Relativistic kinematics

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a nice man
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#1
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I've been working through 'Introduction to Elementary Particles' by Griffiths and found something that has really confused me.

First of all he gives the neutrino zero momentum and therefore p^2=0. I'm assuming that when the book was written the neutrino may have been thought of as massless?

Secondly for the pion he writes that p^2=m^2c^2=E/c and produces a similar expression for the muon.

I have read that this expression is only true for massless particles, and neither the pion or the muon are massless.

Finally, as the pion is at rest, shouldn't its momentum be zero?

Is this something to do with four-vector notation being used?
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Deranging
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(Original post by a nice man)
I've been working through 'Introduction to Elementary Particles' by Griffiths and found something that has really confused me.

First of all he gives the neutrino zero momentum and therefore p^2=0. I'm assuming that when the book was written the neutrino may have been thought of as massless?

Secondly for the pion he writes that p^2=m^2c^2=E/c and produces a similar expression for the muon.

I have read that this expression is only true for massless particles, and neither the pion or the muon are massless.

Finally, as the pion is at rest, shouldn't its momentum be zero?

Is this something to do with four-vector notation being used?
Let us work in Minkowski space with metric
 ds^2 = c^2 dt^2 - d \vec{x} \cdot d \vec{x}

We need to be very careful to distinguish between 4-momentum and 3-momentum.
Let us begin with 3-momentum: The relativistic 3-momentum  \vec{p} = m\frac{ d \vec{x} } {d \tau} = \gamma m \vec{v} , where  \tau is the proper time of the particle and  \vec{v} = \frac{ \vec{d x}}{d t} , the relativistic gamma factor is  \gamma = \frac{1}{\sqrt{1 - \vec{v} \cdot \vec{v} / c^2}} .
Next we define the 4-momentum: This incorporates a 'time-direction' as well and is defined
 p^{\mu} = m \frac{d x^{\mu}}{d \tau} = \left( \begin{array}{c} E/c \\ \vec{p} \end{array} \right) where  E is the energy of the particle.
Why do we do this? Well relativity is a little bit of a he said she said kind of game, what I describe one way from my spaceshuttle another observer in his spaceshuttle will describe differently. So what are we to do? We look for things that everyone will agree upon, Lorentz-invariant objects.
If we are given two four vectors, while different observers will disagree on the individual components of these vectors all observers will agree on the value of the (Minkowski) inner product:  p \cdot q = p^{\mu} q^{\nu} \eta_{\mu \nu} . Here the values  \mu, \nu = 0, 1, 2, 3 are implicitly summed.
So in particular, we can evaluate  p^2 = p \cdot p = p^{\mu} p_{\mu} = p^{\mu} p^{\nu} \eta_{\mu \nu} = E^2/c^2 - \vec{p} \cdot \vec{p} (All the '=' are just to show there are many ways to write the same thing). But, we just argued that all observers agree on this value, so we can choose to evaluate it in a very nice reference frame, the rest frame of the object. In this frame  E = mc^2, \vec{p} = 0 . Thus we find that  p^2 = m^2 c^2 , isn't that neat. In particular, for particles with no mass this vanishes however this does not mean that the momentum of the particle is zero (because of the signature of the Minkowski metric being (+,-,-,-), so Minkowski inner products can be <0, =0, >0>.)

To now turn to your questions, we now know that neutrinos are probably massive, however they have very very very small masses compared to the other particles in the question. Therefore it is an exceedingly excellent assumption to treat them as massless.
No the expression is not just valid for massless particles (wouldn't make sense seeing as the expression features a particle mass... )
Being at rest means no 3-momentum, but you can still have 4-momentum.
Yes, these are all 4-vectors. ( I am not sure about the conventions in Griffiths, but ususally 3-vectors are bold  \mathbf{v} or have arrows on them  \vec{v} , while 4-vectors are not bold.)

Finally, keep at it - particle physics is hard, especially at first the notation can seem funny but it quickly becomes second nature and physics is exceedingly interesting.

Some notes on special relativity to help you along your way:
The last sections of http://www.damtp.cam.ac.uk/user/tong...ity/dynrel.pdf
Section 5 of http://www.damtp.cam.ac.uk/user/tong/em/electro.pdf
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a nice man
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#3
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(Original post by Deranging)
Let us work in Minkowski space with metric
 ds^2 = c^2 dt^2 - d \vec{x} \cdot d \vec{x}

We need to be very careful to distinguish between 4-momentum and 3-momentum.
Let us begin with 3-momentum: The relativistic 3-momentum  \vec{p} = m\frac{ d \vec{x} } {d \tau} = \gamma m \vec{v} , where  \tau is the proper time of the particle and  \vec{v} = \frac{ \vec{d x}}{d t} , the relativistic gamma factor is  \gamma = \frac{1}{\sqrt{1 - \vec{v} \cdot \vec{v} / c^2}} .
Next we define the 4-momentum: This incorporates a 'time-direction' as well and is defined
 p^{\mu} = m \frac{d x^{\mu}}{d \tau} = \left( \begin{array}{c} E/c \\ \vec{p} \end{array} \right) where  E is the energy of the particle.
Why do we do this? Well relativity is a little bit of a he said she said kind of game, what I describe one way from my spaceshuttle another observer in his spaceshuttle will describe differently. So what are we to do? We look for things that everyone will agree upon, Lorentz-invariant objects.
If we are given two four vectors, while different observers will disagree on the individual components of these vectors all observers will agree on the value of the (Minkowski) inner product:  p \cdot q = p^{\mu} q^{\nu} \eta_{\mu \nu} . Here the values  \mu, \nu = 0, 1, 2, 3 are implicitly summed.
So in particular, we can evaluate  p^2 = p \cdot p = p^{\mu} p_{\mu} = p^{\mu} p^{\nu} \eta_{\mu \nu} = E^2/c^2 - \vec{p} \cdot \vec{p} (All the '=' are just to show there are many ways to write the same thing). But, we just argued that all observers agree on this value, so we can choose to evaluate it in a very nice reference frame, the rest frame of the object. In this frame  E = mc^2, \vec{p} = 0 . Thus we find that  p^2 = m^2 c^2 , isn't that neat. In particular, for particles with no mass this vanishes however this does not mean that the momentum of the particle is zero (because of the signature of the Minkowski metric being (+,-,-,-), so Minkowski inner products can be <0, =0, >0>.)

To now turn to your questions, we now know that neutrinos are probably massive, however they have very very very small masses compared to the other particles in the question. Therefore it is an exceedingly excellent assumption to treat them as massless.
No the expression is not just valid for massless particles (wouldn't make sense seeing as the expression features a particle mass... )
Being at rest means no 3-momentum, but you can still have 4-momentum.
Yes, these are all 4-vectors. ( I am not sure about the conventions in Griffiths, but ususally 3-vectors are bold  \mathbf{v} or have arrows on them  \vec{v} , while 4-vectors are not bold.)

Finally, keep at it - particle physics is hard, especially at first the notation can seem funny but it quickly becomes second nature and physics is exceedingly interesting.

Some notes on special relativity to help you along your way:
The last sections of http://www.damtp.cam.ac.uk/user/tong...ity/dynrel.pdf
Section 5 of http://www.damtp.cam.ac.uk/user/tong/em/electro.pdf
Thanks for the reply, I appreciate it. I've had a look at those links, the first one looks particularly useful!
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