Step 2, above, is false, for reasons already stated.Physicsguru said:Mv = \frac{m_0v}{\sqrt{1-v^2/c^2}} = \frac{h}{\lambda_0 \sqrt{1-v^2/c^2} }
Obviously, you must end up disagreeing with some step which I have made, so I now ask you, which one?
russ_watters said:Step 2, above, is false, for reasons already stated.
It was already pointed out to you that this equation makes no sense--what is \lambda_0? If you know that p=0 exactly (which in classical mechanics is what is meant by an object's rest frame), then the DeBroglie wavelength \lambda is infinite.Physicsguru said:Here is where I was going with this. Suppose that:
\lambda = \lambda_0 \sqrt{1-v^2/c^2}
JesseM said:It was already pointed out to you that this equation makes no sense--what is \lambda_0? If you know that p=0 exactly (which in classical mechanics is what is meant by an object's rest frame), then the DeBroglie wavelength \lambda is infinite.
Like I said before, we know they are unequal from both theory and experimentation. For the theory, Newton did not intend for momentum to be used that way when he wrote his momentum equation, and Einstein didn't intend for it to be used that way when he wrote his part. You're mixing classical mechanics with Relativity. In addition, it is well known that classical mechanics is flawed.Physicsguru said:Russ, Energy=hf, therefore h has units of Kg m^2/s. Therefore, h divided by a quantity with units of length has units of classical momentum. How can you simply dismiss setting Mv equal to \frac{h}{\lambda}? You have said they are unequal, but how do you know that?
Because putting a little 0 in subscript usually means the quantity is evaluated in the particle's rest frame. If that's not what you mean, then please supply the meaing of \lambda_0.Physicsguru said:How have you drawn the conclusion that the equation makes no sense? I agree with you, that without an interpretation for \lambda_0, the equation will never "make sense."
JesseM said:Because putting a little 0 in subscript usually means the quantity is evaluated in the particle's rest frame. If that's not what you mean, then please supply the meaing of \lambda_0.
Given that people have already answered "no" for the case of any object with nonzero rest mass, I think you can figure out what our answer to this one would be.Physicsguru said:Ok, we are digressing from the central point of this thread. I thought about it last night, and I didn't actually ask question one, as I fully intended. Let me re-ask both questions:
Question 1: Is it possible to accelerate a body with living beings inside, from rest to the speed of light in 24 hours, such that they remain alive?
As I said before:Physicsguru said:Question 2: If the answer to question one is yes, why is it yes; and if the answer to question one is no, why is it no?
Because the energy of an object with rest mass m_0 moving at velocity v is E = m_0 c^2 / \sqrt{1 - v^2 / c^2}...if m_0 is nonzero, then as v approaches c, the energy approaches infinity.
Because if p=0, \lambda is infinite.Physicsguru said:That is exactly how I would interpret \lambda_0 in that formula. So why would that interpretation be meaningless?
Macro-objects have a DeBroglie wavelength too, it's still given by the formula \lambda = h/pPhysicsguru said:Also, please note that I do not wish to digress from the main question, which seems to be happening. I never meant for this question to be about particles, I meant for it to be about large bodies accelerating with living occupants inside.
perhaps you should point it out then.Physicsguru said:JesseM, you are making an error
I don't know it for certain. Likewise, I don't know for certain that the Earth is round. But there is plenty of evidence for both theories. Do you have an alternate theory that can explain all the observations that are used to support relativity, but which predicts a different formula for the relation between energy and velocity?Physicsguru said:Unless you know for certain that the relativistic energy formula is correct
JesseM said:perhaps you should point it out then. I don't know it for certain. Likewise, I don't know for certain that the Earth is round. But there is plenty of evidence for both theories. Do you have an alternate theory that can explain all the observations that are used to support relativity, but which predicts a different formula for the relation between energy and velocity?
