Do we have time because we are moving?

[PeterDonis]

See: http://www.mth.uct.ac.za/omei/gr/chap8/node8.html

This just shows that calling the effect described "length contraction" is a common misstatement. To see why it's a misstatement, compare the situation in a gravitational field to the situation with relative motion.

(A) Relative motion: you and I are moving past each other at some velocity $v$. Each of us carries a meter stick which, to us, is exactly 1 meter long. As we pass each other, you will measure my meter stick to be shorter than 1 meter, and I will measure your meter stick to be shorter than 1 meter. The key points are (1) we each can make direct measurements, because we are spatially co-located as we pass each other; and (2) the length contraction is symmetric: each of us measures the other's meter stick to be shorter.

(B) Gravitational field: you are out in empty space, far away from all gravitating bodies, and I am down in the gravity well of some large mass. Each of us carries a meter stick which, to us, is exactly 1 meter long.

(1) The first problem is that, unlike in the above scenario, we are not spatially co-located, so there is no direct way for either of us to measure the length of the other's meter stick. The equation on the page you linked to implicitly assumes that there is. But if you try to specify the actual measurement procedure, you will run into difficulties. (A question to consider: in the equation on the page you linked to, which of $ds$, $dr$ represents the length of the meter stick as I measure it? As you measure it?)

For example, suppose you take a second meter stick, and put it alongside your original one, and verify that they are both exactly the same length. Then you lower the second meter stick to me, very slowly and carefully so that its length remains constant. When it reaches me, I put it alongside my meter stick, and guess what? I find that it is exactly the same length as my meter stick.

Now of course you could claim that the second meter stick got "shrunk" as it was lowered into the gravitational field; but this kind of "shrinking", since it affects everything, including any possible measuring instrument, is not directly observable; it's an interpretation that you can put on the equations, but only an interpretation, and not one that's necessary for doing any physics.

(2) The second problem is that, unlike in the scenario with relative motion, the "length contraction" due to the gravitational field is not symmetric. My meter stick "appears" shorter to you, but your meter stick "appears" longer to me. So again, whatever is going on here, it's clearly not the same kind of thing as length contraction due to relative motion, which, as above, is symmetric.

Finally, you might ask: wouldn't the same arguments apply to time dilation in a gravitational field? The answer is no, and it's instructive to consider why. Suppose you and I are situated as above, and we are at rest relative to each other and to the gravitating mass; we can verify this, for example, by exchanging light signals and finding that the round-trip travel time of those signals remains constant. But we will also find that that round-trip travel time is shorter for me than for you: i.e., a given light signal takes longer, by your clock, to make a round trip between us than it takes by my clock. This is a direct measurement that each of us can make, and it gives a direct way of showing how my clock "runs slow" relative to yours. (Once again, this is not symmetric--your clock runs fast relative to mine--so it's not the same kind of thing as time dilation due to relative motion.)

 

[Maxila]

Peter, you’ve made way too much of discussion out of the simple fact I pointed out, it is an axiom of the highest order; as time is variable and relative, length (space) is too, regardless of whether they are changed due to relative motion or gravity, they are both changed in unison and proportionally.

Putting that meter stick aside, for gravitational length contraction one could verify the length difference via the time it would take for a reflected light beam to make the round trip between observer's. The observer in the gravitational field would measure the length between them as being shorter.

 

[phinds]

Hello. I have a question I'd like to ask if I may, so basically if you and I have watches and you were to move relative to me with a velocity of 0.9c, I'd see your watch run slower than it would run when you don't move relative to me, and you'd still see your own watch run normal? Did I get it right?

yes

If yes, how would I see it run slow? Doesn't something like a battery of the watch have to be low so that I can see your watch run slower?

The watch is running at one speed in the frame of reference of the watch and at another speed in another frame of reference. What could a battery have to do with that? You are confusing local observations with remote observations. They don't agree, and that's just a fundamental fall-out of the fact that light moves at the same speed in all frames of reference.

 

[ghwellsjr]

Hi ghwellsjr,
sorry for my long absence!
I'm really struggling with making the time for this, and having still enough plasticity of mind to at least think I might understand ;)

What I think to have found out...most of my general problems (besides mathematics, obviously) with this seem to have originated in my failure to understand why I would "see" light always as moving with c...
Of course, if I'm moving faster I percieve time as I do when standing still...so anything theoretically moving slower compared to me than to the still observer (from his perspective) would seem faster to me (from my "slowed down" perspective)...don't ask me how I could overlook that, I have NO IDEA :D

It's really easy to overlook things that are going on with relativity. The only safe thing to do is precisely specify whatever scenario you are interested in according to an Inertial Reference Frame (IRF) and then use the Lorentz Transformation process to see what it looks like in another IRF. So let's do that with what I think you are describing here although since you weren't very precise, I may not be addressing your issue.

I'm assuming that you have started with the IRF of the "still observer" and you are traveling at some high rate of speed (0.8c) while some other thing is traveling slower than you (0.5c), correct? I'll show the still observer in blue, you are in green and that other thing is in red:

Now you have concluded that since your time is dilated in this frame, you will perceive that the time for that other object will be faster than your time according to your rest frame, correct? Well let's transform to your rest frame and see what happens:

As you can see, the time for that other thing as shown by the spacing of the red dots is dilated even in your rest frame. Does this make sense to you? Or did I misunderstand your scenario?

Thank you again for your patience :)

PS: I'm sorry I didn't make it back earlier to thank you properly for investing your time :(

You're welcome.

 

[ghwellsjr]

(A) Relative motion: you and I are moving past each other at some velocity $v$. Each of us carries a meter stick which, to us, is exactly 1 meter long. As we pass each other, you will measure my meter stick to be shorter than 1 meter, and I will measure your meter stick to be shorter than 1 meter. The key points are (1) we each can make direct measurements, because we are spatially co-located as we pass each other; and (2) the length contraction is symmetric: each of us measures the other's meter stick to be shorter.

