Straight line through spacetime.

[Rotormaster]

Since to travel in a straight line through space time requires the subject in question to have zero change to its state of motion, ie maintain constant velocity at all times. Does this mean that the only entity to ever have and that ever will achieve a straight line through space time is light?

 

[Passionflower]

Since to travel in a straight line through space time requires the subject in question to have zero change to its state of motion, ie maintain constant velocity at all times. Does this mean that the only entity to ever have and that ever will achieve a straight line through space time is light?

The path length of a photon in spacetime is zero, it is a point.

 

[Fredrik]

Since to travel in a straight line through space time requires the subject in question to have zero change to its state of motion, ie maintain constant velocity at all times. Does this mean that the only entity to ever have and that ever will achieve a straight line through space time is light?

Any particle can move as described by a "straight line" in spacetime. The technical term is "geodesic". There are three different types of geodesics: Timelike, null and spacelike. A timelike gedesic represents the motion of a massive particle in free fall. A null geodesic represents the motion of a massless particle. (Massless particles are always in free fall if they're moving through a vacuum).

If you meant to ask specifically about null geodesics, then the answer is that any massless particle in free fall would move as described by a null geodesic. The only massless particles in the standard models are photons and gluons, and I don't think gluons are ever in a state that can be described as free fall. So I guess photons are the only particles that will move this way, at least over long distances. But there could exist other (massless) particles that do the same (if they are very rare or interact very weakly with other kinds of matter...otherwise they would have been discovered by now).

 

[Rotormaster]

But aren't photons the only particle with constant velocity? would this not mean that since everything started they would be the only thing to maintain absolute synchronization with time's arrow ie achieve a straight line through space time?

 

[Ich]

Since to travel in a straight line through space time requires the subject in question to have zero change to its state of motion, ie maintain constant velocity at all times.

Moving "in a straight line" in curved spacetime means moving "along" a geodesic. From the moving particles' point of view, there is no change of their "state of motion" as they are alwasy at rest wrt themselves and feel no forces acting on them. (Photons don't have a point of view, btw.)
Measured in some coordinate system, their speed and direction may change. That's true for light also.

 

[Fredrik]

But aren't photons the only particle with constant velocity?

No. Einstein's equation tells us which curves in spacetime are geodesics. We then define zero acceleration as "motion described by a geodesic", and acceleration as a specific measure of much a particle's motion deviates from geodesic motion. So if there are any timelike geodesics at all in spacetime (and there always are), then massive particles can have zero acceleration as well.

"Constant velocity" isn't a very precise term. We can make the velocity anything we want it to be, and depend on the time coordinate in any way we want it to, just by choosing the coordinate system carefully. That's why I'm talking about zero acceleration instead. This is a coordinate-independent concept.

 

[Dale]

The path length of a photon in spacetime is zero, it is a point.

Careful here. The Minkowski norm is degenerate, so the fact that two points have a separation of 0 does not imply that they are the same point like it would with the Euclidean norm.

 

[stevmg]

Careful here. The Minkowski norm is degenerate, so the fact that two points have a separation of 0 does not imply that they are the same point like it would with the Euclidean norm.

Is your statement due to the so-called "null" line or surface of the space-time coordinate system? (In other words, if something is traveling at c or light speed, wouldn't the interval be \Deltax²/c² - \Deltat² = 0) Neither spacelike nor timelike.

Am I on the right track here?

Remember, I am one of those uneducated ones...

 

[stevmg]

Can anyone out there explain the difference between "Aristotilean" and "Galilean" spacetime viewpoints? They look pretty similar to me.

 

[Dale]

Is your statement due to the so-called "null" line or surface of the space-time coordinate system? (In other words, if something is traveling at c or light speed, wouldn't the interval be \Deltax²/c² - \Deltat² = 0) Neither spacelike nor timelike.

Am I on the right track here?

Yes. In Euclidean geometry the set of all points with 0 distance from the origin is a single point, but in Minkowski geometry the set of all points with 0 interval from the origin is a cone called a light cone.

 

[Passionflower]

Yes. In Euclidean geometry the set of all points with 0 distance from the origin is a single point, but in Minkowski geometry the set of all points with 0 interval from the origin is a cone called a light cone.

However, the shortest possible distance between any two points inside the cone is also 0, it is only when we restrict photons to go only forward in time that the points are only on the cone.

