What is the Constant Speed of Movement through Spacetime?

[RahSuh]

TL;DR Summary

Everyone moving through spacetime at speed c - don't get it

Hi

I have been reading Brian Cox/Jeff Forshaw book on Why does E=mc2 (highly recommend it)

One thing I don't get (page 95) is when they say everything moves through spacetime at the same constant speed c?!

I get why a person/object A at rest moves through space time with speed c - but say that A is moving at speed v with respect to B,

Apparently B thinks A is is moving through spacetime at speed C...

I can't follow the logic in the book - I have googled the question - and have come across terms like propert time (which I get everyone thinks is the same) but can't get to the fact that speed through space time is constant

Any help appreciated

ThanksSuhail

 

[PeroK]

I have been reading Brian Cox/Jeff Forshaw book on Why does E=mc2 (highly recommend it)

One thing I don't get (page 95) is when they say everything moves through spacetime at the same constant speed c?!

Strictly speaking, a popular science book like this isn't a valid reference for learning SR. They are really more for entertainment. Your question highlights this.

If we start with a basic definition of motion as change of position with respect to time. And velocity being rate of change of position with respect to time, and speed being the magnitude of velocity, then you can see immediately that "moving" through spacetime is not well defined. $c$, for example, is defined in terms units of $m/s$. What are you differentiating by in the case of this "motion" through spacetime?

To speak about moving through spacetime at $c$ you would need a new definition of "motion" and "speed".

The more precise statement that Cox/Forshaw are alluding to is that the magnitude of a particle's four-velocity is always $c$.

You might interpret that as "moving through spacetime at $c$" or you might not. But, in principle, there is nothing to "get". The four-velocity of a particle is well-defined and its magnitude is $c$. That's all there is to it. See here for example:

https://en.wikipedia.org/wiki/Four-velocity

 

[RahSuh]

(got you re brian cox - though it is a very good book- I learned more from it than my undergrad SR course!)But going back to the problem - is there an intuitive reason why the magnitude of this 4 vector should be the same for everyone?

(if I had to teach myself special relativity - what would you recommend, I am doing maths/physics at the open university here in the UK - say something at the first year undergrad level??)

I appreciate you taking the time to explain stuff to me!

S

 

[PeterDonis]

when they say everything moves through spacetime at the same constant speed c?!

This is a fairly common pop science way of conflating two very different things:

(1) The invariant length of the 4-velocity vector of an object with nonzero rest mass is $c$.

(2) The magnitude of the ordinary 3-velocity vector of a light ray, i.e., an object with zero rest mass, in any inertial frame is $c$.

(Note that in "natural" relativity units, which are commonly used in textbooks, where $c = 1$, both of these magnitudes are $1$.)

Pop science books try to attribute some kind of deep significance to the fact that these two magnitudes are the same (Brian Greene, for example, is notorious for this). Unfortunately, doing this seriously misrepresents the physics and leads to a lot of confusion, because people trying to actually reason out the physics ask perfectly reasonable questions like the one you ask:

I get why a person/object A at rest moves through space time with speed c - but say that A is moving at speed v with respect to B,

Apparently B thinks A is is moving through spacetime at speed C...

This question simply does not have a good answer within the "moving through spacetime at speed c" viewpoint.

The best way to deal with all this is to simply ignore these "moving through spacetime with speed c" statements and learn the actual physics from a textbook. The actual physics will tell you, in a nutshell, that:

(1) For an object with nonzero rest mass, the invariant length of its 4-velocity vector is $c$, and the magnitude of its ordinary 3-velocity vector in an inertial frame is $v$, its speed relative to observers at rest in that frame--so this magnitude varies from frame to frame, i.e., it is not invariant.

(2) For an object with zero rest mass, the invariant length of its 4-momentum vector is zero (we can't properly call it a "4-velocity" because that would imply that it had invariant length $c$, as for the nonzero rest mass case above), and the magnitude of its ordinary 3-velocity vector in any inertial frame is $c$--so this magnitude does not vary from frame to frame, i.e,. it is invariant.

In other words, the two cases above--nonzero rest mass vs. zero rest mass--are fundamentally different physically, and conflating the two just leads to confusion.

 

[PeterDonis]

is there an intuitive reason why the magnitude of this 4 vector should be the same for everyone?

The magnitude of any 4-vector is an invariant in relativity. That's why the 4-vector formalism is so often simpler and easier to use in relativity than the 3-vector plus time formalism.

 

[PeroK]

(got you re brian cox - though it is a very good book- I learned more from it than my undergrad SR course!)But going back to the problem - is there an intuitive reason why the magnitude of this 4 vector should be the same for everyone?

(if I had to teach myself special relativity - what would you recommend, I am doing maths/physics at the open university here in the UK - say something at the first year undergrad level??)

