Superconductivity at room temperature??

[W.D.Clinger]

As explained above, johanfprins and/or his friend have misinterpreted the passage to which they are objecting. That took some doing, because they had to ignore Einstein's parenthetical remark, which I highlighted at the top of this post.

As seen by observer M, observer M' really is "hastening into a light beam and riding ahead of the other light beam." johanfprins and/or his friend appear to be assuming that Einstein's remark applies also to what is seen in the inertial frame of observer M'. That's a rookie mistake.

Posting two minutes later, johanfprins made essentially the same mistake once again:

Thus, Einstein's reasoning is wrong; and this same wrong reasoning led Einstein to derive a length contraction, which does not occur within K, and a time dilation which does not occur within K^/.

As seen by observer K', observer K really is "hastening into a light beam and riding ahead of the other light beam." johanfprins appears to be assuming that Einstein's remark applies also to what is seen in the inertial frame of observer K. That's a rookie mistake.

 

[ben m]

So, according to an observer which finds him/herself "by accident" at the same position as the detector was within K when the lights switched on, the magnitude of light speeds relative to him/her is the same: Namely c from the tail and -c from the nose.

Yep.

He is thus NOT "hastening towards" or "riding ahead of any light beam" as Einstein has reasoned.

Sure he is. It's just that "hastening towards" a light source does NOT change the apparent velocity of the light. That's one of the postulates, AND one of the visible details of the explicit Lorentz-transformation scenario you set up. Everything's fine.

What's the problem? You don't like the word "hastening"?

 

[johanfprins]

Einstein wrote: "hastening towards the beam of light coming from B " thus clearly reasoning that the speed of the light relative to the observer is in this case c+v, and he wrote "riding on ahead of the beam of light coming from A" thus clearly arguing that the speed of this light relative to the observer is c-v: And he then clearly argued that it is these differences in light speeds relative to the observer which is reponsible for the one light switching on before the other light within the reference frame K of the observer.

This cannot be so since light approaches any observer within K from both directions with a speed c. The reason why he/she sees the one light switching on before the other has nothing to to with running into a light beam or running away from a light beam. In fact, the one light will switch on before the other light whether there is an observer in K or not an observer in K. It has NOTHING to do with an observer. When the observer synchronises his/her clock when the lights switch on at the moment when the nose light passes the observer, still demands that the tail light has already switched on before he/she has synchronised his/her clock.

An observer can only hasten towards a light source, and then he/she will see that the light is redder than the light coming from a light source from which he/she is riding on ahead.

 

[ben m]

Einstein wrote: "hastening towards the beam of light coming from B " thus clearly reasoning that the speed of the light relative to the observer is in this case c+v, and he wrote "riding on ahead of the beam of light coming from A" thus clearly arguing that the speed of this light relative to the observer is c-v:

Ah, good old Einstein. Einstein whose 1st postulate of relativity is "the speed of light is constant", who wrote down mathematical transformations under which the speed of light is constant, who constructed the very thought experiment that you're using to illustrate the constancy of the speed of light ...

... but never mind all that. You have discovered that he didn't think the speed of light is constant, because he used the phrasing "hastening towards the beam" rather than "hastening towards the light source" in a popular-level book on relativity he wrote in 1920.

Is that the idea?

Listen to yourself, Johan. This is pathetic.

 

[ben m]

What does "has had" mean? The fact remains that a photon has the same single frequency and value for k within all inertial reference frames:

An observer can only hasten towards a light source, and then he/she will see that the light is redder than the light coming from a light source from which he/she is riding on ahead.

Huh.

 

[johanfprins]

It is amazing how Ben, Sol and Ziggurat will distort anything to defend wrong physics just because they have been "converted and washed in the blood of the holy triumviarate". Why would hastening towards a light-source and riding on ahead of a light-source, cause you to see the light emitted from the light-source (towards which you are rushing) first and the light from the light source (from which you are moving away), unless the speed with which the light front of the light-beam emitted from the light source (you are moving into) is approaching you faster than the front of the light beam from the light source you are riding on ahead? Remember, the two light fronts were emitted at the time Einstein's observer synchronised his/her clock and when the observer was equidistant from the light sources!

