KJW

You are talking about the theoretical/abstract model. I am asking about how the mass actually does that. There is nothing "abstract" about spacetime geometry. Gravitational time dilation is physically real as is the curvature it implies. It seems to me that many people believe that general relativity is just a theoretical model that describes physics in only an abstract manner typical of theoretical models. But no, geometry is very much a physically concrete notion and the formulae from Ricci calculus very much describes this physically concrete notion. The difficulty arises because the spacetime curvature that describes human-scale physics is quite miniscule, so it becomes very difficult to observe and measure the physically real curvature of spacetime. There are two possibilities of how mass causes spacetime curvature: 1, mass doesn't cause spacetime curvature, but that mass is how we physically interpret the spacetime curvature that exists fundamentally; 2, mass causes spacetime curvature through a mechanism associated with something like quantum physics, the fundamental forces of nature, or perhaps something entirely different.

Another way to say this is that, in this kind of spacetime and under geodesic motion, shapes (ie angles) are preserved, whereas volumes and surfaces are not. You and I discussed this in the thread about the cosmological expansion of time as well as space. By coordinate-transforming to the manifestly conformally flat metric, it becomes easier to derive cosmological redshifts due to the simplified geodesic equation for light as a result of the preservation of angles (speeds in spacetime). This is true for homogeneously and isotropically distributed sources. What about other "smeared" sources? The Weyl tensor is not necessarily zero, but the curvature is not away from the source. A non-zero Weyl tensor can coexist at the same location as a non-zero Einstein tensor. But the energy-momentum tensor corresponds only to the Einstein tensor. I felt the need to distinguish between the curvature that is generated by an energy-momentum source and propagated to remote locations, and the curvature that corresponds to the energy-momentum itself. It's worth mentioning that inside a spherically symmetric shell of matter, the spacetime is flat. That is, while gravitational curvature propagates from the shell of matter to outside of the shell, it does not propagate from the shell of matter to inside of the shell.

You might be wondering where the formula [math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] comes from. I've already provided the clue that it comes from the Killing equation. Here's the derivation: Let [math]T u^\alpha = T \dfrac{dx^\alpha}{ds}[/math] be a timelike Killing vector field, where [math]u^\alpha = \dfrac{dx^\alpha}{ds}[/math] is a unit tangent vector field of timelike trajectories of stationary observers or other objects in stationary spacetime. Note that [math]g_{\alpha\beta}\ u^\alpha u^\beta = 1[/math] The Killing equation: [math]\nabla_\mu (Tu_\nu) + \nabla_\nu (Tu_\mu) = 0 \ \ \ \ \ \ \ \ \ \ \text{(}\nabla_\mu \text{ is the covariant differential operator)}[/math] [math]\nabla_\mu T\ u_\nu + T\ \nabla_\mu u_\nu + \nabla_\nu T\ u_\mu + T\ \nabla_\nu u_\mu = 0[/math] [math]u^\nu (\nabla_\mu T\ u_\nu + T\ \nabla_\mu u_\nu + \nabla_\nu T\ u_\mu + T\ \nabla_\nu u_\mu) = 0[/math] [math]\nabla_\mu T\ u^\nu u_\nu + T\ u^\nu \nabla_\mu u_\nu + u^\nu \nabla_\nu T\ u_\mu + T\ u^\nu \nabla_\nu u_\mu = 0[/math] Considering the individual terms: [math]\nabla_\mu T\ u^\nu u_\nu = \nabla_\mu T = \dfrac{\partial T}{\partial x^\mu}[/math] [math]T\ u^\nu \nabla_\mu u_\nu = T\ g_{\nu\sigma} u^\nu \nabla_\mu u^\sigma = T\ \nabla_\mu (g_{\nu\sigma}\ u^\nu u^\sigma) - T\ g_{\nu\sigma} u^\sigma \nabla_\mu u^\nu = -T\ g_{\nu\sigma} u^\nu \nabla_\mu u^\sigma = 0[/math] [math]u^\nu \nabla_\nu T\ u_\mu = \dfrac{\partial T}{\partial s} u_\mu = 0 \ \ \ \ \ \left(\dfrac{\partial T}{\partial s} = 0 \ \ \text{ by definition} \right)[/math] [math]T\ u^\nu \nabla_\nu u_\mu = T\ b_\mu[/math] Therefore: [math]\dfrac{\partial T}{\partial x^\mu} + T\ b_\mu = 0[/math] [math]b_\mu = -\dfrac{1}{T} \dfrac{\partial T}{\partial x^\mu}[/math] (Note that [math]a^\mu = c^2 g^{\mu\nu}\ b_\nu[/math]) [math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math]

In another thread, I showed that the flat-space Friedmann–Lemaître–Robertson–Walker (FLRW) metric can be coordinate-transformed to a manifestly conformally flat metric. This means that the Weyl conformal tensor is zero, and therefore according to my way of looking at this, there is no gravity (although there is still the curvature associated the Ricci tensor and scalar).

