# Momentum in special relativity

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?

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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• exponent137
vanhees71
Gold Member
2021 Award
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.

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• Sorcerer and exponent137
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.

• 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?

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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}=1## we find
$$(u^0)^2=1+\vec{u}^2.$$
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
$$p^{\mu}=m u^{\mu} = \frac{m c}{\sqrt{1-\vec{v}^2/c^2}} (1,\vec{v}).$$
This implies that for ##|\vec{v}|\ll c##
$$p^0=m c (1-\vec{v}^2/c^2)^{-1/2}=m c (1+\frac{\vec{v}^2}{2 c^2} +\cdots) \; c p^0=E=m c^2 + \frac{m}{2} \vec{v}^2 + \cdots,$$
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.

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.

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@exponent137 please do not quote an entire post in your response.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

[stuff about a momentary situation and agreement with Newton’s forms]

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
$$p^{\mu}=m u^{\mu}.$$

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.

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Cryo
Gold Member
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##

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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• Sorcerer
vanhees71
Gold Member
2021 Award
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.
Thanks for pointing this out, I've corrected it. I've put a comma between the equations rather than a linebreak.

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.
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?

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##.

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.

vanhees71
Gold Member
2021 Award
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!

• protonsarecool, weirdoguy, Cryo and 1 other person
$$\vec{v}=\frac{\mathrm{d}}{\mathrm{d} t} \vec{x}=\frac{c}{u^0} \frac{\mathrm{d}}{\mathrm{d} \tau} \vec{x}=\frac{c}{u^0} \vec{u}.$$
This implies
$$p^{\mu}=m u^{\mu} = \frac{m c}{\sqrt{1-\vec{v}^2/c^2}} (1,\vec{v}).$$
.

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##.

vanhees71
Gold Member
2021 Award
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
\begin{equation}
\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}.
\end{equation}
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
$$\vec{v}=\frac{1}{\gamma} \vec{u}=\frac{c}{u^0} \vec{u}.$$

pervect
Staff Emeritus
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?

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.

stevendaryl
Staff Emeritus
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.

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.

robphy
Homework Helper
Gold Member
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.

It seems to me there should be an analogous motivation/development, without invoking collisions or conservation.

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

robphy
Homework Helper
Gold Member
Without collision is possible (any kind of interaction changing momentum should do the job) but I do not think it is possible without conservation.
But somehow, we can do this in Newtonian physics without first appealing to conservation.
Put another way... in the context of Newton's Laws:
We get impulse and change-in-momentum by using Newton's Second Law.
We get conservation in collisions by using Newton's Third Law.
So, in Special Relativity, I would hope something like that happens.

But somehow, we can do this in Newtonian physics without first appealing to conservation.
Put another way... in the context of Newton's Laws:
We get impulse and change-in-momentum by using Newton's Second Law.
We get conservation in collisions by using Newton's Third Law.
So, in Special Relativity, I would hope something like that happens.
Don't these same rules apply in SR when you replace coordinate time with proper time? I mean, you can certainly get clear from a position coordinate to relativistic energy using proper time and the standard formulus like P = mdx/dτ, F = dP/dt, and W = ∫Fdx.

But somehow, we can do this in Newtonian physics without first appealing to conservation.

I don't know how. All derivations I know include conservation of momentum - either directly or via the third law.

Mister T
Gold Member
We get conservation in collisions by using Newton's Third Law.
So, in Special Relativity, I would hope something like that happens.

We assume the validity of Newton's Third Law, and then show that conservation of momentum follows from that assumption.

In special relativity we assume the validity of conservation of momentum.

But that difference is really just an artifact of the way the knowledge was developed. Physicists used to think that Newton's Third Law was the more fundamental relation. But nowadays physicists regard conservation of momentum as the more fundamental, having rejected Newton's Third Law as invalid because it requires instantaneous action at a distance.

stevendaryl
Staff Emeritus
But somehow, we can do this in Newtonian physics without first appealing to conservation.
Put another way... in the context of Newton's Laws:
We get impulse and change-in-momentum by using Newton's Second Law.
We get conservation in collisions by using Newton's Third Law.
So, in Special Relativity, I would hope something like that happens.

Using conservation of momentum in collisions is equivalent to using the third law, I would think. During a collision, there are unspecified forces acting between the masses. But since the force on one mass due to the other is equal to the opposite of the force on the second mass due to the first, then it follows that the sum of the momenta of the two masses must be constant.

• SiennaTheGr8
robphy
Homework Helper
Gold Member
Yes, Newton's Third is practically the same as conservation of momentum.

But my main point is this:
in PHY 101, we defined momentum using impulse and the Second Law... without using the Third Law or studying a collision.
I am saying that we should be able to do the same thing in Special Relativity:
define relativistic-momentum without using the Third Law or studying a collision.

stevendaryl
Staff Emeritus
Yes, Newton's Third is practically the same as conservation of momentum.

