Originally posted by RamnedIt is correct. A body's relativistic mass is equal to E/c^2. E is the total energy of the body. When the body is at rest, the total energy at rest divided by c^2 is its rest mass. The total energy includes kinetic energy. When a body is moving relative to an observer, then, to the perspective of that observer, it has a higher total energy than when it is at rest, and thus a higher relativistic mass. So, kinetic energy, and all other forms of energy, are always accounted for. In order to accelerate a body at rest and give it kinetic energy, work must be done on it, so its total energy increases; therefore, its relativistic mass increases, also.
Relativity
[b]26. Is the expression E= mc^2 strictly correct? For example, does it accurately account for the Kinetic energy of a moving mass?[/b]
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Originally posted by HolyT'No' does not necessarily mean you are entirely wrong...Are you confident on that answer or is the question confusing in any way?
Well, this is one answer that I'm not changing. So, I'll just stand by for others' input or a rephrasing of the initial question.
"Strictly correct" means the scenarios are all possible.
Originally posted by AThousandYoungE=mc^2/sqrt(1-v^2/c^2)
No, the equation is actually a bit more complex, but I don't want to look it up.
This is the correct equation that accounts for bodies in movement. If you Taylor expand it assuming that v is much smaller than c you'll get the newtonian kinetic energy plus a constant term.
Originally posted by RamnedThe confusing parts are "account for" and "strictly correct." It's a matter of defining terms. If E is the total energy with respect to an observer (which includes kinetic energy), and m is the relativistic mass, then E = mc^2 always does account for the kinetic energy.
'No' does not necessarily mean you are entirely wrong...Are you confident on that answer or is the question confusing in any way?
"Strictly correct" means the scenarios are all possible.
However, if the m is the rest mass, and E is still the total energy, and the body is moving with respect to an observer, then the correction for relative velocity must be thrown in there.
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Originally posted by HolyTDoes the expression account for all scenarios? For example, does it work for light?
The confusing parts are "account for" and "strictly correct." It's a matter of defining terms. If E is the total energy with respect to an observer (which includes kinetic energy), and m is the relativistic mass, then E = mc^2 always does account for the kinetic energy.
However, if the m is the rest mass, and E is still the total energy, and the body is m ...[text shortened]... respect to an observer, then the correction for relative velocity must be thrown in there.
Originally posted by adam warlockThat's not the one I'm thinking about I don't think. I remember some additive term.
E=mc^2/sqrt(1-v^2/c^2)
This is the correct equation that accounts for bodies in movement. If you Taylor expand it assuming that v is much smaller than c you'll get the newtonian kinetic energy plus a constant term.
Ah, found it on Wiki.
E^2 - (pc)^2 = (mc^2)^2
When p = 0, we get the old E = mc^2.
http://en.wikipedia.org/wiki/Special_relativity
Originally posted by AThousandYoungAh. I alwyas think about the equation I gave you even thou this one is more used and in some ways more fundamental.
That's not the one I'm thinking about I don't think. I remember some additive term.
Ah, found it on Wiki.
E^2 - (pc)^2 = (mc^2)^2
When p = 0, we get the old E = mc^2.
http://en.wikipedia.org/wiki/Special_relativity
Originally posted by RamnedYes, it works for light. Photons have relativistic mass but no rest mass. If an observer measures a photon's total energy, its relativistic mass can be calculated as m = E/c^2. If the observer were to accelerate toward the photon's speed, and keep doing the measurement and calculation, the observer would find that the photon's relativistic mass would decrease, until the observer is traveling at the same speed as the photon. At that point, the photon would have zero total energy and a rest mass of zero (mass at zero velocity relative to the observer).
Does the expression account for all scenarios? For example, does it work for light?
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Originally posted by HolyTOk. I think this question is obviously too confusing. So I am going to post a different relativity question to follow. Sorry, I don't always make clear questions 😛
Yes, it works for light. Photons have relativistic mass but no rest mass. If an observer measures a photon's total energy, its relativistic mass can be calculated as m = E/c^2. If the observer were to accelerate toward the photon's speed, and keep doing the measurement and calculation, the observer would find that the photon's relativistic mass would decrea ...[text shortened]... e zero total energy and a rest mass of zero (mass at zero velocity relative to the observer).
Here it is:
26. You are in a speedboat on a lake. You see ahead of you a wave front, caused by the previous passage of another boat, moving away from you. You accelerate, catch up with, and pass the wave front. Is this scenario possible if you are in a rocket and you detect a wave front of light ahead of you?
Originally posted by RamnedNo.
Ok. I think this question is obviously too confusing. So I am going to post a different relativity question to follow. Sorry, I don't always make clear questions 😛
Here it is:
[b]26. You are in a speedboat on a lake. You see ahead of you a wave front, caused by the previous passage of another boat, moving away from you. You accelerate, catch up with, ...[text shortened]... scenario possible if you are in a rocket and you detect a wave front of light ahead of you?[/b]
Originally posted by Ramnedyou can't "detect" a wave of light in front of you that is moving away from you . . .because . . . it is moving at the speed of light away from you.
Ok. I think this question is obviously too confusing. So I am going to post a different relativity question to follow. Sorry, I don't always make clear questions 😛
Here it is:
[b]26. You are in a speedboat on a lake. You see ahead of you a wave front, caused by the previous passage of another boat, moving away from you. You accelerate, catch up with, ...[text shortened]... scenario possible if you are in a rocket and you detect a wave front of light ahead of you?[/b]
"Detecting" is just given in this question to indicate that you are somehow aware that there would be a wave front propogating ahead of the other ship. It could be by pre-arrangement with the other ship; it doesn't really matter.
The rocket and light example is not possible in the same way as the boat and wake example. Relative to you (in the trailing rocket), the other ship's wave front is propogating at the speed of light. You are traveling at zero velocity relative to yourself. You'll never catch up! From the perspective of the lead ship, the wave front is going forward at the speed of light, and your ship is traveling at something less than the speed of light. Still, you'll never catch up.