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Einstein's Theory revisited


ResistETIntervention

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What went wrong in the proof of the theory that nothing can travel faster than the speed of light?

 

In a “normal” size vehicle, let say with the height of the interior of the vehicle (the distance from the floor to the ceiling) is about 3 m (≈3.28 yards), the time it takes the light beam emitted from its source on the floor to hit the ceiling is 10-8 of a second – literally a split bit of a second.

 

The right triangle used to prove in the theory above is obtained under the assumptions that:

  • Light would travel at the constant speed c=3×108 m/s regardless of the frame of reference.
  • From the external observer’s perspective, light would hit the ceiling of the vehicle at the point B on the ceiling directly above the point A where the source of light is on the floor.
  • Two different frames of reference (the passenger’s perspective and the external observer’s perspective) could be related in one single diagram.

 

The first assumption would seem to hold in the case the velocity of the vehicle is insignificant compared to the speed of light: v≪c which is the underlying assumption which leads to the same conclusion. But what if v≈c? From the perspective of the passenger, would the light beam still seem to travel at the constant speed c=3×108 m/s?

 

The second assumption is made under the presumption that the light beam emitted from its source at the point A on the floor will hit the point B directly above it on the ceiling. Once “airborne,” the light beam does not “know” that it is in a moving vehicle. Thus, it will travel in a straight, vertical path. Again, if v≪c, then it would seem that the light beam would hit the point B on the ceiling directly above the source on the floor, as it would if the vehicle were stationary. Indeed, it would seem insignificant where on the ceiling the light beam hits because it would require nanotechnology to measure the infinitesimal distance between the point C on the ceiling where the light beam actually hits and the point B which is directly above the light source at the point A on the floor. However, this insignificant distance is actually the length of the base of the right triangle used in describing Einstein’s thought experiment. So, as insignificant as it may seem, it must be noted that the light beam will not hit the point directly above its source, but towards west (if the vehicle is traveling in the east direction) and distance x away, where x=vt0 is the distance traveled by the vehicle in time t0 it takes the light beam to hit the ceiling, about 10-8 second, as observed by the external observer.

 

Assume, for the moment, the third supposition above – that the scenario can be described in a triangle by combining the two different frames of reference – is justifiable. Then the right triangle used in proving the theory above is as follows, where

 

  • z is the distance the light beam traveled in time t0 that it takes for the light beam to hit the ceiling, as observed by the external observer,
  • x is the distance traveled by the vehicle in time t0, again, as observed by the external observer,
  • y is the distance the light beam traveled in time t1, as observed by the passenger traveling in the vehicle

 

Note that x≪z is in any realistic situation, since in the best of scenarios, even if manmade, supersonic vehicles attain ten times the speed of sound (Mach 10), this right triangle would be so skinny (y≈z) that it would be almost a vertical line segment, rather than a triangle.

 

However, there is no reason to presume that v≪c before anything is proven yet. In that case, there is no reason to make the first two assumptions. For the sake of observing accurately, suppose both the passenger and the external observer can observe the scenario in an extremely slow motion, slow enough that the scenario which occurs within 10-8 second can be viewed over, say, 10 seconds. Then, what would be true is that

 

  • From the perspective of the passenger, the light beam would not travel directly from the point A to the point B directly above it, but to the point C which is x meters to the west away from the point B on the ceiling, where x=vt0 is the distance traveled by the vehicle in time t0 it takes the light beam to hit the ceiling, about 10-8 second, as observed by the external observer. That is, the passenger (traveling in east direction in the vehicle) would see the light beam as traveling northwesterly direction to the point C.
  • On the other hand, from the external observer’s perspective, the light beam which does not “know” that it is in a moving vehicle, would travel in a straight path vertically and hit the ceiling at the point C, rather than at the point B directly above the point A, since the vehicle would have moved x meters.

 

Then, again, assuming that the scenario can be described by combining two different frames of reference, the right right-triangle (or the correct right triangle) would be then one with y as the hypotenuse, and x the base and z the height.

 

Here, this new right triangle (again, very skinny in any practical scenario, almost like a vertical line segment) can be used in two different cases: one in which the distance traveled by the vehicle is from the external observer’s and another in which the distance traveled by the “ground” is from the perspective of the passenger’s (if the vehicle is transparent, say). This is by retaining the third assumption made in proving the theory above.

