no gradient in the fields and the force is zero on both sides of the partition. The only place where the potential energy changes is at the boundary, so the boundary is the only place where the force is not zero. But the force is perpendicular to the boundary and therefore the components of the particle's momentum parallel to the boundary are conserved. This makes it possible to write $v_1 sin\theta_1=v_2 sin\theta_2$. Fascinating. Thank you for your help.

 

[GiorgioPastore]

From my post, you can obtain the quantitative relation between $v_1$ and $v_2$. Without that, it is impossible to say whether $\theta_1<\theta_2$ or the other way around.

The final answer is very simple (it corresponds to equating the mechanical energy in the two half spaces). However, the passage through the family of continuous potentials allows us to get rid of any possible doubt connected with the presence of an impulsive force at the border between two half spaces of constant potential energy.

 

[Ibix]

I don't understand the last post.

The gradient isn't well-defined for the discontinuous potential in the problem, so if you want to consider forces you should really replace it with a smooth function that goes from $U_1$ at $+\infty$ to $U_2$ at $-\infty$ with a parameter ($w$ in the post above) that controls how abrupt the change is, then take a limit as that parameter goes to make the transition more and more abrupt. That's how you handle the issue of "what velocity change is caused by an infinite force applied for zero time?"

The other way to avoid the $\infty\times 0$ issue is to not talk about forces. That's why L&L simply observe that one component of momentum doesn't change. You can see that, whatever the behaviour of the force, it can't have a component parallel to the boundary, so it can only affect the component perpendicular to the boundary, and the energy change is well defined even if the force isn't.

 

[Rick16]

The gradient isn't well-defined for the discontinuous potential in the problem, so if you want to consider forces you should really replace it with a smooth function that goes from $U_1$ at $+\infty$ to $U_2$ at $-\infty$ with a parameter ($w$ in the post above) that controls how abrupt the change is, then take a limit as that parameter goes to make the transition more and more abrupt. That's how you handle the issue of "what velocity change is caused by an infinite force applied for zero time?"

The other way to avoid the $\infty\times 0$ issue is to not talk about forces. That's why L&L simply observe that one component of momentum doesn't change. You can see that, whatever the behaviour of the force, it can't have a component parallel to the boundary, so it can only affect the component perpendicular to the boundary, and the energy change is well defined even if the force isn't.

Thank you for making this clearer. I briefly thought about how I should go about finding the force from this momentary potential energy change, but a decided that in any case the force can only be perpendicular to the boundary, and that is enough for me (for now).

 

[Orodruin]

The gradient isn't well-defined for the discontinuous potential in the problem, so if you want to consider forces you should really replace it with a smooth function that goes from U1 at +∞ to U2 at −∞ with a parameter (w in the post above) that controls how abrupt the change is, then take a limit as that parameter goes to make the transition more and more abrupt. That's how you handle the issue of "what velocity change is caused by an infinite force applied for zero time?"

As a function, no. As a distribution it is absolutely well defined and gives the correct answer. Many times it works out surprisingly well treating the delta distribution as if it were a function - as long as you integrate it at some point. (In this case solving Newton’s equation of motion.)

It is many times much easier than introducing a function with the correct pointwise limit.

 

[Ibix]

Many times it works out surprisingly well treating the delta distribution as if it were a function - as long as you integrate it at some point.

This is what I was getting at with my second paragraph, yes. For example $\int \vec F\cdot d\vec x$ works across the boundary, and I think L&L implicitly use it to equate the potential difference at the step to the kinetic energy change of the particle.

It is many times much easier than introducing a function with the correct pointwise limit.

Definitely agree with this. I only mentioned the limit because modelling the potential as $\lim_{w\rightarrow 0}\tanh(x/w)$ was introduced by another poster, and the OP seemed to want a little more explanation of it.

 

[GiorgioPastore]

As a function, no. As a distribution it is absolutely well defined and gives the correct answer. ...

Although calling the Dirac delta "distribution" is quite widespread in the Physics community, I prefer to call it a generalized function, reserving the term "distribution" for the corresponding functional (as in Schwartz or in the Gel'fand papers and books).