Why are the effects of Time Dilation permanent but Length Contraction is Not?

[nitsuj]

Understood Dalespam.

I appreciate what you are saying.

You've been patient with me in that past with such things & George too to a lesser extent. It's really really helped me in trying to grasp the concepts of SR intuitively.

So it kinda burns me to see a post like I've been commenting on. Where a poster is on the cusp of being correct, but is instead "written off" as being not worthy (the interpretations) of a helpfull reply.

 

[Qzit]

I did not know my thoughts were LET or crackpot ideas Sorry. Please help me understand a little better.

False. Distance is contracted too.

I thought it was only the view that was contracted and not the distance. If your synapsis is slowed than distance would appear contracted but the physical distance remains the same for a photon to travel. How does a spaceship actually contract space in front of it? Please help I am confused. I did not understand that a spaceship could do anything to the space in front of it. If I were mid-point to a destination would I be in a physical contraction?

Your sense of time is contracted. If you were traveling at 0.9C and a photon closing the distance towards you would be 1.9C. A photon traveling parallel with your ship would have a relative departing speed of 0.1C.

True only in the Earth frame.

"C" I thought was a constant. What does not matter where you observe? If there is no observer "C" is always "C". What you observe is an accurate mathematical difference based on relativity. Why do you see it more than an optical affect?

This would be true on the return journey, but false on the outward journey.

If you had a magical telescope (which is impossible to begin with) say your Synapsis on the ship is firing at 16 frames an hour and your twin is firing at16 frames per second on Earth. The Earth twin would always be moving much faster to my way of thinking. This would be real and not just observed I would suspect. If I am wrong please explain.

 

[ghwellsjr]

I still maintain that Qzit has a lot of erroneous ideas. Most basically is his concept expressed in his last sentence that it's only because of your sluggish neurons in your For that explains why observations differ from reality:

Now what is physically happening and what you observe to be happening are two different things because of the speed of your synapsis and the cycling electrons in your FoR.​

In your FoR, your synapses and cycling electrons are perfectly normal. It's the other person's synapses and cycling electrons that are experiencing time dilation. He's got this backwards.

How about this comment:

In fact they can be so slowed that if you had a magic telescope to watch your twin on Earth he would seem to move so fast you would think of him as the comic book hero the Flash. Is the distance contracted? No. Your sense of time is contracted.

He's thinking that because your processes are running slowly, you would see your Earth twin's processes as running fast which is not correct. He's not talking about Relativistic Doppler which is observable with a real telescope, he's talking about a magic telescope that eliminates the Doppler to let you see what's really going on. And he has it wrong, the traveling twin would see the Earth twin's processes as going slow, not fast.

And what do you make of his last comment: "Your sense of time is contracted."? Time is not contracted, it's dilated, but either way, your sense of time is always normal, never contracted or dilated.

Now as to his comments regarding the closing speed of photons which he said is not related to the Twin Paradox:

If you were traveling at 0.9C and a photon closing the distance towards you would be 1.9C. A photon traveling parallel with your ship would have a relative departing speed of 0.1C.

I don't think he understands that the speed of a photon, like the propagation speed of all light, AKA the one-way speed of light, is never observable, it is assigned a speed by a theory, in this case the Theory of Special Relativity, and that speed is always c, not 1.9c or 0.1c. In his later post, he asked:

My other question to you is what is the closing speed between two photons traveling on a collision course?

The answer to that is 2c but it should not be regarded as any measurable or physically real speed, it's just the difference between two speeds in a given FoR. There is nothing traveling at 2c with respect to any FoR or to any other object.

I really want to help Qzit learn SR and if I have misdiagnosed his comments, then I apologize, and he can help me by providing relevant feedback.

 

[nitsuj]

I did not know my thoughts were LET or crackpot ideas Sorry. Please help me understand a little better.
I thought it was only the view that was contracted and not the distance. If your synapsis is slowed than distance would appear contracted but the physical distance remains the same for a photon to travel. How does a spaceship actually contract space in front of it? Please help I am confused. I did not understand that a spaceship could do anything to the space in front of it. If I were mid-point to a destination would I be in a physical contraction?

"C" I thought was a constant. What does not matter where you observe? If there is no observer "C" is always "C". What you observe is an accurate mathematical difference based on relativity. Why do you see it more than an optical affect?

If you had a magical telescope (which is impossible to begin with) say your Synapsis on the ship is firing at 16 frames an hour and your twin is firing at16 frames per second on Earth. The Earth twin would always be moving much faster to my way of thinking. This would be real and not just observed I would suspect. If I am wrong please explain.

I am not skilled enough in SR to debate certain SR concepts.

But I can tell you forget about including biology if you plan on learning SR concepts here. Not supprisingly it shouldn't come up.

