# Basic Energy-Momentum Issue

1. Aug 23, 2008

### mysearch

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] $$E_T = m_o c^2 + 1/2mv^2$$

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

However, relativity also introduces the idea of relativistic momentum:

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

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] $$E_X^2 = m_o^2 c^4 + p^2c^2 = m_o^2 c^4 + m^2v^2c^2$$

Now my initial assumption was that $$[E_X \equiv E_T]$$, 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

Last edited by a moderator: Aug 23, 2008
2. Aug 23, 2008

### Hootenanny

Staff Emeritus
Re: [Something descriptive here will yield more responses]

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.

Note that $E_k = \frac{1}{2}mv^2$ 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 $E_k = \gamma m_0 c^2 - m_0 c^2$ and is derivable from the expression for momentum.

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 $\bold{v} = \left(v_1, v_2, v_3\right)$ and a scalar $a$. Then:

$$a\cdot\bold{v} = a\cdot\left(v_1, v_2, v_3\right) = \left(a\cdot v_1, a\cdot v_2, a\cdot v_3\right)$$

3. Aug 23, 2008

### Fredrik

Staff Emeritus
[1] is wrong. You can't use the non-relativistic formula for kinetic energy. The kinetic energy is $mc^2-m_0c^2$.

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.

Last edited: Aug 23, 2008
4. Aug 23, 2008

### Staff: Mentor

Re: [Something descriptive here will yield more responses]

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. Aug 23, 2008

### mysearch

Many thanks for both quick responses.
Just wanted to say thanks again for the response in #2, #3 & #4. You were all right to point to equation [1], as $$[1/2mv^2]$$ 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 $$E^2 = m_o^2 c^4 + p^2c^2$$ and the concern (mistakenly) that a momentum vector was being added to scalar energy. Anyway, really appreciated the help.