Is it possible to determine absolute speed?

[bkelly]

I started posting to what seemed to be a related thread here:
https://www.physicsforums.com/showthread.php?p=2825641&posted=1#post2825641
After concluding I was hijacking another thread, I am posting here.

The concept is I think that, in theory, one can determine absolute speed. Am I the one moving or is that other object moving past me. I wrote up a theoretical experiment to demonstrate my concept. I decided to post it on a personal website rather than post a bunch of text here. The location is: http://www.bkelly.ws/space_time/index.htm
Click on the link about absolute velocity to view or download the essay.

Ich was kind enough to respond in the first thread and my response is rather general so I will put them in this post. For context, Ich raised the issue of simultaneity and the problems it would cause my experiment.

Hello Ich,

Regarding the trolley and the issue of simultaneity, while the essay stated that it makes the marks as it passed directly in front of Tom, I am claiming that given the goal of the essay it really does not matter if the marks are made when the trolley is a bit to the left of Tom, right in front, a bit to the right, or what the heck, a couple of hundred meters to the right or left. So exactly where the marking event(s) occur is not really important. That should get us down to two events.

However, as I think on this a bit more, we can dispense with simultaneity. Worrying about the simultaneity provides no advantage in discussing this concept. Its sort of like saying we cannot have a theoretical discussion because no one knows how to build a trolley that can move at relativistic speeds. So, given that for our discussion we have a trolley that can move at near the speed of light, the trolley is declared to be 1 meter in width when stationary and has two paint cans, one at each end and they make two marks that are one meter apart, with respect to the trolley if you will, at the appropriate time and place. The concern is not in making and conducting the experiment for real, the concern is the concept of what would happen if we found a way to conduct this experiment.

If "the fence is stationary with Sally and with respect to Tom when Sally stops his motion", it is obviously moving wrt Tom as long as Tom is moving wrt Sally. So Tom sees the trolley mark a moving fence.

Lets go down your path a bit further. Assume that the fence goes past Tom at ½ C while Tom and the trolley are stopped, and the trolley makes its two marks. Then the fence is brought back to a center position and the marks examined. While moving, the fence is length contracted. When stopped then the fence will “un-contract” and the marks will be more than 1 meter apart.

On the other hand, make the fence were stationary and the trolley moves along it at 1/2 C and the marks made. The trolley will be length contracted and the fence will not. The marks will be less than one meter apart. That is in contradiction to the marks being further apart when the fence is moved for the marking then stopped.

That tells me that from the perspective of the trolley, we can detect which was moving. And that is the fundamental purpose of my thoughts.

 

[Doc Al]

Regarding the trolley and the issue of simultaneity, while the essay stated that it makes the marks as it passed directly in front of Tom, I am claiming that given the goal of the essay it really does not matter if the marks are made when the trolley is a bit to the left of Tom, right in front, a bit to the right, or what the heck, a couple of hundred meters to the right or left. So exactly where the marking event(s) occur is not really important. That should get us down to two events.

What's important is when the two marks are made. You seem to think that everyone will agree that they are made simultaneously. Not so. Simultaneity is frame-dependent.

However, as I think on this a bit more, we can dispense with simultaneity. Worrying about the simultaneity provides no advantage in discussing this concept. Its sort of like saying we cannot have a theoretical discussion because no one knows how to build a trolley that can move at relativistic speeds. So, given that for our discussion we have a trolley that can move at near the speed of light, the trolley is declared to be 1 meter in width when stationary and has two paint cans, one at each end and they make two marks that are one meter apart, with respect to the trolley if you will, at the appropriate time and place. The concern is not in making and conducting the experiment for real, the concern is the concept of what would happen if we found a way to conduct this experiment.

If you don't care about when the marks are made and whether they are made simultaneously, then there's no point in discussing the distance between the marks or anything else.

 

[bkelly]

What's important is when the two marks are made. You seem to think that everyone will agree that they are made simultaneously. Not so. Simultaneity is frame-dependent.

If you don't care about when the marks are made and whether they are made simultaneously, then there's no point in discussing the distance between the marks or anything else.

Are you trying to say there is no such thing as two events happening simultaneously? If not then what are you saying? Please frame your answer in the concept of the trolley.

Our theoretical trolley may have a mechanical connection between the two markers one meter apart. It may have an electronic connection with all the relativistic calculation worked out in advance such that the marks are made at the same time. Let's assume we have the technology such that, when the trolley is in front of us, we can see that the two marks are made within one one millionth of a femtosecond of each other and within one billionth of a micrometer of being exactly 1 meter apart. If that is not accurate enough, I will declare it to be as accurate as needed.

The trolley is a single item and moves all together. The trolley, all of the trolley, and all of its parts, should be within a single frame of reference. The two markers always move at the same speed. When I jump on the trolley and ride along at 1/2 C, my measurements continue to record a marking accuracy of a better than billionth of a femtosecond and millimeter.

