# Thread: Do protons rapidly heat up when they collide at CERN?

1. An attempt to calculate the temperature of protons at the instant they collide at CERN, based on the molecular velocity in an ideal gas.

The well-known formula for gas PV and T relationship is PV = nRT
where P is pressure, V is volume (not velocity), n is the number of moles of gas particles, R is the ideal gas constant and T is the absolute temperature in deg K.
Studying how this gas law is combined with the formula for kinetic energy we get
KE = 1/2 m u2 = 3/2 k T. m is the non-relativistic, so called rest mass.
The derivation of this formula can be found on the net in many places.
Appreciate the fact that the velocity for a single gas particle cannot normally be derived from the average temperature throughout a gas. This is because in everyday situations, each particle has a different velocity and so has a different temperature. But my first question is: how different are these temperatures? As a total system, the gas has a temperature that can only be calculated for normal situations but with extra high velocity protons all moving in the same direction as at CERN, I wonder why not. Normally all the hydrogen molecules would be moving randomly but at CERN they are all going in the same direction (which changes due to the accelerator’s circular shape). The protons are very cool when they are injected into the ring because they must not be allowed to expand the beam and the higher their temperature the bigger any component of movement away from the main direction will be, in effect defocussing the beam.
This is where the philosophy or insight becomes difficult because with an everyday gas the temperature is applied to the gas and the molecular speeds generate themselves whereas CERN moves the protons at a given speed and the proton temperature stays cool until the micro pico second while they collide. What I claim is that since the mathematical relationship between molecular speeds and temperature has no preferred direction it should apply to the CERN case. And it seems to me that the only reason for theorists saying that the temperature of any given molecule cannot be calculated is that there will be a distribution of speeds among them in a normal molecular vs temperature situation. But if they all moved at the same speed, they would all have the same temperature. And all moving at the same speed and direction is what happens at CERN.
Also, CERN accelerates not just one proton but very many (1012 ie a tiny fraction of a mole’s worth) of them at the same time, all in a bunch. A mole’s worth is 1022 protons and there are some 2800 bunches going around the ring, one after the other. Let’s concentrate on just one bunch.
Going back and calculating the root-mean-square velocity of gas molecules in normal circumstances from the temperature of gas using the following well known equation (which can be inspected in detail on the net in many places), we have
Vrms = (3RT/M)1/2 In the case of hydrogen atoms (if they could exist as single atoms and which are mostly protons and whose mass is therefore close to that of protons) we can first calculate the temperature of one mole of protons, even though bunches of protons only contain 1012 protons. We must make sure to use units consistently. For example, if the molecular weight is taken to be in grams per mole and the value of the ideal gas constant is in joules per mole per degree Kelvin, and the temperature is in degrees Kelvin, then the ideal gas constant is in joules per mole-degree Kelvin, and the velocity is in .meters per second. Using this scale and the rest mass of protons we get :- V2 = 3RT/M and T = V2M/3R = (0.99999) x3 x108)2 x 1.00794 / 3 x8.314 for one mole’s worth.
T = 8.99982 x 1016 x 1.00794 / 24.942 = 0.3639 x 1016 degrees Kelvin per mole.

