Proving radial properties of particular dimensionless surface plots?

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tade
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surf_101.png


We have a surface function z = f(x,y) ; f(x,y) only contains dimensionless constants, and is itself dimensionless.

If we convert it to cylindrical co-ordinates, z = f(r,θ) , does z only depend on θ?
Meaning we can remove r from the equation, literally.
 
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Discussion

tade said:
z only depend on θ?

Obviously not ! Look at z(r) for e.g. ##\theta=\pi/4##
 
BvU said:
Obviously not ! Look at z(r) for e.g. ##\theta=\pi/4##
what's the equation of f(r,θ) ?
 
You are asking me ?
Or do you mean: if ##\theta=pi/4## then ##f(r,\theta) = f(x,x) ## with ##x=r/\sqrt 2## ?
 
BvU said:
You are asking me ?
I thought you had come up with a counter-example to prove the conjecture false, so I was wondering what that counter-example function was.
 
Conjecture: z is independent of r
Counter example: along the diagonal I see z go up, down, up again and then down again -- clearly not independent of r
 
BvU said:
Conjecture: z is independent of r
Counter example: along the diagonal I see z go up, down, up again and then down again -- clearly not independent of r
sorry, what's the equation of the counter-example function?
 
However, I see an interpretation of
tade said:
f(x,y) only contains dimensionless constants
If x and y do not occur, then f itself is a constant, therefore independent of r, but equall indepndent of ##\theta##
 
BvU said:
However, I see an interpretation of
If x and y do not occur, then f itself is a constant, therefore independent of r, but equall indepndent of ##\theta##
Oh, so just z = c , the surface plot being just a flat plane?
That's a trivial case and the conjecture is that there'll be r-independence in all cases, whichever function you use. As long as the function satisfies the two conditions.
 
In which case your picture is wrongfooting any good-willing helper :rolleyes:
 
BvU said:
In which case your picture is wrongfooting any good-willing helper :rolleyes:
Well, it's certainly not an r-independent function, though I just wanted to make sure people got the idea of "surface plot" immediately.
 
tade said:
As long as the function satisfies the two conditions
I don't see what the dimensionlessness of f or its contained constants has to do with it
 
tade said:
just wanted to make sure people got the idea of "surface plot" immediately
Well, this exercise creates more confusion than it removes
 
BvU said:
Well, this exercise creates more confusion than it removes
Well, no equations are given, the image not referenced, cos its a general conjecture.

I should add some details in the OP though, but its too late now.
 
BvU said:
I don't see what the dimensionlessness of f or its contained constants has to do with it
Its part of the conjecture. Unless its possible to expand the generality of the conjecture even further.