Why Ec2=m and energy x acceleration=force are true

Discussion in 'Pseudoscience Archive' started by ryan2006, Mar 7, 2013.

  1. Prof.Layman totally internally reflected Registered Senior Member

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    Have you collected a lot of experimental evidence that would show that?
     
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  3. Maximum_Planck Registered Member

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    of course. look at all the many posts made by them. so rude and condescending
     
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  5. AlexG Like nailing Jello to a tree Valued Senior Member

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    Nonsense.
     
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  7. AlexG Like nailing Jello to a tree Valued Senior Member

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    Why would the posting of incorrect nonsense deserve any tolerance?
     
  8. ryan2006 Registered Member

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    mass is made up of energy. I understand that energy would have to weigh alot. Does anybody know how much the earth weighs in energy.
     
  9. ryan2006 Registered Member

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    C2 x C2 or a velocity times itself
     
  10. origin Trump is the best argument against a democracy. Valued Senior Member

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    Energy does not have mass. For instance an electron and positron both have mass (about 9 x 10^-28 grams) if they meet with essentially no KE and anihilatle each other (turn to pure energy) the energy of the each of the resulting (2) photons will be 511 keV. Photons do not have a rest mass.

    This is from E=mc^2.

    This equation shows the equivilance of mass and energy, BUT energy is NOT mass and mass is NOT energy!
     
  11. ryan2006 Registered Member

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    Is it fair to say that ec2=m and energy x acceleration=force are hypothesis' with simple observations and without the mathematical proofs? Meaning that maybe Stephan Hawkins would like to provide?
     
  12. ryan2006 Registered Member

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    Radiation or energy from the sun makes the crops grow is another observation
     
  13. origin Trump is the best argument against a democracy. Valued Senior Member

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    No. Taking the equation E=mc^2 and doing some simple algebrac rearrangment gives E/c^2 = m. This is a rather meaningless rearrangement though.

    Nope this is meaningless. It is a simple experiment to show that F= ma. You are saying m=E which is not true. Mass is not energy. As an analogy graphite is not diamond. Graphite may be able to become a diamond but it is not a diamond. If you cannot understand that I would like very much to sell you a million dollars worth of diamonds for only 10,000 dollars. I will supply the diamonds in the form of #2 pencils.
     
  14. origin Trump is the best argument against a democracy. Valued Senior Member

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    Yes, that is a nice random observation. Is that somehow related to the subject.

    I've got one; Howler monkeys have prehensile tails.
     
  15. Prof.Layman totally internally reflected Registered Senior Member

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    You would want to solve for c, so then \( c = {sqrt{E}}/{sqrt{m}} \)
     
  16. origin Trump is the best argument against a democracy. Valued Senior Member

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    2 problems here

    First the equation would be rearranged to this \(c = sqrt{\frac{E}{m}}\) not this \( c = {sqrt{E}}/{sqrt{m}} \) (it makes a difference!).

    Secondly, this is nonsensical because c is a constant so the energy and mass are locked into a ratio dictated by the answer...
     
  17. Pete It's not rocket surgery Moderator

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    Neither of those is a problem.

    For the first, \(\frac{sqrt{A}}{sqrt{B}} = \sqrt{\frac{A}{B}}\) (for all values of A and B)

    For the second, the whole point of the world's most famous equation is that (rest) mass and (rest) energy are equivalent, and related by a constant ratio (c^2).
     
  18. Prof.Layman totally internally reflected Registered Senior Member

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    What Pete said, and maybe you could use it to check your answer, the speed of light is supposed to be a single constant value, so then when someone says, no no no, it must have all of this other junk also behind it, then you could solve for "c" and then find out if \( E = m c^{2} + other junk \) actually gives you the correct speed of light. What is the speed of light squared anyways? I thought everything was supposed to be shown that there is a single constant value of c, not \(c^{2}\), lol.

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    By the way, I think \(c = sqrt{\frac{E}{m}}\) does look a lot better.
     
  19. James R Just this guy, you know? Staff Member

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    Maximum_Planck:

    Please link me to ONE reputable source that backs up your claim.

