# Difference between revisions of "Talk:Counterexamples to Relativity"

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But none of these are counterexamples or "discontinuities": they are just a misinterpretation of the formulas. You don't get the momentum of a photon by taking the momentum formula for a mass and setting m=0 and v=c. That's just not what the formula means, or what they are for. This item should also be removed.--[[User:NgSmith|NgSmith]] Tue Dec 15 10:16:21 EST 2009 | But none of these are counterexamples or "discontinuities": they are just a misinterpretation of the formulas. You don't get the momentum of a photon by taking the momentum formula for a mass and setting m=0 and v=c. That's just not what the formula means, or what they are for. This item should also be removed.--[[User:NgSmith|NgSmith]] Tue Dec 15 10:16:21 EST 2009 | ||

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+ | == Counterexample 9 (Jesus action-at-a-distance) == | ||

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+ | The quoted verse doesn't strongly suggest "action-at-a-distance" in the relativistic sense. Light could travel the distances mentioned in the passage in a fraction of a second, which is well within the precision given in the verse (an hour). The verse and relativity are not in contradiction here. This should be removed. |

## Revision as of 23:40, 4 January 2010

Andy, can you clarify #4 for me? I'm not sure I understand it. JacobB 21:50, 28 November 2009 (EST)

- Sure, I welcome discussion of these important points. As I've said, I have an open mind about this and if something is true, then I accept it. But if something is false, I'll criticize it.

- The theory of relativity has taught for decades that as the velocity of a mass increases, then its (scalar) relativistic mass increases per the Lorentzian transformation. Now apply a force ORTHOGONAL to the velocity. Does that force encounter the increased mass, as relativity says, or encounter the rest mass, as logic would dictate?--Andy Schlafly 22:02, 28 November 2009 (EST)

- Ah, I see what you mean. May I suggest a re-wording? "The logical problem of a force which is applied at a right angle to the velocity of a relativistic mass." I think that might be a little clearer than it is currently stated. Your thoughts? JacobB 22:06, 28 November 2009 (EST)

- Please do. Your edits are always welcome, and you've suggested an improvement here. Thank you for making this change.--Andy Schlafly 22:20, 28 November 2009 (EST)

- Why would logic dictate that? Mass is a scalar, and a force from any direction should encounter the same increased mass, not different masses from different directions.

- I suppose that under Newtonian mechanics, a moving object has a velocity of 0 within the plane perpendicular to its line of motion, and any forces operating in that plane will act on the object as if it is at rest. But that's not what
*logic*dictates, that's what the*previous theory*dictates.

- I suppose that under Newtonian mechanics, a moving object has a velocity of 0 within the plane perpendicular to its line of motion, and any forces operating in that plane will act on the object as if it is at rest. But that's not what

- Essentially your counterexample to relativity is that it makes a prediction that contradicts Newton's laws. This is neithe r a contradiction nor a logical problem, and it is should be edited out.NgSmith

- No, it's a logical problem. If you're suggesting that one force can affect the inertial in an entirely independent, orthogonal direction, that's illogical. One thing cannot affect something else that is entirely independent.--Andy Schlafly 15:40, 12 December 2009 (EST)

- Why is that illogical? What logical principle does it violate?

- See, in relativity, orthogonal doesn't
*mean*independent. In relativity, velocity vectors*do not add.*In relativity, the effect of a new force is not independent of the object's existing momentum. And there is nothing illogical about that; it's just a new theory that contradicts the intuition from the previous theory.--NgSmith

- See, in relativity, orthogonal doesn't

- Ng, something cannot be independent (orthogonal) and yet dependent at the same time. Unfortunately, you're arguing with your own theory at this point. Even most relativity promoters have abandoned the position you take here.--Andy Schlafly 21:37, 12 December 2009 (EST)

- It seems that his point is that something can be orthogonal and dependent. I agree: The cross-product of two vectors is orthogonal to both and yet obviously dependent on both. --EvanW 21:41, 12 December 2009 (EST)

- OK, good point, an orthogonal vector can be a function of other orthogonal vectors. But that's a bit different from what we're discussing. Here it's an orthogonal force that is not dependent on anything else, and yet Ng says it encounters relativistic mass due to a different orthogonal force.

