Spacetime Explained

[lixluke]

An observer has no mass.
As an observer approaches C, time slows down relative to the observer.
When an observer travels at C, time stops relative to the observer.
When an observer is at rest, time is infinite?
When an observer exceeds light speed, time reverses.
An observer traveling anywhere in space at light speed will arrive there instantly in referance to the observer.

Imagine standing on a platfrom in empty space. The platform is accellerating upwards at 9.8m/s2. We will be standing on the platform as if standing on Earth. We are moving upwards at the speed of the platform. If we are standing on Earth, we can pretty much say that we are accellerating upwards at 9.8m/s2. And time flows accordingly.

The faster an observer travels, the slower time moves in reference to the observer. V = D/T. But V and T have an inverse relationship. Let's imagine that V has a direct inverse relationship with time. Thus, V = 1/T.

D/T = 1/T
D = 1

What does this mean? It just means that D is constant. If D is constant and V is constant (C), then T must be constant.

It might seem that the existence of matter bends the flow of spacetime around it. But not matter in a solid sense. Matter in the sense of energy bending the spacetime flow around it. However, it might be more accurate to say that the bending of the spactime flow produces energy (matter)?

Say we have 3 balls in space. They move at different directions and speeds. Each ball observes itself as standing still while the other 2 balls move in space at a particular velocity. Each ball is a ripple of energy. Like a cross section of thread on a fabric. Perhaps there is some point of singualrity that exists as the universal frame of reference for all spactime?

 

[CheskiChips]

Wow, I think you just solved the Unified Field Theory problem. I can't believe no one thought of D=1!

 

[lixluke]

I didn't solve anything. I'm just throwing out facts to lead to a logical solution.

 

[Doreen]

An observer has no mass.
As an observer approaches C, time slows down relative to the observer.

I think there is a problem here, already, but I am not exactly sure what your wording means. The observer does not experience time slowing down, but rather if s/he were to return to the company of those who were not traveling near C, he would find that less time passed for him or her.
But maybe this is what you meant.

 

[Dywyddyr]

I didn't solve anything. I'm just throwing out facts to lead to a logical solution.

Re-check your "facts".
Time doesn't slow down for the moving observer, it slows down as seen by a slower/ non-moving observer.
http://en.wikipedia.org/wiki/Time_dilation

 

[lixluke]

I don't think so. If a ball is traveling at C. And I am right behind it traveling towards it Time must slow down for me in order for me to perceive the ball in front of me traveling at C.

 

[Dywyddyr]

It doesn't matter what you think.
Time remains the same for each observer as seen by themselves.
It doesn't matter how fast I'm going, my clock still runs at the same rate for me, but as seen by observers at different speeds then my clock will vary.

(1) In the case that the observers are in relative uniform motion, and far away from any gravitational mass, the point of view of each will be that the other's (moving) clock is ticking at a slower rate than the local clock. The faster the relative velocity, the more is the rate of time dilation. This case is sometimes called special relativistic time dilation. It is often interpreted as time "slowing down" for the other (moving) clock. But that is only true from the physical point of view of the local observer, and of others at relative rest (i.e. in the local observer's frame of reference). The point of view of the other observer will be that again the local clock (this time the other clock) is correct, and it is the distant moving one that is slow. From a local perspective, time registered by clocks that are at rest with respect to the local frame of reference (and far from any gravitational mass) always appears to pass at the same rate.[1]

From the link above.

 

[lixluke]

Right. From the frame of reference of the observer, spacetime is always at rest. The observer's position in space is always 0. And the observer's perception of time is always constant

For example, there is a ball at some distance in front an observer, and the ball takes off moving away from from the observer at a velocity of C. Becasue the observer is at rest, he will see the ball moving away at C. At the same time, the ball will see the observer moving away in the opposite direction at C while perceiving itself as being at rest.


If they both take off at the same time in the same direction at C, the ball in front will be at rest relative to the observer. And the observer will be at rest relative to the ball.

What they both take off at the same time in the same direction. But the ball takes off at C, and the observer takes off at X < C? Does the observer see the ball in front traveling at C-X?

What if they both take off at C velocity, but in opposite directions? Do each of them see one another traveling at 2C?

 

[Dywyddyr]

Right. From the frame of reference of the observer, spacetime is always at rest. The observer's position in space is always 0. And the observer's perception of time is always constant

Which contradicts your first post.

What if they both take off at C velocity, but in opposite directions? Do each of them see one another traveling at 2C?

Closing speeds

An observer may conclude that two objects are moving faster than the speed of light relative to each other, by adding their velocities according to the principle of Galilean relativity.

