Conservation of relavistic energy

In summary, the conversation discusses the problem of a Kaon particle decaying into two pions and the application of conservation of momentum and energy to determine the momentum of each pion in the Kaon's frame of reference. The equations used in the conversation involve the mass and velocity of each particle, as well as the relativistic formula for kinetic energy. The conversation also suggests using common algebraic tricks to solve for the energy and momentum of the pions.
  • #1
Samuelriesterer
110
0
Problem statement, equations, and work done:

A particle called a Kaon is moving at 0.8c through a detector when it decays into two pions.

Kaon particle: mass = 493.7 MeV/c^2
Pion+: mass = 139.6 MeV/c^2
Pion0: mass = 135.0 MeV/c^2

1) Apply conservation of momentum/energy to determine the momentum of each pion in the kaon frame of reference (where K is at rest).Use the fact that the momentum of the original kaon in its own frame is zero and don't forget to use the relativistic momentum energy conservation equation. Leave the momentum in MeV/c units.

Since the momentum of the kaon is 0 because it is at rest in its own frame, the magnitudes of the momentum of the pions should be equal.

##E_k = m_k c^2 = 493.7 \frac{MeV}{c^2} = (m_{\pi +} + m_{\pi 0}) c^2 + (p_{\pi +} + p_{\pi 0})c##

##493.7 \frac{MeV}{c} = (139.6 + 135.0) \frac{MeV}{c} + 2|p|##

##219.1 \frac{MeV}{c} = 2|p|##

##p = 109.55 \frac{MeV}{c}##

Solving for velocities:

##p{\pi +} = 109.55 \frac{MeV}{c} = \gamma_{\pi +} m_{\pi +} v_{\pi +} = \gamma_{\pi +} 139.6 \frac{MeV}{c^2} v_{\pi +}##

From this:

##v_{\pi +} = .61733 (c), and thus : \gamma_{\pi +} = 1.27113##

And:

##v_{\pi 0} = .63012 (c), and thus: \gamma_{\pi 0} = 1.28783##

From this there is no Conservation of energy:

##E_k = E_{\pi +} + E_{\pi 0} = 493.7 \frac{MeV}{c^2} = \gamma_{\pi +} m_{\pi +} + \gamma_{\pi 0} m_{\pi 0} = doesn't = 493.7!##

(I left units off)

2)Now transform the momentum-energy vectors for each pion and the kaon to the detector frame. Assume that the pion+ is moving in the direction the Kaon was moving in the detector frame.

3) Check to see that the momentum-energy conservation law is followed in the detector frame.

4) If the pions are moving off perpendicular to the Kaon's motion in the detector frame, calculate the momentum energy vectors for the pions in the detector frame and again check to see that momentum-energy is conserved.
 
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  • #2
Samuelriesterer said:
##E_k = m_k c^2 = 493.7 \frac{MeV}{c^2} = (m_{\pi +} + m_{\pi 0}) c^2 + (p_{\pi +} + p_{\pi 0})c##

What's that? What is the relativistic formula for kinetic energy?
 
  • #3
Samuelriesterer said:
Problem statement, equations, and work done:

A particle called a Kaon is moving at 0.8c through a detector when it decays into two pions.

Kaon particle: mass = 493.7 MeV/c^2
Pion+: mass = 139.6 MeV/c^2
Pion0: mass = 135.0 MeV/c^2

1) Apply conservation of momentum/energy to determine the momentum of each pion in the kaon frame of reference (where K is at rest).Use the fact that the momentum of the original kaon in its own frame is zero and don't forget to use the relativistic momentum energy conservation equation. Leave the momentum in MeV/c units.

Since the momentum of the kaon is 0 because it is at rest in its own frame, the magnitudes of the momentum of the pions should be equal.

I would do this for a general decay of a particle of mass M into two particles of masses m_1 and m_2, then plug the numbers into the final equation. You're getting bogged down in numbers. At the level you're at, you should be working more algebraically and not plugging in awkward values at the first opportunity!
 
  • #4
That is E = mc^2 + pc which is total energy = rest energy + KE energy which equals the combined energy totals of the lions. The first part is how my teacher should us, but I can't bring it around to the conservation of energy
 
  • #5
Samuelriesterer said:
That is E = mc^2 + pc which is total energy = rest energy + KE energy which equals the combined energy totals of the lions. The first part is how my teacher should us, but I can't bring it around to the conservation of energy

That's because kinetic energy is not equal to pc for a particle with mass.
 
