Problems with Special Relativity

[JohnMartin]

I have found another problem with satellite orbits.

If you take the equations for orbital velocity around the barycentre in an n-body problem

(i) pv=nrt
(ii) v=ir

which show the relationships between orbital velocity and (i) the orbital radius and time, and (ii) the complex component of the elliptical radius, then

v=ir=nrt/p

so cancelling r

i=nt/p

or

t=pi/n

So for a 3-body interaction such as the sun-earth-moon complex, the orbital time t=pi/3=3.141/3=1.141

but the orbital time of Earth/moon around the earth/moon/sun barycenter is 1 year, not 1.141 years, so G and NM is invalidated.

Am I as incompetent as you think?

JM

[sylas]

Am I as incompetent as you think?

In all seriousness... yes. And you'll never understand how to do basic elementary algebra until you actually recognize that you lack that ability at present. Specifically you need to get past the idea that different equations using the same letters are always using them to represent the same thing. For any given problem, proof or derivation, you need to establish what every one of the variables actually means in that particular problem. At present, you are just relating any two quantities which happen to be represented by the same letter, even though they may be taken from different problems.

Teaching algebra to someone who knows they need to learn is possible. But if you think you understand it, and consider all the published proofs and derivations made over the last century with respect relativity are fundamentally mistaken, and cannot grasp that perhaps the mistakes are yours, you'll continue to be a poster child for invincible ignorance.

Learn the maths first, and with SR it is actually fairly easy (high school level algebra) to derive and prove for yourself all the relations that you are at present thinking you have disproved. You don't need to trust the text books. SR falls out quite easily from first principles once you get the maths right, and use the variables consistently -- as you should for ANY basic maths problem.

[JohnMartin]

A "poster child for invincible ignorance". I wonder if anyone would know who the child is, or would everyone always remain ignorant?

Some good advice in the above post.

JM

[TheLurch]

\I use and rely on special relativity (SR) every day. It works.

Another case where a different aspect of it works with high energy particles:

The muon has a half life of 2.2 milliseconds. Using the speed of light as a limit, they would only rarely travel more than a km without decaying - extremely rarely. Yet we can detect plenty of muons produced by cosmic rays colliding with the atmosphere high above Earth. That's because their high speed allows them to experience time dilation. Time moves more slowly than earth time from their perspective, so their half life gets stretched out, allowing them to travel much further before decaying.

JM, i assume, will take the same approach to this as he did to yours: "I don't understand it, so it must be irrelevant."

[Roy]

Am I as incompetent as you think?

Yes.

I didn't do anything - apart from converting pi to pi - that you haven't done at some point.

[Kbertsche]

I simply do not know anything about the mechanism which you speak of. ...

JM

All that you need to know is that in a particle accelerator, the particles approach a velocity limit of c. If they did not, particle accelerators would not work.

This velocity limit of c is real and verifiable. If you reject SR, how do you explain this?

[37818]

To disprove relativity lets assume it is not true.

We will have one space ship traveling at 1/2 the speed of light traveling between two destinations 1 light year apart.

destination 1 ------------------------------------------------------------------- destination 2.
So they detect each others clocks one year in each other's past.

When the ship > passes destination 1 it will be 1 year before destination 2 sees the ship pass destination 1.
It will take the ship 2 years to reach destination 2.
And destination 1 will not see the ship reach destination 2 for 3 years.

Now when destination 2 sees the ship pass destination 1 it will see the ship arrive at their destination 2 in 1 year, since 1 year has already passed and the ship is half way there.

Now the ship as it passed destination 1 will see destination 1 as only one year latter when it passes destination 2.
And the ship will see 3 years pass as it sees destination 2.

Destination 1 sees the ship's clock after it passes running 2/3 speed. (2 years to travel to 1 year in the observed past. Destination 1 will see the ship arrive at destination 2 in 3 years)
The ship sees destination 1's clock after it passes running at 1/2 speed. (2 years to see what was past 1 year in the past.)

