Does the (non-relativeistic) mass of a stationary object change with its temperature.
For example would a stationary bowling ball have a different mass at 100000 degrees C vs if it were at 0 deg C.
Mass is defined to be an invariant quantity, so no, the mass of a body at higher temperature is still the same. I anticipate someone will try to tell you when you heat something up you are increasing it's internal energy and as a result of the energy going up, the mass goes up. The person in my imagination saying that is wrong. Mass does not increase or decrease. It is invariant.
Either way, that is a relativistic effect. In non relativistic mechanics, mass is generally assumed to be invariant. In thermodynamics when you heat something up it's volume can increase, but it's mass will stay the same.
The trouble is that this effect is rather small. The volume of your bowling ball would not change very much even if you heated it up to just below it's melting point. However, the extension by thermal expansion depends on the natural length of a body so you'll see of bridges and rail lines they leave gaps every so often to leave room for the expansion. If you don't the thermal expansion can actually cause the bridge to collapse or the rails of buckle.
Hope this helps.
Either way, that is a relativistic effect. In non relativistic mechanics, mass is generally assumed to be invariant. In thermodynamics when you heat something up it's volume can increase, but it's mass will stay the same.
The trouble is that this effect is rather small. The volume of your bowling ball would not change very much even if you heated it up to just below it's melting point. However, the extension by thermal expansion depends on the natural length of a body so you'll see of bridges and rail lines they leave gaps every so often to leave room for the expansion. If you don't the thermal expansion can actually cause the bridge to collapse or the rails of buckle.
Hope this helps.
QUOTE (prometheus+Apr 28 2008, 09:09 PM)
I anticipate someone will try to tell you when you heat something up you are increasing it's internal energy and as a result of the energy going up, the mass goes up. The person in my imagination saying that is wrong. Mass does not increase or decrease. It is invariant.
Either way, that is a relativistic effect.
So you're saying that if a hydrogen atom absorbs a photon, it violates Einsteins law of mass-energy equivalence and retains its former mass? I'm not sure I buy that. The rest masses of isolated particles are invariant, but when together in a system they can interact by means of exchanging virtual particles. In an excited hydrogen atom there will be more virtual particles present and thus will have a higher mass than a ground state hydrogen atom.
Either way, that is a relativistic effect.
So you're saying that if a hydrogen atom absorbs a photon, it violates Einsteins law of mass-energy equivalence and retains its former mass? I'm not sure I buy that. The rest masses of isolated particles are invariant, but when together in a system they can interact by means of exchanging virtual particles. In an excited hydrogen atom there will be more virtual particles present and thus will have a higher mass than a ground state hydrogen atom.
QUOTE (barakn+Apr 28 2008, 09:51 PM)
So you're saying that if a hydrogen atom absorbs a photon, it violates Einsteins law of mass-energy equivalence and retains its former mass? I'm not sure I buy that. The rest masses of isolated particles are invariant, but when together in a system they can interact by means of exchanging virtual particles. In an excited hydrogen atom there will be more virtual particles present and thus will have a higher mass than a ground state hydrogen atom.
But a hydrogen atom in the ground state is not just a single electron and a single proton. It's these two particles plus a cloud of photons that transmit the electromagnetic force. Absorbing a single photon hardly changes the amount of particles at all.
Also remember that the particles that make up matter are never isolated and always have a cloud of photons or other gauge bosons associated with them depending on what forces they couple to. What makes you think there will be more virtual particles in excited hydrogen?
The mass energy equivalence is generally taught like this: mass is a form of energy, therefore mass can be converted into energy. While that is almost true it is very misleading. What would be more correct to say is that a particles invariant mass contributes to it's energy. The particles energy can change (like when it's going very fast) but a particles mass cannot change. When a particle has very high energy we get a deviation from the Newtonian kinetic energy = 1/2 m v^2 The KE of a particles is less than you would expect from Newton at a particular velocity. That can be interpreted (poorly) as an increase in mass. What we know now is that the Newtonian expression for KE is only the first approximation to the relativistic expression for KE.