  • #6
PeroK said:
I would do this for a general decay of a particle of mass M into two particles of masses m_1 and m_2, then plug the numbers into the final equation. You're getting bogged down in numbers. At the level you're at, you should be working more algebraically and not plugging in awkward values at the first opportunity!

I second this recommendation.

I also recommend that you always keep m together with c2 as (mc2), and p together with c as (pc). Then the only unit you need to use is MeV. That is, for the kaon, don't think m = 493.7 MeV/c2, but rather (mc2) = 493.7 MeV.

Finally, you should never need to calculate any velocities in a problem like this, unless the problem statement explicitly asks for it. Use conservation of energy, conservation of momentum, and the general relationship between energy, mass and momentum: E2 = (pc)2 + (mc2)2.
 
  • #7
DEvens said:
That's because kinetic energy is not equal to pc for a particle with mass.

I wonder what my fantastic teacher was doing then.

PeroK said:
I would do this for a general decay of a particle of mass M into two particles of masses m_1 and m_2, then plug the numbers into the final equation. You're getting bogged down in numbers. At the level you're at, you should be working more algebraically and not plugging in awkward values at the first opportunity!

I only was trying to calculate the velocities to show it does not fit into conservation of energy.

Can anybody point me in the right direction, I'm only trying to understand this all.
 
  • #8
Samuelriesterer said:
I wonder what my fantastic teacher was doing then.
I only was trying to calculate the velocities to show it does not fit into conservation of energy.

Can anybody point me in the right direction, I'm only trying to understand this all.

Your starting point is the three equations:

##E_1 + E_2 = Mc^2, \ \ E_1^2 = p^2c^2 + m_1^2c^4, \ \ E_2^2 = p^2c^2 + m_2^2c^4##

Can you explain these - at least to yourself?

There are now common ways to work with these sorts of equations: common algebraic tricks. In this case I recommend trying to get an equation for the energy of either pion. Once you have the energy, you can get the momentum from that.
 
  • #9
Samuelriesterer said:
Can anybody point me in the right direction, I'm only trying to understand this all.
I just did.

Kinetic energy for a particle with mass is (gamma - 1) mc^2, total energy (mass plus kinetic) is gamma mc^2.
Momentum is gamma m v.

Alternatively, as suggested by jtbell, E^2 = (pc)^2 + (mc^2)^2
You have E = pc + mc^2, which is wrong.

Your very first equation is wrong. As I have now told you for the third time.
 
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  • #10
DEvens said:
I just did.

Kinetic energy for a particle with mass is (gamma - 1) mc^2, total energy (mass plus kinetic) is gamma mc^2.
Momentum is gamma m v.

Alternatively, as suggested by jtbell, E^2 = (pc)^2 + (mc^2)^2
You have E = pc + mc^2, which is wrong.

Your very first equation is wrong. As I have now told you for the third time.

That equation is an approximation that our teacher said we could use in this situation.

PeroK said:
Your starting point is the three equations:

##E_1 + E_2 = Mc^2, \ \ E_1^2 = p^2c^2 + m_1^2c^4, \ \ E_2^2 = p^2c^2 + m_2^2c^4##

Can you explain these - at least to yourself?

There are now common ways to work with these sorts of equations: common algebraic tricks. In this case I recommend trying to get an equation for the energy of either pion. Once you have the energy, you can get the momentum from that.

Yes I understand those, thank you. Isn't that 3 equations with 4 unknowns? Don't I need: p = p1 + p2 = 0?
 
  • #11
Samuelriesterer said:
Yes I understand those, thank you. Isn't that 3 equations with 4 unknowns? Don't I need: p = p1 + p2 = 0?

I've used p1 = p2 = p, as the momentum of the two pions must be equal.
 
  • #12
Samuelriesterer said:
That equation is an approximation that our teacher said we could use in this situation.

Ok, I think I have discovered the fault.
 
  • #13
PeroK said:
I've used p1 = p2 = p, as the momentum of the two pions must be equal.

Oh I see that now. Thanks.
 
  • #14
I just solved the system of equations using the approximation and got 109.55 for the momentum. I also got the conservation of energy with E1 = 249.15 and E2 = 244.55, so that works.

Now to convert E1 and E2 to the second frame, what equation do I use? If it involves gamma, then what gamma is it and how to calculate it.

