Most likely speed in Maxwell-Boltzmann distribution

In summary, the most likely speed in the Maxwell-Boltzmann distribution can be found by taking the derivative of the natural log of the function f(v) and setting it equal to zero. This is because both the function and its natural log have a maximum at the same value of v. Additionally, if a monotonic function is applied to a function with a maximum, the maximum will remain at the same location. This can be mathematically shown using the chain rule.
  • #1
Vrbic
407
18

Homework Statement


What is the most likely speed in Maxwell-Boltzamann distribution?

Homework Equations


[itex] f(v)dv=4\pi(\frac{m}{2 \pi kT})^{3/2}v^2Exp(-\frac{mv^2}{2kT})dv[/itex]

The Attempt at a Solution


I know I need maximum of f(v) -> [itex]\frac{df}{dv}=0[/itex]. But it is not trivial to do. I found some solution where they said: [itex]\frac{d}{dv^2}(\ln[v^2Exp(-\frac{mv^2}{2kT})])=0[/itex]. But I don't know how they arrive to it. Could somebody advise?
 
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  • #2
Oooou sry, it is easy. I made a mistake I derivate f(v) - function without v2 term. Otherwise, why they can rewrite this problem to logarithm and derivative according to v2.
 
  • #3
They just felt like its easier to differentiate the natural log of f, rather than f itself. Both f and its natural log have a maximum at the same value of v.
 
  • #4
Chestermiller said:
They just felt like its easier to differentiate the natural log of f, rather than f itself. Both f and its natural log have a maximum at the same value of v.
Thank you for your response. So can I say generally that if I apply some monotonic function to other, that extremes stay at same point?
 
  • #5
Vrbic said:
Thank you for your response. So can I say generally that if I apply some monotonic function to other, that extremes stay at same point?
What are your thoughts on this?

Chet
 
  • #6
Chestermiller said:
What are your thoughts on this?

Chet
What are you asking me now? How or my oppinion if it is true?
 
  • #7
Vrbic said:
What are you asking me now? How or my oppinion if it is true?
I'm asking to see if you can reason it out mathematically.
 
  • #8
Chestermiller said:
I'm asking to see if you can reason it out mathematically.
No. I just ask :)
 
  • #9
Suppose g(y) is a monotoically increasing function of y, and y(x) is a function of x with a maximum. Then, by the chain rule,
$$\frac{dg}{dx}=\frac{dg}{dy}\frac{dy}{dx}$$
dg/dy is always positive, so dg/dx has a zero derivative at the same location where dy/dx has a zero derivative.
 
  • #10
Chestermiller said:
Suppose g(y) is a monotoically increasing function of y, and y(x) is a function of x with a maximum. Then, by the chain rule,
$$\frac{dg}{dx}=\frac{dg}{dy}\frac{dy}{dx}$$
dg/dy is always positive, so dg/dx has a zero derivative at the same location where dy/dx has a zero derivative.
Very nice ;) It is true what I said.
 

1. What is the most likely speed in the Maxwell-Boltzmann distribution?

The most likely speed in the Maxwell-Boltzmann distribution is the speed at which the majority of particles in a gas are moving. It is also known as the mode speed and is dependent on the temperature of the gas.

2. How is the most likely speed calculated in the Maxwell-Boltzmann distribution?

The most likely speed in the Maxwell-Boltzmann distribution can be calculated using the formula v mode = √(2kT/m), where k is the Boltzmann constant, T is the temperature in Kelvin, and m is the mass of the gas particles.

3. What factors affect the most likely speed in the Maxwell-Boltzmann distribution?

The most likely speed in the Maxwell-Boltzmann distribution is affected by the temperature of the gas, the mass of the gas particles, and the Boltzmann constant. It also depends on the type of gas and the conditions it is in.

4. How is the most likely speed related to the average speed in the Maxwell-Boltzmann distribution?

The most likely speed in the Maxwell-Boltzmann distribution is not necessarily equal to the average speed. The most likely speed represents the peak of the distribution curve, while the average speed is the average of all speeds in the distribution. However, as the temperature increases, the most likely speed and the average speed get closer to each other.

5. Is the most likely speed the same for all gas particles in the Maxwell-Boltzmann distribution?

No, the most likely speed can vary for different gas particles in the Maxwell-Boltzmann distribution. This is because the distribution takes into account the individual masses of the particles, so lighter particles will have a higher most likely speed compared to heavier particles at the same temperature.

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