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Syl & Der

  1. Feb 28, 2010 #1
    Me and my friend Syl consistently talk about Special Relativity and it's effects in nature when electrodynamic bodies move at relativistic speeds. I wanna make a discussion.

    What is Special Relativity?

    How does it work?

    How much do we understand?
     
    Last edited: Feb 28, 2010
  2. jcsd
  3. Feb 28, 2010 #2

    Fredrik

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    It's a theory of space, time and motion, and it's also a framework in which we can formulate theories of matter and interactions. The exact definition of the theory requires some mathematics, and I get the impression that you don't have a mathematical background, so I'll leave out those details. I'll just mention the two ideas that Einstein used to find the special theory of relativity.

    1. The laws of physics are the same in all inertial coordinate systems.
    2. The speed of light is the same in all inertial coordinate systems.

    The pre-relativistic theory (Newtonian mechanics/Galilean spacetime) says that if you're on a train that's moving with a speed of 100 km/h relative to the ground, and you start walking towards the front of the train with a speed of 5 km/h relative to the floor, your speed relative to the ground is 105 km/h. Special relativity on the other hand says that velocities don't add up like that, and in particular, the second of the two statements above says that if you fire a laser on that train, the light is moving at speed c relative to the floor and relative to the ground.

    This is of course very counterintuitive. It even sounds logically impossible at first, but that's just because we all have with some preconceived ideas about space and time, and it's hard not to let those get in the way. When we let those preconceptions go, it's not hard to find a logically consistent theory that incorporates the two ideas mentioned above. The simplest such theory is special relativity. (General relativity is also consistent with the two statements above).

    The spacetime that this theory describes has some other counterintuitive properties. In particular, two inertial observers with different velocities will disagree about
    • which events are simultaneous.
    • the length of an object that's co-moving with one of them
    • the time between two tics of a clock that's co-moving with one of them.
    The reason why two physical observers can measure the same length or time and both be right is that what they're actually measuring are properties of two different curves in spacetime.

    In addition to the above, velocities don't "add" in the usual way. If they're both in the same direction, the rule is

    [tex]w=\frac{u+v}{1+\frac{uv}{c^2}}[/tex]

    In the example with you walking on a train, we have u=100 km/h, v=5 km/h and w very slightly less than 105 km/h.

    The theory is extremely well understood, and has been tested with an enormous accuracy. It is also perfectly consistent with classical electrodynamics (Maxwell's equations), while the Newtonian/Galilean picture of spacetime isn't. The biggest "problem" I'm aware of is that no one has so far proved, or disproved, that there exists a quantum field theory with interactions in 3+1 spacetime dimensions. This is more of a nuisance to mathematicians than to physicists, since there are theories that physicists happily call "quantum field theories" even though they don't quite live up to the mathematicians' standard, because they can be used to calculate probabilities of possible results of experiments more accurately than we can measure them.
     
  4. Feb 28, 2010 #3
    Hi..

    Those are questions on which one can write review papers..maybe you can start with some specific questions....
     
  5. Feb 28, 2010 #4
    What are the equations that describe simultaneity, and hey don't be afraid to throw some math out there. All though I might not be able to grasp things like the Ricci Curvature just yet, I could still derive the equations for Time Dilation & the Lorentz Contraction, so please show me the math, I want the math
     
    Last edited: Feb 28, 2010
  6. Feb 28, 2010 #5
    A consequence of spacetime.
    RE: Lorentz Tranforms.
    All that has been measured experimentally and postulated theoretically.
     
  7. Feb 28, 2010 #6

    Fredrik

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    Let [itex]\Lambda[/itex] be the Lorentz transformation that does a coordinate change from an inertial frame S to an inertial frame S'.

