Is it possible to determine absolute speed?

[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.

 

[JesseM]

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.

I did address your proposals for "resolving" the simultaneity problem in post #19, which you didn't respond to. I explained why none of them would change the fact that if the marks were made simultaneously in one frame, then they'd be made non-simultaneously in other frames.

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

I asked for clarification of what you meant by "how much difference will it make", if you answer my questions about your meaning I will answer it:

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]

I did address your proposals for "resolving" the simultaneity problem in post #19, which you didn't respond to. I explained why none of them would change the fact that if the marks were made simultaneously in one frame, then they'd be made non-simultaneously in other frames.

I asked for clarification of what you meant by "how much difference will it make", if you answer my questions about your meaning I will answer it:

You said the photon would not strike the fence on either side of the meter stick at exactly the same time. I asked what difference that would make. It is very obvious that I intended to ask that question within the framework of my essay. You know perfectly well the intent of my question.

You said the photons would not arrive at exactly the same time. Ok. So what difference will that make?

If you don't have a valid answer to the question, then: Why did you make the statement?

good night.

 

[JesseM]

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.

A "frame" is just a coordinate system. If Tom is moving at 0.5c relative to Sally, by definition that means any frame in which Tom is at rest (coordinate position not changing with coordinate time) is one where Sally is in motion (coordinate position changing with coordinate time) and vice versa. If by "have the same frame" you mean something different from being at rest in the same frame, please explain. As I told you before, we are free to analyze any combination of objects and events from any frame we choose.

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.

But that would only be true if they were made simultaneously in his frame. Suppose the marks were made simultaneously in Sally's frame, so if the first mark was made at x=0 meters and t=0 seconds in Sally's coordinate system, the next mark was also made at t=0 seconds, but 0.6 meters away at x=0.6 meters in her coordinate system (remember that if the trolley is moving at 0.8c in Sally's frame its length is shrunk by a factor of \sqrt{1 - 0.8^2} = 0.6). Then the coordinates of these marks in Tom's frame are given by the Lorentz transformation:

x'=gamma*(x-vt)
t'=gamma*(t-vx/c^2)

where v=0.5c (or 149896229 meters/second) and gamma=1/sqrt(1-v^2/c^2) = 1/0.866 = 1.1547

So, if the first mark had coordinates (x=0, t=0) in Sally's frame, plugging it into the above gives (x'=0, t'=0) in Tom's frame. Then if the second mark had coordinates (x'=0.6, t'=0) in Sally's frame, plugging it into the above gives:

x'=1.1547*(0.6 - 0.5c*0) = 0.6928 meters
t'=1.1547*(0 - 0.5c*0.6/c^2) = 1.1547*(-0.3/c)=1.1547*(-0.3/299792458) = -0.0000000011555 seconds

So in Tom's frame the marks were actually made 1.1555 nanoseconds apart, and if he was holding a ruler at rest relative to himself next to the trolley and noted the positions on his ruler that were right next to the marks when they happened, he would see that there was a distance of 0.6928 meters between the positions on his ruler that the marks were made. This actually makes sense, because the trolley is 0.866 meters long in his frame, but if the front mark was made 0.0000000011555 seconds before the back mark in his frame, in that time the back of the trolley has moved a distance of 0.5c*0.0000000011555=0.5*299792458*0.0000000011555=0.1732 meters closer to the position on his ruler where the front mark was made, so by the time the back mark is made the back will only be at a distance of 0.866 - 0.1732 = 0.6928 meters from the position where the front mark was made.

