What Is Four-Force in Special Relativity?

[greypilgrim]

Hi.

One of my undergrad professors emphasized explicitly that Newton's second law
$$\mathbf{F}=\frac{d\mathbf{p}}{dt}$$
is not a definition of force, but an empirically found relation between a force law (e.g. Hooke's law) and the change of momentum of an object this force acts upon.

Wikipedia says
The four-force is the four-vector defined as the change in four-momentum over the particle's own time:
$$\mathbf{F}=\frac{d\mathbf{P}}{d\tau} .$$

So what is force in SR, just a definition?

Assume we have a spring with some spring constant that is compressed to half its length and moves at a velocity close to c. How does it act on

1. an object initially in the spring's rest frame (but observed from the lab system)?
2. an object initally at rest in the lab system?

 

[Simon Bridge]

That undergrad prof is incorrect. Newton's law is a definition of what Newton means by "force".
It is very difficult to construct and experiment to show Newton's 2nd law without assuming that definition at least implicitly.

 

[pervect]

Hi.

One of my undergrad professors emphasized explicitly that Newton's second law
$$\mathbf{F}=\frac{d\mathbf{p}}{dt}$$
is not a definition of force, but an empirically found relation between a force law (e.g. Hooke's law) and the change of momentum of an object this force acts upon.

Wikipedia says
The four-force is the four-vector defined as the change in four-momentum over the particle's own time:
$$\mathbf{F}=\frac{d\mathbf{P}}{d\tau} .$$

So what is force in SR, just a definition?

Assume we have a spring with some spring constant that is compressed to half its length and moves at a velocity close to c. How does it act on

1. an object initially in the spring's rest frame (but observed from the lab system)?
2. an object initally at rest in the lab system?

I seem to recall Feynman having vaguely similar ideas. But it's not clear to me what you think a "force" is by your own definition. You've told us what it is not (it's not dp/dt) - but you haven't told us what it is. Offhand, I don't recall Feynman's remarks on the topic.

As far as the law goes, you just add an extra component - the 4-momentum P is just tacks ##P_0## in front of the 3-momentum p, with ##P_0=E##. So the 4-momentum P is just (E,p), E being the energy and p being the 3-momentum. To get the 4-force, you differentiate not with respect to coordinate time t, but proper time ##\tau##. So you write ##dP / d\tau## rather than dp/dt.

 

[bhobba]

That undergrad prof is incorrect. Newton's law is a definition of what Newton means by "force".
It is very difficult to construct and experiment to show Newton's 2nd law without assuming that definition at least implicitly.

Indeed. Newtons first law follows from the second which strictly speaking is just a definition.

The empirical content is in the third law which is equivalent to momentum conservation. However Noethers theorem shows its basically equivalent to spatial symmetry. So what is it's content? Its that natures laws can be expressed as an extremeum principle which is the assumption of Noethers theorem. And that follows from QM. The essence of classical physics is QM.

So what is the content of Newtons laws? Its real content is a paradigm that says - get thee to the forces.

Thanks
Bill

 

[vanhees71]

The four-force, acting on a point particle, is defined as
$$K^{\mu}=\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau},$$
where
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}$$
is the four-momentum of the particle, ##m## is its invariant mass (and one uses exclusively the invariant mass in relativistic physics in the 21st century!), and ##\tau## the proper time of the particle.