 

[stevmg]

Can anyone out there explain the difference between "Aristotilean" and "Galilean" spacetime viewpoints? They look pretty similar to me.

Help...

Someone answer this one...

 

[matheinste]

Can anyone out there explain the difference between "Aristotilean" and "Galilean" spacetime viewpoints? They look pretty similar to me.

There is a lengthy description of both viewpoints in the early chapters of Geroch's General Relativity from A to B.

Matheinste.

 

[stevmg]

There is a lengthy description of both viewpoints in the early chapters of Geroch's General Relativity from A to B.

Matheinste.

Thanks Matheinste, I read Geroch's book and that's why I asked the question. All I know is there are straight lines up, bevelled decks of cards, etc., etc. I don't know if I am coming or going.

Maybe, if I phrase the question this way, at slow non-relativistic speeds as we are in the "real world" what is the most utilitarian approach? Forget about\sqrt{(1 - v^2/c^2)}

 

[Dale]

However, the shortest possible distance between any two points inside the cone is also 0, it is only when we restrict photons to go only forward in time that the points are only on the cone.

None of that statement is true in general. For two arbitrary points inside or even on the light cone the interval may be timelike, spacelike, or null. Also, the interval between two events is purely geometric and has nothing to do with photons nor the arrow of time.

 

[Passionflower]

None of that statement is true in general. For two arbitrary points inside or even on the light cone the interval may be timelike, spacelike, or null. Also, the interval between two events is purely geometric and has nothing to do with photons nor the arrow of time.

Sorry but you are wrong, provided we do not restrict photons to go only forward in time we can always "zig zag" inside the cone to make the distance 0.

 

[matheinste]

Sorry but you are wrong, provided we do not restrict photons to go only forward in time we can always "zig zag" inside the cone to make the distance 0.

If we are discussing the spacetime interval why is it not true that a zig zag path can give a null value if we sum along the path, even if we restrict the motion of the photon to forward pointing in time.

Matheinste.

 

[Passionflower]

If we are discussing the spacetime interval why is it not true that a zig zag path can give a null value if we sum along the path, even if we restrict the motion of the photon to forward pointing in time.

Matheinste.

If we restrict it we could obviously not "zig zag" from a future to a passed event inside the light cone.

 

[matheinste]

If we restrict it we could obviously not "zig zag" from a future to a passed event inside the light cone.

If we are referring to any two events in any time order that may be true. But we can still have a zig zag null path going forward in time between events joined by a timelilke vector within a light cone without either part of the zig zag path being on the light cone.

Matheinste.

 

[matheinste]

Thanks Matheinste, I read Geroch's book and that's why I asked the question. All I know is there are straight lines up, bevelled decks of cards, etc., etc. I don't know if I am coming or going.

I think that in the Aristotlean view there is a priviledged state of absolute rest. In the Galilean view there is no such absolute state, all motion is relative.

Matheinste.

 

[DrGreg]

However, the shortest possible distance between any two points inside the cone is also 0, it is only when we restrict photons to go only forward in time that the points are only on the cone.

So what? The "interval" between two events inside the cone is the longest "distance" (or rather, time) between the events, not the shortest.

 

[stevmg]

I think that in the Aristotlean view there is a priviledged state of absolute rest. In the Galilean view there is no such absolute state, all motion is relative.

Matheinste.

That makes sense in accordance of what I read.

Interesting that relativity is tacked onto Galilean.

 

[stevmg]

Using the equation:

(curvature of space-time geometry) = G * (Mass density of matter in space-time)

How did Einstein know what the mass-density of the Sun was? Was the curvature measured in degrees or radians or whatever?
Does G = 6.67259 x 10^-11 N m²/kg²?

Or did he do it backwards... He measured the curvature angle, used Cavenish's G and calculated the mass-density of the Sun by back substitution.

 

[stevmg]

If we are referring to any two events in any time order that may be true. But we can still have a zig zag null path going forward in time between events joined by a timelilke vector within a light cone without either part of the zig zag path being on the light cone.

Matheinste.

But within each leg of the zig-zag path, wouldn't the path be on the "local" light cone eminating from each "vertex?"

Damn it, I wish I could draw what I'm talking about. It would be so easy with a pencil and paper.

 

[matheinste]

But within each leg of the zig-zag path, wouldn't the path be on the "local" light cone eminating from each "vertex?"