So, if the marks were made on a fence at rest relative to Tom, then the marks are 0.6928 in the rest frame of the fence, so they still will be when the fence is brought to rest in Sally's frame. On the other hand, if the marks were made on a fence at rest relative to Sally, then in Tom's frame the fence is moving at 0.5c in the opposite direction of the trolley, so as soon as the front mark is made it's moving at 0.5c towards the back of the trolley, and in the 0.0000000011555 seconds it takes before a mark is made at the back, the front mark will have gotten closer to the position where the back mark is made by 0.5c*0.0000000011555=0.5*299792458*0.0000000011555=0.1732 meters. In this case, even though the distance between the positions on his ruler where the marks were made is 0.6928 meters, he will measure the distance between the front and back mark as only 0.6928-0.1732=0.5196 meters. This result also makes sense, since if the fence is at rest in Sally's frame the marks should be 0.6 meters apart in her frame, and in Tom's frame the fence is moving at 0.5c so the distance between marks should be shrunk by a factor of 0.866, and 0.6*0.866=0.5196.

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 again, that's only true if the marks are simultaneous in his frame. If they are simultaneous in Sally's frame, he won't measure them as 0.866 meters apart.

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.

Whatever scenario you're considering, we can imagine that besides Sally and Tom, we also have two other observers Sally2 and Tom2, with Sally2 moving at 0.5c relative to Sally (and at rest relative to Tom when Tom is moving at 0.5c relative to Sally), and Tom2 who starts out moving at 0.5c in the opposite direction as Tom, so he's at rest with respect to Sally, along with a second trolley moving at 0.8c relative to Sally2 in the same direction as Tom2, so to Sally it's moving at 0.5c in the opposite direction as Tom and the first trolley. Then whatever happens with the first trolley and Sally and Tom, we replicate it with the second trolley and Sally2 and Tom2...for example, if the first trolley leaves marks on a fence at rest with respect to Tom and simultaneously in the frame of Sally, then the second trolley leaves marks on a fence at rest with respect to Tom2 and simultaneously in the frame of Sally2. Then just as Tom and the marks from the first trolley are brought to rest relative to Sally, Tom2 and the marks from the second trolley are brought to rest relative to Sally2. Then whatever observations Sally and Tom make about their marks, exactly the same observations will be made by Sally2 and Tom2 about their marks, despite the fact that Tom2 was at rest relative to Sally and Tom when the marks were made, and then when he's brought to rest relative to Sally2 that means he's accelerated to 0.5c relative to Sally and Tom. So the situation will be perfectly symmetrical, there'll be no basis for concluding that Tom was moving while Sally was at rest, because exactly the same reasoning would make Tom2 think he was moving (when he was at rest relative to Sally) and Sally2 was at rest (when she was moving at 0.5c relative to Sally just like Tom was before coming to rest relative to her).

 

[JesseM]

You said the photon would not strike the fence on either side of the meter stick at exactly the same time. I asked what difference that would make. It is very obvious that I intended to ask that question within the framework of my essay. You know perfectly well the intent of my question.

No, I really don't, and please don't accuse me of lying. Please just answer my question about whether you want a quantitative calculation of the time difference between the times the photons struck next to either end of the meter-stick in the frame where the meter stick was at rest, or some more qualitative discussion of what impact this non-simultaneity has on your argument, or something else. I'm happy to answer as soon as I know what you're asking, but it's a pretty rude move to refuse to answer this question and then say "you know perfectly well the intent" even though I just told you I wasn't sure and asked politely for a clarification.

 

[Janus]

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.



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.

No. You still have simultaneity issues. When Tom makes his marks (using the trolley as a reference while it is moving at 0.5c relative to him.) he has to make two marks on the wall at the same instant, in other words, simultaneously. (according to him.)
But according to Sally, He will not make both marks simultaneously. He will make one mark before the other. As a result, The trolley will move (according to her) relative to Tom between the moments that he make his marks.

Also, Since I assume that the wall is at rest with respect to Sally, Tom will move with respect to the wall in that same time. As a result, Tom will not make the marks 0.6m apart on the wall according to Sally. Also, according to Tom, the wall is moving at 0.5c relative to him and is length contracted when he makes his (simultaneous) marks on it.

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.

So at the end, Tom will note that the marks on the wall are 1 meter apart. (the wall, and the marks on them un-contract when they are brought into the same frame as he is in.) But this tells him nothing about his own motion, only the relative motion between the wall and himself, as he would get the same result whichever one you consider as "moving".

