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Suppose we have two bodies A and B lying on a horizontal table and the two bodies are going towards each other.

$m_A= m$, $ \vec v_A= -\vec v$

$m_B=2m$, $\vec v_B= \vec {\frac{v}{\sqrt{2}}}$

From the informations given, we see that the two bodies have same initial kinetic energy.

If we find the kinetic energy of the two after an elastic collision, we can notice that the body A with mass $m$ gains some more energy .

Similarly, suppose we have two large bodies with same temperature. If we bring them in contact the molecules will collide elastically with one another.

So why don't their temperature change if the energies of the individual molecules does change ?

Or more specifically

why isn't the same idea of energy exchange true for bodies with same temperature ?


Deriving the fact that energy will transfer (can be ignored)

So let's find the center of mass's velocity : $$\vec v_{cm}=\frac{m_A \vec v_A+m_B\vec v_B}{m_A+m_B}$$. $$\vec v_{cm} = \frac{2m(\vec {\frac{v}{√2}})-m(\vec v)}{3m}\Rightarrow v(\frac{√2-1}{3})$$

In center of mass's frame the velocity of A after collision is given by:

$$\vec v'_A= 2\vec v_{cm} - \vec v_A$$ So, $$\vec v'_A= 2v(\frac{√2-1}{3}) + v \Rightarrow v(\frac{2√2+1}{3})$$

Similarly $$\vec v'_B = {2v(\frac{√2-1}{3})}-{\frac{v}{√2}}\Rightarrow v(\frac{1-2√2}{3√2})$$

So from the above derivation we see that the Energy of each of the molecules change after the head on elastic collision.

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    $\begingroup$ You haven't said this explicitly, but I am guessing the two large bodies are each made up of molecules of different mass? $\endgroup$ May 21, 2021 at 5:46
  • $\begingroup$ @BioPhysicist sorry .. I forgot that . Yeah the two bodies are of different molecules . $\endgroup$
    – Ankit
    May 21, 2021 at 5:58
  • $\begingroup$ In the derivation part the statement In center of mass's frame the velocity of A after collision is given by is a general idea true for elastic collision. $\endgroup$
    – Ankit
    May 21, 2021 at 8:45
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    $\begingroup$ Very good presentation. $\endgroup$
    – Babu
    May 22, 2021 at 14:23

2 Answers 2

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In the center of mass system of two bodies, if the two bodies have a mass equal to each other and scatter elastically, the kinetic energy of each is the same, as observed in the comment.

In the statistical treatment of an ideal gas that you refer to, the masses in order to derive the temperature are all the same. So in both cases, neither the kinetic energy in your example, nor in the temperature of misxing two ideal gases at the same temperature, there is any change.

Here is a discussion about mixing two ideal gasses at the same temperature , which is taken the same after mixing, because of energy conservation . If your two bodies have different masses , they will have different kinetic energies after scatter, but the total energy of the two body system by conservation of energy is the same , before and after. The temperature,is connected to the average kinetic energy, and the average between your two bodies should not change.

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The temperature of a body is its average in the various (spin, magnetization, kinetic...) energy degrees of freedom, not a simple kinetic energy value. We measure temperature by applying a thermometer, and waiting for the thermometer to come to the same temperature as the sample, which happens because of the so-called zeroth law of thermodynamics: heat travels from the hotter of two bodies in contact, to the cooler.

When heat stops making a net increase or decrease in the temperature of the thermometer, that tells us the temperature of the sample is the same as the temperature of the thermometer, and thus the temperature of the sample-and-thermometer-probe is what the thermometer display indicates. We can do this because the probe thermal contact does NOT achieve a temperature difference with the sample, but converges to the same temperature.

The effect of individual interactions is not unimportant; it gives rise to internal fluctuations in any given speck of the system, some of which are observable (Brownian motion is an example; the rushing sound heard in a conch shell is another). The microscopic picture of elements of a gas, for instance, can have a wide range of gas molecule velocities. Single molecules thus do NOT have a measurable statistical property of temperature unless we make a long-time observation and average the various wanderings over many seconds, or apply an ergodic principle (a very useful conceit, ergodicity: the time average is equal to the population-of- other-molecules average).

Ergodic principles are mathematically poorly founded (the theorems are weak ones) but physicists use them (and sometimes call them theorems, if no mathematicians are present). The temperature of a gas particle is inferred routinely from the average that communicates from many particles slowly to a massive thermometer bulb...

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  • $\begingroup$ "which happens because of the so-called zeroth law of thermodynamics: heat travels from the hotter of two bodies in contact, to the cooler." That's not the zeroth law. $\endgroup$ May 21, 2021 at 11:44
  • $\begingroup$ @BioPhysicist - not a restatement of the zeroth law, but 'thermal contact' means the heat flows to make equilibrium, and 'same temperature' means thermal equilibrium is established (which calls for zero heat flow). $\endgroup$
    – Whit3rd
    May 22, 2021 at 5:33
  • $\begingroup$ I don't see how any of that involves the zeroth law. The zeroth law doesn't say anything about heat flow $\endgroup$ May 22, 2021 at 7:20
  • $\begingroup$ @BioPhysicist The zeroth law means that a magnetic liquid and a diatomic gas and a block of marble, which have the same temperature, do not engage in heat energy transfer. Thermal energies aren't always stored similarly in disparate systems. $\endgroup$
    – Whit3rd
    May 22, 2021 at 7:43
  • $\begingroup$ No. The zeroth law is not about heat transfer. It is just a mathematical statement to make thermal equilibrium mathematically well-defined. The fact that two objects at the same temperature do not transfer heat is due to the second law, not the zeroth law. I explain in more detail here. $\endgroup$ May 22, 2021 at 11:17

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