Discussion:
1 mole of protons contains 1022 protons so the mass of just one proton is much smaller than the mass of 1 mole. Since we are discussing the temperature at which just one proton will break up, we have to decide whether to divide mass by 1022 or not. As a guide, reflect that when molecular speeds in a normal gas are calculated, the mass figure used is that of just the one molecule. But if we did that it would reduce the proton temperature dramatically. To counter this, since the protons travel at relativistic speeds, their mass will be much bigger than at normal speeds. Bear in mind also that in the formula for calculating velocity V2 = 3RT/M, M is the weight in grams of 1 mole of gas, not the weight of just one molecule, nor is it the actual number of protons circulated at CERN. And since they all travel at the same speed, each proton travels at the speed calculated for 10 -10 of one mole. The 10 -10 comes in because there are only 10 12 protons in a bunch, not 10 22. So T = 0.3639 x 106 degrees using the rest-mass figure.
Relativistic mass:
The reported speed of protons compared the that of light at CERN is 0.999999991. Call this X, with 7 x1012 volts applied (as reported). So the relativistic mass ratio is (1 – X2) -1/2 This comes to Y, several millions, depending on how many places your calculator can handle. Therefore, the temperature of one bunch of protons at the time of collision would be Y times higher. This comes to anything from about 10 10 to 10 12 degrees.
Summary:
The calculation gives answers which vary enormously depending on what assumptions are made. I think we should use M = 10 -10 x1.00794 grams, and multiply that number by several million to allow for relativity. If we do that the temperature of protons at the moment of impact is around 1012 degrees. But this cannot be right because CERN has only just announced (Dec 2019) its latest temperature to be about 5.5 x 1012 degrees using lead ions crashing into protons to give higher temperatures than protons – protons gives. Either way 1012 is not hot enough to make protons disintegrate just for thermal reasons in a proton-to-proton smash.
So why do they disintegrate? Well, in the report entitled The Big Bang, Protons, Neutrons and Electrons, (see the net), Sister Pat claims that protons were first created when the temperature was 1010 degrees. I am not sure this is correct because in her time-line diagram she shows protons first being created at 1013 degrees. One of these figures is wrong and this causes me to doubt it all. Also, other pages on the net, written by physicists, say that below 1013, protons and neutrons were no longer generated in pairs. Yet other sources say that protons were first produced at 1010 degrees. With such variable information it seems to be impossible to say for certain what the temperature of the protons is. Another aspect of this is that they might not become very hot at all and that they break up for an altogether different reason.

2.

3. Sorry this whole post is ignorant twaddle.. If a little knowledge is a dangerous thing you are fucking lethal...

An attempt to calculate the temperature of protons at the instant they collide at CERN, based on the molecular velocity in an ideal gas.

The well-known formula for gas PV and T relationship is PV = nRT
where P is pressure, V is volume (not velocity), n is the number of moles of gas particles, R is the ideal gas constant and T is the absolute temperature in deg K.
Studying how this gas law is combined with the formula for kinetic energy we get
KE = 1/2 m u2 = 3/2 k T. m is the non-relativistic, so called rest mass.
The derivation of this formula can be found on the net in many places.
Appreciate the fact that the velocity for a single gas particle cannot normally be derived from the average temperature throughout a gas. This is because in everyday situations, each particle has a different velocity and so has a different temperature. But my first question is: how different are these temperatures? As a total system, the gas has a temperature that can only be calculated for normal situations but with extra high velocity protons all moving in the same direction as at CERN, I wonder why not. Normally all the hydrogen molecules would be moving randomly but at CERN they are all going in the same direction (which changes due to the accelerator’s circular shape). The protons are very cool when they are injected into the ring because they must not be allowed to expand the beam and the higher their temperature the bigger any component of movement away from the main direction will be, in effect defocussing the beam.
This is where the philosophy or insight becomes difficult because with an everyday gas the temperature is applied to the gas and the molecular speeds generate themselves whereas CERN moves the protons at a given speed and the proton temperature stays cool until the micro pico second while they collide. What I claim is that since the mathematical relationship between molecular speeds and temperature has no preferred direction it should apply to the CERN case. And it seems to me that the only reason for theorists saying that the temperature of any given molecule cannot be calculated is that there will be a distribution of speeds among them in a normal molecular vs temperature situation. But if they all moved at the same speed, they would all have the same temperature. And all moving at the same speed and direction is what happens at CERN.
Also, CERN accelerates not just one proton but very many (1012 ie a tiny fraction of a mole’s worth) of them at the same time, all in a bunch. A mole’s worth is 1022 protons and there are some 2800 bunches going around the ring, one after the other. Let’s concentrate on just one bunch.
Going back and calculating the root-mean-square velocity of gas molecules in normal circumstances from the temperature of gas using the following well known equation (which can be inspected in detail on the net in many places), we have
Vrms = (3RT/M)1/2 In the case of hydrogen atoms (if they could exist as single atoms and which are mostly protons and whose mass is therefore close to that of protons) we can first calculate the temperature of one mole of protons, even though bunches of protons only contain 1012 protons. We must make sure to use units consistently. For example, if the molecular weight is taken to be in grams per mole and the value of the ideal gas constant is in joules per mole per degree Kelvin, and the temperature is in degrees Kelvin, then the ideal gas constant is in joules per mole-degree Kelvin, and the velocity is in .meters per second. Using this scale and the rest mass of protons we get :- V2 = 3RT/M and T = V2M/3R = (0.99999) x3 x108)2 x 1.00794 / 3 x8.314 for one mole’s worth.
T = 8.99982 x 1016 x 1.00794 / 24.942 = 0.3639 x 1016 degrees Kelvin per mole.

