# Cut The Bullshit In Physics

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### #256 sluggo

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Posted 12 May 2019 - 12:09 PM

MikeBrace#243

This is clearly impossible.   How could we obtain the same fixed valued for C whether the detector is stationary, or when its moving with the lights at any speed, and also even when moving in the exact opposite direction to the light vector at any speed?

When a plane is flying at a constant speed v, in a straight line, without turbulence, Ann can toss a ball across the aisle at 90 deg, to Bob on the opposite side. If the act was recorded on video, and filmed in the plane parked on the ground, there would be no detectable difference. The reason: In the air, all elements in the plane have mass which acquires the plane speed v.
If in the same scenario, Ann directs a light beam at Bob, light has no mass to acquire the speed of the plane, but it does have momentum that directs it forward, while moving at c.
This results in a tiny insignificant delay before reaching Bob, but in a high speed transportation method, the delay becomes more significant.
The propagation speed is c in a vacuum and before SR was known to be slower in denser mediums which can be calculated using a correction factor labeled the index of refraction. You have to know some history.

### #257 sluggo

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Posted 12 May 2019 - 12:17 PM

Sherwood#245

It is also moving at different speeds past various other observers such as observers D,G,Q,M,K, etc., etc., who are probably moving at different speeds with respect to the clock and each other.  Thus the clock has to read different times with respect to each that is at a different velocity.  Difficult for one clock.

[The motion of an observer does not affect the rates of distant clocks or the distance to those clocks.]

'Time dilation' exists only in the mathematics of SR, not in reality.  Light starts out in a spherical direction from a source at speed c.  (If light 'travels at speed c in all reference frames', it certainly travels at speed c from its source.)  Something (object) moves with respect to the light source, call it in the horizontal direction.  Then the 'vertical' (perpendicular) portion of the light has moved distance (I=ct) from the source at speed c for time t.  Meanwhile, the object has moved the horizontal distance x'. So now the light is the vertical distance I plus the horizontal distance x' from the object for a total distance of

H=(x2+I2)1/2

[A sphere of light would be a composite photon emission in all directions. In a lab experiment, light pulses consisting of even single photons can be directed along selected paths to detectors. If a single photon was represented as a sphere, then a group of mirrors on a ceiling would all reflect from an emission from the floor. If the photon energy was E, then the total energy of the reflections could be made arbitrarily large by adding more mirrors. That violates the conservation of energy.]

The light clock shows time dilation in its simplest form.
The clock is an EmitterDetector and a mirror M separated by a distance of 1 unit.
In fig.1 Ann records a photon moving to M, for a half cycle, At=.5.
In fig.2 an identical B-clock moves past Ann at .6c. The photon moves the same distance through space in the same A-time At=.5, but the photon is only 80% to its mirror M.
The reason: the photon has to divide its energy with an x component vt compensating for the clock motion, and a p component ut which becomes the active part of the clock. The geometry results in

u=sqrt(c^2-v^2) or u/c=1/gamma

Light speed within the clock <c, thus the B-clock process runs slower then that of the A-clock, as perceived by A. Adding the fact that Bob's biological clock also runs at 80%, Bob accepts his clock as correct, and concludes c' = x'/t' = (x/g)/(g/t) = x/t = c.

An observer cannot move relative to himself, thus his motion vt masks the x component of light and he is only aware of the p component, as in fig.3. It is obvious from fig.2 that constant and independent light speed (postulate 2) is responsible for the scaling effect, which alters the moving observers perception of his world.
In Newtonian physics, time was universal, independent of motion, therefore the effects of SR cannot be derived from Newton, but variables can always be converted.

### #258 ralfcis

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Posted 12 May 2019 - 02:02 PM

Well why didn't you say so years ago? That is one simple, irrefutable mathematical proof of length contraction . . . except for one tiny problem. It's a circular argument. It is based on the assumption that length contraction is real and is the only possible explanation hence it must be the only possible explanation. Since there can't be another possible explanation, why even look at another possible explanation that is only based on time dilation. All of relativity is based on assumptions that are borne out by the results so relativity must be correct. How else can c remain constant from all perspectives if time dilation doesn't work in tandem with length contraction as Einstein said so. QED ridiculous. Instead of endlessly repeating these circular arguments, show me where my math goes wrong that has an alternate explanation for length contraction. (You can wait until I wrap up the specific proof for the light clock and interferometer.) I say there's no need for an x'-axis (which is the same thing as relativity of simultaneity or  time dilation from the moving perspective). As uncle Al said, a theory must be as simple as possible but no simpler so if everything can be explained without the term length contraction, then the theory is not as simple as possible is it.

