Proving einstein wrong in 2 simple steps

[Jay-qu]

Stereologist is correct Lenvanzanten, I dont know what information you are following in your first post but M&M definitely found that the speed of light is the same no matter which way you measure it. They spent a great deal of time trying to prove themselves wrong, but were unable to.

 

You are right in your second post but have the wrong reasoning. Light of different wavelengths will travel at different velocities in a medium - such as air or a prism. But in vacuum they all travel the same speed - c.

[Haech]

Clearly this is why they should teach modern physics in preschool, before any destructive intuitions about nature develop. That way, we'll always be on the same page when 2 spaceships cross each other.

 

To the OP, that was a very entertaining post. I'm sure you're not being serious.

[Haech]

P.S. I almost crashed into the apparatus that Michelson & Morley used to measure c. True story.

[phillip1882]

actually i am being serious. i honestly don't see how relativity can account for such a distance discrepancy. i have yet to see anyone here actually take a mathematical approach to my second experiment. i would like to see how the difference is accounted for.

to make things simple, let's say the train is 1 light minute long, (impossible on earth i know.)

and traveling at the speed of 0.1*c.

at what time would the train light beam hit the ground receiver from the ground observers point of view?

how much would the train be distance scrunched from the ground observers point of view?

how much time would it take for the train light beam to hit its receiver from the ground observers point of view?

is this data consistent with each-other?

[phillip1882]

here's my guesses to these questions, assuming the second postulate of relativity is true.

1) since both the ground observer and train observer see light traveling the same speed, they should both see the train light beam hit the ground receiver after 1 minute.

2) let's assume light travels at 300 million meters per sec. according to the formula L = L0 *sqrt(1 - v^2/c^2), i get L going from 1.8 billion meters to 1.799 billion meters. already however, i'm seeing a problem. if the train is shortened, wouldn't it have to reach the position of the ground receiver in shorter time as well, given its speed?

3) this part i'm not sure about, i would assume a longer amount of time than 1 minute, but how much longer? and how does it fit in with the shorter distance?

[Haech]

Check out any introduction to modern physics text. Looking for lorentz transforms via wikipedia or google should be fairly helpful. There are plenty of empirical evidence that c is in fact a limit on the transference of information, although you can argue that they are wrong; however, the mathematics is theoretically consistent.

[phillip1882]

but the objects that are transfering information are usually not moving, and when they are its no where need the speed of light. internally yes the speed of light is a limit. that is light can't push itself faster than the speed of light, but like i said i haven't heard it proved that adding velocity to light doesn't increase its speed.

[Haech]

You can't add velocity to light in relativity. From the moving frame, the light goes at c, and from the none moving frame, light goes at c. It's part of the limiting process.

[stereologist]

Phillip you need to read a basic book on this subject.

[modest]

actually i am being serious. i honestly don't see how relativity can account for such a distance discrepancy. i have yet to see anyone here actually take a mathematical approach to my second experiment. i would like to see how the difference is accounted for.

to make things simple, let's say the train is 1 light minute long, (impossible on earth i know.)

and traveling at the speed of 0.1*c.

at what time would the train light beam hit the ground receiver from the ground observers point of view?

Your thought experiment has 2 frames of reference and 4 events. The 4 events are the emission of a photon on the train, the emission on the ground, the detection on the train, and the detection on the ground. Two events are collocated (the two emissions). Both observers can therefore agree on when they happen. The other two events (the two detections), however, are spatially separated. So, the two observers will not agree on when they happen. This is because of the relativity of simultaneity.

 

If the train is going 0.6c then length contraction and time dilation are both a factor of 0.8. Clocks on the train as viewed from the ground will tick 80% as fast as identical clocks on the ground. Meter sticks on the train will be 80% the length as identical ones on the ground. The inverse is true (e.g. rulers on the ground as viewed on the train are .8 times the length of the reference ruler).

 

The two frames will evolve like the following images where the train is moving .6c and the photon detector is 1 light-minute from the emitter as judged from its rest frame. The yellow dot marks the position of the photons. They move from left to right. The "Train-frame" and "Ground frame" represent how things are measured from the observers in those frames. For example, where it says "ground-frame at T=1" you must consider that T=1 is true for the clock on the ground and not on the train. The clock on the train will be running slower from the ground frame's perspective. The left side of the black lines represent the photon emitters and the right side is the detector:

 

 

The solution to the perceived paradox is explained at this website:

The Paradox of Special Relativity

 

To give some specific numerical answers where the train is moving .6c and the detectors are 1 light-minute from the emitters as measured in the rest frame. There are 4 events:

• event A. Train emitter emits a photon
  
• event B. Ground emitter emits a photon
  
• event C. Train detector detects a photon
  
• event D. Ground detector detects a photon
  

We call the train observer T and the ground observer G.

