# Why do we move at the speed of light?

1. Oct 7, 2012

### guitarphysics

Why do we move (including the dimension of time) at the speed of light? I understand that when our velocity increases in a spatial dimension, it will decrease in time, but why is the initial, overall velocity c?

2. Oct 7, 2012

### Simon Bridge

Can you give an example of what you are talking about?

My immediate reaction is: "we don't start out at c."
In our own reference frame we are always stationary.

Certainly nothing can move at the speed of light (except light) - that is kind-of the point.

3. Oct 7, 2012

### Staff: Mentor

This refers to the four-velocity, which is the relativistic four-dimensional spacetime generalization of the velocity of an object. Its magnitude is indeed always c.

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

4. Oct 7, 2012

### Simon Bridge

Well, in that case, the 4-velocity has to be that way because all observers measure the same speed for light.

... we don't "start out" at c, the magnitude of the 4-velocity is always c. Perhaps OP means that a rest-frame has 4-velocity (c,0,0,0)?

remembering that $x_0=ct$ and that, at rest, $\gamma=1$ should help.

5. Oct 7, 2012

### Staff: Mentor

That, and Lorentz transformations keep the magnitude of (proper) 4-vectors invariant.

6. Oct 7, 2012

### Simon Bridge

There's also this:
Well if the 4-velocity is $\mathbf{U}=\gamma(c,\vec{u})^t$ I guess "velocity in the spatial dimension" would be $\gamma\vec{u}$ and the time dimension part is $\gamma c$.

Since $\gamma \geq 1$ - that means that the "our velocity" actually increases "in time" when it "increases in a spatial dimension"? (Though I guess that should be "someone elses velocity" since the observer is always stationary in their own reference frame?)

i.e. if we interpret post #1 in terms of the 4-velocity, there are at least two difficulties with the way it is phrased: you don't "start out" at speed c, and the speed in the time dimension does not decrease. (It sounds a bit like how a pop-science article may phrase things now I think about it...)

@guitarphysics:
A trick for understanding relativity is to get really pedantic about how you talk about it, at least when you are starting out ... everyday language does not do very well. If we are correct in guessing that your understanding of the 4-velocity is as above, then there are two misunderstandings to clear up which probably have lead to the question.
If it isn't, then we are gonna need clarification before we can be of use. Thanks.

7. Oct 7, 2012

### bcrowell

Staff Emeritus
This is something that seems to have propagated by Brian Greene in his popularizations. Physicists in general do describe the four-velocity as having magnitude c, but do not typically describe objects as moving through spacetime with velocity c. The latter is just Greene's way of putting it. It's not wrong, it's just a nontechnical verbal description of an equation that every physicist agrees on. There is a distinction between "moving through space" and "moving through spacetime" (which only Greene talks about).

8. Oct 7, 2012

### guitarphysics

Thanks everyone. Ben, you were spot on! I'm reading The Elegant Universe by Brian Greene. I don't know calculus and I've only been studying physics for the past month or so, but I'm very interested, so I decided to read his book. I've understood most things, but I don't really get WHY the four-velocity has magnitude c. I couldn't really understand a lot of what Simon Bridge and mfb where talking about (as I said, I don't know much in the way of physics or math).

9. Oct 7, 2012

### bcrowell

Staff Emeritus
You might enjoy Relativity Simply Explained, by Gardner.

10. Oct 7, 2012

### Staff: Mentor

If I remember correctly, Greene tacitly describes what would be happening in a 4D Eucidean universe in which an increase in the spatial components of 4 velocity would indeed result in a decrease in the time component. He does this without any, or very little, explanation that, in our actual Minkowski spacetime, this is not actually what happens. Much of if the discussion in his book makes use of the (fictitious) 4D Euclidean universe concept. Nothing is wrong with such an approaach, since it gives readers more of an intuitive feel for what is happening in relativity, but it needs to be accompanied with more extensive caveats.

Chet

11. Oct 7, 2012

### Simon Bridge

Don't worry, that was mostly just me trying to understand the question. Don't be afraid to post the context - it can be very helpful. Had you said "Brian Greene's book" or something earlier the replies would have been more understandable sooner.

I will second bcrowell's book suggestion. iirc it does not have the emphasis on the wierdness that most seem to.

I am trying to improve my ability to describe these things so I'll give it a go, and the others can suggest improvements, then we'll both learn something :)

I'll try and start with a secondary school senior level of physics, and no calculus... that will limit how detailed I can be and you should realize that it is unlikely to do justice to the subject. You will need to know about vectors and coordinates in normal 3D space.

I will be using math symbols a lot - try not to let them intimidate you ;) I'll try to define things as I go so the math will be a kind of short-hand.

