Spacetime Interval & Energy-Momentum 4Vec: Reconciling Hyperbolic Geometry

[MattGeo]

TL;DR Summary

How are the spacetime interval and the energy-momentum four vector related? How is velocity in the space-direction reconciled with the fact that the time-direction component of the energy-momentum four vector contains a kinetic energy term if the object is not at rest?

In a spacetime diagram the spatialized time direction is the vertical y-axis and the pure space direction is the horizontal x-axis, ct and x, respectively.

The faster you go and therefore the more kinetic energy you have, you'll have a greater component of your spacetime vector in the xx-direction. More of your energy and forward "motion" through spacetime is devoted to traveling through space than through time. A consequence is time dilation with which we are familiar.

My question arises because I am confused about how this relates to the energy-momentum 4-vector where the time component of this is mc^2 + 1/2mv^2. The rest mass energy plus the kinetic energy. If the kinetic energy term is rather large, you have a large time component in the energy-momentum 4-vector, but if your kinetic energy is large shouldn't you be traveling "less" through time and "more" through space? There is some subtle disconnect here for me and I would appreciate it if someone could help me to think about this properly.

If the kinetic energy is zero we are left with simply mc^2, the energy that mass has on its own at rest. This tells me that it is the energy that the mass has as it moves through spacetime solely in the time direction. So if you add kinetic energy, the time component becomes larger, and more energy is devoted to travel in the time direction in spacetime. How do we reconcile that with larger energies and velocities means you travel more in the space-direction in the spacetime diagram where we consider the spacetime interval?

Furthermore, would this inability to reconcile this have to do with the hyperbolic geometry of Minkowski spacetime and how it changes the Euclidean Pythagorean relationship to a Non-Euclidean geometry?

 

[PeroK]

I think your post highlights the pointlessness of seeing a particle at rest as "moving solely through the time dimension".

There is a clear relationship between the spacetime coordinates of a particle and its energy-momentum:
$$p^{\mu} = m\frac{dx^{\mu}}{d\tau}$$
Where $\tau$ is the proper time of the particle. That is a better starting point.

 

[MattGeo]

Is there a way you can expand on this with some conceptual elaboration?

 

[PeroK]

Is there a way you can expand on this with some conceptual elaboration?

Try this, for example:

http://home.thep.lu.se/~malin/LectureNotesFYTA12_2016/SR4.pdf

Do you have a textbook on SR?

 

[Ibix]

the energy-momentum 4-vector where the time component of this is mc^2 + 1/2mv^2.

Only approximately - it's actually $\gamma mc^2$.

if your kinetic energy is large shouldn't you be traveling "less" through time and "more" through space?

As @PeroK says, there are better ways to look at this. However, in an inertial frame the spatial components of the energy-momentum tensor have magnitude $\gamma mvc$ while the time component is $\gamma mc^2$ (edit: and not the other way around, as I initially wrote). That means the ratio of the "motion through space" to "motion through time" is $v/c$, which does increase as velocity increases, as you seem to have expected.

 

[MattGeo]

Try this, for example:

http://home.thep.lu.se/~malin/LectureNotesFYTA12_2016/SR4.pdf

Do you have a textbook on SR?

unfortunately I don't have a textbook on Special Relativity. It is discussed in some of my Modern Physics texts and some of my Mechanics texts. I have a beginner's knowledge of tensors and have only ever used them in engineering and continuum mechanics applications. Much of the notation for tensors in relativity is foreign to me.

 

[MattGeo]

Only approximately - it's actually $\gamma mc^2$.
As @PeroK says, there are better ways to look at this. However, in an inertial frame the spatial components of the energy-momentum tensor have magnitude $\gamma mc^2$ while the time component is $\gamma mvc$. That means the ratio of the "motion through space" to "motion through time" is $v/c$, which does increase as velocity increases, as you seem to have expected.

are you sure that the spatial components of the energy-momentum 4 vector have magnitude ϒmc^2? I was fairly sure that was the time component? Unless I have had it backwards all this time.

