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Thread: Zero speed

  1. #1 Zero speed 
    Forum Bachelors Degree Shaderwolf's Avatar
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    What is zero? aren't we jetting through space at about 9/10 the speed of light? (Like other galaxies?) If we are moving realy quick in one direction, wouldn't our mass decrease? Could we find than, by measuring the change of mass figure out what direction we are going? What is direction anyway? Our only true referance point would be the center of the universe. Everything else is moving blindingly fast (and we could never pinpoint the center of the universe anyway, we're probably spiraling out of controle on a massive scale) If you shot an atom at the opposite direction that we were moving, just enough to see it "stop moving" reletive to the center of the universe (because that's all that counts) what would you see?


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  3. #2  
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    "We" are not moving. All other galaxies are moving away from us. The further they are from us, the faster they move.

    ("We" refers to the Local Cluster, for the nitpickers among us.)
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  4. #3  
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    However since it is all relative, then we are moving. In one of those galaxies moving away from us at 0.9c an alien is looking at us and observing that we are moving away at 0.9c, while his neighbour tells him that we are not moving.
    Since everything is believed to have started from the singularity, then we ar all at the centre - our galaxy and the one retreating .9c, and all the others too.
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    So, if we are moving, (relatively) than would the experiment with the photon (atom, whatever) just show the same results in any direction? Even though we have in a sence made it stop, in a sense, it's jetting away from us at a fraction of light speed? There would be no differance in the change of mass, no matter wich direction it was shot?
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    To my simple mind you are only moving IF you are plotting your distance against time, relative to another object, so to say we are 'jetting' through space you really need to specify relative to what. Relative to 'empty' space we are stationary, relative to to our sun we are moving at around 100KM/sec
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    You have to ask yourself the question, relative to which observer. The speeding spaceship travelling at .9c seems to us to have gained mass. Not so to the daring astronauts on board.
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    Quote Originally Posted by Ophiolite
    The speeding spaceship travelling at .9c seems to us to have gained mass.
    Are you sure about this? My sources tell me this is, at best, an outmoded idea, and at worst, gives the wrong results if you try to incorporate Newton's equalities into Relativity in either of its guises.
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    Not at all sure. I'm merely regurgitating what I have picked up from popular science works on the subject, confirmed by my understanding of explanations on forums such as this by people who seemed to know what they are talking about. I'd be happy to hear a better explanation.
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    Sorry that I can't quite think for myself on this subject yet, but the book I've been using to learn physics says the following:

    m = (1 - u²)^-½

    where m is mass of an object when it is stationary to the observer and u is the speed of the object as a fraction of the speed of light (from the reference point of the observer I assume)

    The reason why some of Newton's math was wrong is because it didn't incorperate reletivistic effects.

    Here is some of my own thinking though:
    Lets say one day you decide to see what it is like to jump off a one million story building. You climb to the top of the building, sneak out a window, and manage to find the perfect place to jump off from. Then you jump, and you feel the gravitational force pulling on you, you hit terminal velocity and no longer feel this gravitational force. You could say though that this entire time you have not moved, but that the gravitational force is pulling the Earth toward you. You are not moving at any velocity, but the building is actually passing by you at an incredible speed. Imagine buildings that move at terminal velocity! You have just enough time to ponder a couple more mysteries of the universe before the Earth smashes into you and instantaneously accelerates you away from it very painfully. You and the Earth now travel together again.
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    Quote Originally Posted by Demen Tolden
    the book I've been using to learn physics says the following:

    m = (1 - u²)^-½

    where m is mass of an object when it is stationary to the observer and u is the speed of the object as a fraction of the speed of light (from the reference point of the observer I assume)
    Something missing from this equation, can you see what it is? Hint: you are trying (I assume) to find the change in mass of a body in motion as compared it's mass when stationary. What's missing?

    I am told it will always be, at best, needless, at worst a failed endeavour to find something called relativistc mass

    The reason why some of Newton's math was wrong is because it didn't incorporate relativistic effects.
    Ah no. There's nothing wrong with Newton's maths or his laws: just it's now known they apply only "locally" i.e. when relativistic effects can be ignored (which is usually). That's what you meant, I suppose, just being pedantic.
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    Well if I were to guess what you are getting at I would guess maybe you mean a sort of control variable (i.e., the "real" mass of the object, or the mass of the object at rest), but then that could be found by entering 0 for u, so I guess I am not quite sure what conclusion you are trying to lead me to.

