How can dimensional analysis be carried out correctly?

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1. May 7, 2015

PhysExplorer

Hello!

May I first please direct your attention to Walter Lewin's first Classical Mechanics lecture on Units, Dimensions and Scaling Arguments?

He first carried out a dimensional analysis to determine the relation between the time taken for a ball to fall from a particular height to the height itself, the mass of the ball and gravitational acceleration.

He then mentioned that the dimensional analysis could just as well have been carried out another way, leading to different results.

May anyone please tell me how in this case (and in general) how one can correctly determine all the physical quantities related to the time taken for the ball to fall?

Thank you very much in advance.

2. May 7, 2015

marksman95

From my point of view, there's not such a way. Dimensional analysis just provide you from possible answers to a problem in what dimensions regards.

In the case of the falling ball, you could take into account other variables as the air drag or the distance to the Sun, for example. But, as many times in physics, you idealize the problem. So in this particular case, you suppose that there is no air drag and no other phenomena affecting the fall. What I mean is that your idealized ball is completely described by it's mass, it's height and gravitational acceleration on the Earth' surface.

You could idealize the problem in a different way to take into account other things such air drag, but in that case you should try a different approach also.

Just my point of view.

3. May 7, 2015

ucanihl

I have some problems about this analysis. First, we consider the quantities that may related our problem and then this analysis tells us what that relation is. But we have a dimensionless multiple also. And here is my question, we know, dimensionless multiple does not mean an independent constant multiple. It may change with respect to any quantities that we concern. Maybe it is (mass1+mass2)/( mass1-mass2). If this is possible how this analysis be useful?
One more thing, this problem appears in not only the dim.less multiple. Assume that we found that the power of the mass must be one. But if we have two different masses, how can I know wheter it is "product over sum" or "squared product over cube of sum" or what else??

4. May 7, 2015

Staff: Mentor

As soon as you have two quantities with the same units, dimensional analysis can be tricky to impossible for those quantities. Sometimes you can find the dependence on those quantities in other ways, like reducing the two-body Kepler problem to a one-body problem with the reduced mass, and applying dimensional analysis to this problem with a single mass.

5. May 7, 2015

ucanihl

Considering the symmetricity of two quantities with same units may also be useful. I mean lets say on masses f(m1,m2)=f(m2,m1) must be hold because we choose m1 and m2 arbitrarily. This eliminates much of the possibilities but not all. We cannot write for example (2m1+m2). We must construct our expression from symmetrical terms. But thank you for the reduction idea, it is new to me.

6. May 8, 2015

marksman95

Dimensional analysis is useful in many ways. But I think you are trying to take it too far. What about a law in which an unknown constant has dimensions?

My point is, dimensional analysis is helpful and necessary always. But don't try to take it too far.

7. May 8, 2015

ucanihl

I mean, division of two masses is dimensionless but not constant. It depends on the masses.

8. May 8, 2015

Staff: Mentor

You can have constants like the speed of light, sure, but you can include those in the analysis.

9. May 8, 2015

cnh1995

I'm not an expert but as per my knowledge,dimensional analysis deals with the fundamental physical quantities. So you should first guess the relevant physical quantities for the given situation and break them into fundamental quantities. That's what I remember from my physics class 3 years ago. For example, to calculate the time period of a pendulum, you should assume relevant physical quantities. If I assume it depends on mass, length of pendulum and gravitational acceleration, it would be sufficient. So, I can write
T=kmalbgc.....(1)
[M]0L0T1=k[M]a[L]b[LT-2]c
From the indices' arithmetic, I get a=0, b=1/2,c=-1/2.
Substituting a,b,c in (1), you'll get the relation as
T=k√l/√g..