# I Stanford: objects in spacetime all move at constant speed c?

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1. Apr 17, 2016

### AstroMath

In this Stanford University lecture on Relativity, it is stated:

Likewise, objects in spacetime all move at constant speed c in spacetime but if you change its direction, say by moving at speed v in the x direction, then spatial speed will change and so will the speed along the ct direction. Again, its total speed will still be c through spacetime.

http://web.stanford.edu/~oas/SI/SRGR/notes/SRGRLect6_2007.pdf

Brian Green of Columbia University also says the same thing, as do many others.

Are they right?​

2. Apr 17, 2016

### andrewkirk

Yes they are right, but it doesn't mean what it sounds like it means. In particular it means nothing like the same as moving at speed c through space, which only massless particles like photons can do.

In cases like this, the only exact presentation of the science is a mathematical one, which involve be saying something such as that the four-velocity vector of a massive particle has magnitude c (or magnitude 1, if we're using relativistic units). The linked pdf starts to talk about that around the bottom of p2.

3. Apr 17, 2016

### AstroMath

What do you mean by this, "Yes they are right, but it doesn't mean what it sounds like it means. In particular it means nothing like the same as moving at speed c through space, which only massless particles like photons can do."?

They're not talking about space like you are.

4. Apr 17, 2016

### stevendaryl

Staff Emeritus
The statement that all objects move at speed c through spacetime is equivalent to saying: Every second, you move one second into the future. It's true, but it's not that profound of a statement.

5. Apr 17, 2016

### robphy

A more correct statement would be:
Massive objects in spacetime all move at constant speed c in spacetime (i.e., is a vector with magnitude c) [and spatial speed v (also called $\tanh\theta$ [a slope]) in space]
but if you change its direction, say by moving at nonzero spatial speed v in the x direction,
then spatial component ($\gamma\ v$ or $c\sinh\theta$) will change and so will the temporal component ($\gamma\ c$ or $c\cosh\theta$).
Again, its total speed ($\sqrt{(\gamma\ c)^2- (\gamma\ v)^2}$) will still be c through spacetime.
Massless objects in spacetime all move at constant speed zero in spacetime [and spatial speed c in space].

6. Apr 17, 2016

### AstroMath

So the velocity of light through spacetime is 0? I would argue that the velocity of light is c.

7. Apr 17, 2016

### SiennaTheGr8

Here is the math:

$\left(\dfrac{ds}{d\tau}\right)^2=\left(\dfrac{c \, dt}{d\tau}\right)^2 - \left(\dfrac{d \vec x}{d\tau}\right)^2$,

where $ds$ is the (invariant) infinitesimal spacetime interval, $d\tau$ is the (invariant) infinitesimal proper time interval, $dt$ is the (frame-dependent) infinitesimal coordinate time interval, and $d \vec x$ is the (frame-dependent) infinitesimal displacement.

In plain English, the left side is the (squared) "speed through spacetime" of something with mass. All observers agree on its value. The right side has two terms whose value depends on an observer's velocity relative to the massive object in question: the (squared) "speed through time" and the (squared) "proper velocity."

But this expression can be simplified. First, $ds/d\tau = c$. Second, $dt/d\tau = 1/\sqrt{1-v^2/c^2} = \gamma$. Third, $d \vec x / d \tau = \gamma \vec v$ (where $\vec v$ is normal velocity). So we have:

$c^2=(\gamma c)^2 - (\gamma \vec v)^2$.

This means that everything with mass always moves through spacetime with a "speed" of $c$, where "speed" means the proper-time-derivative of the spacetime interval. As $v$ increases, both terms on the right side increase, and the first of them (the time contribution) is always the bigger of the two. So it's not quite true that "speed through time" decreases as $v$ increases.

What is true is that an object at rest has all of its "motion" through spacetime directed forward in time, whereas moving observers would say that the object moves through both space and time. All observers agree, however, that the difference of the squares of the space and time contributions equals $c^2$, and that the object moves through spacetime at the $\tau$-rate of $c$.

8. Apr 17, 2016

### Staff: Mentor

It's one plausible way of visualizing the relationship between motion in space and paths in spacetime; and it captures the idea that you're always moving forward in time no matter what you do. It's not perfect (for example, either the speed through spacetime is not $c$ but instead $\sqrt{-c^2}$ - note that square root of a negative number - or you have to adopt a sign convention in which your speed through space is the square root of a negative number) and it's no substitute for understanding the underlying math.