 

G figures A and B are simultaneous and collocated. G measures 1 minute between event B and event D and measures 2 minutes between A and C. G measures the length of the train at 0.8 light-minutes, the length between B and D at 1 light-minute and the length between A and C at 2 light-minutes.

 

T figures A and B are simultaneous and collocated. T measures 30 seconds between event B and event D and measures 1 minute between A and C. T measures the length between emitter and detector on the ground at 0.8 light-minutes, the length between A and C at 1 light-minute, and the length between B and D at 30 light-seconds.

 

Hopefully this explains the thought-experiment with enough detail that any objections you still have can be referenced to something very specific.

 

~modest

[phillip1882]

but its not just one paradox, its several.

the train observer would see both light beams reach the position of the train receiver in 1 minute, but the ground observer would see it in 2. the train observer would see the ground as shorter, the ground observer would see the train as shorter. the train observer would see both light beams traveling at 1.6*c toward the stationary receiver, whereas the ground observer would see both light beams traveling at c. i just can't see how you can accept all this as true without having you brain explode from the contradiction.

[modest]

It appears, phillip, that you do not have a working understanding of special relativity. This might be more of a problem than you are accrediting it. Mixing classical intuition with a constant speed of light will give you several paradoxes as you say.

but its not just one paradox, its several.

the train observer would see both light beams reach the position of the train receiver in 1 minute, but the ground observer would see it in 2.

Yes.

 

I recommend this introductory wiki book on special relativity:

Special Relativity - Wikibooks, collection of open-content textbooks

the train observer would see the ground as shorter, the ground observer would see the train as shorter.

That is correct. Length contraction follows the principle of reciprocity (i.e. it is reciprocal)

the train observer would see both light beams traveling at 1.6*c toward the stationary receiver, whereas the ground observer would see both light beams traveling at c.

Nowhere in the above example does any observer see a photon traveling anything but c.

i just can't see how you can accept all this as true without having you brain explode from the contradiction.

You might consider that there are no contradictions in special relativity—that it is an internally consistent system—but that an invariant speed of light causes contradictions without special relativity.

 

In other words: you are trying to apply a constant speed of light to classical physics. That is bound to cause paradoxes which is why special relativity was created in the first place.

 

~modest

[phillip1882]

i'll start form the bottom and work my way up here.

In other words: you are trying to apply a constant speed of light to classical physics. That is bound to cause paradoxes which is why special relativity was created in the first place.

isn't that exactly what Einstein is doing? he assumed classical physics is in general correct, that is space time velocity wouldn't be different for any observer at any speed, but light has its special property.

Nowhere in the above example does any observer see a photon traveling anything but c.

the train observer would see the stationary receiver approach him at 0.6*c yes? he would see his light beam travel away from him at c, yes? he would therefore conclude that the light beams are traveling toward the stationary receiver at 1.6*c yes?

Length contraction follows the principle of reciprocity.

but that doesn't make sense. either the ground observer or the train observer has to be right they cant both be right. that is if the ground observer sees the train as shorter then the train observer should see the ground as longer. or vice versa.

I recommend this introductory wiki book on special relativity

if this is the best that can be offered in the way of explanation i doubt i would want to.

[freeztar]

isn't that exactly what Einstein is doing? he assumed classical physics is in general correct, that is space time velocity wouldn't be different for any observer at any speed, but light has its special property.

No, not at all. He turned classical physics on its head with relativistic physics (well, in fairness, many people had a part of this). Observers travelling at different velocities observe time and space differently.

 

the train observer would see the stationary receiver approach him at 0.6*c yes? he would see his light beam travel away from him at c, yes? he would therefore conclude that the light beams are traveling toward the stationary receiver at 1.6*c yes?

No. In relativity, you can not simply add velocities together like that. It's a bit more than simply a+b (classical physics). I can't remember the exact equation off the top of my head, but I'll go find it and edit it in shortly. EDIT: The equation is

 

[math]w=\frac{u+v}{1-\frac{uv^2}{c}}[/math]

 

Where:

w = combined velocity

u = train's velocity

v = train's light signal velocity ©

 

Here's the page I got it from which has a more in depth explanation: Relativistic Velocities

 

but that doesn't make sense. either the ground observer or the train observer has to be right they cant both be right.

