The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Debate => Topic started by: OccamsRazor on August 15, 2008, 08:22:05 AM
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I'll play along, even though I believe the majority of the people are FE's because they just want the attention, and they do not actually believe this theory.
But for those that believe, here is a simple question I have after browsing your FAQ
Where does this Univeral Acceleration Force come from?
I can't believe that any true FE's know much about physics so I will try to be as simple as possible.
Force = Mass x Accelartion (I hope you all believe in that).
Therefore if the Earth (with mass A is accelerating upwards at 9.8 meter per second squared), there must be a force acting upon it, which I am guess FE's call the Universal Acceleration Force.
Now, in the FAQ, it was asked how comes the earth is not moving at the speed of light.
Quick numbers. The speed of light is 300,000 km/s or 300,000,000 m/s (300 million)
From a standstill, it would take the "flat earth" 30,612,244 seconds or 510,204 minutes or 8503 hours, or 354 days, or roughly 1 year to reach the speed of light.
I am not going debate the age of the earth, but FE's have to agree that we are at the very least, we must be traveling 2008 times the speed of light, and adding speed every second.
Now, the FAQ says that FE'rs believe in Special Relativity, that nothing can reach the speed of light, and that Special Relativity somehow proves we can constantly accelerate and not reach the speed of light.
So I thought, at least for the FAQ sake, I would try to explain special relativity very simply.
E=mc squared. Or Energy = Mass x speed of light squared.
Meaning a lot of energy can create a little mass.
When you break the equation down, the special relativity states that as an object moves faster (approaches the speed of light), the mass increases. Mass being the resistance to acceleration as we saw earlier in Force = Mass x Acceleration.
So this Universal Force Acceleration (is this also infinitely increasing as well?) Because Special Relativity says the Flat Earth and the mass on it is increasing.
So the question I have is where is the energy coming from, and how much is out there.
Especially figuring that there is not enough energy in the Universe to make a mass object reach the speed of light, as the closer you get to c (speed of light) the magnitude of the mass exponentially increases.
FE'rs don't believe in the Earth's Gravity. They must not believe in Special Relativity, because if they do the above starts becoming impossible. Please just discard Special Relativity, I don't know why your FAQ hangs on, if you throw grafity out the window, why not SR?
So the next question is based on throwing out SR?
Do FE'rs believe in the speed of light?
Light (or photons) have no mass. You can't accelerate them. so if the Earch now moving at a minimum of 2008 times the speed of light.
If you took a flashlight, and pointed at your face you would not see the light. The light doesn't move with the Earth, it has no mass you can't speed up something that doesn't have mass. So when the switch is turned on the flashlight, the body, and the earth are moving at 2008 times the speed of light (and moving faster), but the light will be moving at 1 times the speed of light. We will be 2008 times ahead of it you can't see it.
Well, enjoy your rebuttals. This is just one aspect of about 300 that can disprove Flat Earth Theory.
Occam's Razor - All things being equal, the simplest solution is more often the correct solution.
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From a standstill, it would take the "flat earth" 30,612,244 seconds or 510,204 minutes or 8503 hours, or 354 days, or roughly 1 year to reach the speed of light.
What equation did you use in this calculation?
but FE's have to agree that we are at the very least, we must be traveling 2008 times the speed of light
They certainly do not have to agree to this.
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Special Relativity forbids the Earth from surpassing the speed of light.
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9.8 meter per second squared. Means in one second we are traveling 9.8m/s and in 2 seconds we are traveling at 19.6 m/s ...and so on.
speed of light is 300,000 kilometer/second or 300,000,000 meters/second
Take 300 million divided by 9.8 and you get the seconds it would take a constant accelerating body going 9.8 meters per second squared to hit 300,000,000 meters per second,
or
the speed of light.
It is roughly 1 year.
Special Relativity does not prove the earth can not reach the speed of light. Please show me the proof?
Special Relativity PROVES that as a an object approaches the speed of light, the mass increases. Increases exponentially, meaning by a number so large it is unfathomable. Meaning the Force now needed to keep up the "Acceleration" is now unfathomable.
However, FE'rs say the earth is always going 9.8 meters per second squared, that is how they explain that there is no gravity, we just feel the acceleration of the earth.
I hate to break it to all you Flat Earthers but speed is speed. You guys don't want to believe in gravity, so therefore you have to say we are acceleratin up.
