What is time dilation and how does it relate to Einstein's theory?

[MeJennifer]

Yes, and if I look at which events coincide with which markings and which clock-readings on a ruler that is in motion with respect to me, that is a local measurement, because the marking and the clock were right next to the event at the moment it happened. You're confused, all measurements made by looking at an event and then looking at the readings on a ruler-clock system right at the same position as the event as it happened are "local" measurements, and therefore you don't have to worry about light-signal delays (even if it takes a while for the light from the event to reach me, I can still look through my telescope and see which ruler-marking and which clock was right next to the event as it was happening, and the reading on that clock as it was happening--the delay won't change what I see regardless of whether I'm looking at distant readings on a ruler and clock which is at rest relative to me, or distant readings on one that is moving relative to me). In SR, every frame defines the coordinates of events in terms of such local measurements. "Local" in one frame means "local" in every frame, there can be no disagreement between frames about whether two events coincide at the same position and time.

JesseM, when you look through a telescope do you realize that you are simply receiving photons on your retina? You are not actually looking at something away from you. You are simply observing photons that are local.

 

[HallsofIvy]

JesseM, when you look through a telescope do you realize that you are simply receiving photons on your retina? You are not actually looking at something away from you. You are simply observing photons that are local.

While agreeing with the physics here, I would phrase that in exactly the opposite way: you are looking at something far away when photons from that object hit your retina- that's what "looking at" means. You are not "observing" those photons.

 

[MeJennifer]

Furthermore we have to realize when a photon hits our retina it traveled a certain distance and thus some time, since the speed of light is finite. Also the path of this photon is typically not a straight one, since gravity bends spacetime we "see" things often through some kind of a lens and distance is no longer some idea of a straight line in a flat coordinate system.

 

[JesseM]

JesseM, when you look through a telescope do you realize that you are simply receiving photons on your retina? You are not actually looking at something away from you. You are simply observing photons that are local.

You are still misunderstanding the definition of "local"--it doesn't mean the event was measured at my location, it just means it was measured by measuring-devices next to the event itself. If you prefer, instead of inanimate ruler-markings and clocks, you can imagine another experimenter actually sitting on that ruler-marking with a stopwatch and at rest relative to me, who notes the position and time of the event right next to him and then emails the result to me. This is certainly a local measurement, and the fact that there is a delay between the measurement and when I learn of the results isn't going to make the results I see reported in the email any different than the results he measured at the time. Of course, exactly the same is true if I'm looking through a telescope at the ruler-marking and clock-reading right next to the event as it happened, I won't see anything different if I'm 1 light-second away than if I'm 30 light-seconds away, I'll just see the same result sooner or later depending on my distance.

In any case, even if you only wanted to talk about events right next to me, I could equally well note their position and time on a ruler/clock at rest relative to me (the coordinates in my rest frame) or a ruler/clock moving at a constant velocity relative to me (the coordinates in another frame), so your argument still doesn't make sense.

 

[petm1]

SR works, in 4-d space for motion in one direction. I am talking about 3-d space and how every observer always feels like the preferred frame. Twins a and b always see themselves as being in a preferred frame, we know that the one that had an increased motion is the twin that ages slower per SR and experiment, so there must have been a physical change in 3-d that he did not feel, nor see, other than an increase in "g" force.

Everything has a dually so let's set one for SR let's let the equation meters per second (M/S) represent the preferred frame. If I wanted to set an equation to look right with the factors of SR, I would need four factors SR gives me two with length contraction (LC) and time Dilation (TD) to these I would add length dilation (LD) and time constriction (TC). Now I would be able to write out the equation (LC/TC x M/S x LD/TD). Now I can show any changes to the length and the time in two directions either more motion or less motion in a 3-d reference frame.

 

[MeJennifer]

You are still misunderstanding the definition of "local"--it doesn't mean the event was measured at my location, it just means it was measured by measuring-devices next to the event itself.

I am not, and now you are simply condescending.

Nothing changes: A measuring device can only make a local measurement, and that measuring device could be anywere in the universe.

