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This is what I have so far.

We start with the definition that force is the change in momentum with respect to time. In the four dimensions of spacetime, this is:$$\Sigma F_i =\frac{dP}{d\tau}$$Where $\Sigma F_i$ is the sum of 4-forces, both applied and inertial, $P$ is the 4-momentum and $\tau$ is the proper time. For a constant mass system (non-relativistic), this is:$$\Sigma F_i =m\frac{dU}{d\tau}$$Where $U$ is the 4-Velocity. The 4-Velocity is really the product of two tensors, the components and the basis vectors. Expanding, we get:$$\frac{dU}{d\tau}=\frac{d}{d\tau}\left(U^\lambda e_\lambda \right)=\frac{dU^\lambda}{d\tau}e_\lambda +U^\lambda\frac{de_\lambda}{d\tau}$$ The derivative of the basis vector with respect to the affine parameter is represented by the Christoffel Symbols, so the product of the tangent velocity of the fiduciary geodesic and the rate of change in the basis vector is (please provide the derivation if you consider it important):$$U^\lambda\frac{de_\lambda}{d\tau}=\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$ Substituting back in, the expression for the 4-Acceleration becomes:$$\frac{dU}{d\tau}=\frac{dU^\lambda}{d\tau}e_\lambda +\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$$$\frac{dU}{d\tau}=\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$Substituting back into the 4-force expression, we have:$$\Sigma F_i =m\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$If the geometry of spacetime is flat, then the Christoffel symbols disappear and, for a non-relativistic, constant mass system, we have: $$a\equiv\frac{dU}{d\tau}$$ $$\Sigma F_i =ma$$If the spacetime is curved, we have:$$A\equiv\Gamma_{\mu\nu}^\lambda U^\mu U^\nu$$$$\Sigma F_i =m\left(a+A\right)$$Where $a$ is the result the applied force and $A$ is the acceleration of the pseudo (inertial) force of curvature.

I cobbled this together from several papers. I would appreciate any changes that make the notation and the descriptions more standard. In particular, I'm not sure of the treatment of the basis vector, $e^\lambda$ in all the steps.

This is what I have so far.

We start with the definition that force is the change in momentum with respect to time:$$\Sigma F_i =\frac{dP}{d\tau}$$Where $\Sigma F_i$ is the sum of forces, both applied and inertial, $P$ is the momentum and $\tau$ is the proper time. For a constant mass system (non-relativistic), this is:$$\Sigma F_i =m\frac{dU}{d\tau}$$Where $U$ is the velocity. In spacetime, the velocity is a four-velocity and is the product of two tensors, the components and the basis vectors. Expanding, we get:$$\frac{dU}{d\tau}=\frac{d}{d\tau}\left(U^\lambda e_\lambda \right)=\frac{dU^\lambda}{d\tau}e_\lambda +U^\lambda\frac{de_\lambda}{d\tau}$$ The derivative of the basis vector with respect to the affine parameter is represented by the Christoffel Symbols, so the product of the tangent velocity of the fiduciary geodesic and the rate of change in the basis vector is (please provide the derivation if you consider it important):$$U^\lambda\frac{de_\lambda}{d\tau}=\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$ Substituting back in, the expression for the four-acceleration becomes:$$\frac{dU}{d\tau}=\frac{dU^\lambda}{d\tau}e_\lambda +\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$$$\frac{dU}{d\tau}=\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$Substituting back into the sum of forces equation, we have:$$\Sigma F_i =m\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$If the geometry of spacetime is flat, then the Christoffel symbols disappear and, for a non-relativistic, constant mass system, we have: $$a\equiv\frac{dU}{d\tau}$$ $$\Sigma F_i =ma$$If the spacetime is curved, we have:$$A\equiv\Gamma_{\mu\nu}^\lambda U^\mu U^\nu$$$$\Sigma F_i =m\left(a+A\right)$$Where $a$ is the result the applied force and $A$ is the acceleration of the pseudo (inertial) force of curvature.

I cobbled this together from several papers. I would appreciate any changes that make the notation and the descriptions more standard. In particular, I'm not sure of the treatment of the basis vector, $e^\lambda$ in all the steps.

This is what I have so far.

