Neutrinos

[mrjaffa]

Need some help please folks.

Writing an assignment and I've just had to calculate how much energy is created in the first stage of the PPI chain. I think I have this right.

The next part asks to estimate how much energy is carried away by the neutrino. For the above calculation it gave you the masses of the particles except the neutrino as this was negligible and for the neutrino question it asks you to estimate... So I'm guessingit's not a calculation that's needed. But I've gone back through my text and can't see what I'm missing. Can't see how I can make this estimation.

Can anyone point me in the right direction?

[ronin]

0.320 ± 0.081 eV/c2 (sum of 3 flavors) seems to be the accepted, however this seems to have come from Feb this year so there are or will be arguements.

As it is the sum of the 3 flavours I assume the Nature wold news artical saying 0.06eV is average for 1 ??

Stick "neutrino mass" into google and it throws up a number of articals.

[Ptarmigan]

An interesting question !

There is lots of stuff on the interweb about what the neutrino energy _is_ but I cant see anything much about how it is derived or estimated.

Without seeing what you have been given in your notes it is difficult to point !

So presumably you have calculated the input mass and the output mass of particles and the Total energy produced is the mass loss.

So far so good

Have you at any stage had to calculate the _Thermal_ energy produced ( the gamma photon) ?

The energy carried away by the neutrino is the difference between the Total and the Thermal.

and then I get lost, it is above my pay grade, but will be interested in the answer when you find it !

[acey]

Wikipedia says 2% of 26.22 MeV in PPI

http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction#Energy_release

With calculations like that no wonder so much was mis-sold.

http://en.wikipedia.org/wiki/Payment_protection_insurance

[mrjaffa]

An interesting question !

There is lots of stuff on the interweb about what the neutrino energy _is_ but I cant see anything much about how it is derived or estimated.

Without seeing what you have been given in your notes it is difficult to point !

So presumably you have calculated the input mass and the output mass of particles and the Total energy produced is the mass loss.

So far so good

Have you at any stage had to calculate the _Thermal_ energy produced ( the gamma photon) ?

The energy carried away by the neutrino is the difference between the Total and the Thermal.

and then I get lost, it is above my pay grade, but will be interested in the answer when you find it !

Yeah, for the previous question I'd worked out the mass lost and then converted it into energy giving me the energy created at a particular stage during the PPI chain.

The next question is to estimate the percentage of energy carried away by the neutrino. My tutor has told me I need to be thinking about the conservation of energy and the question tells me the entire PPI chain creates 2.67 x 10^7 eV... Hmmmmmm.

[mrjaffa]

Wikipedia says 2% of 26.22 MeV in PPI

http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction#Energy_release

With calculations like that no wonder so much was mis-sold.

http://en.wikipedia.org/wiki/Payment_protection_insurance

Ha! Good one..... I'm not sure I'm given that information in the text so I'm not sure I'm to use those numbers :-/

[mrjaffa]

Being that I am very new to all this. And very new to maths. And very new to scientific calculators. Would any of you kind folk try the following calculations for me to see if I'm inputting right?

Thanks!

Mass of a Hydrogen atom = 1.67262 x 10^-27 kg

So the total mass at the start of this stage is:

(1.67262 x 10^-27) x 2

= 3.34524 x 10^-27 kg

Then, I will minus the mass of the Deuterium and Positron.

(3.34524 x 10^-27) – (3.34358 x 10^-27) – (0.00091 x 10^-27)

= 0.00075 x 10^-27 kg

I will then convert this number to J using E(J)=M(kg)C(m / s)2:

(0.00075 x 10^-27) x 299792458^2

= 4.5 x 10^-22 J

There are 1.6 x 10^-19 eV in 1 J. So to convert J into eV, I will divide by 1.6 x 10^-19:

4.5 x 10^-22 / 1.6 x 10

= 0.0028125 eV

[Ptarmigan]

Hmmmm, I thought I was considering conservation of energy above !
(Energy of neutrino) = (Total energy liberated)-(Thermal Energy carried off by the photon)

where Total energy liberated is the mass loss, ie. the mass of the particles in - mass of particles out.

 my comments after >>s :-

Mass of a Hydrogen atom = 1.67262 x 10^-27 kg
So the total mass at the start of this stage is:
(1.67262 x 10^-27) x 2
= 3.34524 x 10^-27 kg
Then, I will minus the mass of the Deuterium and Positron.
(3.34524 x 10^-27) – (3.34358 x 10^-27) – (0.00091 x 10^-27)
= 0.00075 x 10^-27 kg

>> Yes, so far so good except not "Deuterium",  a "Deuteron"
>> ie. deuterium without its electron , ionised, just the proton+neutron
>> This does not affect your calculation so far though
>> ( as an aside your numbers for the masses of proton, deuteron and positron look ok on a quick google, but later I want to check out if the binding energy of the proton/neutron comes into this at this stage, I think not , , , )

I will then convert this number to J using E(J)=M(kg)C(m / s)2:
>> looks ok so far

(0.00075 x 10^-27) x 299792458^2
= 4.5 x 10^-22 J
>> Nope ! I cant get this !!

