Recent content by elfmotat

You could do a binomial expansion and only use the terms you need. For example, a first order approximation would be: (1+x)^n=1+nx+\mathcal{O}(x^2) \approx 1+nx So: \sqrt{x} \approx 1+\frac{1}{2} (x-1)

Why not? Point particle density is easily and usefully characterized by a delta function all the time in classical physics.

Perhaps varying the action directly instead of using the E-L equations will make more sense to you: S = \frac{1}{2} \int \left [ \eta^{ab} \partial_a \phi \partial_b \phi-m^2 \phi^2 \right ] d^4 x When varying the action, the metric is constant so: \delta S = \frac{1}{2} \int \left [...

It's useful when separating differential equations like the following: a=\frac{k}{r^2} So you can integrate: vdv=\frac{k}{r^2}dr This is used, for example, with finding the time it takes for two masses to come together under gravity.

You could equally well say that "the fundamental idea behind inertial mass is how hard it is to change an object's momentum." You can't relate three-force to three-acceleration without invoking longitudinal and transverse mass, so it has to be a matrix. Because the inner product of two...

If you define inertial mass as "the coefficient relating momentum to velocity", then in SR it's the relativistic mass \gamma m. This is a useful definition because F=dp/dt still holds true in relativity when momentum is defined as p=\gamma mv. If you instead define inertial mass as "the...

The rock has an "acceleration due to gravity" which is -9.8 m/s2. It has an "acceleration due to the ground" which is +9.8 m/s2. Add them together and you get zero.

R(T)=R(T(x,y)) Let's say you're given the point (a,b). Plug that point into T and it returns a number (call it c): T(a,b)=c. Then you plug this number into R to get R(T): R(T(a,b))=R(c)=d. If you're only given x=a but y remains a variable, then T(a,y) is a function dependent only on the...

You're right. Thanks for the correction.

You can calculate it using Jacobi's forumula: http://en.wikipedia.org/wiki/Determinant#Derivative http://en.wikipedia.org/wiki/Jacobi%27s_formula You should get \delta g=gg_{\mu \nu}\delta(g^{\mu \nu})

If you wanted to get hand-wavy, I suppose you could view current as the residue of the magnetic field.

The total force can still be defined by equation (1). What's impossible is the assumption that all particles will have identical accelerations. For that to be possible you need internal forces to offset the tidal forces.

Sorry for derailing your thread a bit further Spinnor, but I felt the need to post this. Unless I messed something up, I believe you can get the following from Maxwell's equations: \nabla_\mu \nabla_\nu \left ( A_\lambda \frac{dx^\lambda}{ds} \right )=\rho g_{\mu \nu} where \rho is charge...

When talking about a "composite" body, it is assumed that there are internal forces which keep the individual particles at the same positions relative to each other. This is known as a "rigid body." So when an external force is applied to the (center of the) composite the force is "transferred"...

Well, that's what we observe from experiment. There's nothing to suggest deviations from f(m,q)=m.