Don't you think saying that we can make direct measurements of the length of a moving object can be misleading? No matter how we do it, we have to make some assumption, usually applying Einstein's second postulate (or equivalent), make some measurements and do some calculations to establish length. It's not like we can just lay a meter stick down next to the moving meter stick and directly see that the moving one is shorter like we could if we were measuring the length of a stationary yardstick with our meter stick.

 

[harrylin]

B [..] .. while I was aware of the difficulty to "see" light in a vacuum I didn't think it would matter for theorization. [..]

Actually, in practice you can see light in a near vacuum by means of scattering; cloud chambers are even how particle physics started!
But that's besides the point. What matters is that, while you can trace the sequence of events, you cannot determine the times in an "absolute" way; it all depends on your clock synchronization. In other words, you typically set your clocks in such a way that they agree with the mirror procedure.

 

[Taragond]

As you can see, the time for that other thing as shown by the spacing of the red dots is dilated even in your rest frame. Does this make sense to you? Or did I misunderstand your scenario?

Actually I think it does, although I'm not sure to what scenario you were referring. In your quotation of my previous post I just wanted to say, that while a standing observer would see the light emanating from my headlights being only .5c faster than me (if I'm at .5c) because of my timedilution these .5c would still seem to be c faster than myself. At least that what I think I had understood...

Anyway, your tables made it clear to me, that from the perspective of anyone, without seeing the timeframe of the others and being in empty space without any reference-points than the tree Observers. To both the blue and green Observers it could seem they themselves are standing still while the others are moving? :D only the red one would know that one is slower while the other is faster than himself? (EDIT: I believe I made an error, as the red one could also believe he's standing still while the others are at the same speed moving in different directions, correct?)

Because in my last scenario I deliberately chose to fly in circles around observer A so the distances between them would always be the same. Unfortunately I didn't do the math how fast B and C would have to go to always have a straight line from A to C with B inbetween, so I chose .4 and .8 c respectively, just as an example.

At last, the scenarios measuring a meter-stick got me thinking. What if the meter-stick would instead be a sqare of 1m²? Wouldn't that be a possibility to measure it? Or am I wrong in assuming the reduction in length would only occur parallel to the movement?

Actually, in practice you can see light in a near vacuum by means of scattering; cloud chambers are even how particle physics started!
But that's besides the point. What matters is that, while you can trace the sequence of events, you cannot determine the times in an "absolute" way; it all depends on your clock synchronization. In other words, you typically set your clocks in such a way that they agree with the mirror procedure.

That's what happens in our atmosphere with laserpointers (especially strong with the green ones or ist that only because our eyes are more succeptible to green light?) right?
The time-determination in absolutes wasn't really what I was going at, only compared to each other. Wouldn't it be possible to observe the difference via a video-signal (alright, probably overkill, as it might be easier to just send the timestamp every (own) second.

For the sake of an argument, I will reduce my last scenario and add to it:

Consider only two observers.
A at rest.
B at .9c on a circular orbit around A sending a typical video-signal at 24 fps of his clock (or whatever really)

Would A only receive said signal with less than 24 fps? I guess so...

 

[PeterDonis]

Don't you think saying that we can make direct measurements of the length of a moving object can be misleading?

Not really. True, we can't lay the two meter sticks side by side; but we could, for example, fire a strobe light and take a very quick picture of the moving stick passing by the stationary one, and then compare their lengths in the picture. Or we could have the moving stick trigger two sets of stationary stop clocks as it passed (one set for the front end of the stick, the other for the rear), and then use the stationary meter stick to measure the distance between two stop clocks that show the same time.

One could always say that these measurements are not "direct" because they involve some kind of calculation or assumption; but you could say the same thing about laying the two meter sticks side by side if both were stationary--you have to assume that the light propagating from the sticks to your eyes behaves in a certain way, so that it's giving you accurate information about how the sticks are positioned spatially. In principle, no measurement is ever "direct" in a pure sense; there are always intervening variables and assumptions.

However, if we compare laying two stationary meter sticks side by side to compare their lengths, with comparing the lengths of two stationary meter sticks that are very far apart, so we can't lay them side by side, it seems obvious that the first measurement is "direct" in a way that the second is not. No matter how we do the second measurement, it is going to involve a measurement of one meter stick that has to be done over a large spatial distance. (I'm assuming we aren't "cheating" by, for example, having someone measure the second meter stick's length locally and then transmit the result by radio--the measurement of the second meter stick's length has to be a real remote measurement, for example by measuring the angle it subtends in the visual field of an observer co-located with the first meter stick.) In other words, the second measurement will necessarily involve extra intervening variables and assumptions, compared to the first, and these extra intervening variables and assumptions will involve, not just the meter sticks or the way the measurements are done, but properties of spacetime itself. (For example, if we measure the length of a spatially distant meter stick by measuring the angle it subtends in our visual field, we have to make an assumption about the geometry of spacetime in between, in order to correctly model the relationship between visual angle and length.)

Perhaps the word "direct" isn't the best word to refer to the difference I just described; but that's the difference I was talking about.

 

[ghwellsjr]

Not really.

I don't disagree with what you say (I've learned not to) but I still think what you added can be misleading and I'll explain why.

True, we can't lay the two meter sticks side by side; but we could, for example, fire a strobe light and take a very quick picture of the moving stick passing by the stationary one, and then compare their lengths in the picture.

Yes, it is possible to do this but only if the strobe fires at a very specific time that isn't at all obvious.

Also, you state that the meter sticks are carried by the observers and you imply that they are taking pictures with their cameras with illumination provided only by a strobe light. This, of course, is impossible. The meter sticks must be passing at some distance from the observers (and their cameras) which means that there will be light delay times that must be taken into consideration. I assume that this is what you had in mind. You also didn't state where the strobe light was in relation to the cameras or if there were two separate strobe lights.