We can't just decrease acceleration as we approach the speed of light, because "Special Relativity" says we can't reach the speed of light. Because if the acceleration decrease the force that the flat earther's claim is gravity, will be less as well.
Bottom line. You Flat Earthers need to completely throw Special Relativity out. You can't believe in it. If you did, then the Earth is not accelerating up.
I SUGGEST YOU FIND ANOTHER WAY to explain the 'phenomenon' we Round Earther's call gravity.
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9.8 meter per second squared. Means in one second we are traveling 9.8m/s and in 2 seconds we are traveling at 19.6 m/s ...and so on.
speed of light is 300,000 kilometer/second or 300,000,000 meters/second
Take 300 million divided by 9.8 and you get the seconds it would take a constant accelerating body going 9.8 meters per second squared to hit 300,000,000 meters per second,
or
the speed of light.
It is roughly 1 year.
Special Relativity does not prove the earth can not reach the speed of light. Please show me the proof?
Special Relativity PROVES that as a an object approaches the speed of light, the mass increases. Increases exponentially, meaning by a number so large it is unfathomable. Meaning the Force now needed to keep up the "Acceleration" is now unfathomable.
However, FE'rs say the earth is always going 9.8 meters per second squared, that is how they explain that there is no gravity, we just feel the acceleration of the earth.
I hate to break it to all you Flat Earthers but speed is speed. You guys don't want to believe in gravity, so therefore you have to say we are acceleratin up.
We can't just decrease acceleration as we approach the speed of light, because "Special Relativity" says we can't reach the speed of light. Because if the acceleration decrease the force that the flat earther's claim is gravity, will be less as well.
Bottom line. You Flat Earthers need to completely throw Special Relativity out. You can't believe in it. If you did, then the Earth is not accelerating up.
I SUGGEST YOU FIND ANOTHER WAY to explain the 'phenomenon' we Round Earther's call gravity.
yeah, i did this one already.http://theflatearthsociety.org/forum/index.php?topic=3152.msg28574#msg28574 (http://theflatearthsociety.org/forum/index.php?topic=3152.msg28574#msg28574)
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Special Relativity does not prove the earth can not reach the speed of light. Please show me the proof?
w=(u+v)/(1+uv/c^2)
Where u is the earth's current velocity, v is 9.8m/s (due to the earth accelerating at 9.8m/s^2 for one second), c is the speed of light, and w is the new velocity.
Start at u=0 and let me know how long it takes to reach the speed of light.
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'u' can't be 0 as it's not possible for anything to be absolutely stationary.
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Use 1 then. Use any number less than the speed of light.
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Fine, start at u=1 then.
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Shouldn't be to much trouble with a relevant formula.
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I gave you the relevant formula.
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How can you use acceleration as a fixed velocity?
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What?
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v makes no sense. v is the figure traveling away from u in this case. It can't be 9.8 as it never could have been 0 at the start of that second, which you seem to have arbitrarily made up. I don't see anywhere in that formula a provision for using acceleration as a fixed velocity on the premise that you've made up a time for it to accelerate.
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It is the addition of velocities. One is the velocity of the Earth, the other is the velocity the Earth would gain from accelerating at 9.8m/s^2 for one second from proper acceleration.
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It is the addition of velocities. One is the velocity of the Earth, the other is the velocity the Earth would gain from accelerating at 9.8m/s^2 for one second from proper acceleration.
and what will be the proper acceleration from that FoR?
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9.8m/s^2, obviously.
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9.8m/s^2, obviously.
So for any FoR the acceleration will be 9.8m/s2
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Uh, no. That is the proper acceleration. Which is what you asked for. ???
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Uh, no. That is the proper acceleration. Which is what you asked for. ???
my mistake I meant proper as in correct not as in proper
so what will the acceleration to be from any FoR?
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my mistake I meant proper as in correct not as in proper
You meant 'proper' not as in 'proper'? Wow.
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my mistake I meant proper as in correct not as in proper
You meant 'proper' not as in 'proper'? Wow.
So what acceleration gets used for any FoR
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Acceleration for what?
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Acceleration for what?
if it is too complicated for you I can slow down a bit
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Well, it would help if you knew how to communicate.