 

[petm1]

SR tells us we will "see" using light I’ll show it as …>…, this is movement between 3-d frames in the 4-d movement of light with the direction indicated by either < or >. Time dilation (TD) and length contract (LC) are the clues that tell us what is happening between our frames and I’ll write it as LC/TD. Let's show this with my equation for the preferred frame using the < to show the direction of light, [LC/TC x M/S x LD/TD]...<...LC/TD. This equation is still in need of one other factor to look more like a well balanced equation and it has to do with a twist, for want of a better term, and that is on the slower, less motion, side of the equation LD/TC. Let’s add this to the equation of my preferred reference frame. LD/TC...>...(LC/TC x M/S x LD/TD)...<...LC/TD. We still use light to see everything we see even those objects within our own frame.

To this equation I would like you to see the view. A sphere always has two sides, inside and out side if we are using light to see we are always on the out side looking in as the light sphere passes our frame, hence the one way view we get with light is showing us a twist of 3-d thru a 4-d medium. But in the 3-d frame we would feel the LC/TC like gravity and the LD/TC as acceleration, when viewing other objects we would see these changes as LD/TC slower with less motion, and LC/TD as faster with more motion.

 

[JesseM]

Nothing changes: A measuring device can only make a local measurement, and that measuring device could be anywere in the universe.

Yes, and if I look at a telescope to see the readings on a measuring device which is measuring the position and time of an event right next to the device, then that qualifies as a "local" measurement of the event according to the definition. Weren't you arguing that I somehow could not call that a "local" measurement in this comment?

In fact we cannot measure anything directly that is not local.

At the most you can calculate something from the perspective of another frame by measuring the light signals you receive (and sent if you are looking for roundtrip times) locally by taking synchronization conventions, Doppler effects and synchronization issues due to the finit speed of light or relativistic effects into account and then perform a Lorentz transformation.

 

[petm1]

Another way to look at this equation is [LC/TC...>...(LD/TC < M/S)] is how I see objects within my frame, while [(M/S < LD/TD)...<...LC/TD] is my view of objects out side my frame.

 

[MeJennifer]

Yes, and if I look at a telescope to see the readings on a measuring device which is measuring the position and time of an event right next to the device, then that qualifies as a "local" measurement of the event according to the definition. Weren't you arguing that I somehow could not call that a "local" measurement in this comment?

In this case you simply get information from a local measurement. I wrote measure anything directly.

Your claim was that we could measure the speed of light in a moving object. I said we could not.

If you measure it in the moving object and then transfer the information you actually measured it in the rest frame not the moving frame.

 

[pess5]

The speed of light is 300,000,000/1 times 1 meter/1 second if you change the meters with length contraction and change the second with time dilation at the same time, you change the speed of light by some factor.

No I think the math is correct for both length contraction and time dilation, but put them into the equation for the speed of light at the same time and you will change the speed of light. The contraction is (length)/gamma. The dilation is t (gamma) let's call gamma .5 now plug it into the speed of light. (300,000,000/1) X 1 meter/1 second X 1/.5 /1 X .5 now I come up with a number of 1,200,000,000 how about you? I don't disagree with any of SR, I just think that I am not seeing the whoe picture.

petm1,

Light speed is the same for all. x/t=c=X/T, where for v>0, x<>X and t<>T.

You mention above that gamma is 0.5. However, gamma must always be greater than 1 for v>0.

At 0.866c, gamma=2. So contractions are 1/gamma=1/2, and dilation is gamma=x2.

You measure light at c. I am moving at 0.866c wrt you. You record t=6hr between 2 events marked by clocks at rest in your frame. Since I am in motion wrt you, you see my clock ticking twice as slow, so I must record T=3hr between those same 2 events. And in fact, I do.

