Recent content by The1337gamer

(\sqrt{3+4i})(\sqrt{3-4i})=\sqrt{(3+4i)(3-4i)} Now how can you simplify the term in the sqrt? Notice we have a complex number multiplied by it's complex conjugate

Is there any way of showing this? I've just read that the tangent/cotangent space is isomorphic to the local Euclidean space so would this imply that Minkowski space is itself locally isomorphic to Euclidean space?

I understand what the tangent and contangent spaces are and how you can find the basis for them. The cotangent space being the dual space corresponding to the tangent space. Tangent vectors transform contravariantly and cotangent vector transform covariantly. I'm not sure why they are used...

So i was considering minkowski space which is a 4-d manifold, why is that we use the tangent and cotangent space, to construct tensors on the space? The definition of a manifold says that the space is locally homeomorphic to Euclidean space. So is the tangent space and cotangent space...

I thought a bilinear form was the tensor product of 2 one-forms/linear functionals, so it would take two vectors to a scalar.

I was looking at this table here: http://en.wikipedia.org/wiki/Tensor#Examples And i didn't understand why a (1,1) tensor is a linear transformation, I was wondering if someone could explain why this is. A (1,1) tensor takes a vector and a one-form to a scalar. But a linear transformation...

For the second term, do you know how to differentiate exponential functions? Can you answer these questions, differentiate with respect to x: e^x 18e^x 4e^2x e^(x^2) e^(x^(1/3)) e^(8x^(-2/3)) For the first term you need to use a substitution, try substituting u=x^1/2

xa^x - a^x/(2lna) isn't correct, xa^x - a^x/(ln(a))^2 is.

It's the bottom line as: 2ln(a) = ln(a^2) What you have is ln(a)ln(a) = (ln(a))^2.

Seems correct to me, you just need to evaluate the 2nd term, the integral, you know how to do integrate a^x as you've already done it, 1/ln(a) is just a constant.

I think if your line integral is path independent, you can use the Fundamental Theorem of Calculus as your vector field is just a gradient field of a scalar function.

Thanks for the feedback. I know a bit about the metric tensor, it is a bilinear form that takes two vectors from the tangent space of our spacetime to a scalar, i think. So i think I am correct in saying the 4x4 matrix representing the metric is the same on all points in flat spacetime...

The interval between two events ds^2 = -(cdt)^2 + x^2 + y^2 + z^2 is invariant in inertial frames. I was wondering, if this same interval still applies and is invariant in non-inertial frames?

The best way to evaluate the derivate would be to use the chain rule. Are you familar with the chain rule?

Try ypn = bn(2^n). b is a constant to be found Notice in the complimentary function, (2^n) is multiplied by just a constant, this is different to the form of the particular solution as the RHS contains a non-zero first degree polynomial multiplying (2^n). But let's say for example your...