Edited by ralfcis, 12 May 2019 - 02:05 PM.

### #259 sluggo

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Posted 14 May 2019 - 11:16 AM

Sherwood#254;

SR transform equation

x' = g(x+ut) = g(u/c)ct + gx

where g is the SR gamma function.

[That is the reciprocal form for x when you know x'.
The plus should be a minus.
If the event is the location of B after 1sec at .8,
Coordinates (x, t) for the rest frame A are (.8, 1).
Coordinates (x, t) for the frame B are (0, .6).

### #260 sluggo

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Posted 14 May 2019 - 11:40 AM

rhertz#243;

tB − tA = t'A − tB

2AB/(t'A − tA) = c,

-

In this way, he is anticipating that c has the same value either in the first lapse when the mirror is hit or in the second lapse, when the light beam bounces back from the mirror to its origin A. Clever?)

[He has just defined the outbound trip time equal to the inbound time for synchronous stationary clocks located at different places. This is what an observer at rest would expect.
-

(Now he includes "v" in the famous round trip, because he needs the factors (c - v) and (c + v) to obtain the famous c2-v2, otherwise impossible to appear. But, doing so, he violates his own 2nd. postulate: speed of light dependent on the motion v of the source or reflexion of the source).

[Not true. The expressions (c±v) are used when two objects are moving and both velocities are required for measurements. It is not propagation speed in space.]

### #261 Sherwood

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Posted 15 May 2019 - 06:18 PM

Sherwood you're confusing Yv with v. v is limited to c but Yv is limitless. v=x/t, Yv =Yx/t =x/t'. At .8c, Y=5/3 so Yv=4/3c but Yc in that frame is 5/3c. It's all a matter of perspective and Yv or Yc is the proper distance travelled over the dilated time of the depicted moving frame from the depicted stationary frame's perspective. See my last posts in the relativity and simple algebra thread for more details.

gu is the ratio of the distance one object moved from another while the light (emanating from either object) traveled the distance I during the time t at speed c according to Special Relativity (SR).  This is the same value you would get in Newtonian Mechanics (NM) by multiplying the speed v=gu by t=I/c.  v is NM speed,  u is SR speed.  They are never the same.

### #262 Sherwood

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Posted 15 May 2019 - 06:24 PM

MikeBrace#243

The propagation speed is c in a vacuum and. . .

This is what Einstein said.  That light always travels at speed c in free space.  The $64 dollar question is "at speed crelative to what !!! ### #263 sluggo sluggo Questioning • Members • 177 posts Posted 16 May 2019 - 10:00 AM This is what Einstein said. That light always travels at speed c in free space. The$64 dollar question is "at speed crelative to what !!!

(The first 3 lines should have been inside the quote box.) Mike was giving the common Newtonian view that the emitter or observer speed should make a difference.  Postulate 1 states c is constant and independent of the source speed. Td and lc are complementary effects that ensure any observer measures the same value c.

### #264 pascal

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Posted 17 May 2019 - 10:48 AM

Its all Super Asymmetry. If you haven't heard of it, you haven't been paying attention!

Jess Tauber

### #265 Sherwood

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Posted 19 May 2019 - 12:20 AM

|                                |                                                  |

|                                |                                                  |

|                                |                                                  |

I      |  300m                     |  300m                                       |  300m

|                                |                                                  |

|__________________|___________________________|

A          400m           B                    500m                   C

A, B, C are passing each other in the horizontal x direction and are momentarily at location A.  Each has a light source at rest with respect to itself. At that moment when they are together, they all turn on their lights.  After a short time, the light from each has traveled at speed c from each in all directions for time t = 1 musec.  I=ct=300m.  Now B is 400m from A, C is 900m from A, and C is 500m from B.  This is the physics.  There is nothing else.  Nothing about energy, momentum, etc.  By Special Relativity (SR):

I2 = (ct)2 - x2 = (ct')2 - x' 2 = (ct'')2 - x'' 2 = . . .

