Using Dimensional Analysis to Guess a Formula

In summary, dimensional analysis is a very important tool in physics and engineering. Units on the left and right side of an equation have to be equal, or the equation would not be valid. The terms of an addition have to have the same units, and it helps to carry units along in your equations to catch mistakes.
  • #1
LBB II
7
0
I'm teaching myself physics by reading notes online posted as an OpenCourseWare physics course at MIT:

http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/

I'm already frustrated in the first module. One major lesson involves guessing a formula based on making the dimensions come out right. One question asks for a formula for the period T of a swinging pendulum. We are told that we should think of some physical quantities on which T might depend. They suggest m, the mass of the bob at the end of the pendulum, θ, the angular amplitude, L, the length of the pendulum, and g, the gravitational acceleration. They say that T cannot depend on m, because none of the other physical quantities have mass units that could cancel out the mass unit of the bob in a formula in order to leave just time units for the period T. Then they talk about how to combine the given quantities to get time units out, and come to the conclusion that T is proportional to the square root of L / g (up to a multiplicative function of θ since the angle is dimensionless).

This is all logical to me except for one thing. Why do the dimensions on each side of the equation have to agree in the first place? For example, the notes also mention Hooke's Law for springs: F = -kx. The dimensions of F are M L / T^2 (M = mass, L = length, T = time) but those for x are just L. The units don't match. They get around this by saying that the dimensions of the spring constant k are
M / T^2. This makes the dimensions on each side of the equation match up. Well, why can't the formula for the period of the pendulum also have some constant with units designed to make the dimensions of each side of the equation match up?

(I know that the formula from the period can be derived and it is indeed proportional to the square root of L / g and it does not depend on m.)

Thanks for any help.
 
Physics news on Phys.org
  • #2
LBB II said:
I'm teaching myself physics by reading notes online posted as an OpenCourseWare physics course at MIT:

http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/

I'm already frustrated in the first module. One major lesson involves guessing a formula based on making the dimensions come out right. One question asks for a formula for the period T of a swinging pendulum. We are told that we should think of some physical quantities on which T might depend. They suggest m, the mass of the bob at the end of the pendulum, θ, the angular amplitude, L, the length of the pendulum, and g, the gravitational acceleration. They say that T cannot depend on m, because none of the other physical quantities have mass units that could cancel out the mass unit of the bob in a formula in order to leave just time units for the period T. Then they talk about how to combine the given quantities to get time units out, and come to the conclusion that T is proportional to the square root of L / g (up to a multiplicative function of θ since the angle is dimensionless).

This is all logical to me except for one thing. Why do the dimensions on each side of the equation have to agree in the first place? For example, the notes also mention Hooke's Law for springs: F = -kx. The dimensions of F are M L / T^2 (M = mass, L = length, T = time) but those for x are just L. The units don't match. They get around this by saying that the dimensions of the spring constant k are
M / T^2. This makes the dimensions on each side of the equation match up. Well, why can't the formula for the period of the pendulum also have some constant with units designed to make the dimensions of each side of the equation match up?

(I know that the formula from the period can be derived and it is indeed proportional to the square root of L / g and it does not depend on m.)

Thanks for any help.

Welcome to the PF.

Dimensional analysis is a very imporatant tool in physics and engineering. The units on the LHS and RHS of any equation have to be equal, or it would not be an equation. You can't have 1 second = 1 hour, right?

Also, each term of an addition has to have the same units. You can't add 1 second + 1 hour directly, without converting into common units for each.

And it helps to carry units along in your equations, to help catch errors early. If you have the same units in the numerator and denominator, you can cancel them out. Like s/s = 1, and s^2/s = seconds.
 
  • #3
One cannot over stress the importance of dimensions. Heed what berkeman wrote above.
 
  • #4
The OP makes a valid point and I don't think either berkeman or LawrenceC has answered the question.

To the OP: In the case of a spring, we intuitively know that the force must also depend on the physical characteristics of the spring, such as stiffness. Perhaps if you take this into account, you will be able to derive the formula for the force exerted by a stretched spring.
 
Last edited:
  • #5
Thanks berkeman and LawrenceC for your advice, but I already understand dimensional analysis (I'm a math professor). My question was about guessing a formula for T in terms of m (for mass), g, L, and θ. The notes I am reading say that the guessing should be done so that the units come out right. For example, T = Lm would be a bad guess because using the MKS system, the units on the left are s and the units on the right are (meters)(kg). The notes I am reading specifically say that based on looking at units alone, we are sure that T does not depend on m. But why can't we have a situation like T = kLm where k is a "pendulum constant" whose units are s/(meters*kg)? This is analogous to the situation with Hooke's Law. I don't think you read my post closely enough before responding. Though I do appreciate the welcome and the attempt at a response.
 
  • #6
alexmahone:

Thank you for addressing my question directly. I'll think more about what you said. I think you've pointed me on the right track. It brings up another question for anyone that cares to answer: Is Hooke's Law only an experimentally verified fact, or has it been derived from more fundamental principles?
 