From the wiki page found here: http://en.wikipedia.org/wiki/Relativity_of_simultaneity

Where an event occurs in a single place—for example, a car crash—all observers will agree that both cars arrived at the point of impact at the same time. But where the events are separated in space, such as one car crash in London and another in New Delhi, the question of whether the events are simultaneous is relative: in some reference frames the two accidents may happen at the same time, in others (in a different state of motion relative to the events) the crash in London may occur first, and in still others the New Delhi crash may occur first.

All parts of the trolley are closer to all the remaining parts of the trolley than those that observe a car crash, and probably even closer that those that participate. The trolley does not have one end in London while the other is in New York City or New Delhi.

To wrap it all up from another perspective, the trolley exists in theory only. If we can presume to discuss a trolley that we can accelerate to 1/2 C, which we can do (discuss the theory that is), then we can presume to have a method to deal with the simultaneity.

So tell me please: Why can we not disregard simultaneity and discuss the concept I am trying to present.

 

[DaveC426913]

Are you trying to say there is no such thing as two events happening simultaneously?

That is correct.

At the very crux of relativity is the relativity of simultaneity. IOW, simultaneity is relative to your frame of reference. It is not simply a wrinkle to be ignored or "dealt with".

It is the very core.

The trolley is a single item and moves all together.

No it does not. Relativity forbids perfectly rigid materials.

If you posit a perfectly rigid material, then you instantly and trivially have something that can move at infinite speed. You could have a radio that communicates with the moon in zero time.

 

[DaveC426913]

Lets assume we have the technology such that, when the trolley is in front of us, we can see that the two marks are made within one one millionth of a femtosecond of each other and within one billionth of a micrometer of being exactly 1 meter apart.

According to whom? Who does this measurement? What is their velocity wrt to the components?

Change the viewpoint and you get a different answer.

Unlike the streetcar that can move at .5c (which can be easily granted as it does not violate any laws of physics), the issues of measurement are not trivial and not technicalities to be waved away by positing advanced technology to circumvent them; they are critical.

 

[bkelly]

My first reply is, ok, they are not simultaneous. But they are close enough to test the theory. Work it that way.

Second, I feel like you are endeavoring to avoid the main crux of my concept. None of the responses have even addressed the core concept much less shot it down.

 

[DaveC426913]

My first reply is, ok, they are not simultaneous. But they are close enough to test the theory. Work it that way.

No, the simultaneity issue will corrupt the effect you are trying to measure.

Second, I feel like you are endeavoring to avoid the main crux of my concept. None of the responses have even addressed the core concept much less shot it down.

The core concept is a non-issue because you are trying to take measurements while ignoring any problems with calibrating them.

As a similar example, I can http://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html" - and by extrapolation that any number equals any other number - as long as you don't worry about the niggling little detail of dividing by zero. If I presented that as my core idea, and insisted that you ignore the detail, how long would you let me go on insisting that 1=2?

 

[bkelly]

So following that concept we can never measure the speed of anything, particularly light because a photon leaving the laser is one event and a photon hitting the receiver is the other and we can never tell if the photon emission really has any relationship to the reception. Heck, we don't really know for sure its the same photon.

Its clear that the respondents don't want to address my concept. I will stop arguing and wait a while to see if anyone else will offer a decent explanation or discuss my concept.

 

[DaveC426913]

So following that concept we can never measure the speed of anything, particularly light because a photon leaving the laser is one event and a photon hitting the receiver is the other and we can never tell if the photon emission really has any relationship to the reception. Heck, we don't really know for sure its the same photon.

Of course we can. If they are comoving - in the same frame of reference - this is a trivial task. Even if they aren't in a comoving FoR, we can still make accurate measurements. But they will be relative measurements - relative to our FoR. They say nothing about any absolute velocity.

Its clear that the respondents don't want to address my concept. I will stop arguing and wait a while to see if anyone else will offer a decent explanation or discuss my concept.

I'm sorry you don't like the answers you are getting. At the risk of sounding sarcastic, are you expecting that the laws of physics will change so that they operate in a manner more to your liking?

You have an opportunity to learn here, to break free from the constraints of your preconceptions about how the universe works. But learning will require you letting go of those preconceptions. Instead of insisting your experiment will work and demanding we discuss it the way you've laid it out, try asking where it is flawed. You will get answers that will enlighten you.

 

[Al68]

Its clear that the respondents don't want to address my concept. I will stop arguing and wait a while to see if anyone else will offer a decent explanation or discuss my concept.

OK, I'll try.

What results would we get if we repeated your experiment on a large spacecraft containing Tom, Sally, trolley, fence, and markers, with identical relative speeds between Tom, the trolley, and Sally, with the spacecraft stationary relative to the sun? Then with the spacecraft moving at 0.5c relative to the sun? at 0.99999c relative to the sun?

The results would be exactly the same.

 

[Dale]

bkelly, let's examine the special case when the trolley is moving at 1 m/s wrt the fence, so that relativistic effects are negligible. Now, let's say that the timing is off such that the front mark is made 0.5 s before the rear mark. In that case the marks will be 0.5 m apart. Will we conclude that the length of the trolley was 0.5 m? Obviously not. Even in Newtonian mechanics the marks must be made simultaneously in order to be a measurement of the length of the trolley.