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Special Relativity does not prove the earth can not reach the speed of light. Please show me the proof?
w=(u+v)/(1+uv/c^2)
Where u is the earth's current velocity, v is 9.8m/s (due to the earth accelerating at 9.8m/s^2 for one second), c is the speed of light, and w is the new velocity.
Start at u=0 and let me know how long it takes to reach the speed of light.
Since V in your equation is based upon the acceleration of the earth what velocity (acceleration) should we use for and FoR
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Since V in your equation is based upon the acceleration of the earth what velocity (acceleration) should we use for and FoR
Well, it would help you knew how to communicate.
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Since V in your equation is based upon the acceleration of the earth what velocity (acceleration) should we use for and FoR
Well, it would help you knew how to communicate.
I love it when some pokes fun of someones grammer with horrible grammer of their own
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I love it when some<one> pokes fun of someones grammer with horrible grammer of their own
Now about that acceleration thing...
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I love it when some<one> pokes fun of someones grammer with horrible grammer of their own
Now about that acceleration thing...
you are the one that is saying that we can use the proper acceleration felt on earth and just use it for any inertial FoR
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Right...that is the point. ???
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Right...that is the point. ???
Since when can you do that?
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Because the acceleration is...proper...
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Because the acceleration is...proper...
So if I am in a different FoR it will always be 9.8m/s/s in relation to that new FoR
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You don't actually know what proper means, do you?
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You don't actually know what proper means, do you?
It looks like you do not know what it means
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So that is a 'no', then?
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So that is a 'no', then?
So your belief is that a0 is always equal to a?
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Where does this Univeral Acceleration Force come from?
Me me me! Pick me! *raises hand*
Its dark matter!
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So your belief is that a0 is always equal to a?
Proper anything is always equal to the quantity measured in the proper frame.
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So your belief is that a0 is always equal to a?
Proper anything is always equal to the quantity measured in the proper frame.
No word games here, I will ask a simple question:
Do you believe that a0 is always equal to a
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The proper acceleration will always be the acceleration measured in the proper frame. That is about as clear as it can get. Do you need me to slow down so you can catch up?
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The proper acceleration will always be the acceleration measured in the proper frame. That is about as clear as it can get. Do you need me to slow down so you can catch up?
So how does that answer my question
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It answers it perfectly.
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It answers it perfectly.
Let me make sure I understand your belief here perfectly
Proper acceleration that we feel on the surface of the earth will always be measured to be the same regardless of the FoR that we are at. Be it on the surface, 100mi away, 3Ly away or looking at it from different directions the acceleration will be always me measured a 9.81m/s/s
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No.
Like I said, you don't know what the definition of proper is.
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No.
Like I said, you don't know what the definition of proper is.
I know what proper is perfectly fine, you are the one making the claim that proper acceleration will be the same for all observers
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No.
Like I said, you don't know what the definition of proper is.
I know what proper is perfectly fine, you are the one making the claim that proper acceleration will be the same for all observers
He's just going to keep avoiding the issue with semantics no matter how you phrase it, it will never be "answered" because they don't know
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No.
Like I said, you don't know what the definition of proper is.
I know what proper is perfectly fine, you are the one making the claim that proper acceleration will be the same for all observers
He's just going to keep avoiding the issue with semantics no matter how you phrase it, it will never be "answered" because they don't know
That is what I am trying to show, that when it comes down to it he does not as much as people give him credit for
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No.
Like I said, you don't know what the definition of proper is.
I know what proper is perfectly fine, you are the one making the claim that proper acceleration will be the same for all observers
Proper is only proper in the proper FoR.
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Since V in your equation is based upon the acceleration of the earth what velocity (acceleration) should we use for and FoR
And just for the record, that's not his equation, that's the equation.
(http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/imgrel/evel1.gif)
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I know what proper is perfectly fine, you are the one making the claim that proper acceleration will be the same for all observers
I have not claimed that at all. Not even once. You are the one having the problem understanding what proper means.
That is what I am trying to show, that when it comes down to it he does not as much as people give him credit for
And in the end, all you are accomplishing is making yourself out to be an idiot. Talk about a plan blowing up in your face...
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I know what proper is perfectly fine, you are the one making the claim that proper acceleration will be the same for all observers
I have not claimed that at all. Not even once. You are the one having the problem understanding what proper means.
