Mystery where propulsion originates

  • #1
Compared how much impulse is potentially created by heating up a gas as compared to the impulse created by the laser for the same amount of energy. Find that the gas can get a million times more impulse per energy increase. If the heat source was a heater that only heats things up by radiating photons, how did the gas end up with a million times more impulse change than the photons from the heater had originally?

Calculation for those that don't take my word for it,

Let us use nitrogen gas and the ideal gas law, PV = nRT
R is the gas constant 8.314 J/(mol K),
We will be using 28 grams for the molecular weight of a mol of N2 molecules.

Volume of 1 m3
4 Mols of Nitrogen
Temperature of 200o K
P V= nRT = (6651.2 Kg )/ (m s2) x 1.0 m3 = 4 x 8.314 J x 200 = 6651.2 J

Now we want to add enough energy to bring the temperature up to 800oK

Volume of 1 m3
4 Mols of Nitrogen
Temperature of 800o K
P V= nRT = (6651.2 Kg )/ (m s2) x 1.0 m3 = 4 x 8.314 J x 800 = 26604.8 J

So in order to raise the temperature of our gas to 800 we added 19953.6 J of energy, and now we want to know how much the impulse was potentially increased by adding that energy.


Knowing that we have 4 mols of N2 nitrogen, the mass is 4x28grams or
.112 Kg and using ½ Mv2 we allow it to equal 26604.8 J we get

½ x .112 Kg x v2 = 26604.8 J v = 974.8 m/s so Mv = 109.2 Kg m/s

We have to subtract the impulse we had that we originally had at 6651.2 J

½ x .112 Kg x v2 = 6651.2 J v = 344.6 m/s so Mv = 38.6 Kg m/s

Adding 19953.6 J to this gas, we gained 109.2 - 38.6 = 70.6 Kg m/s Impulse. At least potentially.

Now we are going to calculate the Impulse created by lasers and of two different types for 19953.6 J output from these lasers.

Speed of light 2.99 e 8 meters

Ruby Laser 694.3 nm = 1.35 x 10^15 Hz
Helium – Silver Laser 224.3 nm = 4.17 x 10^15 Hz

Energy per photon = hf or hc/λ
= 2.86 e-13 u J Ruby laser photon
= 8.86 e-13 µ J Helium-Silver Laser photon

For our example in order to have 19953.6 J in photons we need:

19953.6 J / 2.86 e-19 J = 6.96 e 22 Ruby photons
19953.6 J / 8.86 e-19 J = 2.25 e 22 Helium-Silver Laser photon

To get the momentum for a photon divide the photon's energy by the speed of light

Momentum per photon = 9.57 e -28 kg x m/sec for Ruby laser photon
Momentum per photon = 2.96 e -27 kg x m/sec Helium-Silver photon

For 19953.6 Joules turned into photons for ideal directional momentum = 6.69 e-5 Kg x m/sec for a Ruby laser
= 6.69 e-5 Kg x m/ sec for Helium-Silver laser, so the result is independent of frequency.

For an added 19953.6 Joules we got an impulse increase of 70.6 Kg m/s by heating our gas and only 6.69 e-5 Kg m/s by using laser light.

This result indicates that gas rockets are a million times more efficient at turning energy into impulse.
 

Answers and Replies

  • #2
UltrafastPED
Science Advisor
Gold Member
1,912
216
It all depends upon the efficiency of the conversion ...
 
  • #3
800oK should be read as 800 degrees K, sorry I wrote it in Word and copied it over.
 
  • #4
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,847
4,676
What exactly is the mystery here?

Zz.
 
  • #5
phyzguy
Science Advisor
4,794
1,744
I agree with ZapperZ. If you have a given amount of energy E, and you apply it to a non-relativistic mass m, you get a momentum of :

[tex]p=\sqrt{2mE}[/tex]

Whereas if you apply it to light, you get a momentum of:
[tex]p = \frac{E}{c}[/tex]

In most cases, the first case is a lot bigger than the second case, and the bigger you make m the more different it will be. Where's the question?
 
  • #6
DrClaude
Mentor
7,584
3,946
Volume of 1 m3
4 Mols of Nitrogen
Temperature of 200o K
P V= nRT = (6651.2 Kg )/ (m s2) x 1.0 m3 = 4 x 8.314 J x 200 = 6651.2 J

Now we want to add enough energy to bring the temperature up to 800oK

Volume of 1 m3
4 Mols of Nitrogen
Temperature of 800o K
P V= nRT = (6651.2 Kg )/ (m s2) x 1.0 m3 = 4 x 8.314 J x 800 = 26604.8 J

So in order to raise the temperature of our gas to 800 we added 19953.6 J of energy, and now we want to know how much the impulse was potentially increased by adding that energy.
That's not how you calculate the change in energy of an ideal gas. You have to use the formula
$$
\Delta U = \frac{f}{2} n R \Delta T
$$
where ##f## is the number of degrees of freedom, which we can probably take as ##n=5## for N2 in this temperature range.