Understand that time and length are measurements. It is these measurements that are comparatively different with relative motion. You can as easily works backwards by considering c is constant. c is two measurements, time and length. Simmer that...

For me an easy way to understand this "which perspective (FoR) is "real"" is to consider the simple idea that motion is relative. If you can subscribe to that, the rest of observations made of course have that same principal.

that's about as well as I can explain the concept.

 

[nitsuj]

ghwellsjr, I think I may have taken your post a little differently then your intent.

You made me realize something; I take back my comment the poster was on the cusp of being correct. That, as I found out myself with SR concepts, is far different from understanding.

 

[Qzit]

WOW I see my English is very poor. You are correct to gig me for that. Let’s look at contraction a different way. If you had a spaceship coming towards at ½ the speed of light and there was a mirror on the front and on the back of the spaceship to bounce returning photons we can see contraction as an optical effect of the finite speed of light. Now we shoot a signal to the ship. Half of the signal hits the first mirror and heads back. The second ½ of the signal moves towards the back of the ship. While the second half of signal is moving towards the back the ship has closed the distance towards the signal by 1/3 of the ships length when the second half of the signal bounces off. On return of the signal to the stationary observer the ship appears to be 2/3 its actual size. Moving away it would appear to be 1 and 1/3 its actual size. This is because of the finite speed of lights optical effect not a physical effect to my way of thinking. Am I wrong?

Put a strobe on in the dark to mimic a slowed synapsis firing. You sit down and watch others move. They appear to be moving faster.

 

[Janus]

"C" I thought was a constant. What does not matter where you observe? If there is no observer "C" is always "C". What you observe is an accurate mathematical difference based on relativity. Why do you see it more than an optical affect?

It would be more correct to say that c is invariant. c has the same value relative to any inertial frame as measured from that frame.

So in your ship you will measure the light as traveling at c relative to the ship. In other words, the light coming from in front of you will take the same time to travel from the front of the ship to the back of the ship as the light traveling from the back of you will take to travel from back of the ship to the front.

However, the same light as measured from the Earth travels at 0.1c relative to you going in one direction and 1.9c going in the other. This means that according to the Earth, the light passing you from behind takes longer to pass from back to front of your ship than that light you are meeting takes to pass from front to back of your ship.

 

[ghwellsjr]

WOW I see my English is very poor. You are correct to gig me for that. Let’s look at contraction a different way. If you had a spaceship coming towards at ½ the speed of light and there was a mirror on the front and on the back of the spaceship to bounce returning photons we can see contraction as an optical effect of the finite speed of light. Now we shoot a signal to the ship. Half of the signal hits the first mirror and heads back. The second ½ of the signal moves towards the back of the ship. While the second half of signal is moving towards the back the ship has closed the distance towards the signal by 1/3 of the ships length when the second half of the signal bounces off. On return of the signal to the stationary observer the ship appears to be 2/3 its actual size. Moving away it would appear to be 1 and 1/3 its actual size. This is because of the finite speed of lights optical effect not a physical effect to my way of thinking. Am I wrong?

Let's put it this way, you are half wrong. What you are describing is Doppler but it's not like normal Doppler for sound in air where there is in fact no length contraction. With Relativistic Doppler involving light, there is the effect caused by the finite speed of light which is what you are talking about, but when you subtract that out, there remains an effect caused by the contraction of the length of the spaceship.

 

[Qzit]

Hi ghwellsjr
After you subtract the Doppler how much more contraction is there?

Hi Janus,
How do you measure the one way speed of light in a single direction or is it assumed that inside the ship the speed of light is measured to be the same in both directions?

 

[Dale]

I did not know my thoughts were LET or crackpot ideas

I tried to make it clear. I never said that you were a crackpot, nor did I intend to imply it. I was only mentioning the reason for my aversion to LET.

I thought it was only the view that was contracted and not the distance. If your synapsis is slowed than distance would appear contracted but the physical distance remains the same for a photon to travel. How does a spaceship actually contract space in front of it? Please help I am confused. I did not understand that a spaceship could do anything to the space in front of it. If I were mid-point to a destination would I be in a physical contraction?

Spaceships don't do anything to space. Length contraction is an aspect of the mathematical relationship between different inertial reference frames, just like time dilation. This relationship is called the Lorentz transform. In different reference frames the physical distance is different, even after you correct for all of the view-related and finite-speed-of-light-related effects. You cannot have time dilation without length contraction, they are two sides of the same "coin" (the Lorentz transform).

"C" I thought was a constant. What does not matter where you observe? If there is no observer "C" is always "C". What you observe is an accurate mathematical difference based on relativity. Why do you see it more than an optical affect?

If I am traveling at .9 c in Earth's frame heading towards a pulse of light then in the Earth's frame the distance between the pulse of light and me is closing at 1.9 c. In my frame I am traveling at 0 c by definition, so in my frame the distance between the pulse of light and me is closing at c. So your statement above is only true in the Earth's frame, not in my frame.