    There are some things about reletivity that I still don't understand though. Such as if I was in a rocket that shot off from earth, reletive to myself, the whole universe suddenly changed velocity and mass, but a change in mass should be equivilant to a change of energy right? So from a stricly reletive view point either reletivity needs some adjustment or more likely I just need to learn more about it.

    Maybe saying Newton's math was wrong wasn't the best way to express the idea I was trying to communicate. I should have said that it was less general than general reletivity, which is basically what you are saying i think, so it apears we agree. Thanks for pointing out my mistake.
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    Quote Originally Posted by Demen Tolden
    Well if I were to guess what you are getting at I would guess maybe you mean a sort of control variable (i.e., the "real" mass of the object, or the mass of the object at rest), but then that could be found by entering 0 for u,
    Nah, don't "guess", work it out, it's easy .Look:

    You told me, in effect, that u = v/c, whereby u<sup>2</sup> = v<sup>2</sup>/ c<sup>2</sup>. Ok? You also gave me m = (1 - v<sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup>

    For an object m at rest in its own frame, v = 0, hence v<sup>2</sup>/c<sup>2</sup> = 0 hence m = √ (1 - 0) = 1 so m = 1.

    Is this always true? Can you now see the missing cofactor?
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    I am afraid I can't though I truly wish I could.

    I do understand your/our math just fine, but I still don't know where you are going with it or what you are implying.

    In an effort to figure out your riddle I have been researching applications of "cofactor" used in both science and math for the last hour, but still have not found anything that might apply to this discussion. Thank you for your patience.

    Another guess: Perhaps you mean a unit of measure such as 5 kg or so?
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    OK, maybe cofactor was a poor choice of terminology. Sorry.

    But you can do this, I feel sure:

    Let m = x(1 - v<sup>2</sup>/c<sup>2</sup>)<sup>-1/2</sup>.

    When v = 0, what is x?

    When v ≠ 0 what is x?
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    Well hypothetically if

    m = x(1 - v²/c²)^־½

    and we say v = 0 then

    m = x(1)^־½

    m = x

    To me this says that this hypothetical equation is valid if x is m, the mass of an object when it is stationary to the observer.

    If we again go with the hypothetical

    m = x(1 - v²/c²)^־½

    and we say v = 1 * 10^8 meters per second then

    m = x(1 - 0.33356)^־½

    m = 1.22495x

    To me this says that this can only be true if x is the reciprocal of (1 - v²/c²)^־½ mulitiplied by m, but then I could have just divided by (1 - v²/c²)^־½ to get that. It seems I am going in circles.

    I think what would really make the most sense to me is if m were defined as the mass of an object at its current velocity reletive to the observer and x defined as the mass of the object at rest reletive to the observer. Now that I think about it, I am sure this is the answer you are looking for. That makes sense. Thanks.
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    Quote Originally Posted by Demen Tolden
    Well hypothetically if

    m = x(1 - v²/c²)^־½

    and we say v = 0 then

    m = x(1)^־½

    m = x

    To me this says that this hypothetical equation is valid if x is m, the mass of an object when it is stationary to the observer.
    Yes, good. You showed that the mass of an object at rest relative to an observer is equal to the mass of an object at rest relative to itself. (No, I'm not being sarcastic)

    Let's not get too excited, though

    If we again go with the hypothetical

    m = x(1 - v²/c²)^־½

    and we say v = 1 * 10^8 meters per second then

    m = x(1 - 0.33356)^־½

    m = 1.22495x
    So. You already equated m = x when v =0. So let's give x another name, like m<sub>0</sub>. And now you have some pretty high value for v, so let's rename the m on the LHS of your equation m<sub>v</sub>. Putting all that together, do you (or do you not) find that m<sub>v</sub> is greater that m<sub>0</sub>?

    I think what would really make the most sense to me is if m were defined as the mass of an object at its current velocity reletive to the observer and x defined as the mass of the object at rest reletive to the observer. Now that I think about it, I am sure this is the answer you are looking for. That makes sense. Thanks.
    Bingo, well done!

