Why is the momentum of a particle $\gamma mv$?

Question

I am very new to relativity, and as I was going through Resnick & Halliday, I saw that momentum was no longer $mv$, rather $\gamma mv$. This was the proof:

$$p = mv = m \frac{\delta x}{\delta t_0} $$

where $\delta t_0$ is the proper infinitesimal time interval to cover $\delta x$ distance.

Using the time dilation formula, $\delta t = \gamma \delta t_0$ we can then write $$ p = m \frac{\delta x}{\delta t_0} = m \frac{\delta x}{\delta t} \frac{\delta t}{\delta t_0} = \gamma m \frac{\delta x}{\delta t} $$ However, since $ \frac{\delta x}{\delta t} $ is just the [observed] particle velocity, $$ p = \gamma mv $$

On the other hand, when I went on to Feynmann's book I saw that there he talked about using "the new m" (the relativistic mass) to derive all the equations and said

$$p = mv = \frac{m_0 v}{(1 - \frac{v^2}{c^2})} = \gamma m_0 v $$

My problem is that if both these proofs are true, then why doesn't one consider the other's argument, id est, why aren't both of these factors considered together ? If the mass is really changing, why doesn't Resnick & Halliday account for it? That way the momentum would turn out to be $p = \gamma^2 mv$

Am I missing something here?

Answer 1

They're both saying the same thing: the relativistic momentum is given by $$ \mathbf{p}=\gamma(v)\,m\mathbf{v} $$ The confusion, it seems, is that you are using Feynman's $m=\gamma m_0$ as equivalent to the $m$ in Resnick & Halliday's text; the actual correlation is Feynman's $m_0$ to Resnick's $m$--both of these terms are the (invariant) rest mass. Loosely, Feynamn "attaches" the Lorentz factor ($\gamma$) to the mass while Resnick attaches it to the velocity ($dx^i/d\tau=dx^i/d(t/\gamma)=\gamma v^i$)

Feynman's notation of "relativistic mass," $$ m_{\rm rel}=\gamma(v)m $$ is considered by many to be "outdated" (though there are still some proponents); Resnick & Halliday's notation, on the other hand, is the more modern approach to relativistic velocity. Nonetheless, the two textbooks come out to the same conclusion: $p=\gamma mv$.

Answer 2

The way to define momentum in a special relativistic context is the following:

Start with the trajectory of the particle parametrized by its proper time $x^{\alpha} (\tau)$; define the four-momentum by $p^\alpha = m \frac{\mathrm{d}x^{\alpha}}{\mathrm{d}\tau}$, where $m$ is the mass of the particle (note that I'm only using one mass, not distinguishing between rest mass and mass. As Kyle pointed out in his comment, this is an outdated concept).

If you consider that an inertial observer sees the particle with speed $v$, then you can use the relations between the proper time and the time measured by the observer to write:

$p^i = m \frac{\mathrm{d}t}{\mathrm{d}\tau} \frac{\mathrm{d}x^{i}}{\mathrm{d}t}$

The first derivative is $\gamma$. The second is $v^i$, leading to your expression.

Answer 3

Am I missing something here?

Yes. What you're missing is "the mass of a body is a measure of its energy-content". Read Einstein's original paper, and take note of this: "If a body gives off the energy L in the form of radiation, its mass diminishes by L/c²". Next, imagine your body is a massless photon in a gedanken mirror-box. It isn't actually at rest because it's going round and round at c. But it's effectively at rest, and it therefore adds to the rest-mass of the system. Think of photon momentum p=hf/c as resistance to change-in-motion for a wave propagating linearly at c. Then think of the wave nature of matter and understand this: if that photon is going round and round at c, it still offers resistance to change-in-motion, but we don't call it momentum.

OK, now move the box. You do work on the box to make it move. You add energy to it. As for how much, take a look at the simple inference of time dilation due to relative velocity. See the picture of the light beam bouncing back and forth between the mirrors? That's not totally unlike your mirror-box. And note the mention of Pythagoras's theorem? That's how we derive the Lorentz factor. There's a right-angle triangle, the hypotenuse is the light-path where we say the length = 1 because we're using natural units. The base represents your speed as a fraction of c, and the height gives the Lorentz factor which you see in the expression of proper time:

$$\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$ So you can see how the proper time and the Lorentz factor are related, and you can perhaps get an intuitive feel for the way you have to add more and more energy as the angle flattens. So you'll be happy with the physics of it. (NB: you missed the square root out of your question, see Feynman's book here). The issue is the definition of mass. The "relativistic mass" increased. But like Kyle said, this is a somewhat archaic term. It's a measure of energy, and we tend not to use it any more. See Wikipedia where you can see how Einstein came out against it:

"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."