But if we focus on just one observer who has a strobe light and camera colocated with him and his meter stick which is held out in front of him some distance away so that the camera can capture an image of both ends of the meter stick within its field of view, plus a little more to capture an image of the moving meter stick if necessary. Of course we assume that both ends of the meter stick are equidistant from the strobe light and camera and that the camera is aimed at the center of the meter stick and that the film in the camera is flat so that if we take a picture of his stationary meter stick the markings on it will be equally spaced. This is in contradiction to the notion that we are measuring the angles at which the light is entering the camera. We also assume that the moving meter stick is essentially lined up with the stationary meter stick and will pass by it either slightly above or below it.

Now we are ready to take a picture. How do we do that? First of all, we open the shutter of the camera and fire the strobe and after sufficient time close the shutter. We assume that no light from the strobe can directly enter the camera and only reflected light does. But we can also be sure that no matter when we fire the strobe, the image of the stationary meter stick will appear the same. We don't have to make any assumptions about the speed of the light or whether it is the same in both directions, agreed? We do have to assume that it goes in a straight line after it gets reflected from a target on its way back to the camera and inside the camera.

But it's a different story for the moving meter stick. In order to get the correct image that shows the Length Contraction of the moving meter stick as measured by the stationary meter stick, we must fire the strobe at just one instant in relation to the moving meter stick such that the images of the ends of the moving meter stick are equidistant from the images of the ends of the stationary meter stick. If the strobe is fired earlier than that, the image of the moving meter stick will be too short and if it is fired later, the image of the moving meter stick will be too long, agreed? That's because when it is approaching, the leading edge of the meter stick will be taken at an earlier time than the trailing edge which will have moved closer to the camera. When receding, the trailing edge will be taken at an earlier time than the leading edge which will have moved farther away from the camera. For the same reason, even for the image that shows the correct overall Length Contraction, the image of the center of the moving meter stick will not coincide with the image of the center of the stationary meter stick. It will appear closer to the trailing edge of the moving meter stick. The markings on the image of the moving meter stick will not be equally spaced as they are on the stationary meter stick. So this image will not correctly show the Length Contraction of all parts of the moving meter stick.

Of course, it is easy to tell if we have taken a correct image but it is not easy to take that correct image. And we can just keep firing the strobe hoping to get a correct image, we have to repeat the experiment over and over again until we get lucky. Or else we have to get smart and wire up a trigger and maybe a delay timer to detect when the meter stick is approaching to fire the strobe at just the right time.

But even after we have done all that and gotten our image that shows the moving meter stick symmetrically centered on the stationary meter stick, we must make it clear that we have assumed that the light takes the same amount of time to travel away from the strobe as it does to get back to the camera. This is the assumption that I said we must make in post #35, that of applying Einstein's second postulate.

Or we could have the moving stick trigger two sets of stationary stop clocks as it passed (one set for the front end of the stick, the other for the rear), and then use the stationary meter stick to measure the distance between two stop clocks that show the same time.

Yes, but let's make it clear that we have also applied Einstein's second postulate to synchronize our stop clocks before hand.

One could always say that these measurements are not "direct" because they involve some kind of calculation or assumption; but you could say the same thing about laying the two meter sticks side by side if both were stationary--you have to assume that the light propagating from the sticks to your eyes behaves in a certain way, so that it's giving you accurate information about how the sticks are positioned spatially. In principle, no measurement is ever "direct" in a pure sense; there are always intervening variables and assumptions.

Yes, but one of those assumptions is not Einstein's second postulate, correct? We do have to assume that the light travels in a straight [line] from the ends of the objects to our eyes but we don't care even if the speed is constant, correct? So there is a fundamental difference between making measurements with objects at rest with the observer and objects in motion, correct?

EDIT: inserted "line" above.

However, if we compare laying two stationary meter sticks side by side to compare their lengths, with comparing the lengths of two stationary meter sticks that are very far apart, so we can't lay them side by side, it seems obvious that the first measurement is "direct" in a way that the second is not. No matter how we do the second measurement, it is going to involve a measurement of one meter stick that has to be done over a large spatial distance. (I'm assuming we aren't "cheating" by, for example, having someone measure the second meter stick's length locally and then transmit the result by radio--the measurement of the second meter stick's length has to be a real remote measurement, for example by measuring the angle it subtends in the visual field of an observer co-located with the first meter stick.) In other words, the second measurement will necessarily involve extra intervening variables and assumptions, compared to the first, and these extra intervening variables and assumptions will involve, not just the meter sticks or the way the measurements are done, but properties of spacetime itself. (For example, if we measure the length of a spatially distant meter stick by measuring the angle it subtends in our visual field, we have to make an assumption about the geometry of spacetime in between, in order to correctly model the relationship between visual angle and length.)

Would you consider it cheating if we applied the standard of the length of a meter using a cesium clock to count out a specific number of cycles to time how long it takes for light to travel one meter? If not, how do propose to calibrate these two meter sticks that are so far apart that we can't compare them by transporting a standard meter stick between them (or something equivalent)?

Perhaps the word "direct" isn't the best word to refer to the difference I just described; but that's the difference I was talking about.

To me, the significant difference is whether or not we have to apply Einstein's second postulate.

My favorite way to measure the length of a moving meter stick that is going to pass locally by me is to measure its speed with radar Doppler and then time how long it takes to pass by me but even this assumes Einstein's second postulate. I'm not denying that there are ways to measure Length Contraction, only that it requires an application of Einstein's second postulate along with measurements and calculations (or equivalent).

 

[PeterDonis]

The meter sticks must be passing at some distance from the observers (and their cameras) which means that there will be light delay times that must be taken into consideration. I assume that this is what you had in mind.