Discussion:
1 mole of protons contains 1022 protons so the mass of just one proton is much smaller than the mass of 1 mole. Since we are discussing the temperature at which just one proton will break up, we have to decide whether to divide mass by 1022 or not. As a guide, reflect that when molecular speeds in a normal gas are calculated, the mass figure used is that of just the one molecule. But if we did that it would reduce the proton temperature dramatically. To counter this, since the protons travel at relativistic speeds, their mass will be much bigger than at normal speeds. Bear in mind also that in the formula for calculating velocity V2 = 3RT/M, M is the weight in grams of 1 mole of gas, not the weight of just one molecule, nor is it the actual number of protons circulated at CERN. And since they all travel at the same speed, each proton travels at the speed calculated for 10 -10 of one mole. The 10 -10 comes in because there are only 10 12 protons in a bunch, not 10 22. So T = 0.3639 x 106 degrees using the rest-mass figure.
Relativistic mass:
The reported speed of protons compared the that of light at CERN is 0.999999991. Call this X, with 7 x1012 volts applied (as reported). So the relativistic mass ratio is (1 – X2) -1/2 This comes to Y, several millions, depending on how many places your calculator can handle. Therefore, the temperature of one bunch of protons at the time of collision would be Y times higher. This comes to anything from about 10 10 to 10 12 degrees.
Summary:
The calculation gives answers which vary enormously depending on what assumptions are made. I think we should use M = 10 -10 x1.00794 grams, and multiply that number by several million to allow for relativity. If we do that the temperature of protons at the moment of impact is around 1012 degrees. But this cannot be right because CERN has only just announced (Dec 2019) its latest temperature to be about 5.5 x 1012 degrees using lead ions crashing into protons to give higher temperatures than protons – protons gives. Either way 1012 is not hot enough to make protons disintegrate just for thermal reasons in a proton-to-proton smash.
So why do they disintegrate? Well, in the report entitled The Big Bang, Protons, Neutrons and Electrons, (see the net), Sister Pat claims that protons were first created when the temperature was 1010 degrees. I am not sure this is correct because in her time-line diagram she shows protons first being created at 1013 degrees. One of these figures is wrong and this causes me to doubt it all. Also, other pages on the net, written by physicists, say that below 1013, protons and neutrons were no longer generated in pairs. Yet other sources say that protons were first produced at 1010 degrees. With such variable information it seems to be impossible to say for certain what the temperature of the protons is. Another aspect of this is that they might not become very hot at all and that they break up for an altogether different reason.
By your logic, a train at full speed is hotter than one standing in a station.

5. Originally Posted by exchemist
An attempt to calculate the temperature of protons at the instant they collide at CERN, based on the molecular velocity in an ideal gas.