Sally will also note that the marks are 1 meter apart , due to the relative movement of the wall and trolley during the time between Tom making his marks on the wall( which according to her were not made simultaneously.)

So no, this scenario will not do anything in determining Tom's absolute motion, you would get exactly the same end results in terms of the distance between the marks if you assumed it was Sally and the wall that was moving and not Tom.

And despite you claims otherwise, you have not "resolved" the simultaneity issue. There is no way to design the experiment to get rid of it. It is not a technical issue to be "fixed". It is an integral part of the nature of time and space.

 

[Ich]

bkelly, this is hilarious. I gave you all the answers in the very https://www.physicsforums.com/showthread.php?p=2823558#post2823558".

https://www.physicsforums.com/showpost.php?p=2825668&postcount=1":

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.

Right. They're 1m/.866=1.155m 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.

No. As I said, they're 1m/(1-v²)=1.155m apart.

How much difference will it make?

The difference is 1.155-0.866. Instead of length contraction, you get expansion by the same factor. That's the effect of the simultaneity issue.

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.

You can detect nothing. Because SR only cares about relative speeds. But in order to see how it works, you have to learn SR.

A quote from http://xkcd.com/675/" :

I mean, what's more likely – that I have uncovered fundamental flaws in this field that no one in it has ever thought about, or that I need to read a little more? Hint: it's the one that involves less work.

 

[Austin0]

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.



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.

Hi bkelly I have only gotten the bare gist of your argument and have not followed all the different aspects of simultaneity etc.
But just from your outline it appears that even if you could establish motion with regard to the fence or the troley , or whatever, this still gives you no clue as to absolute motion or any value to put on it.
IMO

 

[Janus]

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.

But you have to go down that path. As I pointed out in my last post, the fact that the marks aren't the distance apart that Tom expected them to be at the end of the experiment has nothing to say about his absolute motion. It can only tell him that there was a relative difference in velocity between Sally and himself when he made them.

In order to establish absolute motion on Tom's part you would have to show that he would get a different end result if it was Sally that was moving and not him. So you have to consider both.

Relativity however, when properly applied, will give identical results no matter who you considered as "moving".

 

[Dale]

Bkelly, let's say that I have a stick moving inertially to the right at speed v in some reference frame of interest. Then the left end of the stick has the worldline:
x_L(t)=vt
And the right end of the worldline has the worldline:
x_R(t)=vt+D
Where D is by definition the length of the stick in the reference frame of interest.

If look at the x positions of a pair of events then we find that:
\Delta x= x_R(t_R)-x_L(t_L)=D+(t_R-t_L)v
This only equals D for the special cases:
t_R=t_L
And
v=0

Therefore, in order to use your marks to determine the length in some frame of interest the marks must generally be made simultaneously in that frame. That is possible in, at most, one frame. Therefore, you simply cannot use this method to compare lengths in multiple frames.

 

[bkelly]

keeping it simple

To All,
A problem that I see in open threads like this is that the thread gets fragmented and there are too many points to address at one time. I am not able to keep track of all these concepts simultaneously. I don’t want to short anyone a deserved reply, but I need to ask about some fundaments for which I don’t understand the importance.

My basic problem is this simultaneity concept. From post 23 by Dave:

Originally Posted by bkelly
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.

I chose that post as it is a good simple place to begin.

From this response I conclude that Dave believes the photons arrive at the fence at sufficiently different times to spoil the test. It is perfectly reasonable to state: It is a given that the light source is stationary wrt to Sally, and wrt to the fence, and the distance from the light source to both ends of the meterstick is identical (at the time the image is created). If you please, let's also presume that all calculations have been done in advance and the light strobes (flashes) at just the right time so that the photons will reach the meterstick and wall and the right time to create the shadow image. Remember I did say that the meter stick is only an atom's width in thickness and only an atom's width distance from the wall.

I say the photons will arrive at both ends of the meter stick as the same time. If Dave says not, then I must ask a few questions.

