Do we have time because we are moving?

[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.

 

[Dale]

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).

In modern cosmology there is no center of the universe, nor is there any "before" or "outside" the universe. It is not that it exists and has no gravity, it simply doesn't exist. We can discuss time dilation within the universe, but not before or outside it.

 

[Taragond]

Thanks Dale,

that's why it is hypothetical I guess? I know all that and I thought I mentioned it, too.
I just felt that if time runs slower the faster you move and we only know our own time personally, then time must run faster if you were slower.
For example:
our galaxy moves in a direction with some million km/h.
what If we could travel off with that speed relative to it in the opposite direction?

We would still have that speed relative to the milky way but relative to the surroundings we would get slower by that factor.
So...the galaxy's time is slowed down compared to it's surrounding space?
We are slowed down compared to the galaxy?
but shouldn't we get closer to the timeframe of the surrounding space by flying in the opposite direction?

Well I guess there just is no "surrounding space" as our galaxy's "speed" is just perceived as such by stretching space? Is that was I was missing? But in that case wouldn't it de facto have no speed?

 

[Dale]

that's why it is hypothetical I guess?

It isn't just hypothetical, it is contradictory to the theory. Since you are asking this question here I presume that you would like an answer which is consistent with the theory of relativity and modern cosmology. It simply is not possible to use a theory to answer a question which presupposes something contradictory to the theory.

I just felt that if time runs slower the faster you move and we only know our own time personally, then time must run faster if you were slower.
For example:
our galaxy moves in a direction with some million km/h.
what If we could travel off with that speed relative to it in the opposite direction?

This is a question which can be answered by the theory. Assuming that you are still close enough to our galaxy to ignore spacetime curvature, then in the galaxy's inertial frame you would be time dilated and in your inertial frame the galaxy would be time dilated.

We would still have that speed relative to the milky way but relative to the surroundings we would get slower by that factor.
So...the galaxy's time is slowed down compared to it's surrounding space?
We are slowed down compared to the galaxy?
but shouldn't we get closer to the timeframe of the surrounding space by flying in the opposite direction?

You would be at rest relative to a local "comoving"* observer. In the comoving observer's frame the galaxy would be time dilated and you would not. However, I don't think that anyone would call the comoving observer's reference frame "surrounding space" nor would anyone speak of speeds relative to "surrounding space".

*by comoving I mean an observer which is at rest relative to the FLRW coordinates, i.e. one that sees no dipole anisotropy in the CMB. See: http://en.wikipedia.org/wiki/Comoving_distance#Comoving_coordinates

 

[Taragond]

This is a question which can be answered by the theory. Assuming that you are still close enough to our galaxy to ignore spacetime curvature, then in the galaxy's inertial frame you would be time dilated and in your inertial frame the galaxy would be time dilated.

That sounds a little like from my perspective, galaxy-time would be slowed down, from the galaxy's perspective, mine would be slowed down.

 

[ghwellsjr]

That sounds a little like from my perspective, galaxy-time would be slowed down, from the galaxy's perspective, mine would be slowed down.

This is only true as long as you understand that "from my perspective" means "from the Inertial Reference Frame" in which I am at rest. It is not true "from the way things look to me". You can't perceive or see or observe Time Dilation either from your rest IRF or from the galaxy's rest IRF. Although things do look slowed down in a galaxy that is moving away from you, that's not just Time Dilation, that's Relativistic Doppler. Time Dilation is Relativistic Doppler with light transit time mathematically removed which requires you to make an assumption about how long it takes for the light to traverse from the Galaxy to you and that requires an assignment of an IRF.

 

[Taragond]

argh always that problem with perspective

proposal:
3 synchronized clocks, one stays at A while two are send with B at .6c away from A
after one year on B, C gets with one of those clocks on B in a lifeboat and heads at .9c away from B in direction of A.
Now the relative speeds are:
A:B => .6c
B:C => .9c
A:C => .3c
how much would the three clocks differentiate after 1, 2, 3 years?

 

[Dale]

That sounds a little like from my perspective, galaxy-time would be slowed down, from the galaxy's perspective, mine would be slowed down.

Yes. (and ghwellsjr is correct in describing what "from my perspective" means in relativity-speak)

 

[Dale]

how much would the three clocks differentiate after 1, 2, 3 years?

In which reference frame?

 

[ghwellsjr]

argh always that problem with perspective

proposal:
3 synchronized clocks, one stays at A while two are send with B at .6c away from A
after one year on B, C gets with one of those clocks on B in a lifeboat and heads at .9c away from B in direction of A.
Now the relative speeds are:
A:B => .6c
B:C => .9c
A:C => .3c
how much would the three clocks differentiate after 1, 2, 3 years?

There's no frame in which your scenario can be carried out. Speeds don't add algebraically. At least I cannot find a way to make your three relative speeds consistent.

I really think you need to focus on very simple scenarios until you get some basic understanding of how Time Dilation and Length Contraction work. Simple scenarios lead to simple explanations.

 

[Taragond]

maybe that's what I am trying to do...

If I had 2 synchronized clocks A&B and send B away with.9c for one year after which it decellerates and flies back. would the clocks still be synchronized? As I understand it, B would show less time has passed?