If you had a magical telescope (which is impossible to begin with) say your Synapsis on the ship is firing at 16 frames an hour and your twin is firing at16 frames per second on Earth. The Earth twin would always be moving much faster to my way of thinking. This would be real and not just observed I would suspect. If I am wrong please explain.

It is impossible to start from a false premise like a magical telescope and obtain any valid conclusions. Can you think about your real question and ask it without any impossible premises?

 

[Qzit]

Hi DaleSpam,

I am beginning to think I think more in line with LET but I do not know the meaning. Between my use of the English language and your replies about LET I am sorry if I was unpleasant. I thought the Lorentz transformations were the same as the optical effect. I did not realize there is a different Lorentz contraction then the one I described with the mirrors. What is the mathematical Lorentz contraction for measuring the length of a ship going ½ “C” to a stationary observer?

 

[Dale]

The Lorentz transform is given here:
http://en.wikipedia.org/wiki/Lorent...ormation_for_frames_in_standard_configuration

All of the usual relativistic features like time dilation, length contraction, and the relativity of simultaneity are contained in it. The Lorentz transformations are not optical effects. They are what remain after all of the optical effects are accounted for. They are not due to the fact that the speed of light is finite, but rather they are due to the fact that the speed of light is invariant.

 

[Qzit]

Hi DaleSpam,

All of the usual relativistic features like time dilation, length contraction, and the relativity of simultaneity are contained in it. The Lorentz transformations are not optical effects. They are what remain after all of the optical effects are accounted for. They are not due to the fact that the speed of light is finite, but rather they are due to the fact that the speed of light is invariant.

Relativity of simultaneity is an explanation of visual contraction not physical contraction. The Lorentz contraction says close to the speed of light the contraction is close to 100%. If a ship could go “C” it would have no length. I can explain relativity of simultaneity that shows it to be a visual effect of the finite speed of light. You are in a ship going near the speed of light. You are in the front of the ship and hit the switch that simultaneously turns on the front and back of a pulse emitter. The front goes off first and the signal travels down the ship at the speed of light. When the signal reaches the back of the ship the back reaches the position the first pulse fired. There is your contraction due to the finite speed of a photon. If you set off the signal in the center of the ship there would be no issue with the simultaneity of relativity and no contraction. Instantaneous and simultaneity of relativity are two different things.


Time dilation on the other hand should be a physical issue. Time slowing down suggests that mass dilates, gets larger and less dense. This is definitely an issue of invariance of “C”. If mass dilates and the path of the electron increases the cycles tick slower to match the longer distance a photon has to travel to maintain its invariance. Every other frame uses their own measuring stick and tick of their clock for measurement. If there measuring stick is shorter like in outer space away from mass there clock ticks faster. In this way the speed of light is measured in every frame to be invariant.

 

[Dale]

Relativity of simultaneity is an explanation of visual contraction not physical contraction.

This is incorrect. The relativity of simultaneity is not an explanation of any kind of contraction.

You are in the front of the ship and hit the switch that simultaneously turns on the front and back of a pulse emitter. The front goes off first

You are contradicting yourself.

When the signal reaches the back of the ship the back reaches the position the first pulse fired.

This is incorrect in every reference frame. There is some time, dt, between when the first pulse is fired and when the signal reaches the back of the ship. In that time the light has moved a distance c dt.

If you set off the signal in the center of the ship there would be no issue with the simultaneity of relativity and no contraction.

This is also incorrect. Length contraction depends on the relative velocity between two frames, not on the location of some light source.

Time slowing down suggests that mass dilates, gets larger and less dense.

Also incorrect. As v->c time slows down, and lengths contract making objects smaller and more dense.

Every other frame uses their own measuring stick and tick of their clock for measurement. ... In this way the speed of light is measured in every frame to be invariant.

At least you got something right.

Qzit. You should stop making assertions and start asking questions. You have some very strangely confused ideas about relativity. I am not sure what the source is. I would like to help, but I can't tell where to start yet.

 

[Qzit]

Hi DaleSpam,

Relativity of simultaneity is an explanation of visual contraction not physical contraction.

This is incorrect. The relativity of simultaneity is not an explanation of any kind of contraction.

You did not understand my thought experiment.

You are in the front of the ship and hit the switch that simultaneously turns on the front and back of a pulse emitter. The front goes off first

You are contradicting yourself.

I am not contradicting myself if you understand the concept of Relativity of Simultaneity. That is why I mentioned instantaneous as different from simultaneity. You have to understand the inherent relativity of the finite speed of light.

When the signal reaches the back of the ship the back reaches the position the first pulse fired.