Thanks guys.
 
  • #15
Samuelriesterer said:
I just solved the system of equations using the approximation and got 109.55 for the momentum. I also got the conservation of energy with E1 = 249.15 and E2 = 244.55, so that works.

Now to convert E1 and E2 to the second frame, what equation do I use? If it involves gamma, then what gamma is it and how to calculate it.

Thanks guys.

The mometum should be nearly double what you've calculated. The energy looks correct. If you have the energy and the mass of a particle, then you can calculate its momentum.
 
  • #16
Well if E1 = m1 + p then p = E1 - m1 = 249.15 - 139.6 = 109.55.

Leaving the c and units out.
 
  • #17
Samuelriesterer said:
Well if E1 = m1 + p then p = E1 - m1 = 249.15 - 139.6 = 109.55.

Leaving the c and units out.

Energy isn't the sum of mass and momentum. It's ##E^2 = p^2 + m^2##
 
  • #18
PeroK said:
Energy isn't the sum of mass and momentum. It's ##E^2 = p^2 + m^2##

I see now. So was it OK to use that approximation to solve the system of equations? Even though I got a different p in that system. Would that throw the values of E off too?
 
  • #19
Samuelriesterer said:
I see now. So was it OK to use that approximation to solve the system of equations? Even though I got a different p in that system. Would that throw the values of E off too?

I haven't seen how you solved the equations. You seem to have got the right energies.

If you look at that approximation mathematically, it can only be valid where one of the terms is very small: where E is approx p or E is approx m. That's not the situation here.
 
  • #20
Basically it was your 3 equations without the squares. <shrugs> when I get the time I'll do the system with the squares and see how it comes out.

Can you point me in the right direction to get the energy and momentum in the second frame?
 
  • #21
Samuelriesterer said:
2)Now transform the momentum-energy vectors for each pion and the kaon to the detector frame

What have you studied on the Lorentz transformation? The LT for momentum and energy is similar to the LT for position and time, because momentum and energy form a four-vector, like position and time form a four-vector.
 
  • #22
PeroK said:
Energy isn't the sum of mass and momentum. It's ##E^2 = p^2 + m^2##

One thing with this equation is that the magnitude of the pion's momentum is not the same:

##p1 = \sqrt{E1^2 -m1^2} = \sqrt{249.15^2 - 139.6^2} = 206.37##
##p2 = \sqrt{E2^2-m2^2} = \sqrt{244.55^2 - 135.0^2} = 203.91##
 
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  • #23
Samuelriesterer said:
Basically it was your 3 equations without the squares. <shrugs> when I get the time I'll do the system with the squares and see how it comes out.

I worked out the 3 equations using the squares and it works out to almost the same:

E1 = 245.57
E2 = 248.13
 
  • #24
Your E1 and E2 are correct for the ##\pi^0## and ##\pi^+## respectively; but the 1 and 2 don't match up with the masses you used in post #22. Hopefully that's just a typo.
 
  • #25
Yes I corrected that, but they still do not equal in magnitude?!
 
  • #26
Why should the energies be equal? The pions have the same magnitude momentum (in this frame), but different masses.
 
  • #27
jtbell said:
Why should the energies be equal? The pions have the same magnitude momentum (in this frame), but different masses.

OK I just recalculated and they do, I made a simple math mistake. :-)

jtbell said:
What have you studied on the Lorentz transformation? The LT for momentum and energy is similar to the LT for position and time, because momentum and energy form a four-vector, like position and time form a four-vector.

I haven't quite gotten to the four vectors thing, but I assumed the new values would be:

##E' = \gamma_k m_k##
##E1' = \gamma_{p1} m_{p1}##
##E2' = \gamma_{p2} m_{p2}##
##P1' = \gamma_{p1} m_{p1} v_{p1}##
##P2' = \gamma_{p2} m_{p2} v_{p2}##

And then solve this system with the same equations as in post #8.
 
  • #28
This approach will work, but it requires you to find the velocities of all the particles in the detector frame. In order to do that, you first have to find the velocities in the kaon's rest frame (where you've found the momenta and energies). It's a lot of work.

Using the Lorentz transformation, you need to know only the relative velocity of the two frames, which is basically the velocity of the kaon in the detector frame, which you're given. The math is a lot simpler.

It's an instructive exercise to do it both ways to see that they agree. However, I suspect from the way the problem is worded, that the author wants you to use the Lorentz transformation.

