eric_g's profile - activity

Use plot(lambda x: EEF(x, 2), x, 0, 3) See the detailed explanations from cell 28 in this tutorial notebook: https:/

Maybe the discussion in this thread is relevant: https://groups.google.com/g/sage-devel/c/GaQHdu0Q6UU

What about something like c_matplotlib = c.matplotlib(figure=fig, sub=ax)

SageMath's covariant derivatives are documented at https://doc.sagemath.org/html/en/reference/manifolds/sage/manifolds/d

SageMath's covariant derivatives are documented here: https://doc.sagemath.org/html/en/reference/manifolds/sage/manifold

Note that the geodesic solver already implemented in SageMath allows one to choose between many ODE solvers, from SciP

Note that the geodesic solver already in SageMath allows one to choose between many ODE solvers, from SciPy or from the

Note that the geodesic solver already implemented in SageMath allows one to choose between many ODE solvers, from SciPy

Note that the geodesic solver already implemented in SageMath allows one to choose between many ODE solvers, from SciPy

Note that the geodesic solver already implemented in SageMath allows one to choose between many ODE solvers, from SciPy

FWIW, it is possible to superpose a SageMath plot to a png image, via Matplotlib. The first step is to convert the plot

You have to access to the tensor components by directly passing the indices to the operator [], not via a string (as @to

Indeed, if the frame argument is not provided, the connection 1-forms are computed for the manifold's default frame, see

Indeed, if the frame argument is not provided, the connection 1-forms are computed for the manifold's default frame, see

You must have the Ubuntu package sagemath-jupyter installed on your system. Then you should launch Jupyter from the Sage

You must have the Ubuntu package sagemath-jupyter installed on your system. Then you should launch Jupyter from the Sage

It works for me with SageMath 10.2. Are you sure you are running a SageMath kernel and not a Python kernel? If yes, plea

Hi, The composition transit_H_to_X = transit_Y_to_X * transit_H_to_Y works well for me (Sage 10.2): it yields a result

It should be diff(f)[i].expr() for $\partial f /\partial x^i$, so I guess commutator_f.expr().collect(diff(f)[0].expr()

You can use str(k) to get a string representation of k.

Actually, since $\Xi := 1 + H_0^2 a^2$ is a constant, one can rescale the $t$ of https://arxiv.org/abs/1805.00243 as $t'

Actually, since $\Xi := 1 + H_0^2 a^2$ is a constant, one can rescale the $t$ of https://arxiv.org/abs/1805.00243 as $t'

Actually, since $\Xi := 1 + H_0^2 a^2$ is a constant, one can rescale the $t$ of https://arxiv.org/abs/1805.00243 as $t'

Actually, since $\Xi := 1 + H_0^2 a^2$ is a constant, one can rescale the $t$ of https://arxiv.org/abs/1805.00243 as $t'

Actually, since $\Xi := 1 + H_0^2 a^2$ is a constant, one can rescale the $t$ of https://arxiv.org/abs/1805.00243 as $t'

Actually, since $\Xi := 1 + H_0^2 a^2$ is a constant, one can rescale the $t$ of https://arxiv.org/abs/1805.00243 as $t'

Actually, since $\Xi := 1 + H_0^2 a^2$ is a constant, one can rescale the $t$ of https://arxiv.org/abs/1805.00243 as $t'

I think that the metric given by Eq. (1) of your reference (https://arxiv.org/abs/1710.00997) is not correct (compare wi

I think that the metric given by Eq. (1) of your reference (https://arxiv.org/abs/1710.00997) is not correct (compare wi

The 5-dimensional Kerr-AdS spacetime has been treated there. The Riemann tensor was computed in a reasonable time (it i

The 5-dimensional Kerr-AdS spacetime has been treated there. The Riemann tensor was computed in a reasonable time (it i

Sorry I read your post too fast. I have updated my answer to the Kerr-de Sitter case.

I have updated my answer to the Kerr-de Sitter case.

The 5-dimensional Kerr-AdS spacetime has been treated there. The Riemann tensor was computed in a reasonable time (it i

[2]: M = Manifold(4, 'M', latex_name=r'\mathcal{M}', structure='Lorentzian') [3]: BL.<t,r,th,ph> = M.chart(r"t r t

The 5-dimensional Kerr-AdS spacetime has been treated there. The Riemann tensor was computed in a reasonable time (it i

It works for me with SageMath 10.1. Which version of SageMath are you using? What is your operating system?

You can define a vector field by providing its components in the default frame like this: v = E.vector_field((function(

You can define a vector field by providing its components in the default frame like this: v = E.vector_field((function(

How to define a vector field over a chart? Hello, I have the following code in sage: E.<x,y,z> = EuclideanSpace()

I have edited my answer to show how to a new vector frame and express the gradient in it.

The answer provided by Sage for the gradient of f is correct: grad(f) = d(F)/dr ∂/∂r + d(F)/dθ/r^2 ∂/∂θ + d(F)/dϕ/(r^2*

The answer provided by Sage for the gradient of f is correct: grad(f) = d(F)/dr ∂/∂r + d(F)/dθ/r^2 ∂/∂θ + d(F)/dϕ/(r^2*

The answer provided by Sage for the gradient of f is correct: grad(f) = d(F)/dr ∂/∂r + d(F)/dθ/r^2 ∂/∂θ + d(F)/dϕ/(r^2*

Actually, parallelism is broken for computations involving symbolic functions (i.e. defined via function as in your case

Your code works for me with SageMath 10.1 (the latest stable version). Which version of SageMath are you using?

The notebook Black hole rendering with SageMath works well on my CoCalc account with the SageMath 10.0 kernel (not the 1

It works for me with SageMath 10.1. Which version of SageMath are you using? What is your operating system?