The well-known formula for gas PV and T relationship is PV = nRT
where P is pressure, V is volume (not velocity), n is the number of moles of gas particles, R is the ideal gas constant and T is the absolute temperature in deg K.
Studying how this gas law is combined with the formula for kinetic energy we get
KE = 1/2 m u2 = 3/2 k T. m is the non-relativistic, so called rest mass.
The derivation of this formula can be found on the net in many places.
Appreciate the fact that the velocity for a single gas particle cannot normally be derived from the average temperature throughout a gas. This is because in everyday situations, each particle has a different velocity and so has a different temperature. But my first question is: how different are these temperatures? As a total system, the gas has a temperature that can only be calculated for normal situations but with extra high velocity protons all moving in the same direction as at CERN, I wonder why not. Normally all the hydrogen molecules would be moving randomly but at CERN they are all going in the same direction (which changes due to the accelerator’s circular shape). The protons are very cool when they are injected into the ring because they must not be allowed to expand the beam and the higher their temperature the bigger any component of movement away from the main direction will be, in effect defocussing the beam.
This is where the philosophy or insight becomes difficult because with an everyday gas the temperature is applied to the gas and the molecular speeds generate themselves whereas CERN moves the protons at a given speed and the proton temperature stays cool until the micro pico second while they collide. What I claim is that since the mathematical relationship between molecular speeds and temperature has no preferred direction it should apply to the CERN case. And it seems to me that the only reason for theorists saying that the temperature of any given molecule cannot be calculated is that there will be a distribution of speeds among them in a normal molecular vs temperature situation. But if they all moved at the same speed, they would all have the same temperature. And all moving at the same speed and direction is what happens at CERN.
Also, CERN accelerates not just one proton but very many (1012 ie a tiny fraction of a mole’s worth) of them at the same time, all in a bunch. A mole’s worth is 1022 protons and there are some 2800 bunches going around the ring, one after the other. Let’s concentrate on just one bunch.
Going back and calculating the root-mean-square velocity of gas molecules in normal circumstances from the temperature of gas using the following well known equation (which can be inspected in detail on the net in many places), we have
Vrms = (3RT/M)1/2 In the case of hydrogen atoms (if they could exist as single atoms and which are mostly protons and whose mass is therefore close to that of protons) we can first calculate the temperature of one mole of protons, even though bunches of protons only contain 1012 protons. We must make sure to use units consistently. For example, if the molecular weight is taken to be in grams per mole and the value of the ideal gas constant is in joules per mole per degree Kelvin, and the temperature is in degrees Kelvin, then the ideal gas constant is in joules per mole-degree Kelvin, and the velocity is in .meters per second. Using this scale and the rest mass of protons we get :- V2 = 3RT/M and T = V2M/3R = (0.99999) x3 x108)2 x 1.00794 / 3 x8.314 for one mole’s worth.
T = 8.99982 x 1016 x 1.00794 / 24.942 = 0.3639 x 1016 degrees Kelvin per mole.

Discussion:
1 mole of protons contains 1022 protons so the mass of just one proton is much smaller than the mass of 1 mole. Since we are discussing the temperature at which just one proton will break up, we have to decide whether to divide mass by 1022 or not. As a guide, reflect that when molecular speeds in a normal gas are calculated, the mass figure used is that of just the one molecule. But if we did that it would reduce the proton temperature dramatically. To counter this, since the protons travel at relativistic speeds, their mass will be much bigger than at normal speeds. Bear in mind also that in the formula for calculating velocity V2 = 3RT/M, M is the weight in grams of 1 mole of gas, not the weight of just one molecule, nor is it the actual number of protons circulated at CERN. And since they all travel at the same speed, each proton travels at the speed calculated for 10 -10 of one mole. The 10 -10 comes in because there are only 10 12 protons in a bunch, not 10 22. So T = 0.3639 x 106 degrees using the rest-mass figure.
Relativistic mass:
The reported speed of protons compared the that of light at CERN is 0.999999991. Call this X, with 7 x1012 volts applied (as reported). So the relativistic mass ratio is (1 – X2) -1/2 This comes to Y, several millions, depending on how many places your calculator can handle. Therefore, the temperature of one bunch of protons at the time of collision would be Y times higher. This comes to anything from about 10 10 to 10 12 degrees.
Summary:
The calculation gives answers which vary enormously depending on what assumptions are made. I think we should use M = 10 -10 x1.00794 grams, and multiply that number by several million to allow for relativity. If we do that the temperature of protons at the moment of impact is around 1012 degrees. But this cannot be right because CERN has only just announced (Dec 2019) its latest temperature to be about 5.5 x 1012 degrees using lead ions crashing into protons to give higher temperatures than protons – protons gives. Either way 1012 is not hot enough to make protons disintegrate just for thermal reasons in a proton-to-proton smash.
So why do they disintegrate? Well, in the report entitled The Big Bang, Protons, Neutrons and Electrons, (see the net), Sister Pat claims that protons were first created when the temperature was 1010 degrees. I am not sure this is correct because in her time-line diagram she shows protons first being created at 1013 degrees. One of these figures is wrong and this causes me to doubt it all. Also, other pages on the net, written by physicists, say that below 1013, protons and neutrons were no longer generated in pairs. Yet other sources say that protons were first produced at 1010 degrees. With such variable information it seems to be impossible to say for certain what the temperature of the protons is. Another aspect of this is that they might not become very hot at all and that they break up for an altogether different reason.
By your logic, a train at full speed is hotter than one standing in a station.

