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I think it is wrong to define the temperature by the average energy of the molecule in all frames of reference. The reason for that is clear: take all of your particles and send them at $100 m/s$ to the north. This won't make the gas hotter, just like the fan does not cool/heat the air (another great mystery!). The organized movement does not participate in the notion of temperature. Seemingly, it has to be defined in a rest frame of gas to make sense.

A simple calculation in mind shows a contradiction. If you define a temperature through the energy, you have to conclude that temperature transforms like an energy (which is a part of a momentum 4-vector). But if you try to expand it through velocity, you will immediately see that velocity squared transforms in a rather ugly way and won't simplify to the vector component transformation. First of all, I believe that the formula for energy you take is not correct in relativistic case.

Next, to the measurement! I think this answeranswer is correct in distinguishing the observation of the temperature from its statistical definition. You can judge about the temperature of a body from the spectrum of black-body radiation it produces, but this is not the measurement of the temperature, but the measurement of the radiation which is subject to relativistic redshift.

I think it is wrong to define the temperature by the average energy of the molecule in all frames of reference. The reason for that is clear: take all of your particles and send them at $100 m/s$ to the north. This won't make the gas hotter, just like the fan does not cool/heat the air (another great mystery!). The organized movement does not participate in the notion of temperature. Seemingly, it has to be defined in a rest frame of gas to make sense.

A simple calculation in mind shows a contradiction. If you define a temperature through the energy, you have to conclude that temperature transforms like an energy (which is a part of a momentum 4-vector). But if you try to expand it through velocity, you will immediately see that velocity squared transforms in a rather ugly way and won't simplify to the vector component transformation. First of all, I believe that the formula for energy you take is not correct in relativistic case.

Next, to the measurement! I think this answer is correct in distinguishing the observation of the temperature from its statistical definition. You can judge about the temperature of a body from the spectrum of black-body radiation it produces, but this is not the measurement of the temperature, but the measurement of the radiation which is subject to relativistic redshift.

I think it is wrong to define the temperature by the average energy of the molecule in all frames of reference. The reason for that is clear: take all of your particles and send them at $100 m/s$ to the north. This won't make the gas hotter, just like the fan does not cool/heat the air (another great mystery!). The organized movement does not participate in the notion of temperature. Seemingly, it has to be defined in a rest frame of gas to make sense.

A simple calculation in mind shows a contradiction. If you define a temperature through the energy, you have to conclude that temperature transforms like an energy (which is a part of a momentum 4-vector). But if you try to expand it through velocity, you will immediately see that velocity squared transforms in a rather ugly way and won't simplify to the vector component transformation. First of all, I believe that the formula for energy you take is not correct in relativistic case.

Next, to the measurement! I think this answer is correct in distinguishing the observation of the temperature from its statistical definition. You can judge about the temperature of a body from the spectrum of black-body radiation it produces, but this is not the measurement of the temperature, but the measurement of the radiation which is subject to relativistic redshift.

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Andrii Magalich
Andrii Magalich
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I think it is wrong to define the temperature by the average energy of the molecule in all frames of reference. The reason for that is clear: take all of your particles and send them at $100 m/s$ to the north. This won't make the gas hotter, just like the fan does not cool/heat the air (another great mystery!). The organized movement does not participate in the notion of temperature. Seemingly, it has to be defined in a rest frame of gas to make sense.

A simple calculation in mind shows a contradiction. If you define a temperature through the energy, you have to conclude that temperature transforms like an energy (which is a part of a momentum 4-vector). But if you try to expand it through velocity, you will immediately see that velocity squared transforms in a rather ugly way and won't simplify to the vector component transformation. First of all, I believe that the formula for energy you take is not correct in relativistic case.

Next, to the measurement! I think this answer is correct in distinguishing the observation of the temperature from its statistical definition. You can judge about the temperature of a body from the spectrum of black-body radiation it produces, but this is not the measurement of the temperature, but the measurement of the radiation which is subject to relativistic redshift.