Now, in relativity there will only be one frame in which the marks are simultaneous. In all other frames not only will they disagree about the length of the trolley, they will disagree about whether or not the marks are related to the length.

 

[Ich]

bkelly,

it appears that you're under the impression that you have a valid argument here, and that all participants here just want to dodge your logic.

If so, you better drop that attitude immediately. PF is a place where you can learn about physics. If you think you don't have to learn, you're wrong here.

On the other hand, make the fence were stationary and the trolley moves along it at 1/2 C and the marks made. The trolley will be length contracted and the fence will not. The marks will be less than one meter apart. That is in contradiction to the marks being further apart when the fence is moved for the marking then stopped.

One more time I offer to help you learn why this statement is wrong. It has to do with - surprise - relativity of simultaneity. If you're not interested, just say so, and we're done. No need wasting anyone's time.

 

[bkelly]

Hello,
You guys are saying that simultaneity will make the events of my theoretical experiment unusable. I don't understand why so allow me to ask a few pertinent questions.

Given the description of the trolley, how much do you expect the problem of simultaneity to change the marks made by the trolley? When I calculate the expected distance of 0.866 meters at a velocity of 1/2 C, what do you say the results will be? Can you give me a tolerance? Do you say the results will change from test to test or will the results be predictable?

In post #3 above, I quoted from a Wiki page about simultaneity and provided a link. In that quote the page clearly implied that observers to a car wreck would all be within a single frame of reference. It referenced two crashes in London and New Delhi indicating that they would be in separate frames of reference. This further implies the immediate wittiness to the crash would all be within the same reference.

Two witnesses to a car wreck can be on opposite sides of the street and be in the same frame of reference. Why is it that two markers on a 1 meter long trolley, with a direct physical connection between them, cannot be in the same reference?

How close must two events be in distance or maybe time in order to be considered one frame of reference or to be one event?

 

[JesseM]

Given the description of the trolley, how much do you expect the problem of simultaneity to change the marks made by the trolley? When I calculate the expected distance of 0.866 meters at a velocity of 1/2 C, what do you say the results will be?

You haven't specified what mechanism the trolley uses to decide how to drop the marks "simultaneously". For example, suppose a flash is set off at the midpoint of the trolley, and at each end there is a device programmed to drop a marker as soon as it detects light from the flash. From the perspective of the trolley observer, if light is assumed to travel at the same speed in both directions in the frame where the trolley is at rest, this procedure should guarantee that both ends drop their markers simultaneously. However, from the point of view of the observer on the track, if they assume light travels at the same speed in both directions in their own frame, then since the back of the trolley is moving towards the position on the tracks where the flash went off while the front is moving away from that position, this observer will say the light catches up with the back before it catches up with the front, so the markers are dropped off non-simultaneously, and thus the distance between the markers is not the same as the length of the trolley in the track frame.

On the other hand, we could set off a flash at a position on the track closer to the front of the trolley than the back, at just the right distance so the track observer would calculate the light should reach both ends simultaneously in the track frame. In this case, the distance between the markers is equal to the length of the trolley in the track frame, but then the trolley observer will just say that this procedure caused the light to reach the two ends non-simultaneously in his own rest frame.

In post #3 above, I quoted from a Wiki page about simultaneity and provided a link. In that quote the page clearly implied that observers to a car wreck would all be within a single frame of reference. It referenced two crashes in London and New Delhi indicating that they would be in separate frames of reference. This further implies the immediate wittiness to the crash would all be within the same reference.

Two witnesses to a car wreck can be on opposite sides of the street and be in the same frame of reference. Why is it that two markers on a 1 meter long trolley, with a direct physical connection between them, cannot be in the same reference?

Unclear what you mean by "in" the same reference frame. Each reference frame is just a coordinate system which can assign coordinates to events throughout space and time, so all events in the universe can be analyzed from the perspective of the same single reference frame. Sometimes authors may talk about observers being "in" a given frame as shorthand for "at rest in", but this only makes sense for objects that exist for an extended period of time, instantaneous events don't have a rest frame since you can't talk about their position at different times in order to determine whether the position is changing or constant.

 

[inko1nsiderat]

Two witnesses to a car wreck can be on opposite sides of the street and be in the same frame of reference. Why is it that two markers on a 1 meter long trolley, with a direct physical connection between them, cannot be in the same reference?

How close must two events be in distance or maybe time in order to be considered one frame of reference or to be one event?

I think the wikipedia article was getting at that when a car crashes into another car, it doesn't really matter what reference frame you're in, the cars were at the same place at the same time because it is a single event in spacetime. What might have confused you about the New Dehli part of the wiki article is that the article talks about two car crashes happening instead of just one. In one frame they might appear to happen simultaneously, but because they are space-like separated another frame can disagree as to which crash happened when. If two car crashes happen on the same block it might appear to happen at the same time to all the people on the block, but the two crashes wouldn't appear to be simultaneous to an observer in a frame that is within a block of both crashes (as seen from the frame of those same pedestrians who witnessed the crash) but was moving significant fractions of the speed of light (like if a cosmic ray happened to be within a block of the crashes and could actually observe anything). It isn't really a matter of how close you are to the events that determines if they appear to be simultaneous or not, it matters how you are moving relative to another frame of reference, thus simultaneity is frame dependent and we call that the relativity of simultaneity.