This result indicates that gas rockets are a million times more efficient at turning energy into impulse.
And what do you get with the laser? A bunch of electronically excited molecules! What about that?

The idea is that if you put in a certain amount of energy into a gas, at equilibrium the gas will be in an identical macrostate whatever the mechanism you used to put the energy in. You concentrated only on the momentum of the photons, but they carry more energy, which eventually, if the system gas+photons is kept in a closed box, will lead to the same result as heating the gas.

And by the way, the unit is kelvin. There are no degrees in there.
 
  • #7
Phyzguy you restated my question basically. The Nitrogen molecule gets hit by something, photons supposedly, that have energy and momentum. The momentum component of the photons is only one millionth of the momentum component that the Nitrogen molecule will pick up, so how does adding energy to the molecule give it momentum exactly. I am not talking formulas I want the exact mechanism described.
 
  • #8
DrClaude
Mentor
7,584
3,946
If the molecule absorbs a photon and re-emits it, nothing much will happen. But you are going to have processes where, before emission the state of the molecule will change, for instance through collisions, or internal conversion (some of the electronic energy will convert to vibrational energy, for instance), and you can get emission of a photon of lower energy than the one absorbed. If everything is enclosed in a box from which beither molecules nor photons can escape, given sufficient time these processes will lead to the conversion of the photons' energy into thermal energy for the molecule. That is, of course, assuming that absorption is significant to start with.

I should point out also that one has to be careful about what we call momentum here: the vector quantity or its magnitude. The vector quantity is of course conserved: if you start with a gas that is not, on average, moving in one direction, and add the laser in a given direction, then after complete absorption of the photons, the gas will be drifting in the direction the laser was going. When converting the energy of the photon into thermal energy, it is the average magnitude of the momentum of the molecules that will have increased.
 
  • #9
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
686
The Nitrogen molecule gets hit by something, photons supposedly, that have energy and momentum.
No. The nitrogen molecules get hit by another nitrogen molecule. Nitrogen is close to an ideal gas. Ideal gases (which don't exist) are perfectly transparent and don't interact with light.
 
  • #10
phyzguy
Science Advisor
4,794
1,744
Phyzguy you restated my question basically. The Nitrogen molecule gets hit by something, photons supposedly, that have energy and momentum. The momentum component of the photons is only one millionth of the momentum component that the Nitrogen molecule will pick up, so how does adding energy to the molecule give it momentum exactly. I am not talking formulas I want the exact mechanism described.

Are you assuming that all momentum transfer is mediated by photons? This may be where your problem lies. When I throw a baseball, the momentum is not transferred from my body to the baseball by photons.
 
  • #11
Dr. Claude, I was confused on how many degrees of freedom to use for N2 also. I did a calculation for N2 using change in specific heat to get energy. Then to get the KE from that I had to assume that it was 3/7 of the total energy to get an answer that almost corresponded to an answer using v^2RMS= 3RT/Molar mass, 995 compared to 960 Joules for what I was calculating. I actually knew PV = nRT was not the right formula except I left it in to see who would find it. I also knew the real ratio of potential impulse for gas would be higher than one million if I used 7 degree of freedom and it did not really change what I wanted to know. Tell you more in next post.
 
  • #12
What I am actually studying is molecular chaos. Energy is a scaler and impulse is a vector and to cause a molecule to increase its rotational and vibrational movements takes force. dE/dx has units of force but I do not know how that works exactly. If it is as I expect an unknown then I suspect that any photon gained by a molecule then re radiated could be in any direction and the rotational and vibrational degrees of freedom will be changed for it, so that if that molecule encounters another molecule, it will push off in maybe a much different direction than it would have had it not had the photon encounter.
 
  • #13
35,392
11,744
dE/dx has units of force but I do not know how that works exactly
I don't see where you get any dE/dx in interactions between molecules and photons.

If it is as I expect an unknown then I suspect that any photon gained by a molecule then re radiated could be in any direction and the rotational and vibrational degrees of freedom will be changed for it
Not in general, but I think those distributions are well-known for many molecules. I just don't know where to look for that.
 
  • Like
Likes 1 person
  • #14
339
15
Are you assuming that all momentum transfer is mediated by photons? This may be where your problem lies. When I throw a baseball, the momentum is not transferred from my body to the baseball by photons.

