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My teacher gives me an equation for the position of a particle to analyse the equation by the dimensional analysis method:

The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position:

$$ s = ka^mt^n $$

I replace the position $s$ (represent of the position) by $[L]^3$ because I think we can't describe a particle position by just one dimension because I assume the particle has a free movement, but my teacher told me that is wrong and now I just want to know why?

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    $\begingroup$ $[L]^3$ is length cubed...otherwise known as volume. $\endgroup$ – Triatticus Sep 7 '18 at 16:38
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Position is indeed 3-dimensions. But you have some problems.

Your teacher has clearly only looked at one dimension. There is nothing wrong with that. Single dimensional motion is possible for a particle. It can be moving along a line, so only one dimension of the position is changing.

You wrote $[L]^3$. If you want to write a 3-dimensional position, you need to write it as a vector. Something like $(s_x,s_y,s_z)$. This would be 3 numbers arranged in order and giving the x, y, and z locations of the particle. Then you would need three equations for the acceleration of the particle in each direction. Your teacher's equation could then be rewritten so.

$$ s_x = k a^m t^n $$ $$ s_y = 0 $$ $$ s_z = 0 $$

And you now have to look at the very same problem, only you have made things more complicated by adding the subscripts. Might as well ignore everything except $s_x$ and just call it $s$.

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The equation is for the distance moved by the particle, and the distance moved is measured in metres. So the dimensions of the left hand side are just metres i.e. $[L]$ not $[L]^3$.

You are quite correct that position is a vector. Suppose we are using $x$, $y$ and $z$ axes, then we define $\mathbf i$, $\mathbf j$ and $\mathbf k$ as the unit vectors along the $x$, $y$ and $z$ axes respectively ($\mathbf i$, $\mathbf j$ and $\mathbf k$ are the standard symbols for these three vectors). In that case the position vector $\mathbf s$ would be given by:

$$ \mathbf s = s_x \mathbf i + s_y \mathbf j + s_z \mathbf k $$

where $s_x$, $s_y$ and $s_z$ are the components of the vector along the three axes. This is usually abbreviated to:

$$ \mathbf s = (s_x, s_y, s_z) $$

which is the way we normally write the coordinates of a point on a 3D graph.

Since $\mathbf s$ is the sum of the three vectors $s_x \mathbf i$, $s_y \mathbf j$ and $s_z \mathbf k$ the dimensions of $\mathbf s$ are the same as the dimensions of the three vectors. But if we take for example $s_x\mathbf i$, the magnitude of this is just the distance measured along the $x$ axis. So the dimensions of $s_x\mathbf i$ are just $[L]$. Likewise the dimensions of $s_y\mathbf j$ and $s_z\mathbf k$ are also just $[L]$. Since $\mathbf s$ is the sum of the three that means the dimensions of $\mathbf s$ are:

$$ \mathrm{Dim}(\mathbf s) = [L] + [L] + [L] = [L] $$

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