Well, lay it on me, baby! Do you think your theory could make correct quantitative predictions about all the experiments listed here, for example? Could you predict the number of http://www.prestoncoll.ac.uk/cosmic/muoncalctext.htm ?Physicsguru said:Yes
Kind regards,
Guru
No, it is impossible to be certain of anything in science, including the roundness of the earth. But if you want to answer the question with a high level of confidence, just do lots of experiments to test that the energy formula (along with other basic formulas in relativity like the time dilation formula) is in fact correct.Physicsguru said:My question in this thread is meant to be taken as, "what if you don't know for certain that E=Mc^2, but you want to try to answer this question with certainty, can you do it?"
g-forces only depend on acceleration, not on velocity (this is true in Newtonian mechanics as well as relativity). So if I accelerate at 9.8 m/s^2 throughout the trip, I will feel earth-type-gravity the whole time, even as I get arbitrarily close to the speed of light.Physicsguru said:You are so entangled in relativistic effects, you have forgotten about the greatest impediment to the answer being yes, which is that they will be crushed by the g-forces long before coming even close to c. I would think you must deal with that issue at some point.
JesseM said:g-forces only depend on acceleration, not on velocity. So if I accelerate at 9.8 m/s^2 throughout the trip, I will feel earth-type-gravity the whole time, even as I get arbitrarily close to the speed of light.
Infinite time. Acceleration doesn't work the same way in relativity that it does in Newtonian mechanics, your velocity at time t in a given frame won't just be (acceleration rate)*(time since your velocity was zero in that frame). See Acceleration in Special Relativity.Physicsguru said:Ok, well this is a start. Suppose, as you say, that you accelerate at 9.8 m/s^2 throughout the trip, how long will it take you to reach the speed of light? (Is your answer anywhere near 24 hours?)
russ_watters said:Like I said before, we know they are unequal from both theory and experimentation. For the theory, Newton did not intend for momentum to be used that way when he wrote his momentum equation, and Einstein didn't intend for it to be used that way when he wrote his part. You're mixing classical mechanics with Relativity. In addition, it is well known that classical mechanics is flawed.
russ_watters said:Why do you think that simply having the same units makes them equal?
JesseM said:Infinite time. Acceleration doesn't work the same way in relativity that it does in Newtonian mechanics, your velocity at time t won't just be (acceleration rate)*(time since your velocity was zero).
24 hours according to the onboard clock, or according to clocks in the inertial frame that the ship started out at rest in?Physicsguru said:Let me try this a different way JesseM. Suppose that you are in a ship which is accelerating at 9.8 m/s^2. If you start from rest, how fast will your ship be moving after 24 hours?
JesseM said:24 hours according to the onboard clock, or according to clocks in the inertial frame that the ship started out at rest in?
jcsd said:Physics guru the answer to you question s have been given, is this thread going anywhere? The only challenge I see are your misconceptions onn relativity.
As for accelartion if you mean extrinsic acceleration (i.e. the coordinate accelartion as measured form some inertial frame) then it is impossible to maintain an extrinsic accelartion of 9.8 m/s^2 indefintely as you soon find the force required to keep up that accelartion is infinite. If you mean intrinsic acceleration, then yes you can maintin an instrinsic accelartion of 9.81 m/s^2 indefintely, but you will never reach a velocity of c or greater in any inertial frame.
Physicsguru said:JCSD the answer is not as simple as the question appears.
Physicsguru said:To put this another way, no one is addressing the question in an epistemologically correct manner, since no one is actually certain that E=Mc^2.
This is supposed to be a hard question, which eventually leads to the Meissner effect, but this thread is nowhere near that point yet.
OK, you can see the equations for velocity as a function of time (both proper time and coordinate time) for constant acceleration according to relativity on http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html page. For onboard time T, the equation is:Physicsguru said:Give the answer in both frames.
(I would also suggest that you cover both cases case 1) relativity correct, case 2) relativity incorrect).
jcsd said:No it's a rather simplistic question that has alreday been answered and as for the relevance of the Meissner effect I really can't see where you're going. If you think there is something being missed it is best you actually say what it is rather than leading everyone on a wild goose chase as the answer already given is undeniably corrcet (within the context of relativty, if you're talking baout any other context then why post it in this forum?) so if you've got another answer you have made a mistake.