Understanding Time Dilation: Exploring Einstein's Theory with Basic Math

[petm1]

yogi says "how does the oberver in the moving frame measure the distance traveled - and he would say - it must have been vt - where v is the relative velocity and t is the number of seconds logged by the clock in the moving frame -"

This is true as long as he just uses his velocity relative to the resting frame, but if he physically mesureses the distance he has traveled, with his contracted ruler, over his dilated second he will get a dilated velocity.

 

[petm1]

The problem arises because of misleading statements as to what is really taking place in a particular experiment - in his 1905 paper Einstein himself says a clock at the equator will run slower than one at the pole (it doesn't for other reasons - but this is not in issue).

yogi, could you give me a link to this, I don't think I have read about it before.

 

[robphy]

My question is why does relativity get to break the rules of math? You change the second, which in the speed of light is the divisor, by the Lorentz transformation. Then SR changes the dividend by the inverse of the Lorentz transformation. I would think that to prove a theroy you must hold to the rules of math while representing reality, if your math does not hold to the rules you have to keep looking. That is why I have been looking for a way that the math works out while still representing what we see.

As far as I know, [Special] relativity does not "break the rules of math". Rather, it suggests to that the "Galilean/Newtonian mathematical model of kinematics and dynamics, together with its physical intuition, definitions, formulas, etc..." is not the appropriate model of the world. So, applying such a model to understand Special Relativity will lead to apparent inconsistencies. A new mathematical model is needed, together with its physical intuition, definitions, formulas, etc... Of course, in the end, the results of experiments will decide which model is more appropriate for our world...and these results suggest that the Galilean/Newtonian model loses this contest.

 

[petm1]

As far as I know, [Special] relativity does not "break the rules of math". Rather, it suggests to that the "Galilean/Newtonian mathematical model of kinematics and dynamics, together with its physical intuition, definitions, formulas, etc..." is not the appropriate model of the world. So, applying such a model to understand Special Relativity will lead to apparent inconsistencies. A new mathematical model is needed, together with its physical intuition, definitions, formulas, etc... Of course, in the end, the results of experiments will decide which model is more appropriate for our world...and these results suggest that the Galilean/Newtonian model loses this contest.

In a nut shell that is what we are talking about. Part of the problem is the fact that the Galilean/Newtonisn's mathematical model works so well for our perfered frame of reference, Earth, and it is used to show the outline for SR which leads to the apparent inconsistencies.

 

[robphy]

In a nut shell that is what we are talking about. Part of the problem is the fact that the Galilean/Newtonisn's mathematical model works so well for our perfered frame of reference, Earth, and it is used to show the outline for SR which leads to the apparent inconsistencies.

Maybe there is a discrepancy in definitions and terminology here.
By Galilean/Newtonian mathematical model, I am referring to the Galilean/Newtonian spacetime model (spatial 3D hyperplanes, stacked to reflect its absolute time structure) together with the Galilean transformations. In this model, there is no preferred frame of reference. (Mathematically, the Galilean transformation has no timelike eigenvectors.) So, the Earth is not a preferred frame of reference in the model above.

The Galilean/Newtonian model does work well for speeds small compared to the speed of light. By work "well", I mean that our measurements are not precise enough to distinguish deviations that suggest the SR model. With a precise enough clock, one need not involve large speeds to see the effects of SR.

 

[petm1]

Maybe there is a discrepancy in definitions and terminology here.
By Galilean/Newtonian mathematical model, I am referring to the Galilean/Newtonian spacetime model (spatial 3D hyperplanes, stacked to reflect its absolute time structure) together with the Galilean transformations. In this model, there is no preferred frame of reference. (Mathematically, the Galilean transformation has no timelike eigenvectors.) So, the Earth is not a preferred frame of reference in the model above.

I believe the Earth is a preferred frame in so far that we are all moving relative to it, granted we all have clocks that have to be synchronize every "time" we get together, but this is what we are talking about also. By the way I also believe that the Earth is the center of my visible universe, and that every point in my "universe" has the potential of being a singularity.

