What Is Four-Force in Special Relativity?

[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.

 

[greypilgrim]

Thank you! That clarifies a lot.

However, now I get the paradox I tried to construct in #15 with rotation-free boosts perpendicular to the force. I'm adapting your notation now:
Let a spring's work direction be $x'$. It moves with $v=0.9\cdot c$ in $y$-direction 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 in $x'$-direction. Hence a force gauge in the spring's rest frame $I'$ measures a restoring force of $\vec{F}'=\vec{K}'=(1\,\textrm{N},0,0)^T$ or $K'=(0,1\,\textrm{N},0,0)^T$.
Now we boost to the lab frame. Since $\vec{\beta}\perp\vec{K}'$, we get
$$K=\Lambda^{-1}K'=(0,\vec{K}')^T\enspace ,$$
hence $\vec{K}=\vec{K}'$. So in the lab frame, the non-covariant three-force is $\vec{F}=\frac{1}{\gamma}\vec{K}\approx (2.3\,\textrm{N},0,0)^T$.

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.

So the situation is asymmetric, the spring resting in the lab is longer. But why? Couldn't we have started with the lab spring, boost to the frame $I'$ and conclude that the other spring must be longer?

 

[Dale]

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.

Again, I have no idea what your reasoning is here. Are you perhaps thinking that Hookes law is a relativistically covariant law? It is not. Hookes law is only valid for v<< c

 

[greypilgrim]

Well I only used springs as an example. Let's take any kind of bathroom-scale-like force gauge. If we use two of them and push them together (top on top), they indicate the same force by Newton 3rd.

Now we assume one of the scales (B) to be moving (fast) in the plane parallel to the scales, such that it passes the resting scales (A). At that precise moment, the scales are pushed together and the forces on both scales are being read off.

Let's say it says 1 N on the resting scales A. By Newton 3rd, this causes an equal force of 1 N directed to the moving scales B. But this is seen in the rest system of A! To know what B actually measures, we need to transform this force to B's rest system. That's what I did in #36. Hence, in B's rest system, the force exerted by A transforms to 2.3 N! And again, by Newton 3rd, this force must be the same as displayed by the scales B.

So gauge B indicates a greater force than gauge A. But here's the catch: We could just as well have started in B's rest frame and used the same line of argument to come to the conclusion that gauge A measures a bigger force.

Clearly those two conclusions contradict each other. So what's wrong?

 

[jartsa]

Well I only used springs as an example. Let's take any kind of bathroom-scale-like force gauge. If we use two of them and push them together (top on top), they indicate the same force by Newton 3rd.

Now we assume one of the scales (B) to be moving (fast) in the plane parallel to the scales, such that it passes the resting scales (A). At that precise moment, the scales are pushed together and the forces on both scales are being read off.

Let's say it says 1 N on the resting scales A. By Newton 3rd, this causes an equal force of 1 N directed to the moving scales B. But this is seen in the rest system of A! To know what B actually measures, we need to transform this force to B's rest system. That's what I did in #36. Hence, in B's rest system, the force exerted by A transforms to 2.3 N! And again, by Newton 3rd, this force must be the same as displayed by the scales B.

So gauge B indicates a greater force than gauge A. But here's the catch: We could just as well have started in B's rest frame and used the same line of argument to come to the conclusion that gauge A measures a bigger force.

Clearly those two conclusions contradict each other. So what's wrong?

Let's say two spaceships have such devices that deliver an impulse of magnitude X, mounted on their sidewalls. Let's call the devices "pushers".

These spaceships pass each other at great speed. When they are next to each other they engage the pushers.

Now both spaceships say that the other pusher pushes with a small force for a long time, making a shallow but long dent on the the spaceship wall. And the length contraction of the other spaceship is the reason why the dent on its wall spans a large part of the wall, and that dent is shallow because the contracted wall material is dense, because of the contraction.It seems quite easy to convert this scenario to a such one where there is a continuous force. I leave it for later time.

 

[Dale]

By Newton 3rd, this causes an equal force of 1 N directed to the moving scales B.

Newton's third law would be in terms of the four vectors, not the three vectors.

A functioning force gauge like what you have been describing (producing a single valued output) measures the norm of the four force, which is equal to the magnitude of the three force in the gauge's rest frame. Since the four forces are equal and opposite then their norms are the same and therefore the readings are the same.

The important thing to remember is that all measured values are scalars, so they are invariant. Often that scalar will have special meaning in the rest frame of the apparatus, and often the scalar will be a contraction with the four velocity of the apparatus, but it is always a scalar.