A "normal" 3D position would be $\vec{q}=(x,y,z)$
a 4D position would be $\mathbf{q}=(ct,x,y,z)$ we also write $\mathbf{q}=(q_0,q_1,q_2,q_3)$ all by convention.
You can just match up the same positions so $q_0=ct,q_1=x,\cdots$

Notice that we use ct instead of just t for the "time" coordinate - this is to be consistent: the units of position have to be some length and you cannot mix units up in one vector.

To turn time into a length-like thingy, we have to multiply it by some speed (because "speed = distance over time" right?) In the past the speed we picked to use for this may have been the speed of the King's favorite racehorse on a fine day timed on the King's stopwatch. This would be problematic - for one thing, the King would have to do all our measurements and he's a busy man.

We pick the speed of light in a vacuum, instead, because it has the special property that everyone measures it to be the same no matter what and so everyone can agree what it is.

This fact that the speed of light in a vacuum is the same for everybody is the cornerstone of relativity. The consequences of special relativity follow from this. Because it is the same for everyone it is considered a fundamental property of the Universe.

The magnitude of $\vec{q}$ would be $q=\sqrt{x^2+y^2+z^2}$ - that's the rule for finding out the distance to a position. The rule follows from the geometry of Euclid (look him up) which most of us think of as "normal" geometry. It's basically Pythagoras.

You'd expect the rule for a 4-vector to be the same, only in 4D - like this: $|\mathbf{q}| = \sqrt{q_0^2+q_1^2+q_2^2+q_3^2}$ and this is the kind of rule that Brian Greene uses as a teaching guide in his book. But he's telling the general public one thing and he's telling his students another thing ... what he tells his students is this:

$|\mathbf{q}| = \sqrt{-q_0^2+q_1^2+q_2^2+q_3^2}$ [1]
... spot the minus sign? This is what happens in our Universe.
Since it obeys a different rule to Euclids, we tend to talk about these things being "in 4-space" rather than "in 4D" but it is usually clear by context which kind of 4D is intended.

If something zips by you in, say, the z direction, then you can write it's velocity, u, like this: $\vec{u} = (0,0,u_z)$ and $u_z=\Delta z/\Delta t$.

In 4-space you'd have to write something like: $\mathbf{u} = (u_0,0,0,u_3)$ since the object is clearly moving in time as well as space - but is not moving in the x or y directions.

How do we work it out?

By analogy, it seems we'd want to do something like $u_3 = du_3/dq_0$ since $q_0$ is what we think of as the 4-space equivalent of the time-axis. But notice that the time we divided by in the 3D version was the time measured by someone who is not relativistic, and the speed has to be non-relativistic as well (or the classical picture would not work).

In relativity we call this the "proper time" and give it the symbol $\tau$. Proper time is related to "regular" time by $\Delta t=\gamma \Delta\tau$ where the $\gamma$ is a factor that depends on the 3D relative speed. This is the "time dilation" effect you have heard about. [3]

So we want to define the 4-velocity more like: $$\mathbf{u} = \frac{\Delta \mathbf{q}}{\Delta\tau} = \left (\frac{\Delta q_0}{\Delta\tau},0,0, \frac{\Delta q_3}{\Delta\tau} \right )$$... and $\tau$ is the time as measured on a clock carried by the object.

But we know that $\Delta q_0=c\Delta t$ and $\Delta t = \gamma \Delta\tau$ ... so $$u_0=\frac{\Delta q_0}{\Delta\tau}=\frac{c\gamma\Delta\tau}{\Delta \tau}=\gamma c$$ ... if the speed were zero (object at rest) then $\gamma=1$, $u_3=0$ so the 4-velocity will be $\mathbf{u}=(c,0,0,0)$ pretty much automatically.

Back to the moving object ... we also know that $q_3=z$ and using $\Delta t = \gamma\Delta\tau \Rightarrow \Delta\tau = \Delta t/\gamma$ we can write: $$u_3=\frac{\Delta z}{\Delta\tau}=\frac{\Delta z}{\Delta t/\gamma}=\gamma\frac{\Delta z}{\Delta t} = \gamma u_z$$... so, the object has a 4-velocity vector: $\mathbf{u}=\gamma(c,0,0,u_z)$

All this just drops out automatically from the math.
The key to everything here is understanding $\gamma$ - which is given by: $$\gamma = \frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$$... in the above case, $u=u_z$. This factor is a direct consequence of all observers measuring the same speed for light in a vacuum.

And that is a whole lecture by itself. [4]

(There are some bits in there that I'm not sure if I should have done a bit more, or a bit differently. No doubt someone will tell me.)

---- footnotes --------------------

[1] there is another rule that goes like this $|\mathbf{q}| = \sqrt{q_0^2-(q_1^2+q_2^2+q_3^2)}$ but they are equivalent.

[2] If we use a very small change, we write: $u_z=\delta z/\delta t$ and for an infinitesimally small change we get $u_z=dz/dt$ ... which is the calculus you see in places like wikipedia.