 

[bkelly]

...Unclear what you mean by "in" the same reference frame. ...

That is getting to the core of the present discussion. I cannot find any reference or reply that says two events must be within X and Y parameters in order to be in the same frame of reference. (You pick the quantities noted by X and Y. Let it be distance, time, or what ever you know to be blocking this discussion.)

The respondents are saying that two markers one meter apart are not in the same frame of reference therefor my theoretical experiment is invalid. So I ask again:
Question 1: What is needed?

Regarding how the two markers are triggered: They are only one meter apart. They are not kilometers apart, not event tens of meters apart. Just one. Hence the question just above, what is the threshold?

So if that question goes unanswered try this.

configuration A: There is a solid rod between the two markers. At some point something in the middle of the rod tweaks it and the rod activates the markers.
Question 2: Are the markers in the same reference? Can the marks be considered to be made simultaneously. if not, why not?

configuration B: Let's make the marks closer together. There is a single spray can. It sends out paint, or light, or what ever your might like, in the usual format that we expect from a spray can. In between the spray can and the fence there is a somethings that blocks the spray. I declare that that something causes a precisely one centimeter wide blockage when the trolley is at rest. There is one event of one pulse out of the spray can, and one centimeter of the spray is blocked when at rest.

Now send that trolley flying by at 1/2 c and have it make a mark. I calculate that the mark will be 0.866 centimeters wide.

Questions 3: Is that all in a single frame of reference? Will that work for my theoretical experiment?

Condition C: Let's go crazy with this. The trolley is a meter stick of standard proportions. Let's make it 2 centimeters wide and one meter long at rest. The meterstick goes flying by right next to the fence such that an observer sees the 1 meter length as along the path of travel, and the 2 centimeter width as a vertical dimension. Make the distance between the meter stick and the fence to be microscopic. Its as close as it can be without touching. The meterstick is also microscopically thin. Considering the past discussions, say it is one atom thick, yet remains optically opaque. And it is within two atoms distance from the fence. As the meter stick goes by at its 1/2 C velocity, a stationary light flashes. The fence has a photo sensitive surface. An image of the meterstick is made on the wall and persists long enough to observe and measure. One light source sent out. One beam of light from a single point in space. One event. The meterstick is close enough to the fence and thin enough, and the light is far enough away that there is no paralax problem of diverging light rays. I don't care if the image is to the left of exactly where we want it, or to the right. Its just there.

Now: My understanding is that the image will be 0.886 meters long. Is that okay for a theoretical discussion?

If not, then I don't understand what your are trying to say. Please explain why not?

One last request: Please excuse any typos. My fingers are getting old and sometimes type what they want and not what I want. Let's not get hung up on a truly minor detail such as that.

Thank you.

 

[DaveC426913]

bkelly, let me ask a question back.


Let's say we set up your experiment exactly as you describe, and it all works out tehnically perfectly. We get the measurements and we determine a length contraction of, say, x, leading to conclusion y (whatever conclusion you want to make).

We now take the entire experiment - trolley, observers and all - and put a rocket behind it and accerelate the entire thing at 1g for a few days. The entire experiment is now moving away from Earth at relativistic speeds.

We now perform the exact same experiment as before.

What do you think we will observe? Do you think we will get different results for x and/or y?

 

[Mentz114]

Hi bkelly,

I read what is posted on your website and I'm baffled. What I don't get is, what is the absolute velocity with respect to ? In the experiments you seem to be measuring the velocity of the metre stick wrt the fence.

 

[JesseM]

That is getting to the core of the present discussion. I cannot find any reference or reply that says two events must be within X and Y parameters in order to be in the same frame of reference. (You pick the quantities noted by X and Y. Let it be distance, time, or what ever you know to be blocking this discussion.)

Again, two events can be analyzed from the perspective of any frame of reference--in that sense they are "in" all of them. However, objects (not events) are sometimes said to be "in" a frame of reference if they are at rest in that frame, so we might say two objects are "in the same frame" if they are at rest relative to one another.

The respondents are saying that two markers one meter apart are not in the same frame of reference

Who said that? I don't see any comments on this thread saying such a thing, can you quote the comment you're talking about so I can see if maybe you're misunderstanding?

Regarding how the two markers are triggered: They are only one meter apart. They are not kilometers apart, not event tens of meters apart. Just one. Hence the question just above, what is the threshold?

