Return to Answer

You may notice that the equations don't pass the test of dimensional analysis. Some factors are missing.

However, let me answer your question:

The reason why the acceleration never exceeds $g$ is that the dome is actually finite, it is truncated at the bottom. For too high values of $r$, your initial formula for $h(r)$ will actually exceed $r$ itself, and you won't be able to find points that are "deeper" below the summit than the total length from the summit along the dome. Well, the dome is actually truncated earlier than that.

See e.g. this presentation

http://philsci-archive.pitt.edu/3195/1/NortonDome.pdf

of the problem. Note that Norton's goal was study the behavior near $h=0$ and $r=0$ which he called an "example of indeterminism in Newtonian classical physics" because the particle may sit at the top for any amount of time, and suddenly freely decide and get rolling. That's why the truncation of the dome isn't important.

My more general comments about Norton's dome and its harmlessness in quantum physics:

http://motls.blogspot.com/2012/10/classical-physics-is-sometimes-more.html?m=1

In that article, I also calculated that the dome has to end up at the point where $dh/dr=1$ because it's the sine of an angle which implies $r_{\rm max}=(9/4)g^2=h_{\rm max}$; I also use an additional coefficient $K$ to make the formulae dimensionally correct.

You may notice that the equations don't pass the test of dimensional analysis. Some factors are missing.

However, let me answer your question:

The reason why the acceleration never exceeds $g$ is that the dome is actually finite, it is truncated at the bottom. For too high values of $r$, your initial formula for $h(r)$ will actually exceed $r$ itself, and you won't be able to find points that are "deeper" below the summit than the total length from the summit along the dome. Well, the dome is actually truncated earlier than that.

See e.g. this presentation of the problem. Note that Norton's goal was study the behavior near $h=0$ and $r=0$ which he called an "example of indeterminism in Newtonian classical physics" because the particle may sit at the top for any amount of time, and suddenly freely decide and get rolling. That's why the truncation of the dome isn't important.

Here are my more general comments about Norton's dome and its harmlessness in quantum physics.

In that article, I also calculated that the dome has to end up at the point where $dh/dr=1$ because it's the sine of an angle which implies $r_{\rm max}=(9/4)g^2=h_{\rm max}$; I also use an additional coefficient $K$ to make the formulae dimensionally correct.

You may notice that the equations don't pass the test of dimensional analysis. Some factors are missing.

However, let me answer your question:

The reason why the acceleration never exceeds $g$ is that the dome is actually finite, it is truncated at the bottom. For too high values of $r$, your initial formula for $h(r)$ will actually exceed $r$ itself, and you won't be able to find points that are "deeper" below the summit than the total length from the summit along the dome. Well, the dome is actually truncated earlier than that.

See e.g. this presentation

http://philsci-archive.pitt.edu/3195/1/NortonDome.pdf

of the problem. Note that Norton's goal was study the behavior near $h=0$ and $r=0$ which he called an "example of indeterminism in Newtonian classical physics" because the particle may sit at the top for any amount of time, and suddenly freely decide and get rolling. That's why the truncation of the dome isn't important.

My more general comments about Norton's dome and its harmlessness in quantum physics:

http://motls.blogspot.com/2012/10/classical-physics-is-sometimes-more.html?m=1

You may notice that the equations don't pass the test of dimensional analysis. Some factors are missing.

However, let me answer your question:

The reason why the acceleration never exceeds $g$ is that the dome is actually finite, it is truncated at the bottom. For too high values of $r$, your initial formula for $h(r)$ will actually exceed $r$ itself, and you won't be able to find points that are "deeper" below the summit than the total length from the summit along the dome. Well, the dome is actually truncated earlier than that.

See e.g. this presentation

http://philsci-archive.pitt.edu/3195/1/NortonDome.pdf

of the problem. Note that Norton's goal was study the behavior near $h=0$ and $r=0$ which he called an "example of indeterminism in Newtonian classical physics" because the particle may sit at the top for any amount of time, and suddenly freely decide and get rolling. That's why the truncation of the dome isn't important.

My more general comments about Norton's dome and its harmlessness in quantum physics:

http://motls.blogspot.com/2012/10/classical-physics-is-sometimes-more.html?m=1

In that article, I also calculated that the dome has to end up at the point where $dh/dr=1$ because it's the sine of an angle which implies $r_{\rm max}=(9/4)g^2=h_{\rm max}$; I also use an additional coefficient $K$ to make the formulae dimensionally correct.

You may notice that the equations don't pass the test of dimensional analysis. Some factors are missing.

However, let me answer your question:

The reason why the acceleration never exceeds $g$ is that the dome is actually finite, it is truncated at the bottom. For too high values of $r$, your initial formula for $h(r)$ will actually exceed $r$ itself, and you won't be able to find points that are "deeper" below the summit than the total length from the summit along the dome. Well, the dome is actually truncated earlier than that.

See e.g. this presentation

http://philsci-archive.pitt.edu/3195/1/NortonDome.pdf

of the problem. Note that Norton's goal was study the behavior near $h=0$ and $r=0$ which he called an "example of indeterminism in Newtonian classical physics" because the particle may sit at the top for any amount of time, and suddenly freely decide and get rolling. That's why the truncation of the dome isn't important.

You may notice that the equations don't pass the test of dimensional analysis. Some factors are missing.

However, let me answer your question:

The reason why the acceleration never exceeds $g$ is that the dome is actually finite, it is truncated at the bottom. For too high values of $r$, your initial formula for $h(r)$ will actually exceed $r$ itself, and you won't be able to find points that are "deeper" below the summit than the total length from the summit along the dome. Well, the dome is actually truncated earlier than that.

See e.g. this presentation

http://philsci-archive.pitt.edu/3195/1/NortonDome.pdf

of the problem. Note that Norton's goal was study the behavior near $h=0$ and $r=0$ which he called an "example of indeterminism in Newtonian classical physics" because the particle may sit at the top for any amount of time, and suddenly freely decide and get rolling. That's why the truncation of the dome isn't important.

My more general comments about Norton's dome and its harmlessness in quantum physics:

http://motls.blogspot.com/2012/10/classical-physics-is-sometimes-more.html?m=1