For example, two fast-moving particles approaching each other from opposite sides of a particle accelerator will appear to be moving at slightly less than twice the speed of light, relative to each other, from the point of view of an observer standing at rest relative to the accelerator. This correctly reflects the rate at which the distance between the two particles is decreasing, from the observer's point of view and is called the closing speed. However, it is not the same as the velocity of one of the particles as would be measured by a hypothetical fast-moving observer traveling alongside the other particle. To obtain this, the calculation must be done according to the principle of special relativity. If the two particles are moving at velocities v and −v, or expressed in units of c, β and −β, where

then this relative velocity (again in units of the speed of light c) is

which will always turn out to be less than the speed of light, regardless of the velocities of the two particles.

http://en.wikipedia.org/wiki/Faster-than-light

 

[lixluke]

When an observer travels at C, time stops relative to the observer.

 

[Dywyddyr]

When an observer travels at C, time stops relative to the observer.

No.
The clock rate stays the same.
As posted in #7 and as you yourself said:

From the frame of reference of the observer, spacetime is always at rest. The observer's position in space is always 0. And the observer's perception of time is always constant

 

[kmguru]

I thought time moves at the same rate while you are inside a spaceship that is traveling at the speed of light - or better yet, if your space you are in is moving faster than the speed of light. Or did I miss something that is new?

 

[lixluke]

If an observer departs from a position moving at C, not a split second will pass for the observer when he arrives at his destination no matter how far away it is.

 

[CheskiChips]

Fine lixluke, I'll be nice, Dywyddyr is either dense or pulling your leg.

It's a matter of semantics. Since neither clock is definite...if you're the one traveling and you want act as if the people not traveling have the clock of perspective then you're "slowing down". If you you have the clock of perspective then you're speeding up. If you have no perspective, you'd sense no change - which is Dywyddyr's unrefined point.

 

[AlphaNumeric]

An observer has no mass.
As an observer approaches C, time slows down relative to the observer.

The only velocity an object with no mass can move at is the speed of light. It follows from $$-m^{2} = p^{\mu}p^{\nu}\eta_{\mu\nu} = 0$$. So either an observer has mass and moves slower than light or has no mass and moves at the speed of light.

Let's imagine that V has a direct inverse relationship with time. Thus, V = 1/T.

No, thus $$V \propto \frac{1}{T}$$ and if the constant of proportionality is k then $$V = \frac{k}{T}$$. Please tell me you're being ironic and not illustrating how you would fail high school physics....

 

[lixluke]

I don't think Dywyddyr is pulling anybody's leg. He's using Wikipedia.

Still, the problem is the observer moving at C in relation to what?

Say there is a ball at a distance from the observer, and the observer travels towards that ball at C. It doesn't matter if the ball is 8 light minutes away or a million lightyears away, he will observe himself hitting that ball instantly.

But what if the observer was at rest, and the ball was traveling towards him at C? Is that not the same thing? Thus, when the ball departs, not a moment would go by for the observer when the ball hits him.

 

[lixluke]

The only velocity an object with no mass can move at is the speed of light. It follows from $$-m^{2} = p^{\mu}p^{\nu}\eta_{\mu\nu} = 0$$. So either an observer has mass and moves slower than light or has no mass and moves at the speed of light.

No, thus $$V \propto \frac{1}{T}$$ and if the constant of proportionality is k then $$V = \frac{k}{T}$$. Please tell me you're being ironic and not illustrating how you would fail high school physics....

The inverse of X is 1/X. In such a case, V = 1/T and T = 1/V.

 

[Dywyddyr]

Fine lixluke, I'll be nice, Dywyddyr is either dense or pulling your leg.

Neither.
How fast does MY clock go according to ME if I am travelling at C?

Re-read his post: he has the observer as the mover.

 

[AlphaNumeric]

No, it means $$V \propto \frac{1}{T}$$. If you drive twice as fast you complete a set journey in half the time. If you take triple the amount of time you drove one third the speed. BUT that doesn't mean [tec]V = \frac{1}{T}[/tex]. Otherwise distance would not come into the definition of velocity, would it? D = VT so $$V = \frac{D}{T}$$. If you have to drive twice as far in the same time you need to drive twice the speed.

The units don't even match. The units of time is 'seconds'. The units of velocity are metres per second. The units of 1/T is 'per second'. So $$V= \frac{1}{T}$$ is wrong by dimensional analysis. An inverse relationship between A and B means $$A = \frac{k}{B}$$ for some quantity k which may or may not have units and other dependencies. $$A = \frac{10}{B}$$ results in A and B having an inverse relationship yet they do not satisfy $$A = \frac{1}{B}$$.

It would seem you did sleep through physics class.

Still, the problem is the observer moving at C in relation to what?

Anything with mass.

 

[lixluke]

Neither.
How fast does MY clock go according to ME if I am travelling at C?

Re-read his post: he has the observer as the mover.

It stands still.