Before asking "why," I want to stick to my knitting so to speak. I ask for your indulgence and skip the why for the moment and concentrate on the results.
How much difference will there be in the timing of the photons hitting the fence?
I expect the answer to that will provide the answer to “How much difference will it make to my test?”

I will be reading other posts and see what I can learn from them. In the meantime, I don’t want to short any post, I just need to concentrate on a few narrow topics at one time. Thanks for your patience.

 

[JesseM]

Before asking "why," I want to stick to my knitting so to speak. I ask for your indulgence and skip the why for the moment and concentrate on the results.
How much difference will there be in the timing of the photons hitting the fence?
I expect the answer to that will provide the answer to “How much difference will it make to my test?”

You can use the Lorentz transformation to answer questions like this. If we know the difference in x-coordinate dx and the difference in time dt between two events in one frame, then if you have another frame moving at speed v along the x-axis of the first, the distance dx' and dt' between the same pair of events is:

dx' = gamma*(dx - v*dt)
dt' = gamma*(dt - v*dx/c^2)

with gamma=1/sqrt(1 - v^2/c^2)

So, if the events of the photons striking the fence right next to either side of the stick are simultaneous in Sally's frame, then in her frame dt=0 seconds for this pair of events. And the dx between them is just the length of the meter stick in her frame--if the stick is moving at 0.5c (with c=299792458 meters/second), then the length will be 0.866 meters, so dx=0.866 meters. And if v=0.5c then gamma=1/sqrt(1-v^2/c^2) = 1/0.866 = 1.1547. So, plugging that into the above you can find dx' and dt' in the frame of the meter-stick:

dx' = 1.1547*(0.866 - 0.5c*0) = 1.1547*0.866 meters = 1 meter
dt' = 1.1547*(0 - 0.5c*0.866/c^2) = -1.1547*(0.5*0.866/c) = -1.1547*(0.5*0.866/299792458) = -0.0000000016678 seconds

So, in the meter stick's frame the distance between the photons striking near either end of it is 1 meter (naturally, since the meter stick is at rest and 1 meter long in its own frame), and the time between them striking is only 1.6678 nanoseconds. But if the wall is moving at 0.5c, in this time it will have moved a distance of 0.5*299792458*0.0000000016678=0.25 meters in this short time! So that means that although the events of the photons hitting at either end happen 1 meter apart in the frame of the meter-stick, the length of the "shadow" on the wall (created by the area that photons are blocked from hitting) is only 1-0.25=0.75 meters in the frame of the meter-stick. Which makes sense, since anything which is 0.866 meters in the frame of the wall should be shrunk to 0.866*0.866=0.75 meters in the frame of the meter-stick.

 

[Mentz114]

bkelly, you're wasting your time. There's no such thing as 'who's really moving'. Uniform motion is observer dependent and all one can ever say is that one thing is moving relative to another or not. This has been understood since aether theories were superceded by SR. But you won't accept this. Why ? Do you really think you know better than everyone who understands SR ?

 

[bkelly]

You can use the Lorentz transformation to answer questions like this. If we know the difference in x-coordinate dx and the difference in time dt between two events in one frame, then if you have another frame moving at speed v along the x-axis of the first, the distance dx' and dt' between the same pair of events is:

dx' = gamma*(dx - v*dt)
dt' = gamma*(dt - v*dx/c^2)

with gamma=1/sqrt(1 - v^2/c^2)

I cannot follow your point here. I am rather good at the logic of writing code and making systems work, but struggled mightily in Calculus. Please elaborate a few points for me.

Please explain the term: the x-coordinate dx. I understand two and three dimensional Cartesian coordinates and understand that dx stands for delta in x or a difference. But I don't know what delta you are referring to. The light is not moving, the fence is not moving, and the distance from the light to to the left side of the meter stick and the right side of the meter stick is identical at the time the image is created. So why is there a dx, which I presume leads to the dt.

