1
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 10, 2013 at 16:45
  • $\begingroup$ @JohnRennie. Yes, that was stupid of me, I've edited the question. $\endgroup$
    – user27182
    Apr 10, 2013 at 17:00

2 Answers 2

2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 other problems could need three or more with only $x,\,t$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.