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

• rwooduk
In summary, the question asks you to find the energy, momentum, and direction of an unknown particle. You calculate the energy by taking the magnitude of the momentum.
rwooduk
[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 components 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 components 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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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.

1 person
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.

1 person
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:
$p^{\mu}p_{\mu}$
working for $c=1$
that is a lorentz invariant quantity and it's indeed the squared rest mass...
$p^{\mu}p_{\mu}=g^{\mu \nu} p_{\nu}p_{\mu}=m^{2}$
by taking $g=diag(+---)$
you get:
$E^{2} - p_{x}p_{x}-p_{y}p_{y}-p_{z}p_{z}=E^{2} - \vec{p} \cdot \vec{p} = m^{2}$
Or you end up with:
$E^{2} = \vec{p} \cdot \vec{p} + m^{2}= \left|{\vec{p}}\right|^{2} + m^{2}$

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:
$E^{2} = \left|{\vec{p}}\right|^{2} c^{2}+ m^{2} c^{4}$

If your particle moves on a plane, you can calculate the energy without needing the z-component...

1 person
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.

1 person
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 components of momentum of the resultant particles?

Thanks again!

You can always use the rule for each particle i:
$E_{i}^{2}= |\vec{p_{i}}|^{2}+m_{i}^{2}$
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 $\vec{p_{1}} \cdot \vec{p_{3}}= |\vec{p_{1}}| |\vec{p_{3}}| cosθ_{13}$ 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...

1 person
ChrisVer said:
You can always use the rule for each particle i:
$E_{i}^{2}= |\vec{p_{i}}|^{2}+m_{i}^{2}$
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 $\vec{p_{1}} \cdot \vec{p_{3}}= |\vec{p_{1}}| |\vec{p_{3}}| cosθ_{13}$ 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!

## 1. What is momentum?

Momentum is a physical quantity that describes the motion of an object. It is the product of an object's mass and velocity.

## 2. How is momentum calculated?

Momentum is calculated by multiplying an object's mass by its velocity. It can also be calculated by multiplying an object's velocity by its momentum.

## 3. What are the units for momentum?

The units for momentum are kilogram-meters per second (kg·m/s) in the SI system and gram-centimeters per second (g·cm/s) in the CGS system.

## 4. How do the x and y components of momentum affect overall momentum?

The x and y components of momentum affect the overall momentum by contributing to its magnitude and direction. The total momentum of an object is the vector sum of its x and y components.

## 5. Can the momentum of a particle change?

Yes, the momentum of a particle can change. It can change in magnitude and direction if there is an external force acting on the particle, according to Newton's Second Law of Motion (F=ma).

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