I appreciate you taking the time to explain stuff to me!

S

A book like that will give you the impression that you've covered a lot of ground, but it will not give you the thorough foundation for the subject. Most popular science books (must) avoid mathematics, for example.

The book I like is by Helliwell:

https://www.goodreads.com/book/show/6453378-special-relativity

Also, Morin has the first chapter of his book online:

https://scholar.harvard.edu/files/david-morin/files/cmchap11.pdf

Re four-vectors, it depends what you mean by "intuitive". Your intuition changes significantly if you know the mathematical basis of something.

One fundamantal aspect of SR is that the spacetime interval between two events is invariant between inertial reference frames. This leads to the concept of a four-vector, comprising four components: one time and three spatial. (This is analagous to the magnitude of an ordinary 3D vector being invariant under rotations.)

The proper time of a particle is also invariant.

The four-velocity, being the derivative of a particle's spacetime coordinates by its proper time is therefore of invariant magnitude.

Note that the magnitude of any four-vector is invariant.

(Although that may be better explained in a textbook with all the required development of these concepts.)

 

[Vanadium 50]

got you re brian cox - though it is a very good book- I learned more from it than my undergrad SR course!

And yet here you are, arguing a point that is incorrect: a point you yourself say you don't get.

Maybe it's not as great a book as you think. Maybe Peter and Ibix are right.

 

[PeterDonis]

I learned more from it than my undergrad SR course!

You might not have learned as much from it as you think. Consider: you spotted one thing that doesn't make sense (the thing you're asking about in this thread). But you're not an expert. How many other things might there be in the book that also don't make sense, but you're unable to tell, so you think they do make sense, but somewhere down the line you're going to find out they don't?

 

[Ibix]

Ibix

You're either psychic or you meant PeroK.

is there an intuitive reason why the magnitude of this 4 vector should be the same for everyone?

I was just going to comment that, in your own rest frame, the rate of change of time with time is, unsurprisingly, 1. And the rate of change of position with time is zero because you are at rest. Thus your four-velocity must be parallel to (1,0,0,0). We want all the components to have the same units so we need to multiply the time component by $c$ (this is hand-wavy - more rigorous justification is available), so your four velocity in your own frame is $(c,0,0,0)$. The magnitude is obviously $c$ and the magnitude is invariant. Thus your (and anybody else's) four velocity has magnitude $c$.

 

[PeroK]

You're either psychic or you meant PeroK.I was just going to comment that, in your own rest frame, the rate of change of time with time is, unsurprisingly, 1. And the rate of change of position with time is zero because you are at rest. Thus your four-velocity must be parallel to (1,0,0,0). We want all the components to have the same units so we need to multiply the time component by $c$ (this is hand-wavy - more rigorous justification is available), so your four velocity in your own frame is $(c,0,0,0)$. The magnitude is obviously $c$ and the magnitude is invariant. Thus your (and anybody else's) four velocity has magnitude $c$.

Yes, and $c$ is just the conversion factor!

 

[Vanadium 50]

You're either psychic or you meant PeroK.

Um...couldn't both be true?

 

[vanhees71]

(got you re brian cox - though it is a very good book- I learned more from it than my undergrad SR course!)

Really? Then I'd try to get my tuition fees back ;-)). SCNR.

 

[Dale]

Summary:: Everyone moving through spacetime at speed c - don't get it

One thing I don't get (page 95) is when they say everything moves through spacetime at the same constant speed c?!

The whole concept of moving through spacetime is suspect. Movement is a change in position with respect to time. But time is part of spacetime already, so moving through spacetime doesn’t happen in that sense.

When you are considering spacetime you are considering geometry where time is part of an object’s shape. What was previously a point particle is now a line.

If the particle is at rest in space then the spacetime line is parallel to the time axis. If the particle is moving through space then the spacetime line is at an angle to the time axis. If the particle is accelerating in space then the spacetime line curves. But in all cases the spacetime line has a geometric shape that doesn’t move in the sense that things move in space.

Instead, what we can do is describe geometric features of the shape. We can, for example, give the unit tangent vector to the curve. To calculate the unit tangent vector we use the equation $\frac{dx^{\mu}}{d\tau}$ which looks like a relativistic generalization of velocity and is therefore called the four velocity.

Since the four velocity is a unit tangent vector it necessarily has unit length. In relativity the unit of speed is $c$. It is not a big revelation that a unit vector has unit length, it is a definition.

 

[PAllen]

To add one thing to @Dale, the reason one wants to use a unit vector for tangent is that the only physical meaning is the direction of the tangent (this gives velocity, and changes in tangent direction give acceleration). In fact, speed in an inertial frame is often most conveniently represented as a hyperbolic angle. Thus magnitude is normalized precisely because it adds no physical content. Thus making it absurd to turn around and ask what this arbitrarily chosen normalized magnitude means (making it c is purely a convention; some authors make it 1, by convention even when not setting c=1).

