A Question About The Speed Of Light

[FBM]

OK, what about tachyons? What if they do exist, after all?

[Xamonas Chegwé]

OK, what about tachyons? What if they do exist, after all?

If they exist, I will already answered this question.

[FBM]

OK, what about tachyons? What if they do exist, after all?

If they exist, I will already answered this question.

Edit: Double post.

[JimC]

The speculated existence of tachyons comes from physicists exploring symmetry issues pertaining to the mathematics behind particle theory. Since the positron was predicted by Dirac, and then later confirmed, such speculations became fashionable. This, of course, is no guarantee of their actual existence...

[Tero]

Your wording was confusing. Tachyons are debatable.

New research has dramatically increased the quantity of positrons that experimentalists can produce. Physicists at the Lawrence Livermore National Laboratory in California have used a short, ultra-intense laser to irradiate a millimetre-thick gold target and produce more than 100 billion positrons

[Seth]

Your wording was confusing. Tachyons are debatable.

How do you debate a tachyon, they won't stand still long enough to ask a question?

[Xamonas Chegwé]

Your wording was confusing. Tachyons are debatable.

How do you debate a tachyon, they won't stand still long enough to ask a question?

That question will be already answered.

[JimC]

If we regard the speed of light (c) as an asymptote, ordinary particles approach it (but can never reach it) as they increase their speed from a point below c, whereas our hypothetical tachyons approach (but can never reach) c as they slow down from a speed above c.

[Calilasseia]

Quite simply, inertial frames of reference are frames of reference within which no acceleration is taking place. The coordinate transformations between any two such frames of reference are symmetric in the Lorentz gamma factor terms, and so, an observer in frame #1, taking measurements of entities in frame #2, moving at speed v relative to frame #1, observes entities in frame #2 experiencing time dilation, mass dilation and length contraction from that observer's viewpoint. Likewise, an observer in frame #2, observes the same time dilation, mass dilation and length contraction, from that observer's viewpoint, taking place within frame #1.

Let's see what happens, shall we? I'll refer to this diagram I've constructed handily for the purpose:

By Pythagoras, we have that x2 = L2 + (½vt)2 = L2 + ¾v2t2

However, from the above diagram, we also have that L = ½ct0, therefore we have:

x2 = ¾c2t02 + ¾v2t2

Multiplying throughout by 4, and dividing throughout by c2, we have:

4x2/c2 = t02 + v2t2

But since 2x/c = t, the left hand side is simply t2. Therefore we have:

t2 = t02 + v2t2

Rearranging, we have:

t2 [1-(v2/c2)] = t02

Which with a little more rearrangement, gives us the final formula:

t = t0[1-(v2/c2)]-½

The term [1-(v2/c2)]-½ is the Lorentz Gamma Factor, a function of v.

Because this term is symmetric in v, and is the same for both observers in both frames of reference, then observer #1 in frame #1 will see observer #2's clock running slow, and vice versa. The equations involved for inertial frames of reference involve coordinate transformations that are symmetric in terms of the Lorentz Gamma Factor, and hence, any observer in any arbitrarily chosen inertial frame of reference, will observe the same time dilation in any other frame of reference moving at v relative to the first, and the situation will be symmetric when the reverse choice of viewpoint is made.

The problem with non-inertial frames of reference, those involving acceleration, is that the above simple algebra is no longer sufficient. The four-dimensional coordinate transformations become rather more involved, to the point of requiring sixteen equations of motion to be handled simultaneously. At this point, one has to move to tensor analysis to handle the situation adequately, and the compactness of the notation allows all 16 equations of motion to be treated as a cohesive unit, that unit obeying certain laws of tensor transformation, again expressible in compact notation. What has to be remembered, however, is that any tensor equation with free indices actually represents a system of equations. If the number of free indices in the equation is m, and the number of dimensions of the space is n, then that tensor equation compactly represents a total of m×n equations in the one symbolic representation. And this, ladies and gentlemen, is why general relativity tends to be bloody hard. That's before you get into the nuts and bolts of the Ricci Calculus, which has its own conventions to simplify certain matters, some of which only make sense after you've spent two years or so mastering the intricacies of the underlying tensor analysis.

Oh, and none of the above covers orientation entanglement, a feature of spaces with dimension greater than 2. To add that into one's understanding of the system, one then has to introduce spinor terms. Spinor calculus is not for those with a weak grasp of mathematics.

[JimC]

I have shamelessly copied the above post to assist my brighter students to understand relativity. What would be even better is to have Cali as a guest lecturer for my lads, if you ever pop over to Oz (I can guarantee some decent Aussie Shiraz as a reward...)

[mistermack]

Quite simply,

You started well, but after that, it was downhill all the way.

[JimC]

Quite simply,

You started well, but after that, it was downhill all the way.

[Hermit]

Quite simply,

You started well, but after that, it was downhill all the way.

Was not! It's all relative, you know?

[mistermack]

Quite simply,

You started well, but after that, it was downhill all the way.

Was not! It's all relative, you know?

You could try and treat the subject with a little more gravity.

[Hermit]

Quite simply,

You started well, but after that, it was downhill all the way.

Was not! It's all relative, you know?

You could try and treat the subject with a little more gravity.

How do you know I've not? You don't. At least not until you open the box.