  • #7
LBB II said:
Thanks berkeman and LawrenceC for your advice, but I already understand dimensional analysis (I'm a math professor). My question was about guessing a formula for T in terms of m (for mass), g, L, and θ. The notes I am reading say that the guessing should be done so that the units come out right. For example, T = Lm would be a bad guess because using the MKS system, the units on the left are s and the units on the right are (meters)(kg). The notes I am reading specifically say that based on looking at units alone, we are sure that T does not depend on m. But why can't we have a situation like T = kLm where k is a "pendulum constant" whose units are s/(meters*kg)? This is analogous to the situation with Hooke's Law. I don't think you read my post closely enough before responding. Though I do appreciate the welcome and the attempt at a response.

Guessing is not required. There are algorithms for determining the correct unit groups. What is required is to be able to identify the physical quantities (variables) that might be involved in a given system. After that you can turn the crank to produce the desired result.

An example might be your pendulum and its period. You might suspect that the period T depends in some way on the length L, mass M, and the acceleration due to gravity g. So write the variables as a multiplied group. Assign an exponent to each variable, giving the first one in the list the exponent 1, followed by unknowns b, c, d, ...

T*LaMbgc

Expand into their base units. (Unfortunately there's name duplication here. For example, T represents both the period of the pendulum and the time unit [T]).

[tex] [T][L]^a[M]^b \left( \frac{[L]}{[T]^2} \right)^c [/tex]

This group of units, taken as a whole, should be unitless (if you've included enough variables to describe the system). So we can construct a set of equations based upon the exponents of the individual units:

For T: 1 - 2c = 0

For L: a + c = 0

For M: b = 0

Solving we find: a = -1/2 ; b = 0; c = 1/2

We can clear the fractions by multiplying through by 2 (this won't alter the equations, just scale them). Then: a = -1, b = 0 ; c = 1, and the exponent of T goes from unity to 2.

Note that mass has dropped out, its exponent being zero.

Rewrite the variable group and include the exponents. Equate the group to a constant.

T2*L-1*g = const

Now you can solve this group for anyone of the variables and see the form of the relationship it has with the others. So for period T:

[tex] T = \sqrt{const \frac{L}{g}} [/tex]
 
  • #8
gneill:

Sorry I didn't make myself clear. (This was my first post here, and I apologize to everyone if I'm being ignorant on this issue and less than clear about my question.) I know how to do the math to find the formula that you arrived at. The "guessing" I was referring to was guessing which physical quantities might affect the period T. Note also that the angular amplitude θ is dimensionless and so T might actually be proportional to f(θ)√(L/g) for some function f. So, we would have to guess the function f. (I know that it turns out that for small angles θ that the period essentially doesn't depend on θ at all.)

My original point (which I'm apparently not making clear, because most responders aren't addressing it) is that there might be some sort of constant k with unknown dimensions that renders your expression T*LaMbgc incomplete. For example, T = k*M*L is dimensionally correct if k has units like s kg-1m-1. I know this is far-fetched and have no physical reason for supposing such a constant exists, but If I had to come up with a dimensionally correct formula for the restorative force, F, of a stretched spring based on physical quantities, I would not think to use a constant k with dimensions of N/m as in F = -k*x. So, I'm wondering what the use is of coming up with a formula just based on getting the units to come out right for the physical quantities involved, which is a major theme and source of exam questions for the first module in the MIT first course in physics I mentioned.

Also note that I understand how to actually find the period of a pendulum by solving a 2nd order linear differential equation (under an approximating assumption involving a small angular amplitude).
 
  • #9
LBB II:

I think that one of the reasons why unit analysis is stressed is that it hammers home the fact that in physics formulas must be consistent in their units or they are wrong (or at best incomplete). Physics is not just a grab bag of rules of thumb. It is also a means of exploring possible relationships that may not be obvious when limited empirical evidence is available, such as at the beginnings of an investigation into uncharted territory.

I don't think that unit analysis is promoted as being a panacea for all of physics research. We're not going to close the labs and stop all experimental research and replace it with "searches of unit space". It's a guide, not a result.

Hooke's law is one of those formulas that came into being through empirical investigation. It was determined that the restoring force was proportional to the displacement. k is the constant of proportionality. It's given units to balance that proportion. It may be that the "law" is in fact the result of much more complex underlying processes and relationships, and a full mathematical expression of them might be daunting. But I have no doubt that at the end of the day, the units would balance.
 
  • #10
Thank you, gneill, for your insightful remarks. I agree with them and I understand that the units always must balance. But I'm still perplexed.

