Refutation of Special Relativity for Dummies

[Darwin123]

Therefore, physicists DON'T drop the first order effect in either distance (x) or time (t). They only drop effects that are second order or higher when they take the Galilean limit. However, you apparently haven't thought out which effect is which. With what expansion are the effects in any order?

The expansions are usually made in terms of 'v/c'. In your units where 'c=1', 'v'. Okay, let 'c=1'. The Galilean limit neglects terms like v^2 and (vx). We do not ignore first order terms like 'v' or 'vx'.

The expansions that you are obviously thinking of are Taylor expansions in terms of 'v'. The other quantities in the Lorentz transformation of independent of 'v'. So both 'v' and 'vx' are first order, not second order.

What disappears is terms like 'v^2'. So the Galilean approximation is really,
sqrt(1-v^2)=1.

The reason one that the 'vx/c^2' term disappears is because for the observer on earth looking at the rocket, 'x=vt'. This is subjective. So in this case:

xv=(vt)t
(vt)t=t v^2
xv =t v^2

Thus, 'xv' is second order in 'v' for this observation only.

Let me elaborate a little bit farther.

You chose units where c=1. Fine. Therefore, v<1 because a signal can't travel faster than the speed of light. So:
v<1
v^2<v
t v^2 < t v

Now, if the observer of earth (x=0) is looking at the rocket after a time t,
X=vt where X is the turn around distance. Substitution into the last inequality expression shows that
vx <vt.
The earth observer is at x=0

The complete Lorentz transform with c=1 is
x'=(-vt)/sqrt(1-v^2)
t'=(t-vx)/sqrt(1-v^2)

Suppose the speed of the rocket is MUCH less than the speed of light, so v<<<1. Because v is much less than 1, we can use the approximation v^2<<v.

x'=-vt.
t'=t.

So, at turn around, the earth twin sees.
x'=X
t'=t

These are the Galilean equations. So if v<<1, the Lorentz transformation becomes the Galilean equations. There is no inconsistency in the Galilean limit.


If you want to know how the 'a' term survives, then we have start with the calculus. We have to subtract quantities very close together in time and space. This has nothing to do with the Galilean limit of the Lorentz transformation.

Your concrete error is still the same. You are hypothesizing that the proper acceleration of the rocket and earth is negligible. The proper acceleration is not negligible even if it is intense for only a very short time.

 

[wogoga]

The vx/c² time-shift as a first-order effect also exposes as a myth the claim that the Lorentz transformation turns into the Galilei Transformation, if v << c:

"For relative speeds much less than the speed of light, the Lorentz transformations reduce to the Galilean transformation in accordance with the correspondence principle." (Wikipedia)​

By dropping the Lorentz-factor as a second-order effect, the Lorentz transformation does not reduce to the Galilean transformation:

x' = x - v t . . . t' = t​

Instead, the Lorentz transformation (here with c = 1) reduces to:

x' = x - v t . . . t' = t - v x​

If we drop the first-order effect vx of the time transformation, then we must also drop the first-order effect vt of the x-transformation, and as result we get the "zero-order" transformation:

x' = x . . . t' = t​

The reason one that the vx/c² term disappears is because for the observer on earth looking at the rocket, x=vt.

It is difficult to make full sense of your posts #119 and #121, but it is clear that you try to rescue the derivability of the Galilean transformation from the Lorentz transformation, by finding a justification for x = 0 in the "reduced" time transformation t' = t - v/c² x. The center x' = 0 of the moving reference-frame F' obviously remains at the center of itself (despite moving according to x = v t in rest-frame F). So you get t' = t from t' = t - v/c² x by replacing x with x' = 0. This resolves the problem at most in a very superficial way, and in a very special case.

In modern physics, the epistemological principle that one single counter-example is enough to refute a theory has been reversed into its opposite: Theories get generally accepted if they are confirmed in very special cases.

