# Is thermal noise a statistical uncertainty?

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• kelly0303
In summary, the conversation discussed a system described by ##y=ax##, where a is the parameter to be extracted and y is the measured quantity. The parameter x is experimentally controlled but has associated uncertainty. It was clarified that x represents the position of a particle in contact with a thermal bath at temperature T, with the particle's energy being ##kx^2/2##. The probability of a given x was stated to follow a Boltzmann distribution. It was questioned whether this uncertainty is of a statistical nature and the concept of "Gaussian and not thermal noise" was brought up. The conversation ended with a discussion on the nature of the thermal probability distribution and the meaning of the "true" value.
kelly0303
Hello! I have a system described by ##y=ax##, where a is the parameter I want to extract and y is the stuff I measure (we can assume that I can measure one instance of y without any uncertainty). x is a parameter I can control experimentally but it has an uncertainty associated to it. In a simplified form (but enough for my question), x is the position of a particle (classically) in contact with a thermal bath at temperature T. For example we can assume that the energy of the particle is ##kx^2/2##, where k is a known constant and for each measurement of y, x has a different x, where the probability of an x is given by the probability of having that given energy based on a Boltzman distribution at temperature T. I am not sure if this is a statistical uncertainty or not. I would say it is, because if I measure many y values, I can narrow down the true y value (if I assume I have Gaussian and not thermal noise, that would go down as ##1/\sqrt{N}##, where N is the number of measurements, right?), but I wanted to make sure this makes sense.

What do you mean by "Gaussian and not thermal noise"? The thermal probability distribution in your case is proportional to
$$e^{-\beta E}=e^{-\beta kx^2/2}$$
which is also Gaussian. Besides, by the "true" value, do you mean the average value?

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