There's no "threshold", events at different locations which are simultaneous in one frame are always non-simultaneous in others. If two events occurred simultaneously in a frame where the distance between them was L, then in another frame moving at speed v relative to the first frame, there will be a difference in time between the two events of gamma*v*L/c^2, where gamma = 1/sqrt(1 - v^2/c^2). So for example, if the second frame is moving at v=0.6c relative to the first, then the time between the events in the second frame is 1.25*0.6c*L/c^2 = 0.75*L/c. So if L=1 meter, then with c=299792458 meters/second, the time between the events in the second frame is about 2.50173 nanoseconds (0.00000000250173 seconds). Small, but if we're talking about a trolley dropping markers it makes a difference, since if the trolley is moving at 0.6c, it will have moved forward a distance of 0.6*299792458*0.00000000250173=0.45 meters in that time. So if the trolley is length-contracted to 0.8 meters in the track frame, then if the back marker is dropped 0.0000000025 seconds before the front one in the track frame, the front marker will be dropped at a distance of 0.8 + 0.45 = 1.25 meters from the back one in the track frame, which is greater than the trolley's rest length, not shorter.

configuration A: There is a solid rod between the two markers. At some point something in the middle of the rod tweaks it and the rod activates the markers.

In relativity it's impossible to have perfectly rigid rods, you can only have more realistic rods where if you push on one end, the other end doesn't move until a sound wave moving through the material of the rod has had time to travel from one end to the other. If the rod is at rest in the trolley frame, then presumably if you jiggle the middle the sound waves going in either direction will travel at the same speed in this frame, so the markers will be dropped simultaneously in the trolley frame. But again, this means the markers are not dropped simultaneously in the track frame.

Question 2: Are the markers in the same reference? Can the marks be considered to be made simultaneously. if not, why not?

Again I don't know what you mean by "in the same reference". Are you just asking if the markers are dropped simultaneously in the track frame? If that is what you're asking, the answer is no.

configuration B: Let's make the marks closer together. There is a single spray can. It sends out paint, or light, or what ever your might like, in the usual format that we expect from a spray can. In between the spray can and the fence there is a somethings that blocks the spray. I declare that that something causes a precisely one centimeter wide blockage when the trolley is at rest.

Is the "something" at rest relative to the trolley, or at rest relative to the track? If at rest relative to the trolley, isn't it in front of the spray can at all times? If at rest relative to the track, then won't it always leave a "shadow" of the same size on the fence regardless of how fast the spray can moves past it?

Condition C: Let's go crazy with this. The trolley is a meter stick of standard proportions. Let's make it 2 centimeters wide and one meter long at rest. The meterstick goes flying by right next to the fence such that an observer sees the 1 meter length as along the path of travel, and the 2 centimeter width as a vertical dimension. Make the distance between the meter stick and the fence to be microscopic. Its as close as it can be without touching. The meterstick is also microscopically thin. Considering the past discussions, say it is one atom thick, yet remains optically opaque. And it is within two atoms distance from the fence. As the meter stick goes by at its 1/2 C velocity, a stationary light flashes. The fence has a photo sensitive surface. An image of the meterstick is made on the wall and persists long enough to observe and measure. One light source sent out. One beam of light from a single point in space. One event. The meterstick is close enough to the fence and thin enough, and the light is far enough away that there is no paralax problem of diverging light rays.

OK, so all photons are treated as traveling at a right angle to the plane of the fence in the fence's rest frame, and they all reach the fence at the same instant in the fence's rest frame? In this case, the image of the meterstick will indeed be 0.866 meters in the fence rest frame. But keep in mind, in the rest frame of the meterstick the same photons are not traveling at a right angle to the plane of the fence and they do not all hit the surface of the fence simultaneously, so you can explain this result in the frame of the meterstick too.

 

[bkelly]

bkelly, let me ask a question back.
Let's say we set up your experiment exactly as you describe, and it all works out tehnically perfectly. We get the measurements and we determine a length contraction of, say, x, leading to conclusion y (whatever conclusion you want to make).

I will answer your question, but first: Now that you have me all worked up about simultaneity and the impossibility of my experiment, and now that I have provided some conditions that I think resolve the simultaneity problem, please respond to the my last post and we can work our way back to the original concept.

Start with my condition C last. Will that suffice to run a theoretical test of contraction due to relativistic speeds?

You baited the hook by refusing to consider the merits of my theory using simultaneity. Now that I have bit the hook and resolved that problem, after considerable effort, it would be rude of you to just say "oh never mind about that." Let's layout a solid conclusion to this problem of simultaneity.

 

[JesseM]

You baited the hook by refusing to consider the merits of my theory using simultaneity. Now that I have bit the hook and resolved that problem, after considerable effort, it would be rude of you to just say "oh never mind about that." Let's layout a solid conclusion to this problem of simultaneity.

Did you read my response? Your condition C doesn't solve the simultaneity issue, since if the photons hit the fence simultaneously in the frame of the fence, they don't hit it simultaneously in the frame of the meter stick.

 

[bkelly]

OK, so all photons are treated as traveling at a right angle to the plane of the fence in the fence's rest frame, and they all reach the fence at the same instant in the fence's rest frame? In this case, the image of the meterstick will indeed be 0.866 meters in the fence rest frame. But keep in mind, in the rest frame of the meterstick the same photons are not traveling at a right angle to the plane of the fence and they do not all hit the surface of the fence simultaneously, so you can explain this result in the frame of the meterstick too.