(ct)2 = I2 + x2;    (ct')2 = I2 + x' 2 ;    (ct'')2 = I2 + x'' 2 ;    . . .

u/c = x/ct;    u'/c = x'/ct';    u''/c = x''/ct'';    . . .

g' = (1/(- (u'/c)2))1/2 = ( (ct' 2) / ((ct' 2) - (ct' 2)(u'/c)2) )1/2 =

((ct' 2) / ((ct' 2) - x' 2) )1/2 = ((ct' 2)/I 2)1/2 = (ct’)/ I

x' = g(x + ut) = g(u/c)ct + gx

Since A has its light source at rest in its reference frame, x = 0 when comparing another reference frame moving with respect to A.  Same applies to B and C for their light sources.  Then the preceding equation becomes

x' = g(u/c) I

For A with respect to B,

xAB = 400m,   = 300m,   tA= I/c = 1 musec,    (ct)B = 500m,    tB = (5/3) musec,    u/c = 400/500,    = 500/300,   xAB = g(u/c) I = (500/300) (400/500) 300 = 400m as illustrated below.

|\

|   \    ct'

I   |      \                            I2 = (ct')2 - x';      g = (ct')/ I;      (u'/c) = x'/(ct');     x' = g(u/c)I

|____ \

x'

For A with respect to C,

xAC = 900m,   I = 300m,    tA = 1 musec,    (ct)C = 948.683298m,    tC = 3.162228 musec,    u/c = 0.948683,    g = 3.162278,    xAC = g(u/c) I = 3.162278 • 0.948683 • 300 = 900m

For  B with respect to C,

xBC = 500m,   I = 300m,    tB = 1 musec,    (ct)C = 583.095189m,    tC = 1.943651 musec,    u/c = 0.857492,    g = 1.943651,    xBC = g(u/c) I = 1.943651 • 0.857492 • 300 = 500m

For  B with respect to A,

xBA = -400m,   = 300m,   tB = 1 musec,   (ct)A = 500m,   tA = (5/3) musec,   u/c = (-400/500),   = (500/300),   xBA = g(u/c) I = (500/300) (- 400/500) 300 = - 400m

For  C with respect to B,

xCB = -500m,   I = 300m,   tC = 1 musec,   (ct)B = 583.095189m,   tB = 1.943651 musec,   u/c = - 0.857492,   g = 1.943651,   xCB = g(u/c) I = 1.943651 • - 0.857492 • 300 = - 500m

For  C with respect to A,

xCA = -900m,   I = 300m,   tC = 1 musec,   (ct)A= 948.683298m,   tA = 3.162228 musec,   u/c = - 0.948683,   g = 3.162278,   xCA = g(u/c) I = 3.162278 • - 0.948683 • 300 = - 900m

The physics here is only about objects and light ‘wavefronts’ which have changed their location over time.  That’s all.  There is nothing about collisions, momentum changes, energy, etc.  The quantities are location and time.  Additionally, changes of location are called distances and changes of location (distances) during a given time are velocity (or speed).  So we have distances, times, and the ratio of distance to time called speed or velocity.  The above showed how these distances, times, and velocity are related in SR.

Notice that in “For A wrt to B” tA=1.0 musec and tB=(5/3) musec while in “For B wrt to A” tA=(5/3) musec and tB=1.0 musec.  Each object has several different ‘times’ and if we had considered more objects traveling different distances we could have many more ‘times’.

There are two key equations in SR which are very similar to equivalent equations which can be derived for Newtonian Mechanics (NM).

The SR versions are

I2 = (ct')2 - x' 2

g = (ct’/I)

The NM versions are

I2 = (c’t)2 - x' 2

g = (c’t/I)

The quantity (ct’) equals exactly the quantity (c’t).  Therefore, I2, x’2, and g will have the same values in both SR and NM.

However, t does not equal t’ and c does not equal c’ and this is where SR and NM differ.

In both SR and NM I equals the distance the light has traveled from its source, possibly spherically, at speed c for time t, i.e., I=ct. So t=I/c.  This makes the light essentially a timing signal.  The light was emitted from a source not moving in its reference frame.  It moved a certain distance I at speed c during time t in all directions in that frame.  From the point of view of another reference frame moving with respect to that reference frame, the light moving perpendicularly (call it vertically) moved not only the ‘vertical’ distance I but also the perpendicular distance x’ for a total distance of

H = (I2 + x’2)1/2

In NM, this obviously increased distance, H during the time t would be called something like c’, so the distance H would be

H = (c’t)

and

I2 = (c’t)2 - x' 2

the NM version of I.

SR insists (postulates, assumes, etc.) that the speed of the light be the same in both reference frames, i.e., that the ratio H/t be the same as I/t.  Since H is greater than I, this is impossible physically.  However, you can mathematically arbitrarily increase t (and call it t’ to differentiate it from t).  But now this is mathematics, not reality.  This is why the SR version is

I2 = (ct’)2 - x' 2

I2 = (c’t)2 - x' 2

It also accounts for the two different forms of the gamma function (g), although the numeric value is the same in both SR and NM, c’t=ct’.  Of course, c≠c’ and t≠t’.

