# B Momentum in special relativity

1. Oct 9, 2018

### exponent137

Although I thought that I understand special relativity enough, I cannot now clearly answer on the following question:

What is the most direct derivation, why momentum in special relativity is $p=\gamma m v$, where $v$ is velocity of the rocket? Let us assume that Lorentz equations are known, so it is not necessary to derive them. Is the postulate of conservation of momentum enough here?

2. Oct 9, 2018

### stevendaryl

Staff Emeritus
I'm not exactly sure whether there is a reason, other than that is the only choice for momentum that is consistent with Special Relativity and has the right non-relativistic limit.

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3. Oct 9, 2018

### vanhees71

The most simple idea is to argue with relativistic covariance and the Newtonian limit. In Newtonian physics the equation of motion reads
$$\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=\vec{F}, \quad \vec{p}=m \vec{v}.$$
This equation holds true in a momentary inertial frame, where the particle is at rest. In order to generalize it to relativistic speeds, we try to write it in a covariant form first.

In the momentary inertial restframe an invariant way to express time is the proper time $\tau$ since its the time measured by an observer at rest in the particle's rest frame. The proper time is defined by
$$\mathrm{d} \tau=\frac{1}{c} \sqrt{\mathrm{d} x^{\mu} \mathrm{d} x^{\nu} \eta_{\mu \nu}}$$
with the components of the Minkowski fundamental form, $(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)$.

Then we'd like to have four-vectors for our mechanical equations of motion, and the most simple way is to use the space-time fourvector and parametrize the worldline of the particle with its proper time. Then instead of $\vec{v}$ we use the four-velocity
$$u^{\mu}=\frac{\mathrm{d}}{\mathrm{d} \tau} x^{\mu}.$$
Then mass as an intrinsic property of the particle should be defined in its rest frame as a scalar, and that's why this socalled invariant mass is the same as in Newtonian mechanics, and we thus keep it named in the same way, $m$. Then we define the four-momentum vector as
$$p^{\mu}=m u^{\mu}.$$
Then we define the four-force vector by
$$\frac{\mathrm{d}}{\mathrm{d} \tau} p^{\mu}=K^{\mu}.$$
From the definition of proper time the definition of four-momentum implies the "on-shell condition"
$$p_{\mu} p^{\mu} = m^2 c^2,$$
and taking the derivative of this equation wrt. proper time gives
$$p_{\mu} \frac{\mathrm{d}}{\mathrm{d} \tau} p^{\mu}=p_{\mu} K^{\mu}=0,$$
which implies that the Lorentz four-force is usually a function of time and space coordinates as well as of momentum. This "on-shell constraint" implies that there are only three independent equations of motion, i.e., the equations of motion for $\vec{p}$, while the fourth equation follows from the constraint equation.

To see the physical meaning of four-momentum, we reexpress it in terms of non-covariant velocity
$$\vec{v}=\frac{\mathrm{d}}{\mathrm{d} t} \vec{x}=\frac{c}{u^0} \frac{\mathrm{d}}{\mathrm{d} \tau} \vec{x}=\frac{c}{u^0} \vec{u}.$$
From $u_{\mu} u^{\mu}=c^2$ we find
$$(u^0)^2=c^2+\vec{u}^2.$$
On the other hand
$$\vec{v}^2=\frac{c^2}{(u^0)^2}{\vec{u}^2}=\frac{c^2}{(u^0)^2}[(u^0)^2-1] \; \Rightarrow \; u^0=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}.$$
This implies
$$p^{\mu}=m u^{\mu} = \frac{m}{\sqrt{1-\vec{v}^2/c^2}} (c,\vec{v}).$$
This implies that for $|\vec{v}|\ll c$
$$p^0=m c (1-\vec{v}^2/c^2)^{-1/2}=m c (1+\frac{\vec{v}^2}{2 c^2} +\cdots), \quad c p^0=E=m c^2 + \frac{m}{2} \vec{v}^2 + \cdots,$$
wich implies that up to the additive constant
$$E_0=m c^2$$
$p^0 c$ is the kinetic energy of Newtonian mechanics if $|\vec{v}|\ll c$. The fact that in this way $E$ becomes the temporal component of a four-vector leads to the conclusion that it is convenient to include the "rest energy" $E_0=m c^2$ in the energy expression. In relativity we thus have
$$E=p^0 c=\frac{m c^2}{\sqrt{1-\vec{v}^2/c^2}}=c \sqrt{m^2 c^2+\vec{p}^2}.$$
The spatial part of the momentum-four vector becomes the Newtonian expression in leading order of powers of $|\vec{v}|/c$ too.