Because of the definition of proper time you have the constraint
$$p_{\mu} p^{\mu}=m^2 c^2.$$
Taking the derivative of this equation with respect to ##\tau## you find
$$p_{\mu} \frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=p_{\mu} K^{\mu}=0.$$
Thus the time component of the four-force is given by
$$K^{0}=\frac{\vec{p} \cdot \vec{K}}{p^0}=\frac{\vec{p}}{E/c},$$
where
$$E=c \sqrt{m^2 c^2+\vec{p}^2}$$
is the energy of the particle (kinetic energy plus rest energy ##m c^2##). Thus, in fact, with a given force ##K^{\mu}## this consistency condition tells you that you have in fact only 3 2nd-order equations of motion as in Newtonian mechanics, i.e., you only need the three spatial equations
$$m \frac{\mathrm{d}^2 \vec{x}}{\mathrm{d} \tau^2}=\vec{K}.$$
An example for a four-force is the motion of a charged particle in the electromagnetic field, which is given by
$$K^{\mu}=q F^{\mu \nu} \frac{p_{\nu}}{m c}=q F^{\mu \nu} u_{\nu},$$
where
$$u^{\nu} = \frac{1}{m c} p^{\nu}=\frac{1}{c} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}$$
is the four-velocity of the particle. Expressing the field-strength tensor ##F_{\mu \nu}## in terms of the usual electromagnetic field components ##\vec{E}## and ##\vec{B}## you get
$$\vec{K}=q u^0 \left (\vec{E}+ \frac{\vec{v}}{c} \times \vec{B} \right), \quad \vec{v}=\frac{\vec{u}}{u^0}.$$
Now
$$u^0=\frac{1}{c} \frac{\mathrm{d}x^0}{\mathrm{d} \tau} = \frac{\mathrm{d} t}{\mathrm{d} \tau}=\frac{1}{\sqrt{1-\vec{v}^2}}, \quad \vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}.$$
Thus you finally get
$$m \frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=\frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau} \frac{\mathrm{d} \tau}{\mathrm{d} t} = \frac{1}{u^0} \frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau} =q \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right)=\frac{1}{u^0} \vec{K}=\vec{F},$$
where ##\vec{F}## is the usual force in the sense of Newton, which is of course always defined by ##\mathrm{d} \vec{p}/\mathrm{d} t##. I don't understand the emphasis of your professors. I'd need more context. It's just Newton's 2nd postulate, and it's on the foundations of all mechanics, including special relativistic one. The only difference between Newtonian and special-relativistic mechanics is the relation between momentum and three-velocity. In the relativistic case
$$\vec{p}=m \frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau}=m u^0 \frac{\mathrm{d} \vec{x}}{\mathrm{d} t}=\frac{m \vec{v}}{\sqrt{1-\vec{v}^2/c^2}},$$
while the Newtonian expression ##\vec{p}=m \vec{v}## is an approximation for ##\vec{v}^2/c^2 \ll 1##.

For some details see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[bhobba]

IOffhand, I don't recall Feynman's remarks on the topic.

He carefully explains it in the lectures which everyone interested in physics should devour.

Basically force is a definition, it physical content being that's how you should analyse problems in classical/relativistic mechanics. When I first encountered this I was fooled into thinking Newtons first two laws were vacuous. John Baez however was kind enough to explain what really is going on. Its not trivial - establishing such a paradigm is vital as centuries of applying it to many diverse areas shows - its just not what beginning texts says it is - it's not laws in the usual sense.

Thanks
Bill

 

[jartsa]

Assume we have a spring with some spring constant that is compressed to half its length and moves at a velocity close to c. How does it act on

1. an object initially in the spring's rest frame (but observed from the lab system)?
2. an object initally at rest in the lab system?

Let's consider number 2:

4-momentum is conserved and 3-momentum is conserved. When we push an object that is already moving fast, we give that object 3-momentum that is the opposite of the 3-momentum that we give to ourselves. We lose energy, the object gains energy. As energy and 3-momentum are the components of 4-momentum, we are giving the object the opposite 4-momentum that we are giving to ourselves.

The pushed object says that it is not receiving any energy or any 3-momentum associated with that energy. That's why 3-force and 4-force are smaller at the pushed object's end, compared to the other end.

EDIT: Let's assume we are pushing a massive object using a small force. Now the pushed object can say that 100% of the energy we use for pushing becomes our kinetic energy.

 

[greypilgrim]

But it's not clear to me what you think a "force" is by your own definition. You've told us what it is not (it's not dp/dt) - but you haven't told us what it is.

According to this prof (and actually, not only this one, at least at my university):
"Force is, what a force gauge measures."

I guess this requires the force gauge to be non-accelerated, and the whole system being in equilibrium, such that the force gauge exerts a force opposite the force to be measured, but of same magnitude (Newton's 3rd law). Also, for every force there is a force law such as Hooke's law, the law of gravitation, the Lorentz force law. Careful: Newton's 2nd law is not a force law according to this definition.

Newton's second law establishes the connection between forces and kinematics, i.e. how a force affects the trajectory of a particle. And the fact that this connection is given by ##\mathbf{F}=\frac{d\mathbf{p}}{dt}## and not by, say ##\mathbf{F}=C\cdot\frac{d^2\mathbf{p}}{dt^2}##, is the empirical part.

 

[PeterDonis]

According to this prof (and actually, not only this one, at least at my university):
"Force is, what a force gauge measures."

