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SporadicFire

Special Relativity

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SporadicFire    162
Ok, so I have been looking at special relativity in past few weeks, and I have noticed some weird things. Don''t worry, nothing to do with homework. so, we have Lorentz transforms: x'' = lambda(v)*(x - v*t) t'' = lambda(v)*(t - v*x/(c*c)) so, I consider the time dilation aspect, and I find this on the net. It all seems to make sense in terms of what it says: x'' = ct'' x''^2 + v^2*t^2 = c^2*t^2 (Pythagorean theorem) c^2*t''^2 = c^2*t^2 - v^2*t^2 t'' = t*lambda(v) which seems well and good except: t'' = lambda(v) * (1 - v/c)*t which means, 1-v/c =1 => v=0 which can''t be! So whats the problem? Am I misunderstanding something? Other than that, if two objects are moving relative to each other at a velocity, then without using any "external" knowledge we know only 2 things: 1. The magnitude of their relative velocity. 2. A sign indicating weither they are heading towards or away from each other. We do not know the DIFFERENCE between their velocities. The traditional relative velocity calculation involving subtraction of two velocities is not available in the vacuum. Because, both the velocities have to be computed relative to the ONLY other object. And, this is one thing that special relativity does not seem to take into account. IMHO, if this were taken into account, there is no possible way to say which object faces time dilation. Please correct me if I am wrong (which I probably am, considering I am not a genius). SporadicFire.

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SporadicFire    162
From what I gather, it is one of Einstein''s assumptions
actually... the speed
of light is the same in all frames of reference. This
implies both:

x=ct
and x=ct''

SporadicFire

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szinkopa    198
But I think x is the position of an object, not the light, and v is the velocity of the same object. And the light''s velocity is c. So I think x=c*t is wrong.
But ask the physicist (13rd post) in the gravity topic.

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SporadicFire    162
Actually, I am quite certain of this. x is the position of
the light wave with respect to the first frame of reference,
and x'' is the position of the light wave with respect to
the second frame of reference. The definition is the
same for both the lorentz transform and the stuff that
I found on the web. Hopefully the physicists will soon be
attracted to this topic.

SporadicFire

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Guest Anonymous Poster   
Guest Anonymous Poster
check your work more carefully, make sure you''re not making a <i><b>stupid</i></b> assumption

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Mastaba    761
x is the position of an event in a linear frame of reference which has velocity v. It is NOT the position of light. x'' is the position of the same event in a different linear frame of reference which has velocity v''.

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Codexus    118
I recommend this book:
"Introduction to the Relativity Principle"
by G. Barton


It's really easy to understand and it changed my view of the universe. The math is quite simple and the book will guide you to the same conclusions Einstein discovered. Once you understand special relativity it's amazing how simple and logical it is.


[edited by - Codexus on May 25, 2004 9:33:38 AM]

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SporadicFire    162
Its
unfortunate that a lot of the material I have run into
so far is more interested in teaching the theory then providing
its mathematical foundations (like proper use and meaning of
Lorentz transforms).

Thanks for the book recommendations/site recommendations.
It will greatly help my persuit for the understanding.

But, just out of curiosity, how does Einstein resolve the other
issue of which frame of reference is moving and which one
is still? and you know, in same line of questioning, what is
meant by rest mass?

SporadicFire.

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Codexus    118
The whole point is that you can't tell which frame of reference is "moving". That is the law of physics stay the same no matter which inertial frame of reference you use. That was already true with classic galilean motion. But then the problem of the observed (and calculated from Maxwell's laws of electromagnetism) fixed speed of light was threatening that notion. As it seemed, it would have to be relative to some universal reference frame.

Einstein's special relativity manages to show that the relativity principle is compatible with a fixed speed of light. But that means the universe works a bit differently than thought before ^^
Rest mass is simply the mass of an object as observed in an inertial frame of reference where it speed is zero.


[edited by - Codexus on May 25, 2004 10:17:06 AM]

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SporadicFire    162
Thanks for all the help. It has made the concepts a little
clearer.

I am still caught in this paradox:

If you cannot tell which frame of reference is moving...then
how do you tell which one is effected by time dilation/length
contraction? Is there a difference between the two frames that
I dont see that causes this?

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Codexus    118
Both frames of reference are equally affected.
You confusion probably come from what I call the space traveller problem. Some guy leaves earth to go explore our neighbour stars travelling at near the speed of light. When he comes back he''s only 1 year older while everyone on earth is 20 years older.
Could we not see this problem relative to the space traveler? The earth is the one moving away while he stays still and as such he should be the one 20 years older?
There is a catch here, special relativity only applies to inertial frames, frames of reference that are not accelerating. This hypothical situation isn''t symetrical as it is the traveller which accelerates. While he''s moving at constant velocity, time seems to pass slower on earth from his point of view, just the same as time seems to slow down on his spaceship from the point of view of an earth observer. But when he starts decelerating to go back to earth, from his point of view time on earth seems to go extra fast.
General relativity has to be used to solve problems involving accelerating bodies.

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SporadicFire    162
Ok...thanks alot for all your help.

So, let me bounce this off you. The effects of special relativity are witnessed when looking at one observer
from the other. So, it works both ways. But, it does
not apply to an accelerating body.

I think I understand it alot better
now. So, since accelaration is beyond the domain of special
relativity, all effects of special rel are temporary? (as in,
when relative velocity becomes zero, every thing seems normal
again, and all clocks show the same time, and all rods that
were same length measure the same).

SporadicFire.

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Squirm    481
One of the problems with special relativity is that the maths is far easier than the understanding, hence the huge number of ''proofs'' that it is wrong, based on misunderstandings, and usually written in a fit of pure rage ...

You are a breath of fresh air when you say you think your wrong and just don''t know why :o)

I agree with the other posters : x'' = ct'' ? The coordinate is the speed of light times time? huh?

You might try sci.physics.relativity rather than a games forum, although if it hasn''t changed since I was last there expect 2,000 new articles every day, 90% of which fit into the category at the top of my post ...

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SporadicFire    162
Oh...sorry, lost my common sense there. Since it doesnt
apply to accelerating bodies, there is no way to actually
change the relative velocity between two frames of reference.

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