I have made that very same assumption.

So how else is momentum transfered other than by photons?
 
  • #15
35,392
11,744
Collisions between particles.

If you look closer at these interactions, they are electromagnetic, but they are not the exchange of radiation.
 
  • #16
Seven degrees of freedom lowered the ratio of gas impulse/energy to photon impulse/energy, calculate the ratio is 309340 for original conditions, less than a third of what I originally thought. Sorry.

Is the electric field photons and explained by eE soft coupling, we are getting into the infrared divergence problem
 
  • #17
339
15
Collisions between particles.

If you look closer at these interactions, they are electromagnetic, but they are not the exchange of radiation.

My thinking was that momentum is transferred either by real photons or virtual photons (in the case of contact forces). I think I see my mistake now, in that contact forces are due more to nuclear repulsion than EM, and therefore this process involves more than virtual photons. Correct?

@Jedi_Sawyer: Sorry. Didn’t mean to get off topic from your original OP. I would start a new thread for my question but I think my question is reasonably resolved.
 
  • #18
35,392
11,744
Contact forces are mainly electromagnetic and related to the Pauli exclusion principle for the electrons - it needs a lot of energy to squeeze electron orbitals together.
Nuclear contact forces are relevant in neutron stars, but you don't produce neutron stars when two objects touch on earth.
 
  • #19
phyzguy
Science Advisor
4,794
1,744
My thinking was that momentum is transferred either by real photons or virtual photons (in the case of contact forces). I think I see my mistake now, in that contact forces are due more to nuclear repulsion than EM, and therefore this process involves more than virtual photons. Correct?

@Jedi_Sawyer: Sorry. Didn’t mean to get off topic from your original OP. I would start a new thread for my question but I think my question is reasonably resolved.

If you view the electromagnetic interaction as due to virtual photons, you need to remember that they are not necessarily on shell, so need not satisfy E = p c.
 
  • #20
339
15
Contact forces are mainly electromagnetic and related to the Pauli exclusion principle for the electrons - it needs a lot of energy to squeeze electron orbitals together.
Nuclear contact forces are relevant in neutron stars, but you don't produce neutron stars when two objects touch on earth.

Hmm. Maybe I have misunderstood this excellent post from Gadong regarding contact forces?

eightsquare,

It is not correct to attribute repulsive interatomic forces to electronic effects. For example, the forces between two atoms are due to energy changes in the entire two-atom system. The dominant contribution is the repulsive interaction between the two nuclei. I will work through a simple example to demonstrate this - two hydrogen atoms approaching.

I make use of the following ionization potential data:

H #1: 13.6 eV (H+)
He #1: 24.6 eV (He+); #2: 54.4 eV (He2+).

Consider the situation when two H atoms are far apart. The nuclei do not interact significantly with each other, nor do the electrons, due to their large separation. The energy stored up in the electron-nuclear interactions is the same for each atom, or in total = -2*13.6 = -27.2 eV. This is the system energy at large internuclear separations.

Now bring the nuclei quite close together. Two things happen: (a) the Coulombic potential energy due to nuclear repulsion becomes significant; (b) the energy stored up in electron-nuclear interactions changes, since each electron can now interact with both nuclei; (c) there are also repulsive interactions between electrons.

For very small internuclear separations (R), the contribution from (a) goes up towards +infinity, while the contribution from (b) and (c) approaches a limiting value, which we can easily calculate: note that as R tends to zero, the system transforms into a helium atom (that is, a neutral atom with a +2 nuclear charge). So, the limiting value of (b) + (c) can be deduced from the sum of first and second ionization energies for He, or -79 eV.

From this, one can see that the energy associated with electronic effects does fall by up to 51.8 eV [i.e. from -27.2 to -79.0] as the nuclei approach. However, it is not difficult to see that this is overwhelmed by the repulsive internuclear interaction which rises much more rapidly.

At intermediate distances the overall interaction results in a negative energy change relative to large separations, so H atoms can form a stable molecule.

So, to sum up, the repulsive forces between atoms are mainly due to nuclear repulsion, which at small R is not counterbalanced by the more attractive electronic contributions.
 
  • #21
35,392
11,744
The key point here is "For very small internuclear separations". You don't get this in contact forces. And hydrogen is a bit special in that respect anyway, as it has just one shell. Apart from hydrogen and helium, all other atoms have more shells, and binding lengths are dominated by the size of the outer shells - the nuclei don't come close to each other.
 

Related Threads on Mystery where propulsion originates

Replies
9
Views
1K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
10
Views
4K
  • Last Post
Replies
6
Views
10K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
2
Replies
33
Views
7K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
11
Views
2K
Top