Feynman II, equation 26.11, off by a factor

[learn.steadfast]

I'm reviewing relativity and am working simple problems to test my understanding. I am reviewing Feynman's HTML book on physics. I don't get the same answer as him for the shape of an E field in a moving particle in one dimension, equation 26.11, but I do get his answer for equation 26.10; and I'm not sure what I'm doing wrong.

http://www.feynmanlectures.caltech.edu/II_26.html

How I worked the problem:
My understanding is that when an object moves at a uniform velocity in the x direction with observers seeing the object move at that velocity; The size of the object and it's fields are supposed to be contracted in the direction of motion; (Eg: according Einstein/Leopold/Infield in another book I read.) The contraction is ONLY in the direction of motion. However, one must also account for the total energy of the object in motion which is a scalar factor determined by the objects speed relative to the measuring frame of reference. So, I looked for all energies and masses that need to be scaled, and I found one reference to energy.

An Electric field is measured by checking the force on a test charge as it's moved in the field. Classical physics would show that force is proportional to the change in energy of the test charge with respect to it's position in the field. $F \propto { {d U} \over { d r }}$ Since the energy falls off as $1 \over r$ the force field must fall off as $1 \over r^2$

The rest frame equation for the magnitude of a vector electric field (force field) is just:
$|E| ={q \over { 4 \pi \epsilon_0 } } \cdot { 1 \over x^2 + y^2 + z^2 }$ . because . $r^2 = x^2 + y^2 + z^2$

Modifying the rest frame equation by accounting for energy scaling and Lorentz transformations ought to allow me to write down down (by inspection) an equation for a moving charge's E field as seen by an outside observer.

In Feynman's example, the point charge is uniformly moving in the x direction with velocity v (as a fraction of the speed of light, v=1 is the same as v=c).
From what I learned from Einstein's book, I think the magnitude of the E field of the moving charge should be the rest frame equation scaled by the motion energy to rest energy factor: $1 \over \sqrt { 1-v^2 }$ . The only other change needed is the Lorentz transformation for x. $x' = { { x - v t } \over \sqrt ( 1-v^2 ) }$

Therefore, a moving charge's E-field as measured by a rest frame observer ought to be:
$|E| ={{ q \cdot { 1 \over \sqrt {1-v^2} }} \over { 4 \pi \epsilon_0 } } \cdot { 1 \over ({ x-v t \over \sqrt {1 - v^2}})^2 + y^2 + z^2 }$

When measuring along axii having origin on the moving charge, the direction of the vector E is purely pointing along the axii. (see Feynman's drawings).

Therefore, I should be able to get Feynman's equation 26.10 by setting x = v t in my equation.
$E ={{ q \cdot { 1 \over \sqrt {1-v^2} }} \over { 4 \pi \epsilon_0 } } \cdot { 1 \over { y^2 + z^2 }}$
Which is clearly the same as Feynman's result.

However, if I set y and z to zero, to try and get equation 26.11, I get:
$E ={{ q \cdot { 1 \over \sqrt {1-v^2} }} \over { 4 \pi \epsilon_0 } } \cdot { 1 \over ({ x-v t \over \sqrt {1 - v^2}})^2 } ={{ q \cdot { 1 \over \sqrt {1-v^2} }} \over { 4 \pi \epsilon_0 } } \cdot { \sqrt {1 - v^2} ^2 \over ({ x-v t })^2 } = { q \over { 4 \pi \epsilon_0 } } \cdot { \sqrt {1 - v^2} \over ({ x-v t })^2 }$

And that looks different than this screen-shot, February 2018, of the Feynman lectures.

Is there some reason that the Lorentz transformation cancels out the total energy multiplication factor? Or is this just a typo in the online transcription of Feynman's book?

http://www.feynmanlectures.caltech.edu/II_26.html

 

[Andrew Mason]

I'm reviewing relativity and am working simple problems to test my understanding. I am reviewing Feynman's HTML book on physics. I don't get the same answer as him for the shape of an E field in a moving particle in one dimension, equation 26.11, but I do get his answer for equation 26.10; and I'm not sure what I'm doing wrong.

http://www.feynmanlectures.caltech.edu/II_26.html

...

Is there some reason that the Lorentz transformation cancels out the total energy multiplication factor? Or is this just a typo in the online transcription of Feynman's book?

http://www.feynmanlectures.caltech.edu/II_26.html

Equation 26.11 appears to me to be correctly derived from Feynman's equation 26.6.