 

Yes, they can. They are both correct from their respective reference frames.

 

I know it doesn't make sense intuitively. It's a very hard concept to grasp for many, myself included. I can't pretend to explain to you the reasons for why time dilation and length contraction occur, much like I can't explain to you what exactly "time" or "space" are, but the effects are observable. Space and time are dependent upon velocity and distance to massive objects. It's very bizzarre, I know, but if you stick with the exanatory power of the math right now and focus on the effects of relativity, you'll be better prepared to tackle the philosophical aspects, including paradoxes.

 

that is if the ground observer sees the train as shorter then the train observer should see the ground as longer.

 

That's exactly what Modest said above.

if this is the best that can be offered in the way of explanation i doubt i would want to.

 

The link Modest gave is great. I highly recommend you heed his advice.

Otherwise, this attitude of "I can't be bothered" is likely to become reciprocal.

[modest]

I recommend this introductory wiki book on special relativity

if this is the best that can be offered in the way of explanation i doubt i would want to.

Regardless of how you feel about special relativity, these concepts have been experimentally confirmed. Special relativity enjoys some of the most extremely accurate predictive confirmations in physics. Everything I'm telling you has been experimentally tested and found true. If your intention is to say "that doesn't make sense so it must be wrong" then let me know and I won't waste more effort explaining it. If your intention is to say "why doesn't this seem to make sense" then I think that's a far-more useful and scientific approach.

 

 

i'll start form the bottom and work my way up here.

In other words: you are trying to apply a constant speed of light to classical physics. That is bound to cause paradoxes which is why special relativity was created in the first place.

isn't that exactly what Einstein is doing? he assumed classical physics is in general correct, that is space time velocity wouldn't be different for any observer at any speed, but light has its special property.

Relativistic mechanics affects much more than the velocity of light. In Newtonian mechanics velocities are added. Think of the train in the example we formed above. It is traveling at 0.6c. Now imagine the person on the train observed another train going in the same direction and measured it's velocity to be 0.6c. In other words, the person on the ground measured the speed of train A to be 0.6c relative to the ground and the person in train A measured the velocity of train B to be 0.6c relative to train A.

 

We might ask: at what velocity is the person on the ground going to measure train B? Newtonian mechanics (before Einstein) would say you add the velocities 0.6c + 0.6c = 1.2c. This makes intuitive sense. If a person is in a car going 80 km/hour and shoots an arrow at 100 km/hour in the same direction that the car is traveling then how fast does the arrow go compared to the road? You add the velocities: 80 + 100 = 180 km/hour.

 

But, if special relativity is correct then a different formula for velocity addition arises from the derivation:

[math]w=\frac{u+v}{1+\left( \dfrac{uv}{c^2}\right)}[/math]

It's explained at this link:

How Do You Add Velocities in Special Relativity?

To answer the train example where the ground observer measures u=0.6c and the train-A observer measures v=0.6c we get:

[math]w=\frac{u+v}{1+\left(\dfrac{uv}{c^2} \right) }=\frac{0.6+0.6}{1+\left(\dfrac{ 0.6*0.6}{1^2} \right) } = \mathbf{0.882c}[/math]

According to SR (special relativity) the ground observer will measure the speed of train-B at 0.882c. So, it's not just the speed of light that is special. All frames of reference which are going fast-enough to reveal relativistic effects must be treated differently. When speeds are very low compared to the speed of light the answers are the same as Newtonian mechanics. If we use the SR formula for the car / arrow situation for example:

[math]w=\frac{u+v}{1+\left(\dfrac{uv}{c^2} \right) }=\frac{80+100}{1+\left(\dfrac{80*100}{1079252850^2} \right) }=\mathbf{180 \ km/hour}[/math]

At slow speeds the u+v method works. But, it's just an approximation. In order for all observers to measure the speed of light the same we need to use relativistic mechanics. This means time dilation and length contraction leading to the Einstein velocity addition formula.

Nowhere in the above example does any observer see a photon traveling anything but c.

the train observer would see the stationary receiver approach him at 0.6*c yes?

Yes. The ground detector moves to the left at .6c in the train frame.

he would see his light beam travel away from him at c, yes?

Yes. Light moves to the right at c in the train frame.

he would therefore conclude that the light beams are traveling toward the stationary receiver at 1.6*c yes?

