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Walter Lewin's first lecture (at 22:16) analyzes the time $t$ for an apple to fall to the ground, using dimensional analysis. His reasoning goes like this:

It's natural to suppose that height of the apple to the ground ($h$), mass of the apple ($m$), and the acceleration due to gravity ($g$) may impact (pardon the pun) the time it takes for the apple to reach the ground. Then $$t \propto h^\alpha m^\beta g^\gamma.$$ On both sides, the units must be equivalent, so $$[T] = [L]^\alpha [M]^\beta \left[\frac{L}{T^2}\right]^\gamma = [L]^{\alpha + \gamma} [M]^\beta [T]^{-2\gamma}.$$ Therefore, $$1 = -2\gamma, \quad \alpha + \gamma = 0, \quad \beta = 0.$$ Solving, we have $$\gamma = -\frac{1}{2}, \quad \alpha = \frac{1}{2}, \quad \beta = 0.$$ Then we conclude $t = k\sqrt{\frac{h}{g}}$, where $k$ is some unit-less constant.

Lewin concludes that the apple falls independently of its mass, as proved in his thought experiment and verified in real-life. But I don't agree with his reasoning.

Lewin made the assumption that $k$ is unit-less. Why could he come to this conclusion? After all, some constants have units, like the gravitational constant ($G$).

Why isn't the following reasoning correct? The constant ($k$) has the unit $[M]^{-z}$; therefore, to match both sides of the equation, $\beta = z$. So indeed, the mass of the apple does impact its fall time.

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    $\begingroup$ No units are needed on "k" to maintain proper dimensions. Arbitrarily adding units to "k" means that you must adjust a variable on the other side of the equal sign to compensate. This process could proceed ad infinitum, and it violates Occam's razor because it it not the simplest answer that matches the problem. In addition, it has been experimentally verified that in the absence of drag forces (e.g., air drag), mass does not affect how fast an object falls. $\endgroup$ – David White Nov 8 '16 at 4:02
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    $\begingroup$ There already is a dimensionful constant. It's called $g$. $\endgroup$ – Emilio Pisanty Nov 8 '16 at 14:07
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    $\begingroup$ The mass of the apple affects the value of the fall time, not its unit. $\endgroup$ – chepner Nov 8 '16 at 14:37
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    $\begingroup$ The key word in the story is "suppose". Its right that you ask why those particular variables were chosen (and by implication others ignored). But, given that this is the first lecture, it should be obvious that it's a set up job, and these are the right variables to use, and like a magician you are being directed to look at the fun parts of balancing dimensional powers... $\endgroup$ – Philip Oakley Nov 8 '16 at 23:51
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    $\begingroup$ Please stop writing clickbait titles and instead focus on describing the question accurately. There is no call to mention Lewin in the title in this instance. $\endgroup$ – Emilio Pisanty Nov 12 '16 at 3:37
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Your example shows a fundamental idea: even though the units agree, this does not mean that the resulting equation is a law of physics. This is why physicists only 'accept' laws that have been tested experimentally.

This idea is nicely explained in the following XKCD comic:
My hobby: abusing dimensional analysis

Here, we get a more extreme example than just 'changing the units of $k$'. It turns out we could arbitrarily add different quantities to an equation, and end up with a new equation that is completely valid. This does not mean that this actually makes any sense! Your new 'law' needs to be validated with experiment, and as you can see in the comic, a single experiment may not be enough.

Dimensional analysis, then, is not used to derive new laws of physics through pure reasoning alone. Instead, your professor already knew, through whatever reason, that $t\propto h^\alpha m^\beta g^\gamma$. Even that is already a bit of a leap of faith - there is nothing+ that keeps you from assuming $t\propto \ln h$. To quote your own post:

[...] as proved in his thought experiment and verified in real-life.

Instead, then, you should see dimensional analysis as a very useful tool, part of a larger toolset to derive certain laws and equations. For example, if you derive an equation through whatever other means, you can use dimension analysis to check whether your new equation is possible at all. Or, in the case of this professor, you may have a general idea where your equation should be going, and you can then use dimensional analysis to get a reasonable idea of the final form of that equation. (Note: this may save you in a closed-book exam one day).

+That's not actually true. There are good reasons why you will probably never see $\ln h$, but that's for some other time.

