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Suppose we have some function of time and space (1-D for simplicity) $G(x, t)$ which, by considering some equation relating $G$ to other quantities we know to be dimensionless. We now conclude that the argument of $G$ must be dimensionless, else we have a contradiction. So let us suppose that $G(x,t) = G(\zeta)$ for $\zeta$ being some dimensionless combination of $x$ and $t$. My question is, is there, in general, a further restriction on the form of $\zeta$? Does $\zeta$ have to be some particular dimensionless combination of $x$ and $t$, or can we just (judiciously, given what we will then do with $G$) pick any $\zeta$ that works?

EDIT

silly of me, $\zeta$ contains some constant to make it dimensionless. My question is, basically, is $\zeta = \lambda \frac{x}{t}$ any better than $\zeta' = \mu \frac 1{\sqrt{xt}}$, or any other $\zeta$ of this style?

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There is no dimensionless combination of x and t. Well, not unless you include constants that have dimensions, but if you do that the argument(s) to $G$ don't have to be dimensionless. –  John Rennie Apr 10 '13 at 16:45
    
@JohnRennie. Yes, that was stupid of me, I've edited the question. –  user27182 Apr 10 '13 at 17:00
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2 Answers 2

As long as the resulting expression is dimensionless and there is no physical restriction on its form, you can pick whatever you like.

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In general there aren't restrictions.

Furthermore:

$G(x,\,t)$ dimensionless does not involves $G(x,\,t)=G(\zeta)$.

For example the probability density of the Stern-Gerlach experiment using Pauli spinors:

$\rho(x,\,t) = K_1 (e^{-(k_2 x - k_3 t^2)^2} + e^{-(k_2 x + k_3 t^2)^2})$ see

It needs two dimensionless parameters and others problems could need three or more with only $x,\,t$.

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