Letting $k = {1 \over { 4 \pi \epsilon_0 }}$ and $\gamma = {1 \over \sqrt {1-v^2}}$ for simplicity, Feynman's equation 26.6 (where y = z = 0) can be written:

$E_x = {{kq \gamma (x-vt)} \over {[(\gamma^2 (x-vt)^2]^{3/2}}}$

which reduces to:

$E_x = {{kq \gamma (x-vt)} \over {[(\gamma^3 (x-vt)^3]}} = {kq \over {\gamma^2(x-vt)^2}}$

Substituting ${1 \over {1-v^2}}$ for $\gamma^2$ this becomes:

$E_x = {kq(1-v^2) \over {(x-vt)^2}}$ which is Feynman equation 26.11

As Feynman points out in his explanation below 26.11, the relativistic effect (speed approaching c - ie. v approaching 1) is a reduction in the magnitude of the electric field in the x direction.

AM

 

[learn.steadfast]

Equation 26.11 appears to me to be correctly derived from Feynman's equation 26.6.

Thank you for checking it. Your adding in gamma and k were very helpful in me seeing the equations are the same. The issue is definitely not a typo, then. The issue has to be in my thinking, or in my math.

As Feynman points out in his explanation below 26.11, the relativistic effect (speed approaching c - ie. v approaching 1) is a reduction in the magnitude of the electric field in the x direction.

That's true, but look at my equation ... it also has $\sqrt { 1 - v^2 }$ in the numerator. So, I'm going to guess that you reworked my math and detected a mistake ... and that when I correct my math, the magnitude will not decrease... ?

Any math mistake pretty much has to be that the Lorentz transform in x changes the x derivative by the chain rule.

The rest frame potential of a test charge should be:
$U = { {k qQ } \over r }$ Where Q is the test charge, which I'll set to 1 for simplicity.
Therefore I need to modify the charge's position with the Lorentz transform $\gamma(x - v t)$ , and also multiply U field by another gamma factor because U is energy.

$U_v = { { kq \gamma } \over \sqrt { \gamma^2 (x-v t)^2 + y^2 + z^2 } }$
Feynman calls this phi, in equation 26.1
Now if I set y and z to zero, and take the spatial derivative in x, I get:
$-E_x = {dU_v \over dx } = k q \gamma \cdot { \partial \over \partial x } ( \gamma^{-2} ( x-v t )^{-2})^{1/2} = k q { \gamma \over \gamma } \cdot { \partial \over \partial x } ( x-v t )^{-1} = { - k q \over { (x-v t)^{2}} }$

Well, my original math was wrong; Putting the Lorentz transform into the energy field equation is not quite the same as taking a Lorentz transformed potential's derivative.
However, this outcome is just as bad in the other direction. Now the field does NOT decrease like it ought to. Feynman's equation 26.4 isn't in simplified form... I'm still working on that. (Edit: I just verified that gamma in 26.4 also totally cancels out) I don't see any more mistakes in my math, so obviously I missed a concept. I think it's Maxwell's equation which says a changing magnetic field creates an electric field.

One last question, to make sure I'm not going in logical circles:

In Feynman's example, I could add a test charge Q at x=1,y=0, z=0,t=0 ( Idon't know or care about earlier times, but it's not moving at t=0). If I give you that test charge (Q=1 Coulomb) and Feynman's description of a charge q @ x=v*t,y=0,z=0; then would you understand Feynman's derivation of equation 26.11 to imply that the total force field experienced by that Q charge is calculated by 26.11 with x=1,t=0,y=0,z=0, alone? eg: Feynman is not expecting us to add additional electric field to 26.11 to account for a changing magnetic field affecting the test charge? (That's already handled by 26.5, I'm guessing, which is added to 26.4 to make 26.6 and finally 26.11 ?)

 

[Andrew Mason]

From what I learned from Einstein's book, I think the magnitude of the E field of the moving charge should be the rest frame equation scaled by the motion energy to rest energy factor: $1 \over \sqrt { 1-v^2 }$ . The only other change needed is the Lorentz transformation for x. $x' = { { x - v t } \over \sqrt ( 1-v^2 ) }$

This is where you may be going wrong conceptually. The potential of the moving charge increases by the $\gamma$ factor, not the field. The field is given by Feynman:

1: $\vec{E} = -∇ \phi - \frac{\partial \vec{A}}{\partial dt}$

Therefore, a moving charge's E-field as measured by a rest frame observer ought to be:
$|E| ={{ q \cdot { 1 \over \sqrt {1-v^2} }} \over { 4 \pi \epsilon_0 } } \cdot { 1 \over ({ x-v t \over \sqrt {1 - v^2}})^2 + y^2 + z^2 }$

The correct equation is Feynman's 26.6. , which follows by adding 26.4 and 26.5

Well, my original math was wrong; Putting the Lorentz transform into the energy field equation is not quite the same as taking a Lorentz transformed potential's derivative.

