That's why the formulas are incorrect. The faster the observer moves, the slower time flows. Thus, at C, time must be 0. Thus D/0 = C which is a paradox. But not really because the whole concept of V=D/T is ineffective. It doesn't work. Furthermore, we will never figure out how spacetime works as long as we appreach matters using ineffective models.
Spacetime Explained
- Thread starter lixluke
- Start date
Sigh.
The only things that can move at the speed of light are massless elementary particles such as photons. Photons are not "observers".
Observers have non-zero mass. They cannot travel at the speed of light, which is an absolute (and the only absolute) in special relativity.
Suppose a crewmember on a non-accelerating spaceship who looks at the spaceship's clock. That person will see the clock as moving at normal speed, regardless of the spaceship's velocity with respect to something else.
Suppose two spaceships start at relative rest. Each spacecraft transmits the time registered on that spacecraft's clock to the other spacecraft. The spacecraft accelerate away from each other for a bit and then drift apart at a constant velocity. Crewmembers on each spacecraft will now perceive the other spacecraft's clock as running slower than their own. It doesn't matter which spacecraft did the accelerating. Each perceives the other's clock as running slow.
The only things that can move at the speed of light are massless elementary particles such as photons. Photons are not "observers".
Observers have non-zero mass. They cannot travel at the speed of light, which is an absolute (and the only absolute) in special relativity.
Suppose a crewmember on a non-accelerating spaceship who looks at the spaceship's clock. That person will see the clock as moving at normal speed, regardless of the spaceship's velocity with respect to something else.
Suppose two spaceships start at relative rest. Each spacecraft transmits the time registered on that spacecraft's clock to the other spacecraft. The spacecraft accelerate away from each other for a bit and then drift apart at a constant velocity. Crewmembers on each spacecraft will now perceive the other spacecraft's clock as running slower than their own. It doesn't matter which spacecraft did the accelerating. Each perceives the other's clock as running slow.
Dywyddyr, I was giving him the benefit of the doubt and assuming the error to be semantics. Turns out he is a crack pot, no offense to you intended.
BARP! EPIC FAIL!
If the ball is moving at C, well, how do you know which direction you are moving in? The total sum of velocity between you and it can be C, and you can only detect your own acceleration, not your speed.
This is the whole point about reference frames, which you don't seem to grasp.
Of course, you can't move towards it, if it is moving at C, as that would require you to exceed C. You can maintain your distance from it, but not move towards it. Unless you meant moving in the same direction?
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How does Dywyddyr's clock know it is moving at C, and not that other objects in the vicinity are moving away from it at C?
How do you reconcile that comment with this one?
My clock always reads the same to me.
Oh dear. Why should D=1? Why not 2 or 3.5 or $$\pi$$?
D must equal 1 because T and V have an inverse relationship.
From the frame of reference of the observer, the observer is standing still. Thus, from the frame of reference of the clock, the clock is standing still.
When the observer is traveling at C, he observers no change in time. There is no observation of any sort of chronological motion. This means if the observer departs at C speed, 0 seconds pass for him when he hits his destination. In fact, in his FOR, he is not moving at all. Thus, it took 0 seconds for his destination to travel to him at C speed.
T and V have a relationship depending on D.
Wrong.
The observer sees no difference at all in the rate of passage of time. His clock runs normally as far as he's concerned: an hour is an hour is an hour.
The observer always sees time flow at the same pace. But when traveling at C, Time isn't flowing, so the observer doesn't see a second pass. He simply arrives at his destination the moment he left.
Yes. It's circular.
Sheer nonsense.
At what point does "time stop flowing" for him?
Where do the equations show this?
At C. What exactly do you think is happening to an observer's clock when he is traveling at a vast distance through space at C? If I travel from earth at C speed, and land 100 light years away, not a second will have passed for me.
Why? How?
According to the traveller it's going at the usual rate.
Wrong.
One more time:
I'm standing still now (in my own reference frame), why isn't my clock?
What usual rate? When time slows down, the traveler cannot notice, so it's always going at its usual rate. At C, time is 0. The observer perceives no movement. At what other rate will the observer perceive his clock movie when traveling at C? Are proclaiming that a year will go by for the observer if he has travelled 1 lightyear away?
So is it slowing down or going at its usual rate?
Wrong.
One second per second... My clock runs normally for me.
One more time:
I'm stationary now, why is my clock still running?
In refeference to the observer, the clock is always moving at its usual rate. So if the observer is traveling faster, he notices his clock ticking the same rate. But in relation to another clock that isn't moving, his clock is moving slower. At C speed, he notices no movement of time because he is chronologically frozen. Meanwhile, the clock outside has moved 1 year when he travels 1 lightyear away.
Doesn't agree with -
Or are you claiming that his clock has the same rate until exactly C?
Why?
Which clock?
No. I am saying that when he travels at C, his own clock is completely stopped. There is no timeflow for an observer moving at C. In one moment, I am departing, in the next moment, I have arrived.
That's what I said you're suggesting.
You've already stated that he sees time running regularly while moving fast
And then you claim it stops completely at C.
Why?
How?
Where's the maths to show this?
(By the way, you won't find any, because you're wrong).