In actuality, I was thinking of a bulk substance when writing my first reply so this is a bit of a digression. The relativistic mass increase myth is an axe that I like to grind as often as possible though, so thanks for giving me the opportunity.
But a hydrogen atom in the ground state is not just a single electron and a single proton. It's these two particles plus a cloud of photons that transmit the electromagnetic force. Absorbing a single photon hardly changes the amount of particles at all.
Also remember that the particles that make up matter are never isolated and always have a cloud of photons or other gauge bosons associated with them depending on what forces they couple to. What makes you think there will be more virtual particles in excited hydrogen?
The mass energy equivalence is generally taught like this: mass is a form of energy, therefore mass can be converted into energy. While that is almost true it is very misleading. What would be more correct to say is that a particles invariant mass contributes to it's energy. The particles energy can change (like when it's going very fast) but a particles mass cannot change. When a particle has very high energy we get a deviation from the Newtonian kinetic energy = 1/2 m v^2 The KE of a particles is less than you would expect from Newton at a particular velocity. That can be interpreted (poorly) as an increase in mass. What we know now is that the Newtonian expression for KE is only the first approximation to the relativistic expression for KE.
In actuality, I was thinking of a bulk substance when writing my first reply so this is a bit of a digression. The relativistic mass increase myth is an axe that I like to grind as often as possible though, so thanks for giving me the opportunity.
QUOTE
So you're saying that if a hydrogen atom absorbs a photon,
See the problem there is that that is different from the question being asked.
An increase in temp is an increase in molecular motion. Since mass = density * volume the idea with heating something up is that it will expand increasing in volume but as a result there is also a direct proportional decrease in density. This results in the mass remaining unchanged.
QUOTE (Precursor562+Apr 30 2008, 12:39 AM)
See the problem there is that that is different from the question being asked.
An increase in temp is an increase in molecular motion. Since mass = density * volume the idea with heating something up is that it will expand increasing in volume but as a result there is also a direct proportional decrease in density. This results in the mass remaining unchanged.
How is it different? The original question was about heating an object. You could heat an object by bathing it in electromagnetic radiation, like toasting bread by having it absorb infrared wavelength photons emitted by hot coils. Heat up a box if hydrogen atoms at low pressure. There - I have basically boiled the original question down to one of the simplest systems possible.
Imagine a photon of energy hc/λ and momentum h/λ being absorbed a hydrogen atom with rest mass m₀ at rest (photon moving in positive direction). The assertion is that the rest mass m₁ of the excited hydrogen atom m₁ is equal to m₀
The relativistic formula for E (with rest mass m):
E = γmc²
E₀ = γ₀m₀m₁ + hc/λ = m₀c² + hc/λ
E₁ = γ₁m₁c²
E₀ = E₁ (conservation of Energy)
γ₁m₁c² = m₀c² + hc/λ
γ₁m₁ = m₀ + h/λc
Relativistic momentum p = γmv (with rest mass m and velocity v):
p₀ = γ₀m₀v₀ +h/λ = h/λ
p₁ = γ₁m₁v₁
p₀ = p₁ (conservation of momentum)
γ₁m₁v₁ = h/λ
Substituting into the energy equation:
γ₁m₁ = m₀ + h/λc
= m₀ + γ₁m₁v₁/c
m₀ = γ₁m₁(1 - v₁/c)
For 0<v₁<c, it can be shown that γ₁(1 - v₁/c)>1, so m₀ < m₁.
Heat up a box of hydrogen and you increase the number of hydrogen atoms in an excited state and hence its mass. Now I could be wrong, but I expect more than a hand waving argument.
Precursor, why would a change in volume mean anything to do with mass? A ton of feathers takes up more volume than a ton of bricks but it's still a ton.
Besides, not all substances always increase in volume with an increase in temperature. Water is densest at 4C. And thankfully too.
