Mechanical Similarity: Solving Equation 10.2

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In summary, the authors state that we are only interested in potentials which are homogeneous functions of the coordinates (homogeneous of degree k), and that the equations of motion permit a series of geometrically similar paths. They then state that the times of the motion between corresponding points are in the ratio t'/t = (l'/l)1-k/2.
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msbell1
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Hi,

I have recently embarked on a quest to read through (and hopefully to understand) all 10 volumes of the Course of Theoretical Physics by Landau and Lifgarbagez. Well, I have already found myself hung up on section 10 in Volume I (Mechanics, 3rd Edition). Specifically, I am stuck on how equation 10.2 was derived. For those of you that do not have the book available (and for those that don't feel like pulling it off the shelf), here's my problem.

The authors first state that we are only interested in potentials which are homogeneous functions of the coordinates (homogeneous of degree k). They then say that we will carry out a transformation in which the coordinates are transformed by a factor A: ra -> Ara, and time is transformed by a factor B: t -> Bt. They then state that velocities will be transformed by A/B, and the kinetic energy by A2/B2. That makes sense to me. Then, since the potential U is homogeneous of degree k, in the Lagrangian U will have a factor of Ak in front of it. Since the Lagrangian can be multiplied by a constant without changing the equations of motion, we need for the kinetic energy T to also have a factor of Ak in front of it. For this to happen, we have

A2/B2 = Ak

Solving for B gives B = A1-k/2

So far so good.

Now the authors claim that the equations of motion permit a series of geometrically similar paths, and the times of the motion between corresponding points are in the ratio

t'/t = (l'/l)1-k/2

where l'/l is the ratio of the linear dimensions of the two paths.

OK, how do they find this expression for t'/t? Thanks a lot for the insight!
 
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  • #2
Hi, msbell1!

This thing
[tex]\frac{t'}{t} = \frac{l'}{l}^{1-k/2}[/tex]
comes from
[tex]B = A^{1-k/2}[/tex] where [tex]t' = Bt[/tex] and [tex]l' = A l[/tex].

P.S. Good choice of the book! ;)
 
  • #3
Thanks physicsworks! That helped a lot, and it makes sense now. I really appreciate your help.
 

Related to Mechanical Similarity: Solving Equation 10.2

1. What is mechanical similarity?

Mechanical similarity is a concept in engineering and physics that describes the relationship between two systems that have the same geometric and dynamic properties. This means that they have similar shapes and move in similar ways.

2. Why is mechanical similarity important?

Mechanical similarity allows us to make accurate predictions about the behavior of a system based on the behavior of a similar system. This is especially useful in engineering, where testing different prototypes can be expensive and time-consuming. By using mechanical similarity, we can make predictions and design more efficient and effective systems.

3. How is mechanical similarity calculated?

Mechanical similarity is typically calculated using the Buckingham Pi Theorem, which states that if two systems have the same dimensionless parameters, they will exhibit the same behavior. These dimensionless parameters are known as Pi terms and are calculated using the system's relevant physical properties.

4. What is Equation 10.2 in mechanical similarity?

Equation 10.2 is a specific equation that is used to solve for the dimensionless parameters (Pi terms) in a system. It is based on the Buckingham Pi Theorem and is often used in engineering and physics to determine the mechanical similarity of two systems.

5. What are some real-world applications of mechanical similarity?

Mechanical similarity has many practical applications in various fields, such as aerodynamics, fluid mechanics, and structural engineering. It is used to design and test aircraft, ships, cars, and buildings. It is also used in biomechanics to study the movement and behavior of living organisms and in chemical engineering to optimize industrial processes.

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