    [tex]\Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}[/tex]

    [tex]\gamma=\frac{1}{\sqrt{1-v^2}}[/tex]

    Let [tex]\begin{pmatrix}t\\ x\end{pmatrix}[/tex] be the coordinates in S of an event that is simultaneous in S' with the origin. (Note that this is the x' axis). We have

    [tex]\Lambda\begin{pmatrix}t\\ x\end{pmatrix}=\begin{pmatrix}0\\ x'\end{pmatrix}[/tex]

    The lower component is irrelevant. The upper one is

    [tex]\gamma(t-vx)=0[/tex]

    and since [itex]\gamma\neq 0[/itex], we have t=vx. Note that this equation defines a line with slope v in a spacetime diagram (with t increasing in the up direction and x increasing to the right).

    This implies that we have the following rule: If the velocity of the spatial origin of S' in S is v, then the t' axis is drawn with slope 1/v, and the x' axis with slope v. And it's not much harder to show any set of events that are simultaneous in S' is a line with slope v in the diagram. Those lines are called simultaneity lines.

    The explicit formula for a Lorentz transformation that I used above follows from a few very naural assumptions about how to make mathematical sense of the ideas in the numbered list in my previous post. Let's start by assuming that a coordinate change between inertial coordinate systems will be a linear function [itex]\Lambda[/itex]. (This is very natural. If these functions don't take straight lines to straight lines, an object that's moving with a constant velocity forever in one inertial coordinate system would be accelerating in another). A linear function can be represented by a matrix, so let's write

    [tex]\Lambda=\begin{pmatrix}a & b\\ c & d\end{pmatrix}[/tex]

    The lines t=x and t=-x represent light rays. #2 on the numbered list in my previous post suggests that we should assume that any point on the line t=x is taken to a point on the same line, and similarly for t=-x. These assumptions imply that there exist numbers [itex]\alpha[/itex] and [itex]\beta[/itex], such that

    [tex]\Lambda\begin{pmatrix}1\\ 1\end{pmatrix}=\alpha\begin{pmatrix}1\\ 1\end{pmatrix}[/tex]
    [tex] \Lambda\begin{pmatrix}1\\ -1\end{pmatrix}=\beta\begin{pmatrix}1\\ -1\end{pmatrix} [/tex]

    It's easy to show that these equations imply a=d, b=c, so that

    [tex]\Lambda=\begin{pmatrix}a & b\\ b & a\end{pmatrix}[/tex]

    Now note that the t' axis, expressed in the coordinates of S, is the line x=vt. This combined with the equivalence of different inertial coordinate systems alluded to in #1 on the numbered list, suggests that we should assume that the t axis expressed in the coordinates of S' is the line x'=-vt'. This means that there's a number [itex]\gamma[/itex] such that

    [tex] \Lambda\begin{pmatrix}1\\ 0\end{pmatrix}=\gamma\begin{pmatrix}1\\ -v\end{pmatrix} [/tex]

    This equation implies [itex]b=-av=-\gamma v[/itex]. We have [itex]a=\gamma[/itex], but this might be a different number for a different v, so I'm redefining [itex]\gamma[/itex] to be a function instead of a number, and I'll write [itex]\gamma(v)[/itex] where I would previously have written [itex]a[/itex] or [itex]\gamma[/itex].

    [tex]\Lambda=\gamma(v)\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}[/tex]

    #1 also suggests that we should assume that we can obtain the inverse just by changing the sign of the velocity.

    [tex]\Lambda^{-1}=\gamma(-v)\begin{pmatrix}1 & v\\ v & 1\end{pmatrix}[/tex]

    So

    [tex]I=\Lambda\Lambda^{-1}=\gamma(v)\gamma(-v)\begin{pmatrix}1-v^2 & 0\\ 0 & 1-v^2\end{pmatrix}[/tex]

    [tex]\gamma(v)\gamma(-v)(1-v^2)=1[/tex]

    But #1 also suggests that we should assume that [itex]\gamma[/itex] is an even function. (Without this assumption, statements about time dilation that two inertial observers make about each other won't be symmetrical). And that gives us the final result

    [tex]\Lambda=\frac{1}{\sqrt{1-v^2}}\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}[/tex]

    If any of this looks difficult because you're unfamiliar with matrices or spacetime diagrams, then I strongly urge you too study those things right away. What I said here can be rephrased without using matrices, but I think it's a terrible idea to go for that option.
     
    Last edited: Feb 28, 2010
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