Besides, not all substances always increase in volume with an increase in temperature. Water is densest at 4C. And thankfully too.
Here is the non hand waving argument:
This equation is pure and complete bunk. E = mc² for a particle at rest. E²=(pc)² + (mc²)² for a particles with momentum p. Even if the equation wasn't bunk which it is, the gamma factor for p = 0 is simply 1 so your bunk formula reduces to the correct one anyway.
There's an error in the first step, which doesn't look good to me. I'm going to say this again: In relativity mass is defined as the invariant length of the energy momentum 4 vector. There is no difference between rest mass and dynamical mass.
QUOTE
The relativistic formula for E (with rest mass m):
E = γmc²
E = γmc²
This equation is pure and complete bunk. E = mc² for a particle at rest. E²=(pc)² + (mc²)² for a particles with momentum p. Even if the equation wasn't bunk which it is, the gamma factor for p = 0 is simply 1 so your bunk formula reduces to the correct one anyway.
There's an error in the first step, which doesn't look good to me. I'm going to say this again: In relativity mass is defined as the invariant length of the energy momentum 4 vector. There is no difference between rest mass and dynamical mass.
Another reason why the calculation of the mass of an atom + a photon is not a very good model is that atoms and photons are quantum particles. Relativity on it's own doesn't include quantum effects like the cloud of virtual photons I mentioned earlier.
QUOTE (prometheus+Apr 30 2008, 08:35 AM)
Here is the non hand waving argument:
This equation is pure and complete bunk. E = mc² for a particle at rest. E²=(pc)² + (mc²)² for a particles with momentum p. Even if the equation wasn't bunk which it is, the gamma factor for p = 0 is simply 1 so your bunk formula reduces to the correct one anyway.
There's an error in the first step, which doesn't look good to me. I'm going to say this again: In relativity mass is defined as the invariant length of the energy momentum 4 vector. There is no difference between rest mass and dynamical mass.
E = γmc² where E is the total energy, and if you don't believe me then you're going to have to start editing some wiki pages. I suggest starting with this one: http://en.wikipedia.org/wiki/Mass-energy_equivalence
You're going to have to change the sentence "If the mass is the relativistic mass, then the energy is the total energy."
We'll see how long your edits stick around. You'll also have to hack into hyperphysics.phy-astr.gsu.edu and change some of their pages too.
This equation is pure and complete bunk. E = mc² for a particle at rest. E²=(pc)² + (mc²)² for a particles with momentum p. Even if the equation wasn't bunk which it is, the gamma factor for p = 0 is simply 1 so your bunk formula reduces to the correct one anyway.
There's an error in the first step, which doesn't look good to me. I'm going to say this again: In relativity mass is defined as the invariant length of the energy momentum 4 vector. There is no difference between rest mass and dynamical mass.
E = γmc² where E is the total energy, and if you don't believe me then you're going to have to start editing some wiki pages. I suggest starting with this one: http://en.wikipedia.org/wiki/Mass-energy_equivalence
You're going to have to change the sentence "If the mass is the relativistic mass, then the energy is the total energy."
We'll see how long your edits stick around. You'll also have to hack into hyperphysics.phy-astr.gsu.edu and change some of their pages too.
Heat is a form of energy, so it does have a mass, of course. Just as chemical or fall height energy, or any other, does.
When writing the energy of a particle, one shouldn't forget terms. Kinetic energy and rest mass are just some of them. Electromagnetic or baryonic interactions would be some other ones, and me may still ignore some of them.
If you want a more detailed insight of heat's mass, at the particle level, it translates to kinetic and electromagnetic energies, which in turn have a mass.
So, divide the number of joules of heat by (3e8)^2 and you get the mass increase in kg. Quite simple scalar computation.
Which won't be much mass.
When writing the energy of a particle, one shouldn't forget terms. Kinetic energy and rest mass are just some of them. Electromagnetic or baryonic interactions would be some other ones, and me may still ignore some of them.