Light in my frame will travel across 3lh in 3hr over that spacetime segment. However in my vantage, your 6hr segment takes 12hr, since your clock ticks twice as slow as mine. This is time dilation. Although your 6hr takes 12hr in my vantage, only 3hr of that 12hr encompasses the common spacetime interval as I experience it. Our perspectives are rotated wrt one another in the spacetime. If I could see your clock, it would tick off 6hr to my 12hr. Light in my frame travels for 12lh across that 12hr segment. So c is always 1lh/hr. By the way, lh means light-hr.

pess

 

[JesseM]

In this case you simply get information from a local measurement. I wrote measure anything directly.

A local measurement isn't "direct"? Is "direct" supposed to mean I made the measurement in my own local neighborhood, so if my partner makes a local measurement in a different region and then sends me the results, that is somehow not "direct"? If that's what you mean, then you're devising completely new terminology and acting like you expect me to understand it without your defining it explicitly--certainly I've never heard any physicist make this sort of distinction between "direct" and "indirect" measurements, it's a totally idiosyncratic use of terminology.

Also, I still can't make sense of what you were talking about earlier about how measurements in other frames would require me to worry about doppler shifts and time delays and the like--if my friend makes a local measurement of an event in his immediate neighborhood, and then sends me the results by email, clearly doppler shifts and time delays aren't going to somehow change the text of the email! And plus, even to assign coordinates to events in my own frame which aren't in my own immediate region, I have to rely on local measurements made elsewhere than where I am, so the issues are exactly the same.

Your claim was that we could measure the speed of light in a moving object.

No, a moving frame. You understand that "frames" in SR are coordinate systems which fill up all of space and time, and thus it's not true that the measurements made to assign coordinates to an event in an observer's own rest frame are all going to be in his immediate area, right?

I said we could not.

Your argument still makes zero sense. If you want to measure the speed of a moving object in either your own frame or in a moving frame, in both cases there must be two measurements of the object at different points on its path, so that you have a change in position which can be divided by the time difference between these two measurements. And regardless of whether I'm making a measurement in my own frame or in a frame in motion relative to me, at least one of the measurements must be somewhere other than my own immediate neighborhood, so it would not be "direct" according to how I'm understanding your terminology, although the two measurements would both be "local" in the standard terminology of physicists. So the issues of "directness" and locality work exactly the same way whether I'm using a ruler at rest relative to me or a ruler in motion relative to me--whatever you'd say about the first case, there's no logical reason for saying anything different about the second case.

Of course, if I am just measuring instantaneous velocity, then the two measurements may be an infinitesimal distance apart so I guess they could both be said to take place in my local neighborhood. But then, I can also make two measurements an infinitesimal distance apart on a moving ruler/clock system which can also be said to take place in my local neighborhood.

If you measure it in the moving object and then transfer the information you actually measured it in the rest frame not the moving frame.

This sentence doesn't make sense to me--a measurement is not "in" an object, nor is it "in" a frame, it's just a pair of events which all observers agree took place in the same local region of spacetime, like "at the moment the explosion happened it was next to the 16-meter mark of this ruler, and the clock mounted on the 16-meter mark read 25 seconds". Of course there is only one frame where the ruler/clock system is at rest, and only in that frame will the local position and time measurements of that ruler/clock system correspond to the coordinates of the frame, but the measurement itself is just a frame-independent fact about events which coincide in the same local region of spacetime.

 

[MeJennifer]

Sorry JesseM there seems to be no point in continuing this discussion, it seems what I write does not make much sense to you.

 

[JesseM]

Sorry JesseM there seems to be no point in continuing this discussion, it seems what I write does not make much sense to you.

Yes, but only because what you write objectively doesn't make sense in the context of relativity. Like I said, measuring the speed of an object in your own rest frame vs. another frame both involve exactly the same types of local measurements, and there is also no difference in terms of whether the local measurements take place near you or in another region (if you're measuring change in distance over a finite period, then in both cases at least one measurement must be away from you, and if you're measuring instantaneous velocity with infinitesimally-close measurements of position and time, then in both cases the measurements can be done in your immediate region). If you disagree with any of that, you are misunderstanding some aspect of SR.

