# Invariance of combinations of physical quantities

1. Nov 6, 2007

### bernhard.rothenstein

Please tell me if there are motivated physical reasons to consider that combinations of dimensionless physical quantities that appear at the exponent of e in distribution functions have the same magnitude in all inertial reference frames in relative motion,

2. Nov 6, 2007

### Staff: Mentor

Dimensionless quantities are used in exponential functions because otherwise the units don't make sense. In other words, I know physically what a meter is and what a meter^2 is, but what is a 2^meter?

It has nothing to do with inertial reference frames or the Lorentz transform.

3. Nov 6, 2007

### bernhard.rothenstein

dimensionless combinations of physical quantities

Thanks for your answer. I think that the problem is somehow connected with SR. Consider the exponment E/kT of e in a given distribution function
(E energy, k Boltzmann constant. T Kelvin temperature. It is dimensionsless. k having the same magnitude in all inertial reference frames the result is that E and T should transform via the same transformation factor.
Please consider all I say above as questions and not as a statement.
Kind regards

4. Nov 6, 2007

### clem

E/kT is not a good candidate for Lorentz invariance, because T is only defined in the overall rest system of the gas.
What about the dimensionless ratio v/c, which does change under LT?

5. Nov 6, 2007

### Staff: Mentor

I am sorry I was unclear.

Certainly you can find many examples of exponential functions that would be used in relativistic mechanics. All I intended to say is that the reason that an exponential function always has a dimensionless number as an argument is more general and not limited to relativity. In general e^(some physical unit) makes no sense.

Taking your example, I can tell you exactly what 1 Joule means, but what does e^(1 Joule) mean? It doesn't mean anything, it is nonsense. It doesn't matter if the context of the Joule is in a relativistic calculation or a regular Newtonian calculation. Either way 1 Joule is meaningful but e^(1 Joule) is not.

6. Nov 6, 2007

### shoehorn

If you're unconcerned with mathematical rigour, a trivial way to see that an exponential function applied to mathematical physics must have a dimensionless quantity in the exponent is that the Taylor expansion of, say, $e^x$ about the point $x=0$ is

$$e^x = \sum_{i=0}^\infty \frac{x^i}{i!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$

As you can see from the expansion, if $x$ had dimensions of, say, length, then the dimensions of $e^x$ would be undefined. Therefore, $x$ itself must be dimensionless.

7. Nov 6, 2007

### Staff: Mentor

Nice explanation, shoehorn. The same explanation could be used for other transcendental functions as well. Sin, cos, sqrt, etc. all must take dimensionless arguments for the same reason.

By the way, bernhard, just because a parameter is dimensionless does not imply that it is frame invariant. Consider the dimensionless parameter beta = v/c. The v is clearly frame variant (0 in rest frame by definition and nonzero in all other frames) and c is constant in all frames. So beta must be frame variant even though it is dimensionless.

Last edited: Nov 6, 2007
8. Nov 6, 2007

### bernhard.rothenstein

invariance of combinations of phyhsical quantities

Thank you for your help. The counterexample you give is interesting but what I have in mind are combinations which do not involve invariant phyhsical quantities. Consider as an example
N=E/h$$\nu$$
where N is say the counted number of photons and h$$\nu$$ is the energy of a single photon. N and h have the same magnitude for all inertial observers in reletive motion and so E and $$\nu$$ should transform via the same transformation equation. Many such examples could be invented.

9. Nov 6, 2007

### bernhard.rothenstein

invariance of combinations

Thanks for the nice explanation. I will insert it in a paper I am preparing.
But how could I quote it?

10. Nov 6, 2007

### bernhard.rothenstein

invariance of combinations

Thanks for your help. Considering Planck's distribution T is the tempareture of the blackbody emitting the radiation.

11. Nov 7, 2007

### robphy

Something of the form:
"Shoehorn, Private communication."