>> it is 3/4 after the yardarm  and two G&Ts so I could be missing something

>> I'd best not tell you what I am getting yet, so as not to cloud the issue, but it might be a good idea to convert everything to 1 sig fig and keep the decimal in the exponent, less chance of missing some tens and things.

>> I'll look at it all again tomorrow.

>> Scrub the rest till we sort this out

,,,,,,,,,,,,,,,,,,,,,,,,,

There are 1.6 x 10^-19 eV in 1 J. So to convert J into eV, I will divide by 1.6 x 10^-19:

4.5 x 10^-22 / 1.6 x 10

= 0.0028125 eV

[Ptarmigan]

(Energy of neutrino) = (Total energy liberated)-(Thermal Energy carried off by the photon)

where Total energy liberated is the mass loss, ie. the mass of the particles in - mass of particles out.

Argh !

I have missed something out, the thermal energy (kinetic energy)  of the deuteron

I have no idea how to calculate that nor the photon energy :(

it must be given or derived somewhere else, in some other previous course work ????

Thrice argh !!

[Ptarmigan]

I think I see where the info we need is !

but I cant make the numbers add up,

so I'll wait for you to re-check your calculation and see if it is me or you that is going mad with our calculators !!

[mrjaffa]

Hey man. Thanks very much for this. Just a quick reply till I get chance to have a proper look later as I'm at work.

I figured out I had gone wrong at the point u mentioned. I'm not using the calculator correctly :-/ I started using the wolfram alpha site which is blumming marvellous so do have some different numbers now.

I'm still trying to get my head round the neutrino question. I'm told in the question how many eV are created throughout the whole PPI chain. Think it was 2.67 x10^7. Now this energy is in the form of gamma rays right? So yeah, if I convert the remaining mass into energy, look at the difference between that and the total energy I should have my answer. Right? But when I'm working it out my numbers must still be wrong. I'm only getting an answer of 3100ev when I convert all the remaining mass energy. Head scratcher..... Also, I'm not sure if I'm suppose to be taking into consideration the speed of the neutrinos. Arrrrgh! And the question says to state any assumptions I made. Ugh.....

[Ptarmigan]

Hey man thanks to you as well, my pleasure, my brain doesnt get enough exersize these days :)

Just a quick one from me also, as I have doctor etc shortly

I may have misled you, I should have said the neutrinos and photons in the plural ! And I misled myself by stopping short of the full chain 1 ! ( I said way back that it was all way above my pay grade lol! )

Yes, I think you have the method, we just need to sort the numbers now.

I wondered way back why he was telling us the total energy released when we were busy calculating it !

Then I wondered if he might be telling us the total _thermal_ energy and we were calculating the mass loss energy.

(so all my worry about the kinetic energy of the deuteron (and the positron for that matter ) is a given in effect !)

So that's why I wrote " I think I see where the info we need is ! "

Yep I think you must still be wrong with your numbers somewhere, for the deuteron part I am getting nearer 1/2 an Mev,

which was worryingly close, but not close enough, to that 2% of 26 Mev that acey found in Wikipedia

but now that I am back on track with 2 neutrinos I may be back in business

Gotta go, see you all later, and by later someone of more use than me may be along to guide you propper ! lol!

oh and yes I think you can neglect the mass of the neutrinos, cos nobody knows  it ! (I think)

[mrjaffa]

Right. I've got my numbers right. I was doing something embarrassingly stupid on the calculator which we'll move over ;-)

So.... I now have the energy created in those first two stages where 2 neutrinos and 2 positrons are created. I'm also given the total energy created in the question during the ppi chain. So. The energy created in those two first stages subtracted from the overall energy is the answer right? But what about the pesky positrons? And they energy carried away by neutrinos, is that mostly in the form of kinetic energy?.... I need a rest....

[Ptarmigan]

"embarrassingly stupid on the calculator which we'll move over"

heheee, they would take over the world if you dont keep a sharp eye on them

I give as evidence m'lud them already mutating into computals

"I'm also given the total energy created in the question during the ppi chain. So. The energy created in those two first stages subtracted from the overall energy is the answer right?"

Umm no I dont think so, not just the two first stages, or am I misunnerstanding something ?

Caveat : a bit of good news today has led to the consumption of a little ( ahem!) red wine with the evening meal, so I fear I may not be of much use to you for a while !

( is there a deadline on your assignment ?! )

I think that to make use of the given total energy you will need to do a full inventory  of all the eventual particles at the end of chain1 ie. down to the eventual helium nucleus and the 4 protons.

That is the only way I can see at the mo. of accounting for the unknown energy of the photons +  the kinetic energy of the eventual particles all jiggling about at 15million K !