 

[MeJennifer]

Yes, but only because what you write objectively doesn't make sense in the context of relativity. Like I said, measuring the speed of an object in your own rest frame vs. another frame both involve exactly the same types of local measurements, and there is also no difference in terms of whether the local measurements take place near you or in another region (if you're measuring change in distance over a finite period, then in both cases at least one measurement must be away from you, and if you're measuring instantaneous velocity with infinitesimally-close measurements of position and time, then in both cases the measurements can be done in your immediate region). If you disagree with any of that, you are misunderstanding some aspect of SR.

Perhaps you should take the effort of actually trying to understand what I am talking about because it seems that you are talking about something else here.
And if you actually think that a measurement taken in an object that is moving relative to you and its value consequently tranfered to you is a measurement of a moving frame then it is you who misunderstands some aspect of SR.

 

[JesseM]

Perhaps you should take the effort of actually trying to understand what I am talking about because it seems that you are talking about something else here.

If you think I'm misunderstanding, then take the time to explain it and respond to my questions.

And if you actually think that a measurement taken in an object that is moving relative to you and its value consequently tranfered to you is a measurement of a moving frame then it is you who misunderstands some aspect of SR.

Still makes no sense--what does it mean to "transfer" a measurement to me? I'm not saying that a measurement using a moving ruler/clock system counts as a valid measurement of position and time in my frame, I'm saying that I can use a moving ruler/clock system to measure position and time (and consequently speed) in some other frame, namely the rest frame of that ruler/clock system. If you don't agree with this, then yes, you're misunderstanding the basic idea of how the coordinate systems of different frames are physically defined in terms of measurements on ruler/clock systems, which was laid out in Einstein's original 1905 paper on SR.

 

[MeJennifer]

If you think I'm misunderstanding, then take the time to explain it and respond to my questions. Still makes no sense--what does it mean to "transfer" a measurement to me? I'm not saying that a measurement using a moving ruler/clock system counts as a valid measurement of position and time in my frame, I'm saying that I can use a moving ruler/clock system to measure position and time (and consequently speed) in some other frame, namely the rest frame of that ruler/clock system. If you don't agree with this, then yes, you're misunderstanding the basic idea of how the coordinate systems of different frames are physically defined in terms of measurements on ruler/clock systems, which was laid out in Einstein's original 1905 paper on SR.

Ok you are on the Earth and try to measure the speed of light in spaceship moving relative to you. You send them a digital message to have them test the speed of light in the spaceship, later you receive the digitized message stating the speed. In this case you still did not measure the speed of light in a moving frame, since the measuring device was measuring in a restframe. The speed of light cannot be measured in a moving frame. That is what I am trying to tell you for several postings.

I'm saying that I can use a moving ruler/clock system to measure position and time (and consequently speed) in some other frame, namely the rest frame of that ruler/clock system.

Yes obviously, but you fail to realize that that is no longer a measurement of a moving frame but a measurement in (another) rest frame.

 

[JesseM]

Ok you are on the Earth and try to measure the speed of light in spaceship moving relative to you. You send them a digital message to have them test the speed of light in the spaceship, later you receive the digitized message stating the speed.

Let's focus on measurements which are in both cases made automatically using my own equipment, so that there is no question of one speed measurement being made by me and the other being made by some other people.

1. Suppose I want to measure the speed of light in the rest frame of a measuring-stick which is 1 light-year long in its own rest frame, and which is also at rest relative to me. I can do this by placing clocks at either end of the measuring-stick, synchronizing them in the measuring stick's frame, and having them record the time at the moment the light beam passes through their local region. One of these measurements may also be made in my local region (if I am sitting near one end of the measuring-stick), but they can't both be--if one measurement was made right near me, the other would be made 1 light-year away from me, so the second clock would have to send a digital signal recording the time that the light had passed by it. Once I received this message, I could divide 1 light-year by the difference in times that the two clocks measured (which would of course be 1 year), and that would be the speed of light in the measuring-stick's frame.