So you can use it to form a mental picture to go with the math, but you can't build any deeper understanding on top of it.

9. Apr 18, 2016

### robphy

Let me be more explicit in my post (#5 above).

Massive objects in spacetime all move at constant speed c in spacetime
(i.e., its 4-momentum vector can be normalized to a "unit" 4-velocity vector with Minkowski-norm c)
[and has spatial speed (slope) v in space].

Massless
objects in spacetime all move at constant speed zero in spacetime
(i.e., its 4-momentum vector has Minkowski-norm zero [and thus can't be normalized])
[and has spatial speed (slope) c in space].

10. Apr 18, 2016

### Staff: Mentor

It depends. Brian Greene, at least, when he makes this pitch in his pop science books, says something that isn't right. He says that all objects, including light, "move at c" in spacetime. But as robphy has pointed out, this isn't correct; it switches between two different interpretations of the term "speed through spacetime"--the norm of the 4-velocity in the case of massive objects, but the coordinate speed through space in an inertial frame in the case of light. This is a good example of a scientist saying something in a pop science book that he would never get away with in an actual peer-reviewed paper.

11. Apr 18, 2016

### A.T.

It's not about right or wrong , just about geometrical interpretations of the same math. The advance through space-time at const c can be visualized better in a space-propertime diagram:

12. Apr 18, 2016

### robphy

Can the clock effect/twin paradox be displayed on this diagram? Or a uniformly accelerated observer?
Or is it restricted to inertial observers?

13. Apr 18, 2016

### A.T.

I think for a uniformly accelerated observer it would look cone-like, as shown here for a local observer hovering in a gravitational field:
http://demoweb.physics.ucla.edu/content/10-curved-spacetime
Also use gravity slider here:

Here the non-local picture for a radial line through the Schwarzshild-geometry (which is non-inertial).

I guess you can transform any space-coordinate diagram into a space-propetime diagram. Here for the twins:

For the twins with constant acceleration it should look somewhat like this:

The key is that the length of all worllines (in space-propertime) represents the passed coordiante time, so they all have equal length. This is the geometrical interpretation closest to the "all moves at c through space-time" meme, that I know of.

Last edited: Apr 19, 2016
14. Apr 18, 2016

### PAllen

For a non-inertial observer in such a diagram, why would you not construct it so the 'defining observer' world line is vertical??!!

It cannot be good for coincident events to be separated on diagram (this occurs even for the inertial frame twin scenario) ...

15. Apr 18, 2016

### A.T.

His world line is shown curved, because he is non-inertial.

16. Apr 18, 2016

### Staff: Mentor

You are being inconsistent here. If you are talking about "velocity through space" then massive objects have $0\le v < c$ and massless objects have $v=c$. If you are talking about "velocity through spacetime" then massive objects have $"v"=c$ and massless objects have $"v"=0$. In neither case do you get $c$ for both massive and massless objects.

If you don't like $"v"=0$ for light then you may want to move on to other topics besides this "velocity through spacetime".

17. Apr 18, 2016

### Staff: Mentor

And what would you base your argument on? The tangent vector to the worldline of a light ray, which is the only feasible interpretation of the term "velocity through spacetime", is null; that means it has zero norm, so light's "velocity through spacetime" is indeed zero under this interpretation. The tangent vector to the worldline of a massive object, OTOH, has norm $c$; that's what justifies the statement that massive objects have a "velocity through spacetime" of $c$.

Another interpretation would be that the concept of "velocity through spacetime" doesn't make sense at all for light. But there is no interpretation I'm aware of under which light's "velocity through spacetime" is $c$. Light's velocity through space is $c$; but as Dale pointed out, you yourself were saying we should be careful not to confuse space with spacetime.

18. Apr 24, 2016

### Jeronimus

Can you elaborate a bit on this?

If as you say, when considering spacetime, massive objects move at c (always i assume), whereas light moves at 0 then it seems acceleration cannot exist any more in the sense of increasing or decreasing the speed of a massive object.

19. Apr 24, 2016

### Staff: Mentor

In the spacetime model, acceleration does not change the speed of a massive object; it only changes its direction. The norm of the object's 4-velocity never changes; it is always $c$. But the direction in which the 4-velocity points in spacetime can change; that is what 4-acceleration does.

20. Apr 24, 2016

### Jeronimus

Why is something which does not affect the velocity of an object called acceleration? 4-acceleration in this case.