One thing I should mention: I distinguish between the curvature directly associated with the energy-momentum and the curvature that is away from its source energy-momentum, and tend not to use the term "gravity" to describe the curvature directly associated with energy-momentum. The curvature associated with gravity is called the Weyl conformal tensor. It has the same algebraic structure as the Riemann tensor, but its contraction is zero (—> zero Einstein tensor). In the [math]{C_{\alpha\beta\gamma}}^\delta[/math] form, it is invariant to conformal transformations: [math]g_{\mu\nu} \longrightarrow \varphi g_{\mu\nu}[/math] for arbitrary scalar function [math]\varphi[/math]

Yes, [math]T[/math] and [math]\dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] depend on [math]M[/math] and [math]r[/math]. I would prefer to say that the curvature associated with mass causes gravity, although the mass is remote from the gravity, unlike gravitational time dilation which is local to the gravity. Whether or not mass causes the curvature associated with it, or is the curvature associated with it is something I can't answer. @Genady, [math]g^{rr} = \dfrac{1}{g_{rr}}[/math]

[math]T[/math] is the relative magnitude of a timelike Killing vector at a given location in three-dimensional space. In terms of the metric, it is [math]\sqrt{g_{tt}}[/math], but Killing vectors are covariant, and the magnitude of a Killing vector is invariant once its magnitude at some location is defined (if [math]K^\mu[/math] is a Killing vector field, [math]\lambda K^\mu[/math] is also a Killing vector field for arbitrary constant scalar [math]\lambda[/math]).

Do you need to consider the stereochemistry of the hydrogenations?

[math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] For a stationary object in the Schwarzschild metric: [math]ds^2 = \left(1 - \dfrac{2GM}{c^2r}\right) c^2 (dt)^2 - \left(1 - \dfrac{2GM}{c^2r}\right)^{-1} (dr)^2 - r^2 \Bigl((d\theta)^2 + sin^2\theta (d\phi)^2\Bigr)[/math] [math]g_{tt} = \left(1 - \dfrac{2GM}{c^2r}\right) c^2[/math] [math]g_{rr} = -\left(1 - \dfrac{2GM}{c^2r}\right)^{-1}[/math] [math]a^r = -c^2 g^{rr} \dfrac{1}{\sqrt{g_{tt}}} \dfrac{d\sqrt{g_{tt}}}{dr}[/math] [math]a^r = -\dfrac{c^2}{2} \dfrac{1}{g_{rr}\ g_{tt}} \dfrac{dg_{tt}}{dr}[/math] [math]a^r = \dfrac{1}{2}\dfrac{dg_{tt}}{dr}[/math] [math]a^r = \dfrac{1}{2}\dfrac{d\left(c^2 - \dfrac{2GM}{r}\right)}{dr}[/math] [math]a^r = \dfrac{GM}{r^2}[/math] One might be surprised to see that the acceleration of a stationary object in the Schwarzschild metric is exactly Newtonian. But this is due to way the [math]r[/math]-coordinate is defined. The [math]r[/math]-coordinate is defined such that the surface area of a sphere of radius [math]r[/math] is [math]4 \pi r^2[/math]. Thus, for the acceleration field to be a conserved field, the surface integral of this field over the sphere needs to be constant with respect to the [math]r[/math]-coordinate, requiring that the acceleration field obey an [math]r^{-2}[/math] law exactly.

One thing I find intriguing is that by an oxidative process, ammonia can be converted to the strongest known reducing agent (azide ion, N3–). Consider boron trifluoride, BF3: A boron atom has three electrons in its outer shell. The three fluorine atoms each provide one electron, giving the boron atom of boron trifluoride six electrons in total in its outer shell. But boron atoms want to have eight electrons in their outer shell to complete the octet. Thus, boron trifluoride will react with a fluoride ion, F–, which supplies a lone-pair of electrons, to produce the tetrafluoroborate anion, BF4–, which has eight electrons in the outer shell of the boron atom, completing the octet.