Yes it is amazing that Einstein could not see that he violated his own postulate. If only Sol, Ben and Ziggurat were able to do a bit of algebra using the Lorentz transformation, they will see that it has NOTHING to do with Einstein's argument why you see one light switching on before the other, but to do with the fact that when you synchronise the clock in K with the light switching on in the tail, the nose of the spaceship is in the future of K, but when you synchronise a clock in K with the light switching on in the nose, the tail is, within K, in the past.

 

[ben m]

Why would hastening towards a light-source and riding on ahead of a light-source, cause you to see the light emitted from the light-source (towards which you are rushing) first

By moving towards the source, you're reducing the distance d between yourself and the source, and the light-travel-time is t = d/c. Reducing d reduces t.

This isn't even complicated.

 

[Ziggurat]

It is amazing how Ben, Sol and Ziggurat will distort anything to defend wrong physics just because they have been "converted and washed in the blood of the holy triumviarate". Why would hastening towards a light-source and riding on ahead of a light-source, cause you to see the light emitted from the light-source (towards which you are rushing) first and the light from the light source (from which you are moving away), unless the speed with which the light front of the light-beam emitted from the light source (you are moving into) is approaching you faster than the front of the light beam from the light source you are riding on ahead?

You have already provided the answer. In the reference frame of the train, the emissions are not simultaneous. The emission from the source you're moving towards (in the reference frame of the ground) occurred earlier in your reference frame, therefore you receive it first. But in the reference frame of the ground, the emissions are simultaneous, and it's your motion that makes you receive one before the other.

Quite simple, really. Yet it's clear you don't understand what's being claimed.

Remember, the two light fronts were emitted at the time Einstein's observer synchronised his/her clock and when the observer was equidistant from the light sources!

They were emitted simultaneously in the reference frame of the ground, NOT in the reference frame of the observer moving with the train. And in the reference frame of the ground, they are equidistant from where that moving observer is when they were emitted. That's the whole point. That's how he demonstrated that simultaneity is reference-frame dependent. Furthermore, since that observer is moving in the reference frame of the ground, the emission events are not equidistant from the detection events in the ground reference frame. What the emitters do after they emit is actually quite irrelevant to the problem.

Yes it is amazing that Einstein could not see that he violated his own postulate.

It's amazing that you clearly have no clue about what he's actually saying.

If only Sol, Ben and Ziggurat were able to do a bit of algebra using the Lorentz transformation, they will see that it has NOTHING to do with Einstein's argument why you see one light switching on before the other

Oh, I can do the algebra. I doubt you can, though. And I know you won't. Internet cranks never actually do the math.

 

[ben m]

If only Sol, Ben and Ziggurat were able to do a bit of algebra using the Lorentz transformation,

Don't imagine that I can't do the math. I *teach* this stuff, Johan. I just did the intro-special-relativity derivations a few weeks ago. We derive the time dilation from the vertical-bouncing-light experiment, the space contraction from the one-way-light-time-of-flight, and the simultaneity shift from the two-lightning bolts. Introducing the full Lorentz transformations, we treated various thought experiments (twins, ladder-barn, etc.) both algebraically and via Minkowski diaThen we covered the Lorentz transformations. Then we introduced the Minkowski metric, relativistic invariants, and the Lorentz group.

they will see that it has NOTHING to do with Einstein's argument why you see one light switching on before the other, but to do with the fact that when you synchronise the clock in K with the light switching on in the tail, the nose of the spaceship is in the future of K, but when you synchronise a clock in K with the light switching on in the nose, the tail is, within K, in the past.

Still not seeing a problem.

 

[sol invictus]

Edited by LashL: 

Removed breach.