2. Now suppose I want to measure the speed of light in the rest frame of a measuring-stick which is 1 light-year long in its own rest frame, but which is moving at a high constant velocity relative to me. I can do this by placing clocks at either end of the measuring-stick, synchronizing them in the measuring stick's frame, and having them record the time at the moment the light beam passes through their local region. One of these measurements may also be made in my local region (if one end of the measuring-stick is passing me at the same moment the light beam is passing it), but they can't both be--if one measurement was made right near me, the other would be made at some distance from me (the exact distance would depend on the velocity of the measuring-stick relative to me), so the second clock would have to send a digital signal recording the time that the light had passed by it. Once I received this message, I could divide 1 light-year (the length of the measuring-stick in its own rest frame, not in my frame) by the difference in times that the two clocks measured (which would of course be 1 year), and that would be the speed of light in the measuring-stick's frame.

So, what is the relevant difference here? Do you agree that the procedure in each case is basically identical, and that in each case I can only determine the speed after receiving a digital signal about an event at a location far from me?

In this case you still did not measure the speed of light in a moving frame, since the measuring device was measuring in a restframe.

But whether a frame is "moving" or "a rest frame" is relative to the observer! I was using a ruler/clock system that was at rest in a frame A to measure the speed of light in frame A, so of course the ruler/clock system would define frame A as its rest frame, but I would define frame A as a moving frame. Is all you are arguing that I can't measure the speed of light in a frame moving relative to me while using measuring-devices at rest in my own frame (at least without doing a mathematical transformation on the results of the measurements)? If so, does that mean you'd agree that I can measure the speed of light in a frame moving relative to me, as long as I set my measuring-equipment in motion relative to me? Maybe you would say that I am "measuring the speed of light in another rest frame" rather than "measuring the speed of light in a moving frame" in this case, but this would be totally idiosyncratic termilogy, in SR "moving frame" and "rest frame" are understood to be relative to an observer.

 

[MeJennifer]

Again, when a measuring device in a spaceship that is moving relative to me makes a measurement then that measurement is made in a restframe not in a moving frame, that is basic relativity and there is nothing idiosyncratic about it.

 

[JesseM]

Again, when a measuring device in a spaceship that is moving relative to me makes a measurement then that measurement is made in a restframe not in a moving frame, that is basic relativity and there is nothing idiosyncratic about it.

Yes, it is idiosyncratic, because "rest frame" and "moving frame" are always defined in relative terms, so if I measure the speed of light in a frame A that is moving relative to me, I don't somehow stop calling frame A "a moving frame" just because my measuring-equipment was at rest in frame A.

 

[MeJennifer]

...if I measure the speed of light in a frame A that is moving relative to me, I don't somehow stop calling frame A "a moving frame" just because my measuring-equipment was at rest in frame A.

True, but then if you make the claim that you measured the speed of light in a moving frame you would be incorrect. The actual measurement was made in the restframe.
Perhaps you fail to understand that no frame can move relative to light, so one most certainly cannot measure the speed in a moving frame.

Anyway, I would not ever expect anything close to "I see your point" from you, since you always argue your are right 100% and the other person is always wrong.

 

[JesseM]

True, but then if you make the claim that you measured the speed of light in a moving frame you would be incorrect. The actual measurement was made in the restframe.

But you are using "the rest frame" as if it's an absolute term rather than a relative one, which is not how physicists talk. The measurement was made in the rest frame of the measuring-device, but it was made in a frame which I define as "a moving frame", I am not somehow forced to call it "the rest frame" because the measuring-device is at rest in it. And if I'm the one who set up the measuring-device and recorded its readings, then I'm the one who made the measurement, so it makes perfect sense for me to say "I measured the speed of light in a moving frame."

Perhaps you fail to understand that no frame can move relative to light

I didn't bother objecting on the other thread where you said our velocity relative to light is 0, but this is also totally idiosyncratic terminology. Certainly we can never "catch up" with a light beam so that its velocity in our rest frame is any lower than c, but normally when you say "X's velocity relative to Y" you mean "X's velocity in Y's rest frame", and light doesn't have a rest frame to begin with. Even if you try to construct a pseudo-frame for light by imagining what my velocity would be in another observer's frame in the limit as that observer's velocity approached c in my frame, but in that limit the observer would see my own velocity approaching c in his rest frame, not 0.

