# Free electron border between relativity and non-relativity

1. Jan 22, 2007

### Odyssey

"Free" electron...border between relativity and non-relativity

Hello guys...I have a couple of issues that I'm not exactly sure about...
1. The problem statement, all variables and given/known data

a) What does it mean to say that an electron is "free"?
b) To figure out the kinetic energy of a particle...one must first have to check if it's a non-relativistic case, or an ultra-relativistic case...my notes have it one should compute the value of $$T/mc^2$$ (T=kinetic energy)...but why do we have to do that?

2. Relevant equations
$$T/mc^2$$
$$T=p^2/2m$$
$$E=pc$$
$$E=T=pc$$

3. The attempt at a solution

a) I think "free" means the electron is moving under no potential? So...in that case...the total energy equals the kinetic energy?

b) I don't know why we have to compute the value of $$T/mc^2$$...but I know if it's <<1 then it's a non-relativistic case, and we can use the formula $$T=p^2/2m$$ to find the Kinetic energy. If $$T/mc^2$$ is >>1 then it's a relativistic case, and we use $$E=T=pc$$ to solve for the kinetic energy? What is the definition of the border between non-relativistic and relatiticistic cases?

Last edited: Jan 22, 2007
2. Jan 22, 2007

### Gib Z

a) correct, although Total energy may include its mass?
b) Well, Energy of a particle is given by $$m_0c^2\gamma$$ whilst its energy at rest is given by the same expression without the Lorentz factor. So the kinetic energy is the difference of these two. So Replace T with $$m_0c^2(\gamma -1)$$ and see what happens :) Hope it helps, you'll find it makes sense because small v makes gamma equal 1, which is correct.

Last edited: Jan 22, 2007
3. Jan 22, 2007

### Odyssey

Thanks for the help. :)
Yes...I plugged in some numbers for v...makes sense! I got one more question...what does the quantity $$T/mc^2$$ represent? It has no dimensions...I replace T with $$m_0c^2(\gamma -1)$$...so it becomes $$m_0c^2(\gamma -1)/m_0c^2$$...what does it means do have $$\gamma -1 >> 1$$? Why does it represent a relativistic case? $$\gamma$$ can only take on values between 0 and 1...but with $$\gamma -1 >> 1$$...we can actually have $$\gamma$$be >1? I think I'm getting a bit mixed up with gamma. =\

4. Jan 23, 2007

### Gib Z

It represents the ratio of the kinetic energy to rest energy.

$$\gamma=\frac{1}{sqrt{1-\frac{v^2}{c^2}}}$$. Say we subbed v=0 into gamma, gamma = 1. That means our T/mc^2 ratio is equal to zero, which makes sense, because the ratio of zero kinetic energy to rest energy should be zero.

What it means for gamma - 1 to approach one is quite simple. Its the same as for gamma to approach 2. Lets set gamma=2 and solve for v. Some simple algebra will show you that v/c = (the square root of 3)/2, which is a good portion of the speed of light. That is why it represents a relatavistic case :)

Your a little confused with the gamma. Yours thinking about $$sqrt{1-\frac{v^2}{c^2}}$$, which can not exceed 1. But gamma is the recipricol of that, and that can exceed 1, because gamma =$$\frac{1}{sqrt{1-\frac{v^2}{c^2}}}$$. Sub in any case where v>0 you will see that whilst the bottom can not exceed 1, the entire gamma can.

This is exactly what you want to see, since gamma is there to sort of make up for the extra energy.

EG- Rest energy is mc^2, whilst Rest+Kinetic is the same, multiplied by gamma. Since we want REST+KINETC to be more than REST alone, gamma should be more than one.

Sorry if i haven't explained it very well.

5. Jan 23, 2007

### Odyssey

Thanks! =) gamma is the recipricol...now I understand!

6. Jan 23, 2007

Good Work :)