OK, so dv/dt has been redefined specifically for the special relativity argument--No, it hasn't. Because we're talking about Special Relativity, the frame of reference is extremely important. The acceleration is constant at 9.81ms^-2 from a non-inertial frame of reference (in this case, the only frame of reference that's easily observable in the real world, e.g. an observer who had just jumped up). The argument of accelerating to the speed of light, however, only becomes at all interpretable in an inertial frame of reference - hence the need to rephrase the equations.
this is a really disingenuous moveNah, it's just physics.
1. The whole point of this page is to show that the earth's acceleration is constant. If the earth's acceleration is dv/dt=g/ɣ^3, it is very clearly not constant. The fact that no one would notice such an obvious contradiction is somewhat alarming to me.Addressed above, not a contradiction. Your username is very applicable.
2. I don't think this is even the correct equation for calculating relativistic acceleration, but my knowledge of special relativity is fairly elementary so I won't argue that point. However, based on my calculations (using separation of variables), the solution given for this differential equation isn't even correct! You can check it yourself by taking the derivative of the "solution"-- it will not give you the original differential equation!Could you show your workings?
3. Here is the conclusion that is arrived at based on these erroneous calculations: "As you can see, it is impossible for dark energy to accelerate the Earth past the speed of light." This statement actually disproves UA-- if the earth's velocity cannot exceed the speed of light, how can it continue accelerating at a constant rate forever? Based on the solution given, as t approaches infinity and v approaches c, it follows that dv/dt must approach 0, meaning that the earth's acceleration is decreasing.This is simply a restatement of your introduction and of point 1. You have no familiarity with Special Relativity, specifically Lorentz transformations, and you're easily confused by multiple frames of reference. You are not arguing against the Flat Earth Theory here, you are arguing against simple physics, which you should have a good grasp of by the time you finished high school.
OK, so dv/dt has been redefined specifically for the special relativity argument--No, it hasn't. Because we're talking about Special Relativity, the frame of reference is extremely important. The acceleration is constant at 9.81ms^-2 from a non-inertial frame of reference (in this case, the only frame of reference that's easily observable in the real world, e.g. an observer who had just jumped up). The argument of accelerating to the speed of light, however, only becomes at all interpretable in an inertial frame of reference - hence the need to rephrase the equations.
If v is indeed the solution of dv/dt, then taking the derivative of v should give us the original differential equation. However, when taking the derivative of the given v, dv/dt=g/(ɣ^3*t^3). Uh-oh, a t^-3 has appeared!2. I don't think this is even the correct equation for calculating relativistic acceleration, but my knowledge of special relativity is fairly elementary so I won't argue that point. However, based on my calculations (using separation of variables), the solution given for this differential equation isn't even correct! You can check it yourself by taking the derivative of the "solution"-- it will not give you the original differential equation!Could you show your workings?
3. Here is the conclusion that is arrived at based on these erroneous calculations: "As you can see, it is impossible for dark energy to accelerate the Earth past the speed of light." This statement actually disproves UA-- if the earth's velocity cannot exceed the speed of light, how can it continue accelerating at a constant rate forever? Based on the solution given, as t approaches infinity and v approaches c, it follows that dv/dt must approach 0, meaning that the earth's acceleration is decreasing.This is simply a restatement of your introduction and of point 1. You have no familiarity with Special Relativity, specifically Lorentz transformations, and you're easily confused by multiple frames of reference. You are not arguing against the Flat Earth Theory here, you are arguing against simple physics, which you should have a good grasp of by the time you finished high school.
You have no familiarity with Special Relativity, specifically Lorentz transformations, and you're easily confused by multiple frames of reference. You are not arguing against the Flat Earth Theory here, you are arguing against simple physics, which you should have a good grasp of by the time you finished high school.
Your failure here basically boils down to the fact that UA would indeed be impossible under classical mechanics, which is clearly your default state of mind when thinking about mechanics. However, classical mechanics is just a simplification of reality which produces reasonably accurate results as long as we're not dealing with (for example) things that approach the speed of light.
If the acceleration changes with speed (as the equation given for dv/dt suggests) it is not constant, regardless of reference frame.It is constant in the reference frame that matters to us locally. It obviously wouldn't be constant in an inertial frame, because we're talking about SR.
There is no Lorentz transform in the calculations done on the Wiki page, we are not talking about multiple frames of reference.I'm sorry, I can't say anything other than "Yes, we are talking about multiple frames of reference, and the Lorentz transformations are referenced directly in the article."
If v is indeed the solution of dv/dt, then taking the derivative of v should give us the original differential equation. However, when taking the derivative of the given v, dv/dt=g/(ɣ^3*t^3). Uh-oh, a t^-3 has appeared!Ah, okay, you're mistaking ∂ for d. I had originally assumed that your notation is incorrect just for the sake of simplicity (or your inability to use the [tex] tag), but it seems that you're actually mixing the two concepts up. That's pretty silly of you.
Do you even math bro?I'm seemingly good enough at it to tell the difference between partial derivatives and total derivatives. That puts me at an unfair advantage over you.
Tsk, tsk, so many assumptions about my background.I'm not making assumptions, I'm drawing conclusions.
Please point out which part of my argument hinges on a classical assumption.That would be points 1 and 3. Your inability to understand the differences between two frames of reference is perfectly understandable if you operate under classical mechanics. It is not at all understandable if you have a rudimentary understanding of SR. Therefore, it follows that you do not have a fundamental understanding of SR and are thus resorting to Newtonian mechanics.
If v is indeed the solution of dv/dt, then taking the derivative of v should give us the original differential equation. However, when taking the derivative of the given v, dv/dt=g/(ɣ^3*t^3). Uh-oh, a t^-3 has appeared!Ah, okay, you're mistaking ∂ for d. I had originally assumed that your notation is incorrect just for the sake of simplicity (or your inability to use the [tex] tag), but it seems that you're actually mixing the two concepts up. That's pretty silly of you.
That would be points 1 and 3. Your inability to understand the differences between two frames of reference is perfectly understandable if you operate under classical mechanics. It is not at all understandable if you have a rudimentary understanding of SR. Therefore, it follows that you do not have a fundamental understanding of SR and are thus resorting to Newtonian mechanics.Without going through all the points you say, I am quite prepared to say that you are correct and the a number of "confused's" points are not valid.