 

[Ibix]

are you sure that the spatial components of the energy-momentum 4 vector have magnitude ϒmc^2? I was fairly sure that was the time component? Unless I have had it backwards all this time.

Yes - sorry, I stated that backwards. I'll correct it. The last sentence that you quoted is correct as written, however.

 

[MattGeo]

So in a spacetime diagram, for instance, if the velocity of some object initially at increases then its spacetime interval will rotate from being only in the time direction to a combination of the time direction and space direction. It now has more "direction in space" and "less direction in time". But the energy-momentum 4 vector is telling us that a larger kinetic energy means a larger contribution from the time component of the 4 vector. How do these two things make sense?

 

[SiennaTheGr8]

Because "more/less direction in space/time" and "larger/smaller contribution from the space/time component" aren't the same concept.

A four-vector's magnitude is the difference of the squares of the temporal and spatial components. So, e.g., if you rotate the timelike unit vector $(1, \vec 0)$ so that it points "more in space" and "less in time" than it did before the rotation, then both the temporal component and the magnitude of the spatial three-vector component must increase in order to preserve the four-vector's magnitude of $1$.

This is different from Euclidean vectors, where the magnitude is the sum of the squares of the Cartesian components. In that more familiar case, rotating a vector so that it points "more in $x$" and "less in $y$" (say) means that (the absolute value of) the $x$-component is increasing and (the absolute value of) the $y$-component is decreasing. It sounds like you might be trying to apply that Euclidean intuition to Minkowski spacetime.

 

[Ibix]

So in a spacetime diagram, for instance, if the velocity of some object initially at increases then its spacetime interval will rotate from being only in the time direction to a combination of the time direction and space direction.

This makes no sense. Interval is the Minkowski geometry equivalent of distance in Euclidean geometry. Does "its distance will rotate" make any sense?

I think what you are trying to say is that an object that changes speed with respect to some inertial frame has a worldline that points in different directions at different times. That would be correct. Interval is the "distance" along the worldline between two events (SiennaTheGr8's post explains why I put distance in scare quotes). The energy-momentum vector is a vector pointing in the direction of the worldline at any given event. You can talk about the direction of a vector or a worldline - but not the direction of an interval.

But the energy-momentum 4 vector is telling us that a larger kinetic energy means a larger contribution from the time component of the 4 vector. How do these two things make sense?

For the reason I gave in my corrected post #5. The magnitude of the time component of your energy-momentum four vector is proportional to $\gamma$ but the magnitude of the spatial components is proportional to $\gamma v$. Sure the timelike component increases, but the spacelike component increases more.

 

[PeterDonis]

More of your energy and forward "motion" through spacetime is devoted to traveling through space than through time.

You are implicitly assuming that one particular frame is somehow preferred, so that motion relative to that frame counts as "some motion devoted to traveling through space instead of through time".

This is not correct. There is no preferred frame, and there is no sense in which, given two observers in relative motion, one is "traveling only through time" while the other is "traveling through space as well as through time". The only difference between the two observers is in which timelike direction their respective 4-velocity vectors point, and all timelike directions are equivalent.

I am confused about how this relates to the energy-momentum 4-vector where the time component of this is mc^2 + 1/2mv^2

No, the time component is $mc^2 \gamma$. What you have given is just a low velocity approximation.

However, the time component by itself has no physical significance. What has physical significance is (a) the norm of the energy-momentum 4-vector, i.e., the rest mass, and (b) the inner product of the energy-momentum 4-vector with some other observer's 4-velocity, i.e., the energy that other observer would measure the object to have.

You need to reconsider your understanding in the light of the above.

 

[PeterDonis]

in a spacetime diagram, for instance, if the velocity of some object initially at increases then its spacetime interval will rotate from being only in the time direction to a combination of the time direction and space direction.

