# Confuse about the relativistic effect

1. Dec 22, 2008

### KFC

Hi there,
I am reading some materials on relativistic effect but since it is quite abstract and I need to think the situation between moving frame and rest frame, it becomes quite confusing. I setup a situation and some questions with answer (from my understand), could you please tell me if my understanding is correct or not?

A moving frame, call primed frame, attached to a moving train at speed U close to speed of light, you are sitting in the train. I am sitting in the platform, call (unprimed) frame.

1) If you measure the length of the train and get $$L_0$$, in my point of view, the length of train is short than $$L_0$$, right?

2) You carry a watch with you, you pretty sure the minute hand of the watch move every 60 second. However, in my point of view, I see the minute hand of your watch is moving ahead more than 60 second. Namely, in my point of view, your watch is running slower than mine.

3) The most confusing concept is the energy in relativistic limit. First, generally, I wonder if the total energy of a particle should be written as

$$E = mc^2$$

??? And this TOTAL energy should include two parts: kinetic energy and rest energy? This is,

$$\textnormal{total energy} = mc^2 = KE + m_0 c^2$$

There, $$m$$ is the mass when the particle is in motion

$$m = \frac{m_0}{\sqrt{1-u^2/c^2}}$$

What confusing me is: when $$u\ll c$$, then $$m=m_0$$, the total energy becomes

$$m_0c^2 = KE + m_0 c^2$$

this gives kinetic energy go to zero? But I think this should give the non-relativistic case. What's going on?

4) Still about energy. The total energy in relativistic limit can also expressed in terms of relativistic momentum

$$\textnormal{Total energy} = mc^2 = \sqrt{p^2c^2 + m_0^2c^4}$$

Is this result exactly the same as the previous definition (E = KE + m_0c^2) ?

Again, if $$u\ll c$$, $$p \to m_0 v$$, then the total energy becomes $$m_0c^2$$ ??? In this non-relativistic limit, I think total energy should be $$\frac{m_0v^2}{2}$$

5) The last question is: in non-relativistic case, the kinetic energy for particle with mass $$m=m_0$$ be

$$KE = \frac{mv^2}{2}$$

For relativistic case,

$$KE = \frac{m_0c^2}{\sqrt{1-u^2/c^2}} - \m_0c^2 = \m_0c^2\left(\frac{1}{1-u^2/c^2} - 1\right)$$

If $$u=c$$, then the first term in the parentheses go to infinity???

All of these questions about energy are really confusing me. Would you please give me some clues what's going on?

2. Dec 22, 2008

### Staff: Mentor

Both of these look like the same question, so I will try to address them together. You are getting this result because you are looking at the total energy as u<<c. It should be no surprise that as u<<c the total energy is dominated by the rest (mass) energy and the kinetic energy contributes a negligble fraction of the energy.

What you really want to do is to solve the original equation for KE and then take the limit of that as u<<c.
$$KE = \frac{m_0 c^2}{\sqrt{1-\frac{u^2}{c^2}}}-m_0 c^2$$

Now, if you take this expression and do a Taylor series expansion about u=0 you get
$$KE = \frac{ m_0 u^2}{2}+\frac{3 m_0 u^4}{8 c^2}+O\left(u^6\right)$$

Where you immediately recognize the first term as being the Newtonian expression for KE.

3. Dec 22, 2008

### KFC

That make sense. Well, I also have a question, since the energy in relativistic limit have two terms (KE and rest energy), so in collision problem like electron and positron produces two photon, when I apply the conservation of energy, should I include two terms for the energy of each particle? What about the photon? Since photon has no mass, should I just simply drop the rest energy? If so, the momentum for the photon just become E/c ?

4. Dec 22, 2008

### JesseM

Yes.
Yes, in your frame it will take longer than 60 seconds for my watch to reach the 60-second mark.
Assuming m is the "relativistic mass" as you say below, that's correct. Most physicists nowadays prefer to avoid using relativistic mass and just refer to rest mass, though.
Yes, from this you can infer that kinetic energy = $$(\gamma - 1)m_0 c^2$$ where $$\gamma = \frac{1}{\sqrt{1 - u^2/c^2}}$$. That way when you add the rest energy to the kinetic energy you get a total energy of $$\gamma m_0 c^2$$.
That equation can never be exactly true as long as u is larger than 0--when u is nonzero, m will always be slightly larger than m0. It is true that when $$u\ll c$$, $$\gamma$$ will be very close to 1, so the kinetic energy $$(\gamma - 1)m_0 c^2$$ will be very small compared to the rest energy $$m_0 c^2$$, and you can treat it as negligible if you just need an approximate answer.
Yes, it's the same. If you substitute in the equation for relativistic momentum, $$p = \frac{m_0 u}{\sqrt{1 - u^2/c^2}}$$, then the resulting equation can be reduced through some algebra to $$\textnormal{Total energy} = \frac{m_0 c^2}{\sqrt{1 - u^2/c^2}}$$. If you try this and don't see how it works I can show you the steps.
Why are you using both u and v to refer to the same object's velocity? Anyway, in the non-relativistic limit the kinetic energy becomes very small compared to the rest energy, so most of the object's total energy is rest energy, it's just that in non-relativistic dynamics you don't even bother to keep track anything other than the kinetic energy (and potential if applicable) because the rest energy of the interacting objects never changes noticeably (whereas in relativistic collisions objects can lose or gain appreciable amounts of rest energy because of heat gained or lost).
Assuming m0 was nonzero, kinetic energy would go to infinity in the limit as u approached c. But no finite amount of applied energy can ever accelerate an object with nonzero rest mass to c in the first place! Only things with zero rest mass like photons can move at c, and to find their energy you can use the quantum equation E = hf, where h is Planck's constant and f is the frequency. And since the relativistic equation tells you a particle with zero rest mass would have E = pc, you can also substitute to get p = hf/c.

5. Dec 22, 2008

### Staff: Mentor

The safest way to do this is using the four-momentum. I know that some people consider this "overkill", but I find it very convenient. So, say we have the electron and positron at rest before the anhilation. Then the four-momentum of the electron is (511.,0,0,0) keV/c, the four-momentum of the positron is also (511.,0,0,0) keV/c, so the total four-momentum of the system is (1022.,0,0,0) keV/c before the anhilation. Then, if one photon goes off on the x-axis then the resulting photons must be (511.,511.,0,0) keV/c and (511.,-511.,0,0) keV/c so that the four-momentum of the system is also (1022.,0,0,0) keV/c after the anhilation.

6. Dec 23, 2008

### Staff: Mentor

In doing conservation of energy and momentum calculations, I always think in terms of only the total energy E (kinetic plus rest energy). If I want to find the kinetic energy of a particle separately, I do it at the very end.

7. Dec 23, 2008

### Staff: Mentor

That goes well with the four-momentum approach since the timelike component is proportional to the total energy rather than the kinetic energy.