Relativistic Energy

  • #1
aliens123
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In Robert M. Wald's General Relativity he writes on page ##61##:

The energy of a particle as measured by an observer - present at the site of the particle - whose 4-velocity is ##v^a## is defined by
$$E=-p_a v^a$$
Thus, in special relativity, energy is recognized to be the "time component" of the 4-vector ##p^a.##

Is this a typo? We get the following:
$$E=-p_a v^a$$
$$E=-p_a(m \gamma) v^a / (m\gamma)$$
$$E=-p_a p^a /(m \gamma)$$
$$E = - (-m^2) / m \gamma)$$
$$E= m/\gamma \neq m \gamma = p^0.$$
 
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  • #2
Ibix
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I'm not sure I follow your maths.

The result as stated by Wald is correct. It's easiest to see by choosing to work in a frame wherein ##v^a## represents rest - then ##v^t=1## and ##v^x=v^y=v^z=0##, and then your first expression reduces to ##E=-p_t=\gamma m##, where ##\gamma## is the Lorentz gamma factor associated with ##p^a##, as measured in this frame.
 
  • #3
aliens123
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I'm not sure I follow your maths.

The result as stated by Wald is correct. It's easiest to see by choosing to work in a frame wherein ##v^a## represents rest - then ##v^t=1## and ##v^x=v^y=v^z=0##, and then your first expression reduces to ##E=-p_t=\gamma m##, where ##\gamma## is the Lorentz gamma factor associated with ##p^a##, as measured in this frame.
In a frame where the particle is at rest ##\gamma =1.## So of course ##m/\gamma = m\gamma##. How is my math confusing?
 
  • #4
Ibix
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In a frame where the particle is at rest $\gamma =1.$ So of course $m/\gamma = m\gamma$. How is my math confusing?
You seem to be confusing the four velocity of the observer, ##v^a##, with the four velocity of the particle (which does not have a symbol defined, but would be ##p^a/m## if ##m## is the mass of the particle). You also seem to be using ##m## for the masses of both the particle and the observer. And you seem to have tried to get a four-momentum by multiplying a velocity by gamma times mass, where it should be just the mass.

So I'm proposing using the frame where the observer is at rest, not the frame where the particle is at rest.
 
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  • #5
aliens123
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You seem to be confusing the four velocity of the observer, ##v^a##, with the four velocity of the particle (which does not have a symbol defined, but would be ##p^a/m## if ##m## is the mass of the particle. You also seem to be using ##m## for the masses of both the particle and the observer. And you seem to have tried to get a four-momentum by multiplying a velocity by gamma times mass, where it should be just the mass.
Wald defines:

The tangent vector ##u^a## to a timlike curve parametrized by ##\tau## is called the 4-velocity of the curve,
$$u^a u_a=-1$$
The energy momentum 4-vector, ##p^a##, of a particle of mass ##m## is defined by
$$p^a = m u^a.$$

So I was operating under the assumption that
$$v^a = (1, dx/dt, dy/dt, dz/dt)$$
$$u^a = (dt/d\tau, dx/d\tau, dy/d\tau, dz/d\tau) = v^a \cdot dt/d\tau = v^a \gamma$$

Where the ##v^a## is of the particle. But you are saying this is incorrect, the ##v^a## here refers to an observer?
 
  • #6
Ibix
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##v^a## is also a four velocity so, like ##u^a##, it is ##\partial x^\mu/d\tau##. It's just defined along a different curve from ##u^a## - your worldline rather than the particle's worldline

Yes, in this context, Wald is saying that the energy you measure is the inner product of your four velocity with the particle's four momentum. If you think about what an inner product is, and note that your four velocity is parallel to the timelike axis of your instantaneous co-moving frame, he's just saying (in a properly covariant way) that the timelike component of the particle's four momentum in your frame is your measure of its energy.

Incidentally, on this forum you get inline maths with two # symbols before and after, rather than one $ symbol.
 
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  • #7
robphy
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A particle with 4-momentum [itex] p^a [/itex] has 4-velocity [itex] u^a = p^a/(\sqrt{-p_b p^b}) [/itex].