By the way: In the future the question whether 'neutrinos' have 'mass' (in the sense of somehow participating in the God-particle constructed by Higgs) will be considered as relevant as past disputes over Holy Trinity (which I do not disdain). Progress is always unfair to the old ways.
Cheers, Wolfgang
www.pandualism.com

 

[Darwin123]

It is difficult to make full sense of your posts #119 and #121, but it is clear that you try to rescue the derivability of the Galilean transformation from the Lorentz transformation, by finding a justification for x = 0 in the "reduced" time transformation t' = t - v/c² x. The center x' = 0 of the moving reference-frame F' obviously remains at the center of itself (despite moving according to x = v t in rest-frame F). So you get t' = t from t' = t - v/c² x by replacing x with x' = 0. This resolves the problem at most in a very superficial way, and in a very special case.

You brought up this 'very special case'. You claimed that the Lorentz transformation is not equivalent to the Galilean limit in what is usually thought of as the Galilean limit (v<<1). You were claiming that this special case was an 'obvious' counterexample. In fact, you were the one who brought up the so called Twin Paradox' which has always been considered a special case. You provided a series of expressions leading to a supposed error by the Einsteinians.

I pointed out the concrete error in your logic. I wrote out the inequality statements necessary to show that the Galilean transform IS the low velocity limit of the Lorentz transform. I pointed out the logical consequences of choosing a velocity for the rocket that is very small. Furthermore, I have pointed out your concrete error in each special case that you have presented in this thread.

You followed your erroneous proof with ad hominem statements. You followed your erroneous proof claiming that physicists were wrong because they never question their 'religion'. You did not even attempt to show how my analysis was logically wrong.

You present mathematical expressions followed by a demand for the rest of us to avoid mathematics. You consistently make mistakes where ever the concept of limit is used. You didn't know that Newton worked with polygons. BTW: Trapezoids are polygons. You reject all materialist philosophy after psoing a materialist question!

In modern physics, the epistemological principle that one single counter-example is enough to refute a theory has been reversed into its opposite: Theories get generally accepted if they are confirmed in very special cases.

By the way: In the future the question whether 'neutrinos' have 'mass' (in the sense of somehow participating in the God-particle constructed by Higgs) will be considered as relevant as past disputes over Holy Trinity (which I do not disdain). Progress is always unfair to the old ways.
This is an obvious non sequitur. This is just a statement to change the subject without admitting that you are wrong. This is an unfalsifiable statement based on a topic (the Higgs boson) that you know nothing about. This brings up a completely irrelevant issue (the Holy Trinity) that the majority of us are uninterested in.

Your non sequitur shows that you can't find a logical error in what I wrote. I showed that the Galilean limit is numerically a good approximation for the Lorentz transformation under certain conditions which include one observer moving slowly (c=1>>>v) with respect to a second observer.

You have not addressed my concrete rebuttal at all. I presented you with the two horns of a dilemma. You choose to hit between the eyes. This may 'win' a debate, but it doesn't resolve the dilemma.


You were never serious about your gambit. You never intended to explain how rebuttals to your 'concrete errors' are wrong.

Question:

How can you attack Maxwell's physics when you don't even understand Newtonian physics?

I look forward to your attacks on Newton's legacy. Newton is, after all, a hero of physicists everywhere. Newton has been the idol of scientists for a lot longer than both Maxwell and Einstein!

I have my hardcopy of Principia. Someone else posted a link to an online copy of Principia! So please attack Newton! Lets see what you got!

 

[Reality Check]

It is difficult to make full sense of your posts #119 and #121, but it is clear that you try to rescue the derivability of the Galilean transformation from the Lorentz transformation, ...

No, wogoga. It is basic mathematics that shows that the Galilean transformation is a limit of the Lorentz transformation.

1. The Lorentz transformation can be expanded in a power series.
2. When v/c is much less than 1 then the higher powers in the series can be neglected.
3. That gives the Galilean transformation !

A fantasy that the Lorentz transformation does not reduce to the Galilean transformation is wrong. It obviously does in the limit of small v/c.

 

[Reality Check]

By the way: In the future the question whether 'neutrinos' have 'mass' (in the sense of somehow participating in the God-particle constructed by Higgs) ...

By the way, wogoga: In the present

• The Nobel prize has just been given to the people who measured the neutrino oscillations that exist if neutrinos have mass.
• The Higgs boson exists.
• The Higgs boson does not give mass to neutrinos. It gives mass to gauge bosons (photons, W and Z bosons, and gluons).
  It is interaction with the Higgs field that gives mass to fermions. But it is not certain whether neutrinos will interact with the Higgs field.