This is fine as long as you recognize that it isn't a statement about physics; it's a statement about how your prof wants to use ordinary language to describe physics. Mathematically, "what a force gauge measures" is perfectly well-defined whether you call it "force" or something else. See below.

for every force there is a force law such as Hooke's law, the law of gravitation, the Lorentz force law. Careful: Newton's 2nd law is not a force law according to this definition.

Again, this isn't a statement about physics; it's a statement about how your prof wants to use ordinary language to describe physics. Basically he's saying he wants to use the word "force" to describe the "F" that appears in Hooke's law, the law of gravitation, etc., and not to describe the "F" that appears in Newton's 2nd law. But again, mathematically, all of these things are perfectly well-defined and obey the same equations no matter what ordinary language words are used to describe them.

 

[stevendaryl]

This seems to be semantics, and it doesn't really make any difference, but I would say that it is incorrect to say that force is by definition equal to [itex]m a[/itex]. An object under stress (being pulled in different directions, by, say, springs or rubber bands) may have zero acceleration, but it has forces. The net force is zero, and that's the way that I interpret [itex]F = ma[/itex]; it says that the net force on an object is proportional to the acceleration of the object. But to compute net force, you need to know all the forces acting on the object, which are not zero (even though the acceleration is zero).

What is the definition of force? It seems to me that Newton's laws leave it undefined.

 

[greypilgrim]

This is fine as long as you recognize that it isn't a statement about physics; it's a statement about how your prof wants to use ordinary language to describe physics. Mathematically, "what a force gauge measures" is perfectly well-defined whether you call it "force" or something else.

In nonrelativistic physics, I'm fine with all this being questions of semantics. I have difficulties transferring this to SR. I try to explain:

One of the most important statements of Newtonian physics is that all observers in inertial frames measure the same forces. Two observers in different inertial frames might measure different velocities of an object, but they measure the same acceleration, and by ##\mathbf{F}=m\cdot \mathbf{a}## the force causing this acceleration is equal in both frames. So it doesn't matter in which inertial frame a gauge rests in order to measure a force.

Suppose we have a spring that moves at a constant velocity with respect to our lab. Above statement now means: If we expand the spring to double its length and want to measure the restoring force, we can do so with our force gauge either moving with the spring or with it resting in the lab.

Now I read on Wikipedia that for the four-force we have
$$\mathbf{F}=\frac{d\mathbf{P}}{d\tau}=(\gamma\frac{\mathbf{f}\cdot\mathbf{u}}{c},\gamma\mathbf{f})$$
where ##\mathbf{f}=\frac{d}{dt}(\gamma m \mathbf{u})=\frac{d\mathbf{p}}{dt}##.

So now we have lovely gamma factors. So what does that mean for our moving, expanded spring? Does a lab-stationary force gauge measure a different force than a force gauge in the inertial system of the spring?

I think this would actually lead to a paradox. Say we use a simple spring force gauge in the lab system and replace the moving spring with an identical (but expanded to double its rest length) spring force gauge as well. So the situation is symmetric with respect to Lorentz transformations between the lab frame and the frame of the expanded spring.
If we now measure in the lab frame and the lab-stationary force gauge gets expanded to anything different from double its length, this would break this symmetry.

So I conclude that in SR as well, all gauges in inertial frames measure the same forces. But then I can't see why there are gamma factors in the formulae which change the values of the forces.

 

[bhobba]

ISo I conclude that in SR as well, all gauges in inertial frames measure the same forces. But then I can't see why there are gamma factors in the formulae which change the values of the forces.

Hmmmm. Maybe, maybe not.

As usual science adviser John Baez gets to the heart of the matter:
http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

Thanks
Bill

 

[Dale]

So now we have lovely gamma factors.

What's wrong with gamma factors? If you have a 3-force in Newtonian physics and you rotate your coordinate system then you will get the spatial equivalent of gamma factors in the sin and cos terms. I just really don't get how you go from gamma factors to a paradox.

For my part, my personal take is that it is simply a matter of making a self-consistent set of definitions. Yes, you can define "force is what a force gauge measures", we tend to do a similar thing with time. So there is nothing wrong with that approach. You could instead define "force is the rate of change of momentum", along with instructions for measuring momentum. I think that either approach works just fine, you just have to decide which quantity is the more fundamental concept, and then that is the one that is simply measured. Most scientists, since Noether's theorem, probably prefer to list momentum as more fundamental than force so that is why you see resistance to the professor's statement.