That particle has energy [itex] E=-p_a v^a [/itex] according to the observer with 4-velocity [itex] v^a [/itex].
(Note that [itex] v^a v_a = -1 [/itex] [ [itex] v^a [/itex] is a future-directed unit timelike vector].)
 
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  • #8
aliens123
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Imagine we are at rest and there is a particle with momentum ##p^a.## To us, the particle has energy $E = p^0.$ Imagine there is a second observer whose velocity with respect to us is ##v^a.## To them, the particle has energy ##\tilde{E} = p^a v_a = -p^0 v^0 + p^1 v^1 + p^2 v^2 + p^3 v^3.##

Also though, the components of ##p^a## measured in the rest frame of the second observer will be given by
$$p'^a = \Lambda (v^c)^a_{\ b} p^b$$
Where ## \Lambda (v^c)## is the Lorentz boost of velocity ##v^a.## So
$$p'^0 = \tilde{E} = \Lambda (v^c)^0_{\ b} p^b = \Lambda (v^c)^0_{\ 0} p^0 + \Lambda (v^c)^0_{\ 1} p^1 + \Lambda (v^c)^0_{\ 2} p^2 + \Lambda (v^c)^0_{\ 3} p^3$$
Implying that $$ \Lambda (v^c)^0_{\ 0} = -v^0, \Lambda (v^c)^0_{\ 1} = v^1, \Lambda (v^c)^0_{\ 2} = v^2, \Lambda (v^c)^0_{\ 3} = v^3$$
Is this correct?
 
  • #9
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In Robert M. Wald's General Relativity he writes on page ##61##:

The energy of a particle as measured by an observer - present at the site of the particle - whose 4-velocity is ##v^a## is defined by
$$E=-p_a v^a$$
Thus, in special relativity, energy is recognized to be the "time component" of the 4-vector ##p^a.##

Is this a typo? We get the following:
$$E=-p_a v^a$$
$$E=-p_a(m \gamma) v^a / (m\gamma)$$
$$E=-p_a p^a /(m \gamma)$$
Wald is correct, that is not a typo. The last step that I quoted is wrong. The mass of the particle times the four-velocity of the observer is not the four-momentum of the particle.

Edit: I see that has already been pointed out. I am just a little late.
 
  • #10
PeroK
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Imagine we are at rest and there is a particle with momentum ##p^a.## To us, the particle has energy $E = p^0.$ Imagine there is a second observer whose velocity with respect to us is ##v^a.## To them, the particle has energy ##\tilde{E} = p^a v_a = -p^0 v^0 + p^1 v^1 + p^2 v^2 + p^3 v^3.##

Also though, the components of ##p^a## measured in the rest frame of the second observer will be given by
$$p'^a = \Lambda (v^c)^a_{\ b} p^b$$
Where ## \Lambda (v^c)## is the Lorentz boost of velocity ##v^a.## So
$$p'^0 = \tilde{E} = \Lambda (v^c)^0_{\ b} p^b = \Lambda (v^c)^0_{\ 0} p^0 + \Lambda (v^c)^0_{\ 1} p^1 + \Lambda (v^c)^0_{\ 2} p^2 + \Lambda (v^c)^0_{\ 3} p^3$$
Implying that $$ \Lambda (v^c)^0_{\ 0} = -v^0, \Lambda (v^c)^0_{\ 1} = v^1, \Lambda (v^c)^0_{\ 2} = v^2, \Lambda (v^c)^0_{\ 3} = v^3$$
Is this correct?
I'm not sure I follow what you're doing there. In any case, the inner product of two four-vectors is invariant. It's a good exercise to show this for four-vectors ##\mathbf{A, B}## generally:
$$\mathbf{A} \cdot \mathbf{B} = A_aB^a = A^aB_a$$ is invariant.

In this particular case, as ##E = -\mathbf{p} \cdot \mathbf{v}## is the energy of the particle in the observer's rest frame, it is the energy of the particle (as measured by the observer) in all frames. That is, in another frame, both ##\mathbf p## and ##\mathbf v## transform according to the four-vector transformation rules, but the inner product is invariant.

