# Recent content by littleHilbert

1. ### Invariance of the speed of light

OK, that was trivial...I knew it must be easy…if one arranges things the right way. I too was using the pythagorean theorem in order to get rid of some terms. But at the same time I was computing the difference u^2_x+u^2_y-c^2 to get 0…and yeah…somehow fell asleep. Many thanks, George.
2. ### Invariance of the speed of light

Oh yes, that sounds much more promising! I'll try this now…thanks in advance! :-)
3. ### Invariance of the speed of light

If you mean using the angles in u_x=u \cos \theta,\, u_y=u \sin \theta, \, u'_x=u' \cos \theta',\, u'_y=u' \sin \theta' …does it really help? I tried to plug them in, too…but same thing…the computation gets lengthier and seems to be getting nowhere. There is actually no more on that page, except...
4. ### Invariance of the speed of light

Hello! Consider the law of addition of velocities for a particle moving in the x-y plane: u_x=\frac{u'_x+v}{1+u'_xv/c^2},\, u_y=\frac{u'_y}{\gamma(1+u'_xv/c^2)} In the book by Szekeres on mathematical physics on p.238 it is said that if u'=c, then it follows from the above formulae that...
5. ### The Structure of Galilean Space

I think I got now what the message is. The message is simply that the spatial distance is not a well-defined function on non-simultaneous events, i.e. not independent under the Galilean transforms, for that is what we wish it to be -- the (classical) frame reference change must preserve...
6. ### The Structure of Galilean Space

At this stage I actually meant nothing, except that I agreed that what Bill said was in fact exactly what Szekeres said by example, and I pointed to my previous post.
7. ### Contraction of Tensors

To George: Aha, ok, if you put it this way then I agree that it becomes a bit less controversial…but still I would then feel inclined to relabel the indices on initial step and then it would actually look all the same, but with relabeled indices. Alright, perhaps I just need a bit more practice...
8. ### The Structure of Galilean Space

Yes, as a matter of fact I recognized the example in his formulation. But the problem is that I saw a slightly different thing in it.
9. ### The Structure of Galilean Space

OK. Now I think that I somehow didn't get the point. I thought the point is that two non-simultaneous events can be brought by a suitable choice of Galiliean frame to simultaneity, i.e. simply by time shift (adding a constant), so that their distance becomes purely spatial distance. You...
10. ### The Structure of Galilean Space

A Galilean transformation is defined as a transformation that preserves the structure of Galilean space, namely: 1. time intervals; 2. spatial distances between any two simultaneous events; 3. rectilinear motions. Can anyone give a short argument for the fact that only measuring the...
11. ### Contraction of Tensors

Hi all! I've got a short question concerning a minor notational issue about tensor contraction I've run across recently. Let A be an antisymmetric (0,2)-tensor and S a symmetric (2,0)-tensor. Then their total contraction is zero: C_1^1C_2^2\,A \otimes S=0. As a proof one simply computes...
12. ### Bounded sets

Oh my it's getting really funny! Yeah, but the fact is that both sets are in the ball with radius 51, and so their union is bounded. I overlooked the value of the constant K, which should of course be the diameter of the (larger) ball, i.e. 102. I "repaired" the proof. It should be OK now...
13. ### Bounded sets

I should have made it more explicit. Here is the complete version: Let A and B be bounded. Take any point z_0\in X, such that for all x\in A we have d(x,z_0)\ge\sup_{u\in A}d(x,u) and for all y\in B respectively d(y,z_0)\ge\sup_{v\in B}d(y,v). Then with K_A:=\sup_{x\in A}d(x,z_0) and...
14. ### Bounded sets

Well, I realised this at the very beginning, but it's now that I can write it down: Let A and B be bounded. W.l.o.g assume that \delta (A)\le\delta (B). By assumption there are numbers K_A and K_B such that \delta(A)\le K_A and \delta(B)\le K_B. Take K:=\max\{K_A,K_B\}<\infty (i.e. we take a...
15. ### Bounded sets

Suppose \delta (A)<\infty. Let x\in A be arbitrary but fixed. Then d(x,y)\le\sup_{(x,y)\in\{x\}\times A}d(x,y)\le\delta (A). Thus setting r:=\delta (A) we find a (closed) ball \bar{B}_r(x):=\{y\in X:d(x,y)\le r\}. It contains all points of A, since for all x\in A and all y\in\bar{B}_r(x) we have...