Yes. That is the problem.

One last question, to make sure I'm not going in logical circles:

In Feynman's example, I could add a test charge Q at x=1,y=0, z=0,t=0 ( Idon't know or care about earlier times, but it's not moving at t=0). If I give you that test charge (Q=1 Coulomb) and Feynman's description of a charge q @ x=v*t,y=0,z=0; then would you understand Feynman's derivation of equation 26.11 to imply that the total force field experienced by that Q charge is calculated by 26.11 with x=1,t=0,y=0,z=0, alone? eg: Feynman is not expecting us to add additional electric field to 26.11 to account for a changing magnetic field affecting the test charge? (That's already handled by 26.5, I'm guessing, which is added to 26.4 to make 26.6 and finally 26.11 ?)

Yes, 26.6 is the sum of 26.4 and 26.5. This follows from Eq. 1 above. The potential $\vec{A_\mu}$ is a four-vector. The spatial derivative necessarily has a time component as we are dealing with space-time.

AM

 

[learn.steadfast]

This is where you may be going wrong conceptually. The potential of the moving charge increases by the γ\gamma factor, not the field. The field is given by Feynman:

Thank you very much for all your help. I am now able to get the right answer...

My ultimate goal is to write a computer simulator of three arbitrarily charged particles in a single frame of reference. The simulator will be using time-steps, and discreet samples of the potential energy fields. Eg: The potential function is calculated around a particle and then propagated at a simulated speed of "c" numerically diffusing in space appropriately. Each particle only knows the potential energy field in a very small volume around it, and nothing about the velocities of the other particles. I plan to reformulate the equations of motion based on how charge's fields appear, and find ways to approximate/extract the velocity terms in Feynman's equations strictly from the local field around each particle.

Your comment makes me pause. I thought I was using the term "field" like other people, https://arxiv.org/abs/astro-ph/9401006, where "field" just means a function of many variables whether or not it's a vector.

I'd like to explore the concept you're clarifyiung a little more ... I've got a BSEE background, so I'm used to thinking of voltages as "potential differences" but force fields (E fields) are seldom used by me. I'm thinking that at least the scalar potential field $\phi(x,y,z)$ is a representation of distributed energy over space and not just in the particle. I'm curious as to why you speak of $\gamma$ applying conceptually to the moving particle but not the field.

Here's a Gedanken experiment to illustrate my puzzlement:
A physics teacher typically introduces potential energy problems in this way: Imagine an experiment with only two electrons q and Q.

q is fixed at the origin, and Q is infinitely far away. To compute the "potential" of the q in Q's field we move Q to some arbitrary point x,y,z; and add up (integrate) all the force * distance = work, required to move Q to x,y,z with q at 0,0,0. Once done, we no longer need to talk about force but only "potential" energy stored up; potential is a scalar field... etc.

Since this is electrostatics, there is a tacit implication that Q is NO longer moving once it arrives at x,y,z. A teacher will also show mathematically that the total work done is independent of the path taken to get there. (Conservative fields). Only the initial and final positions matter.

There's two things I notice ... absolutely no work is done on the fixed charge q because it never moved; Therefore all the work had to be done on Q. While Q was in motion, Q's total energy would increase. But once Q stops, the mass will again be it's rest mass value. Somehow the masses of Q and q are the same rest value at the start and end of the experiment, but energy was added to the system and is not accounted for in the masses.

Relativity says that adding energy increases mass. Since the path to get to the end point for Q and q do not matter, Q being fixed is irrelevant. All we need to know is that q and Q are stopped 1 meter apart exactly t=0 with nothing holding either of them. eg: They just happen to come to a stop in motion when one meter apart at t=0, and either are, or become free to move for all t>=0;

It's obvious the two charges have only their rest masses at t=0, but wijll accelerate due to the "field?" and will increase in mass/energy and gamma for t>=0.

So, I think the energy has to be conceptually stored in the fields, somehow. What other concept is associated with Q and q at t=0, when they are at "rest", that would always supply the energy needed to accelerate the charges and increase their mass?