(note... It's past bedtime for me now, and I'm going to be gone most of the day tomorrow.)
 
  • #29
OK thanks. Can you link a good website that explains the LT process well?
 
  • #31
jtbell said:
A Google search for "Lorentz transformation for momentum and energy" turns up e.g.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html

Yes I saw this page but I don't understand it much. I've scoured the web and haven't found anything that explains it well. How would I go about solving for the velocities. I would assume:

For the rest frame:

##\frac{v1}{\sqrt{1 - v1^2}} = \frac{p1}{m1}##
<same for v2>

For the detector frame, use the relavistic velocity equation:

##v1' = \frac{.8 + v1}{1+.8v1}##
<same for v2'>
 
  • #32
That approach should work, although as I said it uses more algebra and arithmetic. Be careful about + and - signs on the velocities. Also, this method is sensitive to roundoff errors. Don't round off any numbers until you get to the very end. Ideally, keep intermediate results in your calculator as you go along.
 
  • #33
jtbell said:
That approach should work, although as I said it uses more algebra and arithmetic. Be careful about + and - signs on the velocities. Also, this method is sensitive to roundoff errors. Don't round off any numbers until you get to the very end. Ideally, keep intermediate results in your calculator as you go along.

Boy you were right! I worked out all of this yesterday and was way off with the conservation, I thought I was doing something wrong. But I did what you said and left the figures in my calculator and it all worked out! I think I learn more from this forum then in my class :)

Now for the last bit, the pions move off in a direction perpendicular to the kaon. I would assume that the momentum stays the same in the rest frame but the velocities in the detector frame would have to be converted to the y axis:

##u1'_y = \frac{u_y}{\gamma (1- \frac{u_x v}{c^2})}##
<same for u2'y>

I am having a hard time though identifying the variables:

I have v_kaon = 0, v1, and v2 in the rest frame (which I assume remain the same in this question)
and in the detector frame: v_kaon' = 0.8c, v1', and v2'
 
  • #34
Samuelriesterer said:
Now for the last bit, the pions move off in a direction perpendicular to the kaon. I would assume that the momentum stays the same in the rest frame but the velocities in the detector frame would have to be converted to the y axis:

In the kaon frame, the magnitudes of the pion momenta are the same as before, but their directions are now along the +y and -y directions.

In the detector frame, the pions each come out at some angle between the x and y directions. In order to "boost" a velocity that is "originally" along the y-direction, using a relative velocity along the x-direction, you need to use a "vectorized" version of the velocity-addition formula. I've seen such a thing, but don't remember where offhand. Try Googling for something like "relativistic velocity vector addition."

This is where the Lorentz-transformation method becomes really nice. On the page I linked before, it was written in four-vector-and-matrix format. The boost is still along the x-direction, so the Lorentz transformation matrix is the same here. You just apply it to a different energy-momentum four-vector.

(disclaimer: I haven't actually worked out this part yet...)
 
  • #35
I'll do a little digging and see what I can find. Thanks
 

Related to Conservation of relavistic energy

What is the conservation of relativistic energy?

The conservation of relativistic energy is a fundamental principle in physics that states that the total energy of a closed system remains constant over time. This means that energy can neither be created nor destroyed, but can only be transformed from one form to another.

How does the conservation of relativistic energy relate to Einstein's theory of relativity?

Einstein's theory of relativity states that energy and mass are equivalent and can be converted into one another according to the famous equation E=mc². The conservation of relativistic energy is a consequence of this theory and applies to all forms of energy, including kinetic energy and potential energy.

Why is the conservation of relativistic energy important?

The conservation of relativistic energy is important because it allows us to make accurate predictions about the behavior of physical systems. It also provides a framework for understanding the relationship between energy and mass, which has led to many important scientific discoveries and technological advancements.

How is the conservation of relativistic energy applied in real-world situations?

The conservation of relativistic energy is applied in many real-world situations, such as in the design of nuclear reactors and the development of particle accelerators. It is also used in the study of celestial bodies, such as stars and galaxies, to understand their formation and evolution.

Are there any exceptions to the conservation of relativistic energy?

While the conservation of relativistic energy is a fundamental principle, there are some situations where it may appear to be violated. This is often due to the conversion of energy into other forms, such as heat or sound, which may not be accounted for in the initial energy calculations. However, overall, the principle still holds true and is a crucial concept in understanding the behavior of our universe.

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