Notice that this mathematical ledgerdemain can be applied to any ‘timing signal’ of a known speed in its reference frame, whatever that speed is.  I.e., you could have a Special Relativity where the timing signal travels at speed c=300 m/sec instead of 300 million m/sec and is assumed to also travel that same speed in all reference frames moving with respect to it.

Notice also that although light from a given source (just like anything else) is obviously traveling a different speed in reference frames moving with respect to the source’s, it is assumed (along with respect to everything else according to SR) to be moving at speed c from its source.  This was known as the Ritz theory.

x' = g(u/c) I;     (x’/ I) = g(u/c)

shows that x’, the distance the object travels, can be many times the distance I, that the light travels, even approaching infinity as u approaches c.  All the above objects traveled a greater distance than the three lights did between the time they were together and when their lights had traveled 300m for 1 musec at speed c.

u’/c = x’/ct’ = x’/(x’ 2 + I2)1/2

easily shows why “nothing can go faster than light” in SR.  The denominator will always be bigger so u'/c will always be <1.0.

In NM, (x’/I)=(x’/ct)=(v/c).  In SR, (x’/I)=g(u/c)=(ct’/ct)(x’/ct’)=(x’/ct)=v/c which is why SR math can do these time-distance problems although based on erroneous (or wierdly altered, if you prefer) concepts of time and velocity.  This wierd math is carried over into more complex topics such as momentum and kinetic energy and seems to work somewhat because of its ties to NM as shown here.

### #266 sluggo

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Posted 21 May 2019 - 08:51 AM

sherwood#265;

As the graphic shows, B and C are moving faster than c (blue), so all those calculations are invalid. Units are x100 m.

### #267 Sherwood

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Posted 21 May 2019 - 10:38 PM

sherwood#265;

As the graphic shows, B and C are moving faster than c (blue), so all those calculations are invalid. Units are x100 m.

The calculations are valid.  As explained in the post:

"  u’/c = x’/ct’ = x’/(x’ 2 + I2)1/2

easily shows why “nothing can go faster than light” in SR.  The denominator will always be bigger so u'/c will always be <1.0."

The SR speeds, as shown in the post, and repeated below shown in bold, are all less than c.   x', the distances the objects move away from each other, exceeds the distance the light traveled, I, whenever g(u/c) >1.  SR 'speed' (u/c) is always <1.0 while g(u/c) can be any number.     x' = g(u/c) I;     (x’/ I) = g(u/c)   as posted.

x' = g(u/c) I

For A with respect to B,

xAB = 400m,   = 300m,   tA= I/c = 1 musec,    (ct)= 500m,    tB = (5/3) musec,    u/c = 400/500,    = 500/300,   xAB = g(u/c) I = (500/300) (400/500) 300 = 400m as illustrated below.

|\

|   \    ct'

I   |      \                            I2 = (ct')2 - x';      g = (ct')/ I;      (u'/c) = x'/(ct');     x' = g(u/c)I

|____ \

x'

For A with respect to C,

xAC = 900m,   = 300m,    tA = 1 musec,    (ct)= 948.683298m,    tC = 3.162228 musec,    u/c = 0.948683,    = 3.162278,    xAC = g(u/c) I = 3.162278 • 0.948683 • 300 = 900m

For  B with respect to C,

xBC = 500m,   = 300m,    tB = 1 musec,    (ct)C = 583.095189m,    tC = 1.943651 musec,    u/c = 0.857492,    = 1.943651,    xBC = g(u/c) I = 1.943651 • 0.857492 • 300 = 500m

For  B with respect to A,

xBA = -400m,   = 300m,   tB = 1 musec,   (ct)A = 500m,   tA = (5/3) musec,   u/c = (-400/500),   = (500/300),   xBA = g(u/c) I = (500/300) (- 400/500) 300 = - 400m

For  C with respect to B,

xCB = -500m,   = 300m,   tC = 1 musec,   (ct)B = 583.095189m,   tB = 1.943651 musec,   u/c = - 0.857492,   = 1.943651,   xCB = g(u/c) I = 1.943651 • - 0.857492 • 300 = - 500m

For  C with respect to A,

xCA = -900m,   = 300m,   tC = 1 musec,   (ct)A= 948.683298m,   tA = 3.162228 musec,   u/c = - 0.948683,   = 3.162278,   xCA = g(u/c) I = 3.162278 • - 0.948683 • 300 = - 900m

### #268 sluggo

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Posted 23 May 2019 - 10:39 AM

The calculations are valid.  As explained in the post:

"  u’/c = x’/ct’ = x’/(x’ 2 + I2)1/2

easily shows why “nothing can go faster than light” in SR.