Gamma does affect the formulas for both $\phi (x,y,z,t) , \vec{ E }(x,y,z,t)$, The only difference I see is that the force field $\vec{ E }$ is not an energy and so doesn't scale with gamma, directly. Gamma only shows up in force fields because a force field is a derivative of a potential field.

 

[Andrew Mason]

Your comment makes me pause. I thought I was using the term "field" like other people, https://arxiv.org/abs/astro-ph/9401006, where "field" just means a function of many variables whether or not it's a vector.

You were using it in the sense of the electric field, E(x,y,z,t), which is a vector quantity representing the force per unit charge that a charged particle will experience at a particular position and time. That is how Feynman was using it. That is how I was using it.

I'm curious as to why you speak of $\gamma$ applying conceptually to the moving particle but not the field.

I was just saying that one first multiplies the potential of a moving charge by $\gamma$ because it represents energy (per unit charge). The field is determined by differentiating the potential. That is not the same as differentiating the potential and then multiplying the result by $\gamma$. From your last comment, you appear to understand that:

Gamma does affect the formulas for both $\phi (x,y,z,t) , \vec{ E }(x,y,z,t)$, The only difference I see is that the force field $\vec{ E }$ is not an energy and so doesn't scale with gamma, directly. Gamma only shows up in force fields because a force field is a derivative of a potential field.

AM

 

[learn.steadfast]

From your last comment, you appear to understand that:

Yes. That's why in the original post I said difference of U (potential energy) with respect to radius but didn't mention time.

I thought that the force field of the moving charge could be integrated over space to produce a scalar potential field; but now I realize such a field is not the same as phi which Feynman talks about.

Is there a (already known) general solution for what the total scalar potential of a moving charge is? eg: as a algebraic function of (x,y,z,t) and not as an integral? (Eg: not the Lienard Wieichart approach, but it's actual solution?)

Feynman doesn't show a quasi/ or total potential that would give the same E field as he gets. Feynman only shows a potential field (phi) and a separate vector potential that disturbs phi. Feynman's approach is standard, but that approach makes phi an incomplete description of potential energy (which is purely a scalar).

I realize that many people on physics forums believe that no such (conservative?) scalar potential exists and I am a bit puzzled as to exactly why. However; I've done academic searches for a few few days, and there are scalar potential fields for moving charges.

Mathematically, two scalars were demonstrated mathematically (not physically) by:
"E.T. Whittaker, Vol I. 1904, p. 367-372. Proceedings of the London Mathematical Society."
http://www.cheniere.org/misc/Whittak/whit1904.pdf

I think Whittaker was working directly from Maxwell's equations. I am not sure how much knew about Einstein/Heisenburg as this is 1904, and Einstein published 1905. However, Maxwell's equations were already relativisticly correct. So Whittaker's solution already shows one way two scalar potentials can represent a moving particle's energy. The solution isn't in a form I have been able to use, but at least it exists.The second article I came across, when studying Whittaker, shows (indirectly) that the potential energy of integrated fields must have a physical meaning. eg: The E field integration would not end up with just an "apparent" effect due to information traveling at the speed of light.

The author of this second work used two detectors to detect a single photon of gamma ray light, simultaneously. That appears to violate quantum physics ideas, but not Maxwell's equation. The author seems to show that the photon initially had to be quantised, but was no longer quantised by the time it reached the detectors. (Think about Young's double slit experiment and photoelectric effect based photon detectors. I think this experiment puts both of them partially in doubt.) In any event, the experiment shows that the particle's energy really is spread out in space like Maxwell's equation predicts for wave propagation; detectable by two distinct detectors.

http://www.ptep-online.com/2014/PP-37-06.PDF

For a moving charge; I can guess an extremely close approximate potential function. Gradients of my potential function provide the E fields for up to 10% of the speed of light. There is less than 1% error over all space and zero error on the x,y,z axii (as defined by Feynman.) However my approximation degrades horribly by the time 0.6c is reached. I'll post plots later, as I am still working on them. I hope to improve the approximation in the coming days, but if someone KNOWS an analytical solution, I'm already all ears.

 

[vanhees71]

A somewhat less complicated derivation for the field of a uniformly moving point charge is as follows.

