Using Dimensional Analysis to derive an equation (Walter Lewin video)

In summary, the conversation discusses the use of dimensional analysis in determining the time it takes for an object to drop based on differing heights. The equation used is $$t=k{h}^\alpha{m}^\beta{g}^\gamma$$ where k is a constant, h is the height, m is the mass, and g is the gravity. There is some uncertainty about the accuracy of this equation and how it was derived, but it is considered the simplest and most logical candidate. However, it is important to note that all theories are ultimately guesses and must be tested through experimentation.
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Chenkel
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TL;DR Summary
Mr Lewin's dimensional analysis
Hello everyone. I was watching the Walter Lewin lectures and I noticed in the talk he used something called dimension analysis to study the time it takes for an object to drop based on differing heights.



I'm 22:07 minutes into the video.

With some guess work on what's proportional to what, he wrote
$$t = k{h}^\alpha{m}^\beta{g}^\gamma$$Where k is a constant, h is the height, m is the mass, and g is the gravity.

He argued that the dimensional equivalents are $$[T]=[L]^\alpha[M]^\beta\frac {[L]^\gamma} {[T]^{2\gamma}}$$
He said that ##\alpha + \gamma = 0## and ##\beta = 0## and ##1 = -2\gamma## so ## \gamma = \frac {-1} {2} ## and ##\alpha = \frac {1} {2}##

Plugging this values into the first equation we have $$t=k\frac{\sqrt{h}}{\sqrt{g}}$$
I'm wondering about the theory of this approach, how do we know that the equation ##t=k{h}^\alpha{m}^\beta{g}^\gamma## is right? It seems the approach used by Mr Lewin requires a little bit of guess work, but I'm wondering what the logic behind the guess work is, how do we know for example that there isn't some addition or subtraction of certain multiplications happening on the right side, how can we assume this simple formula before solving for the exponents.

Any feed back on this would be helpful, thank you!
 
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  • #2
I often use dimensional analysis to check my calculations.
What Mr Lewin is suggesting is a bit weak. How he came up with Mass, Gravity, and Height as candidate terms for time-to-fall is certainly very ill-defined.
 
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Other possibly interesting sources:
 
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robphy said:
Other possibly interesting sources:
I followed the compadre example of a ball being throw upward, I think there was a mistake in the example they gave.

They wrote $$[m^a(v^y)g^c] = (M)^a(\frac L T)^b(\frac L {T^2})^c = M^aL^{b-c}T^{-b-2c}$$

I believe they meant $$[m^a(v^y)g^c] = (M)^a(\frac L T)^b(\frac L {T^2})^c = M^aL^{b+c}T^{-b-2c}$$

I suppose they are trying to say ## h = km^a(v^y)g^c ## and solve for the exponents that will create a dimensionality of L

So are they guessing that the following equation is true? ## h = km^a(v^y)g^c ## it seems possible but I don't know how reasonable this assumption is; it feels easy for me to imagine getting wrong answers with dimensional analysis because something could have correct dimensions but incorrect formula, is this a reasonable concern?
 
  • #5
Chenkel said:
Summary: Mr Lewin's dimensional analysis
I'm wondering about the theory of this approach, how do we know that the equation ##t=k{h}^\alpha{m}^\beta{g}^\gamma## is right? It seems the approach used by Mr Lewin requires a little bit of guess work, but I'm wondering what the logic behind the guess work is, how do we know for example that there isn't some addition or subtraction of certain multiplications happening on the right side, how can we assume this simple formula before solving for the exponents.
You can't know that for sure. Dimensional analysis is a very powerful technique; but it is primarily used when designing experiments (and checking the final answer).
That is, using dimensional analysis you can write down a relationship between different variables; based on this you can then figure out which measurements you need to do in order to determine the unknowns.
It could very well be that you need to subtract/add terms but if that is the case it should become obvious as you do the measurements.
Dimensional analysis can quite dramatically reduce the the number of measurements you need to do; but it can't replace experiments.
 
  • #6
Chenkel said:
So are they guessing that the following equation is true? ... it seems possible but I don't know how reasonable this assumption is
Two points:
-It is the "simplest", and hence by Occam's Razor the most compelling, candidate.
-No theory is ever proven "true". They are all guesses, although "seeing " some overarching structure in a theory seems to bolster Occam, they are all subject to scrutiny via experiment. We call it science.
 
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FAQ: Using Dimensional Analysis to derive an equation (Walter Lewin video)

What is dimensional analysis?

Dimensional analysis is a mathematical method used to derive an equation by examining the physical dimensions of the quantities involved. It is based on the principle that the units of measurement must be consistent on both sides of an equation.

Why is dimensional analysis useful in science?

Dimensional analysis is useful in science because it allows us to check the validity of equations and identify any errors. It also helps in converting units and simplifying complex equations.

What are the steps involved in using dimensional analysis to derive an equation?

The steps involved in using dimensional analysis to derive an equation are: 1) Identify the physical quantities involved in the problem, 2) Write down the known values and their units, 3) Determine the units of the desired output, 4) Set up a conversion factor using the given units and the desired output units, 5) Cancel out the units to obtain the desired output.

Can dimensional analysis be used in any type of equation?

Yes, dimensional analysis can be used in any type of equation as long as the quantities involved have consistent units. It is a universal method that can be applied to different branches of science such as physics, chemistry, and biology.

Are there any limitations to using dimensional analysis?

One limitation of dimensional analysis is that it cannot account for dimensionless constants or variables. It also cannot determine the numerical value of a constant in an equation, only the relationship between the quantities. Additionally, it may not be applicable in cases where the physical quantities involved are not well-defined or do not have a clear relationship.

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