 

[FactChecker]

One thing I don't get (page 95) is when they say everything moves through spacetime at the same constant speed c?!

This is a mathematical consequence of the Lorentz transformation. The geometry of spacetime simply has that property. The mathematics is closely related to the Pythagorean theorem. The books are just pointing this fact out. One can easily see that the Lorentz transformations indicate a trade-off between velocity in 3-space and the passage of time. The books are just summarizing that fact in a way that is true and easy to state. Any intuition regarding why this geometry applies to the real world is a different question; it is a physics question.

 

[Dale]

some authors make it 1, by convention even when not setting c=1

Oh, I didn’t know that. I haven’t seen that before.

 

[vanhees71]

It's simply convenient to define
$$u^{\mu}=\frac{1}{c} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} s},$$
because you need this quantity very often to project a four-vector or a tensor in its direction and Minkowki-perpendicular to it.

Just some more remarks: The four-velocity is much more convenient to use than the usual three-velocity which is not a four-vector and thus has a very complicated behavior under Lorentz transformations, while the four-velocity is invariant, because it's a four-vector and its components are easily transformed with Lorentz matrices.

The relation to the usual three-velocity is
$$u^{\mu}=\frac{1}{\sqrt{1-\beta^2}} \begin{pmatrix}1 \\ \vec{\beta} \end{pmatrix},$$
where
$$\vec{\beta}=\frac{1}{c} \frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau}, \quad \vec{v}=c \vec{\beta}.$$

 

[PAllen]

Oh, I didn’t know that. I haven’t seen that before.

One example is Bergmann’s 1942 text on SR and GR, which happens to be the first serious text on relativity I studied.

 

[PAllen]

One example is Bergmann’s 1942 text on SR and GR, which happens to be the first serious text on relativity I studied.

One thing that is striking about this book is how modern it seems compared to some later books:
- no mention of relativistic mass whatsoever
- 4 vectors introduced before any discussion particle kinematics in SR
- all SR dynamics derived from relativistic Lagrangian and Hamiltonion formulations (using 4 vectors and one forms - though called covariant vectors).
- in GR, it already showed how the equations of motion follow from the field equations, making the geodesic hypothesis unnecessary, and discussed why this was true for GR but not for other field theories like EM.

 

[vanhees71]

He has also written the very good entries about special and general relativity in the "Handbuch der Physik" (vol. IV).

https://link.springer.com/book/10.1007/978-3-642-45973-3

 

[MikeeMiracle]

Speaking as a laymen without an understanding of the background maths, the whole "moving through spacetime at the speed of light" helps someone like me to visualise roughly what is going on even, if it technically is inaccurate as you say here. A great video on this comes from the first 1m45s of this Cassiopeia video.

 

[PeterDonis]

Speaking as a laymen without an understanding of the background maths, the whole "moving through spacetime at the speed of light" helps someone like me to visualise roughly what is going on even, if it technically is inaccurate as you say here. A great video on this comes from the first 1m45s of this Cassiopeia video.

This video is perpetuating a related misconception: that when an object is moving instead of at rest, some of its "speed through time" gets converted to "speed through space" to keep its "speed through spacetime" the same. This is misleading and you should not be using it. It will trip you up.

 

[FactChecker]

This video is perpetuating a related misconception: that when an object is moving instead of at rest, some of its "speed through time" gets converted to "speed through space" to keep its "speed through spacetime" the same. This is misleading and you should not be using it. It will trip you up.

Good point. If I am completely at rest (in my reference frame), this is how other observers, who are all be traveling at different speeds, would interpret my relative velocity and my passage of time. They would all see a different mixture, but each would see a mixture that could be described as me "traveling" through spacetime at the velocity c. The fact is that I have done nothing at all, they are just observing me through the glasses of their own relative motion.

 

[RahSuh]

This is a fairly common pop science way of conflating two very different things:

(1) The invariant length of the 4-velocity vector of an object with nonzero rest mass is $c$.

(2) The magnitude of the ordinary 3-velocity vector of a light ray, i.e., an object with zero rest mass, in any inertial frame is $c$.

(Note that in "natural" relativity units, which are commonly used in textbooks, where $c = 1$, both of these magnitudes are $1$.)

Pop science books try to attribute some kind of deep significance to the fact that these two magnitudes are the same (Brian Greene, for example, is notorious for this). Unfortunately, doing this seriously misrepresents the physics and leads to a lot of confusion, because people trying to actually reason out the physics ask perfectly reasonable questions like the one you ask:
This question simply does not have a good answer within the "moving through spacetime at speed c" viewpoint.