Last edited: Oct 12, 2018
4. Oct 9, 2018

5. Oct 9, 2018

### SiennaTheGr8

We define momentum as $\gamma m \vec v$ because experiment shows that it's conserved at all velocities (and it reduces to what we already called momentum when $v \ll c$).

Absent experiment, we can show that $\gamma m \vec v$ is the right candidate for such a conserved quantity (read up on the Lewis/Tolman thought experiment), but whether it really is conserved needs to be verified empirically.

6. Oct 10, 2018

### exponent137

Thanks to all. In some time I will continue with this topic.

Was this calculation a main work of Einstein, because Lorentz equations are not his work?

Last edited: Oct 10, 2018
7. Oct 10, 2018

### exponent137

Vanhees, I found some lapses:
After "On the other hand" one bracket fails in the equation.
After "This implies that for", two equations should be in two rows.
Eq. after "we reexpress it in terms of non-covariant velocity" does not agree with eq. before
"This implies that for". probably, in the last equation 1/c is correct not c on the right side of eq., but this is still not enough for a correct result.

Last edited: Oct 10, 2018
8. Oct 10, 2018

### Staff: Mentor

@exponent137 please do not quote an entire post in your response. Only quote the particular things you are responding to.

9. Oct 10, 2018

### Sorcerer

Could this be summarized (for the student not yet ready for the mathematical intricacies of Minkowski space) by something like:

“To get relativistic momentum, multiply mass by (dx divided by proper time)”?​

Now, I know all physical understanding is lost there, and this definition of “proper time” is not general at all, but...

$$p = m \frac{dx}{dτ} = m \frac{dx}{\frac{dt}{γ}} = γm \frac{dx}{dt} = γmu$$

it seems to work.

Edited to remove some of vanhees71‘s derivation after reading PeterDonis request us not to quote entire posts.

Last edited: Oct 10, 2018
10. Oct 10, 2018

### Cryo

You can start with the Lagrangian for the particle, and obtain the quantity that is conserved due to invariance of the Lagrangian under translations in space. This will be the momentum. It will be conserved precisely because of the way it has been obtained.

Before you can do this you need the actual Lagrangian. Here we can do the following. If the particle is in inertial movement, then we can consider its motion in its rest-frame. There the particle is at rest in space, but is moving in time. Essentially its trajectory is a straight line in space-time. Lorentz transforms are linear, so straight lines remain straight lines. Therefore the world-line of the particle for any observer will be a straight line. The action that generates such world-line is the one that minimizes distance between points.

Thus the action is $S=\alpha\int c d\tau$, where $\tau$ is proper time, $c$ is the speed of light, and $\alpha$ is a constant. From basic kinematics we know how to relate the proper time to time of the observer $cd\tau=\sqrt{c^2-\dot{r}^2}dt$. So the Lagrangian is $L=\alpha\sqrt{c^2-\dot{r}^2}$, where $\boldsymbol{\dot{r}}$ is the velocity of the particle. To determine the constant, note that this Lagrangian must reduce to the conventional one $L_{non.rel.}=\frac{m\dot{r}^2}{2}+const$ in the low-speed limit.