"And they energy carried away by neutrinos, is that mostly in the form of kinetic energy?.... I need a rest...."

Hic !

I wish you had not asked that !

I have googled and googled and all the sources ever talk about is the energy of neutrinos in eV

some of which _may_ be in a few eV of rest mass but not much..

In the case of zero rest mass photons we can fall back on E=hv where h is Planck and v the freq

but I cant find anything like that for neutrinos

[mrjaffa]

Oh dear. Little over my head there. I am way over my head with this. But I'm giving it a go! For fun apparently.....

Deadline is Thursday.

To reiterate. First part of the question was to work out how much energy is created in the stages where neutrinos are created.

So. I took the mass of the original 4 H atoms (I originally only calculated for 2), and then subtract the mass of the 2 Deuteron and 2 Positron as that is what produced at this first part of the PPI. It states the mass of the neutrino is negligible and so does not give its mass..... The difference from the start in mass to the end mass of this first stage is then converted to energy using e=mc2. I've gone over the numbers and I get 843759 eV which seems right.....

The next part of the question states the entire chain created 2.67 x 10^7 eV and to estimate the percentage of this energy carried away by the neutrinos....

So. I presume that the 843750 eV created earlier is carried away by the neutrino (but I'm not sure where the positron comes into this). So. 843750 as a percentage of 2.67 x 10^7 is 3.12%..... Which is almost the 2% already mentioned.

Ugh....

[acey]

You've got energy that can get shared between the positron and the neutrino. You don't know how it's shared so need to estimate. The maximum neutrino energy is if the positron gets none (and the minimum is zero, if the positron gets it all). I haven't done the calculation but the figures I see quoted for the neutrino are 0.420MeV maximum, 0.262MeV average. See for example p211 of this book:

http://books.google.co.uk/books?id=k4XRQpKV9kgC&dq=ppi+energy+estimate+positron+neutrino&source=gbs_navlinks_s

The 2% is explained on p216 of same book: 2 x 0.262/27.31 = 1.9%

Google 0.420MeV (or 0.262MeV) and you'll get more references. The upshot is that it's a small percentage.

[Ptarmigan]

You've got energy that can get shared between the positron and the neutrino. You don't know how it's shared so need to estimate.

Yes,

but has that not already been estimated for him by the energy given in the question ?

Proposition : He can calculate the mass loss and compare that with the energy given,

the difference is the energy carried away by the neutrinos.

Assumption : the energy given in the question is the energy retained within the system to contribute to further fusion.** (all of it - the energy of the positrons is given up on recombination, the helium and other stuff and their kinetics remain, the photons are absorbed by more other stuff ! etc&etc aka the thermal energy )

Unknown : the exact wording of the question and the context !

If he had been given the energy equivalent of the mass loss, which he is calculating anyway,  there would be no point to the whole exersize ?

** The energy carried away by the neutrinos does not contribute further, it is lost, not retained within the system for further use.

[mrjaffa]

You've got energy that can get shared between the positron and the neutrino. You don't know how it's shared so need to estimate. The maximum neutrino energy is if the positron gets none (and the minimum is zero, if the positron gets it all). I haven't done the calculation but the figures I see quoted for the neutrino are 0.420MeV maximum, 0.262MeV average. See for example p211 of this book:

http://books.google.co.uk/books?id=k4XRQpKV9kgC&dq=ppi+energy+estimate+positron+neutrino&source=gbs_navlinks_s

The 2% is explained on p216 of same book: 2 x 0.262/27.31 = 1.9%

Google 0.420MeV (or 0.262MeV) and you'll get more references. The upshot is that it's a small percentage.

Thanks for this. The numbers I have are in the same ball park as those quoted but slightly off. For example, I calculated about 800,000 eV released in those first two stages, where as that text says it is about 1,400,000 eV. So now I'm not sure if I need to take a look at this again....

But I think I understand better now about the neutrino and positron. The question asks to estimate energy carried away by neutrinos and state any assumptions made. So I would assume it's about 2% as some of that 800,000 eV carried away by the positron.

[ollypenrice]

Just quote Eddington if you run out of time;

Just now nuclear physicists are writing a great deal about hypothetical particles called neutrinos supposed to account for certain peculiar facts observed in ß-ray disintegration. We can perhaps best describe the neutrinos as little bits of spin-energy that have got detached. I am not much impressed by the neutrino theory. In an ordinary way I might say that I do not believe in neutrinos... But I have to reflect that a physicist may be an artist, and you never know where you are with artists. My old-fashioned kind of disbelief in neutrinos is scarcely enough. Dare I say that experimental physicists will not have sufficient ingenuity to make neutrinos? Whatever I may think, I am not going to be lured into a wager against the skill of experimenters under the impression that it is a wager against the truth of a theory. If they succeed in making neutrinos, perhaps even in developing industrial applications of them, I suppose I shall have to believe—though I may feel that they have not been playing

quite fair. 

Olly