Damn it, I wish I could draw what I'm talking about. It would be so easy with a pencil and paper.

From the start of each leg yes. And all events along each leg.

An event in spacetime is the vertex of a light cone. The set of events on a particles worldine have associated with them a set of light cones.

I was considering a pair of events linked by a timelike vector within the future light cone of some other event in the past of those two events.

Matheinste.

 

[Dale]

Sorry but you are wrong, provided we do not restrict photons to go only forward in time we can always "zig zag" inside the cone to make the distance 0.

This has nothing to do with it. The distance between two points is not the same thing as the length of an arbitrary path between the two points. This is analogous in both Euclidean and Minkowski geometry. When we say that my door is 5 m from my window we do not mean that every possible path from my door to my window is 5 m long. Again, these are purely geometric statements and have no relationship whatsoever with photons or the arrow of time.

 

[Passionflower]

This has nothing to do with it. The distance between two points is not the same thing as the length of an arbitrary path between the two points. This is analogous in both Euclidean and Minkowski geometry. When we say that my door is 5 m from my window we do not mean that every possible path from my door to my window is 5 m long. Again, these are purely geometric statements and have no relationship whatsoever with photons or the arrow of time.

My original statement which you described as "None of that statement is true in general" was:

However, the shortest possible distance between any two points inside the cone is also 0, it is only when we restrict photons to go only forward in time that the points are only on the cone.

You were wrong and rather than admitting that you now seem to cover it up by twisting what I was writing.

Using two arbitrary points A and B inside a light cone we can construct a set of light cones that connect us from A to B. The total length of such a connection is 0, which is the shortest possible path. That is simply a feature of a Minkowski spacetime.

 

[Dale]

Using two arbitrary points A and B inside a light cone we can construct a set of light cones that connect us from A to B. The total length of such a connection is 0, which is the shortest possible path.

This is correct, but the length of such a path is not the distance between A and B. Therefore this statement is simply not true.

the shortest possible distance between any two points inside the cone is also 0

When you measure a distance between two points you do not use an arbitrary path, you use a geodesic. The path you describe is not a geodesic. And photons going forward in time is not relevant to any of this.

Let's say that we have a light cone at the origin, and we take the events A = (t_A,x_A,y_A,z_A) = (1,0,0,0) and B = (t_B,x_B,y_B,z_B) = (3,0,0,0) which are two points inside the cone. Then the distance AB between them is 2 despite the fact that there is a path ACB with C = (t_C,x_C,y_C,z_C) = (2,1,0,0) where AC+CB = 0 and a path ADB with D = (t_D,x_D,y_D,z_D) = (2,1.41,0,0) where AD+DB = 0+2i. Those other paths are completely irrelevant to what the distance AB is.

 

[Passionflower]

When you measure a distance between two points you do not use an arbitrary path, you use a geodesic. The path you describe is not a geodesic.

I really see no point in continuing this "discussion". Hopefully other people can benefit from my initial statement, it is clear it is lost on you.

 

[Dale]

Do you really not understand the difference between the distance between A and B:
\sqrt{c^2(t_B-t_A)^2 - (x_B-x_A)^2 - (y_B-y_A)^2 - (z_B-z_A)^2}

and the length of a path from A to B:
\int_A^B \sqrt{c^2 \left(\frac{dt}{d\lambda}\right)^2 - \left(\frac{dx}{d\lambda}\right)^2 - \left(\frac{dy}{d\lambda}\right)^2 - \left(\frac{dz}{d\lambda}\right)^2 } d\lambda

 

[stevmg]

Do you really not understand the difference between the distance between A and B:
\sqrt{c^2(t_B-t_A)^2 - (x_B-x_A)^2 - (y_B-y_A)^2 - (z_B-z_A)^2}

and the length of a path from A to B:
\int_A^B \sqrt{c^2 \left(\frac{dt}{d\lambda}\right)^2 - \left(\frac{dx}{d\lambda}\right)^2 - \left(\frac{dy}{d\lambda}\right)^2 - \left(\frac{dz}{d\lambda}\right)^2 } d\lambda

Dale -

The top expression is the "shortest" distance between the two points in space-time

The bottom exprerssion is the actual length of the path that one travels from A to B which may not be the shortest distance

Am I right? Did I get it?

