Relativistic Energy and Lorentz factor

In summary, the question asks for the total energy of one particle in the rest frame of the other particle, which can be found by multiplying the rest mass (m0) by the Lorentz factor. The given clue helps to check the answer for a special case when (v/c)^2 = .5. The hint is to find the velocity of the other particle in the rest frame of one of the particles. The equation for energy in terms of velocity and rest mass is E = y m0 c^2. To apply this equation, the velocity transformation equation must be used first and then the resulting velocity can be plugged into the energy equation to solve for the total energy, which is found to be 3m0c^
  • #1
Kunhee
51
2

Homework Statement


[/B]
Two particles of rest mass m0 approach each other with equal and opposite velocity v, in a laboratory frame. What is the total energy of one particle as measured in the rest frame of the other?

But the question gives a clue which reads "if (v/c)^2 = .5, then E = 3m0c^2."

The Attempt at a Solution


[/B]
My current understanding of this question is that the total energy is simply m0c^2 multiplied by the Lorentz factor. Could you explain to me how the clue is relevant?
 
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  • #2
Maybe the clue is to be used to check your answer for the special case when (v/c)2 = .5
 
  • #3
When (v/c)^2 = .5 then the equation is just that E = (1.414)(m0c^2). Where did the 3 come from?
 
  • #4
Kunhee said:

My current understanding of this question is that the total energy is simply m0c^2 multiplied by the Lorentz factor.
Yes
 
  • #5
Kunhee said:
When (v/c)^2 = .5 then the equation is just that E = (1.414)(m0c^2). Where did the 3 come from?
You are asked to find the energy of one of the particles according to a reference frame moving with the other particle. The statement (v/c)^2 = .5 specifies v for each particle in the lab frame. But you need to work in the frame of one of the particles.
 
  • #6
It's not a hint per se - it's just providing you with a particular test case to enable you to check if your obtained answer is correct.

What would be a hint is: what is the velocity of the other particle as measured in the rest frame of one of the particles?
 
  • #7
Oh I see.
Could you help me set up the equation in the particle's frame of reference?

Before I can plug into E = y m0 c^2
I need to find the m0 when the Lorentz factor is applied first because we are
observing it from the frame of the particle?
 
  • #8
Kunhee said:
Before I can plug into E = y m0 c^2
I need to find the m0 when the Lorentz factor is applied first because we are
observing it from the frame of the particle?
##m_0## is unchanged - it is the rest mass of the particle. What is changed is the Lorentz factor - because the particles move with different velocities in different frames.
 
  • #9
Does that mean I need to first do velocity transformation
u = (u' + v) / (1 + u'v/c^2)

And use the resulting velocity to plug into
E = y m0 c^2 ?
 
  • #10
Yes, that is correct - but it would be good if you try not to view it as doing some stuff and then plugging into a formula, because that often hinders with the understanding of the concepts.
 
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  • #11
I see. Thank you!
 
  • #12
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Last edited:
  • #13
Kunhee said:
During the velocity transformation, the u will depend on u' and v.
But we are given (v/c)^2 only.
Perhaps it would help if you show your final expression for the energy. The transformed velocity ##v'## is a function of ##v##, so your final answer will / should be in terms of ##v##, because well, that is your given initial parameter.
 
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  • #14
u' = [u - (-u)] / [1 - (u)(-u)/c^2] = 2u / [1+(u/c)^2

Plugging the u' into Lorentz factor for the Energy equation:

E = y m0 c^2 = m0 c^2 / [1 - ( [2u / [1+(u/c)^2] / c )^2] ^ -1/2

And with further simplification I get m0 c^2 / (1/3) = 3

Got it. Thanks!
 

Related to Relativistic Energy and Lorentz factor

1. What is Relativistic Energy?

Relativistic Energy is a concept in physics that describes the relationship between an object's energy and its mass. It takes into account the effects of special relativity, which states that the energy of an object is not only determined by its mass and velocity, but also by its position in space and time.

2. How is Relativistic Energy calculated?

The formula for calculating Relativistic Energy is E=mc^2, where E is energy, m is mass, and c is the speed of light. This formula was developed by Albert Einstein as part of his theory of special relativity.

3. What is the Lorentz factor?

The Lorentz factor, also known as the gamma factor, is a term used in special relativity to describe the relationship between an object's velocity and its relativistic mass. It is calculated by taking the reciprocal of the square root of (1-v^2/c^2), where v is the object's velocity and c is the speed of light.

4. How is the Lorentz factor related to time dilation and length contraction?

The Lorentz factor plays a crucial role in time dilation and length contraction, which are both consequences of special relativity. Time dilation refers to the slowing down of time for objects moving at high speeds, and is related to the Lorentz factor through the formula t=t0/gamma, where t is the time experienced by the moving object, t0 is the time experienced by a stationary observer, and gamma is the Lorentz factor. Length contraction, on the other hand, refers to the shortening of an object's length in the direction of motion, and is also related to the Lorentz factor through the formula l=l0/gamma, where l is the length experienced by the moving object, l0 is the length experienced by a stationary observer, and gamma is the Lorentz factor.

5. What are some practical applications of Relativistic Energy and the Lorentz factor?

Relativistic Energy and the Lorentz factor have numerous practical applications, including in nuclear power, particle accelerators, and GPS technology. Understanding these concepts is also crucial for the development of advanced technologies such as space travel, where objects are moving at high speeds and special relativity must be taken into account.

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