But my main point is this:
in PHY 101, we defined momentum using impulse and the Second Law... without using the Third Law or studying a collision.
I am saying that we should be able to do the same thing in Special Relativity:
define relativistic-momentum without using the Third Law or studying a collision.

Well, we can say, in both the relativistic and non-relativistic cases, that by definition:

##\overrightarrow{dp} = \overrightarrow{F} dt## The change in momentum of an object is quantity that results from applying a force over time.

##dT = \overrightarrow{F} \cdot \overrightarrow{dr}## The work-energy theorem: the change in kinetic energy (I'm denoting it by ##T##) of an object is equal to the work on the object done by a force acting over a distance ##\overrightarrow{dr}##

But what you don't have is ##M \overrightarrow{dv} = \overrightarrow{F} dt##, the second law. Because it's not true in relativity. So you need some other relationship between velocity and momentum. Consideration of elastic collisions in different frames provide such a relationship.

Newton could have used that, instead of the second law.

stevendaryl
Staff Emeritus
There are some facts about kinetic energy, momentum and velocity that together imply the form of momentum and kinetic energy, but they are a little mysterious. I'm not exactly sure whether there is some deep reason that they are true.

Let ##T## be an object's kinetic energy, let ##\overrightarrow{v}## its velocity and let ##\overrightarrow{p}## be its momentum. Then the work-energy theorem tells us that:

Fact 1. ##\frac{\partial T}{\partial p^k} = v^k##

(##p^k## and ##v^k## mean the ##k^{th}## components of the corresponding vectors). So velocity tells us how kinetic energy changes as a function of momentum. I don't know if that's obvious, or not. But anyway, the next part is much more mysterious.

Let's look at things in two different frames. Pick a frame ##F_1##. Pick another frame ##F_2## that is moving at velocity ##\overrightarrow{u}## relative to the first. Now let there be an object that is moving at velocity ##\overrightarrow{v}## as measured in frame ##F_2## Then let's define the function ##\overline{T}(\overrightarrow{u}, \overrightarrow{v})## to be the kinetic energy of that object, as measured in frame ##F_1##. In terms of this awkwardly defined function, we have the following amazing fact about momentum:

Fact 2. ##\frac{\partial \overline{T}}{\partial u^k}|_{\overrightarrow{u} = 0} = p^k##

So momentum tells us how kinetic energy changes as a function of change of reference frame.

That fact about momentum, kinetic energy, and reference frames is completely mysterious to me. However, by considering elastic collisions as seen from different reference frames, you can see that ##\frac{\partial \overline{T}}{\partial u^k}|_{\overrightarrow{u} = 0}## is conserved (when you sum it over all masses in an elastic collision) if kinetic energy is. But why that conserved quantity should be equal to (a component of) the momentum is a little mysterious.

But facts 1. and 2. together tell us what the momentum and kinetic energy are, in terms of velocity, in both Newtonian physics and relativistic physics.

Vanhees, according to ##c##, it is now clear.
Here are still some typos, which should be corrected according to ##c##. Probably the reason is switch to units ##c=1##.
From ##u_{\mu} u^{\mu}=1## we find
$$(u^0)^2=1+\vec{u}^2.$$
On the other hand
$$\vec{v}^2=\frac{c^2}{(u^0)^2}{\vec{u}^2}=\frac{c^2}{(u^0)^2}[(u^0)^2-1] \; \Rightarrow \; u^0=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}.$$

vanhees71
Gold Member
2021 Award
I don't know how. All derivations I know include conservation of momentum - either directly or via the third law.
You have to explain first, what you mean by "third law" in relativity. Of course, you need the field concept. The third law a la Newton implies action at a distance which violates causality. I think the approach via symmetry concepts is much clearer. You start with Minkowski space's symmetries (the proper orthochronous Poincare group) and then the Lagrangian for free particles is determined (modulo equivalence of course). This implies all the expressions for the conserved quantities (energy, momentum, angular momentum, center-of-momentum motion).

vanhees71
Gold Member
2021 Award
Vanhees, according to ##c##, it is now clear.
Here are still some typos, which should be corrected according to ##c##. Probably the reason is switch to units ##c=1##.
Argh! I should have set c=1. I'm too used to the conventions of my everyday work. Sorry for that. I'll also correct this obvious typo.

The third law a la Newton implies action at a distance which violates causality.

Includes - not implies. The third law also holds for local interactions.

vanhees71
Gold Member
2021 Award
Then define, what you mean by "third Law". It's clear that in special relativity momentum conservation strictly holds true, because momentum is the "Noether charge" of spatial translation invariance, but for me there's no way to accomodate this with the "third law". The very reason that we have field theories as the natural description in relativistic physics is that it enables momentum conservation without the validity of the "third law".

stevendaryl
Staff Emeritus