Note the invariance of the inner product (and, in fact, all scalars) is an absolutely fundamental result in relativity.
 
  • #11
vanhees71
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In Robert M. Wald's General Relativity he writes on page ##61##:

The energy of a particle as measured by an observer - present at the site of the particle - whose 4-velocity is ##v^a## is defined by
$$E=-p_a v^a$$
Thus, in special relativity, energy is recognized to be the "time component" of the 4-vector ##p^a.##

Is this a typo? We get the following:
$$E=-p_a v^a$$
$$E=-p_a(m \gamma) v^a / (m\gamma)$$
$$E=-p_a p^a /(m \gamma)$$
$$E = - (-m^2) / m \gamma)$$
$$E= m/\gamma \neq m \gamma = p^0.$$
You misread the meaning of ##v^a##. It's the four-velocity (components) of the observer (wrt. to computational frame). What Wald calculates (obviously using the east-coast (-+++) signature and the normalization of the four-velocity ##v_a v^a=-c^2##) is the energy of the particle as measured by the observer i.e., in his (momentary) inertial rest frame.
 
  • #12
aliens123
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I'm not sure I follow what you're doing there. In any case, the inner product of two four-vectors is invariant. It's a good exercise to show this for four-vectors ##\mathbf{A, B}## generally:
$$\mathbf{A} \cdot \mathbf{B} = A_aB^a = A^aB_a$$ is invariant.

In this particular case, as ##E = -\mathbf{p} \cdot \mathbf{v}## is the energy of the particle in the observer's rest frame, it is the energy of the particle (as measured by the observer) in all frames. That is, in another frame, both ##\mathbf p## and ##\mathbf v## transform according to the four-vector transformation rules, but the inner product is invariant.

Note the invariance of the inner product (and, in fact, all scalars) is an absolutely fundamental result in relativity.
Isn't energy not an invariant though?
 
  • #13
PeroK
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Isn't energy not an invariant though?
Energy, generally, is frame dependent. Always has been.
 
  • #14
aliens123
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Energy, generally, is frame dependent. Always has been.
Ah okay, I just realized what you mean. ##-\mathbf{p}\cdot \mathbf{v}## in all frames is the energy an observer with velocity ##\mathbf{v}## in their frame?
 
  • #15
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Energy, generally, is frame dependent.

While this is true, it is also misleading as you state it. The quantity ##E = - \mathbf{p} \cdot \mathbf{v}## is an invariant. This quantity is not "energy", but a much more specific, direct observable: energy of a particular object as measured by a particular observer. Results of direct measurements are invariants; all observers and all frames will agree that that particular observer will measure the value ##E## for the energy of that particular object.

##-\mathbf{p}\cdot \mathbf{v}## in all frames is the energy an observer with velocity ##\mathbf{v}## in their frame?

No. The equation ##E = - \mathbf{p} \cdot \mathbf{v}## is valid in all frames, like all equations involving 4-vectors. In whichever frame you pick, you just compute the dot product of ##\mathbf{p}## and ##\mathbf{v}## using the components of those 4-vectors in that frame. No matter which frame you pick, you wlll get the same numerical result, ##E##.

What might be confusing you is that the word "energy", used to describe ##E##, has a much more specific meaning than just the general concept "energy", as explained above.
 
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  • #16
PeroK
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Ah okay, I just realized what you mean. ##-\mathbf{p}\cdot \mathbf{v}## in all frames is the energy an observer with velocity ##\mathbf{v}## in their frame?
Not quite. You have an observer and a particle. The observer measures the energy of that particle and gets the answer ##E##. They write that down on a piece of paper.

The quantity that is written on that piece of paper is ##E = -\mathbf{p}\cdot \mathbf{v}##, where ##\mathbf{p, v}## are the four momentum of the particle and the four velocity of the observer.

If you ask someone to calculate what is written on that piece of paper, they may use the four momentum and four velocity, as measured in their reference frame, and take the inner product. This does not give the energy of the particle that they measure, but calculates the energy of the particle that the other observer measures.

They could, of course, transform the four-momentum of the particle to the other observer's reference frame. But, they don't need to do that: they can do the calculation using the quantities they measure.
 
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