 

[Dale]

Dale -

The top expression is the [STRIKE]"shortest"[/STRIKE] distance between the two points in space-time

The bottom exprerssion is the actual length of the path that one travels from A to B which may not be the shortest distance

Am I right? Did I get it?

Yes, with one small correction. There is only one distance between two points, but there are an infinite number of paths, each with a different length. The distance is the extremum of the lengths of all possible paths.

 

[stevmg]

Yes, with one small correction. There is only one distance between two points, but there are an infinite number of paths, each with a different length. The distance is the extremum of the lengths of all possible paths.

We are on the same sheet of music, DaleSpam. An analogy to cartesian coordinates would be a quarter circle path with point A at (2, 0) and point B at (0, 2) and the center at the origin and the path is the quarter circle.

The distance from A to B is 2\sqrt{2} = 2.2828 while the path from A to B is \pi = 3.1416 which are clearly different. There are clearly other paths from A to B.

In the time-space coordinates there is a "subtraction" if you will when you deviate from the "straight" path which can actually make the combined interval shorter

The twin paradox is a great example of that.

 

[DrGreg]

In the Euclidean geometry of space, the distance between two points is the length of shortest path between them.

Also, in (flat) spacetime, for two events with spacelike separation, i.e. where one event is outside the light cone of the other, the interval between two events is the length of shortest spacelike path between them.

But for two events with timelike separation, i.e. where one event is inside the light cone of the other, the interval between two events is the length of longest timelike path between them.

That's why the zero-length zig-zag path mentioned earlier is an irrelevance.

 

[stevmg]

In the Euclidean geometry of space, the distance between two points is the length of shortest path between them.

Also, in (flat) spacetime, for two events with spacelike separation, i.e. where one event is outside the light cone of the other, the interval between two events is the length of shortest spacelike path between them.

But for two events with timelike separation, i.e. where one event is inside the light cone of the other, the interval between two events is the length of longest timelike path between them.

That's why the zero-length zig-zag path mentioned earlier is an irrelevance.

Guilty as charged!

 

[stevmg]

So I get this right...

If one follows a zig-zag path (each "zig" and "zag" along the local light cone and going forward in time,) the sum total of all the lengths of these "zigs" and "zags" is 0 (not taking into account GR.) Is that right? Is that called the "interval?"

The distance between the origin of the zigzag path and its destination would be the equation as eumerated above by DaleSpam and that distance would not be zero. Is that right?

 

[stevmg]

More terminology questions:

in the expression m(v) = m₀/SQRT[1 - v²/c²]
m(v) is the inertial mass
m₀ is the rest mass

Am I right?

Finally, in a system of particles which do not interact other than banging together, the sum of all momenta for each particle is the total momentum of the system.

Likewise the total energy of the system is the sum of all the resting energy for each particle + the kinetic energy for each particle added up. Right?

Since momenta are linear operators can one prove that the above is true by induction:

Assume true for n particles and from that show it is true for n+1 particles.

I don't get how they do the energy, which the kinetic energy is not linear because it goes with the square of the velocity. It is non linear. Can't use induction then as it is non linear.

 

[Dale]

So I get this right...

If one follows a zig-zag path (each "zig" and "zag" along the local light cone and going forward in time,) the sum total of all the lengths of these "zigs" and "zags" is 0 (not taking into account GR.) Is that right? Is that called the "interval?"

The distance between the origin of the zigzag path and its destination would be the equation as eumerated above by DaleSpam and that distance would not be zero. Is that right?

That is correct, the length of two different paths between A and B in general may be different from each other and different from the distance between A and B (which is the length of the shortest path). That applies both for Euclidean geometry (spatial distance) and Minkowski geometry (spacetime interval).

 

[stevmg]

That is correct, the length of two different paths between A and B in general may be different from each other and different from the distance between A and B (which is the length of the shortest path). That applies both for Euclidean geometry (spatial distance) and Minkowski geometry (spacetime interval).

Any comment about the energy-momentum. You have already pointed me in the direction of the vectors as you did with twin paradox.

My real concern is, how do you work this - at least an elementary way and is the proof by induction valid?

The total energy of a system is invariant but is not conserved when looking from different frames of reference (I think).

Starthaus has a proof that he worked out for this and I would like to see something like that but if he has copyrighted it, then I can understand.