Destination 2 sees the ship's clock running 2 times faster. (2 years pass in 1 year)
And the ship sees destination 2's clock running 1 1/2 times faster. (3 years pass in 2 years)

Now if we plug in the relativistic equations. Both the destinations and the ship see each other's clocks at the same rates. The clock disparity is gone. The traveler's clock was running slower all along.

JM,

Take the time to work this out not assuming SR. As you will find, the traveler sees two destinations clock rates differently than the two destinations see the travelers clock rate. The only way for the traveler and the destinations see the clock rates the same for each other is SR. BTW in SR the travelers clock runs slower. And the time for the traveler to travel between the two destinations 1.732 years instead of the 2. Even though it still takes 2 years between the destinations for the two destinations views.

[shunyadragon]

Am I as incompetent as you think?

JM

Worse, unbelievably worse. Incompetence beyond belief.

It would help if you got a hobby like collecting stamps.

[Kbertsche]

JM, what alternative to SR do you propose? Below is a quote from http://spacetimecentre.org/vpetkov/courses/reid.html. You'll have to go to the link to see his table properly. You'll see that all of the aether theories are disproved by experiment.

[Leonhard]

This thread will be dedicated to exposing problems with Special Relativity Theory.

Problem 1

In Einstein's Paper entitled - On The Electrodynamics of Moving Bodies

By "if we can show that v > c" I think you mean "if we can show that something can have a v > c", as v is just a quantity, and it can have all sorts of values. I think you meant the latter so I'll go with that.

Everything you say is correct up until this point.

so then

L/t = (L'/t')/(1-v²/c²)^1/2

v = v'/(1-v²/c²)^1/2

You made a mistake here. Its right that the length contraction is given by expression you gave but you forgot to take time dilation into account.



After this is done, there is no contraction.

The second and deeper problem is that you don't calculate what a velocity would be in two different frames of reference this way. You can't transform velocities with the lorentz transformation, because a velocity is not a spacetime coordinate, rather its a quantity that indicates direction, and a rate of change of position. This does change between reference frames, but its handled differently.

[Leonhard]

Problem 2

Let an object move from x₁ to x₂

x₁ (same as x'₂)----------------------------------->x₂ (same as x'₁)

Again I assume that (x₁, x₂) and (x^'₁, x^'₂) are two x coordinates, along the same axis, in two different reference frames... and that you mean to say that there is one reference frame where x₁ < x₂ and one where x^'₂ < x^'₁, with x^'₁ = x₂ and x^'₂ = x₁.

I'm afraid that's impossible. There is no lorentz transformation of coordinates that will swap coordinates like that. You're proceeding on a faulty assumption.

We can prove this can't be done. For simplicity I'll only deal with motion along one axis, but extending the proof isn't difficult (left as an exercise).

For the lorentz transformation we have



We'll do the proof by a classical reducto ad absurdum argument. We assume the theorem is false, and if this leads to contradiction, then the theorem must be true.

Postulate: There is a lorentz transformation such that for x₁ < x₂, their values after the transformation will fulfill x'₁ = x₂ and x'₂ = x₁.

From the identities we get



However this is a contradiction, since our starting assumption was that x₁ < x₂. Ergo no lorentz transformation will cause the flip indicated in your example.

[Leonhard]

Problem 3

Let an object move as

x1 (same as x'2)----------------------------------->x2 (same as x'1)

Answered in the same post dealing with Problem 2.

[JohnMartin]

Worse, unbelievably worse. Incompetence beyond belief.

It would help if you got a hobby like collecting stamps.

I gave up on stamps years ago when I couldn't work out why they wouldn't stick to paper. Science seems so much easier.

JM

[JohnMartin]

So far nobody has attempted to resolved problems 1-4.

Interesting.

JM

[DesertBerean]

So far nobody has attempted to resolved problems 1-4.

Interesting.

JM

Considering that your readers, who rarely if ever agreed on anything else, are agreed on your methods of resolving these problems, the issue just might be your understanding or the application thereof.