PLEASE derive Einstein's train experiment (I just tried to make it simpler for you by suggesting the spaceship) from the Lorentz transformation, if you are able to this simple mathematics, then you will see that it is NOT POSSIBLE for any observer EVER to "hasten towards a beam light coming from B" and to "ride ahead of the beam of light coming from A" It is prohibited by Einstein's own first postulate and is also inherent in the Lorentz transformation that it is impossible.

Edited by LashL: 

Removed breach.

Nowhere have I or my friend said that the two simultaneous events will also be simultaneous within another inertial reference frame. All that is claimed, and which follows directly from the Lorentz transformation is that this is NOT CAUSED by an observer "hastening into a light beam and riding ahead of the other light beam" The latter is not possible since the speed of light is always the same relative to any object no matter how fast the object is moving!!!

Edited by LashL: 

Removed breach.

Nonsense. If in frame F object A and a photon are a distance D apart from each other at time t=0, and if they are moving directly towards each other - object A with constant speed v, and the photon with speed c - they will collide at time T=D/(v+c). If instead object A is moving in the same direction as the photon, the collision will happen at time T=D/(c-v). (Those times are as measured in frame F, of course.)

That's true by definition of distance and speed.

Of course this does not imply that the photon moves at speed v+c or c-v in the rest frame of A. When you change frames, the non-linearity of the Lorentz transforms becomes important. But here there is no need to change frames.

ETA - incidentally, the extreme case of this is two photons flying directly towards each other. What is their relative speed? 2c, of course - meaning that if they are a distance D apart, they will collide after a time D/(2c), not D/c. Does that 2c mean something is moving faster than c? No, it does not. Does that 2c mean that in the some other frame, one of the photons is moving faster than c? No, it does not.

 

[johanfprins]

Let us try and make it very simple since it seems that Ben, Sol and Ziggurat at least understand that speed multiplied by time gives a distance. Now consider two identical spaceships K and K^/; each stationary within its own inertial reference frame where the reference frames K and K^/ are moving with a speed v relative to each other. The identical stationary lengths of the spaceships are L₀. Consider the case where the spaceships pass each other along their lengths and when their noses are at the same position the two captains, both sitting in the noses of their respective spaceships, synchronise their clocks and then measure the time it takes for the tail of the other spaceship to reach the nose of his spaceship so that he/she can calculate the length of the passing spaceship. This means that the captain in K will calculate a length L=v(delta)t and tyhe captain in K^/ will calculate a length of L^/=v(delta)t^/.

Now according to the most sacred principle in Special Relativity, the laws of physics must be the same within both K and K^/. Thus when the captain in K measures the length L of a passing spaceship with stationary length L₀ passing him by and the captain in K^/ measures the length L^/ of a spaceship with stationary lengthj L₀ passing him by, both must obtain the same length: i.e. L must be the same as L^/; even if the measured lengths are contracted (which they are not). This means that since L=L^/ the time measurements must be identical: i.e. that: (delta)t must be exactly the same as (delta)t^/. And this means that their two clocks must keep identical time. There is no real time-dilation. From the equations above if should be clear that L=L^/=L₀. Thus there is also not any real actual length contraction.

I hope that you have been taught that length=time*speed. If not please go and sue your primary school teachers and PLEASE do not think that you are competent enough to teach Special Relativity at university level

 

[Ziggurat]

Let us try and make it very simple since it seems that Ben, Sol and Ziggurat at least understand that speed multiplied by time gives a distance. Now consider two identical spaceships K and K^/; each stationary within its own inertial reference frame where the reference frames K and K^/ are moving with a speed v relative to each other. The identical stationary lengths of the spaceships are L₀. Consider the case where the spaceships pass each other along their lengths and when their noses are at the same position the two captains, both sitting in the noses of their respective spaceships, synchronise their clocks and then measure the time it takes for the tail of the other spaceship to reach the nose of his spaceship so that he/she can calculate the length of the passing spaceship. This means that the captain in K will calculate a length L=v(delta)t and tyhe captain in K^/ will calculate a length of L^/=v(delta)t^/.