Yes, but that's true whether the meter stick is moving or not. There is always some amount of approximation involved. But, in a general curved spacetime where the size of a local inertial frame is limited, there is a big difference between measurements that can be made entirely within a single local inertial frame (which, bear in mind, means local in space and time) and measurements which can't. In the latter type of measurement, it's not just a matter of correcting for light delay time; it's a matter of the correction not being unique (because there is no unique way to transport vectors from one event to another in a general curved spacetime). Within a local inertial frame, the correction is unique.

In order to get the correct image that shows the Length Contraction of the moving meter stick as measured by the stationary meter stick, we must fire the strobe at just one instant in relation to the moving meter stick such that the images of the ends of the moving meter stick are equidistant from the images of the ends of the stationary meter stick. If the strobe is fired earlier than that, the image of the moving meter stick will be too short and if it is fired later, the image of the moving meter stick will be too long, agreed?

I agree that you have to fire the strobe at a particular instant, such that the light from the strobe hits each end of the moving meter stick at events that are simultaneous in the frame of the stationary meter stick. (This may be what you were saying, but you worded it differently so I'm not sure.) However, I don't see any real difficulty with setting this up. For example, you could measure the velocity of the moving meter stick (using Doppler), and use that, plus the exact location of the strobe along the stationary meter stick, to determine when to fire the strobe, relative to the instant that the front end of the moving meter stick passes the end of the stationary meter stick, in order to have the strobe light signals hit the meter sticks at the right instant.

let's make it clear that we have also applied Einstein's second postulate to synchronize our stop clocks before hand.

Yes, agreed.

We do have to assume that the light travels in a straight from the ends of the objects to our eyes but we don't care even if the speed is constant, correct?

Yes, we do care that its speed is constant; otherwise we don't know the relationship in time between the images we get from different parts of the objects. For example, in the meter stick example, if the speed of light isn't constant, we don't know how to determine the right instant to fire the strobe.

Would you consider it cheating if we applied the standard of the length of a meter using a cesium clock to count out a specific number of cycles to time how long it takes for light to travel one meter?

No. I am assuming that we have a reliable local method of determining our unit of length that gives the same answer in any local inertial frame. The cesium clock method counts as such a reliable local method.

 

[ghwellsjr]

We do have to assume that the light travels in a straight [line] from the ends of the objects to our eyes but we don't care even if the speed is constant, correct?

Yes, we do care that its speed is constant; otherwise we don't know the relationship in time between the images we get from different parts of the objects. For example, in the meter stick example, if the speed of light isn't constant, we don't know how to determine the right instant to fire the strobe.

I was commenting on your comment with regard to two stationary meter sticks:

One could always say that these measurements are not "direct" because they involve some kind of calculation or assumption; but you could say the same thing about laying the two meter sticks side by side if both were stationary--you have to assume that the light propagating from the sticks to your eyes behaves in a certain way, so that it's giving you accurate information about how the sticks are positioned spatially. In principle, no measurement is ever "direct" in a pure sense; there are always intervening variables and assumptions.

I don't understand why you think it matters when we fire the strobe or why we have to assume that the speed of light is constant in this specific instance.

Please explain.

 

[PeterDonis]

Please explain.

I didn't realize you were talking about the case of stationary meter sticks. Yes, in that case it doesn't matter when you fire the strobe. However, I still think it matters that the speed of light is constant. Strictly speaking, I guess you could say that the speed of light could vary with time without affecting the result; but it can't vary in space, because the paths of the light beams coming from the two ends of the meter sticks are traversing different positions in space. However, variation in time without variation in space in a particular inertial frame would either become partly variation in space in any other frame, or would violate Lorentz invariance; so really I think the assumption of a constant speed of light is necessary for the stationary case as well.

 

[ghwellsjr]

I didn't realize you were talking about the case of stationary meter sticks. Yes, in that case it doesn't matter when you fire the strobe. However, I still think it matters that the speed of light is constant. Strictly speaking, I guess you could say that the speed of light could vary with time without affecting the result; but it can't vary in space, because the paths of the light beams coming from the two ends of the meter sticks are traversing different positions in space. However, variation in time without variation in space in a particular inertial frame would either become partly variation in space in any other frame, or would violate Lorentz invariance; so really I think the assumption of a constant speed of light is necessary for the stationary case as well.

Well, I was speaking strictly and I did say that the light had to travel in a straight line. (I unfortunately left out the word "line" which I have since inserted.)

But the main point I want to make is that strictly for the stationary case, that is, the observer taking a picture of two side-by-side meter sticks that are not moving with respect to him and just a few meters in front of him, or just looking with his eyes, using a strobe light or constant lighting, it is not necessary to invoke Einstein's second postulate to make what you wanted to distinguish as a "direct" length comparison unlike the case of one moving and one stationary meter stick where it is necessary to invoke Einstein's second postulate. I realize that there are other reasons to assume the speed of light is constant and the same in all directions but not strictly for this specific case. Agreed?

 

[Taragond]

Wow...that was a lot...I am not sure I got it all, but I think a question I posted last time (unfortunately hidden between some rambling that may have bored you. If that is the case I apologize. I know how frustrating it can be if you thought you were done with one thing :D) was aimed at a different approach on the same problem, depending your answer on this:

At last, the scenarios measuring a meter-stick got me thinking. What if the meter-stick would instead be a sqare of 1m²? Wouldn't that be a possibility to measure it? Or am I wrong in assuming the reduction in length would only occur parallel to the movement?

Because if so, it should be possible by ataching a perfectly square lightsource and a wide-field-camera to each observers side to automatically measure the "length" by comparing the height of said light-source with it's width corrected for the perspective calculated by it's y-coordinate in the picture. Of course given the paths of the vessels are parallel, or the trajectory of each is known bei both (the measuring one(s)). But I guess if the first part is plausible, there wouldn't be very many positioning lights necessary to automatically calculate that, too...