The momentum is then given by $\boldsymbol{p}=\partial_{\boldsymbol{\dot{r}}}L$

11. Oct 11, 2018

### stevendaryl

Staff Emeritus
Here's a derivation of the relativistic form of momentum under a few assumptions about momentum:
1. Momentum is proportional to the mass of the object.
2. Momentum is in the same direction as the velocity.
3. The magnitude of the momentum depends only on the magnitude of the object's velocity (not its direction)
This implies that $\overrightarrow{p} = m f(v) \overrightarrow{v}$ for some unknown function $f(v)$ that depends on the magnitude of the velocity. To have the correct nonrelativistic limit, $f(v) \rightarrow 1$ as $v \rightarrow 0$.

I'm also going to assume a few things about elastic collisions between spherical masses
1. In an elastic collision between a small mass $m$ and a much larger mass $M$, the direction of the velocity of the smaller mass changes, but not its magnitude (think of a ball bouncing off a brick wall). (Of course, this won't be precisely true, but will be true in the limit as $m/M \rightarrow 0$
2. In an elastic collision between spherical objects, the momentum changes along the axis connecting their centers, but not in the orthogonal direction.
These assumptions are illustrated by the following picture:

The small mass is initially traveling with a velocity of $v$ in the x-direction, and a velocity of $-u$ in the y-direction. Initially, the large mass is stationary. After the collision, the small mass just flips its velocity in the y-direction, and the velocity in the x-direction remains unchanged. (This is only true in the limit as $m/M \rightarrow 0$. So the following is only approximate reasoning).

We assume that $v$ is relativistic, while $u \ll c$ and $U \ll c$.

Our assumption about momentum is that $\overrightarrow{p} = m f(|\overrightarrow{v}|) \overrightarrow{v}$.

So for the small mass, $v^x = v$, $v^y = -u$ and $|\overrightarrow{v}| = \sqrt{v^2 + u^2} = v + \frac{u^2}{2v} + ...$ (expanding the square root using a Taylor series). Since $u \ll v$, we find that $|\overrightarrow{v}| \approx v$ (ignoring smaller terms).

So the momentum in the x-direction is $m f(v) v$, and the momentum in the y-direction is $-m f(v) u$. After the collision, the momentum in the x-direction remains unchanged and the momentum in the y-direction reverses, to $+m f(v) u$.

For the large mass, initially it is at rest. After the collision, we're assuming that it's moving nonrelativistically, so its momentum is just $-MU$.

Using conservation of momentum in the y-direction, we find approximately: $-m f(v) u = +m f(v) u - M U$. So we have one equation:

$MU = 2m u f(v)$

Now, let's switch to the rest frame that is moving at velocity $v$ relative to the original rest frame. In this frame, we have the following picture:

In this frame, the small mass has no velocity in the x-direction. In the y-direction it has velocity $-u'$ before the collision, and $+u'$ after the collision. $u'$ is nonrelativistic, so we can use, approximately, $p = - m u'$ in the y-direction.

The large mass is initially moving at velocity $-v$ in the x-direction. After the collision, it is moving with a small velocity $-U'$ in the y-direction, as well. As before, we can write, approximately, the momentum in the y-direction after the collision as $-M f(v) U'$.

Using conservation of momentum in the y-direction in this frame, we get:

$-m u' = +m u' - M f(v) U'$

So:

$M U' f(v) = 2m u'$

Now, divide by $MU = 2m u f(v)$, our earlier equation, to get:

$\frac{U'}{U} f(v) = \frac{u'}{f(v) u}$

Rearrange that to get:

$(f(v))^2 = \frac{u'}{u} \frac{U}{U'}$

Okay, finally we use the relativistic velocity transformation formuas: For a Lorentz transformation in the x-direction, the y-component of velocity transforms as follows:

$(V^y)' = \frac{1}{\gamma} \frac{V^y}{1 - \frac{(V^x)^2}{c^2}}$

So for the small mass, in the first frame, originally $V^x = v$ and $V^y = -u$. So in the second frame, $u' = \frac{1}{\gamma} \frac{-u}{1-\frac{v^2}{c^2}} = -\gamma u$. So $\frac{u'}{u} = \gamma$.

For the large mass, in the first frame, after the collision, $V^x = 0$ and $V^y = -u$. So in the second frame, $U' = - \frac{1}{\gamma} U$. So $\frac{U}{U'} = \gamma$.

