Physics Standard Deviation

I am a physics enthusiast and I have a question: Why is it meaningless to express the '$\pm$' (standard deviation) value as a percentage?

Standard deviation adds uncertainties to the measured value: $23.3\pm 0.4\,{\rm m}$. One can quickly look at the error (which has units of ${\rm m}$ in my case) and think, The value could be as low as $22.9\,{\rm m}$ or as high as $23.7\,{\rm m}$ without much thinking. Modifying this to being a percentage of the value would be confusing.

Plus it would be arbitrary. As Martin Beckett points out, since the Celsius and Kelvin scales have identical temperature spacing but differing 0-points, making the error (std. dev.) scale to the value would be ridiculous: $$20.0\pm0.1\,{\rm C} \to 20.0\,{\rm C}\pm0.5\%$$ $$293.0\pm0.1\,{\rm K}\to293.0\,{\rm K}\pm0.034\%$$ Same value in units systems would lead to differing uncertainties; this would bring about more confusion.

Depends what you are measuring.

If you are measuring the temperature of a glass of water to 20C +- 0.1C what would the percentage be?

Now what would it be if you measured it as 293Kelvin +- 0.1K

• Perhaps, could you be more general? Jan 14 '14 at 2:35
• He is telling that there are different units of measurements, and so the percentage in the error would vary.
– user28355
Jan 14 '14 at 14:48