If you want a more detailed insight of heat's mass, at the particle level, it translates to kinetic and electromagnetic energies, which in turn have a mass.
So, divide the number of joules of heat by (3e8)^2 and you get the mass increase in kg. Quite simple scalar computation.
Which won't be much mass.
There was an error in the transcription of my notes. I wrote
E₀ = γ₀m₀m₁ + hc/λ = m₀c² + hc/λ
But it should read
E₀ = γ₀m₀c² + hc/λ = m₀c² + hc/λ
It doesn't change my argument.
And I agree with Enthalpy, it wouldn't be much mass. Still greater than zero.
E₀ = γ₀m₀m₁ + hc/λ = m₀c² + hc/λ
But it should read
E₀ = γ₀m₀c² + hc/λ = m₀c² + hc/λ
It doesn't change my argument.
And I agree with Enthalpy, it wouldn't be much mass. Still greater than zero.
QUOTE (barakn+Apr 30 2008, 11:08 PM)
E = γmc² where E is the total energy, and if you don't believe me then you're going to have to start editing some wiki pages. I suggest starting with this one: http://en.wikipedia.org/wiki/Mass-energy_equivalence
You're going to have to change the sentence "If the mass is the relativistic mass, then the energy is the total energy."
We'll see how long your edits stick around. You'll also have to hack into hyperphysics.phy-astr.gsu.edu and change some of their pages too.
Citing wikipedia as a source is not going to cut it I'm afraid.
The equation E²=(pc)² + (mc²)² is derived by taking the inner product of a general energy momentum 4 vector with itself. Here is a relevant bit (also on wikipedia): link If you know anything about 4 vectors in special relativity you'll know that when you take the inner product of a vector with itself you get a scalar that is lorentz invariant, that is, it is the same for all inertial frames. It's that clear: mass is a Lorentz invariant and there can be no 2 ways about it.
I have to say I was rather disheartened that hyperphysics was putting out such garbage, which is why I was happy to see this page.
If you're used to thinking like Newton, relativistic mass is a nice way of visualizing what happens at relativistic speeds. If you're actually trying to use relativity to calculate something, it's worse than a useless concept. It's confusing and science can be confusing enough.
You're going to have to change the sentence "If the mass is the relativistic mass, then the energy is the total energy."
We'll see how long your edits stick around. You'll also have to hack into hyperphysics.phy-astr.gsu.edu and change some of their pages too.
Citing wikipedia as a source is not going to cut it I'm afraid.
The equation E²=(pc)² + (mc²)² is derived by taking the inner product of a general energy momentum 4 vector with itself. Here is a relevant bit (also on wikipedia): link If you know anything about 4 vectors in special relativity you'll know that when you take the inner product of a vector with itself you get a scalar that is lorentz invariant, that is, it is the same for all inertial frames. It's that clear: mass is a Lorentz invariant and there can be no 2 ways about it.
I have to say I was rather disheartened that hyperphysics was putting out such garbage, which is why I was happy to see this page.
If you're used to thinking like Newton, relativistic mass is a nice way of visualizing what happens at relativistic speeds. If you're actually trying to use relativity to calculate something, it's worse than a useless concept. It's confusing and science can be confusing enough.
Being as I rubbished you (quite rightly) for citing wikipedia, I thought I'd find some better references to support what I'm saying, and lo and behold, what should come up?!
Hyperphysics page on momentum four vectors
If you can't be bothered to read that, here is the relevant section:
Hyperphysics page on momentum four vectors
If you can't be bothered to read that, here is the relevant section:
QUOTE (hyperphysics+)
The length of a 4-vector is invariant, being the same in every inertial frame. This invariance is associated with the constancy of the speed of light.
...
The length of this 4-vector (p_mu) is the rest energy of the particle (mc²) . The invariance is associated with the fact that the rest mass m is the same in any inertial frame of reference.
In the mouth of two or three witnesses let every word be established, so here is another one from Wolfram scienceworld.
...