    BUT.. you should be aware that relativistic mass dilation is considered, at best, an unhelpful notion, and at worst, just downright wrong. I'm not enough of an expert to argue either way.
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    so, we have the senario of a spaceship jeting away from the earth at .9c. The daring pilots shoot a pen back at the earth at .9c. To them, the earth is moving away, and so is the pen. To them, these both gain mass. To the people on earth, the spaceship has a large mass, and the pen has stopped, getting smaller in mass. My first idea was that, while you're seeing the pen from the spaceship, it's gaining mas, and from the earth, it's losing mass. You said that this idea might even be outmoded, and mabey even wrong?

    In the "double slit experiment" I used the dialation of mass for a proof that photon have no mass. Does that invalidate my argument?
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    Or is it just the idea of relitivistic mass that does not reach par. Does that not rely on nothing but a basic idea of relitivity (as described by einstien) and the increse of mass ecause of velocity? I know my argumant was sound, because it has ben proven that electrons gain mass when traveling faster in partical accelrators. Where does relitivistic mass lose its lose its fesability, and what does this say about reletivity?
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    We will use your own scenario in which you leave Earth and accelerate to .9c relative to Earth. You can even accelerate a pen backwards at a speed of .9c if you wish.

    Then, if you observe Earth, and all the other nearby stars and stellar objects, you should deduce that you are the only moving object in the universe. Even the pen is motionless with respect to the nearby stars. Now what is your question about mass increase?
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    You are all forgetting that this phenomena can only be explained using time. Einstein has finished this a long time ago, simply look anywhere on the Internet or textbook and you'll see the answer. Don't forget, our velocity is relative to the Earh-which to us stands still. So anything different in motion would alter the perception of time for whomever, whenever and wherever they were observing. Light speed is constant-that law keeps the universe stable.

    As for the mass of something;

    m=m0y

    Where m is the mass that is given its velocity
    Where m0 is the mass we percieve
    Where y is the Lorentz factor.

    Someone else could use this equation stood on a different planet aroud Rigel or something to note our mass, but there noting of our velocity would be different to ours.

    This is very important in remembering that because of these different reference frames, time gives as a result and we see time dilation etc. In a black hole a spaceship could travel at 0.95c but to an outside observer be seen to travel at 1.3c or something-they'd seem to have an infinite mass, or infinite energy, infinite everything really. But then again this cannot be so as the black hole would de-stabilse under the intense gravitational field of the spaceship (infinite mass), everything would fall into a seemingly all powerful G field which passes all strengths of the other 3 fundemental forces and could destroy the whole universe by pulling it towards it! Of couse we know this cannot happen (for some reason), so the second the spaceship APPEARS to travel faster than light-it disappears, to, of course, the past, but who knows what the spaceship crew would see? Just because an outside observor sees them go back in time, does not mean then that they do, or do they> Only experiments will prove this-and they are a long way off being performed. By the time they can be they probably won't be needed.
    "If you wish to make an apple pie from scratch, you must first invent the universe". - Carl Sagan
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    Quote Originally Posted by SteveF
    Now what is your question about mass increase?
    Well here's mine.

    Why is it generally accepted that relativistic mass is a dead donkey? Is it simple aesthetics, or is there some math that says it must be so?

    And don't just wave and say "it's because of E = mc<sup>2</sup>", which incidentally, I do know how to derive. But I cannot see why this makes relativistic mass useless. What am I missing?
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  23. #22  
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    Why is it generally accepted that relativistic mass is a dead donkey?
    Beats me. I never made such a claim.
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    For years I have been asking this question of the "universal reference" of time and space, the so -called "reference of the one". The one "what"? Exactly. The one "what"!.

    So, I decided to theorise HOW that "one", that universal reference, that stationary entity, presumably, perceive all flux around it of space-time.

    "Why not find the centre of gravity of the space-time system", I though to myself......

    What is the centre of gravity of space-time?

    It is a place where gravity is most noticeable, I would think.

    Without being too Socratic with my deductions, would it be absurd to suggest that t theory of gravity is needed to properly explain space-time?

    Could one actually say that gravity is nothing more than a balancing force though, a system balancing force?

    Am I being way too Socratic with that suggestion?


    For those interested, I know of a theory that explains gravity as that space-time system balancing force (www feature below).
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