 

[ghwellsjr]

Wow...that was a lot...I am not sure I got it all, but I think a question I posted last time (unfortunately hidden between some rambling that may have bored you. If that is the case I apologize. I know how frustrating it can be if you thought you were done with one thing :D) was aimed at a different approach on the same problem, depending your answer on this:

At last, the scenarios measuring a meter-stick got me thinking. What if the meter-stick would instead be a sqare of 1m²? Wouldn't that be a possibility to measure it? Or am I wrong in assuming the reduction in length would only occur parallel to the movement?

Because if so, it should be possible by ataching a perfectly square lightsource and a wide-field-camera to each observers side to automatically measure the "length" by comparing the height of said light-source with it's width corrected for the perspective calculated by it's y-coordinate in the picture. Of course given the paths of the vessels are parallel, or the trajectory of each is known bei both (the measuring one(s)). But I guess if the first part is plausible, there wouldn't be very many positioning lights necessary to automatically calculate that, too...

Your question didn't bore me, it just takes time to prepare answers and I spent a lot of time on PeterDonis's issue that a strobe light could be used to measure Length Contraction.

If you understand what Length Contraction is, you will quit asking questions about how to directly observe it. Length Contraction is the ratio of the length of an object as defined according to an Inertial Reference Frame in which the object is moving inertially compared to the length of the object as defined according to the IRF in which the object is at rest. Key to understanding this is that the length of the object according to any IRF is defined as the difference in the spatial coordinates of the object's endpoints (or any other points on the object) that are determined at the same Coordinate Time. Since coordinates are not observable (they can change when we use different IRF's), Length Contraction also cannot be observable. Or to put it another way, different IRF's do not change anything that is observable. What makes the difference between different IRF's is how the propagation of light is defined in them. Remember that the propagation of light is also not observable. Einstein's second postulate is what defines the propagation of light in any IRF. So it takes the application of Einstein's second postulate to a set of observed measurements to establish Length Contraction.

This application will require a measurement involving a roundtrip propagation of light (or radar) signals, from the observer to the object and back to the observer. That is why PeterDonis used a strobe light. In your suggested scenario, you only have a one-way propagation of light from the illuminated square to the wide-field camera so that cannot fulfill the requirement that the dimensions of the object be taken at the same Coordinate Time. You should also realize that nothing is actually happening to an object when we use different IRF's. It's just our description of it that changes.

Does this help?

 

[PeterDonis]

I realize that there are other reasons to assume the speed of light is constant and the same in all directions but not strictly for this specific case. Agreed?

No. The light reaching your eyes at the same time from the two ends of the meter sticks still has to have left the two ends of the meter sticks at the same time; otherwise you are not seeing a "snapshot" of the two sticks at a single instant of time, so you can't interpret what you see as a "length" at all. But Einstein's second postulate is required to ensure that the light reaching your eyes at the same time, if the distances are the same, left the two ends of the meter sticks at the same time.

(To put it another way, if Einstein's second postulate did not hold, you could have an effect similar to Penrose-Terrell rotation even for objects that were stationary relative to you, so even a stationary length comparison--or even a stationary length measurement of a single meter stick--would not be "direct" by your definition, because you would have to correct for different light travel times from different parts of the object.)

 

[ghwellsjr]

No. The light reaching your eyes at the same time from the two ends of the meter sticks still has to have left the two ends of the meter sticks at the same time; otherwise you are not seeing a "snapshot" of the two sticks at a single instant of time, so you can't interpret what you see as a "length" at all. But Einstein's second postulate is required to ensure that the light reaching your eyes at the same time, if the distances are the same, left the two ends of the meter sticks at the same time.

(To put it another way, if Einstein's second postulate did not hold, you could have an effect similar to Penrose-Terrell rotation even for objects that were stationary relative to you, so even a stationary length comparison--or even a stationary length measurement of a single meter stick--would not be "direct" by your definition, because you would have to correct for different light travel times from different parts of the object.)

How does that comport with Einstein's comments from chapter 2 of his 1920 book on relativity:

"ON the basis of the physical interpretation of distance which has been indicated, we are also in a position to establish the distance between two points on a rigid body by means of measurements. For this purpose we require a “distance” (rod S) which is to be used once and for all, and which we employ as a standard measure. If, now, A and B are two points on a rigid body, we can construct the line joining them according to the rules of geometry; then, starting from A, we can mark off the distance S time after time until we reach B. The number of these operations required is the numerical measure of the distance AB. This is the basis of all measurement of length."

 

[PeterDonis]

How does that comport with Einstein's comments

He describes the distance measurement as "marking off", which is not quite the same as what we've been discussing. To compare the lengths of two meter sticks by his method, you would place them together so that they were flush at one end, and verify that by observation--but that observation only involves one end of each stick, so the issue I described doesn't arise. Then you would hold both sticks in place by some method and move to the other end of the "standard" meter stick and see where it fell on the stick you wanted to measure, and make a mark on the stick being measured accordingly. In the case we've been discussing, you could leave out the mark, because the other end of the stick being measured would also be flush with the other end of the standard stick; but if, for example, the stick being measured were two meters long, you would make a mark on it where the end of the standard stick fell. Then you would move the standard stick until its other end was at the mark you just made, verify that by observation, and then move to the other end of the standard stick again, holding both objects in place, to see where it fell on the stick being measured. But in any case, you would be observing only one end at a time, so the issue I described would not arise.

The issues that would arise are, first, how you guarantee that both sticks are in fact held exactly in place while you move from one end to the other of the standard stick, and second, how you guarantee that the standard stick's length is unchanged after you have moved it.