Worse yet, its temperature would be different depending on whether you were measuring it relative to the ground, relative to the center of the Earth, relative to the Sun, etc.

The point here is that while temperature is a measure of average kinetic energy of the individual components, that measurement is made relative to the center of mass for the collection of components. Ergo, the "temperature" for a packet of protons in an accelerator would be determined by the average KE of the individual protons measured relative to the packet's center of mass, and not measured relative to the accelerator frame.

6. Originally Posted by Janus
Originally Posted by exchemist
An attempt to calculate the temperature of protons at the instant they collide at CERN, based on the molecular velocity in an ideal gas.

The well-known formula for gas PV and T relationship is PV = nRT
where P is pressure, V is volume (not velocity), n is the number of moles of gas particles, R is the ideal gas constant and T is the absolute temperature in deg K.
Studying how this gas law is combined with the formula for kinetic energy we get
KE = 1/2 m u2 = 3/2 k T. m is the non-relativistic, so called rest mass.
The derivation of this formula can be found on the net in many places.
Appreciate the fact that the velocity for a single gas particle cannot normally be derived from the average temperature throughout a gas. This is because in everyday situations, each particle has a different velocity and so has a different temperature. But my first question is: how different are these temperatures? As a total system, the gas has a temperature that can only be calculated for normal situations but with extra high velocity protons all moving in the same direction as at CERN, I wonder why not. Normally all the hydrogen molecules would be moving randomly but at CERN they are all going in the same direction (which changes due to the accelerator’s circular shape). The protons are very cool when they are injected into the ring because they must not be allowed to expand the beam and the higher their temperature the bigger any component of movement away from the main direction will be, in effect defocussing the beam.
This is where the philosophy or insight becomes difficult because with an everyday gas the temperature is applied to the gas and the molecular speeds generate themselves whereas CERN moves the protons at a given speed and the proton temperature stays cool until the micro pico second while they collide. What I claim is that since the mathematical relationship between molecular speeds and temperature has no preferred direction it should apply to the CERN case. And it seems to me that the only reason for theorists saying that the temperature of any given molecule cannot be calculated is that there will be a distribution of speeds among them in a normal molecular vs temperature situation. But if they all moved at the same speed, they would all have the same temperature. And all moving at the same speed and direction is what happens at CERN.
Also, CERN accelerates not just one proton but very many (1012 ie a tiny fraction of a mole’s worth) of them at the same time, all in a bunch. A mole’s worth is 1022 protons and there are some 2800 bunches going around the ring, one after the other. Let’s concentrate on just one bunch.
Going back and calculating the root-mean-square velocity of gas molecules in normal circumstances from the temperature of gas using the following well known equation (which can be inspected in detail on the net in many places), we have
Vrms = (3RT/M)1/2 In the case of hydrogen atoms (if they could exist as single atoms and which are mostly protons and whose mass is therefore close to that of protons) we can first calculate the temperature of one mole of protons, even though bunches of protons only contain 1012 protons. We must make sure to use units consistently. For example, if the molecular weight is taken to be in grams per mole and the value of the ideal gas constant is in joules per mole per degree Kelvin, and the temperature is in degrees Kelvin, then the ideal gas constant is in joules per mole-degree Kelvin, and the velocity is in .meters per second. Using this scale and the rest mass of protons we get :- V2 = 3RT/M and T = V2M/3R = (0.99999) x3 x108)2 x 1.00794 / 3 x8.314 for one mole’s worth.
T = 8.99982 x 1016 x 1.00794 / 24.942 = 0.3639 x 1016 degrees Kelvin per mole.