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    $\begingroup$ I think there is something to prevent, for instance $t \propto \ln h$, and it's exactly dimensional analysis: $\ln h$ doesn't have any sane dimensions (just expand $\ln$ as a power series). So, in fact it would have to be the case that $t \propto \ln (h/h_0)$ for some suitable $h_0$: there must be some secret constant in fact. $\endgroup$ – tfb Nov 8 '16 at 9:25
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    $\begingroup$ @tfb I knew I should have added an asterisk there, but I felt it would clutter the answer. But of course, there are many many reasons why that cannot be done. $\endgroup$ – Sanchises Nov 8 '16 at 9:30
  • $\begingroup$ So, let me get this straight: dimensional analysis can't be used unless you have some initial "bias" (i.e. a baby with Einstein's intellect wouldn't be able to figure this out because he hasn't had enough real-life experience)? $\endgroup$ – Fine Man Nov 9 '16 at 20:29
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    $\begingroup$ @SirJony I don't think real-life experience matters that much - Einstein never sat on a train at near-light-speed. Rather, you need to see dimension analysis in a whole framework of tools: experiments, calculus, cause-effect reasoning, other equations that describe the same phenomenon (like checking energy balances, etc). Dimension analysis in itself 'proves' nothing - it only narrows down the paths you can choose towards a model or physical law. $\endgroup$ – Sanchises Nov 10 '16 at 18:43
  • $\begingroup$ @sanchises -- Good point. :) So, is dimensional analysis a tool more often for verification rather than discovery? $\endgroup$ – Fine Man Nov 10 '16 at 19:29
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To do dimensional analysis, we first list all quantities that could affect a quantity of interest. It's true that this is a tricky business. You're totally right to wonder, for example, why we don't include the gravitational constant $G$, or perhaps the mass of the Earth $M$. In particular, if we include $M$, the argument immediately fails, because we can have any power of $m/M$ in the answer.

Prof. Lewin gets around this by a mild cheat. While the values of $M$ and $G$ do affect the time, we know that they only affect the result through the combination $$g = \frac{GM}{R_E^2}$$ where $R_E$ is the radius of the Earth. However, by assuming this, we're sort of assuming the conclusion, so it's a slight cheat.


However, besides the issue of $M$ and $G$, we can rule out other 'dimensionful constants', so the argument goes through. For example, you could say the constant $k$ could have something to do with the Planck mass, but we know it can't, because that involves $\hbar$ and quantum mechanics is irrelevant here. Similar physical reasoning can pare down the number of quantities enough without cheating in most situations.

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    $\begingroup$ Fantastic answer! I do think calling it "a mild cheat" (while I agree in spirit) may distract from there being an important lesson: using the acceleration due to gravity, observed to be constant, can be more useful than gravitational force---which is phenomenologically more complicated. I think this type of thing comes up a lot... obviously there are numerous paths to solution, but some (e.g. via acceleration) end up simpler (or start nearer the answer already). $\endgroup$ – DilithiumMatrix Nov 8 '16 at 1:01
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    $\begingroup$ @SirJony The resolution is, in reality, there's no such thing as a "dimensionful constant", ever. There are only physical quantities. Then we have to ask, which physical quantities could be relevant. We rule out the mass of the Earth by cheating, and all other masses (like the Planck mass) by physical reasoning. $\endgroup$ – knzhou Nov 8 '16 at 1:16
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    $\begingroup$ @knzhou --But $G$ has the dimensions $[\frac{N m^2}{kg^2}]$, which seems to contradict (my understanding of) "there's no such thing as a 'dimensionful constant'". $\endgroup$ – Fine Man Nov 8 '16 at 2:52
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    $\begingroup$ @knzhou -- OK, makes sense. But, then why can't my $k$ be some physical quantity that we don't know? $\endgroup$ – Fine Man Nov 8 '16 at 2:56
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    $\begingroup$ @knzhou -- Oh, I think I get it! So, essentially it could indeed exist (just like Lewin guessed height, mass, etc. did), but experimental evidence shows otherwise? $\endgroup$ – Fine Man Nov 8 '16 at 3:07
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By accepting the presence of "acceleration due to gravity" $g$, Prof Lewin has already postulated that mass does not come into it. In other words - he has added certain knowledge about the physics of the system (namely - if there is such a thing as acceleration due to gravity, and that number is a constant, that means all things accelerate the same way. And if they accelerate the same way, there is no dependence on mass). This circumvents the question "can mass of the object affect the time to fall?".

If you want to answer that question, you CANNOT, in your dimensional analysis, assume that the answer is "no". Similarly, you cannot assume Newton's law of gravity (because if you did, you would come to the same conclusion).

