Can we calculate kinetic energy by multiplying 1/2mv^2 by 1.15 at C/2 speed?

[bobie]

If an electron is traveling at C/2, the Lorenz formula says that mass is about 1.15 m_e, do we get the value of KE simply multiplying 1/2mv² by 1.15?
Is it as simple as that?

 

[dipole]

The expression from classical mechanics, T = \frac{m}{2}v^2 is only valid for small velocities. The full relativistic expression for the kinetic energy is given by,

T = (\gamma -1)mc^2 where m is the rest mass, and \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.

 

[bobie]

\gamma = \frac{1}{\sqrt{1-\frac{(c/2)^2}{c^2}}}
That is the formula I used and I got 1.1547 shall I multiply this or only .15 by .511 MeV?

 

[Dadface]

You have calculated the value for gamma and dipole has given you the equation to calculate the KE. As dipole stated the equation half mv squared "is only valid for small velocities".
Rather than just plugging numbers into the relativistic equation to get the answer try to get a better understanding of the equation and when to use it. A search of the "hyperphysics" site might be a good place to start.

 

[bobie]

Thanks, I suppose 1-γ, 0.15, should be multiplied by .511.

 

[jartsa]

If an electron is traveling at C/2, the Lorenz formula says that mass is about 1.15 m_e, do we get the value of KE simply multiplying 1/2mv² by 1.15?
Is it as simple as that?

Newtonian:

KE =1/2 * constant mass * v^2

Relativistic:

KE = 1/2 * changing mass * v^2

(Kinetic energy is the energy that we get when we stop an object. Stopping involves a change of mass)

 

[jtbell]

Relativistic:

KE = 1/2 * changing mass * v^2

No, unless you want to invent a new kind of velocity-dependent "relativistic mass" in addition to the one that everybody knows about but almost no physicists use.

 

[jartsa]

Thanks, I suppose 1-γ, 0.15, should be multiplied by .511.

Total energy of electron at velocity v = gamma * rest energy

So kinetic energy at velocity v must be ... (gamma * rest energy) - rest energy

 

[jartsa]

No, unless you want to invent a new kind of velocity-dependent "relativistic mass" in addition to the one that everybody knows about but almost no physicists use.

Oh yes, that mass would be the "longitudinal mass" that is proportional to gamma^3, not the relativistic mass, also known as transverse mass, that is proportional to gamma.

(I thought it was the relativistic mass. This time I was so cautious that I checked that longitudinal mass seems to work correctly with the Newtonian kinetic energy formula.)

 

[PAllen]

Oh yes, that mass would be the "longitudinal mass" that is proportional to gamma^3, not the relativistic mass, also known as transverse mass, that is proportional to gamma.

(I thought it was the relativistic mass. This time I was so cautious that I checked that longitudinal mass seems to work correctly with the Newtonian kinetic energy formula.)

NO, that doesn't work either. You would need yet another flavor of 'relativistic mass'. Why do you bother? Physicists just use relativistic formulas, not multiple weird mass definitions to make Newtonian formulas work.

(You would need M = 2(γ-1)mc^2/v^2 as your new 'kinetic relativistic mass'.)

 

[jtbell]

Oh yes, that mass would be the "longitudinal mass" that is proportional to gamma^3, not the relativistic mass, also known as transverse mass, that is proportional to gamma.

So you're claiming that $\frac{1}{2}m_Lv^2 = (\gamma - 1)mc^2$ where m is the rest mass and $m_L = \gamma^3 m$? With a bit of algebra, this proposed equation becomes
$$\frac{1}{2}(\gamma^3 m) (\beta c)^2 = (\gamma - 1)mc^2\\
\frac{1}{2} \gamma^3 \beta^2 = \gamma - 1$$
where β = v/c.

Calculating the left and right sides for various values of β between 0 and 1 shows that they are not equal.

(PAllen beat me to it!)