- I think relativists have abandoned Ng's position, so he's really arguing with his own side at this point. As a result, I urge him to reconsider his views with an open mind once he confirms that.--Andy Schlafly 21:59, 12 December 2009 (EST)

- First of all, relativity has not "abandoned" the prediction we're talking about. The velocity addition formulas for both parallel and perpendicular velocities have not changed, and they still predict that an orthogonal force will have a harder time accelerating a fast-moving object. Physicists may have changed their informal interpretation of this formula, but not the formula itself, nor its predictions.

- Note also that relativity's prediction can't be all that illogical, because this is what we
*actually observe happening to particles at high speeds.*If you think that fast-moving particles commit some terrible offense against basic logic, take it up with God.

- Note also that relativity's prediction can't be all that illogical, because this is what we

- There is a very simple way to settle this matter: write an encyclopedia article where the material is properly sourced. If this is indeed some counterexample or logical flaw in relativity, then one can easily find a book or paper exposing that flaw, and cite it.--NgSmith Sun Dec 13 17:55:04 EST 2009

- OK, I think I see part of the problem you people are having. The word "independent" has two different meanings. Being
*linearly*independent is a concept from pure mathematics. Being*causally*independent is an unrelated metaphysical concept. Whether a force pushing on something causes it to move, and by how much, is completely, umm, independent of whether the vectors involved are linearly independent (orthogonal). Please try to be very careful about the meanings of the terms. SaraT 17:00, 13 December 2009 (EST)

- OK, I think I see part of the problem you people are having. The word "independent" has two different meanings. Being

- I don't think that's the source of our confusion. I think the main problem is that, according to Newtonian mechanics and thus according to our mechanical intuition, orthogonal things tend to operate independently. Not only that, but a force exerted on an object is usually independent of the object's momentum.

- In relativity, none of these things are true, due to the fact that velocities no longer add like vectors (and thus acceleration no longer incurs a cumulative change in velocity in the usual way.) This is seen as some sort of logical flaw or paradox simply because it contradicts the deeply ingrained intuition that came from the previous theory.--NgSmith Sun Dec 13 18:10:46 EST 2009

## Counterexample 4 (limiting behavior)

For the fourth "counterexample," the author points out that the momentum does not approach the momentum of light as and

Aside from the mathematical sloppiness of taking two independent variables to a limit at the same time, at unspecified rates, these sorts of "discontinuities" can be found in just about any scientific theory. In Newtonian mechanics, for example, take the orbit of a planet as the planet's mass goes to 0. For any nonzero mass the orbit is an ellipse; at m=0 it is suddenly a straight line. Is this a "counterexample" to Newton's laws?

Or in electronics, I=V/R. The limiting case is no voltage, no resistance, no current; but if someone foolishly took V/R as both V and R go to zero, he would get a nonsensical answer. Let them both go at the same rate and you get I=1. Is this a "counterexample" to basic electronics?

Or more to the point, momentum in Newtonian mechanics is , and this also fails to give the momentum of a photon at m=0, v=c. Again, is that a "counterexample" to ? Will we see this entry in a corresponding page of "Counterexamples to Newton's laws?"

But none of these are counterexamples or "discontinuities": they are just a misinterpretation of the formulas. You don't get the momentum of a photon by taking the momentum formula for a mass and setting m=0 and v=c. That's just not what the formula means, or what they are for. This item should also be removed.--NgSmith Tue Dec 15 10:16:21 EST 2009

## Counterexample 9 (Jesus action-at-a-distance)

The quoted verse doesn't strongly suggest "action-at-a-distance" in the relativistic sense. Light could travel the distances mentioned in the passage in a fraction of a second, which is well within the precision given in the verse (an hour). The verse and relativity are not in contradiction here. This should be removed.