The Galilean/Newtonian model does work well for speeds small compared to the speed of light. By work "well", I mean that our measurements are not precise enough to distinguish deviations that suggest the SR model. With a precise enough clock, one need not involve large speeds to see the effects of SR.

I would think that on the faster and larger side of reality SR and a clock are needed to "see", but on the smaller and slower side of reality GR and a clock are what you need.

 

[yogi]

yogi, could you give me a link to this, I don't think I have read about it before.

www.fourmilab.ch/etexts/einstein/specrel/www/ - 86k

At the end of part 4

 

[petm1]

Does SR change the speed of light or is it a math problem? Using the Lorentz transformation, to show how far you are contracting the meter. While using, one over the lorentz transformation as the divisior for Time dilation. This changes the ratio into a positive number larger than one.

 

[pess5]

Does SR change the speed of light or is it a math problem? Using the Lorentz transformation, to show how far you are contracting the meter. While using, one over the lorentz transformation as the divisior for Time dilation. This changes the ratio into a positive number larger than one.

One of SR's 2 postulates is that light speed is c, regardless of the relative speed of the light source. The other postulate requires that the same mechanics hold equally in all frame vantages. Hence, light must be speed c in any and all frames, w/o exception. Also, SR addresses only inertial observers in uniform translatory motion.

It seems you may be confusing the Lorentz/Fitzgerald Contraction formula with the Lorentz Transformation formula ?

The Lorentz/Fitzgerald Contraction formula is ...

. x'=x(1-v^2/c^2)^1/2 ... so moving length x is contracted by (1-v^2/c^2)^1/2, a value <1

The Lorentz Transformations are ...

. T = gamma(t-vx/c^2)
. X = gamma(x-vt)
. Y = y
. Z = z

where gamma=1/(1-v^2/c^2)^1/2, a value >1, ie the reciprocal of the Lorentz/Fitzgerald Contraction formula.

Einstein called gamma "beta" in his 1905 OEMB. Later, Einstein's beta was renamed to gamma, since folks liked to refer to beta as v/c, in which case if c is taken as unity and v a percentage of c, then ...

. T = gamma(t-Beta*x)
. X = gamma(x-Beta*t)
. Y = y
. Z = z

pess

 

[petm1]

I don't disagree with any thing that has to do with SR or GR but I believe that the two are connected through our preferred reference frame.

I see myself as a PRF, I have mass, I am my own bio-clock, and I observe. But for this thought experiment let's talk about me (an observer) with two clocks (a light clock, an atomic clock) and sitting on a 1 cubic meter block of concrete marked down to millimeters. My rest frame to me is seamless and everything that I can feel stays the same and appears to have no motion relative to me. According to SR nothing occupies just 1 cubic meter of space per second even though to the observer inside the prf it appears to have a volume of just one cubic meter per second, which is why all reference frames are preferred. To keep track of my prf we will leave it sitting on Earth with a volume of space occupied over time of about 30,000 meters per second. Volume at rest I will call volume, volume of space I will call "vos".

When moving away at some speed we see the changes between the prfs in the form of a ratio with length contraction (LC) changing the length of the meter and time dilation (TD) changes the length of the second, yet inside the moving prf nothing appears to change except the view. This is true for all motion with speeds relative from one prf to the other prf, when looking at their "vos" though you can see that an increase of speed will give a direct increase in their vos. In our universe the fastest thing we can see would be a light wave with a speed 300,000,000 meters per second relative to all prfs. The slowest thing I can think of is the 1 cubic meter of space at the center of the earth’s gravity well.

To show anything with an increase in motion with regards to speed from my prf or faster I would think that their ratio would be dilation per dilation, to show anything getting slower I would expect to see the ratio of contraction per contraction. With these ratios modifying speed in the form of meters per second, I could see everything staying proportional. But everything I have read about SR so far says we see these two effects as inverted with faster having LC/TD, and slower as LD/TC. By the way SR talks about LC/TD, while in my view GR talks about LD/TC, I've been known to think of the Earth as an eddy of time (motion). If looking out from my preferred reference frame and seeing a prf moving away with some speed changes the length contraction and time dilation then in the opposite direction, going slower than and into my prf I would expect to see just the opposite effect in so far as I would see length dilate and time contraction.