 

In the first case, then

 

  • z is the distance the light beam traveled in time t0 , as observed by the external observer,
  • x is the distance traveled by the vehicle in time t0, again, as observed by the external observer,
  • y is the distance the light beam traveled in time t1, as observed by the passenger traveling in the vehicle

 

In the second case,

 

  • z is the distance the light beam traveled in time t0 , as observed by the external observer,
  • x is the distance traveled by the “ground” in time t1 that it takes for the light beam to hit the ceiling, as observed by the passenger,
  • y is the distance the light beam traveled in time t1, again, as observed by the passenger traveling in the vehicle

 

Since we do not presume that the vehicle cannot travel at the luminal or a superluminal speed, we also do not presume that the passenger sees the light beam as traveling at the speed of c=3×108 m/s. We will label it as s for speed. Then, by applying the Pythagorean Theorem on the right triangle in two different cases above yields the following. In the first case, (vt0)2+(ct0)2=(st1)2, and in the second case, (vt1)2+(ct0)2=(st1)2. Setting the two left-hand-sides equal to each other, we see that t0=t1. That is, there is no time dilation. The time that it takes for the light beam to hit the ceiling is the same as observed by the external observer, as well as by the passenger. And the speed at which the light beam would seem to be moving away from the perspective of the passenger is s is the value that satisfies s2=v2+c2. This is not the actual speed of the light beam, but only the perceived speed of the light beam, as observed by the passenger who thinks the light beam is traveling in a diagonal fashion, rather than in a vertical fashion.

 

This perceived, non-actual speed of the light beam computed is a result of combining two different perspectives in one equation, which is an assumption made in “proving” Einstein’s theory: that the two different frames of reference can be combined to describe the scenario in a single diagram. If we do not retain that assumption, then there is no triangle to describe the scenario, let alone a right triangle. Whether we retain that assumption or not, we do not arrive at a result that precludes any object from traveling at superluminal speeds.

 

In the correct right triangle, what does the distance y actually mean then, if it is to have any meaning in the diagram above without producing a non-actual, perceived quantity due to combining two difference references of frame?

 

  • z is the distance the light beam traveled in time t0 , as observed by the external observer,
  • x is the distance traveled by the vehicle in time t0, as observed by the external observer,
  • y is the distance between the source of the light beam on the floor and the head-end of the light beam at time t0, as observed by the external observer

 

With this diagram which includes only one perspective, that of the external observer in a non-moving frame of reference, there is no non-actual, perceived quantity produced in computations.

 

This scenario may be more easily visualized if we change it by exchanging a small ball with a light beam. Here is a thought experiment.

 

A thought experiment: Consider a vehicle with a ceiling of 100 m, traveling at a speed of v=100 m/s. Assume that in the vehicle is vacuum and with no gravity. A ball is emitted vertically upward at a speed of c=1000 m/s (v<c), 100 m/s (v=c), or 10 m/s (v>c). What path of the ball does a passenger in the vehicle observe and an observer outside the vehicle observe? Once “airborne,” the ball does not “know” that it is in a moving vehicle, and it will travel in a vertical fashion as observed by an external observer, though it will seem to the passenger that it is traveling diagonally to where it hits on the ceiling.

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...This scenario may be more easily visualized if we change it by exchanging a small ball with a light beam. Here is a thought experiment.

 

A thought experiment: Consider a vehicle with a ceiling of 100 m, traveling at a speed of v=100 m/s. Assume that in the vehicle is vacuum and with no gravity. A ball is emitted vertically upward at a speed of c=1000 m/s (v<c), 100 m/s (v=c), or 10 m/s (v>c). What path of the ball does a passenger in the vehicle observe and an observer outside the vehicle observe? Once “airborne,” the ball does not “know” that it is in a moving vehicle, and it will travel in a vertical fashion as observed by an external observer, though it will seem to the passenger that it is traveling diagonally to where it hits on the ceiling.

 

It seems you have the picture messed up - check it again please?

"A ball is emitted vertically upward at a speed of c=1000 m/s (v<c)," who checks to see if it is vertical? if it just not quite vertical for the external observer how is he going to adjust it?

Edited by Robittybob1
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It seems you have the picture messed up - check it again please?

"A ball is emitted vertically upward at a speed of c=1000 m/s (v<c)," who checks to see if it is vertical? if it just not quite vertical for the external observer how is he going to adjust it?