Now according to the most sacred principle in Special Relativity, the laws of physics must be the same within both K and K^/. Thus when the captain in K measures the length L of a passing spaceship with stationary length L₀ passing him by and the captain in K^/ measures the length L^/ of a spaceship with stationary lengthj L₀ passing him by, both must obtain the same length: i.e. L must be the same as L^/; even if the measured lengths are contracted (which they are not). This means that since L=L^/ the time measurements must be identical: i.e. that: (delta)t must be exactly the same as (delta)t^/. And this means that their two clocks must keep identical time. There is no real time-dilation. From the equations above if should be clear that L=L^/=L₀. Thus there is also not any real actual length contraction.

I hope that you have been taught that length=time*speed. If not please go and sue your primary school teachers and PLEASE do not think that you are competent enough to teach Special Relativity at university level

You have contradicted yourself, and you don't even realize it. You have acknowledged that simultaneity is no longer constant for different reference frames. But you have constructed a scenario in which two spatially separated events (the nose of rocket 1 reaching the tail of rocket 2, and the nose of rocket 2 reaching the tail of rocket 1) are simultaneous in two different reference frames. This is inconsistent with your own claims, never mind the Lorentz transformations.

You've lost it, johan. Whatever capacity you once had for physics, it now eludes your grasp. Time is evidently cruel indeed.

 

[sol invictus]

Let us try and make it very simple since it seems that Ben, Sol and Ziggurat at least understand that speed multiplied by time gives a distance. Now consider two identical spaceships K and K^/; each stationary within its own inertial reference frame where the reference frames K and K^/ are moving with a speed v relative to each other. The identical stationary lengths of the spaceships are L₀. Consider the case where the spaceships pass each other along their lengths and when their noses are at the same position the two captains, both sitting in the noses of their respective spaceships, synchronise their clocks and then measure the time it takes for the tail of the other spaceship to reach the nose of his spaceship so that he/she can calculate the length of the passing spaceship. This means that the captain in K will calculate a length L=v(delta)t and tyhe captain in K^/ will calculate a length of L^/=v(delta)t^/.

Now according to the most sacred principle in Special Relativity, the laws of physics must be the same within both K and K^/. Thus when the captain in K measures the length L of a passing spaceship with stationary length L₀ passing him by and the captain in K^/ measures the length L^/ of a spaceship with stationary lengthj L₀ passing him by, both must obtain the same length: i.e. L must be the same as L^/; even if the measured lengths are contracted (which they are not). This means that since L=L^/ the time measurements must be identical: i.e. that: (delta)t must be exactly the same as (delta)t^/.

Yes, (delta)t=(delta)t'. By symmetry, each captain must measure the same time and same length.

And this means that their two clocks must keep identical time.

Wrong. As Zig points out, the two events (nose of 1 passes tail of 2, and nose of 2 passes tail of 1) are not simultaneous in either frame K or K'. Since you've already accepted the relativity of simultaneity, that proves that in frame K, the length of ship K' must be contracted (else those two events would be simultaneous), and vice versa.

But if the events are not simultaneous in frame K and yet the captain of K' obtains a reading (delta)t'=(delta)t, his clock must be running slow.

 

[johanfprins]

I noted this absurd argument: "But you have constructed a scenario in which two spatially separated events (the nose of rocket 1 reaching the tail of rocket 2, and the nose of rocket 2 reaching the tail of rocket 1) are simultaneous in two different reference frames."

Read again what I have written: I have ONE synchronisation when the noses pass and the two captains set their respective clocks to zero. After each one separately later switched off his/her clock, they then each calculate the lengths L^/ and L they measured. Captain K stops his clock when the tail of spaceship K^/ passes him to then obtain L=v(delta)t, and captain K^/ stops his clock when the tail of spaceship K passes him to calculate L^/=v(delta)t^/. Have I at this point assumed for no reason at all that the two switching-off actions of the captains are simultaneous: NO I HAVE NOT!