 

[SiennaTheGr8]

I don't know if this helps, greypilgrim, but:

so long as the accelerating object's mass remains constant, the invariant magnitude of the four-force acting on the object is simply the magnitude of the three-force that the object "feels" (call it "proper force," ##\vec f_0##), and it's equal to the object's mass times its "proper acceleration" ##\vec a_0## (which is the acceleration that the object "feels" in response to the three-force; its magnitude is the magnitude of the four-acceleration):

##\vec f_0 = m \vec a_0##

Inertial observers can calculate the magnitude of the proper acceleration using their own coordinate (frame-dependent) measurements of the object's motion:

##| \vec a_0 | = \gamma^3 |\vec a| / \gamma_\bot##

where ##\gamma_\bot = (1 - v_{\bot}^2/c^2)^{-1/2}##, and ##v_{\bot}## is the component of ##\vec v## that's perpendicular to ##\vec a##.

So different observers disagree on the magnitude of the three-force as a coordinate measurement—that is, ##\frac{d \vec p}{dt}## isn't invariant—but they can still use their coordinate measurements of the object's motion to find the magnitude of the proper force because that magnitude is also the magnitude of a four-vector (again, provided that the mass doesn't change).

Someone correct me if I'm wrong about any of this.

 

[greypilgrim]

What's wrong with gamma factors? If you have a 3-force in Newtonian physics and you rotate your coordinate system then you will get the spatial equivalent of gamma factors in the sin and cos terms. I just really don't get how you go from gamma factors to a paradox.

Ok. Say the spring moves with ##v=0.9\cdot c## with respect to the lab frame ##I##. It has a spring constant of 1 N/m and its rest length is 1 m, but it's expanded to 2 m. Hence a force gauge in the spring's rest frame ##I'## would measure a restoring force of ##\mathbf{f'}=(1\,\textrm{N},0,0)^T## or ##\mathbf{F'}=(0,1\,\textrm{N},0,0)^T##.

Now let's boost this to the lab frame. If I'm not mistaken, the boost matrix is
$$
\Lambda^{-1}=
\begin{pmatrix}
\gamma & \gamma\frac{v}{c} & 0 & 0 \\
\gamma\frac{v}{c} & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$$
so
$$\mathbf{F}=\Lambda^{-1}\mathbf{F'}=(\gamma\frac{v}{c}\cdot 1\,\textrm{N},\gamma\cdot 1\,\textrm{N},0,0)^T$$
and ##\mathbf{f}=(\gamma\cdot 1\,\textrm{N},0,0)^T\approx (2.3\,\textrm{N},0,0)^T##.

So if we measure the restoring force of the moving, expanded spring in the lab frame, we find 2.3 N. Assume now we perform this measurement using an exact copy of the moving spring, i.e. spring constant 1 N/m and rest length 1 m. Then this spring will expand to 3.3 m, more than three times its rest length.

But now the situation is completely symmetric! We could either say we use spring A as a gauge to measure the force on spring B or spring B as a gauge to measure the force on spring A.

Let's assume a 3rd inertial frame that symmetrically sees one spring moving with ##u## to the right and the other with ##-u## to the left. Now at the point of measurement, he sees one spring expanded to twice its rest length, the other to 3.3 times its rest length. This is clearly asymmetric, and that's what I find paradoxical. How can one resolve this?

 

[Dale]

so $$\mathbf{F}=\Lambda^{-1}\mathbf{F'}=(\gamma\frac{v}{c}\cdot 1\,\textrm{N},\gamma\cdot 1\,\textrm{N},0,0)^T$$
and ##\mathbf{f}=(\gamma\cdot 1\,\textrm{N},0,0)^T\approx (2.3\,\textrm{N},0,0)^T##.

So if we measure the restoring force of the moving, expanded spring in the lab frame, we find 2.3 N.

Be careful here. ##\mathbf{F}=(2.1,2.3,0,0)=(\gamma \mathbf{f} \cdot \mathbf{u}, \gamma \mathbf{f})## so ##f=2.3/\gamma=1##.

Assume now we perform this measurement using an exact copy of the moving spring, i.e. spring constant 1 N/m and rest length 1 m. Then this spring will expand to 3.3 m, more than three times its rest length.

I have no idea where you are getting this, but I think that fixing the above mistake resolves it.