 

[Taragond]

Your question didn't bore me, it just takes time to prepare answers

That I understand perfectly well! I just wanted to make sure because we're still on the internet, even though I already was pretty sure, it's "rules" don't apply here ;)
Also because of my little expertise on discussing physics I am aware that my questions and statements could be stupid from your point of knowledge and how hard it is to point that out and still have me learn something to correct that...

Length Contraction also cannot be observable. Or to put it another way, different IRF's do not change anything that is observable.

Does this help?

I'm not sure, because I always understood it that way, a rocket would look shorter to us on it's passing, than it actually is. Actually I'm pretty sure that exactly was explained to me in school but maybe there is something missing now. That's why I had the presumption it was observable.
But If I know understand you correctly:

lets say wie have a stationary camera and the suare is dark but flashes over it's entire front just one photon-burst at precisely the same time towards the camera when passing by.
Assuming it is strong enough so the camera can detect them, it would still see a square in the picture?

 

[ghwellsjr]

Length Contraction also cannot be observable. Or to put it another way, different IRF's do not change anything that is observable.

Does this help?

I'm not sure, because I always understood it that way, a rocket would look shorter to us on it's passing, than it actually is. Actually I'm pretty sure that exactly was explained to me in school but maybe there is something missing now. That's why I had the presumption it was observable.
But If I know understand you correctly:

lets say wie have a stationary camera and the suare is dark but flashes over it's entire front just one photon-burst at precisely the same time towards the camera when passing by.
Assuming it is strong enough so the camera can detect them, it would still see a square in the picture?

No, it would not be a square but I think you are still missing my point. When I said that different IRF's do not change anything that is observable, I didn't mean that something that looks like a square to an observer (or a camera) at rest with respect to it would also look like a square to an observer (or a camera) moving with respect to it. What I meant was that however a moving object appears to an observer (or a camera) as analyzed according to one IRF, will continue to show that the object appears the same to that observer (or camera) according to any other IRF.

So let's analyze your scenario to see what a moving square that simultaneously gives off a flash of light looks like to a stationary camera. I'm assuming you meant "at precisely the same time" in the rest frame of the square. Let's use a pinhole camera with a flat film. To make things especially easy for us to analyze, let's make the distance between the film and the pinhole be the same as the distance between the pinhole and the moving square at its closest approach. That means that whatever the positions of the photons according to the rest frame of the camera but in the plane of the moving square when they are emitted will be similar to the positions of the photons in the plane of the stationary film when they are detected. (I hope that makes sense to you.) So all we have to do is determine what the shape of the illumination is in the camera's rest frame without regard to when those photons were emitted. To do that, we start with the rest frame of the square and transform that to the rest frame of the camera. I'm going to assume a square of 1 meter on each side (10 decimeters) and look at a slice along the direction of motion at 0.6c. Here is a spacetime diagram showing the square in its rest frame. Assume the flash occurs at the Coordinate Time of zero, which is the bottom dots on both worldlines:

Now we transform to the rest frame of the camera where the square is moving at 0.6c:

If we look at the Coordinate Time of 10, we see that the length is contracted to 8 decimeters. Since gamma at 0.6c is 1.25, this is in agreement with 10 decimeters divided by 1.25. However, what is important for our scenario is the relative positions of the flashes at the two ends of the square which you will remember are the bottom two dots and these are separated by 12.5 decimeters. So the image that will be recorded on the film is not a square of 10 decimeters on a side but rather a rectangle that is 10 decimeters high and 12.5 decimeters wide. It won't matter what other IRF we transform this scenario to, they all will show an image of a stretched out square rather than a contracted square.

 

[Taragond]

I'm assuming you meant "at precisely the same time"

is that samer than same? :)

I don't seem to understand the diagrams perfectly, as I still don't get why it is streched rather than contracted, but wouldn't I be able to deduce the relative speed of the square just as easily by "measured length"/"measured height"="gamma"=1.25 and calculate 0.6c out of this gamma? (As I knew before, it is perfectly square in it's rest-frame!?)

 

[ghwellsjr]

I'm assuming you meant "at precisely the same time"

is that samer than same? :)

Do you understand that the phrase "at precisely the same time" is ambiguous unless you specify the frame in which it applies? It refers to the Coordinate Times of two events being the same. I assumed you meant in the rest frame of the square. Is that what you meant? If you don't know why it matters, then that is an important concept for you to learn if you are interested in learning Special Relativity.

For example, if you look at the first diagram above for the rest frame of the square, the bottom blue dot is at a Coordinate Time of 0 and the bottom red dot is also at a Coordinate Time of 0, so that means that those two events are simultaneous ("at precisely the same time") in the rest frame of the square. I'm assuming those two dots represent the time at which the flash occurs across the square but I'm only showing two side end points of the square. Since those two events are at the same time, we can say that the width of the square is the difference in the Coordinate Distances for those two events which is 10 decimeters or 1 meter. That is how we specify lengths in Special Relativity. Does that make sense to you?

Now when we transform the coordinates of all the dots to a frame in which the square is moving at 0.6c, we get the second diagram above. Now those two bottom dots are not "at precisely the same time" which means that we cannot use them to determine the length of the square in the frame in which it is moving. If we didn't know any better and used them anyway, we would incorrectly conclude that the length of the square is 12.5 decimeters. Instead, we should find two events that happen at the same time such as the ones that happen at the Coordinate Time of 10 which is the top blue dot and the second red dot up. Those two dots are separated in distance by 8 decimeters which is the correct width of the "square" in the frame in which it is moving at 0.6c. Do you understand now why it is important to specify which frame the expression "at precisely the same time" applies and why that is important in specifying the length of an object?

