Basic Energy-Momentum Issue

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  • #1
mysearch
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Hi, I am trying to quickly resolve a fairly basic question that cropped when considering relativity. Classically, the total energy of a system is often described in term of 3 components:

Total Energy = Rest Mass + Kinetic + Potential

If I ignore potential energy, i.e. a particle moving in space far from any gravitational mass, then I assume the general form above can be reduced to:

[1] [tex]E_T = m_o c^2 + 1/2mv^2[/tex]

Now [tex]m_o[/tex] is the rest mass, while I assume [m] has to be described as the relative mass as a function of its velocity [v], i.e.

[2] [tex]m = \frac {m_o}{\sqrt{(1-v^2/c^2)}}[/tex]

However, relativity also introduces the idea of relativistic momentum:

[4] [tex] p = mv = \frac {m_o v}{\sqrt{(1-v^2/c^2)}}[/tex]

However, the following link show the definition of `Relativistic Energy in Terms of Momentum’: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c4 which I have expanded to the following form:

[5] [tex]E_X^2 = m_o^2 c^4 + p^2c^2 = m_o^2 c^4 + m^2v^2c^2[/tex]

Now my initial assumption was that [tex][E_X \equiv E_T][/tex], but examination of equations [1] and [5] suggests that this cannot be the case. Could somebody explain my error or the difference in the implied energy of these 2 equations?

As a side issue, energy is a scalar quantity, while momentum is a vector quantity. I see how multiplying [p] by [c] gets us back to the units of energy, but was slightly unsure about the maths of mixing these quantities. Would appreciate any clarification of the issues raised. Thanks
 
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  • #2
Hootenanny
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Whilst you're initial assumption is correct, i.e. the total energy of a body is the sum of it's rest energy, kinetic energy and potential energy; your expression isn't. You're mixing classical and relativistic terms.

[tex]E_T = m_o c^2 + 1/2mv^2[/tex]
Note that [itex]E_k = \frac{1}{2}mv^2[/itex] is a strictly classical definition of kinetic energy, you can't simply substitute the relativistic mass for m. Instead, the relativistic kinetic energy is defined as [itex]E_k = \gamma m_0 c^2 - m_0 c^2[/itex] and is derivable from the expression for momentum.

As a side issue, energy is a scalar quantity, while momentum is a vector quantity. I see how multiplying [p] by [c] gets us back to the units of energy, but was slightly unsure about the maths of mixing these quantities.
It is perfectly acceptable to multiply a vector by a scalar, in fact scalar multiplication is one of the operations which defines a vector space. Simply put, to multiply a vector by a scalar you simply multiply each of the components of the vector by the scalar quantity.

For example suppose we have a vector [itex]\bold{v} = \left(v_1, v_2, v_3\right)[/itex] and a scalar [itex]a[/itex]. Then:

[tex]a\cdot\bold{v} = a\cdot\left(v_1, v_2, v_3\right) = \left(a\cdot v_1, a\cdot v_2, a\cdot v_3\right)[/tex]
 
  • #3
Fredrik
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[1] is wrong. You can't use the non-relativistic formula for kinetic energy. The kinetic energy is [itex]mc^2-m_0c^2[/itex].

Edit: D'oh, I was too slow again.

Mysearch, if you want to see a derivation, check out #15 in this thread. Read the stuff at the end first to see the difference between your notation and mine.
 
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  • #4
Dale
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Note that [itex]E_k = \frac{1}{2}mv^2[/itex] is a strictly classical definition of kinetic energy, you can't simply substitute the relativistic mass for m. Instead, the relativistic kinetic energy is defined as [itex]E_k = \gamma m_0 c^2 - m_0 c^2[/itex] and is derivable from the expression for momentum.
You can also take this expression for kinetic energy and do a Taylor series expansion about v = 0. When you do that you recover the classical definition of kinetic energy as the first term.
 
  • #5
mysearch
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Many thanks for both quick responses.
I will follow up on the clarifications and link provided.
I have another issue, related to the conservation of energy,
but will raise it in a separate thread.
Thanks again.
 
  • #6
mysearch
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Just wanted to say thanks again for the response in #2, #3 & #4. You were all right to point to equation [1], as [tex][1/2mv^2][/tex] is only a low speed approximation.

Thanks for the clarification in #2 about scalars and vectors, I was aware that you could multiply a vector by a scalar, which gives you a vector, but as I was looking for an issue that would explain the 'apparent' discrepancy. The issue that I was worrying about was linked to the equation [tex]E^2 = m_o^2 c^4 + p^2c^2[/tex] and the concern (mistakenly) that a momentum vector was being added to scalar energy. Anyway, really appreciated the help.
 

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