Recent content by boniphacy

This is normal for integrators.The truncation error for method of order 4, means the error is 2^4 = 16 lower than for double step: 2h. But this is not true in general, because the error is proportional to some number in fact, which is not constant in general.For example: in the case of circular...

## E = mc^2 \frac{\sqrt{1-2GM/c^2r}}{\sqrt{1 - (r'^2 + h^2/r^2)/c^2}} ## OK. what is correct approximation of this equation, up to second order? ## E = mc^2 + mr'^2/2 + mh^2/2r^2 - mGM/r + the second order term ## we can see 'the first order part' is perfectly Newtonian version.

This is forbidden. We can use the first terms only. This function is used in the GR fantastic theory, exactly: ## E = mc^2 \frac{\sqrt{1-2GM/c^2r}}{\sqrt{1 - (r'^2 + h^2/r^2)/c^2}} ##

for x = y it is: ## \sqrt{\frac{1-2x}{1-x^2}} \approx 1 - x^2 - x^3 - x^4/2 + ... ## so, the first looks is correct. now try: y^2 = x ## \sqrt{\frac{1-2x}{1-x}} \approx 1 - x/2 - 5/8 x^2 + ... ## ? fantastic.

OK. We assume x ~ y^2, then: what is correct now?

We have a function: ## f(x,y)=\sqrt{\frac{1−2x}{1−y^2}} = \frac{\sqrt{1−2x}}{\sqrt{1−y^2}}## for small x and y, we can use standard approximations: ## 1/\sqrt{1−x}=1+x/2+... ## and ##\sqrt{1−x}=1−x/2−... ## Ok. Now we can approximate the whole function f(x,y) First method: ##...

Irrelevant: I can take derivative wrt time t. dE/dt = 0. The result is still the same.

Do You suggest: the standard math is wrong?

I want to derive an acceleration in the case for a stationary mass in the gravity field. I found the total energy in the GR is provided by a simple equation: https://en.wikipedia.org/wiki/Schwarzschild_geodesics ## E = mc^2\sqrt{1 - rs/r} * \gamma ## So, this is easy to provide acceleration...