No, the object's 4-velocity vector will rotate from pointing in one timelike direction in spacetime, to pointing in another timelike direction in spacetime. All timelike directions are equivalent. And 4-velocity vectors are not the same things as spacetime intervals. You need to be careful not to confuse the two.

 

[PeterDonis]

the timelike component increases, but the spacelike component increases more.

No, the timelike component increases more; $\gamma$ is larger than $\gamma v$ since $v < 1$. The timelike component has to increase more since the difference in the squares of the two has to remain constant at $mc^2$.

 

[Ibix]

No, the timelike component increases more; $\gamma$ is larger than $\gamma v$ since $v < 1$. The timelike component has to increase more since the difference in the squares of the two has to remain constant at $mc^2$.

The ratio of the two components, $p^x/p^t=v/c$, increases with $v$, is the sense I was meaning. I did state it that way in #5.

 

[PeterDonis]

The ratio of the two components, px/pt=v/cp^x/p^t=v/c, increases with vv, is the sense I was meaning. I did state it that way in #5.

Ah, ok, I missed that specification in #5.

 

[Ibix]

Ah, ok, I missed that specification in #5.

Actually, I'm not sure what sense $p^t$ increases more than $p^x$. In any frame and after any acceleration, the tip of the arrow representing the object's four momentum lies on a hyperbola identified by that object's (assumed constant) invariant mass $m$. If, at any stage, $p^t$ were to increase more than $p^x$ then the slope of the hyperbola would be greater than one - but it is everywhere between $\pm$1. (Edit: to put it another way, the implied four-acceleration would be timelike.) So the absolute change in $p^x$ must be larger than in $p^t$, and we've agreed about the ratio. Did you fail to account sufficiently for the large "head start" $p^t$ has over $p^x$ at $v=0$ in your constant modulus argument, or am I missing something?

Aside: a hyperbola is a curve where the straight line interval from some fixed event is constant along the curve. Is there a name for that constant interval? It's analogous to the radius of a circle in Euclidean geometry. Wikipedia offers "semi-major axis", but that seems to be in the context of a hyperbola on a Euclidean plane, rather than a Minkowski plane.

 

[PeterDonis]

I'm not sure what sense $p^t$ increases more than $p^x$.

$p^t$ will always have a higher numerical value than $p^x$. But you are right that the ratio $p^x / p^t$ increases as $v$ increases.

 

[vanhees71]

Have a look at

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[robphy]

Just an fyi...
the OP asked this question earlier at
https://physics.stackexchange.com/q...acetime-interval-in-a-spacetime-diagram-relat

I contributed one of the answers with an energy-momentum diagram.

I think the question is asking
how,
while keeping mass fixed,
that increasing relativistic kinetic energy,
which increases the relativistic energy, the timelike-component of the 4-momentum,
also
increases the spacelike-component of the 4-momentum
and thus leads to a larger velocity (larger rapidity).

 

[Ibix]

I think the question is asking
how,
while keeping mass fixed,
that increasing relativistic kinetic energy,
which increases the relativistic energy, the timelike-component of the 4-momentum,
also
increases the spacelike-component of the 4-momentum
and thus leads to a larger velocity (larger rapidity).

Because you can't increase kinetic energy without also increasing momentum, surely?

 

[robphy]

Because you can't increase kinetic energy without also increasing momentum, surely?

Yes, when the mass is fixed, and likely other cases.

But, I suspect that these various concepts haven't been fully connected by the OP
...and others (from similar questions I've seen in the past, here and elsewhere)...
there are misconceptions...

...hence the energy-momentum diagram.

Note that if we allow the mass to change [decreasing mass],
increasing the kinetic energy (as example "B" above)
can be done
with constant relativistic momentum
and decreasing relativistic energy.

So, I think the diagram helps,
as well as the important idea that one has to be specific
about how one varies quantities (especially when taking limits... like how to get to the massless case).