If that's correct, I would expect that in C time runs slower than in B, but from A's perspective in C time runs less slow than in B...so no...probably not...
But if it is a measurable effect with clocks there must be a solution to this?

I guess it would be too much to clarify here why speeds don't add algebraically? not sure, what to search for...

 

[ghwellsjr]

maybe that's what I am trying to do...

If I had 2 synchronized clocks A&B and send B away with.9c for one year after which it decellerates and flies back. would the clocks still be synchronized? As I understand it, B would show less time has passed?

If that's correct, I would expect that in C time runs slower than in B, but from A's perspective in C time runs less slow than in B...so no...probably not...
But if it is a measurable effect with clocks there must be a solution to this?

I guess it would be too much to clarify here why speeds don't add algebraically? not sure, what to search for...

Of course there's a solution for any problem that's consistently and completely described but you have combined parameters that are inconsistent and left out other important details so that there is not a single interpretation of your scenario.

You could have said:

3 synchronized clocks, one stays at A while two are sent with B at .6c away from A.
After one year according to B's time, C gets with one of those clocks on B in a lifeboat and heads at .9c away from B relative to B in direction of A.

Then you could ask your questions like how fast is C traveling in A's rest frame and what time is on each clock in the different rest frames.

 

[Taragond]

You could have said:

3 synchronized clocks, one stays at A while two are sent with B at .6c away from A.
After one year according to B's time, C gets with one of those clocks on B in a lifeboat and heads at .9c away from B relative to B in direction of A.

Then you could ask your questions like how fast is C traveling in A's rest frame and what time is on each clock in the different rest frames.

well...that one!

I am really sorry for my lack of precision in my questions. Maybe it's because english is not my native language, maybe lack of experience in discussions on these matters. Probably a bit of both...

But what does happen if you remove the difference in rest-frames by decellerating before comparing the clocks?

 

[ghwellsjr]

You could have said:

3 synchronized clocks, one stays at A while two are sent with B at .6c away from A.
After one year according to B's time, C gets with one of those clocks on B in a lifeboat and heads at .9c away from B relative to B in direction of A.

Then you could ask your questions like how fast is C traveling in A's rest frame and what time is on each clock in the different rest frames.

well...that one!

Ok, I'm going to do it from the Inertial Reference Frame (IRF) of the B clock after it leaves the A clock at 0.6c because then I can simply specify the speed of the C clock after 1 year to be -0.9c according to the same IRF. I have made a spacetime diagram showing all the timings you asked about. Note that the dots mark off one-month intervals of Proper Time along each of the three clocks' worldlines. I have marked in some of the Proper Time values to help you determine the intervening ones. I have also annotated significant events along the way. Start at the bottom of the diagram and work your way up:

In the above diagram, there are three significant speeds:

1) All three clocks start out at -0.6c with synchronized clocks up until time zero. The A clock (blue) continues on an inertial path.

2) The B clock (red) along with the C clock accelerates to 0.6c with respect to the A clock so it comes to rest in the IRF that I am calling B's IRF (even though it is only B's IRF after time zero).

3) At B's Proper Time (which is also the Coordinate Time) of twelve months, C accelerates away from B at -0.9c toward A.

I have also annotated the Proper Times along each of the worldlines corresponding to the 1-, 2- and 3-year intervals of Coordinante Time as you requested.

Does that make perfect sense to you?

Now you were also interested in the speed of C relative to A. You had said that it would be 0.3c since C originally departed away form A at 0.6c and then returned at 0.9c. But a simple algebraic solution doesn't work. Instead, we can see the answer by using the Lorentz Transformation on the Coordinates of all the events (dots) in the original diagram to create a new diagram moving at -0.6c with respect to the first one. This now becomes the IRF in which the A clock (blue) is at rest:

We can see that the speed of the C clock (black) as it is approaching the A clock (blue) is about -0.65c. Do you know how to determine this speed from the diagram? We can also determine this speed exactly using the relativistic velocity addition formula with v=0.6 and u=-0.9:

s = (v + u)/(1 + vu) = (0.6-0.9)/(1 + (0.6)(-0.9)) = -0.3/(1-0.54) = -0.3/0.46 = -0.652

I have also marked in the Proper Times along each of the clocks' worldlines corresponding to the 1-, 2-, and 3-year intervals of Coordinate Times as per your request. Note that these Proper Times are completely different than those that were determined according to the first IRF.

Even though the Proper Times at which the Coordinate Times are different in the two IRF's, the Proper Times at which identifiable events occur along the worldlines are the same in both IRF's. For example in both diagrams:

1) the last Proper Times at which all three clocks are together is 0.

2) the Proper Times on clocks B (red) and C (black) when they separate is 12 months.

3) the Proper Times on clocks A (blue) and C (black) when they pass each other is 28.8 months for clock A and 22.46 for clock C.

I am really sorry for my lack of precision in my questions. Maybe it's because english is not my native language, maybe lack of experience in discussions on these matters. Probably a bit of both...

But what does happen if you remove the difference in rest-frames by decellerating before comparing the clocks?

Here's another example of lack of precision: are you wanting the C clock to decelerate to the same speed as the A clock when they reunite? But then what about the B clock? It just keeps on going away from the A clock. When do you want it to decelerate and to what speed in which IRF do you want this to happen?