The crucial argument is that captain K calculated the length L of the spaceship, which when stationary has a length L₀; while captain K^/ calculated the length of spaceship K, which when stationary has a length L₀. Thus, the two experiments are identical within both K and K^/: Therefore according to Einstein's postulate that the outcome of the same physics-experiment done within two different inertial reference frames MUST be exactly the same within each reference frame, mandates that L must be equal to L^/. And from this it follows that the time-intervals must be the same and this then proves impeccably that therefore the two captains must also have switched off their clocks simultaneously.

Why, after synchronisation, can two subsequent events within K and K^/ not occur simultaneously? The only time when simultaneous events are non-simultaneous within another reference frame is when they are seperated in space and occuring at the same instant in time within the first reference frame. In the expereiment discussed above one has firstly, that within each reference frame the spatial coordinates of the clock within that refrence frame does NOT change and secondly the clock is not switched on and off at the same instant in time. So what has this got to do with simultaneity within a single reference frame at two separated positions? REALLY my patience is running thin: It is difficult not to cause another infraction when a person posts such utter nonsense!

 

[sol invictus]

Read again what I have written: I have ONE synchronisation when the noses pass and the two captains set their respective clocks to zero. After each one separately later switched off his/her clock, they then each calculate the lengths L^/ and L they measured. Captain K stops his clock when the tail of spaceship K^/ passes him to then obtain L=v(delta)t, and captain K^/ stops his clock when the tail of spaceship K passes him to calculate L^/=v(delta)t^/. Have I at this point assumed for no reason at all that the two switching-off actions of the captains are simultaneous: NO I HAVE NOT!

Yes, you have - because you went from there to the (incorrect) conclusion that neither space nor time is dilated. The truth is that both are (and must be to preserve the constancy of the speed of light).

The crucial argument is that captain K calculated the length L of the spaceship, which when stationary has a length L₀; while captain K^/ calculated the length of spaceship K, which when stationary has a length L₀. Thus, the two experiments are identical within both K and K^/: Therefore according to Einstein's postulate that the outcome of the same physics-experiment done within two different inertial reference frames MUST be exactly the same within each reference frame, mandates that L must be equal to L^/. And from this it follows that the time-intervals must be the same

The clocks will read the same, yes.

and this then proves impeccably that therefore the two captains must also have switched off their clocks simultaneously.

No, it quite obviously does not prove that. If one clock runs slow but is switched off later, the two clocks may read the same - and that, of course, is precisely what happens (in either frame K or K').

REALLY my patience is running thin: It is difficult not to cause another infraction when a person posts such utter nonsense![/SIZE]

It's not surprising that being wrong all the time makes you angry, Johan.

 

[johanfprins]

It's not surprising that being wrong all the time makes you angry, Johan.

It is clearly you who have proved on this forum that you have exactly the same mentality that a Cardinal in the inquisition had during the time of Galileo. Yes what has been done to Galileo makes me mad and when this mentality still lives within a person at present, as it does within you, I cannot help to see red.

So let us see if you can answer a direct question for a change: Do you agree that when you do the same experiment within two inertial reference frames which are moving with a constant speed relative to each other, you must, excluding experimental error, obtain exactly the same result?

To expand a bit: When a rod with length L moves with a speed v relative to an inertial reference frame and when another rod with exactly the same length L moves with the same speed v relative to another inertial reference frame (which in turn is moving relative to the first inertial reference frame), then identical measurements made on the rods within their respective reference frames relative to which they are moving, must according to Einstein give exactly the same result.

 

[sol invictus]

It is clearly you who have proved on this forum that you have exactly the same mentality that a Cardinal in the inquisition had during the time of Galileo. Yes what has been done to Galileo makes me mad and when this mentality still lives within a person at present, as it does within you, I cannot help to see red.

Physics crackpot index: "40 points for comparing yourself to Galileo, suggesting that a modern-day Inquisition is hard at work on your case, and so on."

So let us see if you can answer a direct question for a change: Do you agree that when you do the same experiment within two inertial reference frames which are moving with a constant speed relative to each other, you must, excluding experimental error, obtain exactly the same result?