In (1+3)-noncovariant notation the four-potential is given by
$$\Phi(t,\vec{x})=\frac{q}{4 \pi |\vec{x}|}, \quad \vec{A}=0.$$
The trick is now to formulate this with physical quantities manifestly covariant. The only physical quantity relevant here is the four-velocity of the charge,
$$(u^{\mu})=\frac{1}{\sqrt{1-\beta^2}} (1,\vec{\beta}). \qquad(*)$$
For the particle at rest $\vec{\beta}=0$ and thus $(u^{\mu})=(1,0,0,0)$. So we can write the four-potential in a manifestly covariant way as
$$(A^{\mu})=\frac{q}{4 \pi \sqrt{(u \cdot x)^2-x\cdot x}} u_{\mu}.$$
Since this is a manifestly covariant vector field, the expression holds true in any frame, and for the moving charge, you just have to put (*) for the four-velocity. Now you can get with some algebraic efford
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} A^0.$$
After this effort, you get the magnetic-field components for free
$$\vec{B}=\vec{\nabla} \times \vec{A}=\vec{\beta} \times \vec{E}.$$

 

[learn.steadfast]

A somewhat less complicated derivation for the field of a uniformly moving point charge is as follows.

It's definitely a more compact derivation; but I'm not sure it's really less complicated.
My experience with four vectors is minimal. Einstein only introduces the idea of a four vector in his popular books, but doesn't develop them in the editions I read.

I do understand that Maxwell's equations can be reformulated into a "wave" form by simultaneous solution and substitution of variables; The wave form shows energy moving with the speed of light from any "event." Maxwell's equation in wave form corresponds to a four vector because one of the four vector "lengths" has been made "ct" the wave propagation distance. This "ct" distance affects all kinds of measurements of length, simultaneity, etc. both apparent and real.

In BSEE course work, often the magnetic field can be taken for granted in certain problems because a propagating (eg: not standing wave's) H and E field strengths are related by c. The information in the two fields is redundant.

Propagating waves are |H/c| = |E|. If you know either field and a direction away from the source, you automatically know the other field magnitude and direction. In the case of a uniformly moving point charge (Feynman), all force field vectors are moving away from a point charge; so they should automatically be related by 'c'. The math you provide appears to be consistent with that idea. That's pretty simple.

But when I look at your notation ... all I am seeing is that you are replacing the three separate components of the gradient (E sub x,y,z) with a single vector and operator. I think it's nice that "ct" is inside the four vector, but I still don't see how to get a numeric value for total potential energy when a static test charge is placed somewhere near the moving charge.

eg: I see two separate operators -- a derivative with respect to time -- and a gradient -- that show up in your expression for E. So, A can't be a total potential scalar by itself.
Therefore A might be a more compact vector notation? eg: It's not a scalar description of potential energy of anything; can you even compute a scalar value from it in a simple way?

I'm not sure how the four potential idea fits in; but it seems to have the same problems as Feynman's derivation from a vector potential. A vector potential does not describe the total energy needed to move a charge to a location in a force field (with the charge not being in motion once it gets to the test point.)

Consider an electric field:

In principle, given an electric field "E", a person can start at "infinity" and add up all the work required to move a charge against the field to some arbitrary location x,y,z,t. (Math is a model and not physical motion, so there is no "time" needed to add stuff up.) The answer is a simple number from a line integral $\int \vec F \cdot dx$ which shows how much energy the particle could in principle absorb from the field when being "accelerated" back out toward infinity (and beyond... couldn't resist.!)

But, I don't see a way to get a number by using relativity and changing a frame of reference to make the energy value (a scalar) pop out. I only see a way to get the force fields; which means I have to take integrals to compute a potentia. Time is not a factor. No one appears to have worked the answers to this basic integral even though Feynman computes the exact field necessary for it. We simply don't know a scalar "potential" equivalent to the force field that Feynman gives, or an easy way to compute a potential from it.

I know from trying to guess a potential energy function that would give the right E field, that the potential is roughly ellipsoidal. All I know how to do is take a numerical gradient of a guess function and compare that to the E fields of Feynman. Using errors, I can attempt to improve my guess by inspection. I simply don't know how to solve the calculus, but I'm working on it. The math is still ... complicated.

I've plotted my work so far as a clear example. I'm plotting equal force contours instead of force lines (gnuplot's smart enough to do it, but not force lines.) Since force magnitude is supposed to drop off as one over square radius in any direction, I am plotting against a normalized $1/r^2$ with equal changes in r. That makes the spacing of the |E| contours will uniform on each plotted axii but to different scales. The elliptical nature of the contours is easy to spot. They don't cross x and y at the same radii. There is hardly any difference between Feynman's exact |E| surfaces (curved green lines) compared to my brown potential's gradient contours even at v=0.3c.