In the future

• Neutrinos will still have mass. As will electrons, protons, etc.
• The Higgs boson will still exist.
• The Higgs boson will still not give mass to neutrinos.

 

[Darwin123]

x = 0[/I][/B] in the "reduced" time transformation t' = t - v/c² x. The center x' = 0 of the moving reference-frame F' obviously remains at the center of itself (despite moving according to x = v t in rest-frame F). So you get t' = t from t' = t - v/c² x by replacing x with x' = 0. This resolves the problem at most in a very superficial way, and in a very special case.

BTW: The Lorentz transform does not include 't'=t-v/c^2'. It includes 't'=t-vx/c^2'.

Also, I thought that you wanted to choose units where 'c=1'. If you still want to do that, then you should write 't'=t-vx'.

In your units, If v<<1 then vx<<vt.

There is nothing 'superficial' about it. You made a fundamental error. You implied that vt>vx. This is 'obviously' not true.

You don't understand what I wrote because you don't understand basic mathematics.

I conjecture that you couldn't follow Newton's Principia because you don't understand Euclidean geometry, either. I would enjoy hearing your critique about Newtonian physics.

Cheers! And have a nice day !

 

[slyjoe]

Have we seen any calculus? Introduction to limits may have been useful to the OP.

 

[Darwin123]

Have we seen any calculus? Introduction to limits may have been useful to the OP.

It depends on what you think are the boundaries of calculus. You probably think of calculus as anything that involves the analytical operations of differentiation and integration. However, these two concepts don't make any sense without the concept of limits.

Newton is often said to have invented calculus. However, he didn't have the formalism that has developed concerning differentiation and integration. His famous book, Principia, does not have symbols for differentiation and integration. Newton uses figures to explain his concepts, not symbols. Newton relies on Euclidean geometry to analyze motion. He introduces the concept of infinitesimal without actually naming it.

Newton represents curved trajectories as polygons. Most of the polygons were trapezoids with infinitesimal faces. The OP objected to Einstein's use of polygons. So basically he was protesting the use of calculus.

We briefly discussed infinitesimals without the convenient formalism. Wogaga pointed out that the Kepler used the concept. Infinitesimals according to many people defines calculus.

The concept of limit is usually taught early in the first semester of calculus. Furthermore, the intermediate and advanced courses in calculus usually emphasize limits. There is less emphasis on the algorithms used Therefore, I think of calculus as anything involving the concept of limit.

Moderator, am I permitted to name the OP? I know the OP names me. I don't mind.

A lot of people have difficulty with the concept of limit. I think the educator and psychologist, Piaget, characterized this sort of difficulty. Piaget char Limit is an abstract concept involving formal logic. Most people get by without using abstract concepts or formal logic. They are very good with concrete concepts but not formal concepts.

Many of the science critics probably have difficult with abstract concepts, including the OP. Some of them get the idea that 'real science' consists ONLY of concrete concepts.

The concrete approach constitutes much of science. The concrete approach is faster than the formal approach. However, the concrete approach is more ambiguous. One picture may be worth a thousand words, but 20% of those words are wrong! So it is worth while to say a few words AFTER one draws the picture.

Complaining about people who can reason both ways (like Einstein and Maxwell) is not constructive. I think this is what the OP is doing. He is having difficulty understanding ALL formal concepts, not just relativity. His Einstein bashing is just a call for help !

There are a lot of people viewing this thread. The ratio of views to posts is close to 55. So a lot of people aren't saying anything but are interested.

My conjecture is that many viewers are interested because they either have the same difficulty as the OP, or they are trying to teach somebody with this difficulty. I think it would be useful for these viewers to see where basic abstract reasoning can clarify a scientific problem.

I think concrete reasoning is the basis of all science. However, advances in science also require some abstract concepts. I want to point out to the viewers the importance of formal concepts.

Like limit. Limit is a very important concept in applied mathematics. It is abstract but key to many physical applications.

Maybe I should start a Piaget thread. However, I am a physicist not a psychologist. Maybe someone working in cognitive psychology should start the discussion !

 

[slyjoe]

...

The concept of limit is usually taught early in the first semester of calculus. Furthermore, the intermediate and advanced courses in calculus usually emphasize limits. There is less emphasis on the algorithms used Therefore, I think of calculus as anything involving the concept of limit.

