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location Austin, TX
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visits member for 3 years, 9 months
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I am a Ph.D. general relativist working as a software engineer. I like to still go and do physics as a hobby, and to keep up my skill and knowledge.


16m
comment find a magnetic field on z axis for a triangle
@user107761: is the magnetic field a vector or a scalar?
20h
comment Who foots the (magnetic) energy bill?
@bobie: the currents inside of the magnetic domains in the iron.
1d
comment Lack of scale in Schrödinger equation for square-inverse potential
We already know from classical mechanics that the potentials $\alpha r^{2}$ and $\alpha/r$ are special, in that they are the only central force potentials with stable closed orbits.
1d
comment Why is there a controversy on whether mass increases with speed?
The modern notion would say that $F^{\mu} = \frac{dp^{\mu}}{d\tau}$, and then you get the mass from the magnitude of the momentum. So, the invariant mass is the fundamental thing.
1d
comment Who foots the (magnetic) energy bill?
@bobie: yes, through the electric force. Not through the magnetic field.
Aug
29
comment Which of these two textbook equations of geodesic deviation is correct?
@Peter4075: that's the best thing to do when something isn't clear. Do it the long way, and then make sense of the short way argument about it.
Aug
29
comment Who foots the (magnetic) energy bill?
@AlanSE: but, of course, you know that's not what's going on -- you move some charges, and then exert a force on those moving charges.
Aug
29
comment Who foots the (magnetic) energy bill?
Also, energy and magnetism is generally a subtle thing, and you seem to be confusing force and work in your analysis. I think this question is too broad and changeable in interpretation for you to get an answer that you will be satisfied with.
Aug
29
comment Who foots the (magnetic) energy bill?
@bobie: because this is chasing a moving target.
Aug
29
comment Who foots the (magnetic) energy bill?
Magnetic forces cannot make stationary objects move. the force is proportional to the velocity of the object.
Aug
29
comment Which of these two textbook equations of geodesic deviation is correct?
@Peter4075: they're two copies of the SAME set of four functions.
Aug
29
answered Work done by gravity on a ball & the ball on earth
Aug
28
comment How far has a black hole to be in order for its tidal forces to disintegrate earth?
In fact, a ten solar mass neutron star would be a lot more destructive, because then, you'd have that NS supernova immanently happening. :)
Aug
28
revised Which of these two textbook equations of geodesic deviation is correct?
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Aug
28
comment Which of these two textbook equations of geodesic deviation is correct?
@Peter4075: all that's going on with that is the fact that the Riemann tensor is defined by $\nabla_{a}\nabla_{b}\omega_{c} - \nabla_{b}\nabla_{a}\omega_{c}$. It's an arbitrary choice whether you put the fourth index that you contract on $\omega$ first or last, so this could be either equal to $R_{abc}{}^{d}\omega_{d}$ or $R^{d}{}_{abc}\omega_{d}$. Using the properties of the Riemann Tensor, you can juggle indices around, and show that these two definitions are equal to each other, except that they differ by a minus sign.
Aug
28
comment Which of these two textbook equations of geodesic deviation is correct?
@JohnRennie: yes, but there is an additional convention on the Riemann tensor, depending on whether $\nabla_{[a}\nabla_{b]}\omega_{c} = R_{abc}{}^{d}\omega_{d}$ or whether $\nabla_{[a}\nabla_{b]}\omega_{c} = R^{d}{}_{abc}{}\omega_{d}$. But looking at these two expressions again, the second one is wrong, as it is always identically zero.
Aug
28
answered Which of these two textbook equations of geodesic deviation is correct?
Aug
28
comment Which of these two textbook equations of geodesic deviation is correct?
There are two different conventions for the Riemann tensor, and the tensors can differ by a minus sign between the two conventions. Without telling us which version the two books are using, this question, as written, is unanswerable, since the two expressions you give also differ by a minus sign.
Aug
26
comment Can a human size object move so fast that it ceases to be observable?
@AdamDavis: well, you could also make an argument that there is a finite recession speed where all incident visible light will be reflected back to a viewer at a redshifted frequency too low to be visible anymore.
Aug
26
revised A curious case of Relativistic Velocity Addition
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