Eienstein Defeated?
- TIGERassault
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Re:
Neo's post is much too overcomplicated!
This is a more detailed definition of E=mc2, as this allows for if the object is moving. The concept here is that as v would be equal to c, ie at the speed of light, it would make the equation n/(1-1), which is n/0, which is said to be infinite. Therefore, it would take an infinite amount of energy to make something move as fast as light.
On the other hand, I find the actual definition itself to also be sketchy in that regards, as photons, even though they don't have mass, move at a speed of c, and 0/0 is just as infinite!
This is a more detailed definition of E=mc2, as this allows for if the object is moving. The concept here is that as v would be equal to c, ie at the speed of light, it would make the equation n/(1-1), which is n/0, which is said to be infinite. Therefore, it would take an infinite amount of energy to make something move as fast as light.
On the other hand, I find the actual definition itself to also be sketchy in that regards, as photons, even though they don't have mass, move at a speed of c, and 0/0 is just as infinite!
Re:
This is because you use the wrong equation.TIGERassault wrote:On the other hand, I find the actual definition itself to also be sketchy in that regards, as photons, even though they don't have mass, move at a speed of c, and 0/0 is just as infinite!
The right equation is:
E^2 = m^2 c^4 + p^2 c^2
where p is the momentum p of the object.
For resting particles there is p=0, thus E = m c^2.
For photons there is m=0, thus E = p c
- TIGERassault
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Re:
According to Wikipedia:Munk wrote:This is because you use the wrong equation.
The right equation is:
E^2 = m^2 c^4 + p^2 c^2
where p is the momentum p of the object.
For resting particles there is p=0, thus E = m c^2.
For photons there is m=0, thus E = p c
"with p being the relativistic momentum (ie. p = γp0 = mrel * v)"
So hey, the point is the exact same! In fact, both equations are the same too!
Re:
The way how to add up velocities lies in the special geometry of spacetime, which can be found by the statement that \"the speed of light is constant for any (unaccelerated) observer\".
The equation for energy as above is also a consequence of the same geometry of spacetime.
Thus the reason for \"always below speed of light\" is somewhat \"hidden\".
The equation for energy as above is also a consequence of the same geometry of spacetime.
Thus the reason for \"always below speed of light\" is somewhat \"hidden\".
- TIGERassault
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Is that your way of saying "I don't know why Einstein's formulas said that something can't go faster than light, but I'll make a complex post anyway to make it seem like I know what I'm talking about", or are you actually going somewhere with this?Munk wrote:Thus the reason for "always below speed of light" is somewhat "hidden".
Everyone can believe what they want. Knowing relativity is nothing crucial for survival ^^
I haven't studied physics to tell you rubbish, but exploring relativity by yourself from the very beginning is a great experience. Because you know actually why things are as they are - and not because someone told so.
Of course relativity can be described in 100 words only _scetchy_ like \"infinite energy barrier\" or \"mass increase\". But if you have the time, it is worth to get the full view.
I haven't studied physics to tell you rubbish, but exploring relativity by yourself from the very beginning is a great experience. Because you know actually why things are as they are - and not because someone told so.
Of course relativity can be described in 100 words only _scetchy_ like \"infinite energy barrier\" or \"mass increase\". But if you have the time, it is worth to get the full view.
- Foil
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No, you misunderstand. By "hidden", he means "indirectly seen". As in, "you have to go through some work involving spacetime geometry to get from the premise (constancy of the speed of light) to the conclusion (velocities must always be below the speed of light)."TIGERassault wrote:Is that your way of saying "I don't know why Einstein's formulas said that something can't go faster than light, but I'll make a complex post anyway to make it seem like I know what I'm talking about", or are you actually going somewhere with this?Munk wrote:Thus the reason for "always below speed of light" is somewhat "hidden".
As he said earlier, Munk is an actively working physicist. Don't let the language barrier (Munk, I think your English is actually quite good) confuse you.
- TIGERassault
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No, but it is pretty crucial for being able to join in in a conversation about it...Munk wrote:Everyone can believe what they want. Knowing relativity is nothing crucial for survival ^^
- Foil
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Re:
A couple of quick mathematical notes:
There's a subtle difference (analogous to the difference between the equations a=b and a^2=b^2). Munk is quite right that the more general, correct form is E^2 = m^2 c^4 + p^2 c^2. Using the other form for scenarios that give you indeterminate forms ("n/0", "0/0") can lead to some misunderstandings.