How can B be at 400, C at 900, while light has only moved 300?

### #269 sluggo

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Posted 23 May 2019 - 11:57 AM

Sherwood#265;

I2 = (ct)2 - x2

Light speed is independent of its source and does not add vectorially like masses do.

Refer to lite clock graphic.

c2=v2+u2

### #270 Sherwood

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Posted 23 May 2019 - 05:28 PM

How can B be at 400, C at 900, while light has only moved 300?

Einstein's transform equation:

x' = g(x + ut) = g(u/c)ct + gx   (= g(u/c) I when x=0).

x=0 because we are talking about the origin point (not, e.g., x= 3400) of the frames.

For (u/c) = 0.8,  g = 5/3.

If I = 300m, t = 1 musec.  I = ct = 300m

x' = (5/3)(4/5)300m = 400m

Note that the speed (velocity) of the object now at x' is still only 0.8c according to SR.

If t = 1 musec  and (u/c)= 0.999, then I, the distance the light traveled, is still 300m but the distance x' traveled is x' = 22.366272 • 0.999 • 300m = 6,703.1717m because g = 22.366272 when (u/c) = 0.999.  No matter how large (x'/I) = g(u/c) is, (u/c), the SR 'speed between objects', will always be <1.0.  g(u/c) 'equals' infinity when (u/c) = 1.0 because the calculation for g in this case is a division by zero.

### #271 Sherwood

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Posted 24 May 2019 - 02:34 PM

Sherwood#265;

I2 = (ct)2 - x2

Light speed is independent of its source and does not add vectorially like masses do.

Refer to lite clock graphic.

c2=v2+u2

Like everything else, light speed has to be relative to something.  Since light speed is the same in all reference frames according to SR, then by SR light has to have speed c relative to its source as well as to everything else.

If you have an object emitting something spherically at s=300 m/sec and another object passing by at 400 m/sec, then the NM Interval equation would be

I= (s't)2 – x'    or     (s't)= I2 + x'

meaning that after 1 sec the portion of the something emitted that moves perpendicular to the relative motion of the objects would be the distance (I) from the emitting object and the distance (s't) from the second object.  The distance (s't) is obviously greater than the distance (I).  So the speed of that portion of the emitted something would be (s't)/t or s' with respect to the second object.  If you insist, postulate, or whatever, that s'=s then you have to change the time to, for example, (t') to keep the distance (s't) the same.  You now have the SR Interval equation I2=(st')2 – x'2 where s is the same in all reference frames because the moving reference frame could be any possible reference frame moving any possible distance with respect to the first during the time (t), not just 400m.  s is the equivalent of the c of SR and all the equations and concepts of SR can be derived from this SR Interval equation with c (the s used in this example) being 300 m/sec instead of 300m/musec or any other speed you would care to pick.

Post #265 explains this fairly well.  Don't concentrate on the listed values for the various variables.  That's just the result of working the SR equations.  Read the text carefully several times and think about what it is saying about the meaning of the equations.

### #272 sluggo

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Posted 25 May 2019 - 12:14 PM

Sherwood;

I2 = (ct)2 - x2 = (ct')2 - x' 2 = (ct'')2 - x'' 2 = . . .

u/c = x/ct;    u'/c = x'/ct';    u''/c = x''/ct'';    . . .

From the point of view of another reference frame moving with respect to that reference frame, the light moving perpendicularly (call it vertically) moved not only the ‘vertical’ distance I but also the perpendicular distance x’ for a total distance of

H = (I2 + x’2)1/2

[If that were true, there would be no time dilation observed.]

[On the left, R moves past A at .5c for a time ct. A describes the R coordinates for event e as (x, ct). Because of time dilation, R describes the coordinates for event e as (x', ct'), i.e. event e'. From the graphic,

(ct)2 - x2 = (ct')2 > (ct')2 - (x')2 = (tr)2

What is constant, x/t = x'/t' = x"/t" = c.

On the right, using u/c=.8, A describes the B coordinates as (2.4, 3.0), while B describes his coordinates as (1.44, 1.8) relative to A.
The reference frame clock has the maximum time, and any clock that moves relative to it, loses time.]

[What you present is your own interpretation of SR, which doesn't agree with experimental evidence.]