To expand a bit: When a rod with length L moves with a speed v relative to an inertial reference frame and when another rod with exactly the same length L moves with the same speed v relative to another inertial reference frame (which in turn is moving relative to the first inertial reference frame), then identical measurements made on the rods within their respective reference frames relative to which they are moving, must according to Einstein give exactly the same result.

Yes, that's true. That's why the two captains both measure the same (delta)t. What is incorrect is your next step. As I just explained to you, equal (delta)t could result because there is no length contraction and no time dilation, or because there is both length contraction and time dilation.

Each captain will interpret the other's result as due to the fact that the other's ship is length contracted (hence (delta)t<L₀/v) and that the other's clock is running slow (hence (delta)t'<L₀/v, and in fact, (delta)t=(delta)t'). Of course there is a perfect symmetry between them - which is all that the constancy of the laws of physics in all inertial frames guarantees.

This is physics that high school students and 18-year-old physics freshman learn without difficulty, Johan. It's not much harder, or indeed much different, than the "physics" of perspective - that rotated objects may look thinner or thicker depending on the angle.

 

[johanfprins]

Yes, that's true. That's why the two captains both measure the same (delta)t. What is incorrect is your next step. As I just explained to you, equal (delta)t could result because there is no length contraction and no time dilation, or because there is both length contraction and time dilation.

So let us assume that there is both length contraction and time dilation then it still means that they both measure the same time-intervals and lengths. This will mean that in one case you will have that: (delta)t_n=(gamma)(delta)t^/, where (delta)t_n is the non-dilated time interval within K and you will have that (delta)t_n^/=(gamma)(delta)t where (delta)t_n^/ is the non-dilated time in K^/. Now set the dilated times in these equations equal, and you obtain that the non-dilated time intervals are also equal. So you are talking nonsense again.

This is physics that high school students and 18-year-old physics freshman learn without difficulty, Johan. It's not much harder, or indeed much different, than the "physics" of perspective - that rotated objects may look thinner or thicker depending on the angle.

Yes this is what they learn from people like you who do not know Artha from Martha when it comes to logic.

 

[sol invictus]

So let us assume that there is both length contraction and time dilation then it still means that they both measure the same time-intervals and lengths. This will mean that in one case you will have that: (delta)t_n=(gamma)(delta)t^/, where (delta)t_n is the non-dilated time interval within K

I have no idea what "the non-dilated time interval within K" is supposed to mean. In frame K, one ship is stationary, one is moving. The captain of the stationary ship measures Δt=L₀/(γv). Naturally, he interprets this as indicating that the moving ship has a length L₀/γ; i.e., that it is length contracted by a factor of γ.

The clock of the moving ship K' will read the same Δt. But the explanation - according to frame K - is different. According to stationary clocks in frame K, the moving clock runs slow by a factor of γ. Therefore since the stationary ship is of length L₀, the moving clock reads Δt'=L₀/(γv) when the moving ship passes it.

As you can see, Δt=Δt', which is all that is required by the invariance of physical law.

 

[ben m]

So let us assume that there is both length contraction and time dilation then it still means that they both measure the same time-intervals and lengths. This will mean that in one case you will have that: (delta)t_n=(gamma)(delta)t^/, where (delta)t_n is the non-dilated time interval within K and you will have that (delta)t_n^/=(gamma)(delta)t where (delta)t_n^/ is the non-dilated time in K^/.

Nope. The formula "dt' = gamma dt" is a special case: it only applies to two events happening in the same place in the unprimed frame. The train car has events happening in three different places (front, back, and center) and this formula is inappropriate.

Similarly, the formula "dx' = gamma dx" is strictly appropriate only for measuring the spatial separation of two events that are simultaneous in the unprimed frame. (This one is much easier to apply, especially in problems involving rigid objects.

Try again and actually use the Lorentz transformation, Johan. Including the off-diagonal terms. Write down the four-vector coordinates, in the ground rest frame, of the two flashes and of the "light arrives at detector" event. Transform these three coordinates into the train rest frame. Look at the result.