How can (1/2)mv^2 and p^2/2m give different results?

[sawer]

I am trying to understand the energy and coenergy concept.

How can
$$Energy = \frac{P^2}{2m}$$
and
$$CoEnergy = \frac{1}{2} mv^2$$
be different?
If the reason is increasing of mass according to relativity, both of them have masses.I don't understand the differences...

 

[Orodruin]

Both of these expressions are classical expressions for the kinetic energy of an object and give the same result in Newtonian physics, where $p = mv$.

In relativity, neither of these expressions equate to the energy or the kinetic energy of an object apart from in the Newtonian limit.

What makes you think the expressions would give different results?

 

[sawer]

Both of these expressions are classical expressions for the kinetic energy of an object and give the same result in Newtonian physics, where $p = mv$.

In relativity, neither of these expressions equate to the energy or the kinetic energy of an object apart from in the Newtonian limit.

What makes you think the expressions would give different results?

In The Mathematical Structure of Classical and Relativistic Physics book page:263

For relativistic cases the two equations give different results, but I don't understand why.
As I said if it is about mass increasing, then both of the equations have mass parameters. Because p=mv.

I am asking how can (1/2)mv^2 and p^2/2m be different? What is the reason?

 

[Ibix]

In the relativistic case, $p=\gamma mv$, and $KE=(\gamma-1)mc^2$, where $\gamma=1/\sqrt{1-(v/c)^2}$. If you Taylor expand the expression for $\gamma$ you can recover the Newtonian expressions where v<<c. But you can see that you would not expect your relationship to hold.

Basically, the equality you are expecting is only ever approximate. When you are at low velocity, the error in that approximation is not important.

 

[Orodruin]

As I said if it is about mass increasing, then both of the equations have mass parameters. Because p=mv.

The mass as it is usually defined in modern theoretical physics is not increasing. See our FAQ: https://www.physicsforums.com/threads/what-is-relativistic-mass-and-why-it-is-not-used-much.796527/
However, the relativistic expression for momentum is not $p = mv$, but $p = m\gamma v$.

 

[sawer]

When you are at low velocity, the error in that approximation is not important.

When I am at high velocity also the same "error" can be applied this equation too: $$CoEnergy = \frac{1}{2} mv^2$$
Because it does have mass parameter as energy equation which is $$Energy = \frac{P^2}{2m}$$

I still don't understand why energy and coenergy formulas aren't equal for high velocity.

 

[Vanadium 50]

I still don't understand why energy and coenergy formulas aren't equal for high velocity.

Because both of them are approximations valid (and equal) only in the low-velocity limit.

 

[Ibix]

When I am at high velocity also the same "error" can be applied this equation too: $$CoEnergy = \frac{1}{2} mv^2$$
Because it does have mass parameter as energy equation which is $$Energy = \frac{P^2}{2m}$$

I still don't understand why energy and coenergy formulas aren't equal for high velocity.

The problem is that $p\simeq mv$ and $KE\simeq mv^2/2$. So $KE\simeq p^2/2m$. The approximation gets worse the faster you go.

As Orodruin noted, the m term in these equations does not change with velocity. Modern physics has largely dropped the concept of mass varying with velocity, primarily because of the confusion it causes in cases like this. The only thing changing in this case is v (and hence $\gamma$). And since the full relativistic equations do not have the same dependence on v that the Newtonian ones, your co-energy expression is only approximate.

 

[sawer]

Hmm...
I think this made me confused because, the books about this topic give relativistic case for energy-coenergy equations. So I thought energy and co-energy is equal for classical mechanics but for relativity, only the "energy" function is valid but co-energy gives wrong result. But this is not true either, because in relativistic case neither energy nor co-energy equation is valid. Relativistic energy equation is different from them.

So why in world, do they give "relativistic case" for these energy-coenergy equations. It makes only confusion.

Thank you all...

 

[PeterDonis]

the books about this topic

Which books? I have never seen the term "coenergy" before, so I'm curious as to which books you are reading.

 

[robphy]

Hmm...
I think this made me confused because, the books about this topic give relativistic case for energy-coenergy equations. So I thought energy and co-energy is equal for classical mechanics but for relativity, only the "energy" function is valid but co-energy gives wrong result. But this is not true either, because in relativistic case neither energy nor co-energy equation is valid. Relativistic energy equation is different from them.

So why in world, do they give "relativistic case" for these energy-coenergy equations. It makes only confusion.

Thank you all...

As opposed to the nonrelativistic case, the relativistic case shows that the two notions are actually distinct.
And this distinction leads to an example where the Lagrangian is not always equal to the kinetic-energy minus the potential energy.

Why? Isn't the title of the book referenced
"Mathematical Structure of Classical and Relativistic Physics"?

 

[sawer]

Which books? I have never seen the term "coenergy" before, so I'm curious as to which books you are reading.

System dynamics, engineering mechanics books.
Just google relativity+coenergy term in the book section.

 

[sawer]

And this distinction leads to an example where the Lagrangian is not always equal to the kinetic-energy minus the potential energy.

What does that mean?

 

[harrylin]

When I am at high velocity also the same "error" can be applied this equation too: $$CoEnergy = \frac{1}{2} mv^2$$
Because it does have mass parameter as energy equation which is $$Energy = \frac{P^2}{2m}$$

I still don't understand why energy and coenergy formulas aren't equal for high velocity.

As you noticed, in their "effort to avoid confusion" they actually caused it; surely you would not have been confused if they used the notation m₀. The reason that they didn't do so is probably philosophical.

 

[robphy]

And this distinction leads to an example where the Lagrangian is not always equal to the kinetic-energy minus the potential energy.

What does that mean?

Look on the next page of Tonti's Mathematical Structure of Classical and Relativistic Physics p.264.
The "kinetic" term of the relativistic lagrangian of a free particle is not its "relativistic kinetic energy" $(\gamma-1)mc^2$.

 

[Doug Carver]

http://www.sfu.ca/phys/100/lectures/Collisions/collisions_old.html