Jump to content

Recommended Posts

Posted

I just want to know is mass dependent upon the temperature i.e. if there is an increase on a temperature of any body then does its mass change or it really effects the mass of that particular body

Posted

The mass of an ensemble will depend on the temperature; it's energy internal to the system and not translational energy of the CoM.

Posted

Hotter objects have more mass than when cooler. Atoms in excited states have more energy than atoms in the ground state. The second case has been experimentally confirmed for a nuclear excited state

Posted

"Hotter objects have more mass than when cooler"

would this also be true for a single atom say hydrogen?

Is this mass increase dm only attibutable to the absorbtion of the photon by the electron?

Posted
"Hotter objects have more mass than when cooler"

would this also be true for a single atom say hydrogen?

Is this mass increase dm only attibutable to the absorbtion of the photon by the electron?

 

well, a single atom can't really have temperature. temerature is really more of an emergent property of a collection of atoms.

 

sure you can consider its theoretical temperature based on its velocity but then thats relative. it won't even emit an IR signal that you can measure while a collection of atoms will.

Posted

"it won't even emit an IR signal that you can measure while a collection of atoms will."

 

is this just because we dont have a sensitive enough measuring apperatus yet

or is there a more fundametal property of an ensemble of atoms that is different from a single atom, s mass.

Posted
"it won't even emit an IR signal that you can measure while a collection of atoms will."

 

is this just because we dont have a sensitive enough measuring apperatus yet

or is there a more fundametal property of an ensemble of atoms that is different from a single atom, s mass.

 

No, it's not a sensitivity issue. There is no mechanism by which the lone atom would emit thermal radiation. You get the radiation in an ensemble because of collisions, which cause accelerations.

Posted
No, it's not a sensitivity issue. There is no mechanism by which the lone atom would emit thermal radiation. You get the radiation in an ensemble because of collisions, which cause accelerations.

 

How many atoms do you need to get temperature? two?

Posted

It's not really well-defined. You need enough so that statistical descriptions (e.g. the Maxwell-Boltzmann distribution) work.

Posted
How many atoms do you need to get temperature? two?

 

I don't think two is enough, but if two is the ensemble, or system of interest, the mass of that system could be considered greater than the sum of it's parts, (as you would consider in a gas cloud which would certainly have a temperature) as you could include the kinetic energies with respect to the center of mass.

Posted

"I don't think two is enough, but if two is the ensemble, or system of interest, the mass of that system could be considered greater than the sum of it's parts, (as you would consider in a gas cloud which would certainly have a temperature) as you could include the kinetic energies with respect to the center of mass."

 

It sounds like temprature is only well defined using Maxwell-Boltzmann statistics,

which implies that temprature is "emergent " . As we increase the number of atoms in a vacuum box the temprature increases using the relation PV= nRT.

Similarly the as the Mass increases gravity "emerges" causing additional potential energy.

 

 

This is how I interpret it , does it make sense?


Merged post follows:

Consecutive posts merged

Similarly the mass the Mass increases gravity "emerges" causing additional potential energy.

 

oops typo error apologies!

Posted

The energy of the system doesn't change, how you account for it does. If you have CoM translation, you have a kinetic energy term. If you have multiple particles such that tracking individual particles is impossible, you have a collective description (temperature), and the energy appears in the mass term.

Posted
How many atoms do you need to get temperature? two?

 

there are various property like temperature, pressure, liquid, gas, solid which have collective property i.e these property are useless until and unless there isn't any collection particles to which they're concerned.

 

but a question really comes in mind as quoted above like how much of collection of particle do we need to have to make it possible-for example consider single water molecule H2O, now can you tell me what is it a -

 

A) Solid

B) Liquid

C) Gas

Posted

i say neither,

 

Now if that molecules electron(s) absorb an photon and jumps to a higher orbital

( if that is correct) has the molecules "mass" increased ?

 

"There is no mechanism by which the lone atom would emit thermal radiation" post #10

 

 

Further when it falls back to the lower orbital emmiting a photon , could that not be accepted as "thermal radiation" or is that something else .

Posted
i say neither,

 

Now if that molecules electron(s) absorb an photon and jumps to a higher orbital

( if that is correct) has the molecules "mass" increased ?

 

"There is no mechanism by which the lone atom would emit thermal radiation" post #10

 

 

Further when it falls back to the lower orbital emmiting a photon , could that not be accepted as "thermal radiation" or is that something else .

 

Excited states are more massive, which has been experimentally confirmed with an isomer of one of the isotopes of iron. Here is more detail:

 

http://blogs.scienceforums.net/swansont/archives/278

 

De-excitation of a lone atom is not thermal radiation.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.