I prefer to think of it this way: Oxidation is the loss of electrons. Reduction is the gain of electrons. Acids accept lone-pairs of electrons. Bases donate lone-pairs of electrons. It would seem that there is a correlation between being an oxidising agent and being an acid, and a correlation between being a reducing agent and being a base. However, they are distinct notions. The hydride anion, H–, is an example of both a base and a reducing agent. Indeed, the reaction: H– + H+ —> H2 is both an acid-base reaction and a redox reaction. By contrast, diborane, B2H6, is a Lewis acid that is a reducing agent. Whereas the hydroxide ion, OH–, is not really much different from any other base, the hydrogen ion H+, is singularly unique as a Lewis acid in having no electrons at all. As a proton, a subatomic particle, its charge density is so great that, within the context of chemistry, it will never be free. It will always be bound to some electron density, regardless of how reluctant that electron density is to bond to anything. Therefore, a Brønsted acid is conceptually distinct from a Lewis acid in that the hydrogen ion is always bound to a base, and the measure of the Brønsted acidity is really a measure of the basicity of that base (the weaker the base, the stronger the Brønsted acid).

Why don't you try an infinitesimal translation to find out: [math]x^\mu \longrightarrow x^\mu + \delta x^\mu[/math] Note that: [math]f \longrightarrow f + \partial_\mu f\ \delta x^\mu[/math]

Thank you! You have created another analogy for how gravitational time dilation causes gravity in the "What is gravity?" thread. In that thread, I created the analogy of a slinky as well providing a formula that relates time dilation to acceleration. This formula is based on a geometric principle that I also applied to a disc. It also applies to your problem. Let [math]D_L[/math] be the large diameter, [math]D_S[/math] be the small diameter, and [math]X[/math] be the length of the axle between them. Then the radius of the curvature of the roll of the large diameter, [math]R_L[/math] is given by: [math]\dfrac{1}{R_L} = \dfrac{1}{D_L} \dfrac{D_L - D_S}{X}[/math]

Natural? That’s rather subjective. Not really. Suppose you have a hole in the bottom of a bucket of water. Water is leaking out of the bottom of the bucket. Surely it is natural to say that the hole is causing the bucket to leak rather than the leak is causing the hole in the bucket. Why does it have to? In the first case you’re explaining what causes the redshift. Nothing else. Gravitational redshift and gravity are concomitant. The question is what is causing what? Is there a principle to decide this? True gravity? The equivalence principle says you can’t distinguish it from other acceleration. The equivalence principle only applies locally. The tidal effect distinguishes true gravity from an artificial gravity. However, given that we are discussing the familiar gravity that we all experience on earth, which is true gravity, I felt it was necessary to point out that it isn't only true gravity that is caused by time dilation. In fact, the formula I gave expresses acceleration (not gravity) in terms of time dilation. But one can make that association. And that’s also an equality one can write down. That's a different question. If one wants to know why there is time dilation surrounding a massive object such as the earth, then one probably needs to understand the Einstein equation. But if given that there is time dilation surrounding a massive object such as the earth, then the above explains why that time dilation produces gravity. I hardly think time dilation can be considered a phenomenon we all experience, considering the sophistication of the equipment necessary to detect it. Gravity is the phenomenon we all experience, time dilation is the property of the surrounding spacetime that causes us to experience gravity, and the slinky analogy is the geometric principle that allows ordinary folk to understand it. The original suggestion was that time causes gravity, and that this was a consensus. What textbooks teach this, as apposed to energy-momentum and curvature? What’s the breakdown in the literature? Not liking an explanation is not really a consideration. There are different aspects to the question of what causes gravity. I chose to explain why time dilation causes objects to fall to the ground. After all, why should an object be attracted to a location with slower time (or repelled from a location with faster time)? Apart from what I said above, does this seem natural? Clocks don't "know" about time dilation. They only measure time at the same intrinsic rate regardless of their location or motion. What determines that a time dilation is gravitational (or accelerational) is that it depends on position. By contrast, the time dilation usually considered in SR depends on speed.