The MIT notes say that the period of the pendulum, T, if it indeed depends only on L, M, and g (up to a constant of proportionality and a multiplicative function of θ), must be a product of powers of these quantities, and once you know this, the exponents can be mathematically solved for (and they are unique in this case). But why cannot T, instead, be a product of powers of L, M, g, and some other constant k that does not correspond to a physical quantity like a length or a mass? For constants k with different units we would get different solutions for all the exponents involved. For example, based on unit analysis alone, we can not rule out a possibility such as T = kLM where the units for k are s*m-1*kg-1. The MIT notes make a point to say, for instance, that it is a priori that a formula for T cannot depend on M (because there are no other quantities in the list L, M, g that are derived from mass units and hence the mass units of the bob will not cancel out). But this seems completely false if physics formulas can have constants of proportionality that have units. Don't a lot of physics formulas have such constants?

For example, saying that T cannot depend on M because the units won't work out is like this:

Suppose we were interested in the attractive force F between two bodies of masses M1 and M2 (measured in kg) that are separated by a distance of r meters. Suppose you believe that F depends only on the physical quantities M1, M2, and r. So, you know that the units of F, which are N (working in the MKS system) must be a product of powers of the units of M1, M2, and r. Mathematically solving for the powers shows that there are no solutions. Hmm, that means that F must depend on other physical quantities. It cannot depend just on those mentioned because Newtons have units of kg*m*s-2, and the masses of the bodies and the distance separated by them do not involve time units. But this would be wrong thinking because the gravitational constant, G, comes to the rescue, which has units of m3*kg-1*s-2, and indeed F = G*M1*M2/r2. (I had to look this up on wikipedia--it's been 20 years since I've had a physics course.)

Are the MIT course notes just wrong? Or, what am I missing?
 
  • #11
By the way, this is not a homework question. So I'm wondering why my post got moved to a forum subtitled "Algebra- and calculus-based general physics homework." Maybe it's because my question is very easy (or stupid?) and there's no other forum for it? Or just because it's related to my reading of some online course notes? I'm not enrolled in a course by the way.
 
  • #12
I can propose two ideas that might address your concerns to some degree (or perhaps make things even worse!).

First, perhaps it is necessary to distinguish between relationships that are propositions or axioms of a theory and those that arises from those axioms. The gravitational force relationship, for example, was proposed by Newton as the basis of gravitational theory. It was not derived from previous axioms. As such it is the embodiment of a statement such as "There is an attractive force between any two masses M and m that is proportional to the product of those masses and inversely proportional to the square of the distance between them". The constant of proportionality is thus forced upon us at the outset, and is wholly dependent upon the choice of unit system for the quantities involved. I believe that Hooke's law in its stated form has a similar provenance.

Secondly, it is clear that the constants of proportionality that appear in the axioms of such theories can be abolished entirely by a suitable choice of unit system! In Natural Units the constant G, for example, becomes 1 with no units attached.

Taking these ideas into consideration, dimensional analysis becomes a tool applicable to relationships derived from the encompassing theory with its underlying assumed relationships already in place. Proportionality constants that arise from such analysis would be an artifact of choice of unit system within the underlying axioms of the physical theory that dictates their interrelationships. Introducing spurious constants of proportionality in order to force a random set of variables into unit balance is equivalent to proposing a new theory which may or may not agree with the tenets of the old one. Caveat emptor!
 
  • #13
LBB II said:
By the way, this is not a homework question. So I'm wondering why my post got moved to a forum subtitled "Algebra- and calculus-based general physics homework." Maybe it's because my question is very easy (or stupid?) and there's no other forum for it? Or just because it's related to my reading of some online course notes? I'm not enrolled in a course by the way.

Yes, we try to keep all schoolwork-type questions in the Homework Help forums. This thread may help to explain why the PF HH rules are the way they are:

https://www.physicsforums.com/showthread.php?t=373889

.
 
  • #14
gneill:

Thanks very much! Your response was very enlightening, and I think it fully answers my question.
 

FAQ: Using Dimensional Analysis to Guess a Formula

What is dimensional analysis?

Dimensional analysis is a mathematical method used to convert units of measurement and determine the relationship between different physical quantities.

Why is dimensional analysis useful?

Dimensional analysis is useful because it allows scientists to make predictions and estimate the behavior of physical systems without conducting actual experiments. It also helps in identifying and correcting errors in equations or formulas.

What is the process of using dimensional analysis to guess a formula?

The process of using dimensional analysis to guess a formula involves identifying the known and unknown variables, converting the units of measurement to a common system, and using algebraic operations to manipulate the quantities until the units of the final result match the desired units.

Can dimensional analysis be used for any type of formula?

Yes, dimensional analysis can be used for any type of formula as long as the variables involved have physical units that can be converted and manipulated.

What are some common applications of dimensional analysis?

Dimensional analysis is commonly used in physics, chemistry, and engineering to solve problems related to units, measurements, and physical quantities. It is also used in fields such as economics and biology to analyze relationships between variables and make predictions.

Similar threads

Replies
3
Views
3K
Replies
4
Views
1K
Replies
2
Views
11K
Replies
7
Views
2K
Replies
9
Views
1K
Back
Top