I don't seem to understand the diagrams perfectly, as I still don't get why it is streched rather than contracted,

The square is not stretched in the rest frame of the camera, it is contracted, but it produces a stretched image on the film. I'm just pointing out that Length Contraction is not something that can be directly observed visually to an observer or a camera.

but wouldn't I be able to deduce the relative speed of the square just as easily by "measured length"/"measured height"="gamma"=1.25 and calculate 0.6c out of this gamma? (As I knew before, it is perfectly square in it's rest-frame!?)

How do you know that all parts of the perfect square flashed "at precisely the same time" in the rest frame of the square? What mechanism makes that happen?

 

[Taragond]

Do you understand that the phrase "at precisely the same time" is ambiguous unless you specify the frame in which it applies? It refers to the Coordinate Times of two events being the same. I assumed you meant in the rest frame of the square. Is that what you meant? If you don't know why it matters, then that is an important concept for you to learn if you are interested in learning Special Relativity.

alright, in that context I understand the difference, I just didn't think as much about the time-frames but more about the imaginary optics.
Also, because I ordered only one photon-burst I assumed we were speaking about a single moment in time where in one timeframe one sheeth of photons is released and travels to an open camera otherwise observing continuously total darkness. in that case at some point this camera suddenly receives this flash and saves an image.
It seems to me it doesn't even matter when it receives that light or when it was sent.

ot use them to determine the length of the square in the frame in which it is moving.

I thought it was clear they knew the length, and I added the same hight to determine the length-contraction by comparison to it's height.

Do you understand now why it is important to specify which frame the expression "at precisely the same time" applies and why that is important in specifying the length of an object?

not sure, see above

The square is not stretched in the rest frame of the camera, it is contracted, but it produces a stretched image on the film. I'm just pointing out that Length Contraction is not something that can be directly observed visually to an observer or a camera.

I'm still not sure why it is stretched rather than contracted on film but it seemed to me that the factor is basically the same?

How do you know that all parts of the perfect square flashed "at precisely the same time" in the rest frame of the square? What mechanism makes that happen?

that was hypothetical :D let's say magic that we don't understand yet.

I'm just trying to get a general idea how the light would leave one timeframe and enter another and I assumed length-contraction meant exactly that it would appear shorter to other timeframe-observers.
That being said, I still would like to understand why it would appear stretched instead of actually being contracted, but I'm still a little bit excited the length-contraction would only occur on the movement-direction and guess nearby a huge mass it would apply to all dimensions?

 

[ghwellsjr]

Do you understand that the phrase "at precisely the same time" is ambiguous unless you specify the frame in which it applies? It refers to the Coordinate Times of two events being the same. I assumed you meant in the rest frame of the square. Is that what you meant? If you don't know why it matters, then that is an important concept for you to learn if you are interested in learning Special Relativity.

alright, in that context I understand the difference, I just didn't think as much about the time-frames

What do you mean by "time-frames"? If you mean the depiction of a frame using a spacetime diagram, then why do you drop the term "space"? Just curious.

but more about the imaginary optics.

What do you mean by "imaginary optics"? Just curious.

Also, because I ordered only one photon-burst I assumed we were speaking about a single moment in time where in one timeframe one sheeth of photons is released and travels to an open camera otherwise observing continuously total darkness. in that case at some point this camera suddenly receives this flash and saves an image.

We can say that all the light was emitted at the same time according to the square's rest frame but we cannot say that the light that the camera receives arrives at the same time. Since the light has different distances to travel from the different parts of the square to the camera and since it all travels at c, it must arrive at the camera at different times. This is true for all frames.

It seems to me it doesn't even matter when it receives that light or when it was sent.

If the camera was at rest with respect to the square, then it wouldn't matter when the light was sent or received, and the image on the film would be in the shape of a square but when there is relative motion between them, Relativity of Simultaneity means that it does matter.

ot use them to determine the length of the square in the frame in which it is moving.

I thought it was clear they knew the length, and I added the same hight to determine the length-contraction by comparison to it's height.

Even though you get the same ratio, that's not Length Contraction. Length Contraction has to do with the Coordinate Distance between one pair of separated points on an object as defined according to two different frames, one in which the object is moving and one in which the object is at rest. The Coordinate Distance between a pair of points on an object must have the same Coordinate Time in order to qualify as a length.

Do you understand now why it is important to specify which frame the expression "at precisely the same time" applies and why that is important in specifying the length of an object?

not sure, see above

Well if you don't specify the Coordinate Distance at the same Coordinate Time, then you can get just about any length you want.

The square is not stretched in the rest frame of the camera, it is contracted, but it produces a stretched image on the film. I'm just pointing out that Length Contraction is not something that can be directly observed visually to an observer or a camera.

I'm still not sure why it is stretched rather than contracted on film but it seemed to me that the factor is basically the same?

It is the same but that has nothing to do with Length Contraction.

How do you know that all parts of the perfect square flashed "at precisely the same time" in the rest frame of the square? What mechanism makes that happen?

that was hypothetical :D let's say magic that we don't understand yet.

It's perfectly understandable and it's not magic.

I'm just trying to get a general idea how the light would leave one timeframe and enter another and I assumed length-contraction meant exactly that it would appear shorter to other timeframe-observers.

Light doesn't leave one timeframe and enter another. Light travels at "c" in each inertial frame. I think maybe it's your concept that light is changing frames that is making this difficult for you to understand.

That being said, I still would like to understand why it would appear stretched instead of actually being contracted, but I'm still a little bit excited the length-contraction would only occur on the movement-direction and guess nearby a huge mass it would apply to all dimensions?

No, Length Contraction does not apply to all dimensions, just one.

Do the diagrams in post #50 make sense to you? Do you see how they show Length Contraction?

 

[ImperialThinker]

Okay, I got what you are trying to ask there.