Yes, it is, in this sense: "This is extremely general. In any kind of gravitational field, as long as it is more or less constant with time, and not doing anything too radically relativistic, the coefficient in front of dt2 in the metric is always one plus twice the gravitational potential." Susskind, Cabannes. General Relativity: The Theoretical Minimum (p. 155). The other aspect of what I wrote is the analogy that enables someone to appreciate the esoteric nature of gravity in terms of the geometry of something as familiar as a slinky or a disc. In fact, for a disc, one has: [math]b_r = -\dfrac{1}{C} \dfrac{dC}{dr} = -\dfrac{1}{2 \pi r} \dfrac{d(2 \pi r)}{dr} = -\dfrac{1}{r}[/math] where: [math]b_r[/math] is the radially directed curvature of the circumference of a disc at radius [math]r[/math] [math]C[/math] is the circumference of the disc at radius [math]r[/math] [math]r[/math] is the radius [math]C[/math] increases with [math]r[/math] [math]\Rightarrow[/math] [math]b_r[/math] is directed inward towards the centre.

That's funny, like the name of this chemical compound:

The gravitation time dilation and the curvature field of a Killing vector field of an arbitrary stationary spacetime are concomitant. But gravitational time dilation is a property of the spacetime geometry whereas gravity is a phenomenon that we observe. It is thus natural to say that gravitation time dilation causes gravity. We might say that gravity causes gravitational redshift, but this doesn't really explain gravity. The crucial point is that anywhere there is time dilation in a stationary spacetime, a stationary observer will accelerate as if in a gravitational field. This is a fact that everyone who proposes a cause of gravity must take into account. That is, we have an explanation of gravity, a second explanation is not only unnecessary, but is too many. It also provides some detail to the understanding of gravity that "spacetime curvature" may not provide. It should be noted that gravitational time dilation does not distinguish between true gravity and artificial gravity produced by an accelerated frame of reference in the absence of true gravity (so gravitational time dilation is accelerational time dilation). Thus, I am not addressing a cause of gravity in terms of an energy-momentum source. One thing I should mention is that gravitational redshift is directly proportional to the frequency of the redshifted radiation. Therefore, any amplitude modulation of the radiation is also redshifted by the same relative amount. This means that any mechanical clock will be observed to tick slower. Thus, gravitational redshift is a gravitational time dilation. One can always obtain the acceleration of an arbitrary worldline in an arbitrary spacetime using the absolute derivative of [math]\dfrac{dx^\mu}{d\tau}[/math] with respect to proper time [math]\tau[/math]. But I specifically considered the familiar gravity that we all experience on earth to provide an understanding of the phenomenon that we all experience without the complication of trying to understand the complexities that GR has to offer. I ignored it because it wasn't relevant to what I was discussing. Time dilation in general is about comparing proper times of different spacetime trajectories. It's also about how to relate the endpoints of the different spacetime trajectories so that any comparison is meaningful. In the case of a stationary spacetime, the time translational symmetry along a Killing vector field provides a natural way to define time dilation. There is no suggestion that just any time dilation will cause gravity. My derivation was specific to stationary spacetimes, which is approximately true for earth (spinning rigid objects are stationary, btw).

The familiar gravity that we all experience on earth is caused by gravitational time dilation. I have devised "the slinky analogy" to provide some insight: Note that as the slinky is bent, the side away from the direction of bending is longer than the side towards the direction of bending. Furthermore, the sharper the bending, the greater the relative difference in the length of the two sides. There is a geometric relationship between the relative change in the length of an arc with respect to the radial distance and the curvature of the arc. Gravitational time dilation is the change in the length in spacetime of a timelike worldline with respect to height and is geometrically related to the curvature of the worldline, which corresponds to the acceleration we experience standing on the ground on earth. [math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] where: [math]a^\mu[/math] is the upwardly directed acceleration of an observer at rest on the ground. [math]T[/math] is the relative spacetime length of a timelike Killing vector at a given location in three-dimensional space. [math]\dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] is the gravitational time dilation. [math]g^{\mu\nu}[/math] has signature [math](1, -1, -1, -1, -1)[/math] [math]c[/math] is the speed of light in a vacuum (used to relate the curvature of a worldline to its acceleration) The rubber sheet analogy that is frequently used to explain gravity is incorrect even as an analogy. Of the various problems it has, the one that is relevant to this post is that the familiar gravity that we all experience is not caused by the curvature of three-dimensional space, as suggested by the rubber sheet analogy. It is specifically time dilation that gives rise to such gravity.