That is what I am trying to show, that when it comes down to it he does not as much as people give him credit for
And in the end, all you are accomplishing is making yourself out to be an idiot. Talk about a plan blowing up in your face...
So how about you clearly answer the question then, for an observer not on the earth what value for acceleration should they use to calculate the velocity of the earth from their FoR
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I wouldn't expect a decent answer. He gave a cop out answer in my thread as well.
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I asked this:
So your belief is that a0 is always equal to a?
Proper anything is always equal to the quantity measured in the proper frame.
No word games here, I will ask a simple question:
Do you believe that a0 is always equal to a
And You gave this as your answer
The proper acceleration will always be the acceleration measured in the proper frame. That is about as clear as it can get. Do you need me to slow down so you can catch up?
I see no misunderstanding on our part, either you did not see the subscript by the first "a" or you did not understand the difference between the 2 terms, and since other people seemed to recognize the term we must figure the second condition is the true statement.
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He isn't saying that a0=a in all reference frames. You can use proper acceleration to obtain observed acceleration in any reference frame given the correct variables.
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He isn't saying that a0=a in all reference frames. You can use proper acceleration to obtain observed acceleration in any reference frame given the correct variables.
I did not ask him if you could relate proper acceleration to an observed acceleration in any other FoR. I asked him if he believed they are equal, but instead of answering the question he just gave me a definition of proper acceleration again which really has nothing to do with the question put to him
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I would take that as a no.
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I love this thread, it's facinating, but it would help the viewing audience if you would list your credentials. MIT? Cal-Tech? CERN? What is your background/education in these fields? I'm dying to know. FE and RE alike.
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I would take that as a no.
Then why does he use it as a constant? for any FoR
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Okay, so can somebody explain to me what "proper acceleration" is and how it differs between FoRs, inertial or non-inertial (I really wouldn't know if they made a difference lol)
Explain it to me the same way you would to anyone who hasn't studied physics past the twelfth grade, FE or RE. Honestly I'm just curious if both sides will give me similar explanations.
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The only constant in that equation is c.
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Proper is what you are personally experiencing in your own FoR. Due to relativistic effects the time, acceleration, what have you, that are observed in a different FoR can change unless they are invariant.
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Very basic:
Proper acceleration is the acceleration you feel ie: you step on the gas in your car but the acceleration that another person measures that is not in the car may not be the same measurement.
Now this is taking place at very high speeds (close to the speed of light, denoted by "c")
To relate to FE theory the earth is experiencing an acceleration of 9.8m/s2 but a person outside of the earth would not measure that acceleration
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The only constant in that equation is c.
Not according to theengineer v is also a constant
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What exactly does v symbolize if TheEngineer is saying it equals 9.8m/s. 9.8 starting from what point, exactly? Would v not change if we started with a different value for u? I've got to assume v increases by t 9.8, then. So now I'm confused about that 1uv part. What's the 1 for?
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v is 9.8m/s (due to the earth accelerating at 9.8m/s^2 for one second
This is him picking an arbitrary point in time to make the calculation easier. You can choose any moment in time you wish. By the way, i'm not trying to defend Engy, he can do that himself. I'm just stating what he probably meant.
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What exactly does v symbolize if TheEngineer is saying it equals 9.8m/s.
velocity
Would v not change if we started with a different value for u?
yes
So now I'm confused about that 1uv part. What's the 1 for?
As far as derivation of the equation, you should find a reliable source like a book to explain it.
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I know what proper is perfectly fine, you are the one making the claim that proper acceleration will be the same for all observers
I have not claimed that at all. Not even once. You are the one having the problem understanding what proper means.
That is what I am trying to show, that when it comes down to it he does not as much as people give him credit for
And in the end, all you are accomplishing is making yourself out to be an idiot. Talk about a plan blowing up in your face...
You're a dick... Yes, keep doing what you do because we all know you are only trying to raise your post count. Grow up.
EDIT: This is to the engineer (BTW).
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w=(u+v)/(1+uv/c^2)
Where u is the earth's current velocity, v is 9.8m/s (due to the earth accelerating at 9.8m/s^2 for one second), c is the speed of light, and w is the new velocity.
Start at u=0 and let me know how long it takes to reach the speed of light.
Looks like he is saying v=9.8m/s, the only thing that should change according to him is u and that is based upon w
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cbarnett, if the earth is accelerating at 9.8ms/s, how long will it take until we reach the speed of light?