What we see is not always what we get, I believe that the way we see the world is always in flux with my prf being the center. Going slower relative to me means going deeper into the earth’s gravity well so that my "vos" occupied decreases, not by much but any other motion increases it. Also If you look at a prf that is increasing it volume of space occupied over time by increasing it speed then the proof for time dilation still works.

The hardest part is thinking of the twist with what we see. Larger and faster as length contraction meters getting shorter, and time dilation second getting longer while at the center of my prf looking inward and seeing the opposite, length dilation, meter getting bigger and time contraction, second getting slower. But this is needed to make the math work out, LD/TC...{LC/TC < meter/second < LD/TD}...LC/TD. Length contraction per time dilation is what I see and length dilation and time constriction are what I feel.

I believe that we live in a state of flux, unlike SR I believe that our volume of space occupied over time is the true reality and that we may see ourselves as a prf and at rest with every thing around us but our size is so seamless with everything that it is only little clues that we see from other prf that help point the way. Inside the gravity well of earth, the effects of time contraction and the time dilation over lap, we see the effects of both at the same time, it is different effects that act alike.

 

[petm1]

Does posting this type of statement mean I need to switch this thread to another place?

 

[petm1]

I think that our meter and our second are the basic measurements of our existence. Something about reading that both the second's and the meter's length can be changed with the same motion, but in opposite directions does not look right to me. After reading that it is in the form of a division problem and that it is breaking the rules of math that I learned in elementary school, I knew that something must be wrong with the way I was looking at the problem. This looked like a good site to look for help, but...

Length Constriction/Time Dilation multiplied by the speed of light which is a ratio of length/time does not keep the speed of light constant it will not even keep it proportional, and it only accounts for one directing of travel. It’s too bad, but I guess I am the only one to think this way, and I guess I will have to go to another place to get some feed back on this problem. Thank for the help. Jeff

 

[pess5]

I believe that we live in a state of flux, unlike SR I believe that our volume of space occupied over time is the true reality and that we may see ourselves as a prf and at rest with every thing around us but our size is so seamless with everything that it is only little clues that we see from other prf that help point the way. Inside the gravity well of earth, the effects of time contraction and the time dilation over lap, we see the effects of both at the same time, it is different effects that act alike.

As you said, with contraction comes dilation, since space & time are a fused continuum. The box is say sqr. Einstein's math teacher, Hermann Minkowski modeled Einstein's Special Theory in terms of fused spacetime and rotations between frame perspectives. The time axis may be thought of as a complex spatial axis, which Minkowski labeled the ict axis. The leading "i" signifying a complex axis at 90 deg wrt 3-space. Basically, Minkowski modeled Einstein's time as a 4th spatial axis with length of ct, where c is taken as unity, and so ict is just called x4. So x,y,z,t is called x1,x2,x3,x4. Modeling time like a 1-space allowed Minkowski to view one's motion thru spacetime as a 4-space line, similarly as considering a line in 3-space. This 4-space line he calls a worldline, and it signifies your path which you carve out through spacetime, generally modeled between any 2 spacetime events. It's length is called the worldline length, and it is the amount of duration you experience per yourself over the defined interval (the proper time). There's much more to Minkowski 4-space, but I'll leave it at that for now...

