Creation of Matter: Solving for Mass with E=mc^2

[ArmanCham]

Lets suppose we have a photon and it will going to create a particle particle mass is $16.10^-29$ kg.I used $E=mc^2$ to find the solution and I found 90 MeV but the answer is 180 where did I made wrong.

Thanks

 

[Sagar Singh]

particle mass not clear

 

[Vanadium 50]

If you don't show us the calculation, it will be very difficult to tell you where you went wrong.

 

[ArmanCham]

16.10^-29 kgxc^2 is equal 16x9x10^-13 joule=144x10^-13 joule I convert this MeV and I found 90 MeV but the answer is 180MeV

 

[ChrisVer]

I guess because you need 2 particles at the end?

 

[Sagar Singh]

you forget momentum conservation

 

[ArmanCham]

I thought we need two photons to create a particle but I am not sure

 

[ArmanCham]

you forget momentum conservation

Me ?

 

[Sagar Singh]

I thought we need two photons to create a particle but I am not sure

yaa momentum conservation is must,

 

[ArmanCham]

So final Answer is we need two photon ??

 

[Sagar Singh]

ye

Me ?

yes

 

[ChrisVer]

I wasn't thinking about 2 photons...rather about 2 particles at the end...
$\gamma \rightarrow 1+2$

 

[ArmanCham]

But there's one particle

 

[ChrisVer]

you can't get 1 massive particle from 1 photon...
Simply because that would violate energy/momentum conservation. For the massive particle you can go to a frame where its momentum vanishes (rest frame)...at that frame, the photon's momentum can't be zero (because momentum=energy for photons, so you would get nothing), and you get a violation of energy/momentum...

 

[Vanadium 50]

We are 15 messages into this, and we don't even have a clear idea on what the question is - people are still trying to guess.

 

[ArmanCham]

The question is we have one photon and it creates a particle we know particle mass but intersting thing is answer says photon energy is twice a particle energy why there's extra energy ?

 

[ArmanCham]

you can't get 1 massive particle from 1 photon...
Simply because that would violate energy/momentum conservation. For the massive particle you can go to a frame where its momentum vanishes (rest frame)...at that frame, the photon's momentum can't be zero (because momentum=energy for photons, so you would get nothing), and you get a violation of energy/momentum...

Thats make sense I thought feymann diagram we need one boson and two fermion to make a correct "feymann diagram" there can't be two boson and one fermion.So you are saying 180Mev will create two 90Mev particle or 180 Mev will create 90 Mev particle 90 Mev photon

 

[ArmanCham]

I found the answer I am so so sorry.

 

[Vanadium 50]

The question is we have one photon and it creates a particle

Since that never happens on its own, that's clearly not the question.

 

[ArmanCham]

Yeah,the solution is photon makes matter and antimatter but for that we need two photon and two matter.Wait ! Can one photon create two particles ? Or we need at least two photon If we need at least two photon then the answer will be 90 MeV

 

[ArmanCham]

Since that never happens on its own, that's clearly not the question.

You are right.

 

[ChrisVer]

You need either 2 real photons, or (if you have 1) it will be virtual...

 

[ArmanCham]

γ→e^-+e⁺ my book says this but I know that there must be 2γ

 

[ChrisVer]

I think my previous post answers this...
As an intermediate virtual photon this process is possible or if there are other particles around which can take some energy as well...
For real photons this is not and you need 2γ

 

[ArmanCham]

There where one photon and it can't be virtual photon
Book has false info

thank you.

 

[ChrisVer]

There where one photon and it can't be virtual photon

It has to be virtual because the relation $p^{\mu (\gamma)}p_\mu^{(\gamma)} \equiv E^2_\gamma- \vec{p}_\gamma \cdot \vec{p}_\gamma =0$ cannot hold. It's a simple relativistic exercise to do.. (Equating the squares of total four momenta before and after - use the squares because you can go to the center of mass frame for the two products to get the result which then will be invariant of frame).

If it's not virtual then there should be something around there which shared some of its energy to the interaction. For example some other charged particle $A$:
$\gamma (A) \rightarrow e^- e^+ (A)$

 

[ArmanCham]

Why should I said its not a virtual photon cause, the textbook is high school textbook and in high school I didnt learn virtual photon.So that's why I told book has wrong info.But you are right virtual photons can do that interaction.

 

[ZapperZ]

Why should I said its not a virtual photon cause, the textbook is high school textbook and in high school I didnt learn virtual photon.So that's why I told book has wrong info.But you are right virtual photons can do that interaction.

You really should pay attention to your posts, because I find this thread to be a jumble of confusion.

You started off with a faulty idea that one photon can produce one particle. Not only can this violate conservation of charge (as if gamma producing an electron), but as has been stated, it violates conservation of momentum (consider a photon having just having the same energy as the rest mass energy of an electron - the electron will be created, but it has no net momentum, which means that the momentum of the photon before has not been conserved.