 

[starthaus]

Any comment about the energy-momentum. You have already pointed me in the direction of the vectors as you did with twin paradox.

My real concern is, how do you work this - at least an elementary way and is the proof by induction valid?

No. I already told you that several times.

The total energy of a system is invariant

No, it is not invariant since it is speed dependent:

E=\gamma(v)m_0c^2
E'=\gamma(v')m_0c^2

Nor is the momentum frame-invariant, since it depends on speed and velocity:

\vec{p}=\gamma(v)m_0\vec{v}

It is the norm of the energy-momentum (E , c\vec{p}) that is invariant. Indeed:

E^2-(c\vec{p})^2=(m_0c^2)^2=E'^2-(c\vec{p'})^2

Wait until you receive Spacetime Physics and you study it.

but is not conserved when looking from different frames of reference (I think).

In a closed system, the total momentum \vec{p} and the total energy E are conserved, each one of them. You managed to get everything backwards.

Starthaus has a proof that he worked out for this and I would like to see something like that but if he has copyrighted it, then I can understand.

I haven't copyrighted anything, this is all easily available in textbooks.I wrote a little piece for Battleimage to help him out understanding the energy-momentum conservation, see here. Why don't you wait for your own copy of Spacetime Physics, it is all there.

 

[DrGreg]

More terminology questions:

in the expression m(v) = m₀/SQRT[1 - v²/c²]
m(v) is the inertial mass
m₀ is the rest mass

Am I right?

Yes. m(v) is also called relativistic mass, although its use is deprecated by most modern physicists who prefer to work with rest mass only.

Finally, in a system of particles which do not interact other than banging together, the sum of all momenta for each particle is the total momentum of the system.

Likewise the total energy of the system is the sum of all the resting energy for each particle + the kinetic energy for each particle added up. Right?

Yes. Usually the energy isn't explicitly separated into "rest energy" and "kinetic energy", we just talk about energy.

Since momenta are linear operators can one prove that the above is true by induction:

Assume true for n particles and from that show it is true for n+1 particles.

I don't get how they do the energy, which the kinetic energy is not linear because it goes with the square of the velocity. It is non linear. Can't use induction then as it is non linear.

It's not the "linearity of momentum" that is used here. In fact the function \textbf{p}(m_0,\textbf{v})=\gamma(v) m_0 \textbf{v} is only linear in m₀ and not linear in v. Induction works simply because addition (adding two momenta to get the total momentum) is linear, and therefore similarly for energy. That is true as long as you are looking at this within anyone frame only.

There is a technical difficulty if you want to go further than this and show that the total "energy-momentum" is a well-behaved 4-vector that transforms from one frame to another the way that 4-vectors are supposed to transform (i.e. via the Lorentz transform). The proof isn't entirely trivial due to the relativity of simultaneity.

The total energy of a system is invariant but is not conserved when looking from different frames of reference (I think).

You got "invariant" and "conserved" the wrong way round there!

 

[stevmg]

As stated my book on Spacetime Physics will get here. It is taking me a while to get the terminology straight (look at all the corrections made to my earlier statements inthe immediately preceding posts) and I still need to bounce things off you folks as I learn them. Had to do that all the time in medicine. I never relied totally on my interpretation when it came to new concepts. Never killed anybody and I wish I can say that about others.

I understand that you (starthaus) have stated that the proof is not an induction proof - just getting a second opinion.

When I saw my pulmonolgist earlier this year and he told me that I had sleep apnea, I said "Hey, I want a second opinion!"

He gave it to me...

"OK, Steve, here it is... You're ugly, too.!"

I never win.

 

[Dale]

Any comment about the energy-momentum. You have already pointed me in the direction of the vectors as you did with twin paradox.

I am not sure what you are asking for here. My comment is that I like them

Oh, one thing to remember is that photons are massless so you always know that the energy is equal to the momentum times c.

My real concern is, how do you work this - at least an elementary way and is the proof by induction valid?

Personally, I wouldn't worry too much about the proof. It will be good to read through since it will really make you think about the relativity of simultaneity and how a system of particles is even defined in different frames. But all proofs require some assumptions which then must be validated experimentally.

 

[starthaus]

There is a technical difficulty if you want to go further than this and show that the total "energy-momentum" is a well-behaved 4-vector that transforms from one frame to another the way that 4-vectors are supposed to transform (i.e. via the Lorentz transform). The proof isn't entirely trivial due to the relativity of simultaneity.