    And you're a picture of humility and do not claim superior knowledge or skill, I assume.
     
  20. James R Just this guy, you know? Staff Member

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    ryan2006:

    Do you now admit you were wrong? Are we done here?
     
  21. James R Just this guy, you know? Staff Member

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    Moderator note

    Several off-topic, insulting posts or otherwise inappropriate posts have been removed.
     
  22. Pete It's not rocket surgery Moderator

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    Well, the full equation is actually
    \(E^2 = p^2c^2 + m^2c^4\)
    (p is momentum). The short version is for the special case of objects at rest in the frame of interest.
    If c is a constant, then c^2 is obviously a constant too.
    There are also other universal constants.
     
  23. AlphaNumeric Fully ionized Moderator

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    The \(c^{2}\) term is necessary for the units to match, just \(c^{1}\) wouldn't be enough. Of course that's just a reason why it couldn't be c but that reasoning doesn't preclude something like \(E = 5mc^{2}\), the units still match. The specific form of the expression follows from the internals of special relativity, specifically the form of the metric, Lorentz invariant and conservation of energy and momentum.

    Except in this case we're actually talking about an equation within physics. Regardless of whether or not reality really does satisfy the dynamics of special relativity the fact remains \(Ec^{2} = m\) is meaningless. Others have tried to explain it to you but I'll give it a shot too.

    We all know speed*time = distance if the speed is constant, so d = vt. Velocity is measured in lengths per unit time, \([v] = LT^{-1}\). Distance is measured in lengths, \([d] = L\) and time is measured in units of time, \([t] = T\). So we have \([vt] = [v][t] = (LT^{-1})T = L = [d]\) so it is at least consistent with units. But what if someone said "But I claim it is v = k*d*t[/tex] where k is some constant". Let's look at a specific example. A car is moving at 36 kilometres per hour. It drives for 2 hours and covers 72 kilometres according to d=vt. I decide I prefer metres, not kilometres and I prefer seconds, not hours. There is 1,000 metres in a kilometre and 3600 seconds in an hour. So d = 72,000 metres. v = 36,000 metres per 3600 seconds = 10 metres per second. The time is t = 2 hours = 2*3600 seconds = 7200 seconds. So what is vt? Well vt = (10 meters per second)*(7200 seconds) = 72,000 metres, which indeed is d. So we have gone from 72 = 36*2 to 72,000 = (3600)*10. Fine. No contradiction.

    But what if we did v = k*d*t? So first we do it in kilometres and hours and work out k. v = 36 km/hr. d = 72km. t = 2 so v=k*d*t becomes 36 = k*72*2 = 144k and so this is only going to match is k = 36/144 = 1/4. So now we have v = d*t/4. So now change to metres and seconds. v = (36,000 metres) per (3600 seconds) = 10 metres per second. t = 2 hours = 7,200 seconds. d = 72 km = 72,000 metres. So what is k*d*t = d*t/4? d*t/4 = 72,000 * 7200/4 = 129600000. Last time I checked 129600000 is not equal to 10. So the formula cannot be right as if it works for one choice of units it can break for another.

    The reason it failed is that the 3,600 factor from changing hours to seconds multiplied together instead of cancelling one another. If t had been on the other side then the change would have cancelled out and it would have been okay. This always happens when you don't get your units right. Even if you allow for a constant multiplying factor in each term you can always find a choice of units where the formula is inconsistent.

    Proving it wrong via units proves it wrong. It's like saying 1 km = 1 hour. Yes 1=1 but if you use metres then you have 1000 metres = 1 hour and obviously 1000 isn't equal to 1.

    Firstly the posters here aren't 'forum elites', don't imagine some cabal of people out to get you. Secondly someone showing they understand units doesn't give them a 'bigger physics dick' or make one 'superior', it shows they paid attention in physics lessons when they were 15. This isn't high brow stuff, it is a concept fundamental to any kind of description of physical things, be it engineering, chemistry, biology or physics.

    You should do something about that chip on your shoulder, it must be giving you balance problems.
     

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