This is incorrect in every reference frame. There is some time, dt, between when the first pulse is fired and when the signal reaches the back of the ship. In that time the light has moved a distance c dt.

Of course from a observer at rest the ship and the distance light traveled is the same so the observer at rest sees the two pulses at the same place. The result is the observer at rest sees no length to the ship.

If you set off the signal in the center of the ship there would be no issue with the simultaneity of relativity and no contraction.

This is also incorrect. Length contraction depends on the relative velocity between two frames, not on the location of some light source.

This merely changes the pulse to instantaneous rather than simultaneous of relativity.

Time slowing down suggests that mass dilates, gets larger and less dense.

Also incorrect. As v->c time slows down, and lengths contract making objects smaller and more dense

If your assertion were true than we would not have an invariant speed of light in every frame. A clock that slows down has a longer length measuring stick. The slowed clock has to have enough time for the photon to cover the increased distance of the dilation. If it were smaller the slowed clock would allow the photon to wiz past the measuring stick early destroying the invariant speed of light.

Do you want me to learn from others that are also mistaken?

 

[Dale]

You did not understand my thought experiment.

Probably not. Can you re-explain your thought experiment, this time carefully specifying which frame each quantity is measured in or in which frame any given temporal ordering is?

I am not contradicting myself if you understand the concept of Relativity of Simultaneity.

I understand it quite well. However, the relativity of simultaneity doesn't apply from a single reference frame. If you want to invoke the relativity of simultaneity you need at least two different frames. It may be that you intended that, but it wasn't clear which measurements were intended for which frames from your description.

You have to understand the inherent relativity of the finite speed of light.

None of the relativistic effects are due to the finite speed of light. They are due to the invariance of the speed of light.

For example, suppose that pre-Lorentz aether theories were correct and Maxwell's equations only held in the aether frame and that other inertial frames were related to the aether frame via the Galilean transform. In that case, the speed of light in other frames would be c+v where v is the speed of the frame wrt the aether, i.e. the speed of light would still be finite but not invariant. Since the Galilean transform holds, there would be no time dilation, nor length contraction, and no relativity of simultaneity.

Therefore the relativistic effects are due to the invariance of c, not the finite-ness of c. All of the features of relativity are effects that remain even after correctly accounting for the finite speed of light.

Of course from a observer at rest the ship and the distance light traveled is the same so the observer at rest sees the two pulses at the same place. The result is the observer at rest sees no length to the ship.

This is not correct. I would encourage you to actually work out the math of the Lorentz transform and see for yourself.

If your assertion were true than we would not have an invariant speed of light in every frame.

I would encourage you to work out the math for yourself for this also. Light and the Lorentz transform simply do not work the way you think they do.

If you have trouble working out the math for either of these then I can show you. But it will be more valuable for you if you can do it yourself. If you get stuck please let me know and I will try to help.

 

[Janus]

Hi DaleSpam,


If you set off the signal in the center of the ship there would be no issue with the simultaneity of relativity and no contraction.

This is incorrect. If you set off a signal in the center of the ship, according to someone in the ship, the signal radiates out from the center of the ship and will reach the ends of the ship at the same time. The point of emission and the center of the ship coincide at all times

However for someone on Earth, the signal radiates out from the emission point at c while the ship moves at 0.9c relative to the emission point. The emission point and the center of the ship only coincide at the moment of emission. Thus the rear of the ship is rushing toward the signal with a closing speed of 1.9c and the front of the ship is running away from the signal at with a closing speed of 0.1c. The signal reaches the rear of the ship before it reaches the front of the ship.

This is not an "optical" effect but a real disagreement as to whether or not event at the ends of the ship are simultaneous or not.

There also has to be a contraction. Let's assume that, upon reaching the ends of the ship, the signal is reflected back to the center. The time that passes between emission and return is noted on a clock at the center. We'll say that 1 microseconds elapsed on the clock between emission and return.( .5 us each way for both signals)

According to someone at rest with respect to the Earth, The signal will take 5 us going one way and 0.263 us going the other way, for a round trip time of 5.263 us assuming no length contraction. By time dilation, this means that 0.436 x 5.263 = 2.294 us passed on the ship's center clock.

However, from above, we know that 1 us passes at the ship's center. The ship and Earth must agree as to the reading on the ship clock at emission and return. Since the speed of the signal relative to the observers is invariant, the only way around this is for the length of the ship to be contracted by a factor of 0.436, which reduces the round trip time according to the Earth to 2.294 us and the time passed on the ship clock according to the Earth to 2.294 x .436 = 1us.

 

[Qzit]

Hi DalSpam

This was a sub note to the reference you gave me.

Note that in this equation it is assumed that the object is parallel with its line of movement. Also note that for the observer in relative movement, the length of the object is measured by subtracting the simultaneously measured distances of both ends of the object. For more general conversions, see the Lorentz transformations. An observer at rest viewing an object traveling very close to the speed of light would observe the length of the object in the direction of motion as very near zero.