 

[jartsa]

@greypilgrim

Light bouncing inside an Einstein light clock exerts a force on the mirrors. The faster the clock moves, the smaller the force. But measuring that force is not as easy as you think. Let me demonstrate:

First we have a moving light clock. Then we stop the mirrors. Then the bouncing light pushes the mirrors apart. Hey, it was not difficult at all. :)

In the original frame of the light clock the spring attached to the mirrors stretches a small amount, just like it does in the final frame of the mirrors, although the spring is weakened in the frame where it is moving.

Why does the spring stretch just slightly in the original frame? That is the question you are asking, right?

 

[David Lewis]

One of my undergrad professors emphasized explicitly that Newton's second law is not a definition of force,

He's partly right. The Second Law is not a definition of force in the general sense, which is a push or a pull.

 

[vanhees71]

As should be very clear from the covariant treatment of force (for me a pretty unintuitive notion anyway; all becomes much clearer in terms of the action principle, which is heavily used in relativity to derive dynamical laws) the Minkowski force ##K^{\mu}## is a four-vector. It's related to the non-covariant force ##\vec{F}## via
$$(K^{\mu})=(\gamma \vec{F} \cdot \vec{v}/c,\gamma \vec{F}).$$
The covariant form of the equations of motion for a particle with rest mass ##m## is
$$m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau}=\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau} =K^{\mu},$$
and it's very clear, how to transform from one inertial frame to another one via Lorentz transformations (particularly Lorentz boosts that discribe the transformation between an inertial fram and another one moving with uniform velocity with respect to the first and keeping the spatial axes parallel between the frames, i.e., it's rotation free). That's the great advantage of this equation, but the temporal component of it is redundant, as shown in #5, because you must have
$$K^{\mu} p_{\mu}=0.$$
So you have to solve only the spatial equations, using this constraint to define ##K^0##.

Now, in the early days for relativity, this quite simple ideas were not yet fully understood, and the physicists wrote the equations of motion in a non-covariant way, using the non-covariant notion of force, using the coordinate time, i.e., a frame dependent quantity as the independent variable. Then the equations of motion read
$$\vec{p}=m \gamma \vec{v}, \quad \frac{\mathrm{d}}{\mathrm{d} t} \vec{p}=\vec{F}, \quad \vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}.$$
So, while ##\vec{p}## are the spatial components of the four-vector ##(p^{\mu})=(E/c,\vec{p})##, the velocity ##\vec{v}## is a frame-dependent quantity with quite complicated transformation properties. Under boosts you get the "velocity-addition law". It's always easiest to derive such properties from manifestly covariant equations. That holds true for the force too: You get the transformation in a simple form by using the covariant four-vector ##K^{\mu}##,
$$K^{\prime \mu}={\Lambda^{\mu}}_{\nu} K^{\nu}$$
and then use that to evaluate the transformation property of ##\vec{F}##, which is a non-covariant frame-dependent quantity.

Last but not least one should stress that Newton did not write ##\vec{F}=m \vec{a}## but ##\vec{F}=\mathrm{d} \vec{p}/\mathrm{d} t##, which is the general equation of motion for the center of mass of any (in general extended) body, while the former equation is only valid for a point particle. The same holds true, in modified form (i.e., it refers to the center of momentum rather than the center of mass) also in special relativity.

 

[greypilgrim]

Be careful here. F=(2.1,2.3,0,0)=(γf⋅u,γf)\mathbf{F}=(2.1,2.3,0,0)=(\gamma \mathbf{f} \cdot \mathbf{u}, \gamma \mathbf{f}) so f=2.3/γ=1f=2.3/\gamma=1.

Ah okay, there's the mistake.

So we have:

##\mathbf{f}=\frac{d}{dt}(\gamma m \mathbf{v})=\frac{d\mathbf{p}}{dt}## is what a force gauge measures. It's the same in all inertial frames. Hence the change of momentum caused by a constant force that acts a finite time is also the same in all inertial frames.

However, the acceleration caused by this force can be different in different frames. If we apply the same force to two objects of identical rest mass of which one is at rest and one is moving, then former will be accelerated more than the latter. This agrees with the velocity-addition formula and the fact that nothing can be accelerated to speeds greater than c.

After all that, I still don't know what four-force actually is. I mean I more or less understand the definition and the nice mathematical property of being a four-vector, but can we measure it in an experiment?

