有否理論描述E=pv?

In Special Relativity, the mass m of a moving object is velocity dependent, and is given by, m = (mo)/SRT(1 - v^2/c^2)
where mo is the rest mass, v is the velocity of the object, and c is the speed of light. (SRT denotes "square-root")

Using E = mc^2, the gain in kinetic energy of 1 moving object
= E - Eo = (mo)c^2/SRT(1 - v^2/c^2) - (mo)c^2
where Eo is the rest mass energy

Using Binominal Theorem to expand the term [(1 - v^2/c^2)^(-1/2)]
(1 - v^2/c^2)^(-1/2) = 1 + (1/2)(v/c)^2 - (3/8)(v/c)^4 + higher order terms in (v/c)
When v << c, the higher order terms from (3/8)(v/c)^4 and beyond can be neglected.
Thus, (1 - v^2/c^2) = 1 + (1/2)(v/c)^2 to a good approximation.

Therefore, gain in kinetic energy
= (mo)c^2.[(1 + (1/2)(v/c)^2] - (mo)c^2 = (mo)c^2[1 + (1/2)(v/c)^2 - 1]
= (1/2)(mo)v^2 which is the Newtonian form of kinetic energy

In fact, the relation between energy E and momentum p is given by,
E^2 = (pc)^2 + [(mo)c^2]^2





2015-03-18 23:34:51 補充：
Q:I agree the approximation, but I want the theory in Newtonian mechanics.
A: The 1/2 in classical mechanics is becuase of the calculation in average speed.
Work done = F.s where F is the force and s the dispalcement

2015-03-18 23:36:52 補充：
(cont'd)...
But F = ma = m(v-u)/t where v and u are the final and initial speed, t is the time, m is the mass and a is the acceleration.
and s = [(u+v)t]/2

2015-03-18 23:39:07 補充：
(cont'd)...
To find the kinetic energy KE possessed by an object at speed v, we need to set u = 0.
Hence, KE = work done = F.s = (mv/t).(vt/2) = (1/2)mv^2

2015-03-19 17:34:52 補充：
Your further question:
As I have said in my answer, the equation E = pv is not correct. The correct equation is
E^2 = (pc)^2 + [(mo)c^2]^2
For particles with no rest mass mo, e.g. photons, the equation reduces to
E = pc

2015-03-19 17:37:50 補充：
In your equation: ps/t=p [(u+v)]/2
note that s/t is NOT equal to v, but v/2, because (s/t) gives the average velocity. The v here is the final velocity.

2015-03-19 17:41:19 補充：
(cont'd)
Therefore, the equaion: ps/t=p [(u+v)]/2
just gives you pv/2 = pv/2 when u is set at zero. The result is obvious as this is an equation.

2015-03-19 17:43:30 補充：
超過字數限制，在[意見]欄繼續

2015-03-22 23:32:48 補充：
Your equation should be dW = v(dp)
hence, dW = mv(dv)
Integrate both sides, W = (1/2)mv^2

2015-03-22 23:34:41 補充：
Be aware that dW = v(dp) doesnot lead to W = vp because p is velocity dependent.

2015-03-23 15:58:30 補充：
p = mv, you can't ingore the integrand v in the integration.

2015-03-23 20:00:24 補充：
If you are interested in integrating (dp), the mathematics is as follows:
dW = v(dp) = (p/m)dp
Integrate both sides with respect to p
W = (1/m)integral [p(dp)] = p^2/(2m) = (mv)^2/(2m) = (1/2)mv^2
The same result is obtained.

2015-03-24 15:43:59 補充：
You could do any mathematical combination of the parameters. But the most important point is whether there are any physical significances at all.

2015-03-24 22:48:43 補充：
I suggest you read the discussion posted above again.
The (1/2) in (1/2)mv^2 is because of the use of average velocity in the calculation of displacement s in evaluating work done W.

2015-03-25 15:39:37 補充：
對不起,字數已達上限,不能回答

2015-03-25 15:40:56 補充：
你開一個新post再討論吧!

土扁, you help us review the knowledge, and thus raises our chance of getting the correct answer.

2015-03-19 18:24:05 補充：
I am confused.

Do we have 1/2pv, 1/2Fs, 1/2PV?

Thanks for your detailed explanation.

If the text is limited, you can send me an email. Thanks.

2015-03-23 13:55:06 補充：
In dW = v(dp), integrate both sides, W = vp.

Why is the p velocity dependent? Why is not it time dependent, as F=dp/dt?

2015-03-23 21:35:34 補充：
So, do we have 1/2pv, 1/2Fs, 1/2PV?

2015-03-24 00:37:41 補充：
According to the integration method, I think we should have them.

2015-03-24 16:36:05 補充：
Agree. So I asked "Newtonian mechanics有無理論講" & "The factor of 1/2何解" at first. Why do people treat Fs, 1/2mv^2, 1/2pv, PV differently? Someone said 1/2 factor is not derivable, but from measurment.

2015-03-24 16:37:18 補充：
Also, I found this: Since the kinetic energy increases with the square of the speed, an object doubling its speed has four times as much kinetic energy. For example, a car traveling twice as fast as another requires four times as much distance to stop, assuming a constant braking force.

2015-03-24 16:37:25 補充：
http://en.wikipedia.org/wiki/Kinetic_energy#Kinetic_energy_of_rigid_bodies

2015-03-25 11:12:01 補充：
Yes, I knew. My focus is on why other equations derived from the use of average velocity or the like do not have the 1/2 factor.
If s = [(u+v)t]/2 , set u = 0, s=vt/2, so, Fs=1/2Fvt=1/2Fx.
Also, integrating Fds, we should get 1/2Fs since there are 2 "s".

2015-03-25 11:13:22 補充：
In fact, the original KE formula was proportional to mv^2, the 1/2 factor was added afterwards. What is the physical meaning of v^2?

Thanks for your detailed explanation.

2015-03-25 11:25:53 補充：
My definition of "s" in previous equations is x1-x0, not [(u+v)t]/2 . So, I cannot get your idea clearly. Are they the same?

2015-03-25 20:29:14 補充：
Continue here.

https://hk.knowledge.yahoo.com/question/question?qid=7015032500123&mode=w&from=question&recommend=0&.crumb=eG8F44o98Bz

I give up.

When I saw the last supplementary statement at 19:27:41 on 2015-03-16, I did believe that I am not able to help you.