This leaves you in a position of starting with a different set of possible inputs. The time to fall may depend on the mass of the earth $M$, the mass of the object $m$, the height $h$, and "something else that brings time into the equation", $X$. I call it $X$ because we don't have any information about its role. All we know is that $X$ must at least include dimensions of $T$ (with unknown power) and $L$ (because the answer has to have units of time - which we don't have right now - and can't have units of length - which we do have right now). And possibly $M$.

And that leaves us with an unsolvable mess - three equations (four, if you assume symmetry in mass) with seven unknowns (we don't know the power of X, and we don't know the relative sizes of the coefficients of $M$,$L$ and $T$ in X). Dimensional analysis doesn't always provide us with the answer to every problem - we do actually have to insert some actual physical knowledge to make progress.

If we just postulate that $G$, the gravitational constant, has "something to do with it" as well, then we have dimensions for $X$ ($L^3 M^{-1} T^{-2}$) and we are left with just four unknowns. We can first try to find the force on the apple:

$$F = k~M^a m^b G^c h^d$$

From which we get this equation:

$$ M^1 L^1 T^{-2} = k~M^a M^b L^{3c} M^{-c} T^{-2c} L^d$$

which leads to equations in $L$, $M$ and $T$:

$$T: -2 = -2c\\ M: 1 = a + b - c\\ L: 1 = 3c + d$$

Finally, we know that the force is symmetrical - so we should get the same answer when we swap $a$ and $b$, leading to a fourth equation

$$a = b$$

We can now solve this to get respectively $$c = 1\\ d = -2\\ a = b = 1$$

In other words - we find Newton's law of gravity (to within a constant).

And once we see that the force is proportional to the mass, we know the acceleration is constant - and then the argument can proceed normally.

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    $\begingroup$ This seems like a similar cheat. You can't derive Newton's law of gravitation this way because the units of $G$ have to come out of nowhere. The logic really goes the opposite direction. $\endgroup$ – knzhou Nov 8 '16 at 16:46
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    $\begingroup$ @knzhou - yes. The postulate "this thing called G has something to do with it" essentially begs the question - that is, it presupposed the answer in order to arrive at the answer. I just wanted to point out that without some physical insight, the problem is indeed not solvable - but that you can get there with this one additional piece of information (which is a bit weaker than postulating the existence of $g$, although one follows easily from the other). $\endgroup$ – Floris Nov 8 '16 at 16:49
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I think your objection is valid. For example, suppose we set $z = 1$, then the formula would look like

$$ t = \frac{m}{m_0} \sqrt{\frac{h}{g}} $$

where $m$ is the mass of the apple and $m_0$ is a fundamental physical constant with units of mass. This is mathematically consistent. If this were a law of nature, it would "just" mean that we have a fundamental constant $m_0$ with units of mass. This seems weird but is not unprecedented; for example in the standard model the quark, lepton and neutrino masses are fundamental (not derived) constants with units of mass. We just don't see a lot of such non-unitless constants in classical physics.

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To do dimensional analysis, we list all the quantities that could affect the result, and write (conventionally) the lowest power terms that make the dimensions balance on both sides of the equation. We also add a non-dimensional constant, as this analysis cannot determine the final scale.

Then we write down all possible non-dimensional combinations of quantities, for example Reynolds Number (which in one definition is (mass.acceleration)/(dynamic viscosity.(velocity/distance).area)) and accept that our 'lowest terms' answer could be multiplied by any power of any dimensionless term. That is as far as dimensional analysis will take us.

If our intuition is good, we can use Occam's razor to hack away at the mass of irrelevant dimensionless terms. Experiment is the final arbiter of what the equation actually is, for which dimensionless terms need to be included, and will also put a value on the constant.

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    $\begingroup$ I'm not a fan of this "Occam's razor". It's been mentioned twice (you, and a commenter DavidWhite), so I decided to briefly research it. It seems as if it is a philosophical excuse without scientific backing (don't get me wrong; I love simplicity, but I've never scientifically proven why it's superior to complexity). Maybe I'm not understanding something about it. $\endgroup$ – Fine Man Nov 9 '16 at 20:49
  • $\begingroup$ Occam's Razor is emperical, and holds generally. That is why we need intuition, aka experience, to use it well, and why experiment is the final arbiter of what the equation actually is. $\endgroup$ – Neil_UK Nov 9 '16 at 22:20
  • $\begingroup$ OK, that makes sense. I always need verification about empirical-ness of a law ever since I learned about the 2nd law of thermodynamics. :) $\endgroup$ – Fine Man Nov 11 '16 at 3:54

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