Yes, the two page essay by Rindler on this subject is very interesting. A mechanical application of the rule "the sum of four-vectors is a four-vector" does not apply in the case of the energy-momentum four-vector due to the very definition of relativistic total energy and relativistic three-momentum. Relativity of simultaneity plays a big role in defining the values that enter the conservation equations. Though Rindler's discussion ends up with the conclusion that the resultant energy-momentum of an isolated system of particles is indeed a four-vector, it is the proof that is worth following.

 

[stevmg]

Let's get this correct (for me)

The sum of all energy in a system of particles in which the total momentum is zero is invariant

If the momentum of the system is not zero then the relation:

E² -(cp)² = E₀²

in which E is the energy of the system, cp is the energy due to the momentum of the system, and E₀ is the invariant energy of the system if the momentum were to be brought down to zero.
This invariant quantity E₀ would be the same form all frames of reference.
The conservation applies to all the frames of reference each frame at a time. Different frames of reference will change the total momentum of the system and the total energy of the system but not the E₀ which is the invariant part.

Is this right?

 

[DrGreg]

Let's get this correct (for me)

The sum of all energy in a system of particles in which the total momentum is zero is invariant

If the momentum of the system is not zero then the relation:

E² -(cp)² = E₀²

in which E is the energy of the system, cp is the energy due to the momentum of the system, and E₀ is the invariant energy of the system if the momentum were to be brought down to zero.
This invariant quantity E₀ would be the same form all frames of reference.
The conservation applies to all the frames of reference each frame at a time. Different frames of reference will change the total momentum of the system and the total energy of the system but not the E₀ which is the invariant part.

Is this right?

I'd agree with all that except for one small detail. I wouldn't describe cp as "the energy due to the momentum" (although I suppose in a sense it is). The equation in which it appears has energy-squared in it. But you could describe (E-E₀) as "kinetic energy" of the system, and E₀/c² is the "invariant mass" of the system (which isn't the sum of the individual particles' rest masses except in the special case when they're all at rest relative to each other).

 

[stevmg]

I'd agree with all that except for one small detail. I wouldn't describe cp as "the energy due to the momentum" (although I suppose in a sense it is). The equation in which it appears has energy-squared in it. But you could describe (E-E₀) as "kinetic energy" of the system, and E₀/c² is the "invariant mass" of the system (which isn't the sum of the individual particles' rest masses except in the special case when they're all at rest relative to each other).

If cp isn't the energy of the particle associated with the momentum as seen from a given frame of reference, then what is it?

Of course the point of what I did write was to make sure my concept of invariant energy etc. was correct and that momentum and energy is conserved for each frame of reference while the E₀ is invariant - across all frames of reference.

 

[Dale]

I also would not describe cp as the energy associated with the momentum. "Energy associated with momentum" sounds like another way to say "kinetic energy", and for v<<c cp reduces to cmv which is not the same as 1/2 mv². While cp has units of energy, I would think of it as momentum, not energy. Or, preferably, you could think of it as what it is: the spacelike component of the energy-momentum four-vector expressed in units of energy. Mentioning that it is the spacelike component of a four-vector ensures that its relationship to energy is properly characterized.

 

[stevmg]

I also would not describe cp as the energy associated with the momentum. "Energy associated with momentum" sounds like another way to say "kinetic energy", and for v<<c cp reduces to cmv which is not the same as 1/2 mv². While cp has units of energy, I would think of it as momentum, not energy. Or, preferably, you could think of it as what it is: the spacelike component of the energy-momentum four-vector expressed in units of energy. Mentioning that it is the spacelike component of a four-vector ensures that its relationship to energy is properly characterized.

Thank you. This is new to me. I knew that total energy was mc² + 1/2mv²

I will treat this as a new entity for me - cmv which I will have to wrap my brain around.

I do understand what "spacelike" means.

This is going to take time.

 

[starthaus]

Thank you. This is new to me. I knew that total energy was mc² + 1/2mv²

Not quite, this is the resultant of doing the Taylor expansion of the correct formula:
E=\frac{m_0c^2}{\sqrt{1-(v/c)^2}}.
Like any Taylor expansion, it is valid only for \frac{v}{c}&lt;&lt;1.
It is not valid at large speeds.