The length of this 4-vector (p_mu) is the rest energy of the particle (mc²) . The invariance is associated with the fact that the rest mass m is the same in any inertial frame of reference.
In the mouth of two or three witnesses let every word be established, so here is another one from Wolfram scienceworld.
QUOTE (scienceworld+)
Thus, since c is constant, the momentum four-vector encodes the fact that the rest mass m is also a constant independent of the reference frame.
To sum up then, and to directly quote the current head of department where I work, an eminent theoretical physicist, m = γm_0 is a load of <expletive deleted>. I couldn't have put it better myself.
To sum up then, and to directly quote the current head of department where I work, an eminent theoretical physicist, m = γm_0 is a load of <expletive deleted>. I couldn't have put it better myself.
You're really hung up on this relativistic momentum thing, aren't you. I think you're wrong on this point, and one of the reasons is because it is an inelastic collision between two particles and thus you only need to consider 2 dimensions, one of space and one of time. The equations I've been using result from simplifying your 4 momentum. You're trying to make this more complicated than it is. The fact that I solved for the rest mass after the collision apparently escaped you. I'm very aware of the un-value of the concept of relativistic mass.
Anyway, the equation E = γmc² appears in the Invariant Mass section of Chapter 2 in Modern Physics by Tipler and Llewellyn. It's a decent undergraduate text and since you've become quite rusty at this sort of thing, I recommend you brush up before making any more asinine assertions.
I do hope AlphaNumeric or similar physics heavyweight wieghs in on this issue.
Anyway, the equation E = γmc² appears in the Invariant Mass section of Chapter 2 in Modern Physics by Tipler and Llewellyn. It's a decent undergraduate text and since you've become quite rusty at this sort of thing, I recommend you brush up before making any more asinine assertions.
I do hope AlphaNumeric or similar physics heavyweight wieghs in on this issue.
i always thought temperature was related to the average kinetic energy of the particles in the system. from our frame, an increase in the particles KE would increase the mass of the particle.
QUOTE (barakn+May 4 2008, 06:15 PM)
You're really hung up on this relativistic momentum thing, aren't you. I think you're wrong on this point, and one of the reasons is because it is an inelastic collision between two particles and thus you only need to consider 2 dimensions, one of space and one of time. The equations I've been using result from simplifying your 4 momentum. You're trying to make this more complicated than it is. The fact that I solved for the rest mass after the collision apparently escaped you. I'm very aware of the un-value of the concept of relativistic mass.
So how exactly is the fact that you're ignoring 2 dimensions of space got anything to do with mass being invariant? I'm not saying you have to consider 4 dimensions because in the example you are considering you can always perform a coordinate transformation such that 2 of the dimensions are unchanging with time, as you have correctly recognized.
I don't understand how you can argue this: a hydrogen atom is an electron plus a proton which we'll ignore for a minute. A photon (massless) with a particular energy excites the electron to a higher energy state. The electron is still the same mass, and always has the same mass.
I don't have a copy of that book so I can't really comment on it. I'm having quite a chuckle at your attempt at a jibe though, seeing as I'm competent enough to be doing cutting edge research in quantum field theory, for which special relativity is pretty much essential. I am scratching my head over your use of the phrase "Invariant Mass," too. I'm inclined to suspect that you've simply misunderstood the book.
I don't have a copy of that book so I can't really comment on it. I'm having quite a chuckle at your attempt at a jibe though, seeing as I'm competent enough to be doing cutting edge research in quantum field theory, for which special relativity is pretty much essential. I am scratching my head over your use of the phrase "Invariant Mass," too. I'm inclined to suspect that you've simply misunderstood the book.
I do hope AlphaNumeric or similar physics heavyweight wieghs in on this issue.
Am I not heavy enough for you then? It's pretty funny that you can only cite a wiki page and a general undergrad book which almost certainly talks for about 1 or 2 chapters about special relativity. Did you even read the references I provided? The seminal text on SR is by A P French. You'll find it a much better more in depth and satisfying read than Tipler and Llewellyn I'll wager.