 

[bkelly]

bkelly, you're wasting your time. There's no such thing as 'who's really moving'. Uniform motion is observer dependent and all one can ever say is that one thing is moving relative to another or not. This has been understood since aether theories were superceded by SR. But you won't accept this. Why ? Do you really think you know better than everyone who understands SR ?

No, I pursue it because I see a contradiction in my understanding and in what others say. Are you trying to tell me "Shut up and follow the crowd?"

 

[Mentz114]

No, I pursue it because I see a contradiction in my understanding and in what others say.

Do you mean "No, I pursue it because I see a contradiction between my understanding and what others say."

Have you abandoned this position ?

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.

If that's the case, good luck to you. You do seem to have a misunderstanding about the nature of uniform motion.

 

[talk2glenn]

No, I pursue it because I see a contradiction in my understanding and in what others say.

What contradiction? Be specific.

Am I the one moving or is that other object moving past me.

You've answered your own question. You cannot describe the motion of the object without the "past me". It has only relative motion; there is no such thing as absolute motion. Forget the calculus for a moment - just as a thought experiment try and describe the motion of an object through space, without an observer the object moves relative to.

This has been understood in some form since Galileo posited the man in the ships hold experiment, and predates both relativity and calculus.

 

[DaveC426913]

bkelly, I ask you again:

if you took the entire experiment - trolley, yardstick, Sally and her friend - and accelerated it to .5c and took the measurements again, do you expect to see any differences in the outcome?

 

[JesseM]

Please explain the term: the x-coordinate dx. I understand two and three dimensional Cartesian coordinates and understand that dx stands for delta in x or a difference. But I don't know what delta you are referring to.

Remember, I referred to two events: the first event was the light striking next to the left side of the meter-stick, the second event was the light striking next to the right side. If the first event had position coordinate x₀ in the cartesian coordinate system where the wall is at rest, and the second event had position coordinate x₁ in this same coordinate system, then dx in this coordinate system is x₁ - x₀, that's why I said "the difference in x-coordinate dx" (i.e. dx is the difference between the x-coordinates of the two events, you took this out of context when you just quoted the second part "x coordinate dx"). Likewise, if the first event had time coordinate t₀ in the rest frame of the wall, and the second event had time coordinate t₁, then dt=t₁ - t₀ (so if the events occurred at the same moment in this frame, dt=0 in this frame)

 

[Janus]

To All,
A problem that I see in open threads like this is that the thread gets fragmented and there are too many points to address at one time. I am not able to keep track of all these concepts simultaneously. I don’t want to short anyone a deserved reply, but I need to ask about some fundaments for which I don’t understand the importance.

My basic problem is this simultaneity concept. From post 23 by Dave:



I chose that post as it is a good simple place to begin.

From this response I conclude that Dave believes the photons arrive at the fence at sufficiently different times to spoil the test. It is perfectly reasonable to state: It is a given that the light source is stationary wrt to Sally, and wrt to the fence, and the distance from the light source to both ends of the meterstick is identical (at the time the image is created). If you please, let's also presume that all calculations have been done in advance and the light strobes (flashes) at just the right time so that the photons will reach the meterstick and wall and the right time to create the shadow image. Remember I did say that the meter stick is only an atom's width in thickness and only an atom's width distance from the wall.

I say the photons will arrive at both ends of the meter stick as the same time. If Dave says not, then I must ask a few questions.

Before asking "why," I want to stick to my knitting so to speak. I ask for your indulgence and skip the why for the moment and concentrate on the results.
How much difference will there be in the timing of the photons hitting the fence?
I expect the answer to that will provide the answer to “How much difference will it make to my test?”

I will be reading other posts and see what I can learn from them. In the meantime, I don’t want to short any post, I just need to concentrate on a few narrow topics at one time. Thanks for your patience.

I think a major problem here is that there is a bit of "talking past each other" going on here. I have a feeling that when we say that there are "simultaneity issues", you are thinking of something different from what we mean. It is a "concept" problem and not a "How much of a difference will it make?" problem.

So, before we deal with the "how much of a difference it will make" issue, let's make sure that we are on the same page conceptually first.