We start with the definition that force is the change in momentum with respect to time. In the four dimensions of spacetime, this is:$$\Sigma F_i^\lambda =\frac{dP^\lambda}{d\tau}$$Where $\Sigma F_i^\lambda$ is the sum of 4-forces, both applied and inertial, $P^\lambda$ is the 4-momentum and $\tau$ is the proper time. For a constant mass system (non-relativistic), this is:$$\Sigma F_i^\lambda =m\frac{dU^\lambda}{d\tau}$$Where $U^\lambda$ is the 4-Velocity. The 4-Velocity is really the product of two tensors, the components and the basis vectors. Expanding, we get:$$\frac{dU^\lambda}{d\tau}=\frac{d}{d\tau}\left(U^\lambda e_\lambda \right)=\frac{dU^\lambda}{d\tau}e_\lambda +U^\lambda\frac{de_\lambda}{d\tau}$$ The derivative of the basis vector with respect to the affine parameter is represented by the Christoffel Symbols, so the product of the tangent velocity of the fiduciary geodesic and the rate of change in the basis vector is (please provide the derivation if you consider it important):$$U^\lambda\frac{de_\lambda}{d\tau}=\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$ Substituting back in, the expression for the 4-Acceleration becomes:$$\frac{dU^\lambda}{d\tau}=\frac{dU^\lambda}{d\tau}e_\lambda +\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$$$\frac{dU^\lambda}{d\tau}=\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$ Substituting back into the 4-force expression, we have:$$\Sigma F_i^\lambda =m\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$If the geometry of spacetime is flat, then the Christoffel symbols disappear and, for a non-relativistic, constant mass system, we have: $$a\equiv\frac{dU^\lambda}{d\tau}$$ $$\Sigma F_i =ma$$If the spacetime is curved, we have:$$A\equiv\Gamma_{\mu\nu}^\lambda U^\mu U^\nu$$$$\Sigma F_i =m\left(a+A\right)$$Where $a$ is the result the applied force and $A$ is the acceleration of the pseudo (inertial) force of curvature.

I cobbled this together from several papers. I would appreciate any changes that make the notation and the descriptions more standard. In particular, I'm not sure of the treatment of the basis vector, $e^\lambda$ in all the steps.

This is what I have so far.

We start with the definition that force is the change in momentum with respect to time. In the four dimensions of spacetime, this is:$$\Sigma F_i =\frac{dP}{d\tau}$$Where $\Sigma F_i$ is the sum of 4-forces, both applied and inertial, $P$ is the 4-momentum and $\tau$ is the proper time. For a constant mass system (non-relativistic), this is:$$\Sigma F_i =m\frac{dU}{d\tau}$$Where $U$ is the 4-Velocity. The 4-Velocity is really the product of two tensors, the components and the basis vectors. Expanding, we get:$$\frac{dU}{d\tau}=\frac{d}{d\tau}\left(U^\lambda e_\lambda \right)=\frac{dU^\lambda}{d\tau}e_\lambda +U^\lambda\frac{de_\lambda}{d\tau}$$ The derivative of the basis vector with respect to the affine parameter is represented by the Christoffel Symbols, so the product of the tangent velocity of the fiduciary geodesic and the rate of change in the basis vector is (please provide the derivation if you consider it important):$$U^\lambda\frac{de_\lambda}{d\tau}=\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$ Substituting back in, the expression for the 4-Acceleration becomes:$$\frac{dU}{d\tau}=\frac{dU^\lambda}{d\tau}e_\lambda +\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$ $$\frac{dU}{d\tau}=\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$Substituting back into the 4-force expression, we have:$$\Sigma F_i =m\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$If the geometry of spacetime is flat, then the Christoffel symbols disappear and, for a non-relativistic, constant mass system, we have: $$a\equiv\frac{dU}{d\tau}$$ $$\Sigma F_i =ma$$If the spacetime is curved, we have:$$A\equiv\Gamma_{\mu\nu}^\lambda U^\mu U^\nu$$$$\Sigma F_i =m\left(a+A\right)$$Where $a$ is the result the applied force and $A$ is the acceleration of the pseudo (inertial) force of curvature.

I cobbled this together from several papers. I would appreciate any changes that make the notation and the descriptions more standard. In particular, I'm not sure of the treatment of the basis vector, $e^\lambda$ in all the steps.

This is what I have so far.