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cbarnett, if the earth is accelerating at 9.8ms/s, how long will it take until we reach the speed of light?
it will never reach the speed of light, if you solve the equation properly you will find that the acceleration does not remain constant
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Along with that, no matter how long we accelerate, light will always be traveling 300,000,000 m/s faster than us.
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w=(u+v)/(1+uv/c^2)
Where u is the earth's current velocity, v is 9.8m/s (due to the earth accelerating at 9.8m/s^2 for one second), c is the speed of light, and w is the new velocity.
Start at u=0 and let me know how long it takes to reach the speed of light.
Looks like he is saying v=9.8m/s, the only thing that should change according to him is u and that is based upon w
That is what I am saying.
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So how about you clearly answer the question then, for an observer not on the earth what value for acceleration should they use to calculate the velocity of the earth from their FoR
To find the velocity of the FE? 9.8m/s^2. But I've said that already.
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w=(u+v)/(1+uv/c^2)
Where u is the earth's current velocity, v is 9.8m/s (due to the earth accelerating at 9.8m/s^2 for one second), c is the speed of light, and w is the new velocity.
Start at u=0 and let me know how long it takes to reach the speed of light.
Looks like he is saying v=9.8m/s, the only thing that should change according to him is u and that is based upon w
That is what I am saying.
so what if I want to use any inertial FoR?
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Go right on ahead.
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Go right on ahead.
Same exact equation then?
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So how does light accelerate at the same rate as us? In order for it to constantly be 300,000,000m/s faster than us with our ever-increasing velocity, naturally light would need to accelerate at 9.8m/s2? Yet a = (V2 - V1) / t. If c is a constant, how does light maintain a constant speed relative to us when our own speed is constantly increasing? By your own admission, even light can not move faster than the speed of light. I can only imagine light seeming to move slower, and along with that, the Flat Earth's "gravity" (or whatever you call your equivalent) decreasing, as we can never reach the speed of light. My scale isn't showing reduced weight, though, nor do I seem to be experiencing slowing time. I guess what I'm really looking for is an explanation to perceived weight and light, keeping in mind that relative to us, a is constant, and in any inertial frame of reference V is aymptotal to c, translating to ever decreasing acceleration.
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My scale isn't showing reduced weight, though, nor do I seem to be experiencing slowing time.
That's because in your FoR it doesn't change.
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If c is a constant, how does light maintain a constant speed relative to us when our own speed is constantly increasing?
Accelerating causes our local space-time to become skewed relative to an outside observer.
(http://casa.colorado.edu/~ajsh/sr/lor5f.gif)
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So you're saying that our space-time gets warped. Like the space expanding between us is angled or something? The interpretation I'm coming up with now is that our vector starts shifting and the y-component of our acceleration decreases with time. I've got to assume that's wrong though, as we're supposed to be accelerating upward. Unless it's that angle that makes it look like we're still accelerating upwards. But once again, that's gotta be wrong because we'd be experiencing a pull off to the side.
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So you're saying that our space-time gets warped. Like the space expanding between us is angled or something? The interpretation I'm coming up with now is that our vector starts shifting and the y-component of our acceleration decreases with time. I've got to assume that's wrong though, as we're supposed to be accelerating upward.
This isn't FET, this is relativity, so if you think it is wrong take it up with Einstein. The direction "upward" is a relative term (this may sound like a cop out, but I promise it isn't). In light cone diagrams the x,y, and z axis are combined into one so direction is irelevant.
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I meant that my interpretation must have been wrong. Was it?
PS: Nice quote on the bottom, there.
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Same exact equation then?
Uh, yea. ???
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Same exact equation then?
Uh, yea. ???
I wonder why particle physics is considered difficult then
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I fail to see why this is such a difficult thing for you to understand. The proper acceleration is 9.8m/s^2. For everyone.
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I fail to see why this is such a difficult thing for you to understand. The proper acceleration is 9.8m/s^2. For everyone.
Uh, no
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Really? How so?
Let me ask you this:
A person traveling at .5c will see the watch on his wrist tick at what speed: Normal, slow or fast?
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I meant that my interpretation must have been wrong. Was it?
This might help.
http://casa.colorado.edu/~ajsh/sr/paradox.html (http://casa.colorado.edu/~ajsh/sr/paradox.html)
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Really? How so?