Hi Janus
Are you saying that within the frame of the spaceship the speed of light is measured to be different heading to the back of the ship than heading to the front? If that is the case then we just have to place the switch in a position where the emitters go off instantaneously and not relatively by simultaneity. Than we will not see a contraction in the rest frame wrt the spaceship frame.
I always thought there was a difference in the forward and backward directions in a moving spaceship. I was always told there was not.
I think DelSpam is suggesting that the visual effect of the Lorentz contraction is a physical contraction of the space ship. What is your opinion? Is the Lorentz contraction just a visual or also a physical contraction?

 

[Passionflower]

From my rudimentary understanding, concerning the twin paradox, if one twin leaves traveling near the speed of light and returns, he will find himself younger than his twin who stayed behind. Hence, the effect of time dilation is permanent.

When their spacetime paths are different both time dilation and length contraction are in place. However when they essentially travel on the same spacetime path it is not.

I am not sure why you think time dilation is permanent.

 

[Dale]

This was a sub note to the reference you gave me.

Yes. So now you agree with me on the above points?

 

[Qzit]

Hi DelSpam,

All of the usual relativistic features like time dilation, length contraction, and the relativity of simultaneity are contained in it. The Lorentz transformations are not optical effects. They are what remain after all of the optical effects are accounted for. They are not due to the fact that the speed of light is finite, but rather they are due to the fact that the speed of light is invariant.

Relativity of simultaneity is an explanation of visual contraction not physical contraction. The Lorentz contraction says close to the speed of light the contraction is close to 100%. If a ship could go “C” it would have no length. I can explain relativity of simultaneity that shows it to be a visual effect of the finite speed of light. You are in a ship going near the speed of light. You are in the front of the ship and hit the switch that simultaneously turns on the front and back of a pulse emitter. The front goes off first and the signal travels down the ship at the speed of light. When the signal reaches the back of the ship the back reaches the position the first pulse fired. There is your contraction due to the finite speed of a photon. If you set off the signal in the center of the ship there would be no issue with the simultaneity of relativity and no contraction. Instantaneous and simultaneity of relativity are two different things.

Note that in this equation it is assumed that the object is parallel with its line of movement. Also note that for the observer in relative movement, the length of the object is measured by subtracting the simultaneously measured distances of both ends of the object. For more general conversions, see the Lorentz transformations. An observer at rest viewing an object traveling very close to the speed of light would observe the length of the object in the direction of motion as very near zero.

This was a sub note to the reference you gave me.

Yes. So now you agree with me on the above points?

So we agree that the transformation would be very near zero but the only disagreement is

You think the contraction is a physical effect on the spaceship and the spaceship is being reduced to very near zero length.

I think it is a visual effect that we see zero length with no physical effect on the space ship. Meaning it remains the same physical length or slightly larger in the very near "C" ship as it is in the resting frame ship.

Do I have that correct?

Hi Janus,

Earth frame observer is watching a spaceship made of glass traveling 0.9C. There is a stationary (same as the Earth frame) photon emitter that will emit a photon parallel to the glass ship forward and backward as the ship passes going 0.9C. There is also one inside the glass spaceship that will emit a photon forward and backward. When the spaceship lines up with the observer and the outside photon emitter they go off in unison. The Earth observer sees the outside photon emitter’s photons departing from each other at 2C or each at C in opposite directions. Now inside the ship the Earths observer, observes the forward emitters photon going the speed of light (relative ship frame= 0.1C=C). He sees the rear emitters photon going 0.8C forward (relative ships frame 0.1C=C backwards).
So to the Earths observer the photons are both going forward at "C" and 0.8C respectively.

Do we give up light is measured to be invariant in every frame or give up the idea that the speed of light is the same in every direction within a frame?
They are mutually incompatible statements for a photon.


Happy Holidays

 

[Janus]

Hi DelSpam,







So we agree that the transformation would be very near zero but the only disagreement is

You think the contraction is a physical effect on the spaceship and the spaceship is being reduced to very near zero length.

I think it is a visual effect that we see zero length with no physical effect on the space ship. Meaning it remains the same physical length or slightly larger in the very near "C" ship as it is in the resting frame ship.

Do I have that correct?

Hi Janus,

Earth frame observer is watching a spaceship made of glass traveling 0.9C. There is a stationary (same as the Earth frame) photon emitter that will emit a photon parallel to the glass ship forward and backward as the ship passes going 0.9C. There is also one inside the glass spaceship that will emit a photon forward and backward. When the spaceship lines up with the observer and the outside photon emitter they go off in unison. The Earth observer sees the outside photon emitter’s photons departing from each other at 2C or each at C in opposite directions. Now inside the ship the Earths observer, observes the forward emitters photon going the speed of light (relative ship frame= 0.1C=C). He sees the rear emitters photon going 0.8C forward (relative ships frame 0.1C=C backwards).
So to the Earths observer the photons are both going forward at "C" and 0.8C respectively.

