Particle energy and momentum in different reference frames

In summary, the question discusses the energy and momentum of a particle in a specific reference frame and asks for the energy in another frame with a different momentum. The solution involves using Lorentz transformations and the invariant m^2 = E^2 - p^2. There may be a discrepancy in the calculated relative velocity between the two frames.
  • #1
warfreak131
188
0

Homework Statement



A particle is observed in a certain reference frame (i.e. observer A) has a total energy of 5GeV and a momentum of 3GeV/c.
(a) What is the energy of the frame in which its momentum is equal to 4GeV/c.

Homework Equations



[tex]E'={\gamma}E-{\gamma}{\beta}(cp)[/tex]
[tex]cp'=-{\beta}{\gamma}E+{\gamma}(cp)[/tex]

The Attempt at a Solution



I plugged in cp'=4GeV, cp=3GeV, and E=5GeV, and solved the second equation for v. Then I plugged in v to the applicable variables in the first equation, and I got 4GeV, but the answer is 5.62GeV. And a different part of the question asks for the relative velocity of the two frames, and for that I get .6 c, as opposed to .19 c.
 
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  • #2
You might find the invariant m2 = E2-p2 useful.

Also, the Lorentz transformations can be written in terms of the hyperbolic trig functions:

[tex]\begin{align*}
E' &= E\cosh \theta + p\sinh \theta \\
p' &= E\sinh \theta + p\cosh \theta
\end{align*}
[/tex]

The velocity is then given by [itex]\beta = \tanh \theta[/itex].
 
  • #3
Is the invariant some other form of the equation

[tex]E^{2}=(cp^{2})+(mc^{2})^{2}[/tex]?
 
  • #4
Yeah, it's the same equation with c=1.
 
  • #5
okay, i managed to get the velocity correct, but i still can't manage to get E' to come out correct
 
  • #6
when the question says "total energy", does it mean E^2=(cp)^2+(mc^2)^2 or does it mean E=ymc^2?
 
  • #7
Either way should give you the same answer. Show us your calculations so we can spot your error.
 
  • #8
warfreak131 said:
when the question says "total energy", does it mean E^2=(cp)^2+(mc^2)^2 or does it mean E=ymc^2?

Both E's are the same since

[tex] \sqrt{ (pc)^2 + (mc^2)^2} = \gamma m c^2. [/tex]

I was actually able to get E' to work out, but found [tex]\beta = 0.92 c[/tex] which disagrees with the relative velocity you quoted.
 
  • #9
I got β=0.186, so you should be able to get it to work out.
 
  • #10
vela said:
I got β=0.186, so you should be able to get it to work out.

i managed to get that for beta as well. but when i plug it into the transformation equations to find E', it doesn't come out to be correct
 

1. What is the relationship between particle energy and momentum in different reference frames?

The relationship between particle energy and momentum in different reference frames is described by the famous equation E=mc2, where E represents energy, m represents mass, and c represents the speed of light. This equation shows that energy and momentum are directly proportional to each other, with the speed of light acting as a conversion factor.

2. How does Special Relativity affect the concept of particle energy and momentum in different reference frames?

Special Relativity, a theory proposed by Albert Einstein, states that the laws of physics are the same for all observers in uniform motion. This means that the concept of particle energy and momentum is relative and depends on the observer's frame of reference. In other words, the energy and momentum of a particle can appear different to different observers depending on their relative motion.

3. Can particle energy and momentum be conserved in different reference frames?

Yes, particle energy and momentum can be conserved in different reference frames. This means that the total energy and momentum of a system of particles will remain the same regardless of the observer's frame of reference. However, the distribution of energy and momentum among individual particles may appear different to different observers.

4. How does the concept of relativistic mass play a role in particle energy and momentum in different reference frames?

Relativistic mass, also known as "apparent mass," is a concept that arises in Special Relativity when considering the energy and momentum of a particle in motion. It takes into account the increase in mass of a particle as it approaches the speed of light. This means that the energy and momentum of a particle in motion will appear different to different observers due to the apparent increase in mass.

5. Can the conservation of particle energy and momentum be violated in different reference frames?

No, the conservation of particle energy and momentum cannot be violated in different reference frames. This is a fundamental principle in physics that holds true in all reference frames. While the measurements of energy and momentum may appear different to different observers, the total energy and momentum of a system will always remain constant.

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