A better example would have been a microwave since the HOT coils heat the bread by first heating the air between the coils and bread.
However even with a microwave, the heat is molecular agitation. You can microwave the food all you want and it wont increase in mass. Eventually it will lose mass when it catches fire and burns.
A better example would have been a microwave since the HOT coils heat the bread by first heating the air between the coils and bread.
However even with a microwave, the heat is molecular agitation. You can microwave the food all you want and it wont increase in mass. Eventually it will lose mass when it catches fire and burns.
Precursor, why would a change in volume mean anything to do with mass?
I can't believe you are asking that especially after I have shown on here what the relationship between mass, volume and density are.
If you increase the volume but leave the density the same you will get an increase in mass just as if you increase the density and leave the volume the same will also increase the mass.
Was that suppose to prove some point? If water is densest at 4C then it has not increased in volume. It is at its smallest volume for a specific amount where heating that water will cause it to expand (increase in volume) from that smallest volume (densest state). It will slowly increase in volume (very small increase) until 100C where the water boils, increasing in volume and decreasing in density at a much faster rate.
Although there are cases where a decrease in temperature will also cause an increase in volume (water being the example for that when it turns to ice) but there is a decrease in density that is proportional to the increase in volume as to keep the mass the same.
Was that suppose to prove some point? If water is densest at 4C then it has not increased in volume. It is at its smallest volume for a specific amount where heating that water will cause it to expand (increase in volume) from that smallest volume (densest state). It will slowly increase in volume (very small increase) until 100C where the water boils, increasing in volume and decreasing in density at a much faster rate.
Although there are cases where a decrease in temperature will also cause an increase in volume (water being the example for that when it turns to ice) but there is a decrease in density that is proportional to the increase in volume as to keep the mass the same.
In relativity mass is defined as the invariant length of the energy momentum 4 vector. There is no difference between rest mass and dynamical mass.
Although I am sure this is true, this simply isn't a quantum matter.
Not exactly. The mass you talk of are the molecules. When you talk of E=mc^2 and E=hv the energy (E) is what the existing matter possesses. This energy is simply motion (spin, vibration, etc.).
hv = mc^2
h = De Broglie's constant
V = frequency
m = mass
c^2 = speed of light squared (another constant)
Rewritten as
m = (hv)/c^2
v = (mc^2)/h
This refers to photons. By knowing the frequency you can find the mass or by knowing the mass you can find the frequency.
What I believe is that not all photons are of the same volume/density (I'm more inclined to go with density) and this results in a difference in mass. The frequency imposed onto a photon when 'created' depends upon the mass of the photon particle.
So will an atom absorbing a photon add to the mass of the atom? Absolutely but on a much larger scale (dealing with molecules) the amount of mass that is added when accompanied with the amount of mass given off as photons does not warrant a significant difference in this case.
For a correct answer to the poster's question you don't have to go any further than the relationship between mass, density and volume and heat simply being molecular motion.
So how exactly is the fact that you're ignoring 2 dimensions of space got anything to do with mass being invariant? I'm not saying you have to consider 4 dimensions because in the example you are considering you can always perform a coordinate transformation such that 2 of the dimensions are unchanging with time, as you have correctly recognized.
I don't understand how you can argue this: a hydrogen atom is an electron plus a proton which we'll ignore for a minute. A photon (massless) with a particular energy excites the electron to a higher energy state. The electron is still the same mass, and always has the same mass.
QUOTE
Anyway, the equation E = γmc² appears in the Invariant Mass section of Chapter 2 in Modern Physics by Tipler and Llewellyn. It's a decent undergraduate text and since you've become quite rusty at this sort of thing, I recommend you brush up before making any more asinine assertions.
I don't have a copy of that book so I can't really comment on it. I'm having quite a chuckle at your attempt at a jibe though, seeing as I'm competent enough to be doing cutting edge research in quantum field theory, for which special relativity is pretty much essential. I am scratching my head over your use of the phrase "Invariant Mass," too. I'm inclined to suspect that you've simply misunderstood the book.