Do we give up light is measured to be invariant in every frame or give up the idea that the speed of light is the same in every direction within a frame?
They are mutually incompatible statements for a photon.



Happy Holidays

The Earth observer sees both sets of photons ( the ones emitted by the outside emitter and the ones emitted by the ship emitter) travel at c in opposite directions, with the ones headed towards the rear of the ship arriving at the rear before the ones headed for the the front arrive at the front of the ship. It makes no difference from which emitter the photons came from.

For the Observer on the ship, both sets of photons reach the ends of the ship simultaneously. That is what light speed being invariant and the speed of light being the same in every direction in every frame means. They are not incompatible.

 

[Qzit]

Hi Janus,

The Earth observer sees both sets of photons ( the ones emitted by the outside emitter and the ones emitted by the ship emitter) travel at c in opposite directions, with the ones headed towards the rear of the ship arriving at the rear before the ones headed for the the front arrive at the front of the ship. It makes no difference from which emitter the photons came from.

For the Observer on the ship, both sets of photons reach the ends of the ship simultaneously. That is what light speed being invariant and the speed of light being the same in every direction in every frame means. They are not incompatible.

You are suggesting there is a different reality for each frames observer.

If you are suggesting it’s because of the limit of not being able to measure the one way speed of light that I would understand. The round trip is the same for each I agree.

If it is a different reality I would like to know the reason. (Unless it is just your understanding of a postulate). If there is a physical explanation please in that case I would be very appreciative.


If there is no explanation than we must conclude that the one way speed of light is not measured to be the same in all frames.

 

[Janus]

Hi Janus,



You are suggesting there is a different reality for each frames observer.

No,just different perspectives of the same "reality"

If you are suggesting it’s because of the limit of not being able to measure the one way speed of light that I would understand. The round trip is the same for each I agree.

If it is a different reality I would like to know the reason. (Unless it is just your understanding of a postulate). If there is a physical explanation please in that case I would be very appreciative.


If there is no explanation than we must conclude that the one way speed of light is not measured to be the same in all frames.

The best simple explanation is that measurements of time and space are frame dependent. To use an analogy, imagine a group of people standing in an open field all facing in different directions. If you were to ask them to point North, they will all point in the same direction (assuming they all know the direction of North). If you place two objects on that field and ask them what the North-South distance between them is, they will all give you the same answer. This is an example of frame independence. No matter what direction the people are facing with respect to each other, they will agree on these measurements.

If you now ask them all to point left, they will all point in different directions, as left is defined by the direction they are facing. If you ask them the left-right distance between the two objects, they will give different answers. Some will even say that object one is to the left of object two while other will say that it is the other way around. This is an example of frame dependence. the measurements do depend on how the people face with respect to each other.

What Relativity says is that measurements of space and time are more akin to the second example.

In Relativity, there is something akin to the straight line distance between the two objects in the example. This is frame independent. Everyone agrees to this even though they might not agree as to the Left-Right and Front-Back components of this distance. In Relativity, it is called the space-time interval and represents the separation of event in space and time. This interval is something everyone will agree on. What different frames will disagree upon (such as in your spaceship example) How much of the separation is in time and how much is in space.

One way to look at it is that we measure time and space we are only measuring "slices" of a larger reality, and that different frames are looking at different slices.

 

[Dale]

You think the contraction is a physical effect on the spaceship and the spaceship is being reduced to very near zero length.

I think it is a visual effect that we see zero length with no physical effect on the space ship. Meaning it remains the same physical length or slightly larger in the very near "C" ship as it is in the resting frame ship.

I would certainly say it is a physical effect since "physical" means "of or pertaining to physics" and length contraction is part of the Lorentz transform which definitely pertains to physics.

It is not, however, a visual effect. There are several relativistic visual effects, such as relativistic abberation, relativistic Doppler shift, and Terrell-Penrose rotation. Length contraction is what is measured after properly accounting for these visual effects in a given frame. It is not what is seen visually nor with a high speed camera.

I would again recommend that you work out the two calculations I suggested above. I think that you will find it educational.

 

[Qzit]

One way to look at it is that we measure time and space we are only measuring "slices" of a larger reality, and that different frames are looking at different slices.