Note: The straight green line is separate asymptote. That's where Feynman's green |E| plot intersects with a reference non-moving particle's |E| field (blue circles). So the green line represents a place where the moving charge's gradient has the same magnitude as a non moving charge with same origin. The fact that such an intersection exists and is never at the halfway point between x and y-axis is very interesting. $m = { \Delta y \over \Delta x } = +-\sqrt { {1-(1-v^2)^{2/3}} \over {(1-v^2)^{1/3} -1 } }$

On all axii, my approximation exactly converges to Feynman's solution for all values of v. But, for really fast particles, say v=0.6c, there is significant error when not near the x,y,or z axis. See second plot.

 

[learn.steadfast]

I've been looking at other threads, especially trying to understand "transverse" mass, vs. longitudinal mass. These are concepts I am familiar with as "effective masses" in semiconductors, but not as relativistic distinctions based on direction.

At the same time, I came across a post you made where you call a four-vector a "scalar".

For mass, of course SR is sufficient, and it's "defined" very clearly as a Casimir operator of the Poincare group. With the total four-momentum of your system, it's given by

m2=1c2PμPμ,​

m^2 = \frac{1}{c^2} P_{\mu} P^{\mu},
and since (Pμ)(P^{\mu}) is a four-vector it is a scalar. In the center-momentum frame, where ⃗P=0\vec{P}=0 you have, because of P0=P0=ECM/cP^0=P_0=E_{\text{CM}}/c, you get the famous formula

ECM=mc2,​

Do you have a link to a simple explanation about scalar nature of four vectors?
I think I must be missing something.

 

[PeterDonis]

I came across a post you made where you call a four-vector a "scalar".

He didn't say a four-vector was a scalar. He said the length (more precisely the squared length) of a four-vector was a scalar. If $P^\mu$ is a four-vector, then $P^\mu P_\mu$, which is the squared length of the four-vector, is a scalar. If $P^\mu$ is the 4-momentum, we call its squared length $m^2$, and its square root $m$ is called the invariant mass.

 

[learn.steadfast]

Error correction: I got a sign flipped in the power of the computation of the slope.
It should be:
$m= \sqrt { { 1-(1-v^2)^{-2/3}} \over {1-v^2}^{1/3} }$

Also, since y and z axii are equivalent; this asymptote rotates in three dimensional space to create a "cone" that opens in x.
So, there are three volumetric regions in a unit sphere divided up by this cone. There are two spherical caps, and a band down the center (like a striped billiards/pool ball where r is defined by the slope from origin along either y or z).

Note, $\lim_{v \to 0 } m = \sqrt {2}$ eg: it's not 1.
So, for very tiny velocities mm/year ... the asymptote and cone is still not symeterically dividing up space. The volume of the two spherical caps and cone is not 50% of the total spherical volume around the charged particle. Nor is the surface area of the two caps 50% of the total surface area of the spreading force field's spherical surface.

Light energy spreads out with square law (surface area) diminishing. Light is similar to a certain amount of 'substance' stretching out on the surface of a spherical balloon as it expands in space. I supposed, originally, that the potential energy generating a force field ought to behave approximately the same way.

However, the force field (and hence potential energy) even for extremely slowly moving objects does not spread out equally in the direction of motion vs. perpendicular to that direction. There is a strong asymmetry in how it spreads out no matter how small the motion. This was very surprising to me.

 

[learn.steadfast]

He said the length (more precisely the squared length) of a four-vector was a scalar.

I see. That makes sense. Is there a simple way to calculate a length of a four potential, then, which corresponds to the potential energy of a non-moving charge in it's rest frame? Not the potential phi, or A, which Feynman starts from; but the actual scalar value of the line integral of force from infinity to a point x,y,z near to a uniformly moving particle at say, t=0 ?

Or does the line integral still have to be explicitly computed even with four vectors ?

 

[PeterDonis]

does the line integral still have to be explicitly computed even with four vectors ?

Yes. Four-vectors, scalars, etc. are attached to particular points in spacetime, so if you need a value of something over an extended region, you still have to compute an integral involving the values of those quantities over the entire region. It might be that a particular scalar, such as the invariant mass of a particle, does not change over the entire region of spacetime, but that won't always be true; you have to look at the specific case. (There are even cases where the invariant mass of a particle can change, for example if it emits radiation.)