QUOTE (->
| QUOTE |
| Anyway, the equation E = γmc² appears in the Invariant Mass section of Chapter 2 in Modern Physics by Tipler and Llewellyn. It's a decent undergraduate text and since you've become quite rusty at this sort of thing, I recommend you brush up before making any more asinine assertions. |
I don't have a copy of that book so I can't really comment on it. I'm having quite a chuckle at your attempt at a jibe though, seeing as I'm competent enough to be doing cutting edge research in quantum field theory, for which special relativity is pretty much essential. I am scratching my head over your use of the phrase "Invariant Mass," too. I'm inclined to suspect that you've simply misunderstood the book.
I do hope AlphaNumeric or similar physics heavyweight wieghs in on this issue.
Am I not heavy enough for you then? It's pretty funny that you can only cite a wiki page and a general undergrad book which almost certainly talks for about 1 or 2 chapters about special relativity. Did you even read the references I provided? The seminal text on SR is by A P French. You'll find it a much better more in depth and satisfying read than Tipler and Llewellyn I'll wager.
QUOTE
How is it different? The original question was about heating an object. You could heat an object by bathing it in electromagnetic radiation, like toasting bread by having it absorb infrared wavelength photons emitted by hot coils.
A better example would have been a microwave since the HOT coils heat the bread by first heating the air between the coils and bread.
However even with a microwave, the heat is molecular agitation. You can microwave the food all you want and it wont increase in mass. Eventually it will lose mass when it catches fire and burns.
QUOTE (->
| QUOTE |
| How is it different? The original question was about heating an object. You could heat an object by bathing it in electromagnetic radiation, like toasting bread by having it absorb infrared wavelength photons emitted by hot coils. |
A better example would have been a microwave since the HOT coils heat the bread by first heating the air between the coils and bread.
However even with a microwave, the heat is molecular agitation. You can microwave the food all you want and it wont increase in mass. Eventually it will lose mass when it catches fire and burns.
Precursor, why would a change in volume mean anything to do with mass?
I can't believe you are asking that especially after I have shown on here what the relationship between mass, volume and density are.
If you increase the volume but leave the density the same you will get an increase in mass just as if you increase the density and leave the volume the same will also increase the mass.
QUOTE
Besides, not all substances always increase in volume with an increase in temperature. Water is densest at 4C. And thankfully too.
Was that suppose to prove some point? If water is densest at 4C then it has not increased in volume. It is at its smallest volume for a specific amount where heating that water will cause it to expand (increase in volume) from that smallest volume (densest state). It will slowly increase in volume (very small increase) until 100C where the water boils, increasing in volume and decreasing in density at a much faster rate.
Although there are cases where a decrease in temperature will also cause an increase in volume (water being the example for that when it turns to ice) but there is a decrease in density that is proportional to the increase in volume as to keep the mass the same.
QUOTE (->
| QUOTE |
| Besides, not all substances always increase in volume with an increase in temperature. Water is densest at 4C. And thankfully too. |
Was that suppose to prove some point? If water is densest at 4C then it has not increased in volume. It is at its smallest volume for a specific amount where heating that water will cause it to expand (increase in volume) from that smallest volume (densest state). It will slowly increase in volume (very small increase) until 100C where the water boils, increasing in volume and decreasing in density at a much faster rate.
Although there are cases where a decrease in temperature will also cause an increase in volume (water being the example for that when it turns to ice) but there is a decrease in density that is proportional to the increase in volume as to keep the mass the same.
In relativity mass is defined as the invariant length of the energy momentum 4 vector. There is no difference between rest mass and dynamical mass.
Although I am sure this is true, this simply isn't a quantum matter.
