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Posted

Ralfcis has a point he is the only anti-relativity person to survive on this forum, mainly due to diligence and hard work. Others have tried and failed but Ralfcis may actually have a decent claim that relativity is not as correct as we would think.

@VictorMevilVictorMedvil

I'd like to hear your explanation of the story.

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Posted (edited)

@VictorMevilVictorMedvil

I'd like to hear your explanation of the story.

In their own frames with a differing velocities with respect to the Gamma function would be the same depending on their velocities that they had traveled from the Earth, Meaning once Gamma was taken in account they would all match up if another frame of reference was taken such as a star or Mars or something, though all 4 spaceships would have a different time passed depending on their velocity, once taken across the gamma function and a new object's gamma function they would all align to be the same time passed for that object and once taking in account gamma for each other too.

 

3-s2-0-B9781455731411500101-f10-01-97814

Edited by VictorMedvil
Posted

Ok like everyone who has come before you, you are not reading what I'm writing. I've already explained your story. You are stuck on your ideas which are based on almost total ignorance and you're neither willing to admit this or do anything about it. 

 

P.S. Maybe you should stick to just one thread for your stuff.

Because you are not a supporter of relativity, I did not read your article very carefully. But I've emailed you my thoughts last time. Of course, I will study your mathematical derivation in detail when I have time.

Posted (edited)

In their own frames with a differing velocities with respect to the Gamma function would be the same depending on their velocities that they had traveled from the Earth, Meaning once Gamma was taken in account they would all match up if another frame of reference was taken such as a star or Mars or something, though all 4 spaceships would have a different time passed depending on their velocity, once taken across the gamma function and a new object's gamma function they would all align to be the same time passed for that object.

Such a simple scene, but the explanation is so difficult to understand. Use more assumptions to make up for the loopholes caused by previous assumptions.

 

If Newton's classical physics can't explain these phenomena, I will try to believe these hypotheses, but classical physics can completely explain the Morley experiment, light bending, derivation of mass energy equation and so on. Moreover, Einstein himself disagreed with the fact that the speed of light does not change, and the special theory of relativity based on this assumption can imagine how incorrect it is.

Edited by TonyYuan2020
Posted

Such a simple scene, but the explanation is so difficult to understand. Use more assumptions to make up for the loopholes caused by previous assumptions.

 

If Newton's classical physics can't explain these phenomena, I will try to believe these hypotheses, but classical physics can completely explain the Morley experiment, light bending, derivation of mass energy equation and so on. Moreover, Einstein himself disagreed with the fact that the speed of light does not change, and the special theory of relativity based on this assumption can imagine how incorrect it is.

Here is a picture example to make to easier to understand.

New-Bitmap-Image.png

Posted

In their own frames with a differing velocities with respect to the Gamma function would be the same depending on their velocities that they had traveled from the Earth, Meaning once Gamma was taken in account they would all match up if another frame of reference was taken such as a star or Mars or something, though all 4 spaceships would have a different time passed depending on their velocity, once taken across the gamma function and a new object's gamma function they would all align to be the same time passed for that object and once taking in account gamma for each other too.

 

3-s2-0-B9781455731411500101-f10-01-97814

Back to this story, I would like to ask a question:
Before the destruction of the earth, will the time on the four ships be the same all the time?
Posted (edited)

 

Back to this story, I would like to ask a question:
Before the destruction of the earth, will the time on the four ships be the same all the time?

 

No, they will be different by a factor of their velocity with respect to gamma or γ(v).

Edited by VictorMedvil
Posted

No, they will be different by a factor of their velocity with respect to gamma or γ(v).

But special relativity tells us that t '= t * sqrt (1-V ^ 2 / C ^ 2), as if there is no gamma or γ (V) you said.
 
 
V in γ (V) is the velocity relative to what reference? 
Posted (edited)

 

But special relativity tells us that t '= t * sqrt (1-V ^ 2 / C ^ 2), as if there is no gamma or γ (V) you said.
 
 
V in γ (V) is the velocity relative to what reference? 

 

To another object then each other, the gamma function is

codecogseqn-5.gif

 

hqdefault.jpg

Edited by VictorMedvil
Posted

To another object then each other the gamma function is

codecogseqn-5.gif

 

hqdefault.jpg

Let's continue to simplify. Suppose there is nothing else in this space except the earth and the four spacecraft. Will the time of the four spacecraft remain the same?

Posted

Let's continue to simplify. Suppose there is nothing else in this space except the earth and the four spacecraft. Will the time of the four spacecraft remain the same?

No, they will still differ by their velocity as a function of gamma.

Posted (edited)

No, they will still differ by their velocity as a function of gamma.

It's very contradictory. Isn't the factor relative to the earth the same?   γ (V) 
 
 
Now there's only earth, spaceship A, spaceship B, nothing else.
Edited by TonyYuan2020
Posted (edited)

 

It's very contradictory. Isn't the factor relative to the earth the same?   γ (V) 
 
 
Now there's only earth, spaceship A, spaceship B, nothing else.

 

That is a calculus function meaning it is variable upon the velocity of the objects but it all uses the same equation, depending on the value of V, γ will change.

Edited by VictorMedvil
Posted (edited)

That is a calculus function meaning it is variable upon the velocity of the objects but it all uses the same equation, depending on the value of V, γ will change.

But the speed hasn't changed.
 
Can you explain how it has changed?  γ (V)
 
 
As you said,  γ (V)  ,it's a function of V, so V doesn't change. How can it change? Is there any other factor of change at work? Who are the factors of this change?
Edited by TonyYuan2020
Posted (edited)

 

But the speed hasn't changed.
 
Can you explain how it has changed?  γ (V)
 
 
As you said,  γ (V)  ,it's a function of V, so V doesn't change. How can it change? Is there any other factor of change at work? Who are the factors of this change?

 

Take Calculus and get back with me you don't have the reference knowledge to understand special relativity, link = https://www.khanacademy.org/math/calculus-1

Edited by VictorMedvil
Posted (edited)

Take Calculus and get back with me you don't have the reference knowledge to understand special relativity, link = https://www.khanacademy.org/math/calculus-1

It is often said that the displacement can be obtained by (Calculus ) integrating the velocity function v (t) in the time domain.

 

So I want to know here is that there's something (Calculus ) integrating in the velocity domain? Or is speed (Calculus ) integral on something domain else?
 
 
γ (V)  = Calculus(x*dv)?   or    γ (V)  = Calculus(v*dx)? what is x?
 
first γ (V) =0, because of V is a constant, so dv = 0. So what is the x in the second?
Edited by TonyYuan2020

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