On this we agree but it still does not explain why the Earth observer sees the photon hit the back of the ship first and the front second while the ship observer sees them hit simultaneously. If the Earth frame sees a difference in the relative time (to hit the back and front of the ship) than inside of the spaceship there has to be a difference in the one way speed of light. There is no way to get around that. From an observer in the center of the ship the one way speed of light is different. Not that the spaceship frame could measure it but the reality is the speed within the ship is c+v and c-v. The round trip is “C” in both directions. The Earth observer is the only one that can measure the one way direction of light. That measurement is c+v and c-v for closing and departing distances. So in a moving frame light can cover a different distance in the same amount of time. This would explain why atomic clocks lose Nano seconds going from east to west.

 

[Dale]

On this we agree but it still does not explain why the Earth observer sees the photon hit the back of the ship first and the front second while the ship observer sees them hit simultaneously. If the Earth frame sees a difference in the relative time (to hit the back and front of the ship) than inside of the spaceship there has to be a difference in the one way speed of light. There is no way to get around that.

This is incorrect. The distance traveled by the light is different for the forward and backward paths in the Earth frame, and that distance is different than the distance traveled by both paths in the ship frame. The speed of light is the same, and the various distances are different. Again, you should work out the math to convince yourself of this. If you need help there are many on this form who will be willing.

The Earth observer is the only one that can measure the one way direction of light.

What makes the Earth observer so special?

 

[Qzit]

Hi DaleSpam,

This is incorrect. The distance traveled by the light is different for the forward and backward paths in the Earth frame, and that distance is different than the distance traveled by both paths in the ship frame. The speed of light is the same, and the various distances are different. Again, you should work out the math to convince yourself of this. If you need help there are many on this form who will be willing.

Let's see you are saying the photon physically hits the back of the ship before the one in front hits in the Earth frame. Then they physically hit at the same time in the ships frame. Do you understand how absurd that sounds? The photon has to be in two places at the same time.

Math is no guarantee that you are correct. While the math does contract the visual object the same as your view. You actually believe the contraction is physical. Does the math prove it’s physical?

 

[Dale]

Let's see you are saying the photon physically hits the back of the ship before the one in front hits in the Earth frame. Then they physically hit at the same time in the ships frame. Do you understand how absurd that sounds?

Yes, I do. That is why the relativity of simultaneity is the single most difficult concept for students to grasp in learning special relativity. It goes very strongly against our non-relativistic intuition.

Math is no guarantee that you are correct ... Does the math prove it’s physical?

The math is a guarantee that what I am saying is consistent with SR and that what you are saying is not consistent with SR. Then experiments prove that SR is correct in the domain where it is tested. The math is necessary, but not sufficient, which is why we do experiments too.

In any case, my repeated suggestions that you do the math are not for the purpose of proving my point, they are simply to help you learn. If I were only interested in proving my point then I would do the math myself. But you will learn more and faster if you do it.

 

[nitsuj]

Hi DaleSpam,
Let's see you are saying the photon physically hits the back of the ship before the one in front hits in the Earth frame. Then they physically hit at the same time in the ships frame. Do you understand how absurd that sounds? The photon has to be in two places at the same time.

Math is no guarantee that you are correct. While the math does contract the visual object the same as your view. You actually believe the contraction is physical. Does the math prove it’s physical?

The first postulate is the laws of physics are always the same no matter the observer, as long as they're inertial. Consider that time is merely a measurement, like length is a measurement. Your intuition is right, just got to tune it a bit more physicsy. It would be absurd for someone right next to (inertial with) the experiment to believe the photons didn't hit simultaneously. (im assuming this is the scenario) That observer is in the "same" spacetime, specifically frame of reference as the experiment.

Said differently, considering another observers FoR doesn't make it "reality". Only your own observation does. I think causality is the final nail in that.

In another, more crude way, conscious observers are special in that we can in-vision what would happen in the other FoR. That cannot "physically" be considered a/the "reality". I know nothing of LET but this conscious observer reasoning would be getting off on the wrong foot (with either theory).

I couldn't think of an equation for that.

 

[Dale]

The first postulate is the laws of physics are always the same no matter the observer, as long as they're inertial.

I have never seen the first postulate written that way. It is always written in terms of reference frames, not observers.

 

[Dmytry]

This question is in regard to special relativity.

From my rudimentary understanding, concerning the twin paradox, if one twin leaves traveling near the speed of light and returns, he will find himself younger than his twin who stayed behind. Hence, the effect of time dilation is permanent.

However, I have never read anywhere that the traveling twin's length will also be permanently adjusted due to length contraction.

How is it that one Lorentz transformed aspect remains while the other one vanishes upon the traveling twin's return?

I apologize if this question has been asked before. If so, and you know where to find the responses, please point me in the right direction.

Thank you!

~Dylan

The twin that returns, however, will - upon return - be aging at the same rate as the twin that stayed home. The clock that the twin took with him in a voyage will be counting off time length of 1 second, in a second, when he's back home. The length of the traveling twin's sleep, or the length of his heartbeat (ignoring the senile insomnia, ha ha, and other effects of aging on the staying twin) will not be affected permanently, just as the length of his body was not.

