Kinetic energy, special relativity

In summary, the student is trying to solve a problem where they don't know the value of v. They are told that gamma=\frac{1}{\sqrt {1-\frac{v^2}{c^2}}}. After finding the relationship between \lambda and v, they are told that m_0=\frac{P}{v}\sqrt {1-\frac{v^2}{c^2}}. They are also told that they need to look up the rest mass of the electron. After doing so, they are able to solve the problem.
  • #1
fluidistic
Gold Member
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Homework Statement



Calculate the kinetic energy of an electron whose momentum is 2MeV/c.

Homework Equations


[tex]P=\gamma m_0 v =mv[/tex].
[tex]E_K=(m-m_0)c^2[/tex].

The Attempt at a Solution


I'm told that [tex]\gamma m_0 v=\frac{2MeV}{c}[/tex].
If only I had the mass at rest of the electron (it isn't given in the problem), I could calculate its velocity with the first formula I gave. Then I could calculate its mass (not its rest mass, its apparent mass or whatever it's called). And then I could apply the third formula and this would solve the problem. Am I right?
So, should I look for the electron's rest mass in some book? Is there a missing data, or can I solve the problem without this info?
 
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  • #2
You can do that. But you shouldn't have to look anything up. Do you know the relationship between [tex] \lambda[/tex] (the lorentz factor) and v? if so, solve for the rest mass. Similarly, you can find the relationship between [tex] m [/tex] and [tex] m_0 [/tex]. I hope that helps!
 
Last edited:
  • #3
aq1q said:
You can do that. But you shouldn't have to look anything up. Do you know the relationship between [tex] \lambda[/tex] (the lorentz factor) and v? if so, solve for the rest mass. Similarly, you can find the relationship between [tex] m [/tex] and [tex] m_0 [/tex]. I hope that helps!

I appreciate very much your help.
What I know is [tex]\gamma =\frac{1}{\sqrt {1-\frac{v^2}{c^2}}}[/tex]. I don't know how to solve for the rest mass since v is unknown. I get [tex]m_0=\frac{P}{v}\sqrt {1-\frac{v^2}{c^2}}[/tex] where P and c are known but not v...
 
  • #4
ahh what was i thinking. you're right! you need to look up the rest mass. I'm really sorry, at a quick glance I thought this was just algebra.
 
  • #5
aq1q said:
ahh what was i thinking. you're right! you need to look up the rest mass. I'm really sorry, at a quick glance I thought this was just algebra.
Ok thanks for your help. Problem solved!
 
  • #6
fluidistic said:
Ok thanks for your help. Problem solved!

great!
 

1. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is defined as one-half of an object's mass multiplied by the square of its velocity.

2. How is kinetic energy related to special relativity?

In special relativity, energy and mass are equivalent and can be converted into one another. This means that an object's kinetic energy can be converted into its equivalent mass and vice versa. The famous equation E=mc², where E is energy, m is mass, and c is the speed of light, illustrates this relationship.

3. Does kinetic energy change in different reference frames?

Yes, according to special relativity, the kinetic energy of an object is relative to the observer's frame of reference. This means that different observers in different frames of reference will measure different values for an object's kinetic energy.

4. Can an object have negative kinetic energy?

No, an object cannot have negative kinetic energy. Kinetic energy is always a positive quantity, as it is the result of an object's motion. However, in some cases, the direction of an object's velocity can be negative, resulting in a negative value for its kinetic energy formula. This simply indicates that the object is moving in the opposite direction of the chosen frame of reference.

5. How does special relativity affect the calculation of kinetic energy?

Special relativity introduces the concept of relativistic mass, which takes into account the increase in an object's mass as it approaches the speed of light. This means that the calculation of an object's kinetic energy must also take into account its relativistic mass, resulting in a slightly different formula for high-speed objects.

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