You could say that, but in fact I was making a point. The subscripted form of the partial differential operator is the natural form of this operator, whereas the superscripted form is obtained from the subscripted form by raising the index with the inverse of the metric tensor. The superscripted form of the partial differential operator therefore implicitly contains the inverse of the metric tensor which has to be explicitly taken into account in [math]\partial_\mu \partial^\mu \phi[/math]. Thus: [math]\partial_\mu \partial^\mu \phi = \partial_\mu (g^{\mu\nu} \partial_\nu \phi) = \partial_\nu (g^{\nu\mu} \partial_\mu \phi) = \partial_\nu g^{\nu\mu}\, \partial_\mu \phi + g^{\nu\mu}\partial_\nu \partial_\mu \phi = \partial_\nu g^{\nu\mu}\, \partial_\mu \phi + \partial^\mu \partial_\mu \phi[/math] A similar problem occurs with [math]dx_\mu[/math]. Is [math]dx_\mu = g_{\mu\nu} dx^\nu[/math]? Or is [math]dx_\mu = d(g_{\mu\nu} x^\nu)[/math]? I generally avoid this problem by not using [math]dx_\mu[/math]. However, in the case of: [math]u^\nu = \dfrac{dx^\nu}{ds}[/math] it seems unambiguous to me that: [math]u_\mu = g_{\mu\nu} u^\nu = g_{\mu\nu} \dfrac{dx^\nu}{ds}[/math] and not: [math]u_\mu = \dfrac{dx_\mu}{ds}[/math] The point is that the various subscripted/superscripted forms of a given quantity are not created equally and that there is often one form that is the natural form from which all the other forms are derived, and sometimes it does require one to know which is the natural form.

Hmmm, I nearly misread this. But it seems you are not giving me the credit to know that the identity matrix can only be a tensor if it has one superscript and one subscript. But the metric tensor and its inverse are not only frame-dependent quantities, but they also depend on the intrinsic manifold. By contrast, the Kronecker delta is not only frame-independent, but also independent of the intrinsic manifold. That is, the metric tensor or its inverse tells you both what intrinsic manifold you have, and also the frame within that intrinsic manifold. The Kronecker delta tells you neither. Quantities that are frame-independent but tell you what intrinsic manifold you have would be very useful, but the Kronecker delta is not that quantity. Neither is the metric tensor, but the Riemann tensor is frame-independent on a flat manifold. Only the metric tensor and its inverse can tell you about orthogonality. The ϵ tensor can only tell you about the weaker condition of linear-independence. I suspect you know this, but are underestimating me. I am fully aware that the concept of a tensor is dependent on the group of transformations. In my own research, I explored tensors within the context of the general linear group. This requires that the definition of the partial differential operator be extended beyond the chain rule. It introduces a new quantity to deal with the anholonomy. At the time, it was unclear to me about how to deal with the oxymoronic "anholonomic coordinate systems". Only later did I realise that they were a generalisation of the vierbein. What is special about the differentials of the coordinates is that they are differentials of the coordinates. What more needs to be said? I'm perfectly aware of the modern trend of getting rid of the entire notion of coordinate systems. But I reject this trend, regarding it as throwing the baby out with the bathwater. The notion of covariance is about putting all coordinate systems on equal footing by ensuring that equalities are the same in all coordinate systems. But somehow, even the notion of a coordinate system is an anathema. Even though tensors can be reformulated without coordinate systems such as I mentioned above, there is still an underlying aspect of the notion of coordinate systems within this formulation. It underlies the concept of curvature, regardless of how curvature is defined in the modern formalism.

[Expression 2] is incorrect. [Expression 1], although not incorrect (but see below), is not how I would choose to write it. I generally write partial differential operators in covariant (subscripted) form. Thus, I would use [Expression 2] in the corrected form: [math]\partial_\mu \left( \displaystyle \sum_{n} \dfrac{\partial\mathcal{L}}{\partial(\partial_\mu \phi_n)} \partial_\nu \phi_n - \delta^\mu_\nu \mathcal{L} \right) = 0[/math] At this point, I must admit that I have made an error. I took something you said out of context and placed it into another context in which I said you made an error. I apologise to you and Genady for this. However, I will ask you this: In general, does [math]\partial_\mu \partial^\mu \phi = \partial^\mu \partial_\mu \phi[/math] ?

I'm not aware of any experiment where charge is seen to radiate in some frames but not others. I don't think it requires the absence of electromagnetic acceleration, only the presence of an acceleration due to a non-electromagnetic force. Then one can examine if the radiation is accounted for by the total force or just the electromagnetic force. A proton-proton interaction experiment might be a suitable if the energy is large enough to overcome the electromagnetic repulsion so that the strong force is felt, but not too large to cause a transmutation. 👍 The problem I see is not a disagreement but an inability to see the principle behind the electromagnetic energy-momentum. It seems to come out of thin air... AFAICT, it does not come from Maxwell's equations. Actually, it has occurred to me that it cannot come from Maxwell's equations for dimensional reasons (no mass), but if it can be shown that the quadratic expression in terms of the electromagnetic tensor has energy-momentum-like properties, then this may satisfy me.