Now, in the early days for relativity, this quite simple ideas were not yet fully understood, and the physicists wrote the equations of motion in a non-covariant way, using the non-covariant notion of force, using the coordinate time, i.e., a frame dependent quantity as the independent variable. Then the equations of motion read

⃗p=mγ⃗v,ddt⃗p=⃗F,⃗v=d⃗xdt.​

It might be non-covariant, but the force ##\mathbf{f}=\frac{d}{dt}(\gamma m \mathbf{v})=\frac{d\mathbf{p}}{dt}## is even the same in all inertial frames, isn't that an even better property?

 

[SiennaTheGr8]

But the three-force ##\mathbf f = \frac{d \mathbf p}{dt}## is not the same in all inertial frames. It isn't invariant. That's what I was trying to explain in my post above.

The magnitude of the proper three-force is indeed invariant (provided ##\dot{m} = 0##):

##| \mathbf{f}_0 | = m| \mathbf{a}_0 | = m\dfrac{\gamma^3 }{\gamma_{\bot}}|\mathbf a|##

It's the magnitude of the four-force (when mass remains constant).

Don't think of the four-force as an observable (in the sense of something that we humans could directly measure). Its components power and three-force are the observables, and their values depend on one's frame of reference. The four-force is a convenient way to express mathematically the fact that power and three-force transform between frames together in the same way that space and time do. They "mix." Lorentz transformation.

If we could step "outside" of time and behold spacetime in its full glory, then maybe we'd intuitively understand four-vectors as "things."

 

[greypilgrim]

Ok now I'm hopelessly confused. Didn't Dale point out in #16 that the three-force is the same in both inertial frames?

 

[SiennaTheGr8]

However, the acceleration caused by this force can be different in different frames. If we apply the same force to two objects of identical rest mass of which one is at rest and one is moving, then former will be accelerated more than the latter. This agrees with the velocity-addition formula and the fact that nothing can be accelerated to speeds greater than c.

It occurs to me that there may be a misunderstanding here.

So we define three-force like you did:

##\mathbf f = \frac{d \mathbf p}{dt} = \frac{d}{dt}(\gamma m \mathbf v)##.

But then you say "If we apply the same force to two objects of identical rest mass of which one is at rest and one is moving..."

Well, the point is that if the forces are really the same here, all that means is (assuming constant masses):

##m_1(\dot{\gamma}_1 \mathbf v_1 + \gamma_1 \mathbf a_1) = m_2(\dot{\gamma}_2 \mathbf v_2 + \gamma_2 \mathbf a_2)##

If object 1 starts at rest, then ##\mathbf v_1 = \mathbf 0## and ##\gamma_1 = 1##, and so at that moment (I'm assuming that the forces are applied simultaneously in the "pusher's" frame):

##m_1 \mathbf a_1 = m_2(\dot{\gamma}_2 \mathbf v_2 + \gamma_2 \mathbf a_2)##

What I'm getting at is that we haven't defined force vaguely as amount of push/pull, or even as anything "external" to the object, but rather as the time-derivative of a property of the object that's acted upon. And forgive me if I've misread you, but it seems like maybe you're letting yourself revert to that intuitive Newtonian push/pull stuff.

 

[SiennaTheGr8]

Quick edit:

Since the objects have equal mass (##m_1 = m_2##), they cancel, and:

##\mathbf a_1 = \dot{\gamma}_2 \mathbf v_2 + \gamma_2 \mathbf a_2##

 

[greypilgrim]

Exactly. And in #20 I tried to conclude from that, at least if ##\mathbf{f}## points in the same direction as ##\mathbf{v}_2##, that ##|\mathbf{a}_2|<|\mathbf{a}_1|##. However I forgot about the first term ond the right hand side, which messes things up.

Is ##\mathbf{a}_2##, that ##|\mathbf{a}_2|<|\mathbf{a}_1|## still correct, and how can this be shown?

 

[Dale]

I would never say that a vector is the same in all frames (neither 3 vectors nor 4 vectors). Different frames include spatial rotations, not just boosts, and they change under rotations. So the most you can say is that a vector's magnitude is the same in all frames.

 

[greypilgrim]

and they change under rotations

Well my understanding of linear algebra is that vectors stay the same no matter what basis we use to represent them. Of course the components change, but not the vectors themselves.

Let's put it this way: If we stick to linear boosts, will all three components of ##\mathbf{f}## be the same in all frames?

 

[Dale]

Well my understanding of linear algebra is that vectors stay the same no matter what basis we use to represent them.

