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I'm going through A P French's special relativity. In one chapter (6) the following is set up:

Suppose that a stationary particle of mass $M_0$ is struck by a photon of energy $Q$, which is completely absorbed. The combined system will have mass $M'$ and will recoil with a velocity $v=\beta c$.

It then goes on to say that conservation of energy implies: $$E=M_0 c^2+Q=M' c^2 $$ (1)

and conservation of momentum implies: $$p=Q/c=M' \beta c$$ (2)

Using the first equation to solve for $M'$, and plugging $M'$ into the second equation to solve for $\beta$, we wind up with: $$\beta=\frac{Q}{M_0 c^2+Q}$$

However, if I use what was developed earlier in the chapter, $M'=\gamma M_0$, and assume equation (1) [ignoring (2)], I get a completely different answer. Likewise, if I assume equation (2), I get yet a different answer, for $\beta$.

(So, the first solution would be solving $M_0 c^2+Q=\gamma M_0 c^2$ for $\beta$, with $\gamma=(1-\beta^2)^{-1/2}$. The second solution would be solving $Q/c=\gamma M_0 \beta c$ for $\beta$. I didn't include the solutions I got because they're [from what I gather] incorrect, and because the equations for them aren't too interesting/enlightening/short to write)

I figure the only way to reconcile these problems is if $M'$ doesn't only depend on $\beta$, but if instead the rest mass actually increases from before the collision, so that I have $M_0 c^2+Q=\gamma M_0' c^2$ and $Q/c=\gamma M_0 ' \beta c$. Then, I'd have two equations and two unknowns, so the result in the book would hold, and I can calculate the new rest mass.

My question is: Is this a real effect? If so, is there a name for it? It just seems a bit remarkable, and it's something that if I had ignored momentum and only looked at energy (or vice versa), I would have missed completely.

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When you say you "get a completely different answer", what is the "answer" you're speaking of? Is the "answer" beta? If so, then how exactly are you solving for beta ignoring equation 2? –  Ataraxia Feb 9 '13 at 3:18
    
@phoenixheart6 clarified in the post. Yes, I was solving for $\beta$. –  NeuroFuzzy Feb 9 '13 at 3:41
    
from Wikipedia: It is not good to introduce the concept of the mass M = m/\sqrt{1 - v^2/c^2} of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion. –  Andre Holzner Feb 9 '13 at 20:10
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1 Answer

up vote 3 down vote accepted

Yes there must be an increase in rest mass. Here's why:

Let's start from a position of maximal ignorance. If we assume nothing about the rest mass before and after, then conservation of energy and momentum read \begin{align} M_0c^2 + pc &= \gamma M c^2 \\ pc &= \gamma Mv \end{align} with $p\neq 0$. With some manipulation, this reduces to $$ \frac{M_0}{M} = 1- \frac{v}{c} $$ Now if $M=M_0$, meaning rest mass is conserved, then we get $v=0$, but this contradicts the assumption $p\neq 0$. So we see that rest mass cannot be conserved in this process.

Addendum. I've been thinking about this, and I realized that the increase in rest mass upon absorbing a photon is clear from some basic examples. Here's one:

Suppose that the particle of mass $M_0$ in this problem is a hydrogen atom in its ground state, and imagine that the photon has enough precisely energy to kick the atom's electron into the first excited state while also satisfying energy and momentum conservation. In this case, it is clear that the rest mass of the hydrogen at will have increased since the binding energy of the electron with the nucleus will have increased.

The main idea, then, behind the original though experiment is that in order for the photon to be absorbed, there must be some mechanism for the absorption (like in the Hydrogen atom example I just gave), and whatever that mechanism is, relativistic energy-momentum conservation will guarantee that it must result in an increase of the rest mass of the original stationary particle.

Hope that helps!

Cheers!

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Thanks! Especially for narrowing it down to a clear contradiction (having, $v=0$ be the only solution when $M=M_0$) –  NeuroFuzzy Feb 14 '13 at 0:21
    
Sure thing! Nice question. –  joshphysics Feb 14 '13 at 2:08
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