But that ain't the half of it. See this from Einstein's paper: "If the theory corresponds to the facts, radiation conveys inertia between the emitting and absorbing bodies". A photon conveys inertia. It has a non-zero "inertial mass". And because any concentration of energy causes gravity rather than mass per se, a photon has an "active gravitational mass". And a "passive gravitational mass". Mass is dripping with ambiguity. And for the cherry on top, invariant mass... varies! Mass is a mess.

Answer 4

Am I missing something here?

Well, you (among others) seem to be missing that to measure "momentum" is defined through the application of the gradient of the translation operator $\nabla \hat T_{\mathbf r}[~] := \frac{d}{d \mathbf r_{\mathcal S} }[~]$
to what's given through observational data (e.g. concerning a particular object $A$ under consideration)
by members of a suitable (inertial) system $\mathcal S$.
See also for instance this question: ("Momentum as Generator of Translations"; PSE/q/45067) , and questions linked there.

Arguably, sensibly, this operator should be applied to the quantity which directly characterizes the object $A$ under consideration: its duration (a.k.a. its "proper time") $\Delta \tau_A$ throughout the course of a (suitable, short) experimental trial.

Expressing explicitly $$ \Delta \tau_A := \sqrt{(\Delta \tau_{\mathcal S}[~A~])^2 - (\Delta \mathbf r_{\mathcal S}[~A~] / c)^2} $$

therefore $$ \eqalign{ \mathbf p_{\mathcal S}[~A~] & \simeq \frac{d}{d \mathbf r_{\mathcal S} }[~\sqrt{(\Delta \tau_{\mathcal S}[~A~])^2 - (\Delta \mathbf r_{\mathcal S}[~A~] / c)^2}~] \\ & = \frac{-\Delta \mathbf r_{\mathcal S}[~A~] / c^2}{\sqrt{(\Delta \tau_{\mathcal S}[~A~])^2 - (\Delta \mathbf r_{\mathcal S}[~A~] / c)^2}} \\ & = -\vec \beta_{\mathcal S}[~A~]~\gamma_{\mathcal S}[~A~] / c }.$$

Certain proportionality coefficients can and should be factored in:

• the "invariant mass" $m_A$, as an extensive quantity characterizing object $A$ (in the trial under consideration), and

• $c^2$ for backward compatibility (in terms of dimensionality) with earlier, preliminary conceptions of "momentum" (e.g. in Newtonian physics), and

• a minus sign.

Together, the proportionality coefficient $m_A~c^2$ constitutes the "invariant energy" ("rest energy", "center-of-mass energy") characterizing object $A$ (in the trial under consideration).

Consequently: $$ \mathbf p_{\mathcal S}[~A~] = m_A~c~\vec \beta_{\mathcal S}[~A~]~\gamma_{\mathcal S}[~A~] := m_A~\mathbf v_{\mathcal S}[~A~]~\gamma_{\mathcal S}[~A~].$$

Finally, expressing the relevant geometric data ($\Delta \tau_A$) together with suitable propotionality coefficients as a "phase" $\phi := \text{Exp}[~-i~\frac{m_A~c^2}{\hbar}~\Delta \tau_A~]$,
the appropriate operator to evaluate the momentum of object $A$ with respect to system $\mathcal S$ is: $$ \mathbf{ \hat p}_{\mathcal S} := -i~\hbar~\frac{d}{d \mathbf r_{\mathcal S} }[~].$$

Similarly, of course, for the evaluation of $A$'s energy with respect to system $\mathcal S$ by applying the operator $\hat E_{\mathcal S} := i~\hbar~\frac{d}{d \mathbf \tau_{\mathcal S} }[~]$:

$$ \eqalign{ E_{\mathcal S}[~A~] & := m_A~c^2~\frac{d}{d \mathbf \tau_{\mathcal S} }[~\sqrt{(\Delta \tau_{\mathcal S}[~A~])^2 - (\Delta \mathbf r_{\mathcal S}[~A~] / c)^2}~] \\ & = m_A~c^2~\frac{\Delta \tau_{\mathcal S}[~A~]}{\sqrt{(\Delta \tau_{\mathcal S}[~A~])^2 - (\Delta \mathbf r_{\mathcal S}[~A~] / c)^2}} \\ & = m_A~c^2~\gamma_{\mathcal S}[~A~] },$$

such that $A$'s "invariant energy" can also be expressed as $$ m_A~c^2 = \sqrt{ (E_{\mathcal S}[~A~])^2 - c^2~(\mathbf p_{\mathcal S}[~A~])^2 }.$$