The concept that you aren't getting is called the Relativity of Simultaneity. One of the postulates of Relativity is that the speed of light is the same for all inertial frames.

Using that, we can show that two observers moving relative to each other will disagree at to whether two events are simultaneous or not.

Consider the following scenario:

You have an observer standing beside a railroad track. Two flashes of light occur to either side of him, both an equal distance from him. He see the flashes at the same instant, and knowing that they originated equal distances from him, he knows that they originated simultaneously.

At the same instant that he sees the flashes, a second observer riding on a railway car passes him. The second observer also sees the flashes at the same time.

The following animation show events as they occur according to the embankment observer.

[URL]http://home.earthlink.net/~jparvey/sitebuildercontent/sitebuilderpictures/train1.gif[/URL]

Now let's consider things according to the rail car observer. We already know that he sees that both flashes at the same time, but what does this tell him about when they originated. Remember, the speed of light is a constant for him. He cannot measure the light light of one flash as traveling faster than the other relative to himself. He also knows that between the time the flashes originated his relative distance from the origins has changed. since he was an equal distance from the origins when he saw the flashes, it stands to reason that he was when the flashes originated, he was closer to one than the other. But, if the flashes originated simultaneously, and he was closer to one than the other when that happened, given a constant speed for both light flashes, he should have seen one flash before the other. Instead, he saw both flashes at the same time. The only way that this could be is if the flashes originated at different times.

this Animation shows the same events as the first animation but according to the railway observer.

[URL]http://home.earthlink.net/~jparvey/sitebuildercontent/sitebuilderpictures/train2.gif[/URL]

He sees the flashes simultaneously, but determines that they originated at different times.

The two observer will disagree as to whether or not the flashes originated at the same time or not.

Are you with me so far?

 

[bkelly]

I think a major problem here is that there is a bit of "talking past each other" going on here. I have a feeling that when we say that there are "simultaneity issues", you are thinking of something different from what we mean. It is a "concept" problem and not a "How much of a difference will it make?" problem.

That may well be.

The two observer will disagree as to whether or not the flashes originated at the same time or not.
Are you with me so far?

I don't think I am with you as I don't see anything to do with relativity. Two lights separated by some distance flash. At some point between the two lights the the light waves cross. Anyone that is in the location will see both lights at the same time. It does not matter if they were standing there when the lights flashed, or if they were standing somewhere else and moved there in time to be there when the light waves intersected each other.

However, I presume that you will follow this up and will wait for that post.

Thanks for taking the time to post.

 

[DaveC426913]

bkelly, I ask you again:

Hrmf. You said you would answer me.

I think you're avoiding answering questions that don't jive with what you want to be true.

I think you know that, the moment you tried to answer my question, you'd realize your experiment will fall apart (i.e. by not falling apart).

Accelerating the entire experiment to relativistic speeds and then taking the measurements again will have absolutely no effect on the results. If "being stationary" and "moving at .5c" results in no change to the experiment, one need go no further than that to conclude that there cannot be an absolute velocity.

QED.

 

[bkelly]

Remember, I referred to two events: the first event was the light striking next to the left side of the meter-stick, the second event was the light striking next to the right side. If the first event had position coordinate x₀ in the cartesian coordinate system where the wall is at rest, and the second event had position coordinate x₁ in this same coordinate system, then dx in this coordinate system is x₁ - x₀, that's why I said "the difference in x-coordinate dx" (i.e. dx is the difference between the x-coordinates of the two events, you took this out of context when you just quoted the second part "x coordinate dx"). Likewise, if the first event had time coordinate t₀ in the rest frame of the wall, and the second event had time coordinate t₁, then dt=t₁ - t₀ (so if the events occurred at the same moment in this frame, dt=0 in this frame)

Lets simplify a little bit and go one step at a time. Change the scenario and have the meterstick hovering next to the fence, not moving relative to the fence. The light flashes. The distance between the light and each end of the meterstick is identical. Will the photons at each end of the meterstick reach the fence at the same time?

I say yes.