Discussion:
1 mole of protons contains 1022 protons so the mass of just one proton is much smaller than the mass of 1 mole. Since we are discussing the temperature at which just one proton will break up, we have to decide whether to divide mass by 1022 or not. As a guide, reflect that when molecular speeds in a normal gas are calculated, the mass figure used is that of just the one molecule. But if we did that it would reduce the proton temperature dramatically. To counter this, since the protons travel at relativistic speeds, their mass will be much bigger than at normal speeds. Bear in mind also that in the formula for calculating velocity V2 = 3RT/M, M is the weight in grams of 1 mole of gas, not the weight of just one molecule, nor is it the actual number of protons circulated at CERN. And since they all travel at the same speed, each proton travels at the speed calculated for 10 -10 of one mole. The 10 -10 comes in because there are only 10 12 protons in a bunch, not 10 22. So T = 0.3639 x 106 degrees using the rest-mass figure.
Relativistic mass:
The reported speed of protons compared the that of light at CERN is 0.999999991. Call this X, with 7 x1012 volts applied (as reported). So the relativistic mass ratio is (1 – X2) -1/2 This comes to Y, several millions, depending on how many places your calculator can handle. Therefore, the temperature of one bunch of protons at the time of collision would be Y times higher. This comes to anything from about 10 10 to 10 12 degrees.
Summary:
The calculation gives answers which vary enormously depending on what assumptions are made. I think we should use M = 10 -10 x1.00794 grams, and multiply that number by several million to allow for relativity. If we do that the temperature of protons at the moment of impact is around 1012 degrees. But this cannot be right because CERN has only just announced (Dec 2019) its latest temperature to be about 5.5 x 1012 degrees using lead ions crashing into protons to give higher temperatures than protons – protons gives. Either way 1012 is not hot enough to make protons disintegrate just for thermal reasons in a proton-to-proton smash.
So why do they disintegrate? Well, in the report entitled The Big Bang, Protons, Neutrons and Electrons, (see the net), Sister Pat claims that protons were first created when the temperature was 1010 degrees. I am not sure this is correct because in her time-line diagram she shows protons first being created at 1013 degrees. One of these figures is wrong and this causes me to doubt it all. Also, other pages on the net, written by physicists, say that below 1013, protons and neutrons were no longer generated in pairs. Yet other sources say that protons were first produced at 1010 degrees. With such variable information it seems to be impossible to say for certain what the temperature of the protons is. Another aspect of this is that they might not become very hot at all and that they break up for an altogether different reason.
By your logic, a train at full speed is hotter than one standing in a station.

Worse yet, its temperature would be different depending on whether you were measuring it relative to the ground, relative to the center of the Earth, relative to the Sun, etc.

The point here is that while temperature is a measure of average kinetic energy of the individual components, that measurement is made relative to the center of mass for the collection of components. Ergo, the "temperature" for a packet of protons in an accelerator would be determined by the average KE of the individual protons measured relative to the packet's center of mass, and not measured relative to the accelerator frame.
Indeed.

But since this poster has only ever made one other post, and that was 9 years ago, I'm not counting on him reading this and responding.

7. "Wake up, you wooly head
Put on some clothes, shake up your bed"

8. Great post, you have pointed out some excellent points. This is actually the kind of information i have been trying to find. Thank you for writing this information.

9. The V in PV=nRT refers to average speed of molecules in a gas, which is not moving, not as a group moving together. Winds in tornadoes and hurricanes are not heating up.

10. Originally Posted by mathman
The V in PV=nRT refers to average speed of molecules in a gas, which is not moving, not as a group moving together. Winds in tornadoes and hurricanes are not heating up.
V is volume.

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