I was typing and posting at the same time as JesseM and saw this only after I posted.

In response: So... I have hit upon a method of conducting my theoretical test. We have dispensed with the issue of simultaneity. The instead of a stick a meter long, we could have a longer stick with holes a meter apart and made marks to measure. I could probably find other methods. For the purpose of this discussion, it does not matter. We really could have skipped all this time spent yammering about simultaneity. The essay said that it makes the marks. You (you in general, not specifically you JesseM) could have just as well addressed my concept to start with.

I will start a new post to continue the discussion and not pollute it with this side issue.

 

[DaveC426913]

The fence has a photo sensitive surface. An image of the meterstick is made on the wall and persists long enough to observe and measure. One light source sent out. One beam of light from a single point in space. One event. The meterstick is close enough to the fence and thin enough, and the light is far enough away that there is no paralax problem of diverging light rays.

No. The light rays do not travel the same distance. Each photon arrives at a different time.

Again, this problem of simultaneity is noty simply going to go away. It is not a detail. It is the very crux of relativity.

It is the thing that is relative in "the theory of relativity".

Now, I've addressed your condition C. What will happen to the experiment when it is receding at a relativistic speed away from Earth?

 

[bkelly]

Again, JesseM and I were posting at the some time. I will let that issue ride for a bit and return to the concept of my essay.

Hi bkelly,
I read what is posted on your website and I'm baffled. What I don't get is, what is the absolute velocity with respect to ? In the experiments you seem to be measuring the velocity of the metre stick wrt the fence.

That is the crux of the essay. My understanding of relativity contains the concept that it says, among many things, that there is no such thing as an absolute velocity. There is no such thing as anyone point, point A, being absolutely stationary and some other point, point B being moving relative to to A. Neither A or B can prove that they or stationary and the other moving, or that they are moving and the other stationary.

To the essay, the theory says that if Tom was unable to detect accelerations, it is not possible for him to determine if he was moving or if the fence was moving.

My essay says that I think there is a way to determine if you are moving.

Stage 1: Tom uses the trolley to make marks on a wall that are one meter apart. He measures the marks and finds them to be one meter. Then he causes the trolley to be moving at 1/2 c when he makes the marks. They are now 0.866 meters apart, not one meter. All well and good.

Now for the wrinkle. Tom does his trolley experiment again. To him everything is the same. He watches as the trolley wizzes by and with his impossibly sharp and fast eyes, he sees that the marks are 0.866 meters apart.

But unknown to Tom, Sally has caused Tom to be moving in the same direction as the trolley when it makes the marks, and at 1/2 c with respect to Sally. Then Sally brings Tom back to the second pair of marks. Now he is back in the same reference with Sally, and the same reference where he made the first pair of test marks.

Sally watched as his speed was added relativistically with the trolley speed. Sally with her equally fast and accurate eyes saw the trolley moving at about 239,000,000 meters per second and the marks were 0.6 meters apart.

Sally brings Tom back to the last set of marks and they are both stationary with respect to each other. (To be redundant, Tom was unaware that Sally moved him, and unaware that she had to being him back to the marks. That is a major point in my concept.) The marks are not 0.866 meters apart as Tom expected. They are closer together. Therefore Tom can detect that he was moving at relativistic speed. He also knows that he was moving in the same direction as the trolley. He can also calculate his speed with respect to Sally. That disagrees with the theory.

Jeez, I handn't meant to just write the essay again, but that is about it. If you disagree about some aspect of my test or the results, please let me know exactly where I am wrong. If need be, I can break it up into more paragraphs and smaller sentences.

 

[JesseM]

In response: So... I have hit upon a method of conducting my theoretical test. We have dispensed with the issue of simultaneity. The instead of a stick a meter long, we could have a longer stick with holes a meter apart and made marks to measure. I could probably find other methods. For the purpose of this discussion, it does not matter. We really could have skipped all this time spent yammering about simultaneity. The essay said that it makes the marks. You (you in general, not specifically you JesseM) could have just as well addressed my concept to start with.

There is no possible method to make marks that will not result in some frames saying the marks were made simultaneously while others say the marks were made non-simultaneously. And the distance between the marks in the track frame will depend on whether the marks were made simultaneously in the track frame or non-simultaneously. So, if you want an answer to the question of how far apart the marks are, there is no way to answer that question without bringing up the issue of simultaneity.

 

[bkelly]

There is no possible method to make marks that will not result in some frames saying the marks were made simultaneously while others say the marks were made non-simultaneously. And the distance between the marks in the track frame will depend on whether the marks were made simultaneously in the track frame or non-simultaneously. So, if you want an answer to the question of how far apart the marks are, there is no way to answer that question without bringing up the issue of simultaneity.

Given my condition C with the shadow on the fence, just how much difference will it make?

 

[JesseM]

Stage 1: Tom uses the trolley to make marks on a wall that are one meter apart. He measures the marks and finds them to be one meter. Then he causes the trolley to be moving at 1/2 c when he makes the marks. They are now 0.866 meters apart, not one meter. All well and good.

OK, so that implies the marks are made simultaneously in the frame of Tom. Fine.