Considering a sqr box, it travels through only time per itself, because it is stationary in 3-space. That is, all observers are equally obliged to consider themselves the stationary, since they cannot feel their own inertial motion. So it carves out a volume thru the fused spacetime, ie 4-space, as time flows. Since you are in motion wrt the box, you see it length contracted. However, you are time dilated wrt the box, since it equally sees you in motion at v. Over some defined interval between 2 common spacetime events, the stationary uncontracted box carves out a volume of ... length x time. The box is larger, but the temporal distance (along ict) is shortest per its (proper) frame vantage. You see the box contracted by gamma, but you see it travel thru a much longer dilated temporal path (along ict). The contraction is (length)/gamma. The dilation is t(gamma). 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. This is frame rotation, and it gives rise to contractions & dilations while preserving the volume a body carves out through spacetime as time progresses, a conservation of space & time.

Pess

 

[myoho.renge.kyo]

let's say that you have a clock and i have a clock. your clock and my clock are synchronous. let's also say that you depart from Earth with your clock with a velocity v at a specified direction. let's say that when the sun sets where i am on earth, my clock and your clock display 00:00 universal time. let t be the time that it takes the sun to set again. if i measure the value of t with my clock, and then i measure the value of t again with your clock, the two values of t would be different, but if i measure the value of t using my clock, and you measure the value of t using your clock, the values of t would be equal.

 

[pess5]

let's say that you have a clock and i have a clock. your clock and my clock are synchronous. let's also say that you depart from Earth with your clock with a velocity v at a specified direction. let's say that when the sun sets where i am on earth, my clock and your clock display 00:00 universal time. let t be the time that it takes the sun to set again.

if i measure the value of t with my clock, and then i measure the value of t again with your clock, the two values of duration would be different, ...

but if i measure the value of t using my clock, and you measure the value of t using your clock, the values of t would be equal.

myoho.renge.kyo

Interesting scenario. However, it seems to me that ...

I will record the cycle of an Earth rotation at gamma times faster than you. You'll see a rotation at one every 24 hr. I'll see a rotation every 12 hr. This is because I depart you & Earth at v, some very large velocity near speed c (say). Since Earth is in your frame, you experience no distortion of you or the earth. I am at high speed wrt you and the earth, and so I must experience a distortion of both you & Earth equally. That said, you will record the proper time for earthly rotation. I will not.


pess

 

[petm1]

Considering a sqr box, it travels through only time per itself, because it is stationary in 3-space.

I can see a 1 cubic meter Gold sphere, relative to me it has no movement I can see, but relative with me it has a volume of space of 30,000 meters/second. I feel the 3-d universe but in Minkowski 4-space it has a continuous movement that I see. I think of everything as having a dual nature and just because you add a continuous string of events to my 3d universe, does not mean that the 3-d leaves, it just means that the 4-d overlaps, the dualistic nature of reality means that I have to think of both 3-d and 4-d as being simultaneous. I think, as I feel my motion in 3-d I move in 4-d the only twist is where they intersect.

There's much more to Minkowski 4-space

We talk about the dually of particles yet we deny the same dually of a light wave, by saying that is does not move thru a ether "we say a light wave exists unto itself", yet it is a light wave that exists in 4-d all the time with a 3-d part, just as we exist is 3-d with a 4-d part, both have a equal amount within themselves and both exist at the same time which we limit as a second.

 

[JesseM]

I think that our meter and our second are the basic measurements of our existence. Something about reading that both the second's and the meter's length can be changed with the same motion, but in opposite directions does not look right to me. After reading that it is in the form of a division problem and that it is breaking the rules of math that I learned in elementary school, I knew that something must be wrong with the way I was looking at the problem. This looked like a good site to look for help, but...

What math rules do you think are broken? And what do you mean by "changed with the same motion, but in opposite directions"? Do you just mean that meters are shorter than their rest length, and clock ticks appear longer if the clock is in motion?

Length Constriction/Time Dilation multiplied by the speed of light which is a ratio of length/time does not keep the speed of light constant it will not even keep it proportional, and it only accounts for one directing of travel.

Are you really arguing here that the math is wrong here, and that the combination of length contraction + time dilation + different definitions of simultaneity won't result in each observer measuring the speed of light to be the same in all directions according to his own rulers and clocks? If so, are you willing to go through a numerical example to see why you're wrong about this?