 

The idea of the thought experiment is to get a better visualization of Einstein's thought experiment. The ball is sent off vertically from its source on the floor - the same way a light beam was emitted from its source on the floor of the vehicle in proving that nothing can travel faster than the speed of light.

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The idea of the thought experiment is to get a better visualization of Einstein's thought experiment. The ball is sent off vertically from its source on the floor - the same way a light beam was emitted from its source on the floor of the vehicle in proving that nothing can travel faster than the speed of light.

You still haven't answered my question - if the observation of vertical is made by the outsider how is an adjustment going to be made?

If you really set the experiment up as you propose after the ball hits the roof it will bounce back down at an angle too and you won't have a light clock equivalent.

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You still haven't answered my question - if the observation of vertical is made by the outsider how is an adjustment going to be made?

If you really set the experiment up as you propose after the ball hits the roof it will bounce back down at an angle too and you won't have a light clock equivalent.

Adjustment isn't necessary. You get a right triangle from Einstein's thought experiment with its hypotenuse as the distance that light travels from the floor to the ceiling as observed by the external observer, its height as the distance the light beam travels from the floor to the ceiling as observed by the passenger, and the base as the distance the vehicle travels as observed by the external observer. This right triangle is attained by assuming at least three items which should not be presumed. The other thought experiment is just to visualize it better.

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Adjustment isn't necessary. You get a right triangle from Einstein's thought experiment with its hypotenuse as the distance that light travels from the floor to the ceiling as observed by the external observer, its height as the distance the light beam travels from the floor to the ceiling as observed by the passenger, and the base as the distance the vehicle travels as observed by the external observer. This right triangle is attained by assuming at least three items which should not be presumed. The other thought experiment is just to visualize it better.

Well now you've changed your tune! Compare what you have just said to what you wrote in the OP.

 

This why I have been asking you to check your work.

 

So now the traveling person checks to see if the light clock (ball thrower) is aligned properly.

 

 

Once “airborne,” the ball does not “know” that it is in a moving vehicle, and it will travel in a vertical fashion as observed by an external observer, though it will seem to the passenger that it is traveling diagonally to where it hits on the ceiling.
Edited by Robittybob1
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What went wrong in the proof of the theory that nothing can travel faster than the speed of light?

This perceived, non-actual speed of the light beam computed is a result of combining two different perspectives in one equation, which is an assumption made in “proving” Einstein’s theory: that the two different frames of reference can be combined to describe the scenario in a single diagram.

 

When did you first come to believe, mistakenly, that the thought experiment is what Einstein's conclusions rested on?

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Einstein's conclusion is based on his thought experiment and the computations made by assuming certain suppositions; some of which are outlined in the initial post. The theory was not obtained from the traditional scientific method as no data can be produced to prove that the luminal speed cannot be attained - at the time.

 

Well now you've changed your tune! Compare what you have just said to what you wrote in the OP.

 

This why I have been asking you to check your work.

 

So now the traveling person checks to see if the light clock (ball thrower) is aligned properly.

 

The ball could be shot from a mechanism vertically upwards right in front of the passenger who will not observe the ball hitting at the point on the ceiling directly above the mechanism but a spot slightly next to that point in the opposite direction from that of the vehicle.

Edited by ResistETIntervention
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Einstein's conclusion is based on his thought experiment and the computations made by assuming certain suppositions; some of which are outlined in the initial post. The theory was not obtained from the traditional scientific method as no data can be produced to prove that the luminal speed cannot be attained - at the time.

 

 

The ball could be shot from a mechanism vertically upwards right in front of the passenger who will not observe the ball hitting the ceiling the point directly above the mechanism but a spot slightly next to that point in the opposite direction from that of the vehicle.

Are you just a kid? What you describe will be what happens in an accelerated vehicle. OK you explain to me why it is not directly above?

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What you describe will be what happens in an accelerated vehicle. OK you explain to me why it is not directly above?

 

The vehicle has a velocity. It is moving. It does not need to accelerate. If it did accelerate, the scenario certainly could not be explained in a triangle, for the path of the ball, as observed by the passenger, would not even be a straight line.

 

The ball shot from a mechanism does not "know" that it is in a moving vehicle, once it is "airborne." While it is in midair, the vehicle would have moved a certain distance x meters, say. Thus, the ball will hit the point C on the ceiling that is x meters away from the point B on the ceiling that is directly above the mechanism at the point A on the floor.