Now for the wrinkle. Tom does his trolley experiment again. To him everything is the same. He watches as the trolley wizzes by and with his impossibly sharp and fast eyes, he sees that the marks are 0.866 meters apart.

But unknown to Tom, Sally has caused Tom to be moving in the same direction as the trolley when it makes the marks, and at 1/2 c with respect to Sally.

When you say "the trolley wizzes by", that implies that even though Tom is moving at 0.5c in the same direction as the trolley, the trolley must be moving even faster in that direction? Is it still moving at 0.5c relative to Tom? If so, then according to the relativistic velocity addition formula it must be moving at (0.5c + 0.5c)/(1 + 0.5*0.5) = 1c/1.25 = 0.8c relative to Sally.

Now, in this case are the marks made simultaneously in the rest frame of Sally, or in the rest frame of Tom? Also, are they made on a wall that's at rest relative to Tom, or a wall that's at rest relative to Sally?

Sally brings Tom back to the last set of marks and they are both stationary with respect to each other. (To be redundant, Tom was unaware that Sally moved him, and unaware that she had to being him back to the marks. That is a major point in my concept.) The marks are not 0.866 meters apart as Tom expected. They are closer together.

If the marks were made simultaneously in Sally's frame, on a wall at rest in Sally's frame, then they would be closer together, with the trolley moving at 0.8c in her frame they'd only be 0.6 meters apart. But if that was the case, why did you say "with his impossibly sharp and fast eyes, he sees that the marks are 0.866 meters apart"? If the marks were made non-simultaneously in his frame, then he wouldn't see them being made 0.866 meters apart. On the other hand, if they were made simultaneously in Tom's frame, in Sally's frame the back mark would be made before the front mark, so that would lengthen the distance between the marks in Sally's frame beyond 0.6 meters (how long the distance was would depend on whether the marks were made on a wall at rest in Sally's frame, or if they were made on a wall at rest in Tom's frame which then was brought to rest in Sally's frame along with Tom).

 

[JesseM]

Given my condition C with the shadow on the fence, just how much difference will it make?

It'll be exactly the same as any other method that makes marks simultaneously in the frame of the fence.

 

[bkelly]

It'll be exactly the same as any other method that makes marks simultaneously in the frame of the fence.

In the scheme of my essay, how much difference will it make? Don't dodge the question with any nonsense about how it will compare to something else. You made a issue of this. Now answer the question: How much difference will it make?

 

[JesseM]

In the scheme of my essay, how much difference will it make? Don't dodge the question with any nonsense about how it will compare to something else. You made a issue of this. Now answer the question: How much difference will it make?

It's rather rude to accuse me of "dodging" your question, I gave a simple answer that I thought addressed it, but I can't read your mind to know exactly what kind of answer you want. I'm still not sure what you want--are you asking for a quantitative answer to how much time will pass between each mark being made in the frame where they are made non-simultaneously, like what I did in post #19 (and you didn't comment on), or a qualitative discussion of how much difference it makes to your argument (in which case, please address my questions about the details of your scenario from post #27), or something else?

 

[bkelly]

OK, so that implies the marks are made simultaneously in the frame of Tom. Fine.

When you say "the trolley wizzes by", that implies that even though Tom is moving at 0.5c in the same direction as the trolley, the trolley must be moving even faster in that direction? Is it still moving at 0.5c relative to Tom? If so, then according to the relativistic velocity addition formula it must be moving at (0.5c + 0.5c)/(1 + 0.5*0.5) = 1c/1.25 = 0.8c relative to Sally.

Yes, that is my intent.

Now, in this case are the marks made simultaneously in the rest frame of Sally, or in the rest frame of Tom? Also, are they made on a wall that's at rest relative to Tom, or a wall that's at rest relative to Sally?

This is the key to my concept. I do not know how to say this correctly, but I think the reply is to the effect: We switch frames from Tom to Sally, or, Tom and Sally really have the same frame, but maybe the don't know or cannot prove it. Maybe this proves it one way or the other.

If the marks were made simultaneously in Sally's frame, on a wall at rest in Sally's frame, then they would be closer together, with the trolley moving at 0.8c in her frame they'd only be 0.6 meters apart. But if that was the case, why did you say "with his impossibly sharp and fast eyes, he sees that the marks are 0.866 meters apart"?

the point of Tom's sharp eyes is that he could, somehow, detect that as the trolley went past him and what appeared to be 1/2 c, the marks, to him, at the time he saw them made, they were 0.866 meters apart. However, unknown to him, he was moving past the fence at a relative speed of 1/2 c under control of Sally. He would need really sharp eyes to measure the marks as they were made and move away from him at 1/2 c. That section is not entirely required for my concept, but I think it helps me better describe the setup. Maybe I should drop it.

If the marks were made non-simultaneously in his frame, then he wouldn't see them being made 0.866 meters apart. On the other hand, if they were made simultaneously in Tom's frame, in Sally's frame the back mark would be made before the front mark, so that would lengthen the distance between the marks in Sally's frame beyond 0.6 meters (how long the distance was would depend on whether the marks were made on a wall at rest in Sally's frame, or if they were made on a wall at rest in Tom's frame which then was brought to rest in Sally's frame along with Tom).