Plugging these into the equation for $f(v)$ gives:

$(f(v))^2 = \frac{u'}{u} \frac{U}{U'} = \gamma^2$

So $f(v) = \gamma$.

That's the only choice that respects the relativistic velocity addition formula (and the other assumptions I gave at the top).

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12. Oct 11, 2018

### vanhees71

Thanks for pointing this out, I've corrected it. I've put a comma between the equations rather than a linebreak.

What's wrong here? Here's the derivation again with a bit more steps in between
$$\vec{v}=\frac{\mathrm{d}}{\mathrm{d} t} \vec{x}=\frac{\mathrm{d} \tau}{\mathrm{d} t} \frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} = \left (\frac{\mathrm{d} t}{\mathrm{d} \tau} \right)^{-1} \vec{u} = \frac{c}{u_0} \vec{u}.$$
Again, what's the problem with that?

13. Oct 11, 2018

### DrStupid

You don't need to assume them. 1 and 2 result from Newton's definition of momentum (if you don't force quantity of matter to be identical with rest mass) and 3 from isotropy. This way I got the same starting equation for p(v) in my derivation.

14. Oct 11, 2018

### vanhees71

But we all insist on using the only one useful definition of mass, which is the same in Newtonian and relativistic physics. It's the invariant mass and the invariant mass only!

15. Oct 11, 2018

### exponent137

Vanhees, this second equation does not follow correctly from the first one. $v=cu/\gamma$ is from the first equation or $u=c\gamma v$ is from the second equation, and these two equations are not consistent regarding $c$.

16. Oct 11, 2018

### vanhees71

In the 2nd equation is a simple typo. It must of course be

$$p^{\mu}=m u^{\mu} = m \gamma \begin{pmatrix} c \\ \vec{v} \end{pmatrix}.$$

Is this the problem? I'll correct this typo also in the original posting.

I still do not understand the problem with the derivation of the first equation. You can as well see it the other way around

\label{1}
u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=\frac{\mathrm{d} t}{\mathrm{d} \tau} \frac{\mathrm{d}}{\mathrm{d} t} \begin{pmatrix}c t \\ \vec{x} \end{pmatrix} = \gamma \begin{pmatrix} c \\ \vec{v} \end{pmatrix}.

Here, I've used
$$\frac{\mathrm{d} \tau}{\mathrm{d} t} = \frac{1}{c} \sqrt{\frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} t}}=\sqrt{1-\vec{v}^2/c^2}=\frac{1}{\gamma}.$$
From (\ref{1}) you immediately read off
$$u^0=c \gamma, \quad \vec{u}=\gamma \vec{v}.$$
From this you get
$$\vec{v}=\frac{1}{\gamma} \vec{u}=\frac{c}{u^0} \vec{u}.$$

17. Oct 11, 2018

### pervect

Staff Emeritus
As I recall Goldstein, "Classical mechanics", has a couple of derivations.

From memory, one consider the relativistic boost of a perfectly elastic collision. The vector version of the relativistic velocity addition gives the velocities, the momenta must be a function of mass and the calculated velocities and must be conserved. I think I recall the "boost" being perpendicular to the direction of motion in the derivation I saw.

Another derviation uses the Lagrangian method. It's probably more direct if one is familar with Lagrangian mechanics, but not everyone is.

I'd have to re-read the section to be sure my memory isn't plying some tricks.

18. Oct 11, 2018

### stevendaryl

Staff Emeritus
I didn't get it from Goldstein, which I've never read, but I think my heuristic argument in https://www.physicsforums.com/threads/momentum-in-special-relativity.957224/#post-6070588 is along those lines.

19. Oct 11, 2018

### robphy

In classical physics, we define momentum via the impulse (akin to how kinetic energy is defined from the net work).
All of the precedes any explicit discussion of conservation laws.
It seems to me there should be an analogous motivation/development, without invoking collisions or conservation.

20. Oct 11, 2018

### DrStupid

Without collision is possible (any kind of interaction changing momentum should do the job) but I do not think it is possible without conservation.