I In what cases does dimensional analysis fail?

AI Thread Summary
Dimensional analysis can fail when it does not provide meaningful insights about the physical relationships between quantities, as it cannot determine specific values or relationships like the circumference of a circle. It is not effective in cases where the nature of opposing quantities is unknown, such as in damping of harmonic oscillations, where it cannot derive equations of motion without additional context. While dimensional analysis can confirm the dimensional consistency of equations, it does not guarantee their physical validity. For instance, quantities with the same dimensions, like torque and energy, are fundamentally different and cannot be directly compared. Thus, while dimensional analysis is a useful tool, its limitations must be recognized in complex physical scenarios.
VVS2000
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In what cases does dimensional analysis fails? And also is there like any preferred situation or an 'ideal' situation in order to use dimensional analysis? Thanks in advance!
It possible a diagram would be really really helpful
 
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Is this a schoolwork question?
 
No no Like they were just going through dimensional analysis in our class and I just thought well obviously this won't work for evert case so I just wanted to know what are those cases.
 
VVS2000 said:
Summary: In what cases does dimensional analysis fails?
Well, a number of quantities are dimensionless, but I'm not sure that represents a failure. Can you think of some dimensionless constants or quantities that are used commonly?

Also, these two PF Insights articles would probably be good background reading for you to help understand dimensional analysis better... :smile:

https://www.physicsforums.com/insights/learn-the-basics-of-dimensional-analysis/
https://www.physicsforums.com/insights/make-units-work/
 
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VVS2000 said:
I just thought well obviously this won't work for evert case so I just wanted to know what are those cases.

That's like asking when addition doesn't work - when it's the wrong tool for the job. Maybe you need to subtract or to multiply.

Dimensional analysis cannot tell you if the circumference of a circle is 6r, 2πr or 7r. It can tell you it is not 1/r or r2.
 
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Vanadium 50 said:
That's like asking when addition doesn't work - when it's the wrong tool for the job. Maybe you need to subtract or to multiply.

Dimensional analysis cannot tell you if the circumference of a circle is 6r, 2πr or 7r. It can tell you it is not 1/r or r2.
No no not like that. What if sometimes we don't know the nature of the opposing quantity. How are you going to proceed then?
 
VVS2000 said:
What if sometimes we don't know the nature of the opposing quantity.

I don't understand.
 
Vanadium 50 said:
I don't understand.
Like in damping of harmonic oscillations.
 
  • #10
VVS2000 said:
Like in damping of harmonic oscillations.
What do the overall dimensions of the quantity in the exponent need to be? So what does that tell you about the dimensions of λ ?

http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html
1565279395512.png
 
  • #11
I still don't understand. Perhaps you could use complete sentences? Maybe even paragraphs?
 
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  • #12
Vanadium 50 said:
I still don't understand. Perhaps you could use complete sentences? Maybe even paragraphs?
For example damping in oscillations. What if we don't know the nature of damping? How do you find the equations of motion?
 
  • #13
VVS2000 said:
How do you find the equations of motion?

What does that have to do with dimensional analysis?
 
  • #14
VVS2000 said:
For example damping in oscillations. What if we don't know the nature of damping? How do you find the equations of motion?
Please see post #10.
 
  • #15
Vanadium 50 said:
What does that have to do with dimensional analysis?
I am asking using dimensional analysis how do you find the eqaution of motion then
 
  • #16
berkeman said:
Please see post #10.
Sorry I am new here. Post 10 where?
 
  • #17
No worries. The post number in each thread is shown in the upper right of each post. This post of mine will likely be #17, since I'm replying to your post #16. So just scroll up the thread to look for my reply in post #10. It was a post showing some equations from Hyperphysics to help you see the typical differential equation and solution for a damped harmonic oscillator. :smile:
 
  • #18
berkeman said:
No worries. The post number in each thread is shown in the upper right of each post. This post of mine will likely be #17, since I'm replying to your post #16. So just scroll up the thread to look for my reply in post #10. It was a post showing some equations from Hyperphysics to help you see the typical differential equation and solution for a damped harmonic oscillator. :smile:
That's actually really helpful. Thanks.
 