As do I. I think the problem of limits in the OP's view was brought clearer to me in post 122 where in the limit you get the Galilean equations. OP didn't seem to get that.

 

[Perpetual Student]

wogoga:

By dropping the Lorentz-factor as a second-order effect, the Lorentz transformation does not reduce to the Galilean transformation:
x' = x - v t . . . t' = t
Instead, the Lorentz transformation (here with c = 1) reduces to:
x' = x - v t . . . t' = t - v x
If we drop the first-order effect vx of the time transformation, then we must also drop the first-order effect vt of the x-transformation, and as result we get the "zero-order" transformation:
x' = x . . . t' = t

This is incorrect. We have:

x' = γ(x - vt) and t' = γ(t - vx/c^2)

With c = 1, the v in x' = x - vt is the actual velocity in the x direction. however, because of the c^2 term in the denominator with vx, where t' = vx/c^2, v is the ratio of the actual velocity to c. If you do a dimensional analysis of the two equations, you will see that MUST be the case. To be clear, vx alone is NOT compatible with t, whereas vt is compatible with x.

 

[Perpetual Student]

The above is a fundamantal error that exposes your lack of understanding. As I have advised before, stop wasting your time like this; a real study of physics is much more rewarding. There are texts, papers, videos -- all available on the internet to assist you.

 

[Darwin123]

By dropping the Lorentz-factor as a second-order effect, the Lorentz transformation does not reduce to the Galilean transformation:

x' = x - v t . . . t' = t​

Instead, the Lorentz transformation (here with c = 1) reduces to:

x' = x - v t . . . t' = t - v x​

If we drop the first-order effect vx of the time transformation, then we must also drop the first-order effect vt of the x-transformation, and as result we get the "zero-order" transformation:

x' = x . . . t' = t​

You are describing the zeroeth order limit. In this limit, the other coordinate system doesn't move at all. The rocket remains stationary with the earth observer.

I will refer to the zeroeth order approximation as the 'Zeno limit'. Basically, you are claiming that the rocket can't move a finite distance because every infinitesimal distance is effectively zero length. In the Zeno limit, all variation associated with the velocity 'v' is ignored. This is approximation is physically unreasonable, as 2400 years of research has shown.

The Zeno paradoxes preceded the twin paradox by about 2400 years. You are confusing the Zeno paradox which doesn't concern relativity with the twin paradox that does involve relativity.

The first order approximation is the Galilean limit. This Galilean transformations describe all Newtonian dynamics. In the first order approximation, all terms of order v are kept. All terms of order v^2 are ignored.

The second order approximation can be called the Lorentzian limit. In this limit, the dynamics of special relativity are valid.


Some others posted your error with regard to units. Basically, you are conflating seconds with light seconds. As c=1, the time unit is seconds and the length unit is light seconds. They are different units.

They can't be treated the same in relativity because time is different from space in the metric.
ds^2=dx^2+dy^2+dz^2-(c dt)^2

If c=1, then
ds^2=dx^2+dy^2+dz^2-dt^2

Note that dt is different from the other increments even when c=1.

The minus sign makes time different. If the minus sign weren't there, then maybe you could get away with using the same units for length and time.


Never the less, your problem doesn't seem to be with relativity per se. You seem to have a problem with infinitesimals.

You said that Einstein made a mistake. Einstein said that you can replace a curve with a polygon with an infinite number of line segments. You blamed scientists of mindlessly agreeing with Einstein on this matter.

This approximation was used a long time before Einstein was even born. It was in fact thought out a few decades before the trial of Galileo. The Catholic Church banned this approximation. However, it was well established by the time Newton wrote Principia. BTW: Newton uses this approximation many, many times in Principia.

So this is a concrete error on your part. Curves can be replaced by polygons with an 'infinite' number of sides. The mathematical manipulation of infinity is referred to as calculus.

You shouldn't blame Einstein for infinitesimals. Actually, you are now making TWO concrete errors.

1) Einstein was not the first person to approximate a curve by a polygon of infinite sides.

2) The replacement of a curve by a polygon of infinite sides is a completely valid hypothesis central to science.

 

[fuelair]

As to the refutation, the dummy's are not the ones who do not accept it for the usual suspect reasons!!!!!