 

[petm1]

What math rules do you think are broken?

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.

And what do you mean by "changed with the same motion, but in opposite directions"?

You are contracting the dividend and dilating the divisor in the moving frame's meter/second ratio, yet you are telling me that the observer in this frame will not notice "because" everything will appear the same. We are discussing the "because". All of SR deals with movement along one line in one direction but the meter and the second react in different directions.

Are you really arguing here that the math is wrong here, and that the combination of length contraction + time dilation + different definitions of simultaneity won't result in each observer measuring the speed of light to be the same in all directions according to his own rulers and clocks? If so, are you willing to go through a numerical example to see why you're wrong about this?

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.

 

[MeJennifer]

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.

Sorry but one can only measure the speed of light in one's local frame.

How could one possibly measure the speed of light in a frame moving relative to you?

 

[JesseM]

Sorry but one can only measure the speed of light in one's local frame.

How could one possibly measure the speed of light in a frame moving relative to you?

Just check the measurements of rulers and clocks at rest in that frame!

 

[JesseM]

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.

Do you understand the concept of the "relativity of simultaneity"? It is assumed that each observer measures the speed of anything, including a light beam, using multiple clocks which are at rest in their frame, and "synchronized". For example, if an object passes by the 1-meter mark on my ruler and at the moment it passes a clock at the 1-meter mark reads "15 seconds", and then later the object passes the 9-meter mark on my ruler, and at the moment it passes a second clock at the 9-meter mark reads "17 seconds", then I'll conclude that in my frame the object moved 9-1=8 meters in 17-15=2 seconds, for a speed of 8/2=4 meters/second.

So, each observer is using multiple synchronized clocks to measure speed. But--and this is a key point--different observers disagree on what it means for clocks to be 'synchronized. Two clocks which are synchronized in the frame of an observer at rest relative to them will be out-of-sync in the frame of an observer moving relative to them. The reason for this is that each clock is synchronized using the assumption that light travels at the same speed in all directions in the clocks' rest frame, so I can synchronize two of my own clocks by setting off a flash at the midpoint between them, and setting them to read the same time at the moment the light from the flash reaches each one. But if in your frame my clocks are moving, then one clock will be moving toward the point where the flash went off, while the other will be moving away from that point...so if you assume the light moves at the same speed in both directions in your frame, then in your frame the light must reach the clocks at different times! This means that if I set the clocks to read the same time when the light hits them, in your frame the clocks will be out-of-sync. It works out that if the two clocks are at a distance of x apart in their own rest frame, and they are synchronized in that frame, then if in your frame the clocks are moving at velocity v parallel to the axis that joins them, in your frame the back clock's time will be ahead of the front clocks' time by a factor of vx/c^2.

Once you take time dilation and length contraction and the relativity of simultaneity into account, you will find that each frame does indeed measure the same speed for a light beam. I've already posted a link to this thread where I gave an example showing how it works out, but perhaps I should just repost the example here to increase the chance that you'll actually read it:

Suppose, for the sake of making the math a bit easier, that we measure distance in units of "fivers", where a fiver is defined to be the distance light travels in 0.2 seconds (i.e. 1 fiver = 0.2 light-seconds), so that light is defined to have a velocity of 5 fivers/second. Suppose you see a ruler which is moving at a velocity of 3 fivers/second along your x-axis (0.6c). In its own rest frame, this ruler is 40 fivers long; so in your frame its length will appear to be:
$$40 * \sqrt{1 - v^2/c^2}$$ = $$40 * \sqrt{1 - 9/25}$$ = 40 * 0.8 = 32 fivers long. Also, at either end of this ruler is placed a clock; using the time dilation formula, we can see that for every second on your clock, you will only see these clocks ticking 0.8 seconds forward.

