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2020-07-24 15:18:24 +0100 commented answer can't solve inequality for independent variable

Unfortunately, it does help demonstrating that Mathematica does no better. I guess this is the answer ("Sage can't do what you want"), but I'll give a few days to see if anyone knows more. Thanks.

2020-07-23 17:26:30 +0100 asked a question can't solve inequality for independent variable

One of the frustrations I'm always having with Sage is how it tries to "solve" inequalities. For a random example:

sage: var('a b x')
sage: f = (a - b * x^2) / (x-1)
sage: solve(f > 0, x)
[[x < 1, b*x^2 - a > 0], [1 < x, -b*x^2 + a > 0]]

Which I knew already (since it's just the numerator). Ok, so various signs and things matter, but the fact that it can't even tell me

x^2 > a/b

is frustrating. Is there a reason, or a way to convince Sage to actually "solve" these in some way?

2020-06-04 21:22:28 +0100 commented answer abstract index notation and differential geometry

Ok that does seem to be a partial solution, but it doesn't allow the metric to be represented with abstract indices. For example, the answer to this problem is generically $C^c_{ab}=2\delta ^c_{(a}\nabla_{b)}\ln \Omega-g_{ab}g^{cd}\nabla_d \ln \Omega$.

2020-06-02 00:43:51 +0100 commented answer abstract index notation and differential geometry

Thanks: I will have to work out getting the latest version on my system before testing/marking correct, but it does seem that you've got a solution here.

2020-05-29 19:40:41 +0100 asked a question abstract index notation and differential geometry

I am wondering if there are ways to use abstract index notation in sage. For example, could I define the tensor:

$$C^c_{ab}=\frac{1}{2}g^{cd}(\tilde{\nabla_a} g_{bd}+\tilde{\nabla_b} g_{ad} - \tilde{\nabla_d} g_{ab})$$

this particular object describes the difference between two connections, $\nabla_a$ and $\tilde{\nabla}_b$. Can we define objects like this an manipulate them in Sage? My confusion comes from the common definition of the connection, a la

nabla = g.connection()

Which is not directly a 1-tensor. Specifically, I would like to define a metric, $g_{ab}$, and a conformal transformation $\tilde{g_{ab}}=\Omega^2 g_{ab}$, the corresponding connections, and determine the tensor $C^a_{bc}$ for this particular case. (and of course, we know the answer because this the standard approach to conformal transformations in GR).

2020-02-19 23:01:01 +0100 commented answer Can't substitute H=dot(a)/a in SageMath

Well, since this is my major frustration with Sage, I know that it is not a simple operation, but I can't for the life of me understand why it should be. Although your answer does the job, I'm going to select the other one as it doesn't require a new class definition. Although please, disagree and convince me!

2020-02-14 21:55:09 +0100 asked a question Can't substitute H=dot(a)/a in SageMath

Hello all, I'm working with the Friedman equations, and I've gotten them down to the common form presented in terms of the scale factor. MWE incoming:

M = Manifold(4, 'M', structure='Lorentzian')
fr.<t,r,th,ph> = M.chart(r't r:[0,+oo) th:[0,pi]:\theta ph:[0,2*pi):\phi')
var('G, Lambda, k', domain='real')
a = M.scalar_field(function('a')(t), name='a')
rho = M.scalar_field(function('rho')(t), name='rho')
p = M.scalar_field(function('p')(t), name='p')
g = M.metric()
g[0,0] = -1
g[1,1] = a*a/(1 - k*r^2)
g[2,2] = a*a*r^2
g[3,3] = a*a*(r*sin(th))^2
nabla = g.connection()
Ricci = nabla.ricci()
Ricci_scalar = g.ricci_scalar()
E1 = Ricci - Ricci_scalar/2*g + Lambda*g
print("First Friedmann equation:\n")
E1[0,0].expr().expand() == 0

With the result being

-Lambda + 3*diff(a(t), t)^2/a(t)^2 + 3*k/a(t)^2 == 0

Which is great, but I really want this to be in terms of the Hubble paramter $H=\dot{a}/a$. I can't make that substitution happen:

var('h')
eq1.substitute(diff(a.expr(),t)/a.expr()==h)

just spits out the same thing,

-Lambda + 3*diff(a(t), t)^2/a(t)^2 + 3*k/a(t)^2 == 0

Any ideas?