QUOTE
Heat is a form of energy, so it does have a mass, of course
Not exactly. The mass you talk of are the molecules. When you talk of E=mc^2 and E=hv the energy (E) is what the existing matter possesses. This energy is simply motion (spin, vibration, etc.).
hv = mc^2
h = De Broglie's constant
V = frequency
m = mass
c^2 = speed of light squared (another constant)
Rewritten as
m = (hv)/c^2
v = (mc^2)/h
This refers to photons. By knowing the frequency you can find the mass or by knowing the mass you can find the frequency.
What I believe is that not all photons are of the same volume/density (I'm more inclined to go with density) and this results in a difference in mass. The frequency imposed onto a photon when 'created' depends upon the mass of the photon particle.
So will an atom absorbing a photon add to the mass of the atom? Absolutely but on a much larger scale (dealing with molecules) the amount of mass that is added when accompanied with the amount of mass given off as photons does not warrant a significant difference in this case.
For a correct answer to the poster's question you don't have to go any further than the relationship between mass, density and volume and heat simply being molecular motion.
QUOTE (Precursor562+May 9 2008, 11:57 PM)
Not exactly. The mass you talk of are the molecules. When you talk of E=mc^2 and E=hv the energy (E) is what the existing matter possesses. This energy is simply motion (spin, vibration, etc.).
hv = mc^2
h = De Broglie's constant
V = frequency
m = mass
c^2 = speed of light squared (another constant)
Rewritten as
m = (hv)/c^2
v = (mc^2)/h
This refers to photons. By knowing the frequency you can find the mass or by knowing the mass you can find the frequency.
What I believe is that not all photons are of the same volume/density (I'm more inclined to go with density) and this results in a difference in mass. The frequency imposed onto a photon when 'created' depends upon the mass of the photon particle.
So will an atom absorbing a photon add to the mass of the atom? Absolutely but on a much larger scale (dealing with molecules) the amount of mass that is added when accompanied with the amount of mass given off as photons does not warrant a significant difference in this case.
For a correct answer to the poster's question you don't have to go any further than the relationship between mass, density and volume and heat simply being molecular motion.
This is not a bad effort to be fair, but there is an error in the first step. E = mc^2 is only valid for particles at rest. Since a photon (in vacuum) is always moving with a speed c with respect to any inertial observer you cannot apply this relation to a photon.
For a photon, the mass is zero so E²=(pc)² + (mc²)² reduces to E = pc and your analysis leads to p = hv/c = h/wavelength which is the De Broglie relation.
hv = mc^2
h = De Broglie's constant
V = frequency
m = mass
c^2 = speed of light squared (another constant)
Rewritten as
m = (hv)/c^2
v = (mc^2)/h
This refers to photons. By knowing the frequency you can find the mass or by knowing the mass you can find the frequency.
What I believe is that not all photons are of the same volume/density (I'm more inclined to go with density) and this results in a difference in mass. The frequency imposed onto a photon when 'created' depends upon the mass of the photon particle.
So will an atom absorbing a photon add to the mass of the atom? Absolutely but on a much larger scale (dealing with molecules) the amount of mass that is added when accompanied with the amount of mass given off as photons does not warrant a significant difference in this case.
For a correct answer to the poster's question you don't have to go any further than the relationship between mass, density and volume and heat simply being molecular motion.
This is not a bad effort to be fair, but there is an error in the first step. E = mc^2 is only valid for particles at rest. Since a photon (in vacuum) is always moving with a speed c with respect to any inertial observer you cannot apply this relation to a photon.
For a photon, the mass is zero so E²=(pc)² + (mc²)² reduces to E = pc and your analysis leads to p = hv/c = h/wavelength which is the De Broglie relation.
Ok, I can pick up a 20kg iron mass @ room temperature .... but when it's heated to 500 degrees centigrade I can't?
...... extract from the forthcoming, and most mind-pulpingly cretinous book; "A Brief History of Scorched Flesh" Co-authored by Zarkov & DavidDipfukker.
...... extract from the forthcoming, and most mind-pulpingly cretinous book; "A Brief History of Scorched Flesh" Co-authored by Zarkov & DavidDipfukker.
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