Now, say that when t=0 according to your clock, the clock on the left end of the ruler also reads t'=0. At that moment, a light is flashed on at the left end of the ruler, and you observe how long the light pulse takes to catch up with the right end. In your frame, the position of the light pulse along the x-axis at time t will be c*t, while the position of the right end of the ruler at time t will be v*t + 32. So, the light will catch up to the right end when c*t = v*t + 32 which if you solve for t means t = 32/(c - v). Plugging in c = 5 and v = 3, you get a time of 16 seconds, in your frame.

Now the key to understanding how the ruler can also measure this pulse to be moving at c is to realize that different frames have different definitions of what it means for a pair of clocks to be "synchronized". In your frame, when the clock on the left reads t'=0, the clock on the right will not read t'=0; in your frame, the clock on the right is always 4.8 seconds behind the clock on the left (since the clocks are 40 fivers apart in their own rest frame and they are moving at 3 fivers/second in your frame, then plugging that into the formula $$vx/c^2$$ gives (3*40)/(5^2) = 120/25 = 4.8), so it will read t'=-4.8. This means that after 16 seconds have passed according to your clocks, only 16*0.8 = 12.8 seconds will have elapsed on the ruler's clocks, which means the clock on the right reads -4.8 + 12.8 = 8 at the moment that the light reaches the right end. So remembering that light was emitted when the clock on the left read t'=0, the light must have taken 8 seconds to cross the ruler in the ruler's own frame; and remembering that the ruler is 40 fivers long in its own frame, the speed of the light pulse is measured to be 40/8 = 5 fivers/second. So, light does indeed have the same speed in both frames.[/quote]

 

[MeJennifer]

Just check the measurements of rulers and clocks at rest in that frame!

I am not sure what you are saying but, if I were to test the speed of a light beam comming from an object that is at relative motion with me it will still be c, the speed will be independent of the relative speed between me and the object.

In other words, we can only measure light from our own restframe!

In relativity emitted light does not care about the relative velocity, it always escapes with a speed of c, and this applies to all inertial frames.

 

[JesseM]

I am not sure what you are saying but, if I were to test the speed of a light beam comming from an object that is at relative motion with me it will still be c, the speed will be independent of the relative speed between me and the object.

Yes, of course.

In other words, we can only measure light from our own restframe!

No, that's not the same statement at all. In relativity, you assume that the "speed" of any object is being measured by local measurements on a system of rulers and clocks, with the clocks all at rest with respect to each other and synchronized in their own rest frame using the Einstein synchronization convention. So if a light beam passes a mark on a ruler labeled "1 light-second" and the clock at that mark reads "5 seconds", then later it passes a mark on a ruler labeled "3 light-seconds" and the clock at that mark reads "7 seconds", then according to that ruler-clock system the light beam traveled a distance of 3-1=2 light seconds in a time of 7-5=2 seconds, so its speed is 2/2 = 1 light-second/second according to that ruler clock system.

Now, it's true that every ruler-clock system constructed according to Einstein's procedure will measure light to have the same speed of 1 light-second per second according to measurements on their own rulers and clocks, even if the light beam was emitted by an object moving relative to that ruler-clock system, and even if you have two ruler-clock systems moving side-by-side and measuring the speed of the same beam of light. But you are certainly free to take a peek at a ruler-clock system which is in motion relative to you, and see what marks and clocks the light beam is passing on that system! The statement that "every frame will measure the speed of light to be c" is not saying that every observer is somehow only capable of measuring light's speed in their own rest frame, you can certainly construct a system that will measure light-speed in some frame moving relative to you, but as long as you construct it according to Einstein's procedure (and using Einstein's clock synchronization convention), then it will still measure the speed of light to be 1 light-second per second.

 

[MeJennifer]

In relativity, you assume that the "speed" of any object is being measured by local measurements on a system of rulers and clocks, with the clocks all at rest with respect to each other and synchronized in their own rest frame using the Einstein synchronization convention.

You cannot measure it any other way.
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.

 

[JesseM]

You cannot measure it any other way.
In fact we cannot measure anything directly that is not local.

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.

At the most you can calculate something from the perspective of another frame by measuring the light signals you receive (and send if you are looking for roundtrip times)

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.

 

[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.