 

[fuelair]

Thus of course, those who do.............

 

[wogoga]

By dropping the Lorentz-factor as a second-order effect, the Lorentz transformation does not reduce to the Galilean transformation:

x' = x - v t . . . t' = t​

Instead, the Lorentz transformation (here with c = 1) reduces to:

x' = x - v t . . . t' = t - v x​

This is incorrect. We have:

x' = γ(x - vt) and t' = γ(t - vx/c^2)

With c = 1, the v in x' = x - vt is the actual velocity in the x direction. However, because of the c^2 term in the denominator with vx, where t' = vx/c^2, v is the ratio of the actual velocity to c. If you do a dimensional analysis of the two equations, you will see that MUST be the case. To be clear, vx alone is NOT compatible with t, whereas vt is compatible with x.

I agree that "dimensional analysis" is a prerequisite for every reasonable concept in physics. Yet in our case the situation is so clear that no danger can arise from the implicit use of a length-unit and a time-unit connected by c = 1 (as applied in #75)

By using t' = t - v x instead of t' = t - v/c² x I wanted to stress the symmetry between time and length transformation. Exactly this symmetry explains the constancy of c in the Lorentz transformation (see Simple Derivation of the Lorentz Transformation)

Some may even argue that v/c² x can be removed as a term from the first-order limit because the constant c is as high as 3*10⁸ m/s. Yet it is obvious that the numerical value of c only depends on the units meter and second. If we use light-year and second then the value of c becomes as low as 3.17*10^-8 light-year/sec.

Have we seen any calculus? Introduction to limits may have been useful to the OP.

Instead of joining the ranks of clueless pack followers, you could have used software for symbolic computation and tried yourself to check what corresponds to the mathematical reality. Here the problem formulated with Mathematica-syntax:

γ = 1/Sqrt[1–v^2/c^2]
Series [γ (x - v t), {v, 0, n}]
Series [γ (t - v/c^2 x), {v, 0, n}]​

If n = 0, the result is:

x' = x . . . t' = t​

For the first-order series (with n = 1) we get:

x' = x - t v . . . t' = t - x/c² v​

The second-order result:

x - t v + x/2/c² v² . . . t - x/c² v + t/2/c² v²​

The only possibility to get the Galilean transformation is taking the first order Taylor-approximation for the x-coordinate and the zeroth order approximation for the t-coordinate.

Does anybody know who is responsible for spreading the myth that the Lorentz transformation reduces to the Galilean transformation? I assume and hope it was not Einstein.

Cheers, Wolfgang

 

[tusenfem]

Does anybody know who is responsible for spreading the myth that the Lorentz transformation reduces to the Galilean transformation? I assume and hope it was not Einstein.

Seems more that through bad math quailifications YOU are spreading the myth that the Lorentz transform DOES NOT reduced to the Galilean transform.

 

[Darwin123]

Some may even argue that v/c² x can be removed as a term from the first-order limit because the constant c is as high as 3*10⁸ m/s. Yet it is obvious that the numerical value of c only depends on the units meter and second. If we use light-year and second then the value of c becomes as low as 3.17*10^-8 light-year/sec.

The relevant expression is valid independent of the units used. What matters is the inequality expressions valid in the problem.

The hypothesis is:
v<c

Dividing both sides by c, whatever the units, gives us:
v/c <1.

If the above is correct, then one can expand out all physical quantities as a Taylor series in v/c. The series converges even if the number of coefficients is infinity.

The units of 'c' don't effect the units of the coefficients of this Taylor series. It doesn't matter if c is in light seconds per second (ls/s), ly/y, m/s or km/s.

When you said c=1, most of us thought that the units were seconds (s) for time and light seconds (ly) for length. The speed of light, c, in ly/s is very small. Maybe c= 3x10^-8 ly/s? However, this doesn't change the basic fact that v<c. One simply says that
v<3x10^-8 ly/s.

So let me write out a few variations:
v<3x10^8 m/s
v<3x10^5 km/s
v<1 ly/y
v< 1 ls/s
v < 3x10^-8 ly/s

The fact that v/c<1 is what makes the infinite series converge. As long as v<c, then there is a significant difference between zero order, first order and second order terms.