The best way to deal with all this is to simply ignore these "moving through spacetime with speed c" statements and learn the actual physics from a textbook. The actual physics will tell you, in a nutshell, that:

(1) For an object with nonzero rest mass, the invariant length of its 4-velocity vector is $c$, and the magnitude of its ordinary 3-velocity vector in an inertial frame is $v$, its speed relative to observers at rest in that frame--so this magnitude varies from frame to frame, i.e., it is not invariant.

(2) For an object with zero rest mass, the invariant length of its 4-momentum vector is zero (we can't properly call it a "4-velocity" because that would imply that it had invariant length $c$, as for the nonzero rest mass case above), and the magnitude of its ordinary 3-velocity vector in any inertial frame is $c$--so this magnitude does not vary from frame to frame, i.e,. it is invariant.

In other words, the two cases above--nonzero rest mass vs. zero rest mass--are fundamentally different physically, and conflating the two just leads to confusion.

This is a fairly common pop science way of conflating two very different things:

(1) The invariant length of the 4-velocity vector of an object with nonzero rest mass is $c$.

(2) The magnitude of the ordinary 3-velocity vector of a light ray, i.e., an object with zero rest mass, in any inertial frame is $c$.

(Note that in "natural" relativity units, which are commonly used in textbooks, where $c = 1$, both of these magnitudes are $1$.)

Pop science books try to attribute some kind of deep significance to the fact that these two magnitudes are the same (Brian Greene, for example, is notorious for this). Unfortunately, doing this seriously misrepresents the physics and leads to a lot of confusion, because people trying to actually reason out the physics ask perfectly reasonable questions like the one you ask:
This question simply does not have a good answer within the "moving through spacetime at speed c" viewpoint.

The best way to deal with all this is to simply ignore these "moving through spacetime with speed c" statements and learn the actual physics from a textbook. The actual physics will tell you, in a nutshell, that:

(1) For an object with nonzero rest mass, the invariant length of its 4-velocity vector is $c$, and the magnitude of its ordinary 3-velocity vector in an inertial frame is $v$, its speed relative to observers at rest in that frame--so this magnitude varies from frame to frame, i.e., it is not invariant.

(2) For an object with zero rest mass, the invariant length of its 4-momentum vector is zero (we can't properly call it a "4-velocity" because that would imply that it had invariant length $c$, as for the nonzero rest mass case above), and the magnitude of its ordinary 3-velocity vector in any inertial frame is $c$--so this magnitude does not vary from frame to frame, i.e,. it is invariant.

In other words, the two cases above--nonzero rest mass vs. zero rest mass--are fundamentally different physically, and conflating the two just leads to confusion.

Thanks - this helps a lot!

 

[RahSuh]

A book like that will give you the impression that you've covered a lot of ground, but it will not give you the thorough foundation for the subject. Most popular science books (must) avoid mathematics, for example.

The book I like is by Helliwell:

https://www.goodreads.com/book/show/6453378-special-relativity

Also, Morin has the first chapter of his book online:

https://scholar.harvard.edu/files/david-morin/files/cmchap11.pdf

Re four-vectors, it depends what you mean by "intuitive". Your intuition changes significantly if you know the mathematical basis of something.

One fundamantal aspect of SR is that the spacetime interval between two events is invariant between inertial reference frames. This leads to the concept of a four-vector, comprising four components: one time and three spatial. (This is analagous to the magnitude of an ordinary 3D vector being invariant under rotations.)

The proper time of a particle is also invariant.

The four-velocity, being the derivative of a particle's spacetime coordinates by its proper time is therefore of invariant magnitude.

Note that the magnitude of any four-vector is invariant.

(Although that may be better explained in a textbook with all the required development of these concepts.)

Great - I guess in classical physics, for two events, the space interval was an invariant as well as the time interval

in relativistic physics, the invariant is the spacetime interval

 

[RahSuh]

Really? Then I'd try to get my tuition fees back ;-)). SCNR.

I am old enough, that universities in the UK (back then) were free...

 

[PeroK]

Great - I guess in classical physics, for two events, the space interval was an invariant as well as the time interval

in relativistic physics, the invariant is the spacetime interval

Only the time interval. The space interval isn't invariant across inertial reference frames. The invariant quantity in classical physics is the acceleration, as in $\vec F = m \vec a$

 

[PAllen]

Only the time interval. The space interval isn't invariant across inertial reference frames. The invariant quantity in classical physics is the acceleration, as in $\vec F = m \vec a$

The distance between two events is invariant in classical physics, is it not? All inertial frames agree on it.

 

[PeroK]

The distance between two events is invariant in classical physics, is it not? All inertial frames agree on it.

Not with $x' = x - vt$

 

[PAllen]

Not with $x' = x - vt$

I should have said distance between simultaneous events, and, of course, simultaneity is absolute in pre-relativity physics.