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.
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...
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...
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...
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.
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...
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...
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...
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...
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...
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...
Well, I realized 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...
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...