The photons get to each end of the meterstick at the same time and go by it to hit the fence, or get stopped by it. The stationary meterstick cannot change the distance the photons travel and cannot change their velocity. The photons don't care if the meter stick is moving. It still cannot change the photon's travel distance or velocity.

As I understand you, you are saying that is not the case. Where in the above scenario do we have a disagreement?

 

[bkelly]

Hrmf. You said you would answer me.

I think you're avoiding answering questions that don't jive with what you want to be true.

I think you know that, the moment you tried to answer my question, you'd realize your experiment will fall apart (i.e. by not falling apart).

Accelerating the entire experiment to relativistic speeds and then taking the measurements again will have absolutely no effect on the results. If "being stationary" and "moving at .5c" results in no change to the experiment, one need go no further than that to conclude that there cannot be an absolute velocity.

QED.

My apologies, but to create an analogy: I am on the bottom rung of this ladder and you are several steps up. I need to simplify things and understand this bottom rung before I move up. I'll let my recent two posts ride a bit before I continue.
Thanks for your patience.

 

[Janus]

That may well be.
I don't think I am with you as I don't see anything to do with relativity. Two lights separated by some distance flash. At some point between the two lights the the light waves cross. Anyone that is in the location will see both lights at the same time. It does not matter if they were standing there when the lights flashed, or if they were standing somewhere else and moved there in time to be there when the light waves intersected each other.

However, I presume that you will follow this up and will wait for that post.

Thanks for taking the time to post.

The fact that they see the flashes at the same moment in only a part of the example. (In fact, the example is deliberately set up so they do. The issue revolves around when the events that caused the flash they saw occurred.

The first observer was always halfway between the points where these events occurred. Since the light, traveling at a constant speed, had to take an equal amount of time to reach him from each event, he correctly concludes that these events occurred simultaneously.

The second observer is different. He sees the flashes at the same time, and he knows that he is halfway between the two points when he sees the flashes, however, he also knows that the point where the flashes originated are moving with respect to himself. To him, the light originates at the source and expands as a sphere from a point that maintains a constant distance from himself while the source moves on. (Second postulate of relativity: The speed of light is a constant relative to any inertial frame of reference.) Since the event that caused the flash had to happen sometime before he reached the midpoint between the sources, he had to be closer to one source than the other when they occurred. Thus, if the events that caused the flashes happened simultaneously in his frame of reference, he would have to see one flash before the other. But we've already established that he saw them at the same time. The only way this can happen and still hold to the constant speed of light postulate, is for the events the created the flashes to occur at different times from each other as shown in the second animation.

If you can't visualize the above example, here's another:

You have two clock that you want to synchronize. You decide to carefully measure out the distance between them and set off a flash of light from a spot exactly halfway between the two. The clocks are initially set to zero and are designed to start the instant the light hits them:

[URL]http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/synch1.gif[/URL]

Notice how the expanding light(shown by the circle) moves out from the point between the two clocks, reaching them at the same time and starting them simultaneously. From that point on they run at the same rate and show the same time.

Now, however, we add a second observer in a reference frame that is moving with respect to the two clocks.

He also sees the flash start at a point exactly halfway between the two clocks and expand out as a sphere. However, the clocks do not maintain their positions with respect to him or the expanding light. The clocks are moving to the right (in our example). This means that one clock is rushing towards the part of the light heading in its direction, and the other clock is running away from the light headed in its direction:

[URL]http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/synch2.gif[/URL]

As a result, one clock is struck by light before the other and starts ticking before the other. Once both clocks are running, they run at the same rate, but are always offset from each other.

Remember, this are the exact same clocks and the exact same light, just according to two different frames with a relative motion with respect to each other.

And this is the simultaneity issue that we have been talking about. Events that are simultaneous in one frame will not be simultaneous in all frames.

 

[Bussani]

[PLAIN]http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/synch2.gif[/QUOTE]


I have to ask a side question here: with a setup like this, couldn't you conclude that you're moving at, say, 0.9c, even without something else to immediately compare yourself to?