Recent content by mmmboh

Does this only work because we are taking the average? Because if r>1, then dS(y) would be larger than dS(z), right? And so the left integral (without taking the average) would be larger than the right integral (without taking the average)...

\frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)}u(y)\,dS(y)=\frac{1}{|\partial B(0,1)|}\int_{\partial B(0,1)} u(x+rz)\,dS(z) Why does dS(y)\to dS(z) and not dS(y)\to dS(x+rz)? If you want more information, it comes from http://www.stanford.edu/class/math220b/handouts/laplace.pdf on page 8...

Is the answer to my question also because it would be physically irrelevant?

For this to be true, it must be assumed that the x derivative of the wave function is bounded near +\- infinity, or at least grows slower than the wave function goes to zero...how do we know this? Can we not have a wave function that goes to zero but oscillates quicker and quicker as x grows, or...

Wow that's a very interesting derivation, thanks! By the way, has it ever been the case that someone successfully found the laws of a theory based on minimizing the action, without knowing the laws first? Like just by guessing a Lagrangian? I mean when the least action principle came along, the...

But Newton's laws are invariant under Galilean boosts, but kinetic energy is not...does it still make sense to call kinetic energy a scalar if this is an allowable coordinate transformation? Thanks for the replies, good info.

So why wouldn't we have this problem when using Lorentz boosts?

So allowable coordinate transformations just means coordinate transformations that leave the equations the same? What about Galilean boosts? F=ma either way. I mean if Lorentz boosts are allowable in SR, why aren't Galilean boosts allowable boosts allowable in CM?

I've studied classical physics and never heard this before until recently...the allowable coordinate transformations for classical mechanics are rotations and translations. Could someone explain why this is so? What makes these "allowable" (I know they are orthogonal transformations).

I didn't mean it like that, I mean I've seen the derivation of the Lagrange equations without using the action, and that minimizing the action gives the same results.

Ok so you can't use L=T-V in SR, but why does that imply the action is invariant? The action being invariant is the only justification I've read for why the SR Lagrangian is what it is. But in classical mechanics, Galilean transformations are analogous to Lorentz transformations...so if the...

I'm confused. If you launch a ball in a train, and you are in the train, the kinetic energy of the ball will be less than if you were standing outside the train, so the calculated actions will be different, won't they? Which would mean the action isn't invariant. So why would we assume the...

In classical mechanics, isn't kinetic energy not a Galilean scalar? So the action isn't invariant under Galilean transformations, but we can still use it with Galilean transformations. So why must it be a scalar in special relativity? I think I'm missing something...

I know that, but those laws are derived based on the postulate that an object traveling at c in one reference frame travels at c in all reference frames. I want to know if you can get that based on the postulates: 1. All laws of physics are the same in all inertial frames. 2. There is a speed...

How I studied relativity, we postulated that a particle traveling at c in one inertial frame travels at c in all inertial frame. But now looking through a book, I see that they just postulate that all laws of physics are same in all inertial frames, and that there is a speed limit (c). However...