  • #19
VVS2000 said:
That's actually really helpful. Thanks.
So I cannot use dimensional analysis in such cases?
 
  • #20
berkeman said:
No worries. The post number in each thread is shown in the upper right of each post. This post of mine will likely be #17, since I'm replying to your post #16. So just scroll up the thread to look for my reply in post #10. It was a post showing some equations from Hyperphysics to help you see the typical differential equation and solution for a damped harmonic oscillator. :smile:
So I cannot use dimensional analysis in such cases?
 
  • #21
VVS2000 said:
So I cannot use dimensional analysis in such cases?
Like in the Hyperphysics equations? Sure. Can you show the dimensions for each of the terms in the initial differential equation? And then show how that helps to define the dimensions of λ in the end?
 
  • #22
berkeman said:
Like in the Hyperphysics equations? Sure. Can you show the dimensions for each of the terms in the initial differential equation? And then show how that helps to define the dimensions of λ in the end?
I think I can. I can show the complimentary function for x. Then I can By eliminating the constants by plugging in the initial conditions, we can find the dimensions of lambda
 
  • #23
VVS2000 said:
I think I can. I can show the complimentary function for x. Then I can By eliminating the constants by plugging in the initial conditions, we can find the dimensions of lambda
Okay, but what are the dimensions (or units) for each of the terms in these equations?

1565359058696.png
 
  • #24
VVS2000 said:
I think I can. I can show the complimentary function for x. Then I can By eliminating the constants by plugging in the initial conditions, we can find the dimensions of lambda
The dimensions of c are...?
The dimensions of m are...?
The dimensions of ##\lambda## are...?
 
  • #25
The dimensiona of lambda is (time)^-1, time inverse, basically frequency. .M is mass. That is given. I don't or I am not quite sure on c
 
  • #26
VVS2000 said:
The dimensiona of lambda is (time)^-1, time inverse, basically frequency. .M is mass. That is given. I don't or I am not quite sure on c
The starting equation contains forces. What are the units of force in mks? Each term has to have those same units, right? That helps you figure out the units for the other terms that introduced in the later equations. And yes, since λt is in the exponent, and since the exponent needs to be dimensionless, the units of λ need to be 1/s.
 
  • #27
VVS2000 said:
In what cases does dimensional analysis fails?
Not sure what 'fails' means here, but dimensional homogeneity doesn't guarantee that your equation is physically sensible:

https://en.wikipedia.org/wiki/Dimensional_analysis#Dimensional_homogeneity
Even when two physical quantities have identical dimensions, it may nevertheless be meaningless to compare or add them. For example, although torque and energy share the dimension L2MT−2, they are fundamentally different physical quantities.
 
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  • #28
berkeman said:
The starting equation contains forces. What are the units of force in mks? Each term has to have those same units, right? That helps you figure out the units for the other terms that introduced in the later equations. And yes, since λt is in the exponent, and since the exponent needs to be dimensionless, the units of λ need to be 1/s.
Yeah ok so c has units of M(T)^-1
 
  • #29
A.T. said:
Not sure what 'fails' means here, but dimensional homogeneity doesn't guarantee that your equation is physically sensible:

https://en.wikipedia.org/wiki/Dimensional_analysis#Dimensional_homogeneity
Even when two physical quantities have identical dimensions, it may nevertheless be meaningless to compare or add them. For example, although torque and energy share the dimension L2MT−2, they are fundamentally different physical quantities.
Yeah sorry. I think I did'nt frame my question in the proper way. So the math part of the equation can be checked out but it will not be wise to try and infer the physics aspect of it
 
  • #30
VVS2000 said:
The dimensiona of lambda is (time)^-1, time inverse, basically frequency. .M is mass. That is given. I don't or I am not quite sure on c
That’s correct.
 

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