Okay, let us assume c=1 and then drop the units for v. If c=1, then this is equivalent to saying that,
v<1.

If the above expression is correct, then,
v^2<v.

If m<n, then
v^m < v^n

If we decide to take the terms of the expansion to mth order, then we can drop the terms v^n.

This is far more fundamental than relativity.

I recommend that you review the rules of limits and significant figures before you read any more science documents. Who knows, you may enjoy them once you review them!

 

[Darwin123]

The only possibility to get the Galilean transformation is taking the first order Taylor-approximation for the x-coordinate and the zeroth order approximation for the t-coordinate.

If 0<m<n, and if |v|<c, then
|v/c|^m<|v/c|^n.

Note that the units of v cancel out the units of c. So if this is true in one set of units, then it will be true in other units.

If the above expression is true, then the Lorentz transformation reduces to the Galilean transformation to first order. Therefore, if |v|<<c, then the Lorentz transform reduces to the Galilean transform to first order.

This can be verified by anyone who understands the concept of limits and the idea of units. I came up with those steps myself. Anyone who knows the concept of limit could derive this on his own. Strangely, you insist on giving Einstein the credit for these basic concepts.

Now you are giving Einstein far TOO MUCH credit. If Einstein had come up with the idea of limit, then he would have been the greatest mathematician of the last two millennia. I would never give Einstein the credit for either the 'Taylor series' or even for 'infinitesimals'.

Please note that what I call the 'Taylor series'. We do not call it the 'Einstein series'.

Further, my American and American colleagues call Newton the father of calculus. Some of our German friends like to call Liebnetz the father of calculus. However, none of us call Einstein the father of calculus.

Einstein did not invent calculus. He wasn't smart enough to invent calculus. He only used it. Maybe he stole it! If so, he stole something that has proven of great use to a lot of scientists everywhere!

If you want, then you can call Einstein 'the bastard grandson of calculus'. After all, he later helped to develop tensor calculus and differential geometry. Never the less, these are just extensions of that old time differential calculus. Which is what I conjecture you are missing.

What made you decide that Einstein or Maxwell caused materialist philosophy? I would blame 'materialist philosophy' on the linear combination of Galileo, Newton and Leibnitz.

 

[Reality Check]

Does anybody know who is responsible for spreading the myth that the Lorentz transformation reduces to the Galilean transformation?

That would be the maybe millions of students who have derived that Lorentz transformation reduces to the Galilean transformation for v << c using basic mathematics, wogoga.
The "myth" bit needs a

 

[Perpetual Student]

I agree that "dimensional analysis" is a prerequisite for every reasonable concept in physics. Yet in our case the situation is so clear that no danger can arise from the implicit use of a length-unit and a time-unit connected by c = 1 (as applied in #75)

By using t' = t - v x instead of t' = t - v/c² x I wanted to stress the symmetry between time and length transformation. Exactly this symmetry explains the constancy of c in the Lorentz transformation (see Simple Derivation of the Lorentz Transformation)

Some may even argue that v/c² x can be removed as a term from the first-order limit because the constant c is as high as 3*10⁸ m/s. Yet it is obvious that the numerical value of c only depends on the units meter and second. If we use light-year and second then the value of c becomes as low as 3.17*10^-8 light-year/sec.










Cheers, Wolfgang

You're trying too hard to win an argument and not trying hard enough to understand it.
Let's take a different approach, no calculus, just simple algebra and arithmetic:

x' = γ( x - vt)
t' = γ(t - vx/c^2)

In considering the orders of magnitude below, I have (at times) ignored factors of 3 and .6, etc.

Consider that when c is set to equal 1, we are setting 3x10^8 meters to one length unit. (i.e. one meter is approximately 3x10^-8 length units.) So, the fraction vx/c^2 for a slow velocity like .6 kilometers per minute (10 meters/second = about 10^-7 length units/sec.) and for one meter of x equals 10^-7*10^-8/10^16sec., which is about 10^-31 seconds, in other words vanishingly small or infinitesimal. Contrast this to the value of vt in the equation for x', which is about .6x10^-8. The difference is 23 orders of magnitude!

It is a similar consideration that brings us to consider the γ function to be 1 for slow velocities, which in the above example would be about the square root of 1 minus 10^-30.