# Extreme case of relativitic momentum

1. Jan 8, 2009

### KFC

The relativitic momentum is given by

$$p = \frac{mv}{\sqrt{1-v^2/c^2}}$$

or

$$p = \frac{\sqrt{E^2 - m_0^2c^4}}{c}$$
where $$m_0$$ is the rest mass.

I have two questions about the momentum

1) Are the momentum given by these two expression the same?

2) In extreme case (when v->c), the first expression will give infinite large momentum? How would that happen? If we use the second expression to find the momentum in extreme case, what will energy E there look like? Anyway, I just wonder how to find the momentum in extremem case.

Thanks

2. Jan 8, 2009

### bernhard.rothenstein

Start by expressing the momentum of a particle as

p=p'(1+V/u')/(1-VV/cc)^1/2 (1)
In the limi case u'=c
p=p'sqr[(1+V/c)/1-V/c]1/2 (2)
equation (2) performing the transformation of the momentum of a photon.

3. Jan 8, 2009

### KFC

Thanks for your reply. But I have no idea how do you get this. What does u', p' and V represent?

4. Jan 8, 2009

### Staff: Mentor

Small correction: in the equation above, $m$ should be $m_0$.

Yes. To show this, you need to use the following equation for energy:

$$E = \frac{m_0 c^2}{\sqrt{1-v^2/c^2}}$$

As $v \rightarrow c$, $\sqrt{1 - v^2 / c^2} \rightarrow 0$. But this is in the denominator, so the whole expression $\rightarrow \infty$.

The expression I gave for E also has $\sqrt{1 - v^2 / c^2}$ in its denominator, so $E \rightarrow \infty$ also.

5. Jan 8, 2009

### KFC

Thanks. But I think the momentum shouldn't be infinite large even when v->c. For example, a photon is moving at the speed of light, it still have a finite wavelength so the finite momentum, right? My question is how to find the finite momentum for extreme case.

6. Jan 8, 2009

### Staff: Mentor

Your first formula for p, and the formula I gave for E, are not usable for photons, which have $m_0 = 0$ and $v = c$, because they both give 0/0 which is undefined mathematically.

Your second formula for p, on the other hand, is usable for photons, and gives p = E/c.

7. Jan 8, 2009

### clem

The general equation v=pc/E, may clarify things for you.

8. Jan 8, 2009

Staff Emeritus
The equation I find most useful is (E)2 - (pc)2 = (mc2)2. This is valid for all particles at all velocities.

9. Jan 8, 2009

### bernhard.rothenstein

o.k. Consider that the same tardyon (a particle the speed of which never exceeds c) is detected from the inertial reference frames I and I'. I' moves with constant speed V in the positive direction of the x,x' axes/ Let p and E be momentum and the energy of the tardyon when detected from I and I' respectively. The tardyon moves with speed u relative to I and with speed u' relative to I'. The momentum p and the energy of the tardyon are related in I by
p=Eu/cc (1)
and I' by
p'=E'u'/cc (2)
p and p' are related by the transformation equation [1]
p=[p'+VE'/cc]/sqr(1-VV/cc)]= p'[1+V/u']]/sqr(1-VV/cc) (3)
Replacing the tardyon considered so far by a photon (u'=c) (3) becomes
p=p'sqr[(1+V/c)/(1-V/c) (4)
equation (4) performing the transformation of the momentum of photons.
[1] Robert Resnick, Introduction to Special Relativity (John Wiley and Sons New York, 1968) p.144 (Probably there are newer editions)

10. Jan 10, 2009

### JANm

For question 1 my answer is: indeed these two equations are of the same family. Since T=mc^2-m_0c^2=E-m_c^2,
you can refrase the formula for kinetic energy without the huge potential energy.
Perhaps that will help?