Anyway, I would not ever expect anything close to "I see your point" from you, since you always argue your are right 100% and the other person is always wrong.

Well, I'll make one concession, which is that this argument doesn't seem to have much to do with your misunderstanding any theoretical issues in SR (although I still don't know what you meant when you said I would have to correct for doppler shifts and light-signal delays when measuring the speed of light in another frame but not in my own, since even in my own frame at least one measurement has to be made at a distance from me), it's mainly just because you use language and terminology in a very weird and nonstandard way.

 

[MeJennifer]

Certainly we can never "catch up" with a light beam so that its velocity in our rest frame is any lower than c, but normally when you say "X's velocity relative to Y" you mean "X's velocity in Y's rest frame", and light doesn't have a rest frame to begin with.

Exactly my point, light has no rest frame, light always moves at c.

There is no relative motion between a light and us on the contrary the motion of light is absolute, with the light moving and we standing still!
And since we are standing still light always disperses spherically from us.

But of course only when we are in an inertial state.

..even in my own frame at least one measurement has to be made at a distance from me

Sure, but with a modern measurement apparatus we don't have to go that far out.

 

[petm1]

You mention above that gamma is 0.5. However, gamma must always be greater than 1 for v>0.

At 0.866c, gamma=2. So contractions are 1/gamma=1/2, and dilation is gamma=x2.

So plug in the figures you gave, 300,000k meters / second times 1/2 / 2, still changes the speed of light.

 

[JesseM]

So plug in the figures you gave, 300,000k meters / second times 1/2 / 2, still changes the speed of light.

Did you read my post #56? I showed using a numerical example that when you take into account length contraction, time dilation and the relativity of simultaneity, both observers will measure the speed of light to be the same. If you read it, did you disagree with any step in the analysis? If you didn't read it, don't you think you should look into it before repeating this claim that the speed of light will be changed?

edit: If it helps, I came up with another numerical example where all the numbers aside from the velocity and the gamma-factor are integers, so you should be able to follow it without needing a calculator:

Say there's a ruler that's 50 light-seconds long in its own rest frame, moving at 0.6c in my frame. In this case \gamma is 1.25, so in my frame its length is 50/1.25 = 40 light seconds long. At the front and back of the ruler are clocks which are synchronized in the ruler's rest frame; because of the relativity of simultaneity, this means that in my frame they are out-of-sync, with the front clock's time being behind the back clock's time by vx/c^2 = (0.6c)(50 light-seconds)/c^2 = 30 seconds.

Now, when the back end of the moving ruler is lined up with the 0-light-seconds mark of my own ruler (with my own ruler at rest relative to me), I set up a light flash at that position. Let's say at this moment the clock at the back of the moving ruler reads a time of 0 seconds, and since the clock at the front is always behind it by 30 seconds in my frame, then in my frame the clock at the front must read -30 seconds at that moment. 100 seconds later in my frame, the back end will have moved (100 seconds)*(0.6c) = 60 light-seconds along my ruler, and since the ruler is 40 light-seconds long in my frame, this means the front end will be lined up with the 100-light-seconds mark on my ruler. Since 100 seconds have passed, if the light beam is moving at c in my frame it must have moved 100 light-seconds in that time, so it will also be at the 100-light-seconds mark on my ruler, just having caught up with the front end of the moving ruler.

Since 100 seconds passed in my frame, this means 100/1.25 = 80 seconds have passed on the clocks at the front and back of the moving ruler. Since the clock at the back read 0 seconds when the flash was set off, it now reads 80 seconds; and since the clock at the front read -30 seconds, it now reads 50 seconds. And remember, the ruler was 50 light-seconds long in its own rest frame! So in its frame, where the clock at the front is synchronized with the clock at the back, the light flash was set off at the back when the clock there read 0 seconds, and the light beam passed the clock at the front when its time read 50 seconds, so since the ruler is 50-light-seconds long, the beam must have been moving at 50 light-seconds/50 seconds = c as well! So you can see that everything works out--if I measure distances and times with rulers and clocks at rest in my frame, I conclude the light beam moved at 1 c, and if a moving observer measures distance and times with rulers and clocks at rest in his frame, he also concludes the same light beam moved at 1 c.

