# Dimensional analysis seems wrong for this equation: z = -1/(x^2+y^2)

• I
• dyn
In summary, the conversation discusses a question about dimensional analysis in which the equation given is missing a constant with the appropriate dimension. The question is deemed badly formulated and it is suggested that the equation should include the constant to maintain correct physics. The conversation also mentions the possibility of substituting a different equation to check dimensions.
dyn
Hi
I've just done a question regarding a marble moving on a surface given by z = -1/(x2+y2)
In this case what happens with dimensional analysis ? x and y have dimensions of length while z has dimensions of 1/(length)2.
Is this question badly written or do i just accept that i won't be able to perform dimensional analysis on any later terms that are produced ?
Thanks

Last edited by a moderator:
So if z is a distance, and x and y are distances, then there needs to be some implied dimension in the 1 in the numerator, to make the dimension of z be correct.

It's kind of like you can have y = x^2 mathematically, but if it represents something physical, then you might have to multiply by a coefficient (could be 1) with some dimension (or units).

PeroK and berkeman
dyn said:
HI
I've just done a question regarding a marble moving on a surface given by z = -1/(x2+y2)
In this case what happens with dimensional analysis ? x and y have dimensions of length while z has dimensions of 1/(length)2.
Is this question badly written or do i just accept that i won't be able to perform dimensional analysis on any later terms that are produced ?
Thanks
The question is badly formulated. There's some constant with the appropriate dimension missing. If it's in a math textbook it's simply that mathematicians nowadays don't care for correct physics. If it's a physics textbook it's simply a bad book ;-)).

dyn and scottdave
dyn said:
Is this question badly written or do i just accept that i won't be able to perform dimensional analysis on any later terms that are produced ?
It's a badly written question. But you can argue that the real surface is ##z'=Az##, where ##A## has dimensions of length cubed and equals one in whatever units you are using. At any point in your working you can simply substitute ##z'/A## for ##z## and check dimensions.

dyn
just an opinion but I think ##z=-1/(x^2+y^2)## is better written than ##z = -(1\ \mathrm{m^3})/(x^2 + y^2)## or ##(z/\mathrm{m}) = -1/((x/\mathrm{m})^2 + (y/\mathrm{m})^2)## or whatever because the emphasis is clearly on studying the properties of the surface and not confusing the student with weird dimensional factors

Office_Shredder
I'd simply write ##z=-A/(x^2+y^2)##, where ##A=\text{const}##. Of course, ##A## has the dimension ##\text{length}^3##. A dimensionally wrong equation is a nogo in any physics book!

dyn and scottdave
vanhees71 said:
The question is badly formulated. There's some constant with the appropriate dimension missing. If it's in a math textbook it's simply that mathematicians nowadays don't care for correct physics. If it's a physics textbook it's simply a bad book ;-)).
It was on a university maths exam paper !

Well, that explains it. Mathematicians don't care much for physics nowadays anymore :-(.

dyn

## 1. Why does dimensional analysis not seem to work for this equation?

Dimensional analysis is a method used to check the consistency of units in equations. However, it may not work for this particular equation because it involves inverse square terms, which do not have consistent units.

## 2. Can dimensional analysis still be applied to this equation?

Yes, dimensional analysis can still be applied to this equation to check the consistency of units. However, it may not provide a definitive answer due to the presence of inverse square terms.

## 3. How is dimensional analysis used in scientific research?

Dimensional analysis is commonly used in scientific research to check the validity of equations and to identify any errors or inconsistencies in units. It is also used to derive new equations and to simplify complex equations by reducing the number of variables.

## 4. Are there any other methods besides dimensional analysis to check the consistency of units in equations?

Yes, there are other methods such as the Buckingham Pi theorem and the method of repeating variables that can be used to check the consistency of units in equations. These methods are particularly useful for equations involving inverse square terms.

## 5. Can dimensional analysis be used to solve equations?

No, dimensional analysis is not a method for solving equations. It is only used to check the consistency of units in equations and to simplify complex equations. To solve equations, other mathematical methods such as algebra or calculus are used.

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