[3] $\gamma$ is always bigger than 1, and approaches infinity as u approaches c. This is one of the consequences of c being the same for everybody. It means that if the guy on the spaceship (going fast wrt you) has a clock, and you see it's second hand click forward by 1 second, then your clock will tick off $\gamma$ seconds... so he always seems to be slow.

[4] further reading: Relativity and FTL - despite it's title, is a fun and accessible introduction to the fundamentals of relativity for people who are a tad math declined.

12. Oct 7, 2012

### bahamagreen

"... spot the minus sign? This is what happens in our Universe."

It might help to explain why... (and clean up my lack of skills, especially if I got it right)

With distance as light speed times time you have d = c * t.

d = c * t

d^2 = ( c^2 * t^2 )

d^2 - ( c^2 * t^2 ) = 0

SQRT [ d^2 - ( c^2 * t^2 ) ] = 0

Accordingly, the negative term appears in the 4D distance

13. Oct 8, 2012

### Simon Bridge

the reference in footnote [4] should clear that up.

d=ct for light, but then the gamma factor is infinite so all distances are infinitely contracted in the direction of travel. If the object is travelling at speed u<c it should work out differently ...

I considered saying that the minus sign is needed to make time dilation come out the right way ... in the Greene 4D velocity, moving clocks would run fast wouldn't they?

14. Oct 8, 2012

### robphy

Here is an important takeaway message concerning Greene's "moving through spacetime" at the speed of light idea.
(Actually, I think it was Lewis Epstein that first popularized this "moving through [space]time" phrasing.)

If you must use the "speed through spacetime" phrasing....
• A massive particle travels...
through space at speed less than c.,
but through spacetime with *speed* c [by choice of some convention].
• Light travels...
through space at speed c,
but through spacetime with *speed* zero.

The word speed is used in two similar-but-distinct ways:
In space, speed is the magnitude of the ordinary velocity vector (spatial 3-velocity).
In spacetime, *speed* is the [normalized] magnitude of the spacetime 4-velocity vector for a massive particle. This doesn't quite apply for light since light's 4-momentum has zero magnitude (and thus can't be normalized to a 4-velocity)... but we can call the *speed* zero.

Last edited: Oct 8, 2012
15. Oct 8, 2012

### A.T.

There is no such distinction in Epstein's idea. Both, light and massive objects, "advance" at c through space-propertime. The only special thing about light in this picture is that it advances only though space, and not through propertime, like massive objects do. Not sure if Greene meant the same thing.

16. Oct 8, 2012

### bobc2

One of the renouned mathematicians in the early 1900's was Hermann Weyl. He was Einstein's colleague and close friend. Weyl's description of observers moving along their worldlines (4th dimensions) at the speed of light is the earliest such characterization that I am aware of. Many physicists have described it that way. However, as bcrowell has pointed out, Greene as a different twist on it in the way he describes it (...speed along X1 taking away from speed along X4, etc.). The sketches below attempt to provide a picture of how this works in a 4-dimensional universe. One of the fascinating aspects of special relativity is the way an observer's X1 axis rotates so that a photon worldline always bisects the angle between the X1 and X4 axes. This means that different observers moving at relativistic speeds with respect to each other live in different 3-D cross-sections of the universe, and it accounts for all observers measuring the same light speed: c.

Last edited: Oct 8, 2012
17. May 15, 2013

### guitarphysics

Sorry to bring this post back after such a long time, but I've been learning relativity (from Kleppner and Feynman) for the past month or so, and I understand this all a lot more now :).
A doubt I have though, is why Simon stated the four-velocity as $\gamma$(c,U) when (at least according to Kleppner) it should be $\gamma$(ic,U). (That's another doubt I have- why does Minkowski write it like that? Where did he derive it from?).

18. May 15, 2013

### Simon Bridge

I did it that way because of the position 4-vector (ct,x,y,z).

There are two ways to write the 4-space coordinates.
The other one is (x,y,z,ict) - the i is included in an attempt to make learning it easier iirc - it means students can use the normal formula they know for the magnitude for example, and the Lorentz transformation looks a bit like a rotation. Or something.

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

It seems to work OK for flat space-time but gets awkward as you try to use it in curved space-time ... where the minus sign is better kept in the metric.

[edit: @WannabeNewton - "wick rotation" thanks]

Last edited: May 15, 2013
19. May 15, 2013

### WannabeNewton

20. May 15, 2013

### guitarphysics

Ah, OK. I wasn't aware there had been progress in relativity in the past 40 years. Where can I find a more updated account of SR? Also, is there anything else in SR? (As far as I've seen, there's not much more to it besides the Lorentz contractions, the relativistic expressions for momentum, energy, force, etc., various thought experiments, four-vectors, and relativity of simultaneity)
By the way, for a year-long school project, I'm writing a (probably short) book for the layman on relativity (which is why I started learning it- besides it being extremely interesting), so if anybody has any suggestions before I start writing, they'd be greatly appreciated.