First , let's set up some assumptions.
1. Everything is relative.
2. Speed of light is absolute.
3. Universe has a centre.
--
1. Everything is relative. :- Pretty self explanatory . Things are as we perceive them to be. Eg :- Me : Whoa, that turtle is so slow.. Snail : Wait for me turtle bro, y u so fast?
2. Speed of light is absolute. :- Might be argumentative, it might be relative , but for understanding something , we need to set something as reference. So we take it as absolute.
3. Universe has a centre :- I don't know why so many people say that this is bullsh*t when it is quite plausible. As the universe should had been in thermal equilibrium if it was infinite years old. (thermal equilibrium is achieved after infinite time) But as we know, some stars are hotter and some are cooler. Which means , Universe is not infinite years old. If something is 1000 lightyears away from the centre of the universe, then it is atleast 1000 years old. Conversely, If something is 1000 years old, it is Atmost 1000 lightyears away. So, if something is Finite years old , it is atmost Finite lightyears away. Which means, Universe is not infinitely large. Which means, it has a start ie a centre.

So now your question is , what would be the difference between time on Earth , and time at the centre of the universe where there is no movement and no gravitational field, the constraints that affect time. Well, its a good question but you need to define in respect to what you need to measure time.
Many scientists say there has to be a black hole at the centre of the universe. I tend to believe so too, because it makes sense. At the centre of the black hole there would be no gravitational field. So let's consider two men, one standing on Earth and other at the centre of that black hole. Now, just to be clear, both men will experience the same time, with respect to themselves .
For now, I can only imagine the man at the centre of black hole to be frozen in space time, in the sense that even after 1 billion years (earth time) , he would not experience even 1 microsecond .. So I will say time is infinitely faster for him. But i may be wrong.
As we first thought speed of light was infinite when we were to lazy to do the math. but then we found out that infinite was actually 3 x10^8 mps .. I am sure when we will figure out what "max time" as you said is .. it would be a damn large number too.
Your question is legit.

 

[ghwellsjr]

But i may be wrong.

You are wrong. And if you had read the first few posts after the opening post, you would see why.

 

[ImperialThinker]

You are wrong. And if you had read the first few posts after the opening post, you would see why.

I did read those before posting, Thank you and I still don't get your point.

 

[Dale]

Everything is relative

Not everything is relative (aka frame variant). Many things are invariant.

Universe has a centre :- I don't know why so many people say that this is bullsh*t when it is quite plausible

It is certainly plausible in concept, but when you work out the expected observational consequences you find that there is no evidence to support it. The observations we have better fit a model without a center.

 

[ellipsis]

Energy happens to be relative, since velocity is. KE = (1/2)mv^2, remember. Mass is, since things get effectively more massive when they're accelerated a great amount. But not everything is frame variant.

Position is relative, since you can move the entire universe 5 feet to the left and feel nothing. Same for orientation. One thing that seems to be invariant: shape.

By that I mean, the angles between various particles, the shape they make together. I've been reading "The End of Time" by Julian Barbour. It deals a lot with the minimum amount of information required to describe a physical system.

 

[Taragond]

What do you mean by "time-frames"? If you mean the depiction of a frame using a spacetime diagram, then why do you drop the term "space"? Just curious.

I meant I didn't see the whole procedure but only thought about that frame when the square flashes...

What do you mean by "imaginary optics"? Just curious.

I'm trying to imagine the optics...

We can say that all the light was emitted at the same time according to the square's rest frame but we cannot say that the light that the camera receives arrives at the same time. Since the light has different distances to travel from the different parts of the square to the camera and since it all travels at c, it must arrive at the camera at different times. This is true for all frames.

If the camera was at rest with respect to the square, then it wouldn't matter when the light was sent or received, and the image on the film would be in the shape of a square but when there is relative motion between them, Relativity of Simultaneity means that it does matter.

that is logical. But I'm trying to get why there is contraction and how it happens... that's why I'm trying to ask how it would look if the camera doesn't just get a snapshot but is open the whole time with the flash of the square being the only light ever sent in its direction.

Even though you get the same ratio, that's not Length Contraction. Length Contraction has to do with the Coordinate Distance between one pair of separated points on an object as defined according to two different frames, one in which the object is moving and one in which the object is at rest. The Coordinate Distance between a pair of points on an object must have the same Coordinate Time in order to qualify as a length.

but the 4 points of said square do have the same coordinate time, don't they?

Well if you don't specify the Coordinate Distance at the same Coordinate Time, then you can get just about any length you want.

It is the same but that has nothing to do with Length Contraction.

ok :(

It's perfectly understandable and it's not magic.

great, so why do you ask? just curious ;)

Light doesn't leave one timeframe and enter another. Light travels at "c" in each inertial frame. I think maybe it's your concept that light is changing frames that is making this difficult for you to understand.

yes, it doesn't make things easier for me...

No, Length Contraction does not apply to all dimensions, just one.

so, if all 4corners of the square are in the same coordinate time and length contraction only applys to one of those dimensions, shouldn't I be able to determine length contraction by comparing height to width? still sounds that way... :/

Do the diagrams in post #50 make sense to you? Do you see how they show Length Contraction?

They do, but honestly, I'm probably not one to decide how much...

PS: I do understand if you're a little annoyed, I want to make clear that there is no pressure for me whatsoever, just curious interest. Maybe it's better I come back to this when I had some more time to invest in studies...

So now your question is , what would be the difference between time on Earth , and time at the centre of the universe where there is no movement and no gravitational field, the constraints that affect time.

Thank you, that's exactly what I wanted to ask. Not necessarily at a center but the difference between space and before/outside (completely hypothetical interest).
As it was said, time came into existence with space and space came into existence moving from the start.
Because speed effects time I was thinking about if it just came into existence with movement and mass and what it would mean to be "outside".

Because we know of no outside I set it equal to no movementand no gravitation...

If moving faster means, time runs slower, wouldn't that mean if I were "out" it would run faster compared to here?

@ghwellsjr: I know, probably way too unscientific.