Sure, but in the discussion above you are focused on the components. Particularly your concern about the gamma factor is all about the components.

 

[jartsa]

Let's put it this way: If we stick to linear boosts, will all three components of ##\mathbf{f}## be the same in all frames?

No. Two neutral wires with same electric current attract each other because the repulsive Coulomb force between the electrons is reduced. That is the explanation in the frame of the wires.

In the frame of the electrons the explanation is that the Coulomb force between the electrons is normal, while the Coulomb force between the positive charges is reduced ... and some Lorenz-contraction effects in addition to that.

 

[vanhees71]

It is not the same in all inertial frames. How do you come to that conclusion? As I said you have to Lorentz transform ##K^{\mu}## as a four vector and then determine the non-covariant three-vector ##\vec{F}=\frac{1}{\gamma} \vec{K}##.

 

[vanhees71]

I would never say that a vector is the same in all frames (neither 3 vectors nor 4 vectors). Different frames include spatial rotations, not just boosts, and they change under rotations. So the most you can say is that a vector's magnitude is the same in all frames.

But the force is also changing with rotation-free boosts! As I said you have
$$\vec{F}=\frac{1}{\gamma] \vec{K}$$
in any frame of reference, and that's not the same in the boosted frame, where
$$K^{\prime \mu}={\Lambda^{\mu}}_{\nu} K^{\nu}.$$
So the direction of ##\vec{K}## changes under a boost, and thus ##\vec{F}' \neq \vec{F}##.

 

[greypilgrim]

It is not the same in all inertial frames. How do you come to that conclusion? As I said you have to Lorentz transform ##K^{\mu}## as a four vector [...]

But isn't that exactly what I did in #15? The rotation-free boost only multiplied a ##\gamma## to the spatial part of ##\mathbf{F}'##, which disappears again if you do

[...] and then determine the non-covariant three-vector ##\vec{F}=\frac{1}{\gamma} \vec{K}##.

or
$$\mathbf{f}=\frac{1}{\gamma}\cdot\gamma\cdot\mathbf{f}'=\mathbf{f}'$$
in my notation.

 

[SiennaTheGr8]

@greypilgrim:

vanhees's ##\vec F## (three-force) and ##\vec K## (three-force times ##\gamma##) are measurements made in the same frame. Does your prime notation refer to a different frame? If so, you've misunderstood vanhees, I think.

 

[greypilgrim]

I don't think I've misunderstood. I start from the primed frame, which is the rest frame of the spring, hence ##\gamma=1##. Then
$$\mathbf{F}'=(0,\mathbf{f}')\enspace .$$
If we abbreviate the spatial part of ##\mathbf{F}'## with ##\vec{K'}##, then ##\vec{K'}=\mathbf{f}'##.

Now I perform the boost
$$\mathbf{F}=\Lambda^{-1}\mathbf{F'}=(\gamma\frac{\mathbf{f}'\cdot\mathbf{v}}{c},\gamma\cdot\mathbf{f}')^T=(F_0,\vec{K})^T$$
hence ##\vec{K}=\gamma\cdot\mathbf{f}'## and ##\mathbf{f}=\frac{1}{\gamma}\vec{K}=\mathbf{f}'##.

All unprimed quantities are in the same frame, as in vanshees's notation (he uses ##K^\mu## instead of ##\mathbf{F}## for the four-vector and ##\vec{F}## instead of ##\mathbf{f}## for the non-covariant three-vector).

 

[vanhees71]

The rotation free Lorentz boost has the form (in usual 1+3D notation)
$$\Lambda^{-1}=\begin{pmatrix} \gamma & \gamma \vec{\beta}^{T} \\ \gamma \vec{\beta} & \mathbb{1}+(\gamma-1)/\beta^2 \vec{\beta} \vec{\beta}^{T} \end{pmatrix}.$$
Now in my notation you have
$$K'=\begin{pmatrix} 0 \\ \vec{K}' \end{pmatrix} \; \Rightarrow \; K=\Lambda^{-1} K'=\begin{pmatrix} \gamma \vec{\beta} \cdot \vec{K}' \\ \vec{K}'+(\gamma-1)/\beta^2 \vec{\beta} (\vec{\beta} \cdot \vec{K}') \end{pmatrix}.$$
So your equation is only correct, if ##\vec{K}' \parallel \vec{\beta}##, which is generally not the case.