# Is there time at the speed of light?

1. Jun 11, 2009

### Gothican

I been wondering about this question for a long time - If time slows down as your speed gets closer to the speed of light, then is there time at the speed of light?
First of all the graph of the equation of time/speed has an asymptote at v=c which means that if a mass is moving at the speed of light, then it's time is undefined.

But if you think about it, light has to have time - because it takes him a certain amount of time to get through space.

So what is it?
Can anyone give me a specific definition of time at the speed of light?
Thanks

2. Jun 11, 2009

### Fredrik

Staff Emeritus
See my posts in this thread about the "photon's point of view". In particular, #8 and #14.

The concept of "proper time" is only defined along those curves in spacetime that represent speeds <c, so the concept of "proper time" isn't really relevant here. We could of course define the proper time of a null curve (a curve that represents speed c) to be =0, but it's a pretty pointless thing to do for several reasons, one of them being that no clocks can travel at that speed anyway.

The concept of "coordinate time" is too arbitrary to be useful on its own. The coordinate time between two events can be anything we want it to be, since we're the ones choosing what coordinate system to use. We can remove the arbitrariness by choosing to only consider the co-moving co-moving inertial frame, but there's no such thing as a co-moving inertial frame for a photon.

If you're wondering what the difference is between coordinate time and proper time, this post might help.

If the above didn't make it clear, the answer is "no" (unless you choose to define it as zero for no good reason).

Last edited: Jun 11, 2009
3. Jun 11, 2009

### HallsofIvy

Measured in what frame of reference? If you think about it, the fact that it takes about 8 and a half minutes for light to get from the sun to here, measured in our frame of reference, says nothing about the time interval in the photon's frame of reference. I personally don't think it really makes sense to talk about a photon's "frame of reference" of "point of view" but if we do, the equations say, as you do, that time does NOT pass in a photon's frame of reference.

4. Jun 11, 2009

### Thetom

Hi, I'm not really qualified to even be involved in such a discussion but it reminded me of something I read the other day. This maybe a very simplified way to explain it. Of course, I maybe missing the point entirely.

This is what I read... (not a direct quote)

All things move at the speed of light through the 3+1 dimensions. As we move more and more through the spacial dimensions, we move less and less through the time dimension. Its the same for pure spacial dimensions, as I move north more I move east less. And as I move through the spacial dimension more I move through time less. Light travels through the spacial dimensions so fast that it doesn't move through the time dimension at all. Therefore time doesn't pass for light. Conversely (and this bit's my own deduction :) If you don't move through the spacial dimension at all you'll be moving through the time dimension at the speed of light. Relativity may have some things to say about "not moving through the spacial dimensions" but hopefully this illustrates the co-efficiency between the time/space continuum (of which you are obviously aware).

Basically, in answer to your question, and according to this book, light doesn't move through the time dimension at all. Like I said, a very simplified answer.

5. Jun 11, 2009

### skeptic2

Thetom,

Though I'm in agreement with your answer it leaves me with a question. If instead of light which many of us like to think of as particles, we use radio waves which we generally prefer to think of as waves, then there is the problem with the undulating waveform in which the undulations must occur in zero time in it's own frame of reference. Hmmmm.

6. Jun 11, 2009

### Fredrik

Staff Emeritus
I think that what you have in mind is that every massive particle has a four-velocity with magnitude c. I don't think of that as a statement about a property of massive particles. I think of it as a part of the definition of four-velocity.

Also note that this only applies to massive particles.

That's not right. What you have read is just a really strange way of expressing the fact that if we choose units such that c=1, the magnitude of the ratio $\Delta x/\Delta t$ (i.e. the velocity) goes to 1 when the speed goes to c. It's really a statement that contains no information.

It's not quite that simple. The first difficulty is to define what "for light" means, and as I have tried to explain (see my previous post, and my posts in the thread I linked to), there's no natural way to do it. Any definition we choose is going to be just that: A definition...that we have chosen.

7. Jun 11, 2009

### Nisse

Now I'm really confused. I've read many articles saying that when we're at rest, we move through time at speed c, and that when we move through space we move through time correspondingly slower. Are you saying that's incorrect?

8. Jun 11, 2009

### HallsofIvy

Since "speed c" is measured in units of "distance/time" what could "move through time at speed c" possibly mean?

