Thermos aquaticus
Well-Known Member
Coming from someone who can't do Jr. High math, I will take that with a grain of salt.
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Seeing things in their past? You are full of beans!
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Coming from someone who can't do Jr. High math, I will take that with a grain of salt.
james blunt
Well-Known Member
So all's you have really got is that you can repeat what google says ?
Let us do the maths of the origin of 1 second
An earth day was measured by rotation relative to the sun. We measured 24 hrs and we split this into the increments for preferred accuracy.
Now the earths circumference is 24,901 mile so let us split that into segments to match the 24 hrs.
So you have successfully equated the speed of time to equal the earths spin speed , well done .
TrueBeliever37
Well-Known Member
Are you a photon? If not, then what a photon experiences does not equate to what you experience.
Also, we have all explained the math in multiple posts. Here it is again:
Δt = the observer time, or two-position time (s)
Δt0 = the proper time, or one-position time (s)
v = velocity (m/s)
c = speed of light (3.0 x 10^8 m/s)
Plug in the speed of light for v and solve the equation. You will find that you get a divide by zero, which is the limit that Polymath keeps talking about. For v approaching the speed of light the time keeps getting smaller and smaller and smaller. Just do the math. The same math applies to distance as well.
If this is simply about what you will or will not believe, then that is on you. Reality is not forced to conform to what you will or will not believe. Either you believe in reality or you don't.
You guys keep trying to say I am talking about what I experience. I am NOT talking about what I experience.
If two photons A & B are in a race. And they are both emitted at the exact same instant, from the same source. And both travel at the same speed of light = c. If photon A has to travel any physical distance you say is out there in space, and photon B has to travel a physical distance that is 10,000 times the physical distance photon A has to travel. Do they tie in the race?
james blunt
Well-Known Member
Obviously not.
You are comparing time to a velocity here. THEY ARE NOT THE SAME.
james blunt
Well-Known Member
Think about the history of science sir, think what they actually did when making a measurement of time. Making the speed of time a velocity is exactly what they did . I didn't compare anything, I am stating what science have done and their big mistake.
When you state those distances, you are using a reference frame. When you ask if they tie, you are using the same reference frame. In that reference frame, they do not tie. THERE IS NO REFERENCE FRAME FOR THE PHOTON.
Good enough?
Thermos aquaticus
Well-Known Member
From your frame of reference, they do not tie each other.
james blunt
Well-Known Member
The background of space underlying the photons in motion is the frame.
Thermos aquaticus
Well-Known Member
This, coming from someone who says to forget about everything we have learned.
james blunt
Well-Known Member
I don't say forget everything you have learnt, I suggest you open up your minds and see what you have learnt. Half of you don't understand reality.
Subduction Zone
Veteran Member
There is no such background. It looks like you believe in the long discarded idea of an "ether".
Thermos aquaticus
Well-Known Member
In reality, multiplying a time by a speed gives you a distance.
Subduction Zone
Veteran Member
Please, you don't understand even high school math. You are in no position to make such a claim.
When you ask this question, you are asking from some reference frame (space and time coordinates for all events). In that frame, they do not arrive at the same time.
They do not tie.
Time is involved. The coordinate time for your reference frame. A *different* reference frame will give different distances and times for the travel of those photons, but there will not be a tie in any reference frame.
But, and this is important, you keep asking what the 'photon experiences'. That can only mean what happens in a frame of reference where the photon is at rest. That's what it means to be a reference frame for something. But guess what? The photon is not at rest in *any* frame! So your question is not meaningful. Understood?
HOWEVER, we can take the limit as v-->c of the experiences of things going slower than light and see what those limits are. In this limit, all distances are 0 and all times are 0. It is a degenerate situation because, once again, there is no frame for the photon.
So all's you have really got is that you can repeat what google says ?
Let us do the maths of the origin of 1 second
An earth day was measured by rotation relative to the sun. We measured 24 hrs and we split this into the increments for preferred accuracy.
Now the earths circumference is 24,901 mile so let us split that into segments to match the 24 hrs.
= 0.28820601851 mile per second of rotation .
So you have successfully equated the speed of time to equal the earths spin speed , well done .
Time does NOT have to be measured by the rotation of the Earth. Once again, you are using a very particular (angular) velocity here.
There is no such frame.
Subduction Zone
Veteran Member
I am pretty sure this is going way over his head. Please note that he does not even understand concepts that I learned in the ninth grade of units in the sciences and the number line in math. He not only does not understand these very basic concepts. He refuses to even try to learn. This is a servere case of Dunning Kruger, good luck but I am done.
james blunt
Well-Known Member
Really?
Yes there is , are you ignoring a Higg's type field then that photon energy traverses through ?
Are you going to carry on repeating your education rather than thinking about it yourself?
Why aren't you been critical in your thoughts?
james blunt
Well-Known Member
I will be honest with you and tell you that you are not thinking but just accepting your interpretation of time that you learnt which is completely wrong.
Time does NOT have to be measured by the rotation of the Earth. Once again, you are using a very particular (angular) velocity here.