Again, this is the crux of my thoughts. Tom causes the marks to be made and to his perspective, they are 0.866 meters apart because the trolley is moving. But, and this is it, he did not know he was moving. He is stopped and brought back to the marks by Sally. Tom knows something is afoot, but not what. Not yet. He identifies the marks just made, and sees that they are 0.6 meters apart. He now has proof that he was not stationary, but moving at 1/2 C. And the proof shows that he was moving in the same direction as the trolley.

Without knowing that Sally exists, he detected that he had a velocity relative to something else, namely Sally. That is what contradicts my knowledge of relativity.

 

[Janus]

I will answer your question, but first: Now that you have me all worked up about simultaneity and the impossibility of my experiment, and now that I have provided some conditions that I think resolve the simultaneity problem, please respond to the my last post and we can work our way back to the original concept.

Start with my condition C last. Will that suffice to run a theoretical test of contraction due to relativistic speeds?

Okay,

Condition C: Let's go crazy with this. The trolley is a meter stick of standard proportions. Let's make it 2 centimeters wide and one meter long at rest. The meterstick goes flying by right next to the fence such that an observer sees the 1 meter length as along the path of travel, and the 2 centimeter width as a vertical dimension. Make the distance between the meter stick and the fence to be microscopic. Its as close as it can be without touching. The meterstick is also microscopically thin. Considering the past discussions, say it is one atom thick, yet remains optically opaque. And it is within two atoms distance from the fence. As the meter stick goes by at its 1/2 C velocity, a stationary light flashes. The fence has a photo sensitive surface. An image of the meterstick is made on the wall and persists long enough to observe and measure. One light source sent out. One beam of light from a single point in space. One event. The meterstick is close enough to the fence and thin enough, and the light is far enough away that there is no paralax problem of diverging light rays. I don't care if the image is to the left of exactly where we want it, or to the right. Its just there.

Here's the problem. In the fence frame your source comes from an angle 90° to the relative motion of the stick and is stationary relative to the fence. The light hits both ends simultaneously. This will indeed make a mark 0.866m long on the fence in the fence frame. ( for illustration, we'll assume that the fence has posts 1m apart as measured by someone at rest with respect to the fence. Thus the mark will be 0.866 of the distance between two posts.

Now consider what happens if everything else is the same except we assume that it is the stick that is stationary and it is the fence that is moving at 0.5c. For this to be perfectly reciprocal, this means that the light source moves with the fence.

Two things happen:

1. It is the fence that contracts so that there is a 0.866m distance between the posts.

2.The light from the source no longer comes from an angle of 90°, But at an angle tilted towards the direction opposite that in which the fence is moving (this is called relativistic aberration).
This means that the light hits one end of the stick before it hits the other. This also means that the fence moves between the moment the light hits one end of the stick and the other.

The result will be that the mark it makes on the fence will be shorter than 1 meter long. In fact, it will be 0.75 m long. 0.75 is 0.866 of 0.866, so this means it makes a mark on the fence that is 0.866 the distance between the posts. The exact ratio of mark to fence as you got when you assumed it was the stick that was moving.

IOW, you cannot use the mark on the fence to determine absolute motion, as you get the same results whether you consider the fence or stick as moving. Also, you have not resolved the simultaneity issue, because you still have two events (the light striking the individual ends of the stick) which are simultaneous in one frame but not in the other.

 

[bkelly]

It's rather rude to accuse me of "dodging" your question, I gave a simple answer that I thought addressed it, but I can't read your mind to know exactly what kind of answer you want. I'm still not sure what you want--are you asking for a quantitative answer to how much time will pass between each mark being made in the frame where they are made non-simultaneously, like what I did in post #19 (and you didn't comment on), or a qualitative discussion of how much difference it makes to your argument (in which case, please address my questions about the details of your scenario from post #27), or something else?

Do you think its possible I might feel you were rude to me. You went on and on about simultaneity saying it spoiled my theoretical experiment, then when I resolved the problem, you did not recant, did not actually admit I had a point, and did not really address my responses to the simultaneity problem.

You still have not answered the question from that post: How much difference will it make?

Edit: this references post 30. I hope that that number stays fixed.

 

[DaveC426913]

You said you'd answer my question.

I will answer your question, but first:Start with my condition C last.

No. The light rays do not travel the same distance. Each photon arrives at a different time.

Now, I've addressed your condition C. What will happen to the experiment when it is receding at a relativistic speed away from Earth?

 

[bkelly]

Okay,
...
Now consider what happens if everything else is the same except we assume that it is the stick that is stationary and it is the fence that is moving at 0.5c. For this to be perfectly reciprocal, this means that the light source moves with the fence.
...

I do not want to go down that path. I want to stay with the reference that Sally provides. She is stationary and moves Tom. He does not know it and the facts show him that he was moving. Let's stay in that frame of reference. One problem at a time please.

And it is past bedtime for this guy. I will return soon.