The age is like a position, not like length interval.

 

[Qzit]

Hi nitsuj,

The first postulate is the laws of physics are always the same no matter the observer, as long as they're inertial.

Thank you. That is what I am trying to get across. Relativity of simultaneity is just the distance light travels for each observer. When you say the photon hits the back of the ship first for the Earth observer that is the reality and proves that the one way distance in the spaceship frame that light travels is different than the return trip distance light travels. There does not need to be an observer for reality of physical position in space. The reality is the photon hits the back of the ship in both frames by the same physical position in space. The problem is the MMX experiments was the two way distance light travels. No matter what frame you are in the two way distance will make it appear light travels the same distance/time. If the math you use takes the distance traveled and divided by ½ for the one way speed you will be wrong because the travel out is always adjusted by the return travel. An orthogonal way to prove my point would be with Atomic clocks. You are on the equator and have three atomic clocks synchronized. One stays stationary one is taken in a plane to the west and one to the east around the world. When they return the clock that flew west loses time compared to the stationary one and the one that flew east gains time compared to the stationary one. Einstein said atomic clocks can be used to measure the speed of light. This experiment proves the distance light traverses a moving frame is different than a frame at rest. There was a second experiment I remember reading about using four atomic clocks. Four clocks were synchronized in New York three were taken to the North Pole. One was brought back to New York and the other two were taken to San Francisco. The one brought back to New York stayed synchronized. Out of the two that went to San Francisco one went to New York and the one in New York went to San Francisco. New York to San Francisco lost 14 Nano seconds San Francisco to New York gained 14 Nano seconds. This also proves the one way distance traversed/time interval is different. Wake up the MMX experiments were flawed that Einstein used for Relativity. That does not negate Relativity or simultaneity of relativity. It only proves that simultaneity of Relativity is visual and not physical.

The MMX experiments were not flawed just the math used for the one way distance of traversed light that was assumed.

If you took either clock back to their origin they would be synchronized again.

 

[Dale]

Qzit, pleas learn to use paragraphs to separate your thoughts. A little bit of organization will help communication a great deal.

When you say the photon hits the back of the ship first for the Earth observer that is the reality

I have asked this before, but why? What is it about the Earth observer that makes them so special that their reference frame represents "reality"? Do you believe like Copernicus' detractors that the Earth is at the center of the universe or otherwise occupies some priviledged position?

The reality is the photon hits the back of the ship in both frames by the same physical position in space.

Again, this is not true in any frame, we have already discussed this.

Wake up the MMX experiments were flawed that Einstein used for Relativity. That does not negate Relativity or simultaneity of relativity. It only proves that simultaneity of Relativity is visual and not physical.

Can you provide any mainstream scientific reference which supports your claim?

I would like to remind you that this forum is not for anti-relativity rants, it is for learning mainstream physics. When you signed up for your account you agreed to the rules that prohibit overly speculative posts. If you have an anti-relativity agenda you had best look elsewhere.

 

[Qzit]

Hi DaleSpam,

You and I agree on Relativity being correct but we disagree on what that means. In order to stop getting confused let’s look at everything from the photons perspective. Where do you disagree?
1. The photons minimum and maximum speed are the same in every frame and that = invariant speed of light.
2. Our visual perspective depends on where we observe an object.
3. The photon perspective is the only reality of its own position in space.
If we can agree on that we made progress. Now before you read with a challenging attitude try and follow your logic with the three things we agree on not where we disagree.
You are riding a photon and you can know the exact position of another photon instantaneously. This removes relativity of simultaneity.
Lets go through this thought experiment again because the math is different from the reality of the photon.
A spaceship is traveling ½ the speed of light. Two photons side by side are traveling an intersect course with the space ship. When the two photons reach the front of the spaceship one reflects back to an observer at rest in line with the traveling photons and the ship. The second photon reaches the back of the spaceship and reflects back to the same observer at rest. While the photon was moving from the front to the back the ship moved ½ the speed of light forward intersecting a contracted position relative to the length of the ship. We visualize the photon covering 75% and the ship 25%. This is a contraction of the returned photons measurement from the front to the back that the observer at rest sees. (1-v^2/C^2)=1-25=75 just what the observer at rest sees for the returned photon distance. Now because we think light travels the same speed relative to the ships frame we take the square root and get a contraction of 0.866 just about half + the time dilation increase of the ships length by gamma. Once again the MMX experiments are affecting reality of position. The MMX experiment was not wrong but the conclusions of the experiment were incorrect.

The Earth frame is the only place you can measure the one way speed of light in a space ship. The returned light from the front and back of the ship will be the same and if you incorrectly divide that by 1/2 you are mistaken when the photon actually hit the back and front of the ship.