Energy momentum four vector

  • #1
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In Special Relativity, we have the four vector, (E/c, px, py, pz). However, isn't the first term just `p` given that `E=pc` for a photon? Why is it an energy-momentum four vector when the first term isn't really energy but momentum?
 

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  • #2
Orodruin
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In Special Relativity, we have the four vector, (E/c, px, py, pz). However, isn't the first term just `p` given that `E=pc` for a photon? Why is it an energy-momentum four vector when the first term isn't really energy but momentum?
For a photon (or indeed any massless particle), yes, the energy is ##pc##. However, this is not true for a general massive particle.

Edit: Also, typically, it is just called the 4-momentum.
 
  • #3
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For a photon (or indeed any massless particle), yes, the energy is ##pc##. However, this is not true for a general massive particle.
Wouldn't the dimension of that quantity still not be energy? And be momentum instead
 
  • #4
Orodruin
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Wouldn't the dimension of that quantity still not be energy?
Usually we work in units where ##c = 1##. In those units energy and momentum have the same dimension. If you work in units where ##c \neq 1## yes, the dimensions are those of momentum, that does not mean that the interpretation of the time component is not the energy divided by ##c## (which it is).
 
  • #5
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Usually we work in units where ##c = 1##. In those units energy and momentum have the same dimension. If you work in units where ##c \neq 1## yes, the dimensions are those of momentum, that does not mean that the interpretation of the time component is not the energy divided by ##c## (which it is).
I understand the part where if you set c=1 E=p. I don't get the other part? Time component?
 
  • #6
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Every system—massive or massless—has a four-momentum ##\mathbf{P} = (E, \mathbf{p}c)## whose magnitude ##\sqrt{E^2 - (\mathbf{p}c \cdot \mathbf{p}c)} = \sqrt{E^2 - (pc)^2} = mc^2## is the system's rest energy (mass). If the system happens to be massless (like a light wave), then yes, the time component equals the magnitude of the spatial 3-vector component, as it must to make the magnitude zero.
 
  • #7
I understand the part where if you set c=1 E=p. I don't get the other part? Time component?
If you use unit system with c=1 dimensionless, then Energy, momentum and mass have the same units: mass.

There are no contradiction in that, for example in International System energy and torque are the same units, and there are two different magnitudes.
 
  • #8
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I understand the part where if you set c=1 E=p. I don't get the other part? Time component?
Emphasis mine.


As far as I have seen, every four-vector has three spatial components and one time component. This wikipedia article goes into a little bit (use control F and search for "time component"):
https://en.wikipedia.org/wiki/Four-vector


For example, consider this position vector:

X = (ct, x, y, z).

ct is the time component, and x, y, and z are the space components.

Then take the derivative with respect to proper time dτ, and you'll get a 4-velocity:

U = (γc, γux, γuy, γuz)

(for one coordinate example, the x coordinate: $$\frac{d}{dτ} x = \frac{dx}{dτ} = \frac {dx}{dt} \frac{dt}{dτ} = \frac {dx}{dt} γ $$,

because $$\frac{dt}{dτ} = γ$$,

so you'll get γux)

Notice you have four components. γc is the time component. Keep doing this for all of them. There are always four components, one of which is a time component. You can shorten this to just two symbols, as long as you define γu = (γux, γuy, γuz), so you'd have a 4-velocity of (γc, γu) where γc is the time component and γu is the spatial components of the 4-vector. This, as far as I have seen, is the same for all 4-vectors (three spatial components and one time component) For example, you'll sometimes see (E/c, p) where p is the three-momentum and E/c is the time component of 4-momentum.

Basically, the 4-vector for momentum is: P = m0U = m0(γc, γu) = (γm0c, γm0u) = (γm0c2/c, γm0u) = (E/c, p),


if you remember that energy can be written as E = γm0c2 and that relativistic three-momentum is p = γm0u, with m0 being the invariant rest mass and γ the Lorentz factor.





It's certainly mathematically convenient, but it does deeper than that, I think. There are pretty significant symmetry relations between energy and time and then space and momentum. For example, their conservation laws are connected to it (energy conservation is connected to time symmetry of the laws of physics and momentum conservation is connected to the spatial symmetry of the laws of physics).


In any event, as far as I can tell, ALL four vectors have one time component and three spatial components. I could be wrong, but that's all I've seen thus far in my tour of special relativity.
 
  • #9
PeterDonis
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ALL four vectors have one time component and three spatial components
More precisely, all 4-vectors have one time component and three spatial components in a coordinate chart with one timelike and three spacelike coordinates. The components of a 4-vector depend on the coordinate chart; there are charts that do not have one timelike and three spacelike coordinates.
 
  • #10
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More precisely, all 4-vectors have one time component and three spatial components in a coordinate chart with one timelike and three spacelike coordinates. The components of a 4-vector depend on the coordinate chart; there are charts that do not have one timelike and three spacelike coordinates.
That is interesting. As you probably know by now, I’m new to branching out of the basic version of SR and trying to jump into the adult version. Every little by helps.

Seems like my previous post broke btw. Don’t know what happened lol.
 
  • #11
Mister T
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That is interesting. As you probably know by now, I’m new to branching out of the basic version of SR and trying to jump into the adult version. Every little by helps.
Are you familiar with the spacetime interval? And what it means for an interval to be timelike, lightlike, or spacelike? I suggest you start there with the relationship between space and time. Then when going into the relationship between and energy and momentum you will better understand the temporal and spatial components of the energy-momentum 4-vector.
 
  • #12
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Are you familiar with the spacetime interval? And what it means for an interval to be timelike, lightlike, or spacelike? I suggest you start there with the relationship between space and time. Then when going into the relationship between and energy and momentum you will better understand the temporal and spatial components of the energy-momentum 4-vector.
I'm familiar with the concepts but only on the first/second year physics student level. As in, for example, a lightlike spacetime interval means Δs2=0. But I have not contemplated that with respect to other 4-vector relationships, or really any further than that (other than the obvious stuff: there can't be causal relations between events separated by spacelike intervals, etc).

At this point, all I know regarding time and spatial components is that, at least in the things I've seen (PeterDonis pointed out some things I haven't), if you start with a basic vector and use kinematics to derive something, the thing you get from the time coordinate is going to be the time component, and same with the spatial ones. I also read a paper on Einstein where he called energy the "time component," or something to that affect, which reaffirmed that notion to me.

But as I said, there seems to be some exceptions, and don't worry, I'm gonna get into to them, lol. Still working on tensors too.
 

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