Actually, your example of [math]ds^2 = dx^i g_{ij} dx^j[/math], [math]ds^2 = dx_i g^{ij} dx_j[/math], and [math]ds^2 = dx_i \delta^i_j dx^j[/math] only serves to strengthen my point. Of the three metrics, only in the case of [math]ds^2 = dx^i g_{ij} dx^j[/math] are both vectors differentials of the coordinates. The differentials of the coordinates vector is naturally contravariant as a result of the chain rule. The [math]dx_i = (dr \ \ \ \ r^2 d\theta)[/math] vector is not a differentials of the coordinates vector as it is clearly seen to contain components of the metric tensor. Indeed, this vector required lowering the index of [math]dx^i = (dr \ \ \ \ d\theta)[/math] with the metric tensor. One can't lower indices with the Kronecker delta, another difference between the Kronecker delta and the metric tensor. In all three cases, the metrics [math]ds^2 = dr^2 + r^2 d\theta^2[/math] are the same. Note that the coefficients of the differentials of the coordinates quadratics are components of the metric tensor [math]g_{ij}[/math]. There is not a Kronecker delta in sight. Although the rules of index manipulation do allow the three expressions [math]ds^2 = dx^i g_{ij} dx^j[/math], [math]ds^2 = dx_i g^{ij} dx_j[/math], and [math]ds^2 = dx_i \delta^i_j dx^j[/math], [math]ds^2 = dx^i g_{ij} dx^j[/math] is the natural expression. At [math]r = 0[/math], there is a (removable) singularity that is a natural part of the polar coordinate system. If you want to remove this singularity, you change the coordinate system to something like the Euclidean coordinate system. The Kronecker delta tells you nothing. Its invariance guarantees that. You chose a flat space for your example. How about choosing the spacetime of a Schwarzschild black hole, where the central singularity is not removable. I don't think that what I'm talking about is my interpretation of tensors. I see it as more about general mathematical logic.

I am not going to download your file, and I doubt anyone else will. Please provide something that at least initially does not require us to download your file.

I need to think about this one first Have you ever derived the relativistic Lorentz force law from the divergence of the electromagnetic energy-momentum tensor? If you haven't, it may be difficult to appreciate where I'm coming from. I won't do it here, but basically one starts with the energy-momentum tensor on the left side of the equality, and a particular quadratic expression in terms of the electromagnetic field tensor on the right side of the equality. Perform the divergence of both sides. On the left, one has a force, and on the right after a number of steps, the product of the electromagnetic field tensor with the charge current vector (or charge times velocity). If one is willing to accept without question the electromagnetic energy-momentum tensor, then the force on the left becomes mass times acceleration, and we have an expression that says that the source of the particular quadratic expression in terms of the electromagnetic field tensor, divided by the mass of the charge, is the acceleration of the charge. This is analogous to the expression that says that the source of the electromagnetic field tensor, divided by the charge, is the velocity of the charge. Let's consider something more basic: The Coulomb field around a charge drops with [math]\dfrac{1}{r^2}[/math]. Enclose the charge in a sphere and perform a surface integral of the field to obtain the charge as the source of the field, noting that the surface area increases with [math]r^2[/math], cancelling the [math]\dfrac{1}{r^2}[/math] drop of the field. But the radiation field drops with [math]\dfrac{1}{r}[/math], so that one has to square the radiation field to obtain a field that drops with [math]\dfrac{1}{r^2}[/math] and whose surface integral obtains the source of this radiation field by cancelling with the [math]r^2[/math] increase of the surface area. For me personally, I don't accept without question the electromagnetic energy-momentum tensor. Thus, instead of obtaining acceleration of the charge from the left side of the equality as above, I would like to obtain it from the right side of the equality, thereby disconnecting the acceleration of the charge from the energy-momentum tensor. It is worth noting that due to the antisymmetry of the electromagnetic field tensor, the product of the electromagnetic field tensor with the charge current vector is orthogonal to the charge current vector, just as the acceleration vector is orthogonal to the charge current vector. But that doesn't imply that these two vectors that are orthogonal to the charge current vector are parallel to each other. Nevertheless, the orthogonality of the product of the electromagnetic field tensor with the charge current vector to the charge current vector is a right side of the equality result.