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Interesting Thoughtsby kingofxbox99
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#1
May2414, 10:22 PM

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I was thinking, and somehow I had this little "thought experiment" type thing relating to objects and the speed of light. (This is just my thoughts, I don't know if it's correct or the validity of my assumptions, etc)
Imagine you're in a (large) room, and that the effects of friction, air resistance, etc. are nonexistant. You walk up to a box at the end of the room, and apply a constant force to it. If the force remains constant, the acceleration will remain constant. Eventually, the box's velocity will get close to the speed of light. Once it hits the speed of light, it's acceleration will instantly drop to 0, and in order to maintain a constant force, it's mass will approach infinity (F = ma). Is this why people always say that your mass would approach infinity as you approach the speed of light? Just a cool little thought I had. :) 


#2
May2414, 11:29 PM

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You cannot use ##F=ma## here; that's a simplified formula that only works for speeds that are small compared with the speed of light. For your problem, you have to make two changes: First, you have to use the more general ##F=\frac{dp}{dt}## where ##p## is the momentum, instead of ##F=ma##. Second you have to use the relativistic definition of momentum: ##p=\gamma{m}{v}## where ##\gamma## is ##\frac{1}{\sqrt{1v^2}}## (I'm measuring distances in lightyears and times in years so that the speed of light is one in this formula). Note that if ##v## is small these formulas reduce to the familiar ##F=ma##. But in your problem, ##v## is not small. 


#3
May2514, 01:52 AM

P: 2

Thanks for explaining that! I hate how simplified everything is at the highschool level...



#4
May2514, 02:11 AM

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Interesting Thoughts



#5
May2514, 01:58 PM

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Another complication is that many highschool students have not yet had a full year of calculus, so we can't use calculus, just trignometry and algebra. That makes it hard, especially when you consider that when Newton was figuring out classical mechanics, he needed calculus so badly that he had to invent it. ##F=\frac{dp}{dt}## is the most powerful and general expression of Newton's second law; but if you don't have calculus you have to settle for ##F=ma##, which falls out of ##F=\frac{dp}{dt}## if you remember that classically ##p=mv## and assume a constant mass. 


#6
May2514, 06:00 PM

P: 482

[tex]a = \frac{F}{m}\sqrt {1  \frac{{v^2 }}{{c^2 }}} ^3[/tex] F and a can only remain constant if m is decreased with velocity (e.g. by emission of radiation) according to [tex]m = m_0 \cdot \sqrt {1  \frac{{v^2 }}{{c^2 }}} ^3[/tex] That means when the box reaches the speed of light its mass doesn't approaches infinity but zero. Moreover, energy goes zero too: [tex]E = m_0 \cdot \left( {c^2  v^2 } \right)[/tex] Thus there would be nothing left to accelerate or to apply a force to. 


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