Basically, yes, but you should word it a bit differently. In the train frame, the light and the detector are approaching each other at 1.6c. Of course, this is a different thing from saying: "in the ground frame light is approaching the detector at...". It is not a violation of special relativity to have two things in a single frame of reference which appear to approach each other at more than the speed of light. Imagine shining two flashlights in opposite directions. Each light propagates at c, so in the observer's frame of reference the distance between the two photons increases at twice the speed of light. This is not a violation of SR.

 

SR says that no observer can measure something traveling greater than c. In your example, no observer is seeing anything going faster than c. If the train observer sees the photon moving at c toward the detector and the detector moving at .6c toward the train and he wants to know how fast the detector would observe the photon moving toward it he would use the Einstein velocity formula:

[math]v=\frac{w-u}{1-\left(\dfrac{wu}{c^2} \right) }=\frac{-0.6-1}{1-\left(\dfrac{-0.6*1}{1^2} \right) }=\mathbf{-1c}[/math]

The person in the train concludes that the detector on the ground should see the photon have a velocity of -1c. (-1c means approaching at the speed of light). This is perfectly consistent.

 

The reason this seems like a paradox is because you are intuitively adding velocities (and you probably don't even realize you're doing it). You're thinking: the train is going 0.6c and the photon is going 1c compared to the train so the person on the ground must see (0.6 + 1 = 1.6c). But, that's exactly what Einstein fixed. He figured out that rulers and clocks on a moving reference frame will work differently than the reference frame. This leads directly to the conclusion that you can't simply add velocities.

 

Length contraction follows the principle of reciprocity.

but that doesn't make sense. either the ground observer or the train observer has to be right they cant both be right. that is if the ground observer sees the train as shorter then the train observer should see the ground as longer. or vice versa.

Why? If any ruler which is moving relative to my ruler is length contracted and your ruler is moving relative to mine then yours should be length contracted. Of course, you will see me moving relative to you so you'll likewise demand that my ruler is length contracted relative to yours. In my frame of reference it is you that is moving while in your frame of reference it is me that is moving. This is not a contradiction. It is a necessary condition for relativity to be consistent.

 

~modest

 

EDIT: You beat me to it Freezy, we think so much alike—even referenced the same website :)

[stereologist]

You continue to make the same mistakes again and again. It's long overdue for you to read a book on the subject. Modest has gone way above the call of duty here in presenting a detailed example.

 

he assumed classical physics is in general correct, that is space time velocity wouldn't be different for any observer at any speed, but light has its special property.

 

Read a book and find out where you are mistaken here.

 

he would therefore conclude that the light beams are traveling toward the stationary receiver at 1.6*c yes?

 

As we have repeated staated again and again NO. Everyone sees c!

 

but that doesn't make sense. either the ground observer or the train observer has to be right they cant both be right.

Read a book and learn why this must be case.

 

if this is the best that can be offered in the way of explanation i doubt i would want to.

 

Despite the doubts you exhibit I feel confident that you can get through a book. You have all of the basic pieces you need to do this: curiosity, and an understanding of math.

 

You can do it and better yet I know you will do it well.

[phillip1882]

Read a book and find out where you are mistaken here.

:)

okay, i'll once again post exactly what i mean.

einstein's two assumptions are...

"The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of coordinates in uniform translatory motion."

in english, weather you are going the speed of light or 10 miles an hour, you would measure the ticking of a clock, the motion of an object you threw etc. as being the same within your frame of reference. this part i completely agree with.

"As measured in an inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body."

that is light is always the same speed for all observers, so if your going 90% of the speed of light, and a stationary receiver fired a light beam, you would see it pass you at not 10% of the speed of light, but light speed.

this is what i object to, because accepting it as true creates three self contradicting paradoxes.

the first is length, according to you, the train going 60% speed of light would see the ground as shorter, but the ground observer would see the train as shorter. in particular, in the example you provided, you have the back of the train well past the ground receiver when the light hits the train receiver, from the ground perspective. yet from the train perspective, the back of the train has yet to do so.

the second is time. according to you, the ground observer would see both light beams reach the train receiver after 2 minutes, and the train observer would see them both reach the same position after 1 minute. once again i don't see how both can be right if their clocks start at the same time. if light travels at a constant speed, no matter your velocity, then either both should see it reach the same position after 1 minute, or both at 2 minutes. otherwise light is covering the exact same distance in two different periods of time.

the third is velocity. once again, the train observer would conclude his light beam is traveling toward the stationary receiver at 1.6*c, and the ground observer would conclude his light beam is traveling at c toward the same receiver. yet your claiming that each would see the others light beam traveling at the same speed. :)