2020-01-18 01:53:44 +0100 commented answer Unable to use substitute_function in SageManifolds

Tried it both ways and they work. Thanks very much! So the .operator() returns the function, and the .expr() gives the expression of that function?

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2018-05-07 23:30:42 +0100 asked a question Alternating symbols with SageManifolds

I would like to utilize the alternating symbol $\epsilon_{abcd}$ in SageManifolds. I've worked out that you can define the volume form like

epsilon = g.volume_form()
print(epsilon) ; epsilon.display()


4-form eps_g on the 4-dimensional differentiable manifold M
eps_g = r^2*sqrt(-f(t))*abs(a(t))^3*sin(th) dt/\dr/\dth/\dph

And I can turn this into "the alternating 1-form" by dividing by the determinant of the metric. However, if I try to find the totally upper symbol $\epsilon^{abcd}$, which should just have components +1 and -1, I get

epsilon_up=epsilon.up(g)*det*det
print(epsilon_up);epsilon_up.display()

Tensor field of type (4,0) on the 4-dimensional differentiable manifold
M
abs(a(t))^6/a(t)^6 d/dt*d/dr*d/dth*d/dph - abs(a(t))^6/a(t)^6
d/dt*d/dr*d/dph*d/dth - abs(a(t))^6/a(t)^6 d/dt*d/dth*d/dr*d/dph +
abs(a(t))^6/a(t)^6 d/dt*d/dth*d/dph*d/dr + abs(a(t))^6/a(t)^6
d/dt*d/dph*d/dr*d/dth - abs(a(t))^6/a(t)^6 d/dt*d/dph*d/dth*d/dr -
abs(a(t))^6/a(t)^6 d/dr*d/dt*d/dth*d/dph + abs(a(t))^6/a(t)^6
d/dr*d/dt*d/dph*d/dth + abs(a(t))^6/a(t)^6 d/dr*d/dth*d/dt*d/dph -
abs(a(t))^6/a(t)^6 d/dr*d/dth*d/dph*d/dt - abs(a(t))^6/a(t)^6
d/dr*d/dph*d/dt*d/dth + abs(a(t))^6/a(t)^6 d/dr*d/dph*d/dth*d/dt +
abs(a(t))^6/a(t)^6 d/dth*d/dt*d/dr*d/dph - abs(a(t))^6/a(t)^6
d/dth*d/dt*d/dph*d/dr - abs(a(t))^6/a(t)^6 d/dth*d/dr*d/dt*d/dph +
abs(a(t))^6/a(t)^6 d/dth*d/dr*d/dph*d/dt + abs(a(t))^6/a(t)^6
d/dth*d/dph*d/dt*d/dr - abs(a(t))^6/a(t)^6 d/dth*d/dph*d/dr*d/dt -
abs(a(t))^6/a(t)^6 d/dph*d/dt*d/dr*d/dth + abs(a(t))^6/a(t)^6
d/dph*d/dt*d/dth*d/dr + abs(a(t))^6/a(t)^6 d/dph*d/dr*d/dt*d/dth -
abs(a(t))^6/a(t)^6 d/dph*d/dr*d/dth*d/dt - abs(a(t))^6/a(t)^6
d/dph*d/dth*d/dt*d/dr + abs(a(t))^6/a(t)^6 d/dph*d/dth*d/dr*d/dt

Which is just happening because at some point Sage has figured it out needs to keep track of the sign of a. Is there a direct way to define $\epsilon_{abcd}$ without accessing to the volume form at all? Essentially, I want to directly calculate things like the Euler Characteristic (and signature), which look like

$$R_{abcd}R_{efgh}\epsilon^{abef}\epsilon^{cdgh}$$