9. Jun 11, 2009

### atyy

Along a spacetime geodesic, which is the worldline of a particle moving "freely" in a gravitational field, one can assign an "affine parameter" which changes along it, and in which the geodesic equation looks pretty. Along a non-null geodesic that may be the worldline of a massive particle, the proper time is an affine parameter which has a rather direct experimental interpretation. Along a null geodesic that may be the worldline of a photon, one can still assign affine parameters, but, as Fredrik points out above, these do not have the same experimental interpretation as the proper time.

10. Jun 12, 2009

### Gothican

To Fredric; Are you simply saying that when an object's mass is equal to 0, then it is irrelevant to give him any attribution to time?

What I was hoping to achieve was that a photon which leaves the sun, gets to Mercury and Pluto at the same time. Could this be true based on your previous statements?
At the same time in its frame of reference? Yes. Although, as I said before, I don't think talking about a photon's frame of reference makes much sense. After all, it isn't carrying a pocket watch along with it!

Last edited by a moderator: Jun 12, 2009
11. Jun 12, 2009

### ZikZak

What Fredrik is trying to explain is that all this talk about a "photon's frame of reference" is nonsense. A photon doesn't have a frame of reference. The second postulate is that the speed of light is c in all reference frames. Therefore there is no reference frame in which photons are at rest. No matter how fast you go, photons still pass you at c. It makes no sense to talk about the rest frame of photons, because they don't have any.

Last edited: Jun 12, 2009
12. Jun 12, 2009

### Fredrik

Staff Emeritus
No, it's worse than that. A claim has to make sense to be incorrect, and this one doesn't. So I don't think that's what those articles said. I'm assuming that they talked about the fact that the magnitude of the four-velocity (see e.g. this post) of a massive particle is c in all inertial frames.

The four-velocity is defined to be the vector with components (c,0,0,0) in the co-moving inertial frame. Apply a Lorentz transformation to that, and we see that in an inertial frame where the particle is moving with speed v in the positive x direction, the four-velocity is $u=\gamma(c,v,0,0)$. What they're saying is that the larger the velocity, the smaller the ratio between the 0 and 1 components of the four-velocity:

$$\frac{u_0}{u_1}=\frac{\gamma c}{\gamma v}=\frac c v$$

This is =0 when v=0, and goes to 1 (not infinity, which is what you suggested) as v goes to c.

That's one of the things I'm saying, but I also explained why I'm saying it.

You really need to think about what "time" means here. The concept of "proper time" can't be used here, and if the above is a reference to "coordinate time", then the difference between their arrival times can be anything we choose it to be, because there's nothing that forces us to use a particular coordinate system. There isn't even anything that suggests that there's a coordinate system that's more appropriate than others.

If you define "frame of reference" to mean "co-moving inertial frame" and "reference frame"/"rest frame" to mean "co-moving inertial frame", then yes, this is one of the things I've been saying. I also explained several reasons why I'm saying it, but everyone seems to be ignoring the reasons.

13. Jun 12, 2009

### ZikZak

I'm not; I think you have excellent reasons.

14. Jun 12, 2009

### Naty1

regarding posts 4 and 5: yes there is a "problem" but it's the same one as with a "particle"...nothing can exist without time so the description from Thetom has some shortcomings. I can't remember where I first saw that description, I have posted the source in another thread here some time ago, was it Brian Greene??, anyway, I really like it....and despite Fredrik's objections, it can be useful I believe....

BUT, As Fredrik suggests, it has drawbacks and is perhaps best thought of in conjunction with the rubber sheet analogy for gravity...useful in some respects, but only as a conceptual approach for some perspective, not as an absolutely correct detailed technical insight....

A further useful concept: it time is plotted,say vertically, and say x and y space dimensions on the perpendicular horizontal plane , then constant velocity is an inclined straight line in spacetime, constant "straight acceleration in space" (like down the highway) appears as smooth curve in spacetime , and rotational acceleration (say a merry go round) appears as a corkscrew in spacetime.....so spacetime mediates between velocity and acceleration and types of acceleration!!! (same source, Greene I think)

One famous physicst said something like "time is nature's way of keeping everything from happening at once." This also has it's own shortcomings, since nothing ACTUALLY happens instantaneously, in zero time...but it's a nice broadbrush idea...