Edited by ResistETIntervention
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The vehicle has a velocity. It is moving. It does not need to accelerate. If it did accelerate, the scenario certainly could not be explained in a triangle, for the path of the ball, as observed by the passenger, would not even be a straight line.

 

The ball shot from a mechanism does not "know" that it is in a moving vehicle, once it is "airborne." While it is in midair, the vehicle would have moved a certain distance x meters, say. Thus, the ball will hit the point C on the ceiling that is x meters away from the point B on the ceiling that is directly above the mechanism at the point A on the floor.

 

With the ball example, the ball has a horizontal component to its movement - the same horizontal component as the moving vehicle - and so it can hit the same vertical point.

 

It might be clearer to keep to the light example to discuss the issues you have with relativity?

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With the ball example, the ball has a horizontal component to its movement - the same horizontal component as the moving vehicle - and so it can hit the same vertical point.

 

It might be clearer to keep to the light example to discuss the issues you have with relativity?

 

A light beam emitted vertically upward isn't seeking to hit the point B directly above its source at the point A. It would simply travel vertically upward and end up hitting the point C.

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The vehicle has a velocity. It is moving. It does not need to accelerate. If it did accelerate, the scenario certainly could not be explained in a triangle, for the path of the ball, as observed by the passenger, would not even be a straight line.

 

The ball shot from a mechanism does not "know" that it is in a moving vehicle, once it is "airborne." While it is in midair, the vehicle would have moved a certain distance x meters, say. Thus, the ball will hit the point C on the ceiling that is x meters away from the point B on the ceiling that is directly above the mechanism at the point A on the floor.

Have you been in a moving vehicle?

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Einstein's conclusion is based on his thought experiment and the computations made by assuming certain suppositions; some of which are outlined in the initial post.

 

The only "assumption" in your post that is even close to the postulates of relativity is that the speed of light is constant in all frames.

 

This is simply based on the, quite reasonable, assumption that the laws of physics, including Maxwell's equations, are the same whether you are moving or not. (As the fact that all motion is relative had been established by Galileo many centuries earlier, this is pretty obviously the case.)

 

The rest of the theory was derived purely mathematically from this starting point.

 

The thought experiment you mention, and others, are ways of visualising or explaining what the maths says. They are obviously not the basis of the theory because thought experiments.

 

You need to show (mathematically, not by silly analogies) that one or more of Maxwell's equations, Noether's Theorem or Enstein's derivation of SR is incorrect if you want to show that SR is wrong.

 

And then you need to come up with an alternative explanation for the mountains of evidence (including your functioning computer) that shows that SR is a very precisely accurate theory.

 

Good luck with that.

 

(Note that there are so many errors in your OP, I really can't be bothered to go through it in detail. Life is too short.)

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Have you been in a moving vehicle?

 

By your logic, you would be able to throw a ball vertically upward and catch it while on a skateboard, say.

 

(Note that there are so many errors in your OP, I really can't be bothered to go through it in detail. Life is too short.)

 

I already gave an explanation as to what went wrong in the theory.

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By your logic, you would be able to throw a ball vertically upward and catch it while on a skateboard, say.

Which, neglecting air resistance, is exactly what you can do, as long as you do not accelerate, or decelerate the skateboard. The ball already possesses the same forward velocity as you. From your perspective the ball rises and falls vertically. To an observer, stationery on the sidewalk, it follows a parabola. If you wish it to rise vertically relative to the observer you will need to impart a backward velocity to the ball to counter your forward velocity.

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A light beam emitted vertically upward isn't seeking to hit the point B directly above its source at the point A. It would simply travel vertically upward and end up hitting the point C.

 

The issue is what is meant by 'light simply travels vertically upward'. What is vertical is different for different reference frames. Do you mean with respect to the reference frame of the emitting source of the light?

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The second assumption is made under the presumption that the light beam emitted from its source at the point A on the floor will hit the point B directly above it on the ceiling. Once “airborne,” the light beam does not “know” that it is in a moving vehicle. Thus, it will travel in a straight, vertical path. Again, if v≪c, then it would seem that the light beam would hit the point B on the ceiling directly above the source on the floor, as it would if the vehicle were stationary. Indeed, it would seem insignificant where on the ceiling the light beam hits because it would require nanotechnology to measure the infinitesimal distance between the point C on the ceiling where the light beam actually hits and the point B which is directly above the light source at the point A on the floor. However, this insignificant distance is actually the length of the base of the right triangle used in describing Einstein’s thought experiment. So, as insignificant as it may seem, it must be noted that the light beam will not hit the point directly above its source, but towards west (if the vehicle is traveling in the east direction) and distance x away, where x=vt0 is the distance traveled by the vehicle in time t0 it takes the light beam to hit the ceiling, about 10-8 second, as observed by the external observer.