If this is still confusing, I can also draw some diagrams of this scenario so it's easier to keep everything straight, just let me know if you'd like some.

 

[pess5]

So plug in the figures you gave, 300,000k meters / second times 1/2 / 2, still changes the speed of light.

Observer A & observer B at 0.866c relative, so gamma=2. The B frame per the A frame (B is contracted by 50%) ...

300,000 km/sec x [ (length contraction) / (time contraction) ] = 300,000 km/sec

300,000 km/sec x [ (0.5km/1km) / (0.5sec/1sec) ] = 300,000 km/sec

pess

 

[petm1]

Observer A & observer B at 0.866c relative, so gamma=2. The B frame per the A frame (B is contracted by 50%) ...

300,000 km/sec x [(length contraction) / (time contraction)] = 300,000 km/sec

300,000 km/sec x [(0.5km/1km) / (0.5sec/1sec)] = 300,000 km/sec

This would be fine if SR had time contraction, but I believe it is time dilation in SR.

The spacetime volume is (length)/gamma x t(gamma). So the volume which you see the moving box carve out thru 4-space is the same volume the stationary box sees itself carve out, since (length)/gamma x t(gamma) = length x time. Basically, time is rotated partially into 3-space, and 3-space is equally rotated partially into time, per the viewing observer of a moving body.

Einstein talked about time dilating and length contracting with the increase of motion in SR. In GR he talked about seeing time dilation in a gravity well, but I think that GR is talking about time contraction as in a decrease in motion. So in SR you need to use time dilation which would give you the equation of (length)/gamma x time(gamma) or 1/.5 x .5(1).

I believe that we exist is 3-d space while looking at every thing through the 4-d movement of a light wave. You have to be careful with what you call time contraction because Einstein did not talk of this at all, he talked about an increase in motion causing time to dilate because of the longer length that a light wave must travel between the mirrors of a light clock, and he talked about time dilation in gravity.

This must not be confused with what I call time contraction which would be the slowing of a clock through a slowing of its motion as in gravity well, or while changing its motion. I never read of time contraction nor length dilation, I use these terms just to show an effect opposite of what Einstein talks about, and to show the twist between what we feel and what we see.

 

[JesseM]

This would be fine if SR had time contraction, but I believe it is time dilation in SR.

It depends what you mean by these terms. A moving clock will take longer to tick a single second, but that also means that in a second of time in our frame, the moving clock ticks less time than a second. At 0.866c a moving 1-meter stick only appears to be 0.5 meters relative to our own rulers, and a moving clock only ticks 0.5 seconds in 1 second of our time, so the numbers pess5 gave are correct, although to really understand why each frame measures the same speed of light in all directions you also must include the relativity of simultaneity, as I did in my example above. Are you ever going to look at that example and tell me if you see any problems with it, or if you need additional explanation or diagrams?

 

[MeJennifer]

to really understand why each frame measures the same speed of light in all directions you also must include the relativity of simultaneity

That is like putting the horse behind the carriage.

The relativity of simultaneity is a consequence of the postulate that the speed of light is c in all inertial frames not the other way around.

 

[JesseM]

That is like putting the horse behind the carriage.

The relativity of simultaneity is a consequence of the postulate that the speed of light is c in all inertial frames not the other way around.

Strictly speaking, you could replace Einstein's first postulate by the postulate that the speed of light is isotropic in every inertial frame (ie every inertial frame will measure the speed of light to be the same in all directions in that frame, without assuming from the start that different frames will measure the value of that speed to be the same), and still derive the Lorentz transform and the rest of SR from that.

Anyway, I'm just trying to show petm1 how the math works out if you take for granted the basic features of relativity like length contraction and time dilation and the relativity of simultaneity, not to explain how these features were actually derived.