 

[pervect]

A correct formulation of relativistic dynamics for point particles is:

F = dp/dt
p = $\gamma$ m v

where:
F is the force (3-force)
p is the momentum
t is time (coordinate time)
m is the invariant mass

Note that one must use the chain rule to find dp/dt. In the usual case, the invariant mass of the point particle isn't changing, and one would get something like:

F = \frac{d \gamma}{dt} m v + \gamma m \frac{dv}{dt} = \frac{d \gamma}{d|v|}\frac{d|v|}{dt} m v + \gamma m \frac{dv}{dt}

where v is the 3-velocity vector and |v| is the magnitude of the 3-velocity vector

The chain rule expansion is a bit ugly , if at all possible it's less error prone to stick with F = dp/dt

 

[PAllen]

Thanks, I suppose 1-γ, 0.15, should be multiplied by .511.

Yes, for KE in Mev.

 

[jartsa]

Let's consider following example:

A bike cyclists accelerates from 0 to 0.86 c, while taking good care of his energy balance. (Assistants fire food portions for the cyclist to catch)

Longitudinal mass of the cyclist becomes 8 times larger.

Then there's a short service break. From 0.86 c to 0.87 c the longitudinal mass stays constant.

The change of kinetic energy when velocity goes from 0.86 c to 0.87 c is:

Delta KE = 1/2 * longitudinal mass * (0.87 c)^2 - 1/2 * longitudinal mass * (0.86 c)^2

 

[PAllen]

Let's consider following example:

A bike cyclists accelerates from 0 to 0.86 c, while taking good care of his energy balance. (Assistants fire food portions for the cyclist to catch)

Longitudinal mass of the cyclist becomes 8 times larger.

Then there's a short service break. From 0.86 c to 0.87 c the longitudinal mass stays constant.

The change of kinetic energy when velocity goes from 0.86 c to 0.87 c is:

Delta KE = 1/2 * longitudinal mass * (0.87 c)^2 - 1/2 * longitudinal mass * (0.86 c)^2

This has more mistakes than lines written. I am not going to bother correcting them all, because you just ignore valid responses and make up new nonsense. Please stop.

 

[PAllen]

To bring Pervect's #12 to work and energy:

With F and P as Pervect defined them (relativistic 3 vectors rather than 4 vectors), some natural results follow:

∫F dot dx is change in energy, and will agree with change of E = mγc^2

 

[PAllen]

Let's consider following example:

A bike cyclists accelerates from 0 to 0.86 c, while taking good care of his energy balance. (Assistants fire food portions for the cyclist to catch)

irrelevant

Longitudinal mass of the cyclist becomes 8 times larger.
uq...

irrelevant

Then there's a short service break. From 0.86 c to 0.87 c the longitudinal mass stays constant.

impossible

The change of kinetic energy when velocity goes from 0.86 c to 0.87 c is:

Delta KE = 1/2 * longitudinal mass * (0.87 c)^2 - 1/2 * longitudinal mass * (0.86 c)^2

flat out wrong

 

[jartsa]

irrelevant

irrelevant

impossible


flat out wrong

Excuse me, but it's correct. At speed 0.86 c velocity change is as large as Newton predicts, if we use 8 times more force than what Newton's says to be the correct force for that velocity change. Right?

So eight times more work will be done in this velocity change than what Newton says. E = F*d

 

[HomogenousCow]

Excuse me, but it's correct. At speed 0.86 c velocity change is as large as Newton predicts, if we use 8 times more force than what Newton's says to be the correct force for that velocity change. Right?

So eight times more work will be done in this velocity change than what Newton says. E = F*d

Can anyone decipher what he's trying to say

 

[PAllen]

Excuse me, but it's correct. At speed 0.86 c velocity change is as large as Newton predicts, if we use 8 times more force than what Newton's says to be the correct force for that velocity change. Right?

So eight times more work will be done in this velocity change than what Newton says. E = F*d

If γ is exactly 2, then, at the moment, 3-force is 8 times Newtonian. But integrating over any finite distance, however small, will not be the same 1/2 (γ^3)m v^2. Note, in particular, that:

∫Fdx \propto v^2/2

requires that dp/dt \propto dv/dt, which is false for relativity. You can't pretend gamma is constant for the integration.

So, I repeat, every part of your persistent argument that you can rescue 1/2 mv^2 using any standard form of relativistic mass is just wrong. See post #12, where Pervect kindly wrote the correct formulas.

 

[Dadface]

Jartsa, may I suggest that you look again at relativity and try to understand it better. If you then get stuck you could start a new thread to ask for clarification. Your comments here have been misleading and are not helpful to the education of bobie, the person who asked the original question on this thread.