 

[Dale]

To know what B actually measures, we need to transform this force to B's rest system.

Expanding on what I mentioned above, you never need to transform to B's rest frame provided that you use the correct covariant law.

The standard force gauge I think measures the norm of the four force $\sqrt{F^{\mu}F_{\mu}}$. However, you could think of a four force device which measures not the norm of the four force but rather one of its components, for example say ${e_1}^{\mu} F_{\mu}$ where e1 is the first spacelike vector in a tetrad describing the frame of the gauge.

With that, you could arrange a situation where one device measures a value of 1 N and another measures a value of 2.3 N, but the devices in that situation would not be symmetrical. All frames would agree which device measures what number.

 

[Mark Harder]

I was taught that a force is the negative gradient of its potential. In the case of the spring, the harmonic potential (as a function of distance) is the applicable potential. I find this an attractive definition, since it forces you to specify what the potential in a system of interest is before you can find an expression for the force. If you want to quantitate f(x) according to Newton's #2, you need some measurements, like the position of the mass on a spring and Hooke's constant and the displacement (I think that covers it.) Thus, it appears to me to be a more phenomenological principle, which is another way of saying you enter the numbers on one side of Newton's #2 and "force" comes out the other end.
I know that variational (integral as distinguished from differential) principles generate a lot of QM, including field theories, and I vaguely recall that the definition of force in terms of potentials fits in nicely with physical variational principles. However, I don't know diddly-squat about SR and GR, so I can't shed any light on the Four-force. Nevertheless, beginning with the potential and, I believe, applying relativistic Lagrangian or Hamiltonian dynamics, you might be able to come up with a better "definition" of 4-force. "Better" in the sense that it's a more fertile approach to doing relativistic or quantum physics.

 

[Mr-R]

I was taught that a force is the negative gradient of its potential.

I think that this is only possible for a conservative force. Got nothing against the rest of your post :P

 

[vanhees71]

Concerning posting #38: Newton's 3rd law is equivalent to momentum conservation. Note that Noether's theorem is a far better approach to clarify the basic laws, particularly in the relatistic realm. This immediately tells you that Newton's 3rd law doesn't hold in the relativistic context as it holds in action-at-a-distance models as, e.g., Newton's Law of gravity. Rather you can also have fields like the electromagnetic fields that carry momentum.

Of course you can describe a harmonic oscillator. One realization, which is easy treatable fully relativistically is the somewhat academic situation of the interior of a homogeneously charged spherical ball at rest. In its rest frame you have a harmonic electrostatic force. Now you can think about, what you get in another inertial frame of reference, using the Lorentz transformations. This example is, as unrealistic it may be from a practical point of view, very instructive since you can fully understand, how the force transforms in this case from one reference frame to another since you get the full electromagnetic field in any frame from the exactly calculable electrostatic field in the restframe of the ball. So you can understand the transformation properties of the force under Lorentz transformations as well from the formal application of the Lorentz transform to the Minkwoski four-force as well as from the application of the non-covariant version in any given inertial frame by evaluating the electromagnetic field in this frame. Note that in any frame that's moving wrt. the restframe of the ball, in addition to the electrostatic you have also magnetostatic field components!

 

[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).

I'd like to emend this statement I made earlier.

The caveat I attached to the first part (that the magnitude of the proper force is invariant only when the mass remains constant) is inaccurate. I'm pretty sure that $| \mathbf{f}_0 | = m| \mathbf{a}_0 | = m\frac{\gamma^3 }{\gamma_{\bot}}|\mathbf a|$ holds even if the mass changes with time. You just have to "update" the equation with the new mass every moment, like you'd already do with the other factors on the RHS ($\gamma, \gamma_\bot, |\mathbf a|$).

The caveat is true for the second part of my post: the magnitude of the four-force equals the magnitude of the proper force only when the mass remains constant. If the mass is changing, then the magnitude of the four-force is:

$\sqrt{\left( \frac{1}{c} \frac{d E_0}{d \tau} \right)^2 - | \mathbf{f}_0 |^2}$

or equivalently

$\sqrt{\left( c \, \frac{dm}{d \tau} \right)^2 - | \mathbf{f}_0 |^2}$

($\tau$ is proper time). The first term $\frac{d E_0}{d \tau} = \gamma \frac{d E_0}{dt}$ could be called the "proper power," as it's the invariant rate of change of energy that a traveler "feels," or that an inertial observer in an instantaneously comoving frame would measure.