 

 

No, this is wrong. The others that have responded to this are right.

 

One of the assumptions of relativity is that the physics works exactly the same in all inertial frames of reference. Put another way, any inertial observer can assume s/he is at rest. What happens in a moving car looks no different from what happens in a stationary car.

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By your logic, you would be able to throw a ball vertically upward and catch it while on a skateboard, say.

 

I can juggling whilst riding a bike, I have run a half-marathon whilst juggling, I taught a group of children (with no common language) to juggle on a 747 midway over the atlantic (wouldn't want to try that post 9-11); I can absolutely vouch for the fact that whilst travelling at a constant velocity the ball goes straight up and down.

 

The reason that it might appear to not work on your skateboard is that it is quite difficult to master the art of throwing just upwards when you are moving forwards - you compensate even though you do not need to

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By your logic, you would be able to throw a ball vertically upward and catch it while on a skateboard, say.

And by your logic, you shouldn't be able to throw a ball at all, because once the ball left contact with your hand it would not "know" that it was to keep moving upward.

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Fellows, don't get excited. I'm just saying Robittybob1 would be able to catch a ball thrown vertically upward in a moving vehicle.

 

Robbitybob1, in order to disprove Einstein's theory, all I need to do is to provide one counterexample and I already gave one. In case, any of you decides to retaliate on my sighting, consider what you'd observe if you had sighted a brilliant object achieving a superluminal speed from a stationary position. There are many people in the world who have had such sightings, so there are many counterexamples to the theory. The astronaut Edgar Mitchell also gave his testimony on his sightings and stated that Einstein's theory is already known to be false. Given these facts, what we need to do collaboratively is to reconsider the theory and see what went wrong with the proof the theory, rather than upholding it for the sake of preserving it stubbornly. That would be the correct attitude of true scientists.

 

Here is another faulty assumption made in the theory: that the height of the right triangle used in the proof of the theory is y = ct. In reality, it should be ct - 0.5gt2. Though in most practical situations with small velocities (in comparison to that of light) of any manmade vehicles, we may be able to approximate it as y=ct , any infinitesimal difference here could make all the difference in what is considered here and cannot be ignored. Then accordingly, the hypotenuse of the right triangle would also need to be corrected.

 

 

 

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There are many people in the world who have had such sightings, so there are many counterexamples to the theory.

 

!

Moderator Note

That's not scientific evidence, and let me remind you that you've been warned not to turn your threads into a discussion on UFOs.

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Robbitybob1, in order to disprove Einstein's theory, all I need to do is to provide one counterexample and I already gave one.

 

No you didn't. You have made some incoherent arguments about a thought experiments and some claims about something you have seen. None of which does anything to disprove the theory of relativity.

 

1. Your argument is against a thought experiment and therefore cannot prove or disprove anything.

2. You have provided no maths to support your argument.

3. You have provided no evidence that contradicts relativity (personal anecdotes about thigns you may or may not have seen do not count).

4. You have offered no alternative explanation for all the evidence that supports the theory.

 

So I think we can safely dismiss your claims.

 

 

Here is another faulty assumption made in the theory: that the height of the right triangle used in the proof of the theory is y = ct.

 

That is not an assumption of the theory.

 

 

In reality, it should be ct - 0.5gt2.

 

Why?

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If I sighted an object (not mechanisms that can be controlled remotely or otherwise) that fell from a height took a vertically upward path, rather than a vertically downward path, I would say I had a counterexample to F = mg and s(t) = -0.5gt2 + s0. If millions of others had also sighted such objects, the disproof of F = mg and s(t) = -0.5gt2 + s0 would have been verified multiple times.

 

Your dismissing my claim does not make the theory any less false.

 

No, y = ct is not a formal assumption in the statement of the theory, but it is, nevertheless, a critical assumption that was used in proving the theory.

 

The reason that the height of the right triangle used in proving the theory should be ct - 0.5gt2 (rather than y = ct) is the same as that for anything else that is subject to gravity, isn't it? In case, you want to claim that light is not subject to gravity, consider the reason that even light cannot escape black holes.

Edited by ResistETIntervention
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