We start with the definition that force is the change in momentum with respect to time. In the four dimensions of spacetime, this is:$$\Sigma F_i^\lambda =\frac{dP^\lambda}{d\tau}$$Where $\Sigma F_i^\lambda$ is the sum of 4-forces, both applied and inertial, $P^\lambda$ is the 4-momentum and $\tau$ is the proper time. For a constant mass system (non-relativistic), this is:$$\Sigma F_i^\lambda =m\frac{dU^\lambda}{d\tau}$$Where $U^\lambda$ is the 4-Velocity. The 4-Velocity is really the product of two tensors, the components and the basis vectors. Expanding, we get:$$\frac{dU^\lambda}{d\tau}=\frac{d}{d\tau}\left(U^\lambda e_\lambda \right)=\frac{dU^\lambda}{d\tau}e_\lambda +U^\lambda\frac{de_\lambda}{d\tau}$$ The derivative of the basis vector with respect to the affine parameter is represented by the Christoffel Symbols, so the product of the tangent velocity of the fiduciary geodesic and the rate of change in the basis vector is (please provide the derivation if you consider it important):$$U^\lambda\frac{de_\lambda}{d\tau}=\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$ Substituting back in, the expression for the 4-Acceleration becomes:$$\frac{dU^\lambda}{d\tau}=\frac{dU^\lambda}{d\tau}e_\lambda +\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$$$\frac{dU^\lambda}{d\tau}=\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$Substituting back into the 4-force expression, we have:$$\Sigma F_i^\lambda =m\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$If the geometry of spacetime is flat, then the Christoffel symbols disappear and, for a non-relativistic, constant mass system, we have: $$a\equiv\frac{dU^\lambda}{d\tau}$$ $$\Sigma F_i =ma$$If the spacetime is curved, we have:$$A\equiv\Gamma_{\mu\nu}^\lambda U^\mu U^\nu$$$$\Sigma F_i =m\left(a+A\right)$$Where $a$ is the result the applied force and $A$ is the acceleration of the pseudo (inertial) force of curvature.

I cobbled this together from several papers. I would appreciate any changes that make the notation and the descriptions more standard.

This is what I have so far.

We start with the definition that force is the change in momentum with respect to time. In the four dimensions of spacetime, this is:$$\Sigma F_i^\lambda =\frac{dP^\lambda}{d\tau}$$Where $\Sigma F_i^\lambda$ is the sum of 4-forces, both applied and inertial, $P^\lambda$ is the 4-momentum and $\tau$ is the proper time. For a constant mass system (non-relativistic), this is:$$\Sigma F_i^\lambda =m\frac{dU^\lambda}{d\tau}$$Where $U^\lambda$ is the 4-Velocity. The 4-Velocity is really the product of two tensors, the components and the basis vectors. Expanding, we get:$$\frac{dU^\lambda}{d\tau}=\frac{d}{d\tau}\left(U^\lambda e_\lambda \right)=\frac{dU^\lambda}{d\tau}e_\lambda +U^\lambda\frac{de_\lambda}{d\tau}$$ The derivative of the basis vector with respect to the affine parameter is represented by the Christoffel Symbols, so the product of the tangent velocity of the fiduciary geodesic and the rate of change in the basis vector is (please provide the derivation if you consider it important):$$U^\lambda\frac{de_\lambda}{d\tau}=\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$ Substituting back in, the expression for the 4-Acceleration becomes:$$\frac{dU^\lambda}{d\tau}=\frac{dU^\lambda}{d\tau}e_\lambda +\Gamma_{\mu\nu}^\lambda U^\mu U^\nu e_\lambda$$$$\frac{dU^\lambda}{d\tau}=\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$Substituting back into the 4-force expression, we have:$$\Sigma F_i^\lambda =m\left(\frac{dU^\lambda}{d\tau}+\Gamma_{\mu\nu}^\lambda U^\mu U^\nu\right)e_\lambda$$If the geometry of spacetime is flat, then the Christoffel symbols disappear and, for a non-relativistic, constant mass system, we have: $$a\equiv\frac{dU^\lambda}{d\tau}$$ $$\Sigma F_i =ma$$If the spacetime is curved, we have:$$A\equiv\Gamma_{\mu\nu}^\lambda U^\mu U^\nu$$$$\Sigma F_i =m\left(a+A\right)$$Where $a$ is the result the applied force and $A$ is the acceleration of the pseudo (inertial) force of curvature.

I cobbled this together from several papers. I would appreciate any changes that make the notation and the descriptions more standard. In particular, I'm not sure of the treatment of the basis vector, $e^\lambda$ in all the steps.