Do we see objects in their past?

Exactly that, I think you may of understood .

 

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What a waste of time.

 

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Years of looking at time dilation, I agree a waste of time.

 

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Two different inertial frames, A & B, can set up many clocks synchronized in their own frame and one and only one in B can briefly be synchronized with one near it in A. You probably falsely believe all can be always synchronized in both A & B. That identical clocks must tick at one rate, independent of the frame they are in. I. e. deny the dilation of time for both viewing the other's clocks. Just as you deny observers each frame find the other's meter sticks are actually too short (vs their meter sticks).

GR has many predictions, and all that have been tested have been found valid - there is no point in my trying to convince you of what all knowledgeable physicists agree on - Deep in your uneducated heart, you know better. So I will be even less active here than I have been.

 

You clearly are not considering the invariant between two point sources, the subject of relativity is obviously beyond your ability.

 

No, actually you seem like the idiot.

 

I thank you for your reply, you also are not considering the invariant between two points sources as shown in the video simulation. Imagine you have two clocks, they are travelling relative to each other

+ve=c

-ve=c

So now instead of imagining the coloured planes to be photons, imagine them are the clocks in motion relative to each other. You can clearly observe both clocks travelling through the invariant of light length between the two point sources . The length between two set points is an invariant and can not contract. If we send a man to the moon , the moon does not get closer to the earth.

 

Sorry interruption at work.
Going back to the point before history....
Let's give Alice and Bob a pair of identical clocks. They stand together and sychronise the clocks. Then Bob walks (slowly) a distance of (say) 300m. I think the spacetime interval is going to be zero (please check all this). If a photon leaves Bob's clock and is detected by Alice, in the photon frame (if allowed) we'd have an elapsed time of zero and a distance of zero so still spacetime interval of zero.

 

Yes, the net difference is 0,


lets say it takes the photons 1's to travel either direction, 1-1=0 net difference.

t1-t2=o

 

No frames are envisaged traveling at c. There is no rest frame for photons, traveling at c. ...

 

I know objects can not travel at c, it was an example, you say there is no rest frame for photons travelling at c, I disagree , you are not accounting for the relatively observed rest frame of c between point sources. space is not visually opaque, space is not visually observed to be moving, photons propagating through space are not visually opaque , all objects in motion are in motion relative to the visual stationary reference frame of space-time and relative to each other.

 

Here's an easy one. Removing the speed of light altogether...

Alice and Bob synchronise clocks and (slowly) move 300m apart.
I'd like to introduce Colin at this stage.
Colin starts off at Bob's location. At 10:00:00.0000000000 am he starts his stopwatch (0.00000000000 seconds) and walks towards Alice at 1m/s. After 300.0000000000 seconds he meets Alice and records what her clock shows.
Does he record 10:05:00.00000000000000 or something else?

Edit... I don't think this shows what I wanted it to show... maybe someone can think of a way to show what I really wanted to show.

 

Unfortunately this forum prohibits the use of Crystal Balls and Tarot Cards, so I don't think anyone is going to be able to figure what you really wanted to show.

 

Fair comment.

Starting with a plausible geometry s²=x²-c²t² and as little rigour as possible...
Let the distance between Alice and Bob be x, the time in Colin's frame be T and in the AliceBob frame be t...
-c²T²=x²-c²t²
x=vt (< that's just Newton)
-c²T²=(vt)²-c²t²
-c²T²=t²(v²-c²)
so
T²=t²(c²-v²)/c²
T²=t²(1-v²/c²)
or
In the ALiceBob frame elapsed time
t = T/(√(1-v²/c²)) where T is the time Colin records.
So now we can let Colin go a bit faster.

 

reaction time would be ten minutes, so instantaneous reaction wouldn't occur until ten minutes after you signed something withsmoke signals about how bored you are. question: the moon is three seconds away?

 

Quick question: Have we invalidated SR yet?

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Having sat through hours and hours of philosophy of time lectures and read a ton of ink spilled on this issue, it really gets me every time someone completely ignorant of the physics and the history of the topic claims that their idea, "may be beyond your thinking."

There is nothing new in ignorance. Nor is it hard to understand.

 

Let the distance between Alice and Bob be x, the time in Colin's frame be T and in the AliceBob frame be t...
-c²T²=x²-c²t²
Really we want Colin to move between Bob and Alice in zero time and see what t is then
try
x²-c²t²=0
giving
t=x/c as an inescapable time difference.
There r problems with this tho'.

 

Let Colin time the journey by using the constant fps of a camcorder.