You should be more careful with indeterminate forms like "n/0" and "0/0", by actually doing some limit evaluations. Those indeterminate forms don't necessarily represent "infinity". In fact, something which evaluates to "0/0" may actually diverge to infinity, or it might not exist... or it might even represent a finite number!TIGERassault wrote:...n/0, which is said to be infinite.
...
On the other hand, I find the actual definition itself to also be sketchy in that regards, as photons, even though they don't have mass, move at a speed of c, and 0/0 is just as infinite!
Not quite the same.TIGERassault wrote:According to Wikipedia:Munk wrote:This is because you use the wrong equation.
The right equation is:
E^2 = m^2 c^4 + p^2 c^2
where p is the momentum p of the object.
For resting particles there is p=0, thus E = m c^2.
For photons there is m=0, thus E = p c
"with p being the relativistic momentum (ie. p = γp0 = mrel * v)"
So hey, the point is the exact same! In fact, both equations are the same too!
There's a subtle difference (analogous to the difference between the equations a=b and a^2=b^2). Munk is quite right that the more general, correct form is E^2 = m^2 c^4 + p^2 c^2. Using the other form for scenarios that give you indeterminate forms ("n/0", "0/0") can lead to some misunderstandings.
Ok.... When I said infinite \"fuel\" I don't think I said what I really meant. I didn't mean you would constantly have like a full tank of gas. I mean with A fusion Reaction for speed it could possibly have the potential to break said rules of space fabric and all that crap. Thats what I think anyway.
- SuperSheep
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- Foil
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Hehe, that one was fun to teach my calculus class.
Me:\"What's the limit of sin(x)/x as x->0?\"
Various students:
\"Well, sin(0)=0, so we have 0/x, so it's 0!\"
\"No, the denominator goes to zero, so it diverges to infinity!\"
\"Wait... the numerator goes to zero, so it's going to zero... but the denominator goes to zero, so it's going to infinity? The limit must not exist!\"
Me:\"What's the limit of sin(x)/x as x->0?\"
Various students:
\"Well, sin(0)=0, so we have 0/x, so it's 0!\"
\"No, the denominator goes to zero, so it diverges to infinity!\"
\"Wait... the numerator goes to zero, so it's going to zero... but the denominator goes to zero, so it's going to infinity? The limit must not exist!\"
- TIGERassault
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I know, but it's just a lot easier to call it infinity and leave it at that.Foil wrote:You should be more careful with indeterminate forms like "n/0" and "0/0", by actually doing some limit evaluations. Those indeterminate forms don't necessarily represent "infinity". In fact, something which evaluates to "0/0" may actually diverge to infinity, or it might not exist... or it might even represent a finite number!
But trying to calculate p gives the same problem as in the other equation!Foil wrote:There's a subtle difference (analogous to the difference between the equations a=b and a^2=b^2). Munk is quite right that the more general, correct form is E^2 = m^2 c^4 + p^2 c^2. Using the other form for scenarios that give you indeterminate forms ("n/0", "0/0") can lead to some misunderstandings.
- Foil
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The point is that "calling it infinity" isn't necessarily correct. Something which evaluates to "0/0" isn't necessarily infinite (e.g. the sin(x)/x as x->0 example SuperSheep mentioned).TIGERassault wrote:I know, but it's just a lot easier to call it infinity and leave it at that.
How so?TIGERassault wrote:But trying to calculate p gives the same problem in the other equation!Foil wrote:...the more general, correct form is E^2 = m^2 c^4 + p^2 c^2. Using the other form for scenarios that give you indeterminate forms ("n/0", "0/0") can lead to some misunderstandings.
I assume you're talking about your earlier question about photons (rest mass=0)? As Munk said, that simplifies the equation to E=pc ( => p=E/c).
- TIGERassault
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Hmm... I hadn't thought of that as p=E/c...Foil wrote:I assume you're talking about your earlier question about photons (rest mass=0)? As munk said, that simplifies the equation to E=pc ( => p=E/c).
Ok, now I'm just confused.
Re:
Yes, you see ? All those relativistic equations with the velocity v doesn't make sense, as you approach v -> c.TIGERassault wrote:But trying to calculate p gives the same problem in the other equation!
This is because m > 0 and v=c is contradictory.
Working with relativistic theory in this way shows you, that the velocity is not a "good" variable. With velocity involved, you need to seperate equations for massive and massless particles, the transition is somewhat obscure.
Working with momentum p is much simpler and equations holds for both (massive and massless).
This is a strong hint that the momentum p is more fundamental than the velocity, and in fact it is true:
Momentum p shares the spacetime geometry, while the velocity v does not!
Of course for human's life velocity is much more intuitive than momentum, but for relativistic physics it's the other way around ^^
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