Let me ask you this:
A person traveling at .5c will see the watch on his wrist tick at what speed: Normal, slow or fast?
and the person from any outside inertial FoR will see what?
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This is all hearsay anyway as relative physics is just made up by people who want to sound clever. Newtonian physics is the only one that actually works.
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This is all hearsay anyway as relative physics is just made up by people who want to sound clever. Newtonian physics is the only one that actually works.
You, sir, have just been sig'd.
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I'm Honored!
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and the person from any outside inertial FoR will see what?
Please answer my question.
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and the person from any outside inertial FoR will see what?
Please answer my question.
A person in proper time will see his watch tick at normal speed, now how would that person watch tick compared to any other FoR
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A person in proper time will see his watch tick at normal speed
Exactly. I am glad you have seen your error.
now how would that person watch tick compared to any other FoR
Differently.
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A person in proper time will see his watch tick at normal speed
Exactly. I am glad you have seen your error.
now how would that person watch tick compared to any other FoR
Differently.
When did I ever claim that a watch in proper time would not tick normal?
so if the watch will tick differently then how is the acceleration measured to be the same as in the proper FoR?
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When did I ever claim that a watch in proper time would not tick normal?
Applied to acceleration:
I fail to see why this is such a difficult thing for you to understand. The proper acceleration is 9.8m/s^2. For everyone.
Uh, no
so if the watch will tick differently then how is the acceleration measured to be the same as in the proper FoR?
Who said it would?
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When did I ever claim that a watch in proper time would not tick normal?
Applied to acceleration:
I fail to see why this is such a difficult thing for you to understand. The proper acceleration is 9.8m/s^2. For everyone.
Uh, no
so if the watch will tick differently then how is the acceleration measured to be the same as in the proper FoR?
Who said it would?
you did.
How Is acceleration measured?
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you did.
Hmm, that's strange, I don't remember doing such a thing. Perhaps you can show me where I did so.
I am still waiting for you to reconcile this:
I fail to see why this is such a difficult thing for you to understand. The proper acceleration is 9.8m/s^2. For everyone.
Uh, no
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you did.
Hmm, that's strange, I don't remember doing such a thing. Perhaps you can show me where I did so.
I am still waiting for you to reconcile this:
I fail to see why this is such a difficult thing for you to understand. The proper acceleration is 9.8m/s^2. For everyone.
Uh, no
a=a0-a1
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you did.
Hmm, that's strange, I don't remember doing such a thing. Perhaps you can show me where I did so.
I am still waiting for you to reconcile this:
I fail to see why this is such a difficult thing for you to understand. The proper acceleration is 9.8m/s^2. For everyone.
Uh, no
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you did.
Hmm, that's strange, I don't remember doing such a thing. Perhaps you can show me where I did so.
I am still waiting for you to reconcile this:
I fail to see why this is such a difficult thing for you to understand. The proper acceleration is 9.8m/s^2. For everyone.
Uh, no
SO now that I have shown that every acceleration may not always be the same for all observers lwt us assume that the acceleration of the observer is zero. he will then track the acceleration of the body like this:
alpha=1/{1-(v/c)2}
so start at v=9.8m/s and tell me what happens to the acceleration as it approaches the speed of light
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SO now that I have shown that every acceleration may not always be the same for all observers
Sorry, when was that an issue?
so start at v=9.8m/s and tell me what happens to the acceleration as it approaches the speed of light
It decreases. But I've said that already.
You really need to think these arguments out better. All you are doing is making yourself out to be an idiot.
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The proper acceleration is 9.8m/s^2. For everyone.
And when have you said that acceleration decreases
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In your very own thread:
The acceleration decreases over time.
It must suck for you to be proven an idiot all the time.
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In your very own thread:
The acceleration decreases over time.
It must suck for you to be proven an idiot all the time.
In your very own thread:
The acceleration decreases over time.
It must suck for you to be proven an idiot all the time.
So from 2 months ago you found a comment you made that may have absolutely nothing to do with the topic at hand, that is your proof?
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So from 2 months ago you found a comment you made that may have absolutely nothing to do with the topic at hand, that is your proof?
It has everything to do with the topic at hand. :-\
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So we started at 9.8m/s, what are we at now?
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Less than the speed of light.