9
votes
Accepted
Why Curie temperature is bigger for smaller lattice in 2D Ising model
Why numerical simulation gives seemingly wrong result?
The exact expression for $T_c$ is given by
$$T_c=\frac{2J}{\ln\left(\sqrt{2}+1\right)}\approx 2.269\,J,$$
where we assume that $k_B=1$. This ...
- 2,454
6
votes
Accepted
Errors when the quantity is an exponent
The most general rule for uncertainty propagation comes from a derivative of the function. If you sketch a graph of a function, you can see how this works. For a function $f(x)$, if $\frac{\partial ...
- 6,866
3
votes
Bar suspended by three vertical ropes
This is a canonical example of an underdefined problem (the number of unknowns (3 tensions) is less than the number of conditions (torque=0 and total force=0). You've tried to come up with a third ...
- 3,451
2
votes
Accepted
Obtaining Geodesic equation for Massive particles using Schwarzschild metric
The second equation doesn't really come from the Euler-Lagrange equations (though one could probably derive it from them.) It's much easier to see from the definition of the particle's proper time:
$$...
- 47.1k
1
vote
Accepted
Pendulum equation
I am offering two ways of solving such question.
The first method is the Lagrangian equation:
$\frac{d}{dt}(\frac{\partial L}{\partial \dot q})=\frac{\partial L}{\partial q} \ and \ L= T-V$ where q ...
- 36
1
vote
Accepted
How to correctly calculate minimum distance with kinematic equations
I think your answer is correct.
An alternative approach, without calculus, is to note that the minimum separation will occur when the velocity of Q equals the constant velocity of P - up until this ...
- 47.4k
1
vote
Elongation of rod in two cases
Assuming Hooke's law holds in every case, what you said about case 1 is right.
Assuming there is no friction, When only 1 force is applied to 1 end of the rod in case 2.
Here the rod must be ...
- 33
1
vote
Accepted
Potential - metal sphere in a uniform electric field
Yes, but your last equation is valid for all $\theta$. You can use this fact and the orthogonality of the Legendre polynomials to conclude that all the coefficients are zero.
Indeed, for “any” ...
- 9,525
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