But now, you are claiming that your book is wrong, which essentially showed that γ→e-+e+, which is essentially CORRECT, because it produced the e-p pair (not sure why you think this is wrong) in the vicinity of a massive object (which is why pair production often occurs when gamma photons are shot at a massive target). Your book never showed one gamma producing one electron or one positron. It showed that that one gamma produced electron AND positron. This is correct. What you calculated in the beginning isn't!

So why is the book wrong? Why did you ignore your text? Why did you calculate a photon being converted to just one particle, rather than the particle-antiparticle pair? From what I've read, I think you made the error, not your text.

Zz.

 

[ArmanCham]

You really should pay attention to your posts, because I find this thread to be a jumble of confusion.

You started off with a faulty idea that one photon can produce one particle. Not only can this violate conservation of charge (as if gamma producing an electron), but as has been stated, it violates conservation of momentum (consider a photon having just having the same energy as the rest mass energy of an electron - the electron will be created, but it has no net momentum, which means that the momentum of the photon before has not been conserved.

But now, you are claiming that your book is wrong, which essentially showed that γ→e-+e+, which is essentially CORRECT, because it produced the e-p pair (not sure why you think this is wrong) in the vicinity of a massive object (which is why pair production often occurs when gamma photons are shot at a massive target). Your book never showed one gamma producing one electron or one positron. It showed that that one gamma produced electron AND positron. This is correct. What you calculated in the beginning isn't!

So why is the book wrong? Why did you ignore your text? Why did you calculate a photon being converted to just one particle, rather than the particle-antiparticle pair? From what I've read, I think you made the error, not your text.

Zz.

1) This is quoted from question " What's the minimum photon energy, to create a particle which its mass is 16.10^-29 kg" I was confused cause the question claims that single particle created by single photon.There is no "s"after particle.
2) After the answer I noticed that one photon cannot be produce one particle cause of momentum violation so I thought I am missing something.Then I asked you guys.Then I get some answers then I noticed in the book there says one photon makes electron and pozitron.I thought it must be wrong cause I learned that 2 photon makes proton and pozitron.
3) Why book is wrong ?
First thing is its talking about annihilation.In annihilation I thing we need 2 photon.If we have virtual photon the equation can be true but this is not that situation.
4) Why you ignore your text ?
I bet If you were in my shoes you would do the same thing.
5) Why did you calculate a photon being converted to just one particle, rather than the particle-antiparticle pair?
I have already give the answer to this question.The book claimed particle not "particles".Thats the point where I am confused.But the book has wrong info.Book claims that annihilation equation is one photon makes e-p pair but all of us know that that's not true.
Virtual particles can do that.But the book is High School book and I didnt learned such a thing in high school unless they want to teach us Advanced Quantum Mechanics

 

[ChrisVer]

annihilation equation is one photon makes e-p pair

It can in a non-empty environment.

 

[vanhees71]

Well, the correct way to explain it is as follows:

A photon hits a very massive particle like an atomic nucleus with mass $m_A$. Let's assume that the photon energy is very small compared to the mass of the nucleus (I'm using natural units with $\hbar=c=1$ so that masses, energies and momenta all have the same units; the use of kg is very inconvenient in particle physics, so let's use electron volts or in this case MeV).

Now if the photon hits the nucleus under these circumstances the nucleus will almost get momentum transferred (consider a ball hitting a wall). Thus the energy-momentum balance for making an electron-positron pair at rest reads
$$E_{\gamma}=2 m_e,$$
and that's the minimum photon energy needed for this reaction.

A single photon cannot decay to an electron positron-pair in the vacuum, because this is forbidden by energy-momentum conservation. That's easy to see by considering the relativistic invariants of the corresponding kinematics. If this would happen, you'd have
$$p_{\gamma}=p_{e^-}+p_{e^+},$$
where the $p_j$ are the four-momenta of the particles. Squaring the equation in the sense of the Minkowski product you'd get
$$p_\gamma^2=0=(p_{e^-}+p_{e^+})^2=2 m_e^2 + p_{e^+} \cdot p_{e^-}.$$
Now
$$p_{e^+} \cdot p_{e^-}=E_{e^+} E_{e^-}-\vec{p}_{e^+} \cdot \vec{p}_{e^-}.$$
Since now $E_{e^{+}}=\sqrt{\vec{p}_{e^+}^2+m_e^2} > E_{E^{+}}$ you can never make the right-hand side vanish in contradiction to the result on the left-hand side.

For a process to occur you must conserve energy and momentum and obey the "on-shell conditions", $p^2=m^2$, for all particles in the initial and the final state! That's not possible for a single photon in empty space, as just shown.

 

[ArmanCham]

Thank you for detailed explanation.