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Momentum of a particle, I have x and y componants of momentum

  • Thread starter rwooduk
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[Note from mentor: this post does not use the standard homework template because it was originally posted in one of the other forums. It was moved here when it became apparent that it is based on a homework-type exercise.]


how do I resolve these componants to find the energy of the particle?

since energy is not a vector how do i calculate the energy of a particle given x and y componants of momentum?

do i simply take the magnitude to find the total momentum and use this to find the energy?

thanks in advance for any suggestions!
 
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  • #2
jtbell
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Correct, find the magnitude of momentum, p, and then use ##E^2 = (pc)^2 + (m_0 c^2)^2## if the particle is relativistic, or ##E = p^2 /(2m_0)## if it's not.
 
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  • #3
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What about the z component? Do you know that as well? is it unknown or is it assumed to be zero? If it is unknown than you cannot calculate the energy.
 
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  • #4
ChrisVer
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Energy is the t-th component of the 4-momentum, while the rest components x-y-z take the normal momentum's compoents.... In special relativity, you can have a lorentz invariant out of the 4-momentum by taking the Lorentz product:
[itex]p^{\mu}p_{\mu}[/itex]
working for [itex]c=1[/itex]
that is a lorentz invariant quantity and it's indeed the squared rest mass...
[itex]p^{\mu}p_{\mu}=g^{\mu \nu} p_{\nu}p_{\mu}=m^{2}[/itex]
by taking [itex]g=diag(+---)[/itex]
you get:
[itex]E^{2} - p_{x}p_{x}-p_{y}p_{y}-p_{z}p_{z}=E^{2} - \vec{p} \cdot \vec{p} = m^{2}[/itex]
Or you end up with:
[itex]E^{2} = \vec{p} \cdot \vec{p} + m^{2}= \left|{\vec{p}}\right|^{2} + m^{2}[/itex]

Now if you want to move into the Standard units (c=1 means that you take [length]=[time] ) you need to make the conventions through a dimensional analysis... this will give you:
[itex]E^{2} = \left|{\vec{p}}\right|^{2} c^{2}+ m^{2} c^{4}[/itex]

If your particle moves on a plane, you can calculate the energy without needing the z-component...
 
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  • #5
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If this is from an actual experiment (with a magnetic field in z-direction to measure x- and y-momentum), you have to reconstruct momentum in z-direction via the flight direction of the particle. The 3-momentum vector points in the same direction as the flight direction.
 
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  • #6
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Thanks for the suggestions!

It's a question where you are given x and y momentum of three particles which are the result of an unknown decaying particle. The question asks you to find the energy, momentum and direction of the unknown particle.

the z componant of momentum of the three particles is zero.

so i'll go with taking the magnitude to find the energy of each particle.

any ideas as to how i would find the direction of the unknown particle?

would i simply sum the x and y componants of momentum of the resultant particles?

Thanks again!!
 
  • #7
ChrisVer
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You can always use the rule for each particle i:
[itex]E_{i}^{2}= |\vec{p_{i}}|^{2}+m_{i}^{2}[/itex]
together with the fact that 4momenta are being conserved....

In that case you will have somewhere the sum of momenta for particle 1,2 and 3... when you square it, you'll get their squared values together with (normal 3D) inner products among them, so terms like [itex]\vec{p_{1}} \cdot \vec{p_{3}}= |\vec{p_{1}}| |\vec{p_{3}}| cosθ_{13}[/itex] and so you will also have a knowledge over the angles... but in most cases, this can also be seen by some conservation...for example if you can see that the z component momentum is zero, you don't have to worry about its angle... and so on...
 
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  • #8
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You can always use the rule for each particle i:
[itex]E_{i}^{2}= |\vec{p_{i}}|^{2}+m_{i}^{2}[/itex]
together with the fact that 4momenta are being conserved....

In that case you will have somewhere the sum of momenta for particle 1,2 and 3... when you square it, you'll get their squared values together with (normal 3D) inner products among them, so terms like [itex]\vec{p_{1}} \cdot \vec{p_{3}}= |\vec{p_{1}}| |\vec{p_{3}}| cosθ_{13}[/itex] and so you will also have a knowledge over the angles... but in most cases, this can also be seen by some conservation...for example if you can see that the z component momentum is zero, you don't have to worry about its angle... and so on...
thats great, will try that!

thanks again, really appreciated!!
 

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