15. Jun 12, 2009

### Thetom

Yes the quote was from Brian Greene's book: The Elegant Universe. I've not read it all so cant recommend it or anything (is a pop-science book for the laymen promoting Greens area of mathematics). He does specifically say that a photon doesn't age at all, though.

Here's a direct quote:
"...Einstein found that precisely this idea - the sharing of motion between different dimensions - underlies all of the remarkable physics of special relativity...
...Einstein proclaimed that all objects in the universe are always traveling through space-time at one fixed speed - that of light...
...If an object dose not move through space all of the objects motion is used to travel through time...
...Something traveling at light speed through space will have no speed left for motion through time. Thus light does not get old; a photon that emerged from the big bang is the same age today as it was then. There is no passage of time at the speed of light."

I cant say whether or not his statements are true. I thought it seemed fitting to the question, though. I'm sorry if I confused the issue at all Gothican.

I know its only a simple explanation and is sure to break down somewhere. And I know there may be some semantic issues about what we mean by 'time'. But if time is take to be literally a dimension, and the passage of time is motion through that dimension, then the idea that something can be sitting still in time (but moving in space) is very easy to grasp.

16. Jun 12, 2009

### Nisse

This is the sort of thing I was referring to, but did not express it as eloquently. I would be grateful for Frederik's comments on this.

17. Jun 12, 2009

### matheinste

Later in the same paragraph we have the clarification----

We are presently talking about an object’s combined speed through ALL four dimensions---three space and one time---and it is the object’s speed in this generalized sense that is equal to that of light.

Matheinste

18. Jun 13, 2009

### Fredrik

Staff Emeritus
Most of what Greene says are things I don't mind, e.g. "through space-time at one fixed speed", "all of the objects motion is used to travel through time", "it is the object’s speed in this generalized sense that is equal to that of light". This is his attempt to say the same things that I said about four-velocity, while consistently avoiding any reference to mathmatics. Unfortunately that last thing just makes it difficult, if not impossible, to understand what he's saying (if you don't know the mathematics already).

This is a part of the Greene quote that I don't like at all, because it's so easy to misinterpret: "Something traveling at light speed through space will have no speed left for motion through time".

Think of a spacetime diagram representing the inertial coordinates in which you are stationary. Time in the "up" direction, space in the "right" direction. (No need to consider more than one spatial dimension for this). What we mean by saying that an object that's stationary (in this frame) is moving only through time (in this frame) is that its four-velocity is a vector pointing straight up. Its components are (c,0). The four-velocity of an object moving with speed v to the right is $\gamma(c,v)$. This vector isn't pointing straight up. It makes an angle arctan(v/c) with the time axis. This angle is in the interval [0,45°). That's what we mean when we say that the object is moving through both space and time. We're talking about the direction of its four-velocity vector in our own rest frame.

Now look at the words "Something traveling at light speed through space will have no speed left for motion through time". He's not talking about a particle with speed c. He's saying that if something has a four-velocity vector that makes a 90 degree angle with the time axis, it's not moving through time. This is an "object" (in some generalized sense) which is moving at infinite speed, not light speed. The words "traveling at light speed" is a reference to the fact that a four-velocity vector always satisfies $-(u^0)^2+\vec u^2=-c^2$, and "through space" specifies that the projection of the vector he's talking about onto the time axis is zero.

Now let's talk about the worst part of the Greene quote: "Thus light does not get old; a photon that emerged from the big bang is the same age today as it was then. There is no passage of time at the speed of light". This looks like complete crazy talk in this context, because he didn't even say that he has now abandoned the description of motion in terms of which way the four-velocity vector is pointing in his rest frame, and is now talking about which way the four-velocity vector (c,0) of a stationary object will point if we apply a Lorentz transformation with speed v and take the limit v→∞. A Lorentz transformation with velocity -v takes (c,0) to $\gamma(c,v)$, i.e. it tilts the time axis by an angle arctan v. It also tilts the x axis in the opposite direction by the same amount. In the limit v→∞, both axes get tilted 45° in opposite directions, which makes them coincide. (They will point in the same direction or opposite directions, depending on the direction of the velocity).

There are several reasons why this inappropriate, and very misleading. The first is that he didn't say that he's no longer considering the four-velocity vector in his own rest frame, and is instead